RICHARDST. ANDRE
&
Core Topics and Section Prerequisites
Chapter 5
Orderings of
Comparability &
Cardinality
Cardinals ————_-
Axiom of Choice
See
5.4
5.5
Chapter 7
tee
ae
Chapter 4
Functions
;
4.1, 4.2, 4.3, 4.4
Renineee.
ho lh, Day ores:
Ordering Relations
3.5
Modular Arithmetic
3.4
Chapter 2
Sets & Induction
DANN 2 3, 2.4, 2.5
Chapter 1
Logic & Proofs
Lheell Ne Peale! Cea a eB
Mathematical Writing
Style
Appendix
Preface to the Student
Core
TE he) 6 A
ai
Limits & Continuity
45
;
Chapter 3
Relations & Partitions
Concepts of Analysis
Principles of
Counting
2.6
Proofs from
Number Theory
1.8
Strategies for
Proofs
Loe
4.7
Chapter 6
Concepts of Algebra
6.1; 6.2; 6:3,.6.4, 6.5
Digitized by the Internet Archive
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A TRANSITION
TO
ADVANCED MATHEMATICS
EIGHTH
EDITION
A TRANSITION
TO
ADVANCED MATHEMATICS
Douglas D. Smith
University of North Carolina Wilmington
Maurice Eggen
Trinity University
Richard St. Andre
Central Michigan University
pee UINGAGE
«~ Learning’
Australia * Brazil
Japan * Korea * Mexico « Singapore * Spain * United Kingdom ¢ United States
oe CENGAGE
aS SLeamines
A Transition to Advanced Mathematics,
8th Edition
Douglas Smith, Maurice Eggen, and
Richard St. Andre
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Printed in the United States of America
Print Number: 01
Print Year: 2014
To our wives
Karen, Karen, and Karen
2
eee
CONTENT S
ee
ee
Ee
eee
Preface
ix
Preface to the Student
cHapteER
1
Logicand Proofs
1.1
1.2
1.3
1.4
1.5
1.6
Propositions and Connectives
1
Conditionals and Biconditionals
10
Quantified Statements
18
Basic Proof MethodsI
28
Basic Proof Methods II 42
Proofs Involving Quantifiers
51
Mathematical Writing Style 64
Strategies for Constructing Proofs 67
1.7
1.8
cHAptTteER
vi
xii
Proofs from Number Theory
1
76
2_
Sets and Induction
2.1
2.2
2.3.
2.4
2.5
2.6
Basic Concepts of Set Theory
85
Set Operations 95
Indexed Families of Sets 104
Mathematical Induction
114
Equivalent Forms of Induction
128
Principles of Counting
137
85
Contents
Cer
A P IER
Cena)
3
Relations and Partitions
ull
S22
33
3.4
Bye)
Relations
153
Equivalence Relations
163
Partitions
175
Modular Arithmetic
182
Ordering Relations
189
153
Functions
I Ae a
203
Functions as Relations
203
Constructions of Functions
212
Functions That Are Onto; One-to-One Functions
Inverse Functions
Set Images
237
251
Cardinality
ER
222
231
Sequences
243
Limits and Continuity of Real Functions
Gh
AS Psi
Vii
261
Equivalent Sets; Finite Sets 261
Infinite Sets 271
Countable Sets 279
The Ordering of Cardinal Numbers
287
Comparability and the Axiom of Choice
295
Gals
As P TER
6
Concepts of Algebra
6.1
6.2
6.3
6.4
6.5
Algebraic Structures
303
Groups
312
Subgroups
320
Operation Preserving Maps
Rings and Fields 334
303
326
Vili
Contents
cHapteR
hapep ek ND
JZ
Concepts of Analysis
7.1
7.2
7.3
7.4
7.5
The Completeness Property 344
The Heine—Borel Theorem
351
The Bolzano—Weierstrass Theorem
363
The Bounded Monotone Sequence Theorem
Equivalents of Completeness
374
NX
343
368
Sets, Number Systems, and Functions
Answers to Selected Exercises
Index
427
387
379
PeR-E
Ff ACC E
his text is intended to bridge the gap between calculus and advanced courses
in at least three ways. First, it guides students to think and to express themselves mathematically—to analyze a situation, extract pertinent facts, and draw
appropriate conclusions. Second, it provides a firm foundation in the major ideas
needed for continued work. Finally, we present introductions to modern algebra
and analysis in sufficient depth to capture some of their spirit and characteristics.
In summary, our main goals in this text are to improve the student’s ability to think
and write in a mature mathematical fashion and to provide a solid understanding of
the material most useful for advanced courses.
Exercises marked with a solid star (« )have complete answers at the back of the
text. Open stars (:) indicate that a hint or a partial answer is provided. “Proofs to
Grade” are a special feature of most of the exercise sets. We present a list of claims
with alleged proofs, and the student is asked to assign a letter grade to each “proof”
and to justify the grade assigned. Spurious proofs are usually built around a single
type of error, which may involve a mistake in logic, a common misunderstanding
of the concepts being studied, or an incorrect symbolic argument. Correct proofs
may be straightforward, or they may present novel or alternate approaches. We
have found these exercises valuable because they reemphasize the theorems and
counterexamples in the text and also provide the student with an experience similar
to grading papers. Thus, the student becomes aware of the variety of possible errors
and develops the ability to read proofs critically.
The eighth edition is based on the same goals as previous editions, with several new or substantially revised sections and many new and revised expositions,
examples, and exercises. One of the new features is a mini-section in Chapter 1 on
mathematical writing style that describes good practices and some of the special
characteristics that distinguish the way mathematics is communicated. In addition
to advice on what to include in a proof and what to leave out, this short section
offer tips on the use of symbols and other details that help in writing clear, readable
proofs.
xX
Preface
A listing of useful preliminary concepts of sets, the number systems, and the
terminology of functions that students have presumably encountered in prior study
is now found in an Appendix. This makes the prerequisite material easy to locate
and keeps the focus of the text on mathematical reasoning and the core content.
An expanded section on strategies for constructing proofs follows the introductory sections on methods of proof and the discussion on writing style. This section
summarizes basic proof methods and includes more than 60 exercises involving
proofs. Proofs from elementary number theory appear in a separate section where
the Division Algorithm is accepted without proof in order to practice basic proof
methods on a coherent set of results about divisibility and the greatest common
divisor. We have deliberately placed this early in the text before any discussion of
inductive proofs or the Well-Ordering Principle. Later, in Chapter 2, students observe
the power of inductive methods to prove the Division Algorithm and other results.
There is a new Section 3.4 on modular arithmetic and a new Section 4.7 on
limits of functions and continuity of real functions. Other sections with the most
substantial revisions are Section 2.6 on combinatorial counting and Section 4.6 on
sequences.
We consider the core material (see the diagram on the inside front cover) to be
the first several sections of Chapters | through 5. Chapter | introduces the propositional and predicate logic required by mathematical arguments, not as formal logic,
but as tools of reasoning for more complete understanding of concepts (including
some ideas of arithmetic, analytic geometry, and calculus with which the student
is already familiar). We.present methods of proof and carefully analyze examples
of each method, giving special attention to the use of definitions and denials. The
techniques in this chapter are used and referred to throughout the text. In Chapters
2, 3, and 4 on sets, relations, and functions, we emphasize writing and understand-
ing proofs that require the student to deal precisely with the concepts of set operations, equivalence relations and partitions, and properties of injective and surjective
functions.
Chapter 5 emphasizes a working knowledge of cardinality: finite and infinite
sets, denumerable sets and the uncountability of the real numbers, and properties
of countable sets. As shown in the diagram on the inside front cover, each of the
first five chapters offers opportunities for further study, including basics of number
theory, modular arithmetic, limits and continuity of real functions, and the ordering
of cardinal numbers.
Chapters 6 and 7 make use of the skills and concepts the student has acquired
from the core—and thus are above the earlier work in terms of level and rigor.
In Chapter 6, we consider properties of algebras with a binary operation, groups,
substructures, and homomorphisms, and relate these concepts to rings and fields.
Chapter 7 considers the completeness property of the real numbers by tracing its
consequences: the Heine—Borel Theorem, the Bolzano—Weierstrass
Theorem, and
the Bounded Monotone Sequence Theorem, and back to completeness.
Preface
xi
We sincerely thank our reviewers for the eighth edition: Dave Feil, Carroll
University; Rita Hibschweiler, University of New Hampshire; William Howland,
University of St. Thomas; Katherine Kelm, California State University, Fresno;
Irene Loomis, University of Tennessee at Chattanooga; Dale Rohm, University of
Wisconsin-Stevens Point; Daniel Seabold, Hofstra University; Boris Shekhtman,
University of South Florida; Beimnet Teclezghi, New Jersey City University; and
David Walnut, George Mason University.
We also wish to thank Elizabeth Jurisich for her suggestions after accuracy
checking the text and the Answers to Selected Exercises, and the staff at Cengage
for their exceptional professional assistance in the development of this edition and
previous editions.
Finally, we note that instructors who adopt this text can sign up for online
access to complete solutions for all exercises via Cengage’s Solution Builder, service at www.cengage.com/solutionbuilder.
Douglas D. Smith
Richard St. Andre
PREFACE
1-0-1
HAE
Soi7u
DEON
ie
——————————————EEE———
“T understand mathematics but I just can’t do proofs.”
Many students approach the study of mathematical reasoning with some apprehension and uncertainty, perhaps expecting that the study of proofs is something they
won’t really have to do or won’t use later. These feelings, expressed in the remark
above, are natural as you move from courses where the goals emphasize performing computations or solving certain equations to more advanced courses where the
goal may be to establish whether a mathematical system has certain properties. This
textbook is written to help ease the transition between these courses. Let’s consider
several questions students commonly have at the beginning of a “transition” course.
Why write proofs?
Mathematicians often collect information and make observations about particular
cases or phenomena in an attempt to form a theory (a model) that describes patterns
or relationships |among quantities and structures. This approach tto the development of a theory uses inductive reasoning. However, the characteristic thinking
of the mathematician is deductive reasoning, in which one uses logic to develop
and extend atheory by drawing conclusions based on statements accepted as true.
Proofs are essential in mathematical reasoning because they demonstrate that the
conclusions are true. Generally speaking, a mathematical explanation for a conclusion has no value if the explanation cannot be backed up by an acceptable proof.
The first goal of this text is to examine standard proof techniques, especially
concentrating on how to get started on a proof, and how to construct correct proofs
using those techniques. You will discover how the logical form of a statement can
serve as a guide to the structure of a proof of the statement. As you study more
advanced courses, it will become apparent that the material in this book is indeed
fundamental and the knowledge gained will help you succeed in those courses.
Moreover, many of the techniques of reasoning and proof that may seem so difficult at first will become completely natural with practice. In fact, the reasoning
xii
Preface to the Student
xiii
that you will study is the essence of advanced mathematics, and the ability to reason
abstractly is a primary reason why applicants trained in mathematics are valuable
to employers.
Why not just test and repeat enough examples to confirm
a theory?
After all, as is typically done in natural and social sciences, the test for truth of a
theory is that the results of an experiment conform to predictions and that when
the experiment is repeated under the same circumstances, the result is always the
same. One major difference is that in mathematics we often need to know whether
a given statement is always true, so while the statement may be true for many (even
infinitely many) examples, we would never know whether another example might
show the statement to be false. By studying examples, we might conclude that the
statement “x? — 3x + 43 is a prime number’ is true for all positive integers x. We
could reach this conclusion testing the first 10 or 20 or even the first 42 integers 1,
2, 3, ..., 42. In each of these cases and others, such as 44, 45, 47, 48, 49, 50, and
more, x” — 3x + 43 is a prime number. But the statement is not always true because
43° — 3(43) + 43 = 1763, which is 41 - 43. Checking examples is helpful in gaining insight for understanding concepts and relationships in mathematics, but is not
a valid proof technique unless we can somehow check all examples.
Why not just rely on proofs that someone else has done?
One answer follows from the statement above that deductive reasoning characterizes the way mathematicians think. In the sciences, a new observation may force a
complete rethinking of what was thought to be true; in mathematics what we know
to be true (by proof) is true forever unless there was a flaw in the reasoning. By
learning the techniques of reasoning and proof, you are learning the tools of the
trade. A proof is the ultimate test of your understanding of the subject matter and
of mathematical reasoning.
What should I know before beginning Chapter 1?
The usual prerequisite for a transition course is at least one semester of calculus.
We will sometimes refer to topics that come from calculus and earlier courses (for
example, differentiable functions or the graph of a parabola), but we won’t be solving equations or finding derivatives.
We assume that you have encountered the basic concepts of sets and subsets;
that you are familiar with the natural number system and the integers and rational,
real, and complex numbers; and that you have worked with functions—especially
with functions defined on sets of real numbers with real number images. See the
Appendix for a quick review of these essential ideas and notations, which will be
used throughout the text.
Xiv
Preface to the Student
What am I allowed to assume for a proof?
You may be given specific instructions for some proof-writing exercises, but generally the idea is that you may use what someone studying the topic of your proof
would know. That is, when we prove something about intersecting lines, we might
use facts about the slope of a line, but we probably would not use properties of
derivatives. This really is not much of a problem, except for several of our earliest examples in which we prove well-known facts about even and odd integers. In
those few examples we construct proofs using other properties of number systems,
but not what we already know about evenness and oddness. This is done so that we
can study the structure of proofs in a familiar setting.
Remember
We don’t expect you to become an expert at proving theorems overnight. With
practice—studying lots of examples and exercises—the skills will come. Our goal
is to help you write and think as mathematicians do, and to present a solid foundation in material that is useful in advanced courses. We hope you enjoy it.
Douglas D. Smith
Richard St. Andre
CHAPTER
1
Logic and Proofs
We strongly recommend that you read the entire Preface to the Student before
beginning this first chapter. As described there, mathematics
is concerned with
the formation of a theory (a collection of true statements called theorems) that
describes patterns or relationships among quantities and structures. It is characterized by deductive reasoning, in which one uses logic to develop and extend
a theory. A proof of a theorem is a justification (a deduction) of the truth of the
theorem. A proof is obtained by drawing conclusions based on statements initially
accepted as true (the axioms) and statements previously proved. How one puts
together a sequence of statements to build an acceptable justification is at the heart
of this text. Thus, this chapter begins with the basic logic underlying proofs. It
introduces the essential methods used to construct correct proofs.
Writing a proof of a theorem requires thorough understanding of the theorem’s
mathematical concepts. So that you will be familiar with terminology and notations
that will be used throughout this book, we recommend that you review the material
in the Appendix before beginning this chapter.
1.1
Propositions and Connectives
Our goal in this section and the next is to examine the structure of sentences used
in making logical conclusions. Most sentences, such as “z > 3” and “Earth is the
closest planet to the sun,” have a truth value. That is, they are either true or false.
We call these sentences propositions. Other sentences, such as “What time is it?”
(an interrogatory sentence) and “Look out!” (an exclamatory sentence) express
complete thoughts
but have no truth value.
DEFINITION A proposition is a sentence that has exactly one truth
value. It is either true, which we denote by T, or false, which we denote by F.
2
CHAPTER 1
Logic and Proofs
Some propositions, such as “7? = 60,” have easily determined truth values.
By contrast, it will take years to determine the truth value of the proposition “The
North Pacific right whale will be an extinct species before the year 2525.” Other
statements, such as “Euclid was left-handed,” are propositions whose truth values
may never be known.
Sentences like “She lives in New York City” and “x? = 36” are not propositions because each could be true or false depending on the person to. whom “she”
refers and what numerical value is assigned to x. We will deal with sentences like
these in Section 1.3; until then, when we say that a sentence like “x > 6” is a proposition, we are assuming that the variable x has been assigned some specific value.
The statement “This sentence is false” is not a proposition because it is neither
true nor false. It is an example of a paradox—a situation in which, from premises
that look reasonable, one uses apparently acceptable reasoning to derive a conclusion that seems to be contradictory. If the statement “This sentence is false” is true,
then by its meaning it must be false. On the other hand, if the given statement is
false, then what it claims is false, so it must be true. The study of paradoxes such
as this has played a key role in the development of modern mathematical logic.
A famous example of a paradox formulated in 1901 by Bertand Russell” is discussed in Section 2.1.
By applying logical connectives to propositions, we can form new propositions.
3
=f
ig
Ae
Tiucze so
DEFINITION
Thé negation ifa proposition P, denoted’XP yis the proposition Cot P? The proposition ~P is true exactly when P is false.
The truth value of the negation of a proposition is the opposite of the truth
value of the proposition. For example, the negation of the false proposition “7 is
divisible by 2” is the true statement “It is not the case that 7 is divisible by 2,” or
“7 is not divisible by 2.”
DEFINI
N_
Given propositions
P and Q, the conjunction of P and Q,
Qis true exactly whenothy
epepo kien CP and
-_ o P 4
ee
and Q are true>)
(P
The English words but, while, and although are usually translated symbolically
with the conjunction connective, because they have the same effect on truth value
as and.
wd
Examples.
Let C be the proposition “19 is composite” and M be “45 is a multiple of 3.” Then C is false and M is true. Thus the proposition C A M is false.
* Bertrand Russell (1872-1970) was a British philosopher, mathematician, and advocate for social
reform. He was a strong voice for precision and clarity of arguments in mathematics and logic. He
coauthored Principia Mathematica (1910-1913), a monumental effort to derive all of mathematics from
a specific set of axioms and well-defined rules of inference.
1.1
Propositions and Connectives
3
We read C A M as “19 is composite and 45 is a multiple of 3.” On the other hand,
if we let C be “Copenhagen is the capital of Denmark” and M be “Madrid is the
capital of Spain,” then the statement “Copenhagen is the capital of Denmark
while Madrid is the capital of Spain” is a true proposition with the same symbolic
form, C A M.
Oo
The examples above illustrate an important distinction between a statement
and the form of a statement. The form P A Q itself has no truth value. Only when
the components P and Q are assigned to be specific propositions does P \ Q have
the value T or F. Those combinations of truth values for P and Q that yield true and
those that yield false can be displayed in a truth table for P A Q.
P
Q
PASO
a
i
T
F
ey
i
iF
F
tT
F
IF
Ie
DEFINITION — Given propositions P and Q, the@isjunction of P and Q,
denoted P v Q, is the proposition CP or OOP
PVVvQis trueexactly when at
i“least one ot P or Q is true.
The truth table for P V O is
Ip
Q
ae)
AL
IF
il
lel
ap
T
iE
ls)
ay
W
le
le
Example.
IfR is the proposition “12 is a prime number” and S is “16 is an integer power of 2,” we know Ris false and S is true. Thus, “12 is a prime number or
16 is an integer power of 2,” which has the form R v S, is true. The false proposition “Either 12 is a prime number or 161 1snotan integer power of 2” has the form
RV AS.
——
_
o
The statement “Either 7 is prime and 9 is even, or else 11 is not less than 3”
may be symbolized by (P A Q) V ~R, where P is “7 is prime,” Q is “9 is even,”
and R is “11 is less than 3.” Because the propositional form (P A Q) V~R has three
components (P, Q, and R), it follows that there are 2* = 8 possible combinations
of truth values in its truth table. The two main components are P A Q and ~R. We
make truth tables for these and combine them by using the truth table for Vv.
4
CHAPTER 1
Logic and Proofs
P
O
R
PAO
~R
(PA Q)V~R
7
F
rT
F
a
F
ii
F
u
i
F
F
Tl
7
F
F
A
T
7
T
F
F
F
F
‘i
F
F
F
Hf
F
F
F
F
F
F
F
ak
T
8
T
z
F
F
F
T
ag
HE
T
In practice, we don’t make a complete truth table to determine the truth value
of a specific statement such as “Either 7 is prime and 9 is even, or else 11 is not less
than 3.” We can conclude that this statement is true because
We know P is true, Q is false, and R is false.
Therefore, P A Q is false and ~R is true.
Thus (P A Q) V ~Riis true.
The reasoning here follows the steps necessary to create line 7 of the table.
Some compound forms always yield the value true (or false) just because of
the way they are formed.
DEFINITIONS _ A tautology is a propositional form thatis true for every
assignment of truth values to its components.
a
a
ee
~A contradiction is a propositional form that is false for every assignment
of truth valuesto its components.
The Law of Excluded Middle,
art
:
P V ~P, is an example of a tautology because
P \ ~P is true when P is true and true when P is false. We know that statements like
“20341 is a prime number or 20341 is not a prime number” and
“The absolute value function is continuous or it is not continuous”
must be true because both have the form of this tautology.
Example.
Proof.
Prove that (P Vv Q) v (~P A ~Q) is a tautology.
The truth table for this propositional form is
Q
a
i
F
|
domdd
F
ENO)
ma44
~P
~Q
~PK~O
4mm 4407
sles
ole
ie
(PV
ON
(APA)
Haas
Because the last column is all true, (P V Q) v (~P A ~Q) is a tautology.
|
1.1
Propositions and Connectives
5
Both ~(P v ~P) and Q A ~Q are examples of contradictions. The negation of
a contradiction is, of course, a tautology.
Particularly important in writing proofs will be the ability to recognize or write
a statement equivalent to another. Sometimes, knowledge of the mathematical content enables us to write an equivalent statement. For instance, if one step in a proof
is the statement “The ones digit of the integer x is zero,” a later step could be the
equivalent statement “The integer x is divisible by 10.”
In other cases, the meaning of a statement does not come into play; it is the
form
of the statement that may be used to find a useful equivalent. We say two
propositional forms are equivalent if they have the same truth tables.
Some of the most commonly used equivalences are presented in the follow- ing theorem. You may wish to make truth tables for each pair of forms to verify
that they are equivalent, but in each case you should understand the equivalences
by examining their meanings. For example, in part (h), negation is applied to a
conjunction. The form ~(P A Q) is true precisely when P A Q is false. This happens
when one of P or Q is false, or, in other words, when one of ~P or ~Q is true. Thus,
~(P A Q) is equivalent to ~P V ~@Q. That is, to say “We don’t have both P and Q”
is the same as saying “We don’t have P or we don’t have Q.”
All parts of Theorem 1.1.1 may be verified by constructing truth tables for each
pair of propositional forms. (See Exercise 5.)
Theorem 1.1.1
For propositions P, Q, and R, the following are equivalent:
(2)
(b)
(C)
(d)
(C)
(f)
ee
3P:'v0
eee NO
RV (ON ER
ea OWN)
PA(O.VR)
and
~(~P)
puavel
=O) \Y IP
gual
Ql
eae
eO)) Val
ean ian (2) OE:
and
WPA) V(PAR)
(g)
(h)
PV(QAR)
~(PAQ)
and
and
(PPVQ)A(PYVR)
~PVv~Q
ji)
~(PvQ)
and
~PA~Q
Double Negation Law
Commutative Laws
Associative Laws
Distributive Laws
DeMorgan’s Laws
Sy
“Sy
Say?
Se
As an example of how this theorem might be useful, suppose that for some
integer x we have determined that the statement “x is even and x > 10” is not true.
Then its negation,
“Tt is not the case that the integer x is even and x > 10,”
is true and has the form ~(P A Q), where P is “x is even” and Q is “x > 10.” By part
(h) of Theorem 1.1.1, this is equivalent to ~P V ~Q, which is
“Tt is not the case that x is even or it is not the case that x > 10.”
* Augustus DeMorgan (1806-1871) was an English logician and mathematician whose contributions
include his notational system for symbolic logic. He also introduced the term mathematical induction
(see Section 2.4) and developed a rigorous foundation for that proof technique.
6
CHAPTER 1
Logic and Proofs
An easier way to say this is
x is not even or x is not greater than
CC
i
10
10,
”
which may be restated as
oy iS OGdOL vi
LO.
A denial of a proposition P is any proposition equivalent to ~P. A proposition
has only one negation, ~P, but always has any denials, including ~P,—~~P,
~~~x«-~-P, etc. Some denials of “x is odd” are “x is not odd,” “ xis even,” and “x is
divisible by 2.” DeMorgan’s Laws provide other ways to construct useful denials.
Example.
A denial of “Either the defendant paid a fine or the judge declared a
mistrial” is
“The judge did not declare a mistrial and the defendant did not pay a fine.”
This can be verified by first writing the two sentences symbolically as P V J and
(~J) A (~P), respectively. Then we observe that P Vv J is equivalent to J Vv P, so
a denial of PVv J is equivalent to ~ (J V B), which we know by DeMorgan’s Laws
is equivalent to (~/) A (~P). We could also verify that the sentence is a denial
by checking that the truth tables for P V J and (~/) A (~P)) have exactly opposite
values.
Oo
Example. Suppose L, and L, are two lines in a coordinate system. Find a denial
of the statement
“L, and L, have the same slope or L, and L, are vertical lines.”
The mathematical concepts expressed determine the form of the statement. The
component “L, and L, have the same slope” cannot mean “L, has the same slope”
and “L, has the same slope.” However, “L, and L, are vertical lines” must mean
“L, 18 a vertical line” and “L, is a vertical line.” The correct symbolization is
S V (P A Q), where S is “L, and L, have the same slope,”, P is “L, is a vertical
line,” and Q is “L, is a vertical line.”
The negation of the statement is ~[S Vv (P A Q)], which is equivalent to
~S A ~(P A Q). This form, in turn, is equivalent to ~S A (~P Vv ~Q). The denial
we seek is
“L, and L, do not have the same slope, and
either L, is not a vertical line or L, is not a vertical line.”
Oo
Does someone who says, “Not P or Q” mean “Neither P nor Q” or “Either not
P or else Q”? That is, should the symbolic translation be ~(P v Q) or (~P) v Q?
The two translations are not equivalent, so the English sentence needs further
explanation. Ambiguities like this can be tolerated in casual conversation but not
in situations where precision matters—for example, in mathematics and in legal
documents. To avoid ambiguities in symbolic statements, we use parentheses ( ),
square brackets [ ], or braces { }.
1.1
Propositions and Connectives
7
Propositional forms are often written without all the parentheses you might
expect. To correctly understand such a form, use these rules:
First, ~ always is applied to the smallest proposition following it.
Then A connects the smallest propositions surrounding it.
Next, V connects the smallest propositions surrounding it.
Also, when the same connective is used two or more times in succession, parentheses are restored from the left. Thus, ~P Vv Q is an abbreviation for (~P) Vv Q,
but ~(P Vv Q) is the only way to write the negation of P V Q. Here are some other
examples:
~P V ~Q abbreviates (~P) Vv (~Q)
PV OVX Rabbreviates
PV (O A R)
PA ~QV
~R abbreviates [P A (~Q)] V (~R)
RAP AS <A Q abbreviates [((R A P) A S]
AQ
There is no requirement to leave out as many parentheses as possible. For
example, ~P \ ~R V ~P A Ris an abbreviation for [(~P) A (~R)] Vv [(~P) A R],
but for most readers the form (~P A ~R) V (~P A R) is easier to read.
Exercises
1.1
1.
*
*
Which of the following are propositions? Give the truth value of each proposition.
(a)
(b)
(c)
(d)
What time is dinner?
Itis not the case that z is not a rational number.
x/2 1s arational number.
2x + 3y is areal number.
(e)
Either 7 is rational and 17 is a prime, or 7 < 13 and 81 is a perfect
square.
Either 2 is rational and 7 is irrational, or 27 is rational.
Either 57 is rational and 4.9 is rational, or there are exactly four primes
less than 10.
—3.7 is rational, and either 37 < 10 or 37 > 15.
It is not the case that 39 is prime, or that 64 is a power of 2.
There are more than three false statements in this book, and this statement is one of them.
(f)
(g)
(h)
(i)
(j)
2.
*
For each pair of statements, determine whether the conjunction P A Q and the
disjunction P V Q are true.
(a) Pis “V2 <2” and Q is “97 is a prime number.”
(b) Pis “The moon is larger than Earth” and Q is “The prime divisors of 12
( © Pare 2and 3.7
(ey Ps 57 + 127= 137 and Os “V2 + V3 = V2 + 3.”
*
(d)
P is “France is south of Italy” and Q is “New Zealand is in Europe.”
8
CHAPTER 1
Logic and Proofs
(e)
P is “0,5, and 10 are all natural numbers” and Q is “98 has two prime
divisors.”
(f)
Pis “Hexagons have 5 sides” and Q is “V2 * 3 = DEE
Make a truth table for each of the following propositional forms.
(a)
(ec)
*+ W (el
(2)
(i)
(k)
PA ~P
PA~O
NO)
20
Oy
PaO? RK)
PAP
(b) PV ~P
(d) PROVO)
yey. OC)
Ce
Gj) PPAQDVCPAR)
() (PAQ)V(RA~S)
If P, QO, and R are true while S and K are false, which of the following are
true?+
(a) QARAS) *
(> (PV Q)A(RVS) 1
(ec) ~Pv~O |
(g)
(Pv S)A (PV
K)
(b) OVRAS)
(d)) CPM Onin
*® (“OVHAQVS)
(h)
KA~GSV
=!
Q),
Use truth tables to verify each part of Theorem 1.1.1.
Which of the following pairs of propositional forms are equivalent?
(a)
eS
Vy (OES
CRP
ae
Oe)
OVV RP AIO YR)
(bik
(dd)
CP i (=O) Be Pave?)
(WP AO)
P AO
O)WARSP
(f).
APA
O} NEE.
(O17)
Determine the propositional form and truth value for each of the following:
(a) Itis not the case that gold is not a metal.
(b)
(c)
19 and 79 are prime, but 119 is not.
Julius Caesar was born in 1492 or 1493 and died in 1776.
(d)
(e)
Perth or Panama City or Pisa is located in Europe.
Although 51 divides 153, it is neither prime nor a divisor of 409.
(f)
(g)
While the number z is greater than 3, the sum 1 + 27 is less than 8.
It is not the case that both —5 and 13 are elements of N, but 4 is in the
set of rational numbers.
Suppose P, Q, and R are propositional forms. Explain why each is true.
(a) If P is equivalent to Q, then Q is equivalent to P.
(b) If P is equivalent to Q, and Q is equivalent to R, then P is equivalent to R.
(c) If P is equivalent to Q, then ~P is equivalent to ~Q.
(d) If Q is equivalent to R, then P A Q is equivalent to P A R.
(e) If QO is equivalent to R, then P V Q is equivalent to P Vv R.
Suppose P, Q, S, and R are propositional forms, P is equivalent to Q, and S is
equivalent to R. For each pair of forms, determine whether they are necessarily equivalent. If they are equivalent, explain why.
(a)
PandR
(b)
Pand~~Q
(c)
(e)
PASandQAR
~(PAS)and~Qv
~R
(d)
(f)
PVSandQvR
PAQandSAR
1.1.
10.
Propositions and Connectives
9
Use a truth table to determine whether each of the following is a tautology,
acontradiction, or neither,
-
(ao
A OMe PAe~O)
(DHE P)
(CHM CE" OPAC
PO)
(ROY
ON EIN OV MCR TRON
(e)
(QA ~P)A
ON)
NCS.
I) SAO LORY
(PEO)
~(P AR)
Give a useful denial of each statement. Assume that each variable is some
fixed object so that each statement is a proposition.
.
a)
xis a positive integer.
(b) Cleveland will win the first game or the second game.
(Gio:
(d) 641,371 is a composite integer.
(e) Roses are red and violets are blue.
(f)
K is not bounded or K is compact.
(g) Mis odd and one-to-one.
(h)
(i)
(j)
(k)
12.
The matrix M is diagonal and invertible.
The function g has a relative maximum at x = 2 or x = 4 and a relative
minimum at x = 3.
Neither z < s nor z < fis true.
R is transitive but not symmetric.
Restore parentheses to these abbreviated propositional forms.
(a) PN
AO
eS
(DONS
(C)
(d)
VASA S Es a)
ee ic O Ny
NN
~PV
OA ~~PAOVR
ee
13-14. Other logical connectives between two propositions P and Q are possible.
13.
The word or is used in two different ways in English. We have presented
the truth table for V, the inclusive or, whose meaning is “one or the other or
both.” The exclusive or, meaning “one or the other but not both” and denoted
@, has its uses in English, as in “She will marry Heckle or she will marry
Jeckle.” The “inclusive or” is much more useful in mathematics and is the
accepted meaning unless there is a statement to the contrary.
(a) Make a truth table for the “exclusive or” connective \.
(b) Show that A~B is equivalent to (A V B) A ~(A A B).
14.
“NAND” and “NOR” circuits are commonly used as a basis for flash memory
chips. A NAND B is defined to be the negation of “A and B.” A NOR B is
defined to be the negation of “A or B.”
(a)
(b)
(c)
Write truth tables for the NAND and NOR connectives.
Show that (A NAND B) v (A NOR B) is equivalent to (A NAND B).
Show that (A NAND B) A (A NOR B) is equivalent to (A NOR B).
10
1.2
CHAPTER 1
Logic and Proofs
Conditionals and Biconditionals
Sentences of the form “If P, then Q” are the most important kinds of propositions in mathematics. You have seen many examples of such statements in previous study: from precalculus, “If two lines in a plane have the same slope, then
the lines are parallel”; from trigonometry, “If sec 6 = 2, then sin@ = ae from
calculus, “If
fis differentiable at x) and f(x) is a relative minimum for f, then
Ff") = 0.”
DEFINITIONS _ For propositions P and Q, the conditional sentence? > Q
is the proposition “If P, then Q. Proposition P is called the antecedent and
Q is the consequent. The sentence P > Q is true if and only if P is false or
QO is true.
The truth table for P > Q is
The only case where
IP
Q
(P= 0)
AD
le
W
le
Ti
IW
Iz
In
a
a
Ie
T
P > Q is a false statement occurs on line 3 of its truth table,
when P is true and Q is false. This agrees with the way we understand promises.
Example.
Suppose someone makes this promise to a friend:
“Tf the weather is warm, we will go hiking.”
The antecedent is “The weather is warm” and the consequent is “We will go hiking.”
This promise would be broken if the weather turned out to be warm and the friends
did not go hiking (line 3 of the table.) In every other situation, the statement is
true. When the weather was warm and the friends went hiking (line 1 of the table),
the promise was kept. Whether the friends go hiking or not, in the event that the
weather is not warm, we wouldn’t say the promise was broken. (These are lines 2
and 4 of the table.)
Our truth table definition for
P= Q captures the same meaning for “If...,
then...” that you have always used in mathematics. For example, if we think of x
as some fixed real number, we all know that
oli
8s then
oe
is a true statement, no matter what number x we have in mind. Let’s examine why
we say this sentence is true for some specific values of x, where the antecedent P is
“x > S”-and the consequent O is “x5.”
1.2
Conditionals and Biconditionals
11
When x is a number greater than both 8 and 5 (for example, 11), both P and Q
are true, as in line | of the truth table. When x is between 5 and 8 (for example, 7),
P is false and Q is true, as in the second line of the table. When x is less than both
8 and 5 (for example, 2), we have the situation in line 4. In all three cases, P > O
is true. In fact, “If x > 8, then x > 5” is always true because there can be no case
corresponding to line 3 of the truth table. We are not claiming that either P or Q is
true. What we do say is that no matter what number we think of, if it is larger than
8, then it is also larger than 5.
One curious consequence of the truth table for P => Q is that a conditional
sentence may be true even when there is no connection between the antecedent and
the consequent. The reason for this is that the truth value of P > Q depends only on
the truth value of components P and Q, not on their interpretation. For this reason,
all of the following are true:
“If sin z = 1, then 6 is prime.” (line 4 of the truth table)
13 > 7 => 2
3e— 57. lines kof thetruthitablie)
“7” = 3 => Paris is the capital of France.” (line 2 of the truth table)
and both of these are false by line 3 of the truth table:
“If Saturn has rings, then (2 + 3)? = 2? + 32.”
“If 47 > 10, then | is a prime number.”
Other consequences of the truth table for P => Q are worth noting.
¢
When Pis false (lines 2 and 4), it doesn’t matter what truth value Q has:
will be true.
P=> Q
¢
When Q is true (lines | and 2), it doesn’t matter what truth value P has:
will be true.
P=> Q
e
When P and P > Q are both true (on line 1), O must also be true.
Two propositions associated with P > Q are its converse and contrapositive.
DEFINITION
Let P and Q be propositions.
The converse of P= QisO=> P.
The contrapositive of P= Q is (~Q) > (~P).
For the conditional sentence “If z is an integer, then 14 is even,” the converse
of the sentence is “If 14 is even, then z is an integer” and the contrapositive is “If
14 is not even, then z is not an integer.” The sentence and its contrapositive are
true, but the converse is false.
For the sentence “If 1+ 1= 2, then V10 > 3,” the converse and
contrapositive are, respectively, “If V/10 > 3, then 1 + 1 = 2” and “If V10 is
not greater than 3, then 1 + 1 is not equal to 2.” In this example, all three sentences are true.
These two examples show that a conditional sentence and its converse are
not always equivalent. Thus, the truth value of P = Q cannot be inferred from its
12
CHAPTER 1
Logic and Proofs
converse Q => P. However, a statement and its contrapositive are equivalent, as the
following theorem shows.
Theorem 1.2.1
For propositions P and Q, P > Q is equivalent to its contrapositive (~Q) = (~P).
Proof.
The proof is carried out by examination of the truth table.
POO
9) Oe
ogee
ee ee Cr
are)
ea
ipeers
Hi ame
Eaperr
0
T
F
i
F
‘i
F
if
EF
F
44
ai
i
‘
F
i
P => Q is equivalent to (~Q)
identical to the sixth column.
> (~P) because the third column in the truth table is
|
The biconditional connective, defined next, is symbolized with a double arrow
<>, which reminds one of both <— and >. This is no accident because P +> Q is
equivalent to (P > Q) A (Q=> P).
DEFINITION — For propositionsP andQ, the biconditional sentence
CP <> Qis the proposition (P if and only if Q.”)The sentence P <> Q is true
exactly when P and Q have the same truth values.
Mathematicians often abbreviate “P if and only if Q” as “P iff Q.” The truth table
for P< Q is
i
|
ay
A
Sh
Sh
sah
nal
1S)
Examples. The proposition “2? = 8 iff 49 is a perfect square” is true because
both components are true. The proposition “7 = 22/7 if and only if V2 is a rational
number” is also true. The proposition “6 + 1 = 7 iff Argentina is north of the equator” is false because the truth values of the components differ.
Oo
Definitions are important examples of biconditional sentences because they
describe exactly the condition(s) needed to satisfy the definition. Be aware that
definitions in mathematics, however, are not like definitions in ordinary English,
which are based on how words are typically used. For example, for a period of at
most a few dozen years, the standard meaning of the word wireless was a broadcast
radio receiver. Definitions in mathematics have precise meanings that stay fixed
1.2
Conditionals and Biconditionals
13
over time. The definition of an “odd” integer in the Appendix tells you exactly what
that word always means. You may form a helpful mental image or concept, but the
idea that an odd integer ends in 1, 3, 5, 7, or 9 is a consequence of the definition,
not the definition.
Definitions may be stated with the “if and only if” wording, but it’s also common practice to state a formal definition using the word “if.” For example, we could
say that “A function
fis continuous at a number a if...,” leaving the “only if” part
understood. Either way it’s worded, biconditionality provides the test of whether a
statement could serve as a definition or is just a description.
Example. The statement “Horizontal lines have slope 0” could be used as a definition, because a line is horizontal if and only if its slope is 0. However, “A quadratic
function is a polynomial” is not a definition, because the sentence “A function is
quadratic if and only if it is a polynomial” is false.
Because the biconditional sentence P <> Q is true exactly when the truth values of P and Q agree, the2 propositional forms P and Q are equivalent precisely
when P => Qisa tautology. This means all of the statements in Theorem 1.1.1 may
be restated using the <> connective. For example, the first of DeMorgan’s Laws
(Theorem 1.1.1(h)) may be written ~ (P A Q) & (~P V ~Q).
The next theorem contains several additional important pairs of equivalent propositional forms involving implication. They will be used often to construct proofs.
Theorem 1.2.2
For propositions P, Q, and R, the following are equivalent:
(ayaeP=—0
and
~PVQ
(bD)
eee
ain wan (loa)
Nn Opal)
(c)
P=)
and
PA~@Q
(d)
~(PAQ)
and
P>~@Q
(e)
~(P A Q)
and aan Ow?
(f)
P=>(O=Ree ands
(SO)
(9))
P]S(O7 RK) = and
OO
h)
(h) Oe G2
40S RR and
aeRO)
Exercise 8 asks you to prove each part of Theorem 1.2.2. The natural way to
proceed is by constructing and then comparing truth tables, but you should also think
about the meaning of both sides of each statement of equivalence. With part (a), for
example, we reason as follows: P > Q is false exactly when P is true and Q is false,
which happens exactly when both ~P and @Q are false. Since this happens exactly
when ~P V Qis false, the truth tables for P= Q and ~P Vv Q are identical.
Note that many of the statements in Theorems 1.1.1 and 1.2.2 are related. For
example, once we have established Theorems
1.1.1 and 1.2.2(a), we reason that
part (c) is correct as follows:
~(P => Q) is equivalent, by part (a), to
~(~P V Q), which is equivalent, by Theorem 1.1.1(i), to
~(~P) A ~Q, which is equivalent, by Theorem 1.1.1(a), to
PK
~@.
14
CHAPTER 1
Logic and Proofs
Recognizing the structure of a sentence and translating the sentence into symbolic form using logical connectives are aids in determining its truth value. The
translation of sentences into propositional symbols is sometimes very complicated
because some natural languages (such as English) are rich and powerful with many
nuances. The ambiguities that we tolerate in English would destroy structure and
usefulness if we allowed them in mathematics.
Even the translations of simple sentences can present special problems.
Suppose a teacher says to a student,
“If you score 74% or higher on the next test, you will pass this course.”
This sentence clearly has the form of a conditional sentence, although almost everyone will interpret the meaning as a biconditional.
Contrast this with the situation in mathematics where “If x = 2, then x is a
solution to x7 = 2x” must have only the meaning of the connective =, because
x? = 2x does not imply that x is 2.
Here are some phrases in English that are ordinarily translated using the conNectives => Or <=.
Use P = Q to interpret:
li, then Q.
P is sufficient for Q.
P only if Q.
Ova P,
Q whenever P.
Q is necessary for P.
Use P <> Q to interpret:
- P if and only if Q.
. P if, but only if, Q.
= __-P'is equivalent to Q.
:
/ P is necessary and sufficient for Q.
\__ P implies Q, and conversely,
QO, when P.
The word unless is one of those connective words in English that poses special
problems because it has so many different interpretations. See Exercise 11.
Examples. Translate each of these statements into symbols. Think of a as a fixed
real number.
Statement:
In symbols:
a > 5 is sufficient for a > 3.
a > 3 is necessary fora > 5.
G.>.S.only it a = 3.
|a| = —a whenevera < 0.
ip
= a= 3
GfSei
— (0a)
Oe
=d > 3
a0 = )\a|\= =a
|a| = 2 is necessary and sufficient for a* = 4.
d=2Sa7=4
oO
It is not always necessary to know the meaning of all the words in a statement
to determine a correct translation. When you see “S is compact is sufficient for
S to be bounded,” you understand that the interpretation is “S is compact = S is
bounded,” even if you don’t know what compact and bounded mean.
There will be more than one way to translate a sentence symbolically. For
example, if we let C denote the proposition “S is compact” and B denote the
1.2
Conditionals and Biconditionals
15
proposition “S is bounded,” the statement “If S is compact, then S is bounded” may
be translated as C > B or as ~B > ~C or as ~C Vv B, because all these forms are
equivalent.
On the other hand, the sentence “17 and 35 have no common divisors” shows
that the meaning of words must be considered. The translation “17 has no common
divisors A 35 has no common divisors” makes no mathematical sense. Compare
this to the proposition “17 and 35 have digits totaling 8,” which can be written as
a conjunction.
Example.
Suppose b is a fixed real number. The form of the sentence “If b is an
integer, then b is either even or odd” is P => (Q v R), where P is “b is an integer,”
QO is “b is even,” and R is “b is odd.”
Example.
oO
Suppose a, b, and p are fixed integers. “If p is a prime number that
divides ab, then p divides a or b” has the form (P A Q) > (R V S), where P is “p
is a prime,” Q is “p divides ab,” R is “p divides a,” and S is “p divides b.”
o
The rules presented at the end of the previous section that allow us to sometimes reduce the number of, or restore omitted parentheses to, a propositional form
can be extended to the connectives = and <>:
The connectives ~, A, V, =>, and <> are always applied in the order listed.
Thus, ~ applies to the smallest possible proposition, then A is applied with the
next smallest scope, and so forth. For example,
P>~QvVR#
S isan abbreviation for (P > [(~Q) V R]) S S,
PV ~O —& R='S is an abbreviationtor [Pv.(~Q)]
= R=
S),
and
P= Q=
Ris an abbreviation for (P > Q) > R.
Whenever you abbreviate a form by eliminating some parentheses, be sure to leave
enough to make the form easy to read.
Exercises 1.2
1.
Identify the antecedent and the consequent for each of the following conditional sentences. Assume that a, b, and f represent some fixed sequence,
integer, or function, respectively.
x ((a) If squares have three sides, then triangles have four sides.
“(by If the moon is made of cheese, then 8 is an irrational number.
(c)) b divides 3 only if b divides 9.
*
(d)
The differentiability of fis sufficient for fto be continuous.
16
CHAPTER 1
Logic and Proofs
(e)
A sequence a is bounded whenever a is convergent.
((f))
A function fis bounded if fis integrable.
(g)
1+2
((h)
The fish bite only when the moon is full.
= 3 is necessary for 1 -- 1 = 2:
(i)
A time of 3 minutes, 48 seconds or less is necessary to qualify for the
Olympic team.
Write the converse
Exercise 1:
and contrapositive
of each conditional
sentence
in
What can be said about the truth value of Q when
(a) PisfalseandP=>Qistrue?
(b)
Pis true and
P= Qis true?
(c) PistrueandP=>Qisfalse?
(d)
Pis false and P <> Q is true?
(e) Pistrue and P = Q is false?
Identify the antecedent and the consequent for each conditional sentence in
the following statements from this book.
(a)
(c)
(e)
Exercise 3 of Section 1.6
The PMI, Section 2.4
Theorem 4.7.2
(b)
(d)
(f)
Theorem 2.1.1(c)
Theorem 3.3.1
Corollary 5.3.6
Which of the following conditional sentences are true?
((a) If triangles have three sides, then squares have four sides.
(b) If hexagons have six sides, then the moon is made of cheese.
* (©) If7 +6 = 14, then5 +5 = 10.
\
(d)
The Nile River flows east only if 64 is a perfect square.
(e)
Earth
=
North Sea.
has
one
moon
only
if the
Amazon
River
flows
into
the
((f)) If Euclid’s birthday was April 2, then rectangles have four sides.
(g)
5 is prime if \V/2 is not irrational.
(h)
1+ 1 = 2 is sufficient for 3 > 6.
6. Which of the following are true? Assume that x and y are fixed real numbers.
* ((a) Triangles have three sides iff squares have four sides.
(b)
745 = 12 if and only if 1 + 1 = 2.
((c) D5 eOy==00) 4-0 If) + 1 = 10.
((d)) A parallelogram has three sides iff 27 is prime.
(e)
The Eiffel Tower is in Paris if and only if the chemical
symbol for
helium is H.
(Dy
aay
OS
1
ey
N12 if V1
i
0,
(g) x? > O if and only if x > 0.
(h) x* — y = Oiff (x — yx + y) = 0.
(i) x* + y* = 50 if and only if (« + y)? = 50.
7.
Make truth tables for these propositional forms.
(ayer = (OP),
* (c)
(e)
O
*(b)
(~PSO0)VQP).
~QO=>(Q = P).
(PAO)V(QAR)S>PVR.
(d)
(PV Q)>(P AQ).
(CSS)
oO)
SiR):
A0=]
Rial
1.2
Conditionals and Biconditionals
17
Prove Theorem 1.2.2 by constructing truth tables for each equivalence.
Determine whether each statement qualifies as a definition.
(a)
(b)
(c)
y = f(x) isa linear function if its graph is a straight line.
y = f(x) is a quadratic function when it contains an x? term.
A quadrilateral is a square when all its sides have equal length.
(d) A triangle is a right triangle if the sum of two of its interior angles is 90°.
(e) Two lines are parallel when their slopes are the same number.
(f) A quadrilateral is a rectangle if all its interior angles are equal.
10.
Rewrite each of the following sentences using logical connectives. Assume
that each symbol f, x, 1, x, B represents some fixed object.
(a) If fhas a relative minimum at x, and if f is differentiable at x,, then
f'(X) =
0!
(b)
Ifnis prime, then n = 2 orn
(c)
Ris symmetric and transitive whenever R is irreflexive.
is odd.
(d)
(e)
(f)
(g)
B is square and not invertible whenever det B = 0.
fhas a critical point at x, iff f’(x)) = 0 or f’(xq) does not exist.
2<n-—6isa
necessary condition for 2n < 4 orn > 4.
6>n-—3onlyifn
> 4o0rn > 10.
(h)
xis Cauchy implies x is convergent.
(i)
fis continuous at x) whenever
(j)
If fis differentiable at x9 and fis increasing at xo, then f’(x9) > 0.
lim f(x) = f(x9).
5 8G)
11.
Dictionaries indicate that the conditional meaning of unless is preferred, but
there are other interpretations as a converse or a biconditional. Discuss the
translation of each sentence.
(a) I will go to the store unless it is raining.
(b) The Dolphins will not make the playoffs unless the Bears lose all the rest
of their games.
(c) You cannot go to the game unless you do your homework first.
(d) You won’t win the lottery unless you buy a ticket.
12.
Show that the following pairs of statements are equivalent.
(a)! BCP 0)
FB and ~R
(PO).
(b> PAO)
Rand (2k) =O;
(k= (O07 Band HONK) SP:
(ie (Chandi eye hk).
(e)) (P> Q)=>
Rand (PA ~Q) VR.
\@/) P <= Qand(~P v Q) A (~Q V P).
13.
14.
Give, if possible, an example of a true conditional sentence for which
(a)
the converse is true.
(b)
the converse is false.
(c)
the contrapositive is false.
(d)
the contrapositive is true.
Give, if possible, an example of a false conditional sentence for which
(a) the converse is true.
(b) the converse is false.
(c) the contrapositive is false.
(d)
the contrapositive is true.
18
CHAPTER 1
Logic and Proofs
15.
Give the converse and contrapositive of each sentence of Exercises 10(a), (b),
(f) and (g). Decide whether each converse and contrapositive is true or false.
16. Determine whether each of the following is a tautology, a contradiction, or neither.
re ee
(0) — iel —
(D) a PSP
(EaGO);
ORS Pra:
(c) Wh
(dd) oh (Bee =O):
(e) PAQV~OSP.
() [QAP=>QO|>P.
(2)
PSO
~(~PV Q)V (~P AQ).
(Om
sh) LP S(O
1, 2)
CU
QO) Faw.
G)
(k)
(ie
17.
(PV Q>O>P.
[P>(QAR)] >[R>(P>Q)].
OU wi
a
)),
The inverse, or opposite, of the conditional sentence
P= Q is ~P > ~@.
(a)
Show that
(b)
For what values of the propositions P and Q are P => Q and its inverse
both true?
Which is equivalent to the converse of a conditional sentence, the contrapositive of its inverse, or the inverse of its contrapositive?
(c)
1.3
BR) N (RP).
P > Q and its inverse are not equivalent forms.
Quantified Statements
Unless there has been prior agreement about the value of x, the statement “x > 3” is
not a proposition because it is neither true nor false. A sentence that contains variables
is called an open sentence or predicate and becomes a|proposition only when its vari‘ables are assigned specific values. For example, “x > 3” is true when x is given the
value 7 and false when x is 2.
When P is an open sentence with a variable x, the sentence is symbolized by
P(x). If P has n variables x, x5,...,.x,, we write pear ieee anes x,). For example,
if P(x, y, z) represents the open sentence “x + y = z*,” then P(4, 5, 3) is the true
proposition 4 + 5 = 3°, while P(1, 2, 4) is the false proposition | + 2 = 4?.
The collection of objects that may be substituted to make an open sentence a
true proposition is called the truth set of the sentence. Before a truth set can ‘be
determined, we must be given or must decide what objects are available for consideration; that is, we must have specified a universe of discourse. In many cases the
universe will be understood from the context. For the sentence ‘“‘x likes chocolate,”
the universe is presumably the set of all people. We will often use the number systems N, Z, @, R, and C as our universes. (See the Appendix.)
Example. The truth set of the open sentence “x” < 5” depends on the collection of
objects we choose for the universe of discourse. With the universe specified as the
1.3
Quantified Statements
19
set N, the truth set is { 1, 2}. For the universe Z, the truth set is {—2, —1, 0, 1, 2}.
When the universe is R, the truth set is the open interval (— V5, V5).
DEFINITION
With a universe specified, two open sentences P(x) and
Q(x) are equivalent if they have the same truth set.
Examples. The sentences “3x + 2 = 20” and “x = 6” are equivalent open sentences in any of the number systems named above. On the other hand, “x? = 4” and
“x = 2” are not equivalent when the universe is R. They are equivalent when the
universe is N.
Oo
Although words such as truth set, universe, and equivalent open sentence may
be unfamiliar to you, the concepts are not new. The equation (x? + 1)(x — 3) = 0
is an open sentence. Solving the equation is a matter of finding its truth set. For the
universe IR, the only solution is x = 3 and thus the truth set is {3}. But if we choose
the universe to be C, the equation may be replaced by the equivalent open sentence
(x + i)(x — 1)(x — 3) = 0, which has truth set (solutions) {3, 7, —7}.
To determine whether the sentence
“There is a prime number between 5060 and 5090”
is true in the universe N, we might try to individually examine every natural number, checking whether it is a prime and between 5060 and 5090, until we eventually
find any one of the primes 5077, 5081, and 5087 and conclude that the sentence
is true. (A quicker way is to search through a complete list of the first thousand
primes.) The key idea here is that although the open sentence “‘x is a prime number
between 5060 and 5090” is not a proposition, the sentence
“There is a number x such that x is a prime number between 5060 and 5090”
does have a truth value. This sentence is formed from the original open sentence by
applying a quantifier.
DEFINITION _ The symbol 4 is called the existential quantifier. For an
open sentence\P (x), the sentence((4.x) P(x))is read “There exists x such that
P(x)’ or “For some x, P(x).” The sentence (4.x) P(x) is true if the truth set
of P(x) is nonempty.
An open sentence P(x) does not have a truth value, but the quantified sentence
(4x) P(x) does. One way to show that (4.x) P(x) is true for a particular universe is to
identify an object a in the universe such that the proposition P(qa) is true. To show
that (4x) P(x) is false, we must show that the truth set of P(x) is empty.
20
CHAPTER 1
Logic and Proofs
Let’s examine the truth values of these statements for the universe R:
Examples.
Ele
yee)
(b)
GaG? = 0)
(c)
GINGS3h7=——1)
(d)
(Ad
> 3 Vx? =—1)
Statements (a) and (d) are true because 3, 7.02, and many other real numbers are in
the truth set of x > 3, and therefore in the truth set of x > 3 v x* = —1. Statement (b)
is true because the truth set of x* = 0 is precisely {0} and therefore is nonempty.
Because the open sentence x* = —1 is never true for real numbers, the truth set of
x >3 Ax? =-—1 is empty. Statement (c) is false in the universe R.
All four statements are false in the universe {1, 2}, and in the universe {—3, 0}
only statement (b) is true.
Sometimes we can say (4x) P(x) is true even when we do not know a specific
object in the universe in the truth set of P(x), only that there (at least) is one.
Example.
Show that (Sx)(x? — 12x° + 16x — 3 = 0) is true in the universe of
real numbers.
For the polynomial f(x) = x’ — 12x* + 16x — 3, we see that f(0) = —3 and
fC.) = 2. From calculus, we know that fis continuous on [0, 1]. The Intermediate Value Theorem tells us there is a zero for fbetween 0 and 1. Even if we don’t
know the exact value of the zero, we know it exists. Therefore, the truth set of
x’ — 12x? + 16x — 3 = 0 is nonempty. Hence (Sx)(x’ — 12x73 + 16x — 3 = 0)
is true.
-
o
The sentence “Every number x is greater than 0” needs a different quantifier
because it is not enough to find at least one value for x for which “x > 0” is true.
The open sentence “x > 0” must always be true—that is, true for every object in
the universe. The sentence “Every x is greater than 0” is true when the universe is
N but is false when the universe is the integers.
DEFINITION
The symbol V \s called the universal quantifier. For an
open sentence P(x), the sentence (Vx) P(x) is read “For all x, P(x)” or “For
every x, P(x).” The sentence (Vx) P(x) is true if the truth set of P(x) is the
entire universe.
Examples.
In the universe of natural numbers, the sentences (Vx)(x + 2 > 1) and
(Vx)(2x is an integer) are true. However, the sentence (Vx)(x + 2 > 1) is false in
the universe of real numbers because —5 + 2 > 1 is false, and (Vx)(2x is an integer)
is false in IR because 0.6 is not in the truth set.
The sentence (Vx)(2x + 1 < 6) is false in N because 3 is not in the truth set,
and (Wx)(2x — 1 > x) is false in N because 1 is not in the truth set. The two sentences are false in the universe of real numbers for the same reasons.
1.3
Quantified Statements
21
Some universally quantified sentences that are true in R are
Wx)
= 10 orn = 0 or me10), V2
0))
andeWn@-+
2 > x).
Oo
There are many ways to express a quantified sentence in English. Look for key
words such as for all, for every, for each, and similar words that require universal
quantifiers. Look for phrases such as some, at least one, there exist(s), there is
(are), and others that indicate existential quantifiers.
You should also be alert for hidden quantifiers because natural languages allow for
imprecise quantified statements where the words for all and there exists are not present.
Someone who says “Polynomial functions are continuous” means that “All polynomial
functions are continuous,” but someone who says “Rational functions have vertical
asymptotes” must mean “Some rational functions have vertical asymptotes.”
How should the sentence “All apples have spots” be written in symbolic
form? If we limit the universe to just apples, a correct symbolization would be
(Vx)(x has spots). But if the universe is all fruits, we need to be more careful. Let
A(x) be “x is an apple” and S(x) be “‘x has spots.” Should we write the sentence as
(Vx)[A(x) A S(x)] or (Vx)[A(@x) => S(x)]?
The first quantified form, (Vx)[A(x) A S(x)], says “For all objects x in the universe, x is an apple andx has spots.” Since we don’t really intend to say that all fruits are
spotted apples, this is not the meaning we want. Our other choice, (Vx)[A(x)
> S(x)],
is the correct one because it says “For all objects x in the universe, if x is an apple,
then x has spots.” In other words, “If a fruit is an apple, then it has spots.”
Should the symbolic translation of “Some apples have spots” be
(Ax) [A@) A S(x)] or (Sx) [A(@) > S(x)]? The first form says “There is an object
x such that it is an apple and it has spots,” which is correct. On the other hand,
(Sx)[A(x) > S(x)] reads “There is an object x such that, if it is an apple, then it
has spots,” which does not ensure the existence of apples with spots. The sentence
(4x)[A(x) > S(x)] is true in every universe for which there is an object x such that
either x is not an apple or x has spots, which is not the meaning we want.
In general,
“All P(x) are Q(x)” should be symbolized (Vx)(P(x) > Q(x),
and
“Some P(x) are Q(x)” should be symbolized (4x)(P(x) A Q(x)).
Here are several examples.
Examples.
(a)
Translate each of these sentences using quantifiers.
“For every odd prime x less than 10, x° + 4 is prime.”
The sentence means that if x is prime, and odd, and less than 10, then x7 +4
is prime. It is written symbolically as
(Vx)(x is prime A xis odd A x < 10=
(b)
x? + 4 is prime).
“Some functions defined at 0 are not continuous at 0.”
22
CHAPTER 1
Logic and Proofs
This is translated as
(Af )(f is defined at 0 A f is not continuous at 0).
(c)
“Some real numbers have a multiplicative inverse.”
This statement could be symbolized
4x)(x is areal number A x has a real multiplicative inverse).
—
However, ‘“‘x has an inverse” means there is some number that is an inverse for x
(hidden quantifier), so a more complete symbolic translation is
(4x)[x is a real number A (Sy)(y is areal number A xy =
1)].
One correct translation of “Some integers are even and some integers are odd” is
dx)(x is even) A (4x)(x is odd)
—
because the first quantifier (4x) extends only as far as “even.” After that, any variable (even x again) may be used to express “Some integers are odd.” It would be
equally correct and sometimes preferable to write
(4x)(x is even) A (4y)(y is odd),
but it would be incorrect to write
(4x)(* is even A x is odd),
because there is no integer that is both even and odd.
Definitions in their symbolic forms often use multiple quantifiers. For example,
the definition of a rational number may be symbolized as follows:
ris arational number if (Sp)(Sq)(p EZAGELAGFVAr=
r).
Statements of the form “Every element of the set A has the property P” and
“Some element of the set A has property P” occur so frequently that abbreviated
symbolic forms are used:
“Every element of the set Ahas the property P” may be restated as “If x € A,
then...” and symbolized by
we
(Vx'e A) P(x).
“Some element of the set A has property P” is abbreviated by
(Axe) P(x).
Thus the definition of a rational number given above may be written as
ris a rational number if (4p € Z)\(4q € Z)(q#OAr = r).
1.3
Quantified Statements
23
Example. The statement “For every rational number there is a larger integer” may
be symbolized by
(Vx)[x € Q => (Azz € Zand z > x)}
or
(¥xe.O)(E742 Z)@ Sex):
As was noted with propositional forms, it is necessary to make a distinction
between a quantified sentence and its logical form. With the universe all integers,
the sentence “All integers are odd” is an instance of the logical form (Vx) P(x),
where P(x) is “x is odd.” The form itself, (Vx) P(x), is neither true nor false, but
becomes false when “‘x is odd” is substituted for P(x) and the universe is all integers.
DEFINITION
Two quantified sentences are equivalent in a given
universe if they have the same truth value in that universe. Two quantified
sentences are equivalent if they are equivalent in every universe.
Example.
(Vx)(x > 3) and (Vx)(x > 4) are equivalent in the universe of integers
(because both are false), in the universe of natural numbers greater than 10 (because
both are true), and in many other universes. However, if we choose the universe
W torbe thesintervall (3:7, co), them y(Vx) Ces 3) Misi truewanda(V 50) (Conen-b mis stalse
in U. The sentences are not equivalent in this universe, so they are not equivalent
sentences.
Oo
We can construct equivalent quantified statements using the theorems in
Sections 1.1 and 1.2. For example, the statement (Vx)(P(x) => Q(x)) is equivalent
to (Vx)(~Q(x) > ~P(x)) by Theorem 1.2.1(a). The two equivalences in the next
theorem are essential for building proofs that involve quantifiers.
Theorem 1.3.1
If A(x) is an open sentence with variable x, then
(a)
(b)
~(Vx)A (x) is equivalent to (Sx) ~A(2).
~(4x)A(x) is equivalent to (Vx) ~A(x).
Proof.
(a)
Let Ube any universe.
The sentence ~(Vx)A
(x) is true in U
iff (Vx)A (x) is false in U
iff the truth set of A(x) is not the universe
iff the truth set of ~ A(x) is nonempty
iff (4x) ~A() is true in U.
(b)
The proof of this part is Exercise 7.
=
24
CHAPTER 1
Logic and Proofs
Theorem 1.3.1 is helpful for finding a useful denial (that is, a simplified form
negation) of a quantified sentence. In the universe of natural numbers, the
the
of
sentence “All primes are odd” is symbolized (Vx)(x is prime = x is odd). The
negation is ~(Vx)(x is prime > x is odd). When we apply Theorem 1.3.1(a), this
becomes (4x)[~(x is prime > x is odd)]. By Theorem 1.2.2(c) this is equivalent
to (4x)[x is prime A ~(x is odd)]. We read this last statement as “There exists a
number that is prime and is not odd” or “Some prime number is even.”
Example. For the universe of all real numbers, find a denial of “Every positive
real number has a multiplicative inverse.”
The sentence is symbolized (Vx)[x > 0 => (Sy)(xy =
1)]. The negation and
successively rewritten equivalents are
=Vx)ix0=
Eyer =1)]
(4x) ~[x > 0 = (yay = 1]
(Ax)[x > 0A ~(y)ay = DI]
(dx)[x > 0 A (Wy) ~@y = 1)]
(Ax)[x > 0A (Wy)ay ¥ DJ
This last sentence may be translated as “There is a positive real number that has no
multiplicative inverse.”
Oo
Example. For the universe of living things, find a denial of “Some children do not
like clowns.”
The sentence is (4x) [x is achild A (Vy)(y is
aclown => x does not like y)]. Its
negation and several equivalents are
~ (4x) [xis a child A (Vy)(y is a clown > x does not like y)]
(Vx) ~[x is a child A (Vy)(y is a clown => x does not like y)]
(Vx) [xis a child => ~(Vy)(y is a clown => x does not like y)]
(Vx) [xis a child = (dy) ~(y is a clown => x does not like y)]
(Vx) [xis a child = (dy)(yis aclown
A ~ x does not like y)]
(Vx) [xis a child = (Sy)(y is aclown A x likesy)]
The denial we seek is “Every child has some clown that he/she likes.”
Oo
Example.
To find a simplified denial of (Vx)(4y)(Az)(Vu)(Az)(x + y +z>2u+v),
we begin with its negation and apply Theorem 1.3.1 five times in succession, working inward from the outermost quantifier (Vx). Each use of the theorem moves the
negation symbol across a quantifier and changes that quantifier to another, and the
1.3
Quantified Statements
25
last use also negates the open sentence with five variables. The result is the simplified form
(Ax)(Vy)(¥z)(suyVv)x + y +z < 2u + v).
oO
Example.
The symbolic form of “All Australians play soccer” is (Vx € A)P(x),
where A is the set of all Australians and P(x) is “x plays soccer.” Determine whether
(Ax)@ € A A ~P(x))
is a symbolic form of a denial of the sentence.
Begin by listing several denials of (Vx © A)P(x). Start with its negation and its
equivalent obtained by using Theorem 1.3.1(a). Then add to the list the unabbreviated form of the denial and some of its equivalents derived from Theorem 1.1.2.
Thus we find these denials:
~(Vx € A)P(X)
(dx € A) ~P(x)
~(Vx)\(x
~(Vx)(x
Gx)
6 A= Pi(x))
(Ax)\a« € A A ~P(@))
€ A> P(r))
€ A V P(x))
To determine whether (4x)(x
¢ A A ~P(x)) is a denial, make a second list
consisting of this sentence and some of its equivalents:
Goin
Ga evAn Vee)
~(Yx)\(oS AV
PG)
We suspect that (4x) (x € A A ~P(x)) is not a denial because we haven’t found
any logically equivalent form from one list in the other list. We can be sure it’s not
a denial by examining the two forms in blue. The form (4x) (x € A A ~P(x)) says
“Some Australian does not play soccer.” However, the form (ax) (x € A A ~P(x))
says “Some non-Australian does not play soccer.” These certainly have different
meanings.
Oo
We sometimes hear statements like the complaint one fan had after a great Little
League baseball game. “The game was fine,” he said, “but everybody didn’t get to
play.” We easily understand that the fan did not mean this literally, because otherwise
there would have been no game. The meaning we understand is “Not everyone got to
play” or “Some team members did not play.” Such misuse of quantifiers, while tolerated in casual conversations, is always to be avoided in mathematics.
A special case of the existential quantifier is defined next.
DEFINITION
The symbol 3!is called the unique existential quanti-
fier. For an open sentence P(x), the sentence (4!x) P(x) is read “There is a
Q unique x such that P(x).” The sentence (4!x)P(x) is true if the truth set of
P(x) has exactly one element.
=
26
CHAPTER 1
Logic and Proofs
Recall that for (4x) P(x) to be true it is unimportant how many elements are
in the truth set of P(x), as long as there is at least one. For (4!) P(x) to be true, the
number of elements in the truth set of P(x) is crucial—there must be exactly one.
In the universe N, (4!x) (x is even and x is prime) is true because the truth set of
“x is even and x is prime” contains only the number 2. The sentence (Alx)? = 4)
is true in the universe N, but false in Z.
Theorem 1.3.2
If A(x) is an open sentence with variable x, then
(a)
Gly) AWSXEnAG)
(b)
(A!x)A(x) is equivalent to (Ax)A(x) A (Vy)(Vz)[AQ) A A(z) > y = ZI.
Part (a) of Theorem 1.3.2 says that 4! is indeed a special case of the quantifier
+. Part (b) says that “There exists a unique x such that A(x)” is equivalent to “There
is an x such that A(x) and if both A(y) and A(z), then y = z.” The proofs are left to
Bxercise: Ly.
Exercises 1.3
1..
Translate the following English sentences into symbolic sentences with quantifiers. The universe for each is given in parentheses.
*
(a)
Not all precious stones are beautiful. (All stones)
*
(b)
(c)
All precious stones are not beautiful. (All stones)
Some isosceles triangle is a right triangle. (All triangles)
(d)
No right triangle is isosceles. (All triangles)
(e)
(f)
(g)
(h)
Every triangle that is not isosceles is a right triangle.
All people are honest or no one is honest. (All people)
Some people are honest and some people are not honest. (All people)
Every nonzero real number is positive or negative. (Real numbers)
*
(i)
Every integer is greater than —4 or less than 6. (Real numbers)
*
(j)
(k)
(1)
Every integer is greater than some integer. (Integers)
No integer is greater than every other integer. (Integers)
Between any integer and any larger integer, there is a real number. (Real
numbers)
There is a smallest positive integer. (Real numbers)
No one loves everybody. (All people)
Everybody loves someone. (All people)
For every positive real number x, there is a unique real number y such
that 2” = x. (Real numbers)
*
*
(m)
(n)
(0)
(p)
2.
For each of the propositions in Exercise 1, write a useful denial, and give a
translation into ordinary English.
3.
Translate these definitions from the Appendix into quantified sentences.
(a) The natural number a divides the natural number b.
(b) The natural number n is prime.
(c) The natural number n is composite.
1.3
(d)
The sets A and B are equal.
(e)
(f)
The set A is a subset of B.
The set A is not a subset of B.
Quantified Statements
27
Write symbolic translations using quantifiers for each of the five important
properties of Z listed in the Appendix under the heading The Integers.
The sentence “People dislike taxes” might be interpreted to mean “All people
dislike all taxes,” “All people dislike some taxes,” “Some people dislike all
taxes,” or “Some people dislike some taxes.” Give a symbolic translation for
each of these interpretations.
Leta 7s C="
SV 04 and We 20327, 20}
these four different universes is the statement true?
(a) Gs) Gis.odd =x5=-8);
(b) (Sx)(wis odd A x > 8).
(CPR) Gasioddi=
a8):
(d) (Vx)(xis odd A x > 8).
(a)
ain which, of
Complete the following proof of Theorem 1.3.1(b).
Proof:
Let U be any universe.
The sentence ~(4x)A(x) is true in U
ie,
iff (Vx) ~ A(X) is true in U.
(b)
Give a proof of part (b) of Theorem 1.3.1 that uses part (a) of that theorem.
Which of the following are true? The universe for each statement is given in
parentheses.
-
Al
(a)
(b)
(Wx)(x + x > x). (R)
(Vx)(x + x => x). (N)
(c)
(Ax)(2Qx + 3 = 6x + 7). (N)
(d)
(e)l
(4x)G* = x’). (R)
Gy6* =o -R)
g4s__
4
~
)
8h x) ES) i
(£) pn(SCO) =
(g) (Vx)Q? + 6x + 5 > 0). (R)
(h)) (Vx)? + 4x + 5 > 0). (R)
(i,
(Ax)(x? — x + 41 is prime). (N)
(j)
(Vx)? — x + 41 is prime). (N)
(k)
WM)
(Vx)(x3 + 17x? + 6x + 100 > 0). (R)
(¥x(Vy)Le < y > (Aw) < w < y)].(R)
Give an English translation for each. The universe is given in parentheses.
(a)
(Wx)(x = 1). (N)
(b)
(c)
(Alxy(x => 0 Ax < 0). (R)
(Vx)(x is prime A x # 2 = x is odd). (N)
(4) (Alx(log,x = 1).(®)
(e)
~(A00? < 0). (R)
(f)
(Al? = 0). (R)
(g)
(Vx)(x is odd = x? is odd). (N)
28
CHAPTER 1
Logic and Proofs
10.
(4.
*
Which of the following are true in the universe of all real numbers?
(a)
(V¥x\(dy)@ + y = 0).
(b)
(Ax)(Vy)( + y = 0).
OMG nEGyGc ty).
>} Gy\y=
0 ney >0)|
x
(d)
Wolxne0
x
(e)
ff)
(g)
(Vy)(Ax)(Vz)(xy = xz).
(Cy GS y).
(Vy)(Ax)@ < y).
(th)
Gly)o < 0A y+ 3 = 0).
x @
(jf)
(k)
11.
*
*
12.
(CWVye= y’).
(Vy\(AL)@= y’).
(Alx)\(Gy) (Ww)? > x — y).
Let A(x) be an open sentence with variable x.
(a) Prove Theorem 1.3.2 (a).
(b)
(c)
(d)
Show that the converse of Theorem 1.3.2 (a) is false.
Prove Theorem 1.3.2 (b).
Prove that (S!x)A (x) is equivalent to (4x)[A(@) A (Vy)(AQ) > x = y)].
(e)
Finda
useful denial for (4!x)A (x).
Suppose the polynomials a,x” + a,_,x""! +--+ aj and b,x" + b,x! +-++4+
by are not equal. Which of the following must be true?
Gs ene
ED)
13.
*
*
1.4
14.
(b)
(c)
(d)
(e)
(f)
(g)
a; #5; whenever 0 <i <n.
a; #5, for every i such that 0 <i <n.
a, #b, for some i such that 0 <i <n.
Itis not the case that a;= b; for all i such that 0 <i <n.
tis not the case that a; = b,; for some i such that 0 <i <n.
There is ani such that 0 <i <n anda; #b,.
(h)
Ifa,= 5; for alli such thatO <i<n-—
1, thena, #b,,.
Which of the following are denials of (4!x) P(x)?
(a) (x) Pav (Wx) ~PG@).
(b)
(c)
(d)
(¥x) ~P(x) Vv (Ay)(G2)0 4 z A PQ) A PC).
(Vx)[PQ) => (Gy)(PO) Ax # y)].
~(¥x)(Vy)L(P@) A PO) > x = y).
Riddle: What is the English translation of the symbolic statement V3 3V?
Basic Proof Methods |
A theorem in mathematics is astatement that describespattern
a
or relationship
among quantities or structures. A proof of a theorem is a justification of the truth
of the theorenr that follows the principles of logic.
We cannot define all terms or prove all statements from previous ones. We
begin with an initial set of statements, called axioms
(or postulates), that are
assumed to be true. We then derive theorems that are true in any situation where
1.4
Basic Proof Methods I
29
the axioms are true. The Pythagorean’ Theorem, for example, is a theorem whose
proof is ultimately based on the five axioms of Euclidean’ geometry. In a situation
where the Euclidean axioms are not all true (which can happen), the Pythagorean
Theorem may not be true.
There must also be an initial set of undefined terms—concepts fundamental
to the context of study. In geometry, the concept of a point is an undefined term. In
this text the real numbers are not formally defined. Instead, they are described in
Appendix as the decimal numbers along the number line. While a precise definition
of a real number could be given,* doing so would take us far from our intended
goals.
From the axioms and undefined terms, new concepts (new definitions) can be
introduced. And finally, new theorems can be proved. The structure of a proof for
a particular theorem depends greatly on the logical form of the theorem. Proofs
may require some ingenuity or insightfulness to put together the right statements to
build the justification. Nevertheless, much can be gained in the beginning by studying the fundamental components found in proofs and examples that exhibit them.
The rules that follow provide guidance about what statements are allowed in
a proof, and when. The first four rules enable us to replace a statement with an
equivalent or to state something that is always true or is assumed to be true.
\
In any proof at any time you may:
State an axiom, an assumption, or a previously proved result.
The statement of an axiom is usually easily identified as such by the reader
because it is a statement about a very fundamental fact assumed about the theory.
Sometimes the axiom is so well known that its statement is omitted from proofs, but
there are cases (such as the Axiom of Choice in Chapter 5) for which it is prudent
to mention the axiom in every proof employing it.
The statement of an assumption generally takes the form “Assume P” to alert
the reader that the statement is not derived from a previous step or steps. We must
be careful about making assumptions, because it is only when all the assumptions
are true that we can be certain that what we proved will be true. The most common
assumptions are hypotheses given as components in the statement of the theorem to
be proved. We will discuss assumptions in more detail later in this section.
Proof steps that use previously proven results help build a rich theory from the
basic assumptions. In calculus, for example, before one proves that the derivative
sin Ax
of sin x is cos x, one usually proves that Jim
= |. It is easier to prove this
x20
Ny
* Pythagoras, latter half of the 6th century B.c.E., was a Greek mathematician and philosopher who
founded a secretive religious society based on mathematical and metaphysical thought. Although
Pythagoras is regularly given credit for the theorem named for him, the result was known to Babylonian
and Indian mathematicians centuries earlier.
+ Euclid of Alexandria, circa 300 B.c.£., made his immortal contribution to mathematics with his famous
text on geometry and number theory. His Elements sets forth a small number of axioms from which
additional definitions and many familiar geometric results were developed in a rigorous way. Other
geometries, based on different sets of axioms, did not begin to appear until the 1800s.
+ See the references cited in Section 7.5.
30
CHAPTER 1
Logic and Proofs
result first, and then cite the result in the proof of the fact that the derivative of
sin x is cos x. A result that serves as a preliminary step is often called a lemma.
.) In any proof at any time you may use the tautology rule:
State a sentence whose symbolic translation is a tautology.
Example.
“Kither
PAD
Suppose a proof involves a real number x. You may at any time state
x > 0 or x < 0” because this statement
is an instance of the tautology
IP.
An important skill for writing proofs is the ability to rewrite a statement in an
equivalent form that is more useful or helps clarify its meaning.
In any proof at any time you may use the replacement rule:
State a sentence equivalent to any statement earlier in the proof.
For example, if a proof contains the statement “It is not the case that x is even
and prime,” we may deduce that “x is not even or x is not prime.” This step is valid
because the first statement has the form ~(P A Q) and the second has the form
~P Vv ~@. We have applied the replacement rule, using one of DeMorgan’s Laws.
A thorough knowledge of the logical equivalences of Theorems 1.1.1 and 1.2.2
is essential when one uses the replacement rule because these replacements are
done routinely, without mentioning the relevant rules of logic.
It is impossible to read or write proofs in advanced mathematics without using
definitions. Because understanding and using definitions is so crucial, we restate
the replacement rule specifically for definitions.
In any proof at any time you may:
Use a definition to state an equivalent to a statement earlier in the proof.
The precise definition of “divides” given in the Appendix makes it possible to
build proofs involving divisibility properties of N that are among the first examples
we do. An explanation of why a divides b that uses an inexact definition of divisibility, such as “a divides b because it goes in evenly,” is practically useless in writing a proof. The key idea is that divisibility is defined as it is, so that in a proof we
can replace the statement “a divides b” with the equivalent statement “b = a - k for
some integer k.” Conversely, if we want to deduce that a divides b, the definition
tells us that we need a step that says there is an integer k such that b = a- k.
The most fundamental rule of reasoning is modus ponens, which is based on
the tautology P A [P > Q)] => Q. In Section 1.2 we showed that this tautology
means that whenever P and P => Q are both true, we may deduce that Q is also
true. This rule allows us to make a connection so that we can get from statement P
to a different statement Q.
In any proof at any time you may use the modus ponens rule:
After statements P and P = Q appear in a proof, state Q.
1.4
Basic Proof Methods |
31
When we use modus ponens to deduce Q from statements P and P > Q, the
statement P could be an assumption, a compound proposition whose components
are hypotheses or previously proved results, or any other statement known to be
true at this point. The conditional sentence P > Q could also be a previous theorem
or tautology or any other statement that appears earlier in the proof.
You have used these proof rules, at least informally, possibly to answer questions like the one in the next example, which comes from a calculus exam. Notice
how the solution (1) states the assumption, (2) replaces the assumption using the
definition of differentiability, (3) uses a known result, (4) applies modus ponens to
two previous statements, and (5) uses the definition of continuity to deduce the last
Statement.
Example.
Suppose f is a function defined on an interval containing 2, and we
know that the derivative of fat 2 exists. Find Jim f(x) = f(2). Explain your answer.
We are given that f’ (2) exists. Thus fis differentiable at x = 2. A theorem from
calculus says that if fis differentiable at x, then fis continuous at x. Therefore,
fis
continuous at x = 2. We conclude that lim, fx) = f(2).
Oo
x
The next example comes from outside mathematics and shows that it may be
the form of propositions, and not the meaning, that enables us to make a deduction.
Example.
You are at a crime scene and have established the following facts:
(1) If the crime did not take place in the billiard room, then Colonel Mustard is guilty.
(2) The lead pipe is not the weapon.
(3) Either Colonel Mustard is not guilty or the weapon used was a lead pipe.
From these facts and modus ponens, you may construct a proof that shows the
crime took place in the billiard room:
Proof.
Statement (1)
Statement (2)
Statement (3)
Statements (1) and (2) and (3)
Statements (1), (2), and (3)
imply the crime took place
in the billiard room.
Therefore, the crime took place
in the billiard room.
~Bi— VI
PAL;
~MvL
(SB=M)
GF
KHL A (AMV £L)
Nad SO I EN Bo
3
is a tautology (see Exercise 2).
6B
a
The last three statements above are an application of the modus ponens rule:
We deduced Q from the statements P and P= Q, where Q is B and P is
(~B=>M)A~LA(~M VL).
Because our proofs are always about mathematical phenomena, we also need
to understand the subject matter of the proof—the concepts involved and how they
32
CHAPTER 1
Logic and Proofs
are related. Therefore, when you develop a strategy to construct a proof, keep in
mind both the logical form of the theorem’s statement and the mathematical concepts involved.
You won’t find truth tables displayed or referred to in proofs that you encounter in mathematics: It is expected that readers are familiar with the rules of logic and
correct forms of proof. As a general rule, when you write a step in a proof, ask yourself whether deducing that step is valid in the sense that it uses one of the five rules
above. If the step follows as a result of the use of a tautology, it is not necessary to
cite the tautology in your proof. In fact, with practice you should eventually come
to write proofs without purposefully thinking about tautologies. What is necessary
is that every step be justifiable.
The first—and most important—proof method is the direct proof of statement
of the form P > Q, which proceeds in a step-by-step fashion from the antecedent
P to the consequent Q. Since P > Q is false only when P is true and Q is false, it
suffices to show that this situation cannot happen. The direct way to proceed is to
assume that P is true and show (deduce) that Q is also true. A direct proof of
P= Q
will have the following form:
DIRECT PROOF
Proof.
Assume P.
OF P > @Q
Therefore, Q.
huss
7O.-
|
J
Some of the examples and exercises in this and the next section involve open
statements with variables. The proof techniques to handle open sentences and
quantified statements are discussed in detail in Section 1.6. For now, whenever
we encounter a sentence with a variable, imagine that the variable represents some
fixed object.
You don’t need to see a proof to be convinced that the next number after an odd
number is an even number, but we’ll examine proofs of this and several other obvious results in our first examples. These examples are chosen so that you don’t have
to deal with new concepts at the same time as you are learning how to write proofs.
The important thing to learn from these examples is that a direct proof proceeds
step by step from the antecedent to the consequent.
Example.
Letx be an integer. Prove
that ifx is odd, then x + 1 is even.
The theorem has the form
P => Q,
where P is “xis odd” and Q is “x + 1
is even.”
Proof.
Let x be an integer.
Given. We may assume this hypothesis because it is given in the statement
of the theorem.
1.4
Basic Proof Methods |
33
Suppose x is odd.
Assume the antecedent P is true. The
goal is to derive the consequent Q.
Then x = 2k + 1 for some integer k.
We have replaced P with an equivalent statement—the
definition of
“odd.” We now have a statement we
can work with.
Thenx+1=(2kK+1)+1.
We add | to each side of the equation
to get an equivalent statement.
Then x + 1 =2k+2=2(k=+
1),
We use algebra and the fundamental
so x + | is the product of 2 and the
integer k + 1.
fact that if k is an integer, thenk +1.
is an integer.
Thus x + 1 is even.
We have deduced Q.
Therefore, if x is an odd integer, then
x + 1 is even.
|
We conclude that
P > OQ.
The right-hand column was included to describe how the steps are connected.
Proofs are not usually written this way because, in practice, such a column is
unnecessary. Writing a proof in two-column form can be a good way to begin to
understand its structure, but the sequence of statements on the left is the complete
proof and should be written in shorter form, as follows:
Proof.
Let x be an integer. Suppose x is odd. Then x = 2k + 1 for some integer
k. Then
x + 1 = (2k+ 1) + 1. Because (2k + 1) + 1=2k + 2 =2(k+ 1), we see
that x + 1 is the product of 2 and the integer A + 1. Thus x + 1 is even. Therefore,
if x is an odd integer, then x + 1 is even.
a
This form of the proof assumes that the person reading it knows the relevant
definitions and the basics of logic. Good proofs include enough detail so that readers with the appropriate background can follow the logical steps and fill in computations as necessary. This can be challenging for someone first learning to read and
write proofs, but these skills come with practice. After you’ve written more proofs
using a variety of methods, you’ll find useful advice about mathematical writing
following Section 1.6.
Writing proofs will come much easier to you when you understand the vital
importance of definitions, beginning with this very first example, where the fact
that an odd number x is such that “x = 2k + 1 for some integer k” gave us something specific to work with: an equation that we could manipulate. That’s the first
way a definition was used in the example. The second use of a definition came
when we concluded that x + 1 is even: we determined that x + 1 was even because
it satisfied the condition given in the definition of even.
You'll find that using the definitions often provides a hint about how to begin
a proof. Notice how the next theorem uses the definition of divides. The proof is
34
CHAPTER 1
Logic and Proofs
again given in two-column style. Try covering the right-hand column the first time
you read the proof.
Theorem 1.4.1
Let a, b, and c be integers. Ifadivides b
and b divides c, then a divides c.
The theorem has the form P \Q=> R,
where P is “a divides b,” Q is “b
divides c,” and R is “a divides c.”
Proof.
Let a, b, and c be integers.
Given.
That
is,
we
assume
this
hypothesis is true.
Suppose a divides b and b divides c.
Assume the antecedent P / Q is true.
The antecedent is a conjunction, so
both components are true.
Then b = ak for some integer k, and
c = bm for some integer m.
Replace each assumption by an
equivalent using the definition of
“divides.” Note that we do not assume
that k and m are the same integer.
Then c = bm = (ak)m = a(km).
To show “a divides c,” we must show
c is a multiple of a. We use algebra to
obtain an expression
for c in terms of a.
Since kA and m are integers, km is an
integer.
We use a fundamental property of
the integers.
Thus a divides c.
We have deduced R.
Therefore, if a divides b and b divides
c, then a divides c.
|
We _ state
ROSS
the
conclusion — that
No other proofs in this text will be presented in this two-column format.
However, we will sometimes include stylized parenthetical comments offset by
( ) to help explain how and why a proof is proceeding as it is. These comments
would not be included in a proof written for experienced readers.
Example.
Suppose a, b, and c are integers. Prove that if a divides b and a divides
c, then a divides b — c.
Proof. Suppose a, b, and c are integers and a divides b and a divides c. (Now
use the definition of divides.) Then b = an for some integer n and c = am for
some integer m. Thus, b — c = an — am = a(n — m). (We next use the fact that
the difference of two integers is an integer.) Since n — mis an integer, we have that
a divides b — c.
a
Our next example of a direct proof comes from an exercise in precalculus
mathematics about distances between points in the Cartesian plane and uses algebraic properties available to students in such a class.
Example.
Prove that if x and y are real numbers such that x < —4 and vi
the distance from (x, y) to (1, —2) is greater than 6.
2, then
1.4
Basic Proof Methods I
35
Proof.
Assume that x and y are real numbers such that x < —4 and Vie 2.
(Then (x, y) is a point in the shaded area in Figure 1.4.1.) Then x — 1 < —5, so
(x — 1)? > 25. Also y + 2 > 4, so (y + 2)? > 16. Therefore,
We = Drei
22 = 1/25. 16= 36,
V/
so the distance from (x, y) to (1, —2) is greater than 6.
Lea e(,
a
—2)
Figure 1.4.1
The proofs above exhibit the strategy for developing a direct proof of a conditional sentence:
1.
Determine precisely the hypotheses (if any) and the antecedent and consequent.
2.
Replace (if necessary) the antecedent with a more usable equivalent.
3.
Replace (if necessary) the consequent by something equivalent and more readily shown.
4.
Beginning with the assumption of the antecedent, develop a chain of statements that leads to the consequent. Each statement in the chain must be deducible from its predecessors or other known results.
To discover a chain of statements from the antecedent to the consequent, it
is sometimes useful to work backward from what is to be proved: To show that
a consequent is true, decide what statement could be used to prove it, another
statement that could be used to prove that one, and so forth. Continue until you
reach a hypothesis, the antecedent, or a fact known to be true. After doing such
preliminary work, construct a proof that progresses forward until it ends with the
consequent.
Example.
Let a and b be positive real numbers.
Prove that if a < b, then
b? — a’? > 0.
Before we write the proof, we work backward from the consequent. First
rewrite b? — a2 > 0as(b — a)(b + a) > 0. This inequality will be true when both
factors are positive. The term b — a is positive because we assumed that b > a.
Also, b + a > 0 because of our hypothesis that both a and b are positive. We now
know how to write the proof.
36
CHAPTER 1
Logic and Proofs
Proof.
Assume that a and b are positive real numbers and that a < b. Since both a
becausentbe
andb are positive, b + a > 0. Because a < b, we see that baa
Therefore,
O.
>
a)
+
a)(b
—
(b
positive,
is
product of two positive real numbers
es
Ei
yy,
Sometimes, working both ways—backward from what is to be proved and forward from the hypothesis—until you reach a common statement from each direction will help reveal the structure of a proof.
Example.
Prove that if x? < 1, thenx* — 7x >
— 10.
Working backward from x2 — 7x > —10, we note that this can be deduced
from x2 — 7x + 10 > 0. This can be deduced from (x — 5)(x — 2) > 0, which
could be concluded if we knew that x — 5 and x — 2 were both positive or both
negative.
Working forward from x? = 1, we have —1 <x < 1, so x = 1. Therefore,
x < 5 and x < 2, from which we can conclude that x — 5 < 0 and x — 2 < 0,
which is exactly what we need.
Proof.
Assume that x? < 1. Then —1 < x < 1. Therefore, x < 1. Thus x < 5 and
x < 2,andso we have x — 5 < Oandx — 2 < O. Therefore, (x — 5)(x — 2) > 0.
Thus x- 17x -- 10 > 0. Hence x7 — 7x > —10.
|
When either P or Q is a compound proposition, the steps in proving statements
of the form P => Q depend on the forms of P and Q. We have already constructed
proofs of statements of the form (P A Q) => R. When we give a direct proof of a
statement of this form, we have the advantage of assuming both P and Q at the
beginning of the proof.
A proof of a statement symbolized by P= (Q A R) would probably have two
parts. In one part we prove P => Q and in the other part we prove P > R. We would
use this method to prove the statement “If a circle of radius r is inscribed in a square
that is inscribed in a circle, then the center of the outer circle is the center of the
inner circle, and the radius of the outer circle is rV2.”
To prove a conditional sentence whose consequent is a disjunction—that
is, a sentence
of the form P= (Q.Vv R)—one
often proves
either the equiva-
lent P \ ~Q=>R
or the equivalent P \ ~R => Q. For instance, to prove “If the
polynomial f has degree 4, then f has a real zero or f can be written as the product
of two irreducible quadratics,” we would prove “If f has degree 4 and no real zeros,
then f can be written as the product of two irreducible quadratics.”
A statement of the form (P V Q) => R has the meaning “If either P is true or 0
is true, then R is true.” A natural way to prove such a statement is by cases, so the
proof outline would have the form
Case 1.
Case 2.
Assume P.... Therefore R.
Assume Q.... Therefore R.
1.4
Basic Proof Methods |
37
This method is valid because of the tautology
(PV OSRI
SIPS
RAQHS
RI.
The statement “If a quadrilateral has opposite sides equal or opposite angles equal,
then it is a parallelogram” is proved by showing both “A quadrilateral with opposite
sides equal is a parallelogram” and “A quadrilateral with opposite angles equal is
a parallelogram.”
The two similar statement forms (P=>Q)>R and
P=>(Q=>R) have
remarkably dissimilar direct proof outlines. For (P > Q) > R, we assume P > Q
and deduce R. We cannot assume P; we must assume P > Q. On the other hand,
in a direct proof of P= (Q = R), we do assume P and show QO = R. Furthermore,
after the assumption of P, a direct proof of Q > R begins by assuming Q is true as
well. This is not surprising, because P > (Q => R) is equivalent to (P A Q) > R.
The main lesson to be learned from this discussion is that the method of proof
you choose will depend on the form of the statement to be proved. The outlines we
have given are the most natural, but not the only, ways to construct correct proofs.
Before we begin the proof of the next example, consider integers of the form
n = 2m + 1 for some integer m. A little experimentation shows that when m is
even—for example, when n is 2(—2) + 1, or 2(0) + 1, or 2(2) + 1, 2(4) + 1, etc. —
then n has the form 47 + 1, and otherwise n has the form 4i — 1. The statement
below has the form (P V Q) => (R, V R,), where P is “m is even,” Q is “m is odd,”
R, is “n = 4j + | for some integer j,” and R, is “n = 47 — | for some integer 7.” The
proof method we choose is to show that P = R, and 0 => R>.
Example.
Suppose n is an odd integer. Then n = 4j + 1 for some integer j, or
n = 4i — 1 for some integer 1.
Proof.
Suppose n is odd. Then n = 2m + 1 for some integer m.
Case 1.
If m is even, then m = 27 for some integer j, and so n = 2(27) + 1 =
47 + 1.
Case 2.
If m is odd, then m = 2k + 1 for some integer k. In this case, n =
22k + 1)+ 1=4k +3 =4(k + 1) — 1. Choosing i to be the integer
k + 1, wehaven = 4i — 1.
|
The statement to be aval may have any logical ‘form. For example, to prove
that every number x in the closed interval [0, 5] has a certain property, we might
consider the cases x = 0, 0 < x < 5, and x = 5. The exhaustive method was our
method in the previous example, and in the suggested proof of Theorem 1.1.1,
which requires examination of all combinations of truth values for each pair of
propositions. Naturally, the idea of proof by exhaustion is appealing only when
the number of cases is small or when large numbers of cases can be systematically
handled. Care must be taken to ensure that all possible cases have been considered.
Example.
Let x be a real number. Prove that —|x| < x < |x|.
38
CHAPTER 1
Logic and Proofs
(Because the absolute value of x is defined by cases (\x| =x ifx= 0;
Proof.
|x| = —x ifx < 0), this proof will proceed by cases.)
Case 1.
Case 2.
Suppose x > 0. Then|x|= x. Since x > 0, we have
=K< 0 = x,-which is =|x|.s xi) )|)4\\in this ease.
Suppose x < 0. Then|x|=—x.
XS eS
Thus, in all cases
—K OF =(=5) =
—x <x.
Hence,
Since x < 0, x < —x. Hence, we have
eS Sa AVIChsi ae
we have —|x| <x < |x|.
x =|)
|
There have been instances of truly exhausting proofs involving great numbers
of cases. In 1976, Kenneth Appel and Wolfgang Haken of the University of Illinois
announced a proof of the famous Four-Color Conjecture. The original version of their
proof contains 1,879 cases and took 3 4years to develop.”
Finally, there are proofs by exhaustion with cases so similar in reasoning that
we may simply present a single case and alert the reader with the phrase “without
loss of generality” that this case represents the essence of arguments for the other
cases. Here is an example:
Example.
Prove that for the integers m and n, one of which is even and the other
odd, m? + n* has the form 4k + 1 for some integer k.
Proof.
Let mand n be.integers. Without loss of generality, we may assume that m
is even and nis odd. (The case where m is odd and n is even is similar.) Then there
exist integers s and f such that m = 2s and n = 2t + 1. Therefore, m* + n* =
(25) et Oth
== Ag?
Apes Pe (5?
72k
4 1, Since s2. rg
is an integer, m* + n? has the form 4k + 1 for some integer k.
=
Exercises 1.4
1.
*
Analyze the logical form of each of the following statements and construct
just the outline of a proof. Since the statements may contain terms with which
you are not familiar, you should not (and perhaps could not) provide any
details of the proof.
~
(a) Outline a direct proof that if (G, *) is a cyclic group, then (G, *) is
abelian.
* The Four-Color Theorem involves coloring regions or countries on a map in such a way that no two
adjacent countries have the same color. It states that four colors are sufficient, no matter how intertwined
the countries may be. The fact that the proof depended so heavily on the computer for checking cases
raised questions about the nature of proof. Verifying the 1,879 cases required more than 10 billion
calculations. Many people wondered whether there might have been at least one error in a process so
lengthy that it could not be carried out by one human being in a lifetime. Haken and Appel’s proof has
since been improved, and the Four-Color Theorem is accepted; but the debate about the role of computers in proof continues.
1.4
(b)
(c)
(d)
(e)
Basic Proof Methods |
39
Outline a direct proof that if B is a nonsingular matrix, then the determi-
nant of B is not zero.
Suppose A, B, and C are sets. Outline a direct proof that if A is a subset
of B and B is a subset of C, then A is a subset of C.
Outline a direct proof that if the maximum value of the differentiable function f on the closed interval [a, b] occurs at xo, then either
Xo = GOL 4, = Dorf’ Xp) = 0.
Outline a direct proof that if A is a diagonal matrix, then A is invertible
whenever all its diagonal entries are nonzero.
A theorem of linear algebra states that if A and B are invertible matrices, then
the product AB is invertible. As in Exercise 1, outline
(a)
(b)
adirect proof of the theorem.
adirect proof of the converse of the theorem.
Verify that [((~B>M)
example on page 31.
A ~LA(~M
Vv L)) => B is a tautology.
See the
These facts have been established at a crime scene:
(i)
If Professor Plum is not guilty, then the crime took place in the kitchen.
(ii) If the crime took place at midnight, then Professor Plum is guilty.
(iii) Miss Scarlet is innocent if and only if the weapon was not the candlestick.
(iv) Either the weapon was the candlestick or the crime took place in the
library.
(v) Either Miss Scarlet or Professor Plum is guilty.
Use each of the following as a sixth clue to solve the case. Explain your
answer.
(a)
(b)
(c)
The crime took place in the library.
The crime did not take place in the library.
The crime was committed at noon with the revolver.
(d)
The crime took place at midnight in the conservatory. (Give a complete
answer.)
Let x, y, and z be integers. Prove that
rm
(a)
if.x and y are even, then x + y is even.
(b)
(¢)
if xis even, then xy is even.
if x and y are even, then xy is divisible by 4.
((d)
if x and y are even, then 3x — 5y is even.
(e)
if x and y are odd, then x + y is even.
(f)
if x and y are odd, then 3x — 5y is even.
(g)
if x and y are odd, then xy is odd.
(h)
(i)
(j)
if xis even and y is odd, then x + y is odd.
if exactly one of x, y, and z is even, then the sum of x, y, and z is even.
if exactly one of x, y, and z is odd, then xy + yz is even.
Let a and b be real numbers. Prove that
(a)
|ab| = |al|d|.
(b)
|la—b|
= |b—- al.
40
CHAPTER 1__ Logic and Proofs
a
(c)
a
(d)
(e)
(f)
(g)
|a + b| < |a| + |b| (The Triangle Inequality).
if |a| < b, then —b <a <b.
if —b <a <b, then|a| <b.
‘Ff
a
*
b
b|
mapllcdnr erste
|b
(la)
ola pI:
upp ose a, b, c, and d are integers. Prove that
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(I)
(m)
2a — 1 is odd.
if ais even, then a + | is odd.
if ais odd, then a + 2 is odd.
a(a + 1) 1s even.
1 divides a.
a divides a.
if a and b are positive and a divides b, then a < b.
if a divides b, then a divides bc.
if a and b are positive and ab = 1, thena=b=
1.
if a and b are positive, a divides b, and b divides a, then a = b.
if a divides b and c divides d, then ac divides bd.
if ab divides c, then a divides c.
if ac divides bc, then a divides b.
Give two proofs that if n is a natural number, then n? + n + 3 is odd.
(a)
(b)
Use two cases.
Use Exercises 7(d) and 5(h).
Let a, b, and c be integers and x, y, and z be real numbers. Use the technique
of working backward from the desired conclusion to prove that
(a)
if x and y are nonnegative, then #0.z SN Ky:
Where in the proof do we use the fact that x and y are nonnegative?
if a divides b and a divides b + c, then a divides 3c.
if ab > O and be < 0, then ax? + bx + c = 0 has two real solutions.
if
=2x% = Osthen Qe ets 5-el 1.
if an isosceles triangle has sides of length x, y, and z, where
z = V2kxy, then it is a right triangle.
x = y and
if x and y are positive and xy? — 25x? — 9y* + 225 < 0, then x < 3 and
ye
10.
orelsex 3) and ye= 5,
Recall that except for degenerate cases, the graph of Ax? + Bxy + Cy? +
Dy oe Ee yer
01s
an ellipse iff B*? — 4AC < 0,
a parabola iff B* — 4AC = 0,
a hyperbola iff B7 — 4AC > 0.
(a)
Prove that whenever
ellipse.
A > C > B > 0, the graph of the equation is an
1.4
Proofs to Grade
11.
Basic Proof Methods |
Al
(b)
Prove that the graph of the equation is a hyperbola if AC < 0 or
Bie
4A
0.
(c)
Prove that if the graph is a parabola, then BC = 0 or A = B?/(4C).
Exercises throughout the text with this title ask you to examine “Proofs to
Grade.” These are claims alleged to be true and supposed “proofs” of the
claims. You should decide the merit of the claim and the validity of the proof,
and then assign a grade of
A (correct), if the claim and proof are correct, even if the proof is not the
simplest or the proof you would have given.
C (partially correct), if the claim is correct and the proof is largely correct. The proof may contain one or two incorrect statements or justifications, but the errors are easily correctable.
F (failure), if the claim is incorrect, or the main idea of the proof is incor-
rect, or there are too many errors.
You must justify assignments of grades other than A, and if the proof is
incorrect, you must explain what is incorrect and why.
(a) Suppose a is an integer.
Claim.
(b)
*
(c)
If a is odd, then a* + 1 is even.
“Proof.”
Let a. Then, by squaring an odd we get an odd. An odd plus
an odd is even. So a? + | is even.
a
Suppose a, b, and c are integers.
Claim.
If a divides b and a divides c, then a divides b + c.
‘“‘Proof.”’ Suppose a divides b and a divides c. Then for some integer
q, b = aq, and for some integer g, c = aq. Then b + c = aq + aq=
2aq = a(2q), so a divides b + c.
a
Suppose x is a positive real number.
Claim.
The sum of x and its reciprocal is greater than or equal to 2.
That is,
ee
“Proof.”
Multiplying
Wo = oct
oles OF Thustcaeel
1
x
by x,
Se,
we
get x7 +1 > 2x.
:
*
(d)
greater than or equal to zero, so x + — > 2 is true.
5G
algebra,
|
Suppose
m is an integer.
Claim.
(e)
By
Ar: Any real number squared is
If m7 is odd, then m is odd.
‘Proof.’
Assumemisodd.Thenm = 2k + 1 forsomeintegerk. Therefore, m2 = (2k + 1)? = 4k? + 4k + 1 = 2(2k? + 2k) + 1, which is
odd. Therefore, if m7 is odd, then m is odd.
a
Suppose a is an integer.
Claim. a> + a’ is even.
“Proof.” a> + a* = a*(a + 1), which is always an odd number times
an even number. Therefore, a* + a? is even.
a
42
1.5
CHAPTER 1
Logic and Proofs
Basic Proof Methods Il
The method of proof we considered in the previous section, the direct proof, is the
most basic form of proof for P > Q. It proceeds as a chain of statements from P to
Q. In this section we consider indirect proof techniques for statements of the form
P = Qas well as proof methods for statements of the form P <= Q.
The indirect method called contrapositive proof or proof by contraposition
of P > Q makes use of the fact that P > Q is equivalent to its contrapositive,
~Q=
~P.To use this method, we give a direct proof of ~Q => ~P and conclude
by replacement that P > Q.
PROOF OF P => @ BY CONTRAPOSITION
Proof.
Assume ~Q.
Therefore, ~P.
SUS as ON
heretores2— 0);
r
sell
This method works well when the connection between denials of P and Q are easier
to understand than the connection between P and Q themselves, or when the statement of either P and Q is itself a negation.
In the following examples of proof by contraposition we use familiar properties of inequalities and the property that every integer is either even or odd, but not
both. As in the last section, we assume that variables represent fixed quantities.
Example.
Let m be an integer. Prove that if m* is even, then m is even.
Proof.
(The antecedent is P, “‘m* is even” and the consequent is QO, “m is even.”’)
Suppose that the integer m is not even. (Suppose ~Q.) Then m is odd, so
m = 2k + | for some integer k. Then
m
= (dh
1)
Ak? Ak
l= 2(2k?
+ 2k) + 1.
Since m? is twice an integer, plus 1, m? is odd. (Since k is an integer, 2k? + 2k is
an integer.) Therefore, m? is not even. (We have concluded ~P.)
Thus, if m is not even, then m? is not even. By contraposition, if m is even,
then m is even.
Example.
P|
Let x and y be real numbers
such
that x < 2y. Prove
that
if
Txy < 3x? + 2y?, then 3x < y.
Proof.
Suppose x and y are real numbers and x < 2y. (Let P be Txy < 3x? + 2y?
and Q be 3x <y.)
Suppose
3x > y. (We assume
~Q.)
Then
2y —x>0
1.5
Basic Proof Methods Il
43
and 3x — y > 0. Therefore, (2y — x)(3x — y) = 7xy — 3x” — 2y? > 0. Hence,
Tay > 3x? -Qy*
We have shown that if 3x > y, then 7xy > 3x? + 2y. Therefore, by contraposition, if 7xy < 3x” + 2y?, then 3x < y.
]
An indirect proof of a statement P by the method of contradiction uses the
logic that if P cannot be false, then P must be true. To use this method, we temporarily assume that P is false and then see what would happen. If what happens is
impossible—that is, if we deduce some contradiction Q A ~Q—then we know that
P must be true. The correctness of this reasoning may be verified by checking that
if Q is any proposition, then [(~P) > (Q A ~Q)] is equivalent to P.
PROOF OF P BY CONTRADICTION
Proof.
Suppose ~P.
Therefore, QO.
Therefore, ~Q.
Hence, Q A ~@Q a contradiction. (With Q and ~@ established, this step
may be omitted.)
TMS, 12,
.
Example.
Prove that the graphs of y = x* + x + 2and y = x — 2 donotintersect.
Proof. Suppose the graphs of y = x” + x + 2 and y = x — 2 intersect at some
point (a, b). (Suppose ~P.) Since (a, b) is a point on both graphs, b = a” + a + 2
and
b = a — 2. Therefore,
a - 2 =a? +a+2,
so a = —4.
Thus,
a? < 0.
(The proposition Q is “a* < 0.”) But a is a real number, so a” > 0. This is a
contradiction. (The statement a* < 0 A a* > 0 is a contradiction.) Therefore, the
graphs do not intersect.
a
Two aspects of proofs by contradiction are especially noteworthy. First, this
method of proof can be applied to any proposition P, whereas direct proofs and
proofs by contraposition can be used only for conditional sentences. Second, the
strategy of proving P by proving ~P > (Q A ~Q) has the disadvantage that when
we set out to prove P, we may have no idea what proposition to use as Q. Thus
writing a proof by contradiction may require special insight, or years of work, or
possibly even a well-chosen guess to find a useful Q. The advantage of this method
is that there may be many propositions Q such that ~P implies both Q and ~Q, and
any such proposition may be used to construct the proof.
The next proof by contradiction is a classical result whose proof can be traced
back to Hippasus, a disciple of Pythagoras, circa 500 B.c.£. One of several legends
44
CHAPTER 1
Logic and Proofs
has it that Hippasus proved that \V/2 is not a rational number while traveling by
ship with his Pythagorean colleagues. The Pythagoreans, steadfast believers that all
numbers are rational, supposedly threw him into the sea to drown.
The proof relies on the definition of a rational number: r is rational if r = ;for
some integers a and b, with b # 0. We may assume that a and b have no common
factors, because otherwise we would simply reduce ; by canceling all common
factors.
Example.
V2 is an irrational number.
a
Assume that V2 is a rational number. (We assume ~P.) Then V2 = 5
Proof.
for some integers a and b, where b #0 and a and b have no common factors.
a
(The statement O is “a and b have no common factors.”) From V2 = b we have
a
Ue 2 which implies that 257 = a. Therefore a? is even and so a is even. (Recall
the first proof in this section.) It follows that there exists an integer k such that
a = 2k and therefore 2b? = a® = (2k)* = 4k?. Thus b* = 2k?, which shows that
b* is even. Therefore b is even. Since both a and b are even, a and b have a com-
mon factor of 2. (We have deduced the statement ~Q.) This is a contradiction. We
conclude that V2 is irrational.
=
Recall that a natural number greater than | is prime if its only positive divisors
are | and itself. The next proof by contradiction, attributed to Euclid, shows that
there are infinitely many primes. By this we mean that it is impossible to list all of the
prime numbers from the first to the Ath (last) one, where k is a natural number. It uses
the fundamental result that every natural number greater than | has a prime divisor.
Example.
The set of primes is infinite.
Proof. Suppose the set of primes is finite. (Suppose ~P. This means that the set
of primes has k elements for some natural number k. Then the set of all primes can
be listed, from the first one to the kth (last) one.) Let pj, p>, P3,---, Py, be all those
primes. Let n be one more than the product of all of them: n = (p, pp3...p,) + 1.
(We made up a number n that will not have any of the p; as prime factors.) Then nis a
natural number, son has a prime divisor q. Since gis prime, g > 1. (The QO statement
is“q > 1.”) Since q is a prime and pj, p>, P3,---, p; are all the primes, q is one of the
p; in the list. Thus, g divides the product p, p> p3...p,. Since q divides n, g divides
the difference n — (pp) p3...p,). But this difference is 1, so g = 1. (This is ~Q.)
* You may wonder why V2 is important or why it should be the first number to be proved irrational.
The ancient Greek geometers constructed numbers (lengths of line segments) using only a compass and
a straightedge. It’s easy to construct a square with sides of length 1, for which the length of a diagonal
is V2. The fundamental Pythagorean belief that all numbers that arise in nature are either integers or
ratios of integers is disproved by the irrationality of V2.
1.5
From the contradiction,
g > 1 and g =
Basic Proof Methods II
45
1, we conclude that the assumption that the
set of primes is finite is false. Therefore, the set of primes is infinite.
a
Example.
to
Prove the square shown
in Figure
1.5.1(a) cannot
be completed
form a “magic square” whose rows, columns, and diagonals all sum to the same
number.
1
2
3
4
5
7
6
8
9
10
(a)
1
2
3
a
b
4
5
6
7 |
Cc
8
d
su OM
of
10
(b)
Figure 1.5.1
Proof.
Suppose the square can be completed with entries a, b, c, d, e, f, as shown
in Figure 1.5.1(b). Since the sums of the second row and second column are the
same, b + 15 = c + 15. Thus, b = c. Comparing the sums of the first column and
the lower-left-to-upper-right diagonal, 1 + b+ 7+e=e+c+5+a. Thus
a = 3, and the first row sums to 9. Thus the “magic sum” is 9. (This is our Q
statement.) But the main diagonal sum (1 + 4 + 8 + 10 = 23) is not 9. (This is
our ~Q statement.) This is a contradiction. We conclude that the square cannot be
completed.
5
Proofs of biconditional sentences often make use of the fact that
equivalent to P= Q A Q=
P. These proofs have the following form:
P <= Q is
aa
TWO-PART PROOF OF P = Q
Proof.
(i) Show
P=} QO.
(ii) Show
O=> P.
herelores a <— 10)
ot
The separate proofs of ‘parts (i) and (ii) may use different methods.
Often
one part is easier to prove than the other part. This is true, for example, of the
proof that
“The natural number p is prime if and only if there is no positive integer
greater than | and less than or equal to Vx that divides x.”
|
46
CHAPTER 1
Logic and Proofs
It immediately follows from the definition of prime that “x is prime” implies “There
is no positive integer greater than | and less than or equal to \/x that divides x.” The
converse requires more thought and is an exercise in Section 1.8.
The parity of an integer isthe attribute of being either odd or even. The integer
31 has odd parity and 42 has even parity. The integers 12 and 15 have opposite parity.
The next example is a proof of a biconditional statement about parity with a two-part
proof. Both parts of the proof have two cases. The proof we give is not the shortest possible, but it does illustrate the two-part approach to proving a biconditional statement.
Example.
Let m and n be integers. Then m and n have the same parity if and only
if m* + n? is even.
Proof.
(i) Suppose m and n have the same parity. We have two cases.
(a)
If both m and n are even, then m = 2k and n = 2) for some integers k
andj.Then m? + n? = (2k)* + (Qj) = 2(2k? + 2j7), which is even.
(b)
If both m
and n are odd, then m=2k+4+
1 and
n=2j+
1 for
some-integers k and j. Then m? + n? = (2k +1) +Qj4+ 1? =
D20ie = 2k E27? = 27
In both cases m* + n? is even.
(ii)
1), which is even.
Suppose m? + n? is even. (To show that n has the same parity as m, we use
some previous examples and exercises about even and odd integers.) Again
we have two cases.
(a)
If mis even, then m? is even. Therefore, since m? + n? is even and m2
(b)
that 7 is even.
If mis odd, then m7 is odd. Therefore, since m2 + n? is even and m? is odd,
is even, n> = (m? + n?) — m7? is even. From n? is even, we conclude
n> = (m? + n?) — m? is odd. From n? is odd, we conclude that n is odd.
Hence, if m is even, then 7 is even, and if m is odd, then n is odd. Therefore,
mand n have the same parity.
a
In some cases it is possible to prove a biconditional sentence P < Q that uses
“if and only if” throughout. This amounts to starting with P and then replacing it
with a sequence of equivalent statements, the last one being Q. With n intermediate
statements R,, R5,..., R,, a biconditional proof of P = Q has the form
BICONDITIONAL PROOF OF P & Q
Proof.
P if and only if R,
if and only if R,
if and only if R,,
if and only if Q.
a
1.5
Basic Proof Methods II
47
Example. The triangle in Figure 1.5.2 has sides of length a, b, and c. Use the Law
of Cosines to prove that the triangle is a right triangle with hypotenuse c if and only
ifa? + b? = c?.
a
Figure 1.5.2
Proof.
By the Law of Cosines, a? + b? = c* — 2abcos 6, where @ is the angle
between the sides of length a and b. Therefore,
e+bh=c
iff 2abcosd=0
litCOStr—n)
iff 6 = 90° (because 0° < 6 < 180°.)
Thus, a? + b? = c? if and only if the triangle is a right triangle with hypotenuse c.
a
When you begin to write the proof of a statement, keep in mind that the logical
form of the statement, the meanings of the concepts involved, and your knowledge
of previous results are all key factors in deciding what proof method to use. Usually
a statement can be proved using any of several different methods. Our last example
in this section gives three different proofs of the same result: a direct proof, a proof
by contraposition, and a proof by contradiction.
Example.
Let x and y be positive real numbers such that x — 4y < y — 3x. Prove
that if 3x > 2y, then 12x” + 10y? < 24xy.
Direct Proof.
Let x and y be positive real numbers such that x — 4y < y — 3x.
Suppose (hate = 2ys Mhen 4y'— oy — O'and ox — 2y > 0, so(4x — Sy)Gx — 2y)'—
12x? — 23xy + 10y? < 0. Thus 12x? + 10y* < 23xy. Since xy is positive, 23xy < 24xy.
Thus 12x? + 10y? < 24xy. Therefore, if 3x > 2y, then 12x? + 10y* < 24xy.
a
Proof by Contraposition. Let x and y be positive real numbers such that x — 4y
< y — 3x. Suppose 12x? + 10y* > 24xy. Because xy is positive, 24xy > 23xy, so
12x? + 10y? > 23xy. Then 12x” — 23xy + 10y? = (4x — 5y)( 3x — 2y) > 0, and
because 4x — 5y < 0, we have 3x — 2y < 0. Therefore, 3x < 2y. We conclude that
if 3x > 2y, then 12x* + 10y* < 24xy.
)
48
CHAPTER 1
Logic and Proofs
Proof by Contradiction.
Let x and y be positive real numbers such that x — 4y <
y — 3x. Suppose 3x > 2y and 12x? + 10y? > 24xy. Then 4x — Sy < Oand 3x — 2y > 0,
so (4x — Sy)( 3x — 2y) = 12x” — 23xy + 10y < 0. Then 12x? ENO = 23) But
12x” + 10y? > 24xy > 23xy. This is a contradiction. We conclude that if 3x > 2y,
then 12x” + 10y? < 24xy.
a
By now you may have the impression that, given a set of axioms and definitions of a mathematical system, any properly stated proposition in that system can
be proved true or proved false. This is not the case. There are important examples
in mathematics of consistent axiom systems (structures exist that satisfy all the
axioms) for which there are statements such that neither the statement nor its negation can be proved. It is not a matter of these statements being difficult to prove or
that no one has yet been clever enough to devise a proof; it has been proved that
there can be no proof of either the statement or its negation within the system. Such
statements are called undecidable in the system because their truth is independent
of the truthofthe axioms.
:
The classic case of an undecidable statement involves the fifth of five postulates that Euclid set forth as his basis for plane geometry: “Given a line and a point
not on that line, exactly one line can be drawn through the point parallel to the
line.” For centuries, some thought Euclid’s axioms were not independent, believing that the fifth postulate could be proved from the other four. It was not until
the 19th century that it became clear that the fifth postulate was undecidable. There
are now theories of Euclidean geometry for which the fifth postulate is assumed
true and non-Euclidean geometries for which it is assumed false! Both are perfectly
reasonable subjects for mathematical study and application.
Exercises 1.5
1.
*
Analyze the logical form of each of the following statements, and construct
just the outline of a proof by the given method. Do not provide any details of
the proof.
(a) Outline a proof by contraposition that if (G, *) is a cyclic group, then
(G, *) is abelian.
*
(b)
Outline a proof by contraposition that if B is a nonsingular matrix, then
the determinant of B is not zero.
(c)
Outline a proof by contradiction that the set of natural numbers is not
finite.
(d)
*
(e)
(f)
Outline a proof by contradiction that if x is a nonzero real number, the
multiplicative inverse of x is unique.
Outline a two-part proof that the inverse of the function f from A to B is
a function from B to A if and only if f is one-to-one and fis onto B.
Outline a two-part proof that a subset A of the real numbers is compact
if and only if A is closed and bounded.
1.5
Basic Proof Methods Il
49
A theorem of linear algebra states that if A and B are invertible matrices, then
the product AB is invertible. As in Exercise 1,
(a) outline a proof of the theorem by contraposition.
(b) outline a proof of the converse of the theorem by contraposition.
(c) outline a proof of the theorem by contradiction.
(d) outline a proof of the converse of the theorem by contradiction.
(e)
outline a two-part proof that A and B are invertible matrices if and only
if the product AB is invertible.
Let x, y, and z be integers. Write a proof by contraposition to show that
(a)
(b)
if xis even, then x + | is odd.
if x is odd, then x + 2 is odd.
(c)
if x” is not divisible by 4, then x is odd.
(d)
if xy is even, then either x or y is even.
(e)
if x + yis even, then x and y have the same parity.
(f)
if xy is odd, then both x and y are odd.
(g)
if 8 does not divide x? — 1, then x is even.
(h)
if x does not divide yz, then x does not divide z.
Write a proof by contraposition to show that for any real number x,
(a)
if x? + 2x < 0, then
(b)
if x” = 5%
MeO:
(ec)
ifx? + x = 0; then x — 0.
(d)
if (x+ 1-1)
(e)
if x(x — 4) > —3, then x < 1 orx > 3.
6 = O then
=
=
3.
<0, then
x < 1.
A circle has center (2, 4).
(a) Prove that (—1, 5) and (5, 1) are not both on the circle.
(b) Prove that if the radius of the circle is less than 5, then the circle does not
intersect the line y = x — 6.
(c) Prove that if (0, 3) is not inside the circle, then (3, 1) is not inside the
circle.
Suppose a and b are positive integers. Write a proof by contradiction to
show that
(a)
(b)
(c)
(d)
if a divides b, thena < b.
if ab is odd, then both a and b are odd.
if ais odd, then a + | is even.
ifa — bis odd, then a + bis odd.
(e)
ifa= band ab < 3,thena
= 1.
Suppose a, b, c, and d are positive integers. Prove each biconditional statement.
(a) ac divides bc if and only if a divides b.
(b)
(ec)
(d)
(e)
a-+
1 divides b and b divides b + 3 if and only if a= 2 and b = 3.
a+c=bDand 2b — a=d ifandonly ifa=b—candb+c=d.
a+2c#dorb —aF 2dif and only if b + 2c ¥ 3d or 3a + 4c Fb.
a’,at+b,anda+b+careall odd if and only if ab + bc and b + c¢ are
even and a + c is odd. Use Exercise 5 of Section 1.4.
50
CHAPTER 1
Logic and Proofs
8.
Let a, b, c, and d be positive integers, By an example and an exercise in this
section, (i) a2 + b? is even if and only if a and b have the same parity, and (ii)
a+c=band 2b — a=difand only ifa=b —candb+c =d. Using these
results, which of the following can we conclude, and why? If the statement is
not a consequence of these results, give a counterexample.
(a) a? +b? is odd if and only if a and b have opposite parity.
(b)
(ce)
(d)
If2b—aF#d,thenb+cF#d.
If2b—aFd,thenb+c#dora#b-—c.
If2b—a#d,thenb+c#danda#b—c.
Prove by contradiction that if m is a natural number, then
Proofs to Grade
n
n+
1
>
n
n+2
;
10.
Prove that V5 is not a rational number.
11.
Three real numbers, x, y, and z, are chosen between 0 and |. Suppose that
0O<x<y <z < 1. Prove that at least two of the numbers x, y, and z are
within ;unit from one another.
12.
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
(a) Suppose m is an integer.
Claim. If m? is odd, then m is odd.
“Proof.” Assume that m? is not odd. Then m? is even and m* = 2k for
some integer k. Thus 2k is a perfect square; that is, V2k is an integer.
If V/2k is odd, then V2k = 2n + 1 for some integer n, which means
mm = 2k = (2n + 1)* = 4n? + 4n + 1 = 2(2n* + 2n) + 1. Thus m? is
odd, contrary to our assumption. Therefore, V2k = m must be even.
Thus if m7 is not odd, then m is not odd. Hence if m? is odd, then m
*
(b)
is odd.
a
Suppose ¢ is a real number.
Claim.
If fis irrational, then 5r is irrational.
“Proof.”
Suppose 5r is rational. Then St = p/q, where p and q are
integers and g # 0. Therefore, t = p/(5q), where p and 5q are integers
and 5q # 0, so t is rational. Therefore, if f is irrational, then 5r is irra-
(c)
(d)
tional.
a
Suppose x and y are integers.
Claim.
If x and y are even, then x + y is even.
“Proof.”
Suppose x and y are even but x + y is odd. Then, for some
integer k,x +
y= 2k +71. Therefore, x + y + (—2)k = 1. The left
side of the equation is even because it is the sum of even numbers.
However, the right side, 1, is odd. Because an even cannot equal an odd,
we have a contradiction. Therefore, x + y is even.
a
Suppose a, b, and c are integers.
Claim.
If a divides both b and c, then a divides b + c.
“Proof.”
Assume that a does not divide b + c. Then there is no
integer k such that ak = b + c. However, a divides b, so am = b for
some integer m; and a divides c, so an = c for some integer n. Thus
am + an =a(m+n)=b-+c. Therefore, k = m+n
is an integer
1.6
(e)
Proofs Involving Quantifiers
51
satisfying ak = b + c. Thus the assumption that a does not divide
b + cis false, and a does divide b + c.
|
Suppose m and n are integers.
Claim.
If m? +n? is even, then m and n have the same parity.
“Proof.”
Suppose m? + n? is even, and m and n have opposite parity.
We may assume that m is odd and n is even. Then for some integers j
and k,m = 2j+ 1 andn = 2k, som* +n? = 47? + 47+ 1+ 4k? =2(27? +
(f)
2j + 2k*) + 1, which is odd. This is a contradiction.
a
Suppose a and b are positive integers.
Claim.
If a+ 1 divides b and b divides b + 3, then a = 2 and b = 3.
“Proof.”
Assume a + 1 divides b and b divides b + 3. Then
b=(a-+ 1)kandb + 3 = bj, for some integers k andj.Choose k = | and
p=leThen ves
2).s0 2b — b— 3. Thus b'— 3. From b = (a -- 1)k;
Wie aViei=— i(Gia ilsOm—eulnerelorena =
1.6
rancipi—so.
a
Proofs Involving Quantifiers
In previous sections we proved statements such as “Ifx is odd, then x + 1 is even”
by asking you to think of x as some fixed integer. A more complete version of this
statement includes a universal quantifier with variable x: “For every integer x, if x is
odd, then x + | is even.” The methods for proving statements with quantifiers are
discussed in this section, beginning with the universal quantifier.
A direct proof of (Vx)P(x) begins by letting x represent an arbitrary object
in the universe and then shows that P(x) is true for that object. We may use any
method to prove that P(x) is true, but we are allowed to use only properties of x
that are shared by every element of the universe. Because x represents any element
of the universe, we can then conclude that (Vx) P(x) is true.
DIRECT PROOF OF (Vx)P(x)
Proof.
Let x be an arbitrary object in the universe. (The universe should be
named or its objects described.)
Hence P(x) is true.
Therefore, (Vx) P(x) is true.
a
Many of the proof examples in Sections 1.4 and 1.5 are examples of complete direct proofs for statements of the form (Vx) P(x). Whenever those statements
involved a universally quantified statement, we were careful to begin the proof with
an assumption such as “Let n be an integer” or “Let x and y be real numbers.”
The direct proof that follows shows that the sum of a positive real number and
its reciprocal is always at least 2. The logical form of this statement is (Vx) P(x),
52
CHAPTER 1
Logic and Proofs
where the universe is (0, 00) and P(x) 1s “x + 7 > 2.” The proof was derived by
working backward from x + t =a.
Prove that the sum of a positive real number and its reciprocal is at
Example.
least 2.
Proof.
(Begin by assuming thatxis an element of the universe.) Let xbe a positive
real number. We know that (x —
1)* > 0, which implies that ye
ee
0)
Thus x? + 1 > 2x. (Next we use a property shared by all elements in the universe,
namely that x is positive.) Because x > 0, we conclude that
x+i>2.
x + |
x
= 2 sl nerelore:
=
In our next examples we consider odd functions. By definition, a real function
fis an odd function iff(—x) = —f(x), for every real number x.
Example.
Prove that the function f given by f(x) = x° — 4is an odd function.
Proof.
(To prove that fis odd, we must prove (Vx)P(x), where the universe is R and
P(x) is “f(—x) = —f(x).”) Let x be a real number. Then f(—x) = a
xe + t =
ee — “) = —f(x). Therefore,
fis an odd function.
Be
X
5
Usually, when we attempt to prove (Vx) P(x), we find that the open sentence
P(x) is a combination of other open sentences joined by logical connectives. For
instance, the logical form of the statement in the example about the sum of a number
and its reciprocal is (Vx)(H(x) =
K(x)), where the universe is R, H(x) is “x > 0,”
and K(x) is “x + t > 2.” As always, the logical form of P(x) helps us determine a
method for completing the proof.
Example.
If fis an odd function, then f(0) = 0.
Proof.
(The statement we must prove has the form (Wf)[A( f ) => B(f)], where the
universe is the set of all real functions, A(f )is “fis odd,” and B(f) is “f(O) = 0.” )
Let fbe a real function. (We now give-a direct proof of A( f) => B(f ).) Assume that
fis odd. Then for every real number x, f(—x) = —f(x). Because 0 is a real number,
we have f(—0) = —f(0). But —0 = 0, so f(—0) = f(0). Thus f(0) = —f(0). Adding
f(0) to both sides of this equation, we have 2f(0) = 0, so f(0) = 0.
a
In the proof above, we made two assumptions about f, for different reasons. The
first assumption was that fis a real function—that is, fis in the universe. We did not
assume that f was a polynomial, or a particular function, or anything else about f.
The second assumption was that fis odd, which we made as the first step of a direct
proof of a conditional sentence. Although their roles are different, we could have
combined these assumptions into one sentence: “Let f be a real function that is odd.”
1.6
Proofs Involving Quantifiers
53
It would be a mistake to give an example (or several examples) of odd func-
tions whose image at 0 is 0 and then claim that f(0) = 0 for every odd function f.
Examples may sometimes help us decide whether a statement is true. Examples
can also help guide our thinking about how to proceed with a proof. However, we
cannot prove that a universally quantified statement is true by showing that it’s true
for selected values of the variable.
A linear combination of the integers a and b is an integer of the form ax + by,
where x and y are integers. For example, some linear combinations of 8 and 12 are
4 = 8(—1) + 12(1) and 20 = 8(10) + 12(—5). The next theorem establishes this
important result that we will use later: If the integer c divides both a and b, then
c also divides every linear combination of a and b. It incorporates all our earlier
examples and exercises showing that if c divides two numbers, then it divides the
sum, difference, and multiples of those numbers.
The logical form of the theorem involves five universal quantifiers with universe the integers:
(Va)(Vb) (¥c)((c divides a A c divides b) > (Wx) (Vy)(c divides ax + by))
Reading the form from left to right will indicate the structure for a proof. The opening statement in the proof deals with the quantifiers for a, b, and c. Later in the
proof we consider the quantifiers associated with x and y.
Theorem 1.6.1
If a, b, and c are integers, and c divides a and c divides b, then c divides every linear
combination of a and b.
Proof.
(Introduce the first three quantified variables.) Let a, b, and c be integers.
(Next assume the antecedent.) Suppose that c divides a and c divides b. Then
there exist integers k and m such that a = ck and b = cm. (Introduce the next two
variables.) Let x and y be integers. Then ax + by = (ck)x + (cm)y = c(kx + my).
Because kx + my is an integer, c divides ax + by. Therefore, c divides every linear
combination of a and b.
a
The method of proof by contradiction is often used to prove statements of the
form (Vx) P(x). Because ~(Vx) P(x) is equivalent to (4x) ~ P(x), the form of the
proof is as follows:
PROOF OF (Vx)P(x) BY CONTRADICTION
Proof.
Suppose ~( Vx) P(x).
Then (Sx) ~P(x).
Let t be an object such that ~P(f).
Therefore Q A ~Q, which is a contradiction.
Thus (4x) ~P(x) is false, so (Vx) P(x) is true.
|
54
CHAPTER 1
Logic and Proofs
The following example of a proof by contradiction comes from an exercise
in a trigonometry class. It uses algebraic and trigonometric properties available to
students in the class.
JE.
Example.
Proof.
Prove that for all x € (0, oF sinx + cosx > l.
(The statement has the form (Wx)P(x), where the universe is the open
interval (0, a and P(x) is “sin x + cos x > 1.”) Suppose that the statement is false.
Then there exists a real number
ft, with 0 < t <
> such that sint + cost < l.
(We have deduced (At) ~ P(t).) Because the functions sin x and cos x are positive
for every x € (0; ah sin t > 0 and cos t > 0. Therefore,
O < snt + cost < 1
0 < (sinf
+ cos f)* < 1? = 1
0 < sin? t+ 2 sin ¢cos t + cos? t < 1
0 < 1+ 2sintcost
<1
Sle
(We use the identity sin? t + cos* t =
sin acosi,
= 0:
1.) But2 sint cos t < 0 is impossible because
both sin t and cos ¢are positive. Therefore, if 0 < x < 7, thensinx + cosx>
1. m
There are several ways to prove existence theorems—that is, propositions of
the form (4x) P(x). In a constructive proof we actually name an object a in the
universe such that P(qa) is true, which directly verifies that the truth set of P(x) is
nonempty.
CONSTRUCTIVE PROOF OF (4x)P(x)
Proof.
Specify one particular object a.
If necessary, verify that a is in the universe.
Therefore, P(a) is true.
Thus, (4x)P(x) is true.
z
Some constructive proofs are quite easy to devise. For example, to prove that
“There is an even prime natural number,” we simply draw on our knowledge of
natural numbers to note first that 2 is a natural number and then that 2 is prime and
2 is even.
Constructive proofs often hide all the preliminary work that goes into finding
an object a that makes a proof of P(a) possible. Some constructive proofs have
eluded mathematicians for centuries. The question of whether any nth power is a
1.6
Proofs Involving Quantifiers
55
sum of fewer than n nth powers was raised by Leonard Euler’ in the mid-1700s.
A computer search in 1968 discovered a fifth power that was the sum of four fifth
powers. Here is an example for fourth powers:
Example.
Prove that there exists a natural number whose fourth power is the sum
of three other fourth powers.
Proof.
20,615,673 is one such number because it is a natural number and
20615673* = 26824404 + 15365394 + 18796760+.
|
Another strategy to prove (4x) P(x) is to show that there must be some object
for which P(x) is true, without ever actually producing a particular object.
INDIRECT
PROOF
OF (4x)P(x)
Proof.
Therefore, there must be an object a such that P(q) is true.
Therefore, (4x)P(x) is true.
|
Astronomers use slight variations in the intensity of light from a star to prove
the existence of planets that cannot be observed directly. How do mathematicians
prove that something exists without specifically exhibiting the object? To fill in the
gaps in the proof outline of (Axv)P(x), we look for facts that imply that some element a in the universe is in the truth set of P(x). A thorough understanding of the
concepts expressed in P(x) is essential.
Rolle’s Theorem from calculus is a good example of an existence theorem that
is proved indirectly. Its proof uses properties of continuous functions and properties
of the real numbers. The next example makes use of The Fundamental Theorem of
Algebra and The Complex Root Theorem.
Example.
Prove that the polynomial
FO =e
ae
x
0S
has a real zero.
Proof.
(The universe is R. The statement has the form (At)(r(t) = 0).) By the
Fundamental Theorem of Algebra, r(x) has 71 zeros that are either real or complex.
* Leonard Euler (1707-1783) was a brilliant Swiss mathematician who spent much of his career at the
Imperial Russian Academy of Sciences in St. Petersburg and the Berlin Academy. He made profound
contributions to calculus, number theory, and graph theory, as well as to physics and astronomy. He was
the first to introduce the idea of function and the familiar f(x) notation.
56
CHAPTER 1
Logic and Proofs
Since the polynomial has real coefficients, by the Complex Root Theorem its
nonreal zeros come in pairs. Hence the number of nonreal zeros is even, and that
leaves an odd number of real zeros. Therefore, r(x) has at least one real zero.
a
Example. There are n people (n > 2) at a party, some of whom are friends. Prove
that there exists someone at the party who is friends with the same number of partygoers as another person.
Proof.
(We assume such a person does not exist and look for a contradiction.)
Suppose that for every person, the number of people that person is friends with is
different from the number of people anyone else is friends with. The smallest possible number of friends is 0 and the largest is n — 1, so the number of friends dif-
ferent people have must be 0, 1, 2,..., 2 — 1. Then the person who is friends with
n — | people is friends with everyone else at the party. This is impossible, because
one person at the party has no friends at the party. We conclude that there are two
(or more) people who are friends with the same number of people at the party.
™
Existence theorems may also be proved by contradiction. The proof technique
has the following form:
PROOF OF (3x)P(x) BY CONTRADICTION
Proof.
Suppose ~(4x) P(x).
Then (Vx) ~ P(x)
Therefore, ~Q A Q, which is a contradiction.
Thus ~(4x) P(x) is false.
Therefore, (4%) P(x) is true.
a
The core of a proof of (4x) P(x) by contradiction involves making deductions
from the statement (Vx) ~ P(x).
Example. Let S be a set of 6 positive integers, each less than or equal to 10. Prove
that there exists a pair of integers in S whose sum is 11.
Proof. Suppose the set S has six elements, and there is no pair of numbers in S$
whose sum is 11. Then S may contain at most one element from {1, 10}. Similarly,
S may contain at most one number from each of the sets {2, 9}, {3, 8}, { 4, 7}, and
{5, 6}. Thus S contains at most five elements. This is a contradiction. We conclude
that S contains a pair of numbers whose sum is 11.
]
1.6
Proofs Involving Quantifiers
57
Sometimes a statement to be proved has the form (Sx) P(x) > Q. As a first
step, we assume (4x) P(x). However, the fact that some object x in the universe
has the property P(x) does not give us much to work with. A useful step is to name
some particular object that has the property and use the property of the object to
derive OQ.
Example. The graph of x? + y* = r?, withr > 0, is acircle with center (0, 0) and
radius r. Prove that if one of the x-intercepts of the circle has rational coordinates,
then all four intercepts have rational coordinates.
Proof.
Suppose that there exists an x-intercept (a, 0) of the circle with rational
coordinates. Then ais a rational number and a? + 0? = r2,soa? = r*anda = +r.
Then the other x-intercept is (—a, 0). To find the y-intercepts, we solve 0? + y? = r?
and find y = +r = +a. Therefore, the four intercepts are (a, 0), (—a, 0), (0, a),
and (0,—a), all of which have rational coordinates.
|
The logical forms of many statements in mathematics are a mixture of universal and existential quantifiers. We must deal with each in succession, starting from
the lert,
Theorem 1.6.2
Between any two rational numbers x and y there is a rational number z.
Proof.
Without loss of generality, we may assume thatx< y. (With this assumption,
the statement may be symbolized by (Wx € Q)(Vy € Q)[x <
(ax € Q)
S
(x < z < y)].) Suppose thatx and y are rational numbers. Then x = and y= Ae
‘
x
some integers p, g, 5, t, where neither g nor f is 0. Let z be
y >
us (We must show
that z is rational.) Then
st (tt) aaa)
(ee
Tap Le
Meso eee) a
ee oes Ma;
Both pt + gs and 2qt are integers (because products and sums of integers are
integers) and 2gt # 0, so z is rational. (We must show that z is between x and y.)
Key
Vy
Furthermore,
Example.
x =
5
<
5
a
5
= "ylsOrm<
2 XY.
a
Prove that for every natural number n, there is a natural number M such
that for all natural numbers m > M,
58
CHAPTER 1
Logic and Proofs
Proof.
(The statement may be symbolized by
(Vn
EN)(AM EN)(Vm € N) (nSil=
l
1
me < +)
We begin with the universal quantifier on the left.) Let n be a natural number.
(We must prove the existence of a natural number M with the given property.)
Choose M to be 3n. Let m be a natural number, and suppose that m > M. Then
|
|
m > 3n, and 3mn > 0, so dividing by 3mn we have vi Fas (The choice of 3n
for M is the result of some scratchwork,
working backward from the intended
conclusion — =< -—,)
m
3n
|
Example. Prove that there is a real number with the property that for any two
larger numbers, there is another real number that is larger than the sum of the two
numbers and less than their product.
Proof.
(We must prove (4z)(Vx)(Vy)[(& > z A y > z) > (Aw)(x + y < w < xy)].)
Let z be 2. (This choice later ensures that x + y < xy.) Now suppose that x and y
are real numbers, x > z and y > z. Without loss of generality, we may assume that
x <y. Then x + y < 2y < xy. Now let w be the midpoint between x + y and xy. Then
X+YyY<W<Xy.
,
|
A proof of a statement about unique existence always involves multiple
quantifiers. A common technique for proving a proposition of the form
(S!x)P(x) is based on proving the equivalent statement: (4x) P(x) A (Vy)(Vz)
[P(y) A P(z)=y =z]. Because this propositional form is a conjunction, the
method will have two parts:
PROOF OF (4!x)P(x)
Proof.
(i) Prove that (4x) P(x) is true. Use any method.
(ii) Prove that (Vy)(¥z)[PQ) A P@) => y = Zz].
Assume that y and z are objects in the universe such that P(y) and P(z)
are true:
Therefore) yz.
From (i) and (11) conclude that (4!x) P(x) is true.
|
=<}
Example.
inverse.
Prove that every nonzero real number has a unique multiplicative
1.6
Proofs Involving Quantifiers
59
Proof.
(The statement has the form (Vx € R)\(x
#0 => (Aly € R)(xy = 1).) Let x
be a nonzero real number. (We show that there that is a unique real number y such
that xy = | in two Steps.)
‘
(i)
(ii)
dabnahe
(First give a constructive proof to show that y exists.) Let
lie
y = —. Since
x
:
1
x # 0, yis areal number. Then xy = (+) = |. Therefore, x has a multiplicative inverse.
iy
(Now show that x cannot have two different inverses.) Suppose that w and z
are multiplicative inverses for x. Then xw =
1 and xz =
1, so
XW = KZ
AW ae X
ee)
x(w
— z) = 0.
Because x # 0, w — z = O. Therefore, w = z.
|
Great care must be taken in proofs that contain expressions involving more
than one quantifier. Here are some manipulations of quantifiers that permit valid
deductions:
(Vx) (Vy) P(x, y) <= (Vy) (Vx) P(x, y).
(Ax)(Ay)P(x, y) <> (Ay)(Gx) PG, y).
[(Vx) P(x) V (¥x)O(x)] => (WX) PO) V O(x)].
(Vx)[P@) > OG) => [Vx)P@) => (YH) O@)1(Wx)[ PO) A QO(x)] = [(Vx) P(X) A (¥x) Ox).
ea
Oe
Se
(Ax) (Wy) P(x, y) => (Vy) (Ax) PC, y).
You should convince yourself that each of these is a logically valid conditional or biconditional. For example, the last on the list is always true because
if (4x)(Vy) P(x, y) is true, then there is (at least) one x that makes P(x, y) true no
matter what y is. Therefore, for every y, (4x) P(x, y) is true because this particular
X exists.
The converses of statements 3, 4, and 6 are not valid. These converses and
other common incorrect deductions involving multiple quantifiers are discussed in
Exercise 8.
Finally, there are times when we will want to prove a quantified statement is
false. We know that (Vx) P(x) is false precisely when ~( Vx) P(x) is true and that
~(Vx) P(x) is equivalent to (4x) ~P(x). Therefore, one way to prove that (Vx) P(x)
is false is to prove that (4x) ~ P(x) is true.
A constructive proof of (4x)(~ P(x)) names an object a in the universe such
that P(a) is false. The object a is called a counterexample to (Vx) P(x). The
number 2 is a counterexample to the statement “All primes are odd.” The function
60
CHAPTER 1
Logic and Proofs
f(x) = |x| is a counterexample to “Every function that is continuous at 0 is differentiable at 0.”
Example. Some beginning algebra students believe that (x + y)? = x* + y*. In
symbolic terms, they believe that (Vx)(Vy)[(x + y)? = x? + y?] is true in the universe of real numbers. This mistake could be corrected by providing a counterexample—for instance,
Proving that
x = 3 and y = 4.
~ (4x) P(x), is false is equivalent to proving that (Vx) ~P(x) is
true. This is the logic behind our next example. We proved in Section 1.4 that every
odd integer can be written in the form 47 — 1 or 4k + 1. We now show that there
does not exist an integer that can be written in both of these forms. The proof is by
contradiction.
Example. There is no odd integer that can be expressed in the form 47 — | and in
the form 4k + 1 for integers 7 and k.
Proof. Suppose n is an odd integer, and suppose n = 47 — | andn = 4k + 1 for
integers jand k. Then 4j — 1 = 4k + 1, so 4j — 4k = 2. Therefore, 27 — 2k = 1.
The left side of this equation is 2(j — k), which is even, but | is odd. This is a
contradiction.
Py
Exercises 1.6
1.
Prove that
*
(a)
(b)
(c)
(d)
there exist integers m and n such that 2m + 7n = 1.
there exist integers m and n such that 15m + 12n = 3.
there do not exist integers m and n such that 2m + 4n = 7.
there do not exist integers m and n such that 12m + 15n = 1.
(e)
(h)
forevery integer ¢, if there exist integers m andn such that 15m + 16n = t,
then there exist integers r and s such that 3r + 8s = t.
if there exist integers m and n such that 12m + 15n = 1, then m and n
are both positive.
for every odd integer m, if m has the form 4k + 1 for some integer k,
then m + 2 has the form 47 — 1 for some integer j.
for every integer m, if m is odd, then m? = 8k + 1 for some integer k.
(i)
for all odd integers m and n, if mn = 4k —
*
*
(f)
(g)
| for some integer k, then m
or n is of the form 47 — | for some integer /.
2.
*
Prove that for all integers a, b, and c,
(a) if adivides b — 1 and a divides c — 1, then a divides bc — 1.
(b) if a divides b, then for all natural numbers n, a” divides b”.
(c)
(d)
if ais odd, c > 0, c divides a, and c divides a + 2, thenc = 1.
if there exist integers m and n such that am + bn = 1 andc ¥ +1, then
c does not divide a or c does not divide b.
1.6
3.
Proofs Involving Quantifiers
61
Prove that if every even natural number greater than 2 is the sum of two primes,"
then every odd natural number greater than 5 is the sum of three primes.
4. Provide either a proof or a counterexample for each of these statements.
(La) For all positive integers x, x7 + x + 41 is a prime.
0b) , (Vx)(Sy)(x + y = 0). (Universe of all reals)
(c)
(Vx)(Vy)(x > 1 Ay > 0Sy > x). (Universe of all reals)
(d)
For integers a, b, c, if a divides bc, then either a divides b or a divides c.
(e)
For integers a, b, c, and d, if a divides
a divides b — d.
b — c and a divides c — d, then
) For all positive real numbers x, x7 — x > 0.
For all positive real numbers x, 2° > x + 1.
For every positive real number x, there is a positive real number y less
than x with the property that for all positive real numbers z, yz > z.
For every positive real number x, there is a positive real number y with
the property that if y < x, then for all positive real numbers z, yz > z.
(b)
Prove that the natural number x is prime if and only if x > 1 and there
is no positive integer greater than 1 and less than or equal to Vx that
divides x.
Prove that if p is a prime number and p #3, then 3 divides p* + 2.
(Hint: When p is divided by 3, the remainder is 0, 1, or 2. That is, for
some integer k, p = 3k or p= 3k + Lor pp= 3k +22.)
6.
Prove that
(a)
(b)
*
1
for every natural number n, rs<< Ih,
(c)
there is a natural number M such that for all natural numbers n > M,
1
= << (Oy
n
for every natural number n, there is a natural number M such that
(d)
(e)
(f)
(g)
1
NL,
Dh
there is a natural number M such that for every natural number n, — < M.
there is no largest natural number.
‘
there is no smallest positive real number.
For every integer k there exists an integer m such that for all natural
numbers n, we have
(h)
0
<m+k <n.
For every natural number n there is a real number r such that for all
:
1
natural numbers m and ¢, if tf > m > ms then t +n > 102.
(i)
there is a natural number K such that —
number larger than K.
0.01 whenever r is a real
4
* No one knows whether every even number greater than 2 is the sum of two prime numbers. This
is the famous Goldbach Conjecture, proposed by the Prussian mathematician Christian Goldbach in
1742. You should search the Internet to learn about the million-dollar prize (never claimed) for proving
Goldbach’s Conjecture. Fortunately, you don’t have to prove Goldbach’s Conjecture to do this exercise.
62
CHAPTER 1
Logic and Proofs
(j)
(k)
(1)
(m)
(n)
there exist integers L and G such that L < G and for every real number
hit LesenGthent 403 O22
there exists an odd integer M such that for all real numbers r larger than
1
M, we have — = 0:01.
2r
forevery natural number x, there is an integerksuch that 3.3x + k < 50.
thereexistintegersx < 100andy < 30suchthatx + y < 128 and forall
real numbers r and s, if r > x and s > y, then (r — 50)(s — 20) > 390.
for every pair of positive real numbers x and y where x < y, there exists
a natural number M such that if n is a natural number and n > M, then
fpean!
Starting at 9 a.m. on Monday, a hiker walked at a steady pace from the trailhead up a mountain and reached the summit at exactly 3 p.m. The hiker
camped for the night and then hiked back down the same trail, again starting
at 9 a.m. On this second walk, the hiker walked very slowly for the first two
hours, but walked faster on other parts of the trail and returned to the starting
point in exactly six hours. Prove that there is some point on the trail that the
hiker passed at exactly the same time on the two days.
Show by example that each of the following deductions involving multiple
quantifiers is not valid. Note that parts (b), (c), and (d) are the converses of the
valid deductions 3, 4, and 6, respectively, on page 59.
(a)
(b)
(ec)
(d)
Proofs to Grade
(Av)P(x) = (¥x)P(Q)
(Wx)[PQ) V O@)] = [(¥x)PQ) V (Wx)QQ)]
[(Vx)P(x) = (Vx)Q(x)] => (Vx)[PO) > Q@)]
(Vy)(Ax)P@, y) = (Ax)(Vy)P(, y)
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
(a) Claim.
Every polynomial of degree 3 with real coefficients has a real
zero.
“Proof.”
The polynomial p(x) = x* — 8 has degree 3, real coefficients, and a real zero (x = 2). Thus the statement “Every polynomial
(b)
of degree 3 with real coefficients does not have a real zero” is false, and
hence its denial, “Every polynomial of degree 3 with real coefficients
has areal zero, is true) ~
is
Claim.
There is a unique polynomial whose first derivative is 2x + 3
and which has a zero at x = 1.
“Proof.”
The antiderivative of 2x + 3 is x* ++ 3x+C. If we let
p(x) = x? + 3x — 4, then p’(x) = 2x + 3 and p(1) = 0. So p(x) is the
(c)
desired polynomial.
Claim.
Every prime number greater than 2 is odd.
a
“Proof.”
The prime numbers greater than 2 are 3, 5, 7, 11, 13, 17,
19,.... None of these is even, so all of them are odd.
|
1.6
x(a)
Claim.
Proofs Involving Quantifiers
63
There exists an irrational number r such that rY? is rational.
“Proof.”
If \/3V? is rational, then
r = V3 is the desired example.
Otherwise, \/3¥? is irrational and (V3V2)V2 = (V3) = 3, which is
rational. Therefore, either V3 or V3¥? is an irrational number r such
that rV? is rational.
(e)
Claim.
“Proof.”
Case
a
For every real number x, |x| > 0.
We proceed by three cases: x > 0, x = 0, and x < 0.
ec
0 Choose, srorsexample
mvi="45 [hen
(4) = 4. Thus
|x| = 0.
(f)
Case2.5
Case 3.
x = ON Then! 010s Thus sii| 0:
x < 0. Choose, for example, x = —5. Then |—5| = 5. Thus
[esilleezan
a
Claim.
If x is prime, then x + 7 is composite.
“Proof.”
Letx be a prime number. If x = 2, then x + 7 = 9, which is
composite. If x # 2, then x is odd, so x + 7 is even and greater than 2.
In this case, too, x + 7 is composite. Therefore, ifxis prime, then x + 7
is composite.
=
(g)
Claim.
If fis an irrational number, then t — 8 is irrational.
“Proof.”
Suppose there exists an irrational number ¢ such that t — 8 is
rational. Then t — 8 = s:where p and q are integers and q # 0. Then
=
; +8 =
p + 8q , with p + 8q and q integers and g # 0. This is a
contradiction because fis irrational. Therefore, for all irrational numbers
t,t — 8 is irrational.
(h)
Claim.
|
For real numbers x and y, if xy = 0, then x = 0 or y = 0.
“Proof.”
Case 1.
Case 2.
(i)
“Ifx = 0; then xy = Oy =.0If y= 0; then xy"= x00:
In either case, xy = 0.
i)
Claim.
For every real number x in the interval (3, 6), there is a natural
1
1
number K such that for every real number y, if y > K, then 7 < 70:
“Proof.”?
Assume that x is in the interval (3, 6). Then x > 3. Let K be
x + 7. Then K > 10. Suppose that y is a real number and y > K. Then
|
ie
(j)
LOS SO
vy
Claim.
10
|
For every natural number n,n < n°.
“Proof.”
Let n be a natural number. Since 7 is a natural number,
7) Since
is positive, 7 > =n. Therefore, n = n? for all natural
numbers n.
a
64
CHAPTER 1
Logic and Proofs
Mathematical Writing Style
A proof is nothing more than an explanation why a statement is true. Someone
else will read it, so your proof needs to be clear and correct. Writing correct
proofs requires both knowledge of the subject matter and understanding of correct proof methods. In addition, you should be aware that good mathematical
writing has a certain style—accepted ways to explain your work. Adopting this
style will help you communicate clearly and effectively, especially when the
material is complex or unfamiliar to your audience.
Here are some recommendations that will help you write clear, readable
proofs. The general comments are followed by some specific details.
Grammatically Correct Writing
First, proofs should be written in proper, correctly punctuated English.
Thoughts can be expressed in many different ways, but some guiding principles for clear writing always apply. Style and grammar manuals for
(nonmathematical) English, as well as guides such as The Chicago Manual of
Style,” can be helpful.
Always use complete sentences. A fragment like “Let x.” makes no sense,
but an equation like “x = 4” is a complete sentence with subject “x,” verb
“equals,” and object “4.” Equations do not normally stand alone, except in a
sequence of equations, each derived from its predecessor. Logical symbols such
as ~, A, V, =>, ,
V, and Jd are convenient abbreviations, but it’s best to avoid
using them in the final version of your proof. Although a solution to a calculus
assignment might consist just of a series of equations, writing “steps” consisting
of equations or formulas is not enough for a proof. Generally you need words to
explain the relationships among steps.
Use connecting words such as Therefore, Thus, So, It follows that, This
implies that, Consequently, Hence, and “For this reason.” You should try to
vary the words you choose; sometimes one of the words just sounds better. Reading your proof aloud can help you find places where you need to add connecting
words between steps.
Helping your audience understand your explanation
Readers must be able to follow your line of thinking. Use words to tie together
the steps in your proof. You must keep your reader informed about what your
steps mean, where they come from, and where they lead to (but not all in the
same sentence)."
* The Chicago Manual of Style, 16th edition, University of Chicago Press, 2010.
* We do not mean that you should put in comments (that look like this). Such comments are used in
this text solely as an aid in helping you learn how to structure proofs correctly and to explain some of
the mathematical relationships involved in a proof. They are not part of the proof.
Mathematical Writing Style
In mathematical writing it’s always assumed that the audience is familiar
with the rules of logic, so you don’t have to explain, for example, that your proof
uses a particular method. At times you do need to explain what you are doing.
For example, if you introduce a new variable in a proof, be sure to explain what
the variable represents. “Let y be the....”
Alert your reader about what is coming up next. A mathematician might
announce a new theorem by saying “We prove that...,” using the first personal
plural. Similarly, when you write a proof with several parts, you might want to
use the phrase “Next we prove that...” or “Next we show....”
What mathematics to include
Students often wonder what they may assume when writing a proof. In general,
for an exercise in this text you may use any preceding material: definitions, theorems, and previous exercises, as well as the preliminary material presented in the
Appendix. It’s true that when we did our first proofs about even and odd integers,
we did not use everything we know about integers, but that was only so we could
practice the use of definitions.
Obviously we can’t always include in a proof everything back to the beginning of a subject. In advanced textbooks it is assumed that the audience understands topics such as those covered in the early chapters of this text. Professional
journals assume much more but provide references for the nonspecialist. You
should think of your audience as being other students who have read just as much
of the text as you have. Include just enough information so that another student
should be able to understand your proof.
You may not skip over a part of a proof that you don’t know how to do by
using the words Clearly, Obviously, or It is apparent that.... Those words are
appropriate when the claim really is obvious but tedious to verify, or when the
explanation is too long to include in the proof. The words serve as a cue that the
reader should expect to take some time to go through the details (and mathematicians usually do). Students reading advanced mathematics texts learn that the
time needed to verify what some authors call “obvious” may stretch into days.
Use those words sparingly.
Writing drafts
With practice it is possible to write clear, easily understandable—even elegant—
proofs. You needn’t try for elegant at this stage, but you should expect that your
first draft will need to be rewritten, maybe more than once. If you used the “‘forward and backward” method or otherwise constructed your proof in pieces, put
the pieces together in the proper order.
Go over your proof again, cleaning out steps that aren’t needed, putting your
statements in the proper order, and verifying that your proof is complete. When
possible, seek feedback from an instructor or another student.
65
66
CHAPTER 1 _ Logic and Proofs
Learning to recognize good proofs
One of the best ways to learn to write good proofs is to read as many correctly
written proofs as you can. Make critical thinking your attitude. That is, when reading a proof or listening to someone present a proof, think about the overall strategy
of the proof, make sure you follow the connections between steps, and be confident that you can fill in all omitted details. Develop the habit of checking all proofs
(not just Proofs to Grade exercises) for errors in logic or incomplete arguments.
Finally, here is a collection of tips about details such as capitalization, the
use of symbols, and the use of certain words.
When you refer to a specific theorem, lemma, or the like, capitalize the
name, as in “Theorem 5.3.”
When you give an object a name, be sure to write the name with the object
you’re naming. Thus if you are describing a sequence of real numbers, don’t
say “The sequence of real numbers x is...,” say “The sequence x of real
numbers is....”
The phrase “Since..., it follows that...” may be used as a substitute for
“Tf..., then...,” but in formal writing, use the form “Because
;
ee
;
rather than “Since
Avoid using the word any, which can be ambiguous. When you mean every
or each or some, choose the correct word.
Usually, the words Assume and Suppose should be followed by the word
that. Don’t say “Assume f is continuous.” Instead, say “Assume that f is
continuous.” However, there are times when the word that interrupts the
flow of a sentence. The wording “We have A = B” reads much better than
“We have that A = B.”
Use symbols consistently. In this text, we generally use the symbols P and
Q for propositions; A, B, C for sets; a, b, c for integers; f, g, h for functions;
and x, y, z for real numbers. You may choose other symbols, but don’t use
A, f, and z for functions in the same proof.
Avoid starting a sentence with a variable. Instead of “‘A is a subset of B,” say
“The set A is a subset of B.”
Sometimes symbols should be separated by words to avoid confusing the
reader. Instead of “Because x is an element of A, y, z, and w are not in B,”
write “Because x is in A, we have that y, z, and w are not in B.”
When you state a proposition or theorem, avoid using symbols that are not
needed. Instead of “Every compact set A of real numbers is closed,” say
“Every compact set of real numbers is closed.”
It is generally agreed that the following words are written correctly
without a hyphen: nonzero, nonempty, nonnegative, nonincreasing, and
nondecreasing.
1.7
1.7
Strategies for Constructing Proofs
67
Strategies for Constructing Proofs
For most people, the hardest part of writing a proof is knowing where or how to
start. There is no surefire way to construct a proof, but there are systematic ways
to begin. In particular, we will emphasize the fact that the logical form of the statement to be proved is usually the best guide to the finding the natural form a proof
should take.
Writing a proof is not done by staring at a statement to be proved until a complete proof suddenly appears in your head. Neither should you always think formally, step by step, while you search for a proof. You are allowed to use your
intuition to picture the situation and “see” that a statement is true. Such informal
thinking—piecing together facts and seeing the relationships among definitions and
previous results—can be just what is needed to discover a formal proof.
The most important strategy is to make a start. Almost any start, including a
false start, can give you some ideas that can be useful if you have to begin again
with a new approach. Here’s a summary of strategies you should try when you
begin to write a proof:
1.
Understand the statement to be proved. Make sure you know the definitions
of all the terms that appear in the statement. If there are formulas or equations
in the statement or the definitions, jot them down on scratch paper and be
sure that you understand the relationships the formula or equation expresses.
Often these equations and formulas are algebraically manipulated in steps in
the proof.
2.
Determine the logical form of the statement. It is important to be able to
write (or at least visualize) the complete symbolic translation, with quantifiers,
of the statement to be proved, because the logical form of the statement will
usually offer insight into how to proceed. Each logical form suggests a natural
outline for a proof, as the examples that follow illustrate.
3.
Identify the assumption(s) and the conclusion. Most theorems can be stated
in the form of a conditional sentence. Look for known facts and previous
results that might connect the antecedent with the consequent.
4.
Look for the key ideas. Ask yourself what is needed to reach the conclusion.
Look for relationships among the terms, equations, and formulas involved.
Recall known facts and previous results about the antecedent and consequent. For example, many theorems about right triangles use the Pythagorean
Theorem.
Even though the logical form of the statement usually offers insight into how
to proceed, this does not mean you should be overly concerned with naming different types of proofs or with devising “formulas” for writing proofs of a certain type.
The following list suggests proof strategies to try first for various logical forms,
but it’s not true that if a statement has a certain form you must always use a certain
proof technique. In a sense, any correct proof could be considered a good proof, but
68
CHAPTER 1
Logic and Proofs
it often happens that using one method produces a proof that feels awkward or artificial compared to that produced by another method. Once you have a correct proof,
consider rewriting it, looking for a more natural or elegant approach. Another reason not to be too concerned with naming types of proofs is that several “methods”
may be needed in the same proof. For example, a proof by cases may use a direct
proof in one case and a contradiction argument in another.
The remainder of this section includes specific suggestions about what to try
first once you have determined the logical form of the statement.
“If P, then Q”—Try a direct proof.
Try this method first whenever you encounter a conditional statement. Begin by
assuming that P is true. P may be a conjunction of several statements, so we may
assume all those statements are true. When Q is a conjunction, we must derive each
of its components.
Example.
Let a, b, and c be real numbers and i = V—1. If (a — bi)(c — ci) =
1 —iand
(a+ bi)* = 4, then b = 0 andc
= +%.
Proof.
Leta, b, and c be real numbers. Assume that (a — bi)(c — ci) =
1 — i and
(a + bi)* = 4. Then ac — be — (be + ac)i = 1 — i. It follows that ac — bc = 1 and
—(bc + ac) = —1. Then c # 0, and by adding the two equations we find —2bc = 0,
so b = 0. Therefore, ac = 1. From (a + bi)? = 4 and b = 0 we have a = +2, so
C= sep.
:
|
“If P, then Q”—Try a direct proof, working forward and backward.
Once the hypotheses and conclusion have been identified, write the assumptions,
leave some space, and write the conclusion as the last line. Try to deduce statements
from the hypotheses that are more useful or easier to manipulate. Work “‘backward”
by rewriting the conclusion or trying to find a suitable statement or statements from
which your conclusion follows. The idea is to try to reason forward from your
assumption and backward from your conclusion until you join them.
Example.
Let a, b, and c be real numbers. If a? = ac — ab + bc anda +b#0,
then the lines y = ax + b and y = cx + dare parallel.
Proof.
Let a, b, and c be real numbers. Assume that a” = ac — ab + bc and
a+b #0. (To show that the lines are parallel, we need to show that a = c.) Then
a +ab=ac +be,soa(a+b)=clat+ b). Since a + b # 0, we have a = c. There-
fore, the lines are parallel.
=
“If P, then Q”—Try a proof by contraposition.
When a direct proof fails, consider a proof that ~Q implies ~P. This method is the
best option when Q has the form of a negation. The next example relies on previous
results, including the fact that x* + x is even for every integer x.
1.7
Example.
Strategies for Constructing Proofs
69
If a is an odd integer, then x? + x — a* = 0 has no integer solution.
Proof. Suppose that x” + x — a* = 0 has an integer solution f. By the quadratic
formula,
pag
V1 + 4a2
2
2t=-1+
V1 + 4a
%W+il= tVl 44a
4? 4+ 4¢+1=1+4
4a
rP+t=a
Because f + tis even, we have that a” is even, and therefore a is even. (Note that it
is not always necessary to cite references to results established by earlier exercises
and examples.) We conclude that if a is an odd integer, then x* + x — a* = 0 has no
integer solution.
=
“If P, then Q”—Try a proof by cases.
The method of exhaustion is suggested when P has the form P, V P,. We show that
(i) P; implies Q and (ii) P, implies Q. The cases method may also be used when
the antecedent P does not have the form of a disjunction, so long as we remember
to consider every possibility. For example, to prove that rational functions have a
certain property, we could show that functions have the property in three separate
cases: when the degree of the numerator is less than, is equal to, and is greater than
the degree of the denominator.
Example. Let m and n be integers. If m and n differ by 3 and mn is negative, then
Moan =).
Proof. Let m and n be integers. Because mn is negative, m and n have opposite signs. Without loss of generality, we may assume that m < 0 < n. The only
possibilities are that (1) m = —2 andn = 1 or (ii) m = —1 andn = 2. In either case,
ie ee
g
Any statement—Try a proof by contradiction.
By assuming the negation of the statement to be proved, it is possible to apply the
method of contradiction to any statement. In particular, when the direct proof and
contrapositive methods for proving “If P, then Q” both fail, or when Q has the form
of a negation, try the method of contradiction. Assuming both P and ~Q gives you
two hypotheses to work with as you aim for some contradiction.
Example.
There is no real numberx such that x? — 2x < 0 and x? — 6x — 8 <0.
Proof.
Suppose that there is a real number x such that both x* — 2x < 0 and
x* — 6x — 8 < 0. Then x(x — 2) < Oand (x — 4)(x — 2) < 0. If x — 2 is positive, then
70
CHAPTER 1
Logic and Proofs
from the first inequality, x must be negative. However, x > 2 and x < 0 is impossible. On the other hand, if x — 2 is negative, then from the second inequality, x — 4
must be positive. However, x < 2 and x > 4 is impossible.
fm
Here are some other ideas for beginning proofs of statements with specific forms.
“If P, then Q, Vv Q,”
This statement is equivalent to the implication “If P and ~Q,, then Q,,” so any
of the methods above may be applied. If you aim for a direct proof, you have the
advantage of having both P and ~Q, as hypotheses.
Example.
OL yi
Proof.
If the point (x, y) is inside the circle (x — 6)? + (x — 3)? = 8, thenx > 4
Suppose that the point (x, y) 1s inside the circle (x — 6? +(y- 3)? = 8 and
x < 4. Then (x — 6)? + (y — 3)” < 8 and 6 — x > 2. Because (x — 6)* > 4, we see
that (y — 3)? = 4. Thus 2 >y— 3 > —2,s0 y > 1.
&
“P if and only if Q”
What you should hope for is a proof in which you construct a list of equivalent
statements linking P and Q. But usually, and especially when P and Q are complicated, you will need to prove P > Q and Q = P separately. Rather than worrying
about which proof form to use, a good strategy is to begin by proving either of the
two implications and then checking to see whether each step can be reversed so
that (by modifying the words that connect statements) the proof can be converted
to an “iff” proof.
Example. Suppose that a, b, c, and d are real numbers with a # 0 and c # 0. The
lines y = ax + b and y = cx + d have the same x-intercept if and only if ad = be.
Proof.
Leta, b, c, and d be real numbers with a ¥ 0 and c + 0. The lines have the
same x-intercept if and only if —b/a = —d/c, if and only if —bc = —ad, and if and
only if ad = be.
8
$(Wx)P(x)”
Very often, the statement to be proved begins with the word every or the phrase
for all. There may be more than one universal quantifier, and quantifiers may be
hidden. The first sentence of your proof will almost always have the form “Let x be
an object such that
” or “Suppose x is in
%
Example.
Proof.
For all real numbers s and ¢, if t > 4 and s(s — 4) < 0, then s < t.
Let s and ¢ be real numbers such that ¢ > 4 and s(s — 4) < 0. Then s and
s — 4 have opposite signs, so s — 4 < 0. Thuss <4 <t, sos <t.
a
1.7
Strategies for Constructing Proofs
71
(Ax) P(x)”
You may be able to give a constructive proof by finding an object that has the
desired property. Otherwise, you may be able to show, perhaps by contradiction,
that an object with the property must exist.
Example.
Let & be a natural number. There exists a natural number n such that n,
n+1,n+ 2,...,andn +k are not prime. (Thus, no matter how large k is, some
consecutive primes differ by at least k.)
Proof.
Let k be a natural number. Choose n to be (k + 1)! + 2. Then n is divisible
by 2,n+ 1=(k+ 1)! + 3 is divisible by 3,n + 2 = (k + 1)! + 4 is divisible by
4, and so on, until we get ton + k= (k+ 1)! +k + 1, which is divisible by k + 1.
Therefore, none of these k consecutive numbers is prime.
a
“(Alx) P(x)”
There are two parts to proving that there is a unique object with a given property:
We must prove that there is such an object, and that there is only one such
object. It often happens that the natural first step is to find what the unique
object must be. (This is the usual first step in solving an equation.) Usually it
is then easy—but essential—to show that the discovered object has the desired
property.
Example.
Let a be a fixed real number. Let /, be the line y = ax + 7 and 1, be
the line y = cx + 5. There is a unique real number c such that /, and /, have the
same x-intercept.
Proof.
(i)
(ii)
Suppose that y = ax + 7 and y = cx + 5 have the same x-intercept. Then
—Tla = —5/c, so c = (5/7)a.
Suppose that c = (5/7)a. Then the x-intercept of the line y = cx + 5 is
—5/e = —5/(5/7)a = —7/a, which is the same as the x-intercept of the line
y=ax+7.
a
The other way to prove a statement of the form (4!x)P(x) is to begin with
the existence step. After that, there are three approaches to proving uniqueness. You may (1) prove that any two objects with the property must be equal,
(2) derive a contradiction from the assumption that two distinct objects have
the property, or (3) prove that every object with the property is equal to some
specified object.
Example.
There is exactly one real number x such that (x + 4)(x — 6) < 0 and
(x — 6)\x— 11) <0.
72
CHAPTER 1_
Logic and Proofs
Proof.
(i) There is at least one number, 6, for which both inequalities are true.
(ii) Suppose that (x + 4)(x — 6) < 0, (x — 6)(x — 11) < 0, and x # 6. Then x > 6
is impossible, because for x > 6, x + 4 and x — 6 are both positive. And
x < 6 is impossible, because for x < 6, x — 6 and x — 11 are both negative.
We conclude that 6 is the only number x such that (x + 4)(x — 6) < O and
(x — 6)(x — 11) < O.
|
Exercises 1.7
1
*
(a)
Let a be a negative real number. Prove that if a is a solution to the
equation x* — x — 6 = 0, then ais a solution to x° + 2x7 +x+2=0.
x
*
(b)
(Cc)
*
(d)
x
(e)
Let x be a real number. Prove that 0 < x < 3 impliesx + 1 > (x — 1)2.
Let a and b be positive integers such that a # 0. Prove that if a does
not divide b, then there is no positive integer x that is a solution to
ax? +bx+b—a=0.
Sit 2
Prove that if x is a real number and x > 1, then
|4
Prove that no point inside the circle (x — 3)? + y* = 6 is on the line
Weehie
ged hes
x
(f)
Let x be a real number. Prove that if x7 + 7x = 9x + 15, then x < 2 or
x-—4
> 0.
x—3
x
(g)
x
(h)
Suppose that x is a real number. Prove that —x* + 7x — 10 < 0 if and
onlysifene= 2 or x5!
Prove that if two nonvertical lines are perpendicular, then the product
of their slopes is —1. (Recall that nonvertical lines are those lines in the
plane that have slope.)
x
(i)
Prove that there exists a point (x, y) inside the circle (x — 3)? + (y — 2)?
= 13 such that x < 0 and y > 3.99.
x
(j)
Prove that there is a pair, and only one pair, (x, y) of real numbers such
2.
Prove that if (a + bi)(c + di) = 2, thena 4 0 ord ¥ 0. Give (a) a direct proof,
thatO <x <2,0<y <1, and 3x? 4+ 2y’ > 14,
(b) a proof by contraposition, and (c) a proof by contradiction. Which two of
the three methods produce very similar proofs?
3.
Prove that
(a)
(b)
(c)
(d)
5n* + 3n + 4 is even, for all integers n.
for all integers n, if 5n + 1 is even, then 2n? + 3n + 4 is odd.
the sum of five consecutive integers is always divisible by 5.
n> — nis divisible by 6, for all integers n.
1.7.
(e)
(f)
(g)
(h)
(i)
Strategies for Constructing Proofs
(n> — n)(n + 2) is divisible by 12, for all integers n.
every four-digit palindrome number is divisible by 11. (A palindrome
number is a number that reads the same forward and backward.)
if a, b, and c are real numbers and (a — bi)(c — ci) = 1 — i, then ac = 1.
if p is a prime integer, then p + 19 is composite.
if n is a natural number and r is the remainder when n is divided by 3,
then if = 5 or 11, the sum 7° + r+
(j)
(k)
(I)
73
fis prime.
if m and n are natural numbers, n < 4, and (2 — n)(2 — m) > 2(m — n),
shen 71 — LOL tas 3.
if n > 2 is an even natural number, then 2” — | is not prime.
if S is a set of real number such that a, b are in S, and ifx < aandx <b
for every element x of S, then a = b.
(m)
if a and b be integers and b is odd, then 1 and —1 are not solutions of the
(n)
equation ax + bx?
+a=0.
if two nonvertical lines have slopes whose product is —1, then the lines
are perpendicular.
Let L be the line 2x + ky = 3k. Prove that
(a)
if k is not equal to —6, then L does not have slope i
(b)
for every real number k, line L is not parallel to the x-axis.
(c)
(d)
there is a unique real number k such that L passes through (1, 4).
the x-intercept of L is less than k if and only if k is negative.
Prove that
(a) if x + y is irrational, then either x or y is irrational.
(b) if xis rational and y is irrational, then x + y is irrational.
(c) there exist irrational numbers x and y such that x + y is rational.
(d) for every rational number z, there exist irrational numbers x and y such
that ci yz.
(e)
(f)
(a)
(b)
for every rational number z and every irrational number x, there exists a
unique irrational number y such that x + y = z.
for every positive irrational number x, there is a positive irrational number y such that y < % and y < x.
Prove that except for two points on the circle, if (x, y) is on the circle
with center at the origin and radius r, then the line passing through (x, y)
and (r, 0) is perpendicular to the line passing through (x, y) and (—r, 0).
Which two points are the exceptions?
Let (x, y) be a point inside the circle with center at the origin and radius r.
Prove that the line passing through (x, y) and (r, 0) is not perpendicular
to the line passing through (x, y) and (—r, 0).
Prove that
(a)
every point on the line y = 6 — x is outside the circle with radius 4 and
center (—3, 1).
(b)
there exists a three-digit natural number less than 400 with distinct digits,
such that the sum of the digits is 17 and the product of the digits is 108.
74
CHAPTER 1
Logic and Proofs
(c)
(d)
if f does not have a maximum value on the interval [5, 7], then fis not
differentiable on [5, 7]. Use the Extreme Value Theorem.
the equation x3 + 6x — 1 = Ohas at most one real solution. Use Rolle’s
Theorem.
Prove that
(a)
;
|2x — 1|
for all real numbers x, if x > O, then cae ——
(b)
forallreal numbersx, if —2 < x < lorx > 3,then meena
(c)
(d)
for all real numbers x and y, x < (x + y)/2 if and only if (x + y)/2 < y.
for all real numbers x and y, (x + 1)? = (y + 1)? if and only if x = y or
Gee
Ss Sn
(e)
there is a real number x that is less than 7 and such that for every y > x,
(x — 1)@ + 2)
>
Yay? + y > 24.
Prove or disprove:
(a)
Every point inside the circle (x — 3)? + (y — 2)? = 4 is inside the
circle x- -— y- = 41.
(b)
(c)
If (x, y) is inside the circle (x — 3)? + (y — 2)? = 4, then x — 6 < 3y.
Every point inside the circle (x — 3)? 4 (y — 2)? = 4 is inside the
(d)
eirclere 25 )4
iy 2)
25:
If (x, y) is inside the circle (x + 3)? + (y + 2)? = 4 or the circle (x + 1)”
(e)
If (x, y) is inside the circle (x — 5)? + (y — 1)? = 17, then y < 2orxy> 0.
(f)
There is a unique real number x such that for all real numbers y, if y < x
then x + y is negative and x — y is positive.
+ (y — 3)? =9, thenx + y <3.
10.
Positive real numbers a and b, where a > b, are in the golden ratio’ if
+b
2
ae:
.= e ate This common ratio is a constant denoted @. (The exact value of
TANS
nb
ayes
é
d is Lea ae but we do not need that expression in this exercise.)
(a)
b
Suppose a and b are in the golden ratio. Prove that .=
a
(b)
Proofs to Grade
ie
—
Prove by contradiction that @ is an irrational number.
Assign a grade of A (correct);>C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
(a) Claim.
There is a unique three-digit number whose digits have sum 8
and product 10.
“Proof.”
Let x, y, and z be the digits. Then x +
y+ z= 8 and
xyz = 10. The only factors of 10 are 1, 2, 5, and 10, but since 10 is
* Many famous mathematicians have been intrigued by properties of the golden ratio. For instance, the
golden rectangle is a rectangle with sides of length a and b, and it can be divided into a square and a
smaller rectangle with ¢ (phi) as the ratio of their areas.
1.7
*
(b)
Strategies for Constructing Proofs
75
not a digit, the digits must be 1, 2, and 5. The sum of these digits is 8.
Therefore, 125 is the only three-digit number whose digits have sum 8
and product 10.
=
Claim.
There is a unique set of three consecutive odd numbers that are
all prime.
“Proof.”
The consecutive odd numbers 3, 5, and 7 are all prime. Suppose
that x, y, and z are consecutive odd numbers, all prime, and x # 3. Then
y=x-+2
and z=.x + 4. Since x is prime, when x is divided by 3,
the remainder is | or 2. In case the remainder is 1, then x = 3k + 1
for some integer k > 1. But then y = x + 2 = 3k + 3 = 3(k + 1), so
y is not prime. In case the remainder is 2, then x = 3k + 2 for some
integer
k > 1. But then z=x+4=
3k +2+
4 =
3(k + 2), so z is
not prime. In either case we reach the contradiction that y or z is not
prime, Thus x — 3 and soy —*5 and 27 = 7; Therefore, the only three
consecutive odd primes are 3, 5, and 7.
(c)
|
Claim.
If x is any real number, then either 27 — x is irrational or 7 + x
is irrational.
“Proof.”
It is known that z is an irrational number; that is, 7 cannot
be written in the form 5 for integers a and b. Consider x = z. Then
mz — x = O, whichis rational, but zm + x = 2. If 27 were rational, then
DF
; for some integers a and b. Then z =
as so 7 is rational. This
is impossible, so 27 is irrational. Therefore, either
7 — x ora +x is
irrational.
(d)
Claim.
a
If xis any real number, then either 7 — x is irrational or 7 + x
is irrational.
‘Proof.’
It is known that z is an irrational number; that is, 7 cannot
be written in the form for integers a and b. Let x be any real number.
Suppose both z — x and wz + x are rational. Then, since the sum of
two rational numbers
is always rational, (w — x) + (7 + x) = 2m 1s
Zr 1S
rational. Then 27 = £ for some integers a and b. Then z = ap SO
rational. This is impossible. Therefore, at least one of 7 — x or m + x
is irrational.
(e)
Claim.
=
For all real numbers x and y, x7 — 3x = y? — 3y” if and only if
w= onsets
“Proof.”
Suppose that x? — 3x = y? — 3y. Then x? + xy — 3x — xy —
y? + 3y =0, so (x + y — 3)(x — y) = 0. Therefore, x =
(f)
yorx+y=3.
m
Claim.
For all real numbers x and y, the equality xy = % (x + y)?
holds if and only ifx = y = 0.
“Proof.”
Part (i) Suppose that x = y = 0. Then xy = 0 = % (x + y)’, so the equality holds.
Part (ii) Suppose that x and y are real numbers and xy = %2 (x + y)’.
Then 2xy = x? + 2xy + y’, sox’ + y* = 0. Since the square of
a real number is never negative, x7 = y?=0,sox=y=0.
@
76
CHAPTER 1
Logic and Proofs
(g)
If 7 is prime andn +5 orn + 12 is prime, then n = 2.
Claim.
“Proof.’ Assume that n is a prime number. Then n > 2, so ifn + 5 is
prime, then n + 5 must be odd. Therefore, n must be even. Since 2 is the
5
only even prime, n = 2.
(h)
Claim.
Let a, b, and c be real numbers with a # 0. If ax* + bx
+c =
0 has no rational roots, then cx? + bx + a = 0 has no rational roots.
“Proof.”
Suppose that cx? + bx + a = 0 has a rational root p/q. Then
c(p/qy + b(p/q) + a = 0. Then c + b(q/p) + a(q/p)* = 0, so g/p is a
rational root of the equation ax” + bx
1.8
+c =0.
a
Proofs from Number Theory
This section is devoted to examples of proofs from elementary number theory—that
branch of mathematics concerned with the integers and questions about divisibility,
primes, and factorizations. The term elementary is used not because the subject is
low level, but because no methods from other fields of mathematics are used.
Our primary goal is to gain more experience with the proof methods of previous sections, so we consider only results that we can establish using those methods.
At the same time, we see how a rich theory can be built up from definitions and
previous theorems. We have already made a good beginning in number theory by
proving some results about divisibility, parity, and prime numbers.
The most fundamental theorem about the integers is the Division Algorithm, which
we state here without proof. In Chapter 2, the Division Algorithm will be presented as
Theorem 2.5.1 and proved using a technique that will be introduced in Section 2.5.
The Division Algorithm (See Theorem 2.5.1)
For all integers a and b, with a # 0, there exist unique integers g and r such that
b=ag+rand0 <r < |a|.
The integer a is the divisor, b is the dividend, q is the quotient, and r is
the remainder. For example, 23 divided by 4 gives a quotient of 5 and remainder 3, because 23 = 4-5 + 3. Note, however, that it would be incorrect
to say that —23 divided by 4 has quotient —S and remainder —3, even though
—23 = 4(—5) + (—3). Remainders can never be negative, so when we divide by
4, the only possible remainders are 0, 1, 2, and 3. When —23 is divided by 4, the
quotient is —6 and the remainder is 1, because —23 = 4(—6) + 1.
It is the fact that the remainder must be nonnegative and as small as possible
that makes the quotient and remainder unique. Notice that dividing b by a produces
a remainder of 0 exactly when there is an integer g such that b = ag + 0, which
happens exactly when a divides b.
A common divisor of nonzero integers a and b is an integer that divides both
aand b. For example, by comparing the list of divisors of 18 and the list of divisors
of 24, we see that there are eight common divisors of 18 and 24: —6, —3, —2, —1,
1, 2, 3, and 6. In this case, 6 is the largest (greatest) common divisor.
1.8
Proofs from Number Theory
77
DEFINITION
Let a and b be nonzero integers.
We say the integer d is the greatest common divisor (gcd) of a and b, and
write d = gcd(a, b),’ if
(i)
dis acommon divisor of a and b.
(ii) |every common divisor c of a and b is less than or equal to d.
It follows from the definition that gcd(a, b) is always a positive integer. By
comparing divisors, we find gcd(45, 20) = 5 and gced(84, 70) = 14.
There is no requirement that a and b must be positive. Therefore, we can find
gcd(—5, 20) = 5, ged(21, —35) = 7, and gcd(—9, —27) = 9. The integers 24 and
35 have no positive common divisors except 1, so gcd(24, 35) = 1.
In Section 1.6 we said that a linear combination of integers a and b is an expression of the form ax + by, where x and y are integers. Theorem 1.6.1 states that if c
is a common divisor of a and b, then c divides every linear combination of a and
b. In particular, because gcd(a, b) is a common divisor of a and b, it follows that
gcd(a, b) divides all linear combinations of a and b.
Example.
For the integers 12 and 18, some linear combinations are
(2G ral ol 6
(21) + I8Q)yr=24
12(10)+ 18(—2)
= 84
12(0) + 18(—4) = =72
Because 2 is a common divisor 2 of 12 and 18, we know that 2 divides 6, 84, 24,
—72, and all other linear combinations of 12 and 18. And because gced(12, 18) = 6,
we have that 6 must also divide all linear combinations of 12 and 18.
Oo
There is much more to be said about linear combinations. Whereas we look
for the greatest common divisor of a and b, we look for the smallest positive linear
combination of a and b. A natural first question to ask is: For every pair a, b of
nonzero integers, is there always a smallest positive linear combination? Yes, there
is, but once again, we need to assume the result here and wait until Chapter 2, where
we have the tools to give the proof. See Theorem 2.5.3.
Still, it is not too soon to see how we can use this result and basic proof tech-
niques to understand the connection between the gcd and linear combinations. We
have just seen that 6 is a linear combination of 12 and 18, so the smallest positive linear combination of 12 and 18 must be 1, 2, 3, 4, 5, or 6. Calculating more
combinations, such as 12(3) + 18(1) = 54 and 12(1) + 18(1) = 30, suggests that
every linear combination of 12 and 18 is a multiple of 6. If this is true, it rules out
1 through 5 as the smallest positive linear combination of 12 and 18. In fact, the
suggestion that all linear combinations of 12 and 18 are multiples of 6 is true, and
a more general result is stated in the next theorem.
“Number theory and other branches of mathematics often employ special notation and abbreviations. To
avoid confusion with other concepts, we have chosen not to use the shorter notation (a, b) for the ged.
Number theorists also often use the abbreviation a |b for “a divides b.”
78
CHAPTER 1
Theorem 1.8.1
Logic and Proofs
Let a and b be nonzero integers. The gcd of a and b is the smallest positive linear
combination of a and b.
Proof. Let d = as + bt be the smallest positive linear combination of a and b.
(To show that d = gcd(a, b), we must show that d satisfies parts (i) and (ti) of the
definition.)
(i)
(Show that d divides a and d divides b.) By the Division Algorithm, there exist
integers g and r such that a = dg + r, where 0 < r < d. Then
r=a—dq=a-—(as+
bt)qg =a — as — btqg=a(l
— s) + b(—t9).
Because | — s and —tgq are integers, we conclude that r is a linear combination ofa
and b. But 0 < r < d, and d is the smallest positive linear combination. Therefore,
r=0. (fr > 0, then r would be a positive linear combination smaller than d, which
is the smallest.) Therefore, a = dq, so d divides a.
Replacing a by b in the foregoing steps, we have that d divides b. Thus d is a
common divisor of a and b.
(ii) (Show that all common divisors are less than or equal to d.) Suppose that c is
a common divisor of a and b. By Theorem 1.6.1, c divides every linear combination of a and b. Thus c divides the linear combination d. Therefore, c < d.
(We have used Exercise 7(g) of Section 1.4.)
By parts (1) and (11), d is the greatest common divisor of a and b.
a
Notice that because every common divisor of a and b divides every linear
combination of a and b, it is an immediate consequence of Theorem 1.8.1 that every
common divisor of a and b divides gcd(a, b).
There are several ways to calculate gcd(a, b). We first used the method
of
listing all divisors of a and all divisors of b, finding their common divisors, and
choosing the largest one. By Theorem 1.8.1, we could choose the smallest positive
combination among all linear combinations of a and b. Other methods involve the
prime factorizations of a and b (see Exercise 17). All these methods work well
for small integers but are impractical for large numbers. Euclid developed another
method that is a very efficient way to find the gcd(a, b) of positive integers a and b.
Note how its proof repeatedly applies the Division Algorithm and Theorem 1.6.1.
Theorem 1.8.2
(Euclid’s Algorithm) Let a and b be positive integers with a < b. Then there are
two lists of positive integers ¢,, qy,..., @y_1, qj, and 1}, T>,..., T4_1, 7, Such that
CL) Oi
Wise
Sar
ae eh
(2) (bSaqr;
a= 1149 115
Tyas, = To Gea
pd
eat eae
1.8
Proofs from Number Theory
79
Te-2 = eid
Tr-1 II= "eI (that is, 7,4, = 0)
Furthermore, gcd(a, b) = r,, the last nonzero remainder in the list.
Proof. Let a and b be positive integers with a < b. By the Division Algorithm,
for 0 < a < b there exist a positive integer g, and a nonnegative integer r, such that
b=aq,+r,and0 <r, <a. Ifr, =0, the lists terminate. Otherwise, for 0 < r, < a,
there exist a positive integer g, and nonnegative integer r, such that a = r,q, +r, and
O<r, <r,.Ifr, =0, the lists terminate. Otherwise, for 0 < r, < r,, there exist a positive integer q; and a nonnegative integer r, such that r, = rq, +r, and 0 < 1; < 15.
Continuing in this fashion, we obtain a decreasing sequence of nonnegative
IME SCLSHara
naar This list must end, so there is an integer k such that r,,, = 0.
Thus we have
CE
aley nee = cone gh
Ade eigen)
210)
We now show that r, = d. (We show first that r, < gcd(a, b) and then show
that r, => gcd(a, b).)
(i) The remainder r, divides r,_, because r,_,; = rq,41. Also, r, divides r,_,
because r, divides both r, and r,_, and hence r, divides their linear combination r,_>»= p_14, + Ty. Likewise, r, divides r,_3 because r, divides both r,_,
and r,_, and hence divides their linear combination r,_3 = ry 24x) + Th_1:Continuing up the list of equations, we see that r, divides both a and b. Therefore,
r, < gcd(a, b).
(ii) We see that gcd(a, b) divides r, because r; = b — aq, is a linear combination
of a and b. From this, gcd(a, b) divides r, because r,; = a — ry, q> is a linear
combination of a and r,. Likewise, gcd(a, b) divides r; because r, = 1, — ry q3
is a linear combination of r, and r,. Continuing down the list of equations, we
see that gcd(a, b) divides r, = ry,» — rp_ dz. Therefore, we have r, > gcd(a, b).
From (i) and (11), 7, = gced(a, b).
Example.
|
Use Euclid’s Algorithm to find gcd(32, 12). We use repeated division.
32 = 12(2)+8
Psy
8 =4(2)+0
Because the last nonzero remainder is 4, we have gcd(32, 12) = 4.
The usefulness of Euclid’s Algorithm is this: by reversing the steps in the algorithm, we can find the coefficients that yield gcd(a, b) as a combination of a and b.
Example.
Write gcd(44, 104) as a combination of 44 and 104.
We use Euclid’s Algorithm to find gcd(44, 104).
104 = 44(2) + 16
44 = 16 (2) + 12
80
CHAPTER 1
Logic and Proofs
16
121)
12
= 4(3)
4
The last nonzero remainder is 4, so gcd(44, 104) = 4. Now apply the sequence of
equations in reverse order.
4 = 16 — 12(1)
12 = 44 — 16(2), so 4 = 16 — [44 — 16(2)](1) = —44 + 16(3)
16 = 104 — 44(2), so 4 = —44 + [104 — 44(2)](3) = 44(—7) + 104(3).
Thus, 4 = 44(—7) + 104(3). Therefore, gcd(44, 104) = 104x + 44y, where x = 3
and y = —7.
Oo
A linear Diophantine’ equation in two variables is an equation of the form
ax + by = c, where we seek integer solutions. The foregoing example shows how
Euclid’s Algorithm may be used to find integer solutions for some of these equations. See Exercises 13 and 14.
The last result of this section is found in Euclid’s Elements. Its proof makes use
of the fact that the gcd of two numbers may be expressed as a linear combination of
the numbers. It also makes use of the concept of relatively prime integers.
DEFINITION
We say that nonzero integers a and b are relatively prime,
or coprime, if gced(a, be
The numbers 12 and 35 are relatively prime. The numbers 15 and 36 are not,
because gcd(15, 36) = 3. The integer 2 is coprime with every odd integer.
Lemma 1.8.3
(Euclid’s Lemma) Let a, b, and p be integers. If p is a prime and p divides ab, then
p divides a or p divides b.
Proof. Suppose that p is prime and p divides ab. Assume that p does not divide
a. Then p and a are relatively prime, so there exist integers s and ft such that
as + pt = 1. Then b = abs + bpt. Because p divides both abs and bpt, we have
that p divides their sum, so p divides b. Thus either p divides a or p divides b.
m
Euclid’s Lemma is frequently used in one of its equivalent forms:
if p divides ab and p does not divide a, then p must divide b,
or
if p does not divide a and p does not divide b, then p does not divide ab.
* Diophantus lived in Alexandria sometime during the third century C.B. It is said he wrote several books
on “arithmetic” —involving various algebraic equations. Some of his works were later translated and
have survived to this day. Other than these writings, little else is known about him.
1.8
Proofs from Number Theory
81
Exercises 1.8
For each given pair a, b of integers, find the unique quotient and remainder
when b is divided by a.
(A) Wi =.85) 3510)
(D) ea,
0 = 36
(C) od = = 36
ae (Cd) =,
D = —36
(ec) fa = 7 b= 44
(ff) a=-—8, b= —52
(2) ia a — Sai
Let a and b be positive integers, and let r be the nonzero remainder when b is
divided by a. Prove that when —b is divided by a, the remainder is a — r.
Let a and b be integers, a # 0, and a > b. Prove that when b is divided by a,
the quotient is 0 if and only if b > 0.
For each pair of integers, list all positive and negative common divisors, and
find gcd(a, b).
(a) @ = 8, b=310
(c) a=18,b= —54
(a)
(b)
(c)
(dD) ira
(d) a=
5. b= 36
—8,b = —52
Write 2 in two different ways as a linear combination of 12 and 22.
Write —4 in two different ways as a linear combination of 12 and 22.
Describe all integers that can be expressed as a linear combination of 12
and 22.
Find d = gcd(a, b) and integers x and y such that d = ax + by.
(CVE: (hea olles Tks,
(Dy a= 26) b 232
(C) ode .04p 9)
(dd)
5b 4A
(e) (rai 005)
=6
(f).6 @=48.b:
= 50
Let a, b, and c be natural numbers and gced(a, b) = d. Prove that
(a)
(b)
adivides b if and only if d = a.
if a divides bc and d = 1, then a divides c.
(c)
if c divides
(d)
gcd(S, *)ale
for every natural number n, gcd(an, bn) = dn.
a and
c divides
b, then
ged(4, 7)= a In particular,
Which elements of the set {3, 6, 10, 63} are relatively prime to 7? to 21? to 30?
Prove that for every prime p and for all natural numbers a,
(a)
gcd(p, a) = p iffp divides a.
(b)
gcd(p, a) = 1 iffp does not divide a.
Let q be a natural number greater than | with the property that g divides a or
q divides b whenever g divides ab. Prove that g is prime.
Let a, b, and c be nonzero integers such that a and b are coprime and a and c
are coprime. Prove that a and be are coprime.
Let a be an integer and p and q be distinct primes such that p divides a and q
divides a. Prove that pq divides a.
82
CHAPTER 1
Logic and Proofs
13:
Let a and b be nonzero integers that are relatively prime, and let c be an integer.
Prove that the equation ax + by = c has an integer solution.
14.
Let a and b be nonzero integers and d = gcd(a, b). Letm = éania? = HtShow
that if x = s and y = ris a solution to ax + by = c, then so is x = s + km
and y = t — kn for every integer k. (This shows how linear combinations
help describe solutions to equations.)
iS:
Let a and b be nonzero integers and c be an integer. Prove that the equation
ax + by = c has an integer solution if and only if gcd(a, b) divides c.
16.
For nonzero integers a and b, the integer n is a common multiple of a and b
if a divides n and b divides n. We say that the positive integer m is the least
common multiple of a and b, written as lem(a, b), if
(i)
mis acommon multiple of a and b, and
(ii)
if mis a positive common multiple of a and b, then m < n.
Using ideas from Section 2.5, it can be proved that lcm(a, b) always exists.
Find Icm(a, b) for
(2) ec — OND 14
(Ded
(OC)
(a) Wd =s1 2.
yey,
Dy 39
Ope
5
= 45
Let a, b, and c be natural numbers, gcd(a, b) = d, and Icm(a, b) = m. Prove
that
(a)
(b)
(C)
adivides b if and only if m = b.
-m < ab.
tiga ale then7
ab:
(d) if c divides a and c divides b, then lem(¢, ®)=".
(e)
for every natural number n, lem(an, bn) = mn.
(f)
gcd(a, b) - lem(a, b) = ab.
Let a and b be integers, and let m = Icm(a, b). Use the Division Algorithm to
prove that if c is acommon multiple of a and b, then m divides c.
The greatest common divisor of positive integers a and b may be calculated
using prime factorizations. Suppose a = p{!p--.pié and b = qj g?--- qs» are
the prime factorizations. Then
gcd(a, b) = wi w3.-- wir,
—
t
2
where each w; is a prime factor of both a and b and whenever w; = p, = 4,
then the exponent /; is the smaller of the corresponding exponents r, and s,.
Use this method to find the gcd of each of the following pairs.
(a) a= $4, b= 30
(D) 5a = 1327
p= 42
(C) aa 0000
= a4)
(d) -@=315)).=
180
20.
The least common multiple of positive integers a and b may be calculated
using prime factorizations. Suppose a = pj! p)--. pj and b = qi gy... g'n are
the prime factorizations. Then
Iem(a,
b) =
wi} w} stefe wr,
1.8
Proofs from Number Theory
83
where each w, is a prime factor of either a or b. Whenever w; = p, = q, then
the exponent /; is the larger of the corresponding two exponents r, and s,,. If
w; = p, but is not a factor of b, then t; = r,. If w; = q,, but is not a factor of a,
(Hem t5—=csy.
Use this method to find the Icm of each pair in Exercise 19.
Proofs to Grade
21.
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
(a)
Claim.
“Proof.”
For all natural numbers n, gcd(n, n + 1) =
(i) | dividesnand
1.
| dividesn + 1. (ii) Suppose that c divides n
and c divides n + 1. Then | divides c. Therefore, gcd(n,n + 1)=1.
(b)
Claim.
“Proof.”
For all natural numbers n, ged(2n —
Obviously
1, 2n + 1) =
@
1.
1 divides both 2n — | and 2n + 1. Suppose
that c divides 2n — 1 and 2n + 1. Then c divides their sum, 4n, so c
also divides 4n*. Furthermore, c divides their product, 4n? — 1. Since
c divides 4n* and 4n* — 1, c divides 4n? — (4n? — 1) = 1. Therefore,
c < |. Thus | is the greatest common divisor.
(c)
Claim.
ie
ON
|
Let 7 be a natural number. If 3 does not divide n, then 3 divides
“Proof.”
Suppose that 3 does not divide the natural number n. By the
Division Algorithm, when n is divided by 3, there is a remainder of | or
2, so for some g, we have n = 3g + 1 orn = 3g +2. Then n? +2 = 9q?
+ 6g +3 or n? + 2 = 9q? + 12q + 6, so 3 divides n? + 2.
g
(d)
Claim.
prime.
If a and bd are relatively prime, then a? and b* are relatively
“Proof.” Suppose that a and b are relatively prime and gcd(a’, b*) > 1.
Then gcd(a’, b*) has a prime factor p. Therefore p divides a’ and b*. By
Euclid’s Lemma, p divides a and b. This is a contradiction. Therefore a
(e)
and b? are relatively prime.
5
Claim.
If a and b are integers such that a > 1, b > 1, and a does not
divide b and b does not divide a, then gcd(a, b) = 1.
“Proof.”
Let a and b be integers such that a > 1 and b > 1. Suppose
that a divides b and b divides a, and gcd(a, b) = 1. We know a divides
a, so from a divides b, we have that a is a common divisor of a and
(f)
b. By the definition of gcd, a < gcd(a, b). By the same reasoning,
b < gcd(a, b). Thus a and b are both less than or equal to 1. This is a
contradiction. We conclude that if a does not divide b and b does not
divide a, then gcd(a, b) = 1.
a
Claim.
If Icm(a, b) = ab, then a and D are relatively prime.
“Proof.”
Suppose that lcm(a, b) = ab. By Exercise 17(f), the product
of gcd(a, b) and Iem(a, b) is ab. Therefore, ged(a, b) =1, so a and b are
relatively prime.
(g)
Claim.
a
If a divides bc but a does not divide b, then c divides a.
“Proof.” Suppose that a divides bc. Therefore, there exists an integer k such
that bc = ka. Since a does not divide b, then k divides b, so b = mk for some
integer m. But then kmc = ak. Therefore, mc = a, so c divides a.
a
84
CHAPTER 1
Logic and Proofs
(h)
Claim.
Ifa is an integer and p is a prime, then either p divides a or p
and a are relatively prime.
“Proof.”
Suppose that p is a prime and p and a are not relatively
prime. Let d= gcd(p, a). Since d > 1, dhas a prime factor g > 1. Then
q divides p. Butp is a prime and g > 1, so g = p. Therefore, p divides a.
|
Gans ASP TeE Re. 2
Sets and Induction
Starting from the theory of sets, one can construct all the number systems, functions, calculus, and other areas of mathematics. Thus, the study of sets is the foun-
dation for the entire structure of mathematics.
A full development of these constructions is beyond the scope of this text,
but we will provide some set-theoretic concepts used throughout the text and
advanced mathematics. Sections 2.1 and 2.2 provide precise definitions for familiar concepts such as union and intersection. In Section 2.3 we extend the union
and intersection operations to collections of sets and discuss how to use indices to
organize a family of sets. Proof methods using forms of mathematical induction
are discussed in Sections 2.4 and 2.5. Basic methods for counting the elements in
a finite set appear in the optional Section 2.6.
ax)|
Basic Concepts of Set Theory
We assume that you have had some experience with sets, set notation, and common
sets of numbers such as the integers and real numbers, as described in the Appendix.
In general, uppercase will be used to denote sets, and lowercase letters to denote the
elements in sets. To designate a set, we use the notation
{x: P(x)},
where the open sentence P(x) describes the property that defines the set. For example, the set A =
{1, 3,5, 7, 9, 11, 13} may be written as
{x:x€ N, x is odd, and x < 14}.
The set of all integer multiples of 3 is the set 3Z = {3z: z € Z}, and this set contains 0, 3, —3, 6, —6, 9, —9, etc.
85
86
CHAPTER 2 _ Sets and Induction
A word of caution: Some sentences P(x) may not be used to define a set. In
1902, when the theory of sets was new, Bertrand Russell” and others pointed out
paradoxes that revealed flaws in the then-common assumption that for every open
sentence P(x), there corresponds a set {x: P(x)}. See Exercise 3 for a version of
the Russell Paradox.
The resolution of Russell’s and other paradoxes involved making a distinction between sets and arbitrary collections of objects. Sets may be defined within
a system of axioms for set theory, first developed by Ernst Zermelo’ and Abraham
Fraenkel.” Their axioms assert, for example, that a collection of two sets constitutes
a set (Axiom of Pairing) and that the collection of all subsets of a set is a set (Axiom
of Powers). Under their system, known paradoxes such as Russell’s are avoided.
It is not our purpose here to carry out a formal study of axiomatic set theory.*
However, all of our discussions of sets are consistent with the Zermelo—Fraenkel
system of axiomatic set theory.
A second word of caution: Recall that the universe of discourse is a collection
of objects understood from the context or specified at the outset of a discussion,
and that all objects under consideration must belong to the universe. Some ambiguity may arise unless the universe is known. For example, membership in the set
A = {x:x? — 6x = 0} depends on an agreed-upon universe. The set A is {0, 6}
for the universe of real numbers, but A is {6} for the universe of natural numbers.
DEFINITION
or null set.
Let @ = {x: x #.x}. Then W is a set, called the empty set
It is an axiom that © is a set. Since for every object x in every universe, x is
equal (identical) to x, there are no elements in the collection ©. That is, the state-
ment x € © is false for every object x. We could define other empty collections,
such as B = {x: x € R and x” < 0}, but we will soon prove that all such collections are equal, so there really is just one empty set.
In the Appendix we said A is a subset of B and wrote A C B if every element
ofA is an element of B. If A is not a subset of B, we write A ¢ B. For X = {2,4},
Vr e( 28345) wand
Z = t)2..3.96)} eXG-Y and
X GZ.
* Bertrand Russell (1872-1970) was a British philosopher and mathematician and a strong proponent of
social reform. He coauthored Principia Mathematica (1910-1913), a monumental effort to derive all of
mathematics from a specific set of axioms and a well-defined set of rules of inference.
¥ Ernst Zermelo (1871-1953) was a German mathematician whose work on the axioms of set theory has
profoundly influenced the foundations of mathematics. In 1905 he discovered a paradox similar to the
Russell paradox. He developed a theory of sets based on seven axioms but was unable to prove that no
new paradoxes could arise in his system.
*** Abraham Fraenkel (1891-1965), born in Germany, spent much of his career in Israel. In the 1920s
he made attempts to improve the set-theoretic axioms of Zermelo to eliminate paradoxes. Within his
system of ten axioms he proved the independence of the Axiom of Choice. (See Section 5.5.)
‘* A complete study of the foundations of set theory from the Zermelo—Fraenkel axioms may be found
in Notes on Set Theory by Y. N. Moschovakis (Springer-Verlag, Berlin, 1994). The study of set theory
is still active today, and many unsolved problems remain.
2.1
Basic Concepts of Set Theory
87
The distinction between “is an element of” and “‘is a subset of” is crucial, especially when we work with sets whose elements are sets.
Example.
Let X = {1, {2, 4}, {5}, 8}. This set has four elements; two of these
elements are natural numbers and two are sets. Thus 4 € {2, 4} and {2, 4} € X, but
ago Also we have (5) e X, {1 45)) Cx and {{51) GX but (5) 2 Xx.
If x € A, the correct terminology is that A contains x. When B C A, we say
A includes B. For instance, let A be the set {1, {2, 3}, 3, {7}, 8}. Then A has five
elements, and two of its elements are sets. We may say that 8 is an element of A or
that A contains 8, but we don’t say that A includes 8. The setA also contains {2, 3}
because {2, 3}€ A. On the other hand, the setA includes the sets {8} and {1, {7}},
because every element of each of these sets is an element of A.
In symbols, we write the definition of A C B as
ACBS(Vx\xEAsSxeE
B).
Because of its logical construction, a proof of the statement A C B is usually a
direct proof, taking the form:
DIRECT PROOF OF A CB
Proof.
Let x be any object.
Suppose x € A.
Thus x EB.
iherelore Aub:
Example.
|
Let A = {2, —3} and B = {x eR: x? + 3x? —4x — 12 = 0}. Prove
that A CB.
Proof.
Suppose x € A. (We show A CB by individually checking each element
of A.) Thenx = 2orx = —3.If x = 2, then 2? + 3(27) — 4(2) —12 = 0. Ifx= —3,
then (—3)° + 3(=3)? — 4(—3) —
Example.
12 = 0. In both cases, xe B.Thus,ACB.
Let a and b be natural numbers, and let aZ and bZ be the sets of all
integer multiples of a and b, respectively. Prove that if a divides b, then bZ C aZ.
Proof. Suppose that a divides b. Then there exists an integer c such that b = ac.
(To show bZ C aZ, we begin with an element from bZ.) Let x € bZ. Then x is a
multiple of b, so there exists an integer d such that x = bd. But then x = bd =
(ac)d = a(cd). Therefore, x is a multiple of a, so x € aZ.
a
88
CHAPTER 2
Theorem 2.1.1
Sets and Induction
(a)
(b)
Forevery set A,@ CA.
Forevery set A, A CA.
(c)
For all sets A, B, and C,if A C
Band BCC, then A CC.
Proof.
(a)
Let A be any set. Let x be any object. Because the antecedent is false, the
(b)
Let A be any set. (To prove A CA, we must show that for all objects x,
if x EA then x EA.) Let x be any object. Then x ©€A>x EA is true.
(Here we use the tautology P => P.) Therefore A C A.
(c)
See Exercise 8.
sentence x € @ > x € A is true. Therefore © C A.
|
Recall that sets A and B are equal iff they have exactly the same elements;
that is,
A = Bifand only if (Vx)(xe
AS xe B).
Thus, one method of proving A = B is to give a sequence of equivalent statements starting with the statement x © A and ending with x € B. However, since
xE€A&xeEB
also say
is equivalent
A=
to ®E€AS>xeEB)A(XEB>x€EA),
we
may
Biff A GC B and BGA.
For this reason, a proof that A = B will typically have the following form:
TWO-PART
PROOF OF A = B
Proof.
(i) Prove that A C B (by any method).
(ii) Prove that B C A (by any method).
Therefore A = B.
Example.
B= ties
Prove that
ye
&
A = B, where A = {n € N: (n — 3)(n + 5) < 2n} and
Proof.
(i)
(ii)
Substituting 1, 2, and 3 forn, we have, respectively, (—2)(6) < 2, (—1)(7) <4,
and (0)(5) < 6. Thus every element ofB is in A, so B C A.
Suppose t € A. Then t € N and (t — 3)(t + 5) < 24, so ? + 2t¢ — 15 < 2t.
Therefore, #7 < 15. Because t is a natural number, t = 1, 2, or 3. Therefore
t € B. We conclude that A C B.
By (i) and (11), A = B.
a.
2.1
Example.
Basic Concepts of Set Theory
89
Prove that
X = Y, where
X = {x ER: ix + 1] > 5 —x} and Y= (2, oo).
Proof.
(i)
Suppose x € X. Then |x + 1| > 5 — x. (To simplify the expression |x + 1|,
consider two cases.)
Casel.
Ifx>
—1, then |x + 1] =x-+
1. Therefore,
x + 1 > 5 — x, which
implies that 2x > 4. Thus x > 2, sox € Y.
Case 2.
If x < —1, then |x + 1| = —x — 1. Therefore, —x — 1 >5—-x.
Then —1 > 5, which is impossible.
In every possible case, ifx€ X, then x € Y. Therefore X C Y.
(ii)
Suppose x € Y. Then x > 2,sox+ 1 > 3. Thus |x + 1| > 3. Also, sincex > 2,
we have 3 > 5 — x. Therefore, |x + 1| > 5 — x, sox € X. We conclude that
YG x.
By (i) and (11), X = Y.
|
The set B is a proper subset of the set A if B C A and A ¥ B. To denote that B
is a proper subset of A, some authors write B C A and others write B ¢ A. The only
improper subset of A is the set A itself.
We are now in a position to prove that there is only one empty set, in the sense
that any two empty sets are equal.
Theorem 2.1.2
IfA and B are sets with no elements, then A = B.
Proof.
Theorem 2.1.3
See Exercise 12.
|
For any sets A and B, if AC Band A # ©, then B 4 ©.
Proof.
Suppose A C B and A # ©. Since A is nonempty, there is an object t such
that t © A. Since t € A, t € B. Therefore,
B ~ ©.
a
We sometimes use Venn’ diagrams to display simple relationships among sets.
For example, suppose we want to find nonempty sets A, B, and C such that A C B,
A#B,C CA, and A¢ C. We begin with three overlapping sets that represent the
sets A, B, and C in Figure 2.1.1(a). Because A C B, there are no elements in the two
regions of A that are outside B. Because C C A, there are no elements in the two
regions of C that are outside A. These four regions are shaded in Figure 2.1.1(b).
Because A is not a subset of C, there is some element x in the remaining region of A
that does not overlap C, and because A # B, there is some element y in B that is not
in A. Finally, C is required to be nonempty, so there is an element z in C. There may
* John Venn (1834-1923) was a British philosopher and logician best known for his use of diagrams to
describe relationships. He developed the idea that the probability of an event is a prediction of the frequency with which that event will occur.
90
CHAPTER 2
Sets and Induction
be other elements in these sets, but the solution we have found is A = {x, z},
B= {x5 2yand'€ =z} s(See Figure 2.1, 1(c).)
Note that although Venn diagrams are often useful for understanding relationships among sets, it may be difficult or impossible to use them when more than
three sets are involved.
[SSE
(a)
(b)
(c)
Figure 2.1.1
One of the axioms of set theory asserts that for every set A, the collection of all
subsets of A is also a set.
DEFINITION
Let A be a set. The power set of A is the set whose elements
are the subsets of A and is denoted
(A). Thus
P(A) = {B: BCA}.
The power set of a setA is a set whose elements are themselves sets, specifically
the subsets ofA.For example, if A =
POS
{a, b, c, d}, then the power set of A is
Gb} sMaeeys |andieb,c}, 4b.
1@, a,@ Wd,
Cad) (De G.a hy.
“)
may D}s tok}
(ao, Ch,
Peery
For every set X we have © C X, © € P(X), {GO} C P(X), and OW C P(X).
Example.
MOG
Let X =
{{1, 2,3}, {4,5}, 6}. Then P(X) =
25 SETA HeaG it 1.25 Sha Sy
At, 2.8), 6}, (14,
53, 6),.X4.
In addition, we have
{{4, 5}} © PCX) and {{ {4, py C. PX),
(because {4,5} xX),
{4,5} ¢ oe (because {4,5} ZX)
{O} EX, {OD} ¢ P(X), and {{D}} 2 P(X)
(because © ¢ X).
oO
Notice that for the set A = {a, b,c, d}, which has four elements, P(A) has
16 = 2+ elements, and for the set X given previously with three elements, P(X) has
8 = 23 elements. These observations illustrate the next theorem.
2.1
Theorem 2.1.4
Basic Concepts of Set Theory
91
If A is a set with n elements, then (A) is a set with 2” elements.”
Proof.
(The elements in P(A) are the subsets of A. Thus to prove this result, we
must count all the subsets of A.) If n = O (that is, if A is the empty set), then
P(D) = {WG}, which is a set with 2° = 1 elements. Thus the theorem is true for
ae
Suppose that A has n elements, for n > 1. Write A as A = {X1, X>,..., X,}To describe a subset B of A, we need to know, for each x, € A, whether the element
is in B. For each x,, there are two possibilities (either x, € B or x, € B). Thus there
ace exactly 2 27 2) =
2 (n factors) different ways of making a subset of A.
(The counting rule used here is called the Product Rule. See Theorem 2.6.5 and the
discussion following that theorem.) Therefore, P(A) has 2” elements.
|
The next theorem is a good example of a biconditional statement for which a
two-part proof is easier than an iff proof.
Theorem 2.1.5
Let A and B be sets. Then A C B if and only if P(A) C P(B).
Proof.
(i)
(ii)
(i) Assume that A C B and suppose that X € Y(A). Then X C A. Because
A CB, we have X C B by Theorem 2.1.1. Therefore X € 9(B). Because
X € P(A) implies X € P(B), P(A) C PB).
Assume that 9(A) C (B). By Theorem 2.1.1,
ACA; so A € P(A). Since
P(A) C P(B), A € P(B). Therefore A © B.
B
The second half of the proof of Theorem 2.1.5 could have been done differently. We could have shown that A C B by giving a direct proof that x € A implies
x € B. A proof that consists of a series of steps beginning with “Assume x € A”
and leads to a conclusion that “Therefore x € B” is often called an element-chasing
proof, and it is the natural way to prove the most basic facts about sets. As we
build our knowledge of set properties in the next section, we may use theorems
already proved, as we just did when we used Theorem 2.1.1, to shorten our proof of
Theorem 2.1.5. An element-chasing proof of part (11) would be just as correct, but
most people prefer a shorter, more elegant proof.
Exercises 2:1
1.
If possible, name an object that is an element of both given sets. Then, if possible, name an object that is an element of the first set but not of the second.
* ‘ (a)) {1, 16, 128} and the set of positive integer powers of 2
(b)) {6, 42, 74} and the set of integer multiples of 6
* This theorem is the reason why some mathematicians use 2 to denote the power set of A.
92
CHAPTER 2
Sets and Induction
x |Co) {0, —6, 7} andN
\(d), {—5,%, 98.6} and@
* Co) N and Z
* ey Z andN
( (f)) RandQ
(h) RandC
* Wy [5, 8) and {5, 6, 7}
2
7) ) (2, 5] and (2, 5)
Levies {x: P(x)}. True or false?
(a) IfaeX, then P(a).
(b) If P(a), thena € xX.
(c)
If ~P(a), thena € X.
3. x(a)
The Russell Paradox A logical difficulty arises from the idea, which
at first appears natural, of calling any collection of objects a set. Let’s
say that set B is ordinary if B € B. For example, if B is the set of all
chairs, then B ¢ B, because B is not a chair. It is only in the case of very
unusual collections that we are tempted to say that a set is a member of
itself. (The collection of all abstract ideas certainly is an abstract idea.)
Let X = {x: x is an ordinary set}. Is X € X? Is X ¢ X? What should we
say about the collection of all ordinary sets?
(b)
In the town of Seville, the (male) barber shaves all the men, and only
the men, who do not shave themselves. Let A be the set of all men in the
town who do not shave themselves. Who shaves the barber? (That is, is
the barber an element of A? Is he not an element of A?)
cs
True or false?
(a) SE&){{5}, {6}}.
x (Cc) {5} ©X5}.
AAC) Moy S19;,0}:
(g) {{5}}©{{{5}}, {6}}.
Gi) BESAN6} }GA5).6):
(b)
(dD
@)
Ch)
GD)
5) (5, {6}}.
eric
te} St e6
Sess
ai hs
PALO} Poi 54 6} }-
True
or false?
(a) Wel©, {O}}.
(ec) {P}EIS, {G}}.
(e) {{D}} ED, {D}}.
(g) For every set A, O@)A.
(b)
(d)
(Ey
(h)
OC {G, {D}}.
(One Io 1):
(Oe:
(OV }s
For every setA, {O} CA.
Co
(kK)
tee
On Oho)
123
tl 23.14}.
MD
1
chen
lag
toda
t ls o).
Ata} Mott,
2,3, (47).
Give an example, if there is one, of sets A, B, and C such that the following
are true. If there is no example, write “Not possible.”
(a) ACB,BEC,and
ACC.
(b) 4 6 BB e Cand Ga!
aire)
ve
ee
(c)) AGB Be Ceand AS Cc
(d) ACB,
BEC, and AEC.
(ec)
AGC B,ACB,andBCC.
(f) "Ae BAS
c Th Prove that if x €
Cand BCC
Band A C B, thenx € A.
2.1
Basic Concepts of Set Theory
93
Prove part (c) of Theorem 2.1.1: For all sets A, B, and C,if AC Band
thenvAni@
BCC,
Prove that if AC B, BC C, and
10.
11.
CCA, then A = BandB
=C.
Let
X = {x€ R: x — 6x? — 19x + 30 > 0}.
(a)
(b)
Prove that {1.2, —2.5} CX.
Prove that {1.2, —2.5} is a proper subset of X.
(c)™
Prove that {0,.2:575,—5 PZ x,
Prove that
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
{xeER:%x-—2>10}
= (16, 00)
{x eR: |x — 4| =2|x| —2} = {-6,2}
{x ER: 2|x + 3] +x=0} = {-6, —2}
{x eR: |x| =6 — |2x|}= {-2, 2}
{x ER: |x +3] < —4x — 2} =(—oo, -1]
{xeER: |x+3| =5 — |x|} = {-4, 1}
{xeER: |x —4| <x} =(2, cw)
{3} is a proper subset of {x € R: |x — 2| = 4 — |x|}
Prove that there is only one empty set. That is, prove that if A and B are sets
with no elements, then A = B. (Theorem 2.1.2.)
For a natural number a, let aZ be the set of all integer multiples of a. Prove
that for all a,b
€ N, a = bif and only if aZ = bZ.
Find the power set, (X), for each of the following sets.
(a)
(ce)
X= {0,A,D}
(b)
X={O
{a}. 1b} {eo 8})} @,
X= {S, {S}}
XS
AS eens
Let A, B, and C be sets and x and y be any objects. True or false?
(a) Ifx eA, then x € P(A).
(b) Ifx eA, then {x} € P(A).
(c) Ifx eA, then {x} C P(A).
(d) If {x, y} © P(A), then x € A andy EA.
(e) If BCA, then {B} € P(A).
(f) IfB CA, then B & P(A).
(g) If Be P(A), then BCA.
(h) If CC Band Be (A), then C € P(A).
List all of the proper subsets for each of the following sets.
(a)
(c)
©
ts
(b)
(d)
{D, {O}}
{0, A, 0}
True
or false?
(a)
De PA{Y, {O}}).
(b)
{OG} © PUY, {P}}).
(c)
(e)
(g)
Gi)
(k)
{{O}}
=~PUPS, {S}})).
{DO} CAPUD, {P}}).
3€Q.
{3} e PQ).
{3} CQ.
@M
(f)
(hy)
@
Q)
PCPS, {O}}).
{{O}} SPUD, {D}}).
{3} C PQ).
{{3}} CPO).
{{3}} © PQ).
94
CHAPTER2
Sets and Induction
18.
Let A and B be sets. Prove that
(a) A=Bifand only if P(A) = PCB).
19.
Assign a grade A (correct), C (partially correct), or F (failure) to each. Justify
(b)
Proofs to Grade
ifA is a proper subset of B, then P(A) is a proper subset of P(B).
assignments of grades other than A.
(a)
Claim) aitX ete Nj x7, 14 and Y= 1, 2,3), thenXx — Y“Proof”? Since 1? =1 < 14,2? = 4 <14,and3*?=9<14,X=Y. »
(b)
Claim.
If A, B, and C are sets, and A C
“Prool.anlctAs={ L. 5, Sia Be Hie,
Band
BCC, then A CC.
8.10 \-and C= fel 2
6,3, LO}. Then A GB, and 6 S.C,,and A © C.
x
(c)
Claim.
If A, B, and C are sets, and
“Proof.”
A C
|
Band B CC, then A CC.
Suppose x is any object. If x€ A, then x € B, since A C B. If
x € B, then x € C, since B € C. Thus, x € C. Therefore, A € C.
(d)
Claim.
a5,
If A, B, and C are sets, and
A C
ia
Band B CC, then A CC.
“Proof.”
If x eC, then, since
BC C, x € B. Since
AC Band x € B,
it follows that x € A. Thus x © Cimplies x € A. Therefore,
ACC.
gm
Claim.
If A,B, and C are sets, and
“Proof.”
ACB.
SOAS
A C
Band B CC, then A CC.
Suppose A C B and B CC. Then x € A and x € B, because
Then x € B and x € C, because B C C. Therefore,
OVAUGSGC:
x © A and
ia]
Claim.
Let
X = {x € R: |x + 4| = |2x|}. Then
X = {4}.
“Proof.” (i) The only element of {4} is 4, and |4+ 4, =8 = |2-4|,so4 eX.
Therefore, {4} CX.
(ili) Suppose x € X. Then |x + 4| = [2x].
Case l:
x>0.Thenx+4>0,so
siheretorews
Case 2:
|x+ 4| =x+4 and |2x| = 2x.
4a 20 SOne—
x< —4.Thenx+ 4 <0Oandx <0,so0 |x+4| = —x-—4
and |2x| = —2x. Therefore, —x — 4 = —2x, sox = 4.
But x = 4 is impossible when x < —4, so X has no elements in this case. Therefore, if x € X, then x = 4.
By (i) and (ii), we conclude that X = {4}.
(g)
Claim.
“Proof.”
Assume
x € P(A). Therefore,
(h)
A is a set. Suppose x¢ A. Then
A C P(A).
Claim. If A isa set, then A C P(A).
“Proof.”
Assume A is a set. Suppose
(i)
x CA.
Thus
BZ
x € A. Then
{x} CA. Thus
|
Claim.
If A and B are sets and (A) C P(B), then A C B.
“Proof.”
x«eA> {x} CA
=> {x} © P(A)
=> {x} © P(B)
{x} © P(A). Therefore, A C P(A).
x
|
If A is aset, then A C P(A).
2.2
Set Operations
95
Snes
=e b,
Therefore,
*
(j)
Claim.
xe A= x € B. Thus A C B.
|
If AC Band BEC, then AEC.
“Proof.”
Suppose AC B and BEC. Then there exists x € B such
that x ¢ C. Since x € B, x € A by definition of subset. Thus x € A and
x € C. Therefore, A ¢ C.
|
222
Set Operations
In this section we give precise definitions of set union, intersection, and difference
and prove some of their familiar properties. The operations are binary operations
because each combines two sets to produce another set.
DEFINITIONS
Let A and B be sets.
The union ofA andB is the sect AUB = {x:xeAorxe B}.
The intersection ofA and B is the set
AM B = {x:x eA andxe B}.
The difference ofA and B is the set A — B = {x:x © Aandx ¢ B}.
The set A U Bis a set formed fromA and B by choosing as elements the objects
contained in at least one of A or B; AM B consists of all objects that appear in both
A and B; and A — B contains exactly those elements of A that are not in B. The
shaded areas in the first three Venn diagrams of Figure 2.2.1 represent, respectively, the result of forming the union, intersection, and difference of two sets.
HOO)
A-B
Figure 2.2.1
Examples.
For A = {1,2,4,5,7} and B = {1,3,5, 9},
AID, a3
AMD r=
A-— B=
B= A
203,43
OF /29}
ib Dh,
{2,4,7},
(3,91.
Disjoint sets A and B
96
CHAPTER 2
Sets and Induction
Examples.
For intervals of real numbers, we have
[3 nOhh (453 se
[3,6] 1 [4,8) =
[3,6] — [4, 8) =
[4, 8) — [3, 6] =
[4, 8) — 6,6] =
s3.05)
[4,6]
[3, 4)
©, 8)
[4,5] UG, 8).
Two sets are said to be disjoint if they have no elements in common.
DEFINITION
Sets A and B are disjoint iff
AM B = ©.
As shown in the last Venn diagram of Figure 2.2.1, when sets A and B are
known to be disjoint, we represent them as non-overlapping regions.
Examples.
The sets {1, 2, b} and {—1, ¢, v, 8} are disjoint. The set of even natural numbers and the set of odd natural numbers are disjoint. The intervals (0, 1) and
[1, 2] are disjoint, but (0, 1] and [1, 2] are not disjoint because they both contain the
element 1.
o
The set operations of union, intersection, and difference obey certain rules
that enable us to simplify our work or replace an expression with an equivalent
one. Some of the 18 relationships in the next theorem seem to be obviously true,
especially if you compare sets using Venn diagrams. For example, the Venn diagrams for
AB and BMA
are exactly the same; see part (j). However, simply
drawing a Venn diagram does not constitute a proof. The properties must be confirmed by using the definitions of set operations, subset, and set equality, or by
using previous theorems. We prove parts (b), (f), (j), (m), and (p) and leave the
others as exercises.
Theorem 2.2.1
For all sets A, B, and C,
((a)) ACAUB.
(b)
ANBCA.
(©)
ANG=@.
(d)
AUS =A.
(e-)
(g)
ANA=A.
A-SG=A.
(f)
(h)
AUA=A
OD-A=@.
@” FAUB=3 UA)
(jj)
ANB=BNA.
(k)
AU(BUC)=(AUB)UC.
(I)
AN(BNC)=(ANB)NC.
(m) AN(BUC)=(ANB)U(ANC).
((m))
AU(BNC)=(AUB)N(AUC).
(0)
AC Bifand
only if AUB =B.
Commutative Laws
\Associative Laws
}Distributive Laws
2.2
_(p)
Set Operations
97
A C Bif and only if ANB =A.
iA eB themAW
CC BOC,
If AGB; then Aim C <.B.G.C,
© (q)
(r)
Proof.
(b)
(We must show that if x €
AM B, then x € A). Suppose
x€A
x € A. (We used the tautology
and x € B. Therefore
x €
AN B. Then
P \
O=> (Q.)
ThusAN BCA.
(We must show that x € A UA if and only if x € A). By the definition of
union, x € A UA if and only if x € A or x € A. This is equivalent to x € A.
(f)
Therefore,
(j)
AUA = A.
(This biconditional proof uses the definition of intersection and the equivalence of statements of the forms P \ Q and Q A P.)
xEANBiffxeAandxEeB
ise Se S Jo} Buoval se eS Al
iff xe BNA.
(As you read this proof, watch for the steps in which the definitions of union
and intersection are used (two for each). Watch also for the use of the equivalence from Theorem 1.1.1 (f).)
(m)
xEAN(BUC)
iff xe Aandxe BUC
iff xe A and (xe BorxeEC)
iff (x € Aandx € B)or(xe€ A andx eC)
iff x EeANBorxEe
ANC
iff xe (ANB)U(ANC).
Therefore ABU) C)
(A OB) UO AC).
(Each part of this proof makes use of earlier parts of this theorem.)
(p)
(i)
Assume that A C B. We must show that
AM B = A. Suppose x € A.
Then from the hypothesis A C B, we have x € B. Therefore, x € A and
x€B,
(ii)
so
x € ANB.
This shows
that
A
CAMB,
which, combined
with A
B CA from part (b) of this theorem, gives
AM B = A.
Assume that
AM B= A. We must show that A C B. By parts (b) and
(h) of this theorem, we have
BN AC B and
BNA =ANB. Therefore,
AM B C B. By hypothesis,
AN B = A, so A CB.
We conclude that A C Bif and only ifAN B=A.
r|
When you suspect that a relationship among sets is not always true, try to construct a counterexample. To find a counterexample for (A
UB) 1C = AU(BNC),
we need sets such that the shaded regions of Figures 2.2.2 (a) and (b) have different elements. That is, we find sets A, B, and C such that A contains at least one
element that is not in C. One counterexample is B = {3,5}, C = {4,5, 6}, and
Ari
se4yernen (Aw Bic
=) 4,5} while AU (BC)
= {2,.3,4, 5}.
98
CHAPTER 2
Sets and Induction
G
E.
(AUB)NC
AU(BNC)
(a)
(b)
Figure 2.2.2
With the universe fixed throughout a discussion, the operation of finding the
complement is a unary operation because it is applied to a single set to produce
another set.
DEFINITION
Let U be the universe and A C U. The complement of A is
the sel As =U —A.
The set A‘ is the set of all elements of the universe that are not in A. (See
Figure 2.2.3.)
Figure 2.2.3
For the set A =
{2, 4, 6, 8}, we have A° =
{10, 12, 14, 16,...} if the universe
is all even natural numbers, whereas A° = {1, 3,5, 7,9, 10, 11, 12, 13,...} if the
universe is N. For the universe R, if B= (0, oo), then BS = (—oo, 0]. If D= {5},
then Ds (oom) Ul(O.100)))
The next theorem includes several results about the relationships between
complementation and the other set operations. We give two proofs of part (e) to
illustrate different proof methods.
Theorem 2.2.2
Let U be the universe, and let A and B be subsets of U. Then
(a)
(A) =A.
(b)
AUAT=U.
(c)
ANA
=2.
2.2
(I)
A—B=ANB.
(e)
A CB
Cy
Ali Br=2 it and/only 1f A'S Bo
(2)
AWB)
(h)
Set Operations
99
if and only if BS CA’.
Be.
(ANB) =A UBS.
De Morgan’s Laws
Proof.
(a)
(e)
By definition of the complement x € (A‘)° if and only if x ¢ A® if and only if
He Anlhererore(AN =A)
First proof. (This is a two-part proof. The first part is an element-chasing
proof. The second part is proved using the first part.)
Part 1.
Assume that A C B. Suppose x € B°. Then x ¢ B. Since A C B and
Part 2.
x € B, we have x € A. Therefore, x € A°. Thus, BS C A°.
Assume that B° C A°. Then by Part 1, we have (A‘)* € (B®).
Therefore, using (a), A C B.
By Parts 1 and 2, we conclude that A C Biff BS C A’.
Second proof.
A C Bif and only if for all x, if
:
xe A thenx eB
if and only if for all x, if x ¢@ Bthenx
€A
if and only if for all x, if x © B° then x € A°
if and only if BS C A®.
(g)
xe (AUB) if and only ifx
¢AUB
if and only if it is not the case that x € A orx € B
if and only if x ¢ A andx € B
if and only if x€ A° and x € B®
if and only if x € ACN BS.
=
The proofs of the remaining parts are left as Exercise 8.
The ordered pair (a, b) is an object formed from the two objects a and J,
where a is called the first coordinate and b the second coordinate. Two ordered
pairs (a, b) and (c, d) are equal whenever their corresponding coordinates are
equal—that is, when a = c and b = d. Thus (3, 7) # (7, 3) even though the sets
{3,7} and {7,3} are equal. A more rigorous definition of an ordered pair as a set
is given in Exercise 18.
In previous study you have dealt with the ambiguity of using the same notation
(3, 7) for the ordered pair that represents a point in the plane and also for the open
interval of real numbers with endpoints 3 and 7. The context in which (3, 7) appears
should always make the meaning clear.
We also say the ordered n-tuples (a), d5,..., a,,) and (C;, C3,..., C,) are equal
iff a; = c; fori = 1, 2,...,n. Thus the ordered 5-tuples (4, 9, 5, 0, 1), (5, 4, 9, 0, 1)
and (0, 1, 4, 5, 9) are all different.
100
CHAPTER 2_
Sets and Induction
Let A and B be sets. The product (or cross product) of
DEFINITION
A and B is
Ax
B=
{(a,b):acAandbeB}.
The set A x B, which is read “A cross B,” is the set of all ordered pairs having
first coordinate in A and second coordinate in B. The cross product is sometimes
called the Cartesian product of A and B, in honor of René Descartes.”
Example.
If A = {1,2} and B= {2, 3, 4}, then
ALB =| (2), l, 5), (1.4), 25-2), (2,3); (2, 4) 4Thus (1,2) eA x B, (2,1) €A
ple, A x BAB
ej
x B, and {(1, 3), (2, 2)} CA
x B. In this exam-
x A since
NE
OLN OI 079 EN
“
GROIN
PARE
ong©
i
The product of three or more sets is defined similarly. For example, for sets A,
B,andC,A x B x C= {(a,b,c):a€A,beEB,andceEC}.
The following theorem presents some useful relationships between the cross
product of sets and the-other set operations.
Theorem 2.2.3
If, A, B, C, and D are sets, then
(a)
(bt)
Ax (BUC)=(A
x B)U(A x C),.
Ax (BNC)=(Ax
BNA x C).
(ie
othe
(d)
(e)
(f)
(Ax BYN(C x D) =(ANC) x (BND).
(Ax B)U(C x D)C (AUC) x (BUD).
(A x B)N(B x A) = (ANB) x (ANB).
op
Proof.
(a)
(Because both A x (BUC) and (A x B)U(A
pairs, their elements have the form (x, y).)
(x,y)€A
x C) are sets of ordered
x (BUC) iff xe AandyE
BUC
iff xe A and (y € BoryeEC)
iff (x € A and y €-B) or (x EA andy €C)
* René Descartes (1596-1650) was a French mathematician, philosopher, and scientist. His work
Discours de la méthode defined analytical geometry by combining the geometric notions of curves and
areas with algebraic equations and computations. Descartes was the first person to use superscripts to
indicate exponential powers of a quantity.
2.2
Mheretore
(e)
Set Operations
101
iff (x,y)
€A x Bor(x,y)EA x C
iff (x,y)
€(A x B)U(A x C).
Al (BiG) =A «x BYU ARC).
If Gy) =
x B)U(C xD), then (yy) EA x B or G, y) EC x D. If
(x,y) €A x B, then xE€A and yeB. Thus
xe AUC
and ye BUD,
because ACAUC
and BCBUD.
Thus (x, y)e(AUC) x (BUD).
If (x, y) € C x D, a similar argument shows (x, y) € (AUC) x (BUD).
This shows that (A x B) U(C x D)
C(AUC) x (BUD).
w
Parts (b), (c), (d), and (f) are proved in Exercise 15. Part (e) of Theorem 2.2.3
cannot be sharpened to equality. See Exercise 16(a).
Exercises 2.2
Bet Ay
DiRT
NS
4
ii 3 coin Oh Bia 0
AG
eS iO, odo 2Oy rand
SC
= 11,2 Ae
(a)
AUB.
(b)
(c)
(ec)
(g)
Gi)
A-B.
A-—B)-C.
(ANC)ND.
(ANB)UANC).
(d)
A—(B-C).
(f) TAL Gia D),
(h) AN(BUC).
Gj) (AUB) -—(CND).
aero
aman
@
ie):
ANB.
Let the universe beall real numbersLet
Av—)[3,.8),
B = )2)61)3
and D = (5, oo). Find
(a)
(c)
(ec)
(2
(i)
2
AUC.
Bac:
B-D.
eae
(AUC) — (BND).
(b)
(d)
(ff)
(h)
(jj)
AUB.
ANB.
A-—B.
D-A.
B-(AUC).
Let the universe be the set Z. Let E, D, Z*, and Z~ be the sets of all even,
odd, positive, and negative integers, respectively. Find
(a) E-Z*.
(Dy Zk.
(d) (Zt).
(C) 25 — ee
(g) E-Z.
(1) CE)
(c) D—E.
ORE
Ome”
Let A, B, C, and D be as in Exercise 1. Which pairs of these four sets are disjoint?
Let A, B, C, and D be as in Exercise 2. Which pairs of these four sets are disjoint?
Give an example of nonempty sets A, B andC such that
BP
Nm
x4¥ (a)
CCAUBandANBEC.
(b)
ACBandCCANB.
(c)
AUB
SG Cand CCB:
(d)
AFBUC,BEZAUC,
(C\pe
(peat
Abe > UO Cn biaeAle CC CAUB AMD — All C, and A 7 B:
SC An CeB Bim C.cA.and A= BUC.
andCCAUB.
102
CHAPTER 2_
Sets and Induction
Prove the remaining parts of Theorem 2.2.1.
Prove the remaining parts of Theorem 2.2.2.
Let A, B, and C be sets. Prove that
x
(a)
AC BifandonlyifA
(by?
(c)
fA CBW Cand ANB =; then A CC.
C CA
Bit and only if-C CA and C CB.
—-B=2©.
(d)
ifACB,thenA
(eg
(A=
(2)
GQGAUB)N
Ce CAUGiNC).
(h)
A — Band B are disjoint.
By =
—-CCB-C.
Cas Ae
Cer (BC),
((f)\- if A-G.Gand BEC, then
10.
AUB:S GC
Let A, B, C, and D be sets. Prove that
(a)) if CCAand DCB,
thenCNDCANB.
(b) ifCCAandDCB,thenCUDCAUB.
°
11.
x
(c)
(d)
(ce)
ifCCA,DCB,andA
and B are disjoint, then C and D are disjoint.
ifC CAandDCB,thenD—ACB—-C.
ff AUBeECUD,
ANB = OG: and CGA, then BC D.
Provide counterexamples for each of the following.
(a)
IEAUCCBUC,
then A CB.
(d)
(e)
(ff)
IAM Ce BG, then A CB.
If(A — B)N(A — C) = ©, then
BN C= ©.
P(A) — A(B) C P(A — B).
A-(B-—C)=(A—-B)—-(A-C).
A—(B-—C)=(A—-B)—-C.
12.
*
LetA and B be sets.
(a) Prove that P(A NM B) = P(A) N P(B). You may use Exercise 9(c).
(b) Prove that P(A) U P(B) C P(A U B).
(c) Show by example that set equality need not be the case in part (b). Under
what conditions on A and B is P(A U B) = P(A) U P(B)?
(d) Determine whether 9(A) and 9(B) are disjoint. Prove your answer.
13.
List the ordered pairs in A x Band B x A in each case.
(A) ee caan als SOD =f dee ket, Ete
x
(b)
(c)
(d)
A= {1,2,.11,2}),B
= (q, {1}, 7}.
A= {P, {D}, {O, {D}}},B = (©, {D}), {SP}, {PD}, O)}.
A= {(@, 4), G, 1)},B = {G 1), @, 3)}.
14.
In general, for nonempty setsA and B, the sets A x Band B x A are not equal.
Find and prove a necessary and sufficient condition for A x B = B x A.
15.
Complete the proof of Theorem 2.2.3 by proving
&
(a) “A XB
(b)
(C),
qd)
GC) =A
Bins
©),
Ax O=©.
“Ase B)IQKG XD) =A
Cis
GD):
GX BOB
A= Gn By) <A MB):
2.2
16.
Proofs to Grade
103
Give an example of nonempty sets A, B, C, and D such that
(a) (A x B)U(C x D)#(AUC)
x (BUD).
(b) (C x C) —(A x B)#(C — A) x (C — B).
(c)
lle
Set Operations
“AS(Bx<
©) F7 (4 xB) C.
Let a, b, c, and d be real numbers.
(a)
Prove that if 0 < a < b, then (a, b) ¢ (2a, 2b).
(b)
Prove that if a, b, c, and d are in the closed interval [0, 1], a < c < dand
b+d>1+a-+c, then the closed intervals [a, b] and [c, d] are not disjoint.
18.
One way to define an ordered pair in terms of sets is to say (a, b) ={ {a}, {a, b}}.
Using this definition, prove that (a, b) = (x, y) iffa=xandb=y.
—
19:
Let A and B be sets. Define the symmetric difference of A and B to be
AA
B=(A
(a)
(c))
AAB=BAA.
AAA]@:
— B)U(B
— A). Prove that
(b)
(d)
AA B=(AUB)
— (ANB).
AAO=A.
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
(a)
Claim.
If A CB,thenA —-CCB—C.
“Proof.”
Assume A C B. Suppose x € A. Then x € B, since A CB.
Let C be any set. Then x € A and x ¢ C. Then x € B and x ¢ C. Thus
x€A — Candxé€B
— C. Therefore,
A -—- CC B—C.
|
(b)
Claim.
If A CB,thenA
“Proof.”
—- CCB—C.
Assume A C B. Suppose A — C. Then x EA
Then x € B, because A C B. Since
(c)
and x € C.
C. Therefore,
A— @GB—G
a
Claim.
If AC B,thenA —-CCB-—C.
‘Proof.’
Assume that
ACB. Then xe€A and xe B. Suppose
xE€A—C.
Then xEA
and x€C.
Since xeB
and x¢€C,
x € B — C. Therefore, A —
(d)
x € B and x € C, B —
CC B — C.
|
Claim.
ACBiffANB=A.
“Proof.”
Assume that ACB.
Suppose
xe AMB. Then xEA
and x € B, so x EA. This shows that
AM B= A. Now assume that
ANMB=A. Suppose xe A. Then x €ANMB, since
A=ANMB; and
therefore, x € B. This shows that x € A implies x € B, so A C B.
(e)
Claim.
IfANBAOand
BNC, thn ANCHE.
‘“Proof.’ Assume
ANB#M and
BNC#S©. Since
there exists x such that x ©€AMB; thus xe A. Since
there exists
dy)
x €
ANBFQ,
BNC#S,
BM C; thus x € C. Hence x € A and x € C. Therefore,
x € ANC, which shows that
AN C# ©.
Claim(A x BB) UC=(4 x1C)
U (Bes ©).
“Proof.”
r
xE(A x B)
UCiffxEA x BorxEeC
iffxeAandxe BorxeC
iffxeA x CorxeEeBxC
ite
Ax COB x):
=
a
104
2.3
CHAPTER 2
Sets and Induction
Indexed Families of Sets
A set of sets is called a family or a collection of sets. In this section, we extend the
definitions and results about the union and intersection of two sets to families of sets.
Throughout this section, we will use script letters, 4, %, 6, ... to denote
families of sets. For example,
etl
eee
ert O a 4
2M) On he, LOTT
is a family consisting of four sets. The set € = {(—x, x):x © R and x > 0} is
an infinite family of open intervals. The sets (—1, 1), (=/2,'V 2), and (—5, 5) are
elements of €. See Figure 2.3.1.
5, 5)
> (V2, V2)
|
5S
|
4
|
3
|
2
—_—__—
(-1, 1)
|
l
|
|
-1
0
I
2
|
3
|
4
|
5
Figure 2.3.1
DEFINITION
The union of the family » (or the union over 4) is
JA = {x:x eA forsomeA
€ SX}.
AEA
Using this definition, for any object x we may write
xe JA
if and only if (GA € A)
A).
AEd
This symbolic statement expresses the direct relationship between the union over
a family and the existential quantifier 4. To show that an object is in the union of a family, we must show the existence of at least one set in the family that contains the object.
Figure 2.3.2(a) is a Venn diagram showing the union over the family 4 = {R, S, T}.
iS
R
di
(A
AEM
(b)
Figure 2.3.2
7
2.3
For the family
of four sets above, L) A =
Indexed Families of Sets
105
{1, 2, 3,4, 5, 6, 7, 9, 10}.The family
AEA
DN
eke St, a)
das) Vis a family of two sets and L) isi Walc Apiaoh yale
Be
Thus, when there are only two sets in a family, the union of the family is the same
as the union of the two sets in the family.
For the family @ =
{(—x, x): x © R and x > 0}, every real number y is an
element of (—|y| — 1, |y| + 1), which is a member of 6. Therefore,
) C = R.
CE€
DEFINITION
SA) is
The intersection of the family « (or the intersection over
()\A = {x:x €A for every A € A}.
AEA
For the intersection over a family 4, we write
x € {)A if and only if (VA € A)(x E A).
AEH
Figure 2.3.2(b) shows the Venn diagram for the intersection of the family
M = {R,S,T}. Using the family sf above again as an example, ()A = {3}
because 3 is the only object contained in all four sets in 4. The aeseatton of the
family € =
{(—x, x): x € R and x > 0} is the set {0} because 0 is the only num-
ber in every set in €.
Properties of unions and intersections for families are similar to the properties
found in Section 2.2. Parts (d) and (e) of the next theorem are De Morgan’s Laws
applied to families of sets.
Theorem 2.3.1
Let & be a family of sets.
(a)
For every set B in the family 4, () A CB.
AEA
(b)
For every set B in the family 4, B C [ JA.
AEA
(c)
If the family f is nonempty, then (]) A C UA.
AEA
(d)
(ma) a Oe
Aes
(e)
Aes
(
()4) = AEH
[1A
AeA
De Morgan’s Laws
AcE
106
CHAPTER 2
Sets and Induction
Proof.
(a)
Let 4 bea
family of sets and B € A. Suppose x € (| A. Then x € A for every
AEd
A € A. (Notice that the setA in the last sentence is a variable. It stands for
any set in the family.) In particular, x € B. Therefore, (iA eB
AEA
(b)
(c)
(d)
See Exercise 3(a).
Let
be a nonempty family. Choose any set C € 4. By parts (a) and (b),
(\AcCCC
UA, and therefore,
(Aaa
AEA
Aca
AEA
AEA
ve (( 4)
AEA
if and only ifx @ {) A.
Aen
if and only if it is not the case that for every A € A, x € A.
if and only if for some B € A, x € B.
if and only if for some B € A, x € B°.
if and only if xe UL) A’.
AEA
(e)
(One proof
of part (e) is very similar to the proof given for part (d) and is left
as Exercise 5(b). Because part (d) has been proved, we may use it to give an
alternate proof. We also use (twice) the fact that (A) = A.)
(Ua) =(Uary =((na)) = 0
It was necessary in part (c) that the family « be nonempty. If
is the empty
family, then the intersection is equal to the universe of discourse. (See Exercise 5.)
This observation is a reason to be cautious about dealing with the empty family of sets.
Theorem 2.3.2
Let 4 be a nonempty family of sets and B be a set.
(a)
IfBCAforallA€ A, thenBC [)A.
(b)
IfA CBforallA es, then
AEA
ACB.
AE
Proof.
(a)
Assume that B CA
for all A € 4. Suppose x € B. Then x € A forall A € Xt.
Therefore x € {] A.
Acs
(b)
See Exercise 3(b).
|
The following example shows how Theorems 2.3.1 and 2.3.2 can be used to
determine the union and intersection of a family.
2.3
Example.
Let € = {{[—r, 7°+ 1):reR andr
Indexed Families of Sets
107
> 0}. Prove that (a) U) C =
and
(Geto):
am
Cee
Proof.
(a)
First, for every xin R, x € [— |x|, |x|? + 1),so RC
LU C. Also, every interval
CE€
in € is included in R, so by Theorem 2.3.2(b),
(b)
U C CR. Thus
ie) Gene
First, [0, 1) is included in every interval in &. Therefore, bye iheorent
Zo O(a (Oo
iS () C. Also, the interval [0, 1) is in €, so by Theorem
Cee
2.3.1(a), {)C © [0, 1). Thus we have (| C = [0, 1).
Cee
#
CE
It is often helpful to associate an identifying tag, or index, with each set in
a family of sets. Consider, for example, the collection
B, = {0, 1, 2,...,n}
2 = {B, :n € N}, where
. Some of the members of93 are the sets B, = {0, 1, 2} and
Be = {0, 1, 2, 3, 4, 5, 6}. In this example, each natural number n corresponds to
a set B,. By specifying the index, as we did when we selected n = 2 or n = 6, we
specified the corresponding set. Furthermore, by specifying a set of indices, such as
the set {2, 6, 10}, we can specify a subfamily of %, namely the family {B,, By, By}.
DEFINITIONS
Let A be a nonempty set such that for each a € A there
is a corresponding set A,. The family {A,: a € A} is an indexed family of
sets. The set A is called the indexing set, and each a € A is an index.
Indexing is a common phenomenon in everyday life. Suppose an apartment
building has six rental units, labeled A through F. At any given time, for each apartment, there is a set of people residing in that apartment. These sets may be indexed
by A = {A, B, C, D, E, F}. Let P, be the set of people living in apartment k. Then
P = {P,:k € A} is an indexed family of sets.
An index is simply a label that provides a convenient way to refer to a certain set.
There is no real difference between a family of sets and an indexed family. Every
family of sets could be indexed by finding a large enough set of indices to label
each set in the family.
Example. ForthesetsA, = {1, 2, 4,5}, A, = {2, 3,5, 6}, andA, = {3, 4, 5, 6},
the index set has been chosen to be A = {1, 2,3}. The family 4 indexed by A
is A = {A,, A>, A} = {A;: ie A}. The family
could be indexed by another
Seta Fon amsiancen it
I=. { 10,210 a
}5,Aj, — (152,45 5},A5, = 12, 3, 5,6}, and
A= (3, 4,5,6) then (A; i] A} = (AzpieT}.
o
Example. For all neN,
let A,={n,n+1,2n}.
Then A, = {1,2},
A, = {2, 3,4},A; = {3,4, 6}, and so forth. The set with index 10 is Ay) =
{10, 11,20}. Except for the set A,, every set in the family {A;:i¢ N} has
108
CHAPTER 2
Sets and Induction
three elements. To describe the family of sets that contains only A,, A3, Ajo, and Aj5,
we write {A;: 7 €{2, 3, 10, 15}}, or we let T = {2, 3, 10, 15} and write {A;: 1 € iffe
Oo
Example. Let A = {0, 1, 2,3, 4}, and let A, = {2x + 4, 8, 12 — 2x} for each
x € A. Then Ay = {4, 8, 12}, A, = {6, 8, 10}, A, = {8}, Az = (6, 8, 10}, and
A, = {4, 8, 12}. The indexing set has five elements, but the indexed family 4 =
{A,: x € A} has only three members because A; = A; and Ay = Ay.
O
As the above examples demonstrate, an indexing family may be finite or infinite, the member sets need not have the same number of elements, and different
indices need not correspond to different sets in the family.
The operations of union and intersection over families of sets apply to indexed
families, although the notation is slightly different. For a family d = {A,: a € A},
we write
[a AS =
UA
aecA
AEA
and
(\A, =
()A.
aed
AEA
Then
xe
UA, if and only if (Ga € A)(x € A,) and
aed
xe (A, if and only if (Va € A)(x € A,).
acl
Example.
For each real number r, define B, = [r?, r? + 1]. Then Bap = E >|,
By = (0, 1], and B,, = [100, 101]. This is another example in which we have dif-
ferent indices representing the same set. For example, B_, = B, = [4,5]. Here the
index set is RR, and we prove that (i) al B. = ©, and (ii) (2 Be
reR
0. co):
reR
Proof.
(i)
Ifthe real number x € ‘a B,, then x € B, for every r € R. However, x € By =
reR
[0, 1] and x € B,,. = [100, 101] is impossible. Therefore () B=
@.
reR
(ii)
Suppose x € [0, 00). Then Vx > 0 and x € By, = |x, x + 1), sox e€ L Be
reR
Now suppose x € {
)B,. Then x € B, = [/?, 7 + 1) for some r € R. Then
0 <r <x,sox€
reR
[0, 00). Therefore JB, = [0, 00).
&
reR
Example.
For neéN, let A, = {n,n + 1, 2n}. Then JA, =N. To show, for
neN
example, that 27 € (J A, we need only point out some index n such that 27 € A,.
neN
2.3
Indexed Families of Sets
109
Either index 26 or 27 will do. Because there is no number x such that x € A, for all
neEN,{)A, = @.
neN
There is a convenient variation on the notation for union and intersection when
the index set is the natural numbers. For an indexed family 4 = {A,:n EN},
(ee)
co
we can write (J A; instead of (J A,. The intersection over of is written {
)A,.
i=l
Example.
4
neN
15
i=l
For eachn EN, let A, = {n,n + 1,n7}. For 4 = {A,:neEN},
4
p=
>i
Tle
iG»
14, 056) 1 16825)36 a (a)Aa |
il
(2,3, 4, 9)
The results of Theorem
Exercise 6.
DEFINITION
Ant
2.3.1 may be restated for indexed families. See
The indexed family 4 = {A,: a € A} of sets is pairwise
disjoint if for all a and £ in A, either A, = Ag or A, MN Ag = ©.
The family {A,,A,,A3} in Figure 2.3.3(a) is pairwise disjoint. However,
the family {B,, B,,B;} in Figure 2.3.3(b) is not pairwise disjoint. Although
B, 1B, = ©, the sets B, and B; are neither identical nor disjoint.
A,
at
(a)
(b)
Figure 2.3.3
Two questions are commonly asked about the concept of pairwise disjoint families.
The first is about why we bother with such a definition when we could simply say a
family is disjoint if and only if ia A, = ©. Having an empty intersection is not the
aeA
110
CHAPTER 2 _ Sets and Induction
same as being pairwise disjoint and not nearly as useful (see Section 3.3). The fam= {a,b}, C, = {b,c}, and C, = {a,c} is notpairwise
ily {C,, C,, C3} with Cis
disjoint, even though nes
=
The second common question asks why the definition says “either A, =
A,
Ag or
Ag= ©” instead of “A, 1 Ag = © whenever a # f.” That is, why not say
“if a # B, then A, and Ag are disjoint”? The family { [n,n + 1): n € N} is pairwise disjoint because whenever n # m, we have [n,n + 1) [m,m-+ 1) = Ds
However, in some families of sets, it happens that different indices correspond to
the same set, so the definition allows for this possibility.
Example.
Suppose 2 = {B,, B,, B3, By, B;, Bg}, where
By =
ayere}
B, = {d,g}
Bes
0,.Gxe}
Ba
B; = {d,g}
By = {b, fh}
Nia
ee}
The family % is pairwise disjoint. Note that B, = B; = B, and B, = B;, so
Exercises 2.3
1. Find the union and intersection of each of the following families.
A
ee
Ls 94) 51 25 Ss Foden
Od's (dp 4acds
Os dba A 45 Oscds
Oa te
(Dect loa) ), 12.4, Or, 1 7,95 LL, Lo tae, LO, [2h}e
* (c) A= {A,:n € N}, where A, = {5n, 5n+ 1, 5n + 2,..., 6n} for each
natural number n.
(2)
ee
IN where BS =
1 2.3). n} for each natural
numbern.
* (e) Ais the family of all sets of integers that contain 10.
Gt
(Aron e115 253,.,-, 10}} whereAj = {1},A, = {2, 3},A, =
(34 Olson hae et LOME Leas ek:
A = {A,:n © N}, whereA, = (0, ) for each natural number n.
(h)
Gj)
o = {A,: r € (0, 00)}, whereA, = [—7, r) for each
r € (0, 00).
d= {Are R}, where
A, = (irl. 2|r| + 1) foreachre R.
m= ine n € N}, where a Sie — oh — Oh
Og, 2 re
(k))
for eachn € N.
tA ee 3}, where A=
(m)
= {C,:n € Z}, where C, = [n,n + 1) for eachn € Z.
4 = {A,:n € Z}, where A, = (n,n + 1) for eachn € Z.
3
E 2+ i)for each n € N — {1, 2}.
—{(DeneNt where D, = (—n, *)forneN.
2.3.
Indexed Families of Sets
111
(0)
sf = {pN: pisa
prime}, where pN = {np: n € N} for each prime p.
(p)
J={T,:neZ},
whereT, =
{(x,y)eR
x R:0<x<1,0<y<x"}
for eachn € Z.
(q) V={V,:neN},whereV, = {(x,y)eER x R:0<x<1a"<y< Vx}
(r)
for eachn EN.
A= {E,:n EN}, where E£, = R — [0, n] for each natural number n.
Which families in Exercise | are pairwise disjoint?
(a)
(b)
Prove part (b) of Theorem 2.3.1.
Prove part (b) of Theorem 2.3.2.
Find the union and intersection of these families, and prove your answers.
fa)
A= {A,: n € N}, where A, = {4n, 4n + 1,..., 5n} for each natural
number n.
(Db) 1 y= Be nieoN} pewhere By = Ne= {n, me 1} for each natural
number n.
(c)
= {C,: ne N}, where C-=~{1 —n, n — 4} for each natural
number n.
(dd),
S=1 D726 7). where Dire
By Theorem 2.3.1,
()A C [JA
AE
O p25)
100 aller
is true for every nonempty family of sets.
AEA
Let the universe be IR, and let W be the empty family of subsets of R . Show
that O) AC
AEA
(b)
Lal A is false in this case by proving that (a) () A=R
AEA
and
AEA
PA =@.
AEA
(a)
(b)
(c)
State the results
(a) — (e) of Theorem
Olia= {eA Poco
ie
for an indexed
family
Prove part (e) of these restated results directly, using the definitions of
intersection and union for indexed families.
Give a proof of part (d) of the restated results that uses part (e) as the key
step of the proof.
Let M = {A,: a € A} bea
(ay
2.3.1
family of sets, and let B be a set. Prove that
BO WA, = UGNA).:
aecA
acd
(b) BU ()4,= [] BVA).
ach
aek
Let
d = {A,:a € A}, and let ® = {Bg: Be I}. Use Exercise 7 to write
(a)
(U A,)a) (
ia)Ba)as a union of intersections.
aeA
(b)
Bel
(( A,)U (a B,)as an intersection of unions.
aed
pel
112
CHAPTER 2_
Sets and Induction
93
family of sets,
= {A,: a € A} bea
Let
A # ©, and B be a set. Prove the
statement is true, or give a counterexample.
(a)
p-(A,)=
acA
ee
Say
acA
(b) p-(UA,) = Le
acd
(c)
A,).
acA
(M4) -8=
acd
(GAs a8).
acA
(d) (U4,)-3= UG,
= B).
acA
10.
If A =
Oy
{A,: a € A} isa
US
ae.
Dy”
12.
family of sets and I C A, prove that
es.
acA
(VT eine
acA
11.
acA
ae.
Let & be a nonempty family of sets.
(a)
What is the largest set X such that X C A for all A € A? That is, find the
set X such that (i) X C A for all A € A and (ii) if V CA for all A € &,
ihengVaSexXc
(b)
What is the smallest set Y such that A C Y for all A € SA? That is, find
the set Y such that (i) A C Y for all A © & and (ii) if A C W for all
Aێ WS, then Y C W.
Let X = {1, 2, 3,4,..., 20}. Give an example of each of the following:
(a)
a family of of subsets ofX such that () A = {1} and [JA =X.
(b)
a family % of four pairwise disjoint subsets of X such that ) B = X.
(c)
a family € of 20 pairwise disjoint subsets of X such that L) C = X.
Aed
Acw
BeB
Cee
13.
Give an example of an indexed collection of sets {A,:n€N}
(i) each A, © (0, 1) and (ii) for all m,n EN, A,,
such that
NA, # ©, but [| A, = ©.
neN
14.
Let A be a family of pairwise disjoint sets. Prove that if 2 C 4, then Bis a
family of pairwise disjoint sets.
15.
Let
and & be two pairwise disjoint families of sets. Let € = 4M % and
R= AUR.
(a) Prove that € is a family of pairwise disjoint sets.
(b) Give an example to show that % need not be pairwise disjoint.
(c)
Prove that if ) A and |) Bare disjoint, then % is pairwise disjoint.
Acsd
BER
2.3
16.
Let
Indexed Families of Sets
113
4 = {A;;i e N} be a family of sets, and let k and m be natural numbers
with k < m. Prove that
k+1
(a)
k
LJ 4,= UA,UA,.,
=!
Pi
ll ee
ll 1
k+l
(b)
x
(c)
(Gs= IN
ll =
>
Ces
3
(d)
ie IN
Si
ll ee
>
=~
Sal
(e)
G=> IN Cs >
si
(f)
17.
>) = IN
=
i
al
>
Suppose 4 = {A,: i € N} is a family of sets such that for alli, j € N, ifi <j,
then A; © Aj. (Such a family is mala a nested family of sets.)
(a)
Prove that for every
(b)
Prove that UJeens
ke N, Q Ana
t=
18.
i=l
Give an example of a nested family {A;:
condition.
ie N} (see Exercise 17) for each
co
*
(a)
Dri a Sl
i=]
(b) [ 1A; = Coo, 1
i=l
()
lo)
()4;=
i=l
{0,1}
Co
dd) (\A;=2
Il
=
Proofs to Grade
19.
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
*
(a)
Claim.
For every indexed family {A,:a@ € A}, (] A, © UA.
acA
“Proof.”
Choose any Ag € (Ayers).
(Apes
and AvenlUAy
acl
(b)
Therefore,
by
Theorem
2.3:2,
2.1.1(c),
acA
(Ane)
*
acA
Lben by Theorem
Ae
acA
acA
Claim.
If A, CB foralla€ A, then
a
JA, CB.
acA
114
CHAPTER 2
Sets and Induction
“Proof.”
Suppose x € |) A,. Then, because A, © B for all a € A,
acd
x € B. Therefore
=
(JA CB.
acd
(c)
’
For every indexed family {A,: a € A},
Claim.
“Proof”?
Let
As— {7 s, fh,An =
) AY.
(NAG
{a,b)e, d},A, =
{bicrd,ej, and
A,={c,d,e,f}.Then () A, = {c,d} € {a,b,c,d,ef} = UA, =
aeA
(d)
Claim.
acd
For every indexed family {A,:@ € A},
{]Ay © WAS
acA
“Proof.”
Assume
(])A,¢ UA,.
aed
x@
UA,. Since
acd
Then,
aeA
for some
xe {)A,,
aed
aeA
x ¢ UA,, it is not the case that x € A, for some
acd
aeéA.
Therefore x ¢ A, for every
a € A. But because x € ( 7a
x€A,
for every ae A. This is a contradiction,
acA
so we conclude
(eee
acA
(e)
Claim.
(oe)
lo.@)
1
Ee = 4] hier) | ot 10= [—3, 10].
iia
“Proof.”
be
acA
n
By Exercise 7,
il
vi
3 a
1
ig
1
1] ere, E en 0]
Asst] Ufe-s)
n=
n
n=1
uf
[.e)
(*\"(=3810)
n=
[==3, 105
2.4
5
Mathematical Induction
In 1889, Giuseppe Peano’ set forth five axioms that provided a complete and rigorous definition of the natural numbers based on the notion of successors. The axioms
assert that
(i)
1 is a natural number;
(ii) |every natural number has a unique successor, which is a natural number;
(iii) |no two natural numbers have the same successor; and
* Giuseppe Peano (1858-1932) was an Italian mathematician who made many contributions to mathematical logic and set theory, especially its language and symbolism. He was the first to use the modern symbols for union and intersection. His ‘“Rormulario Mathematico”
manuscript (1908) contains
4,200 precisely stated mathematical formula and theorems. Other contributions include his “space filling
curve” counterexample, a forerunner of fractal images.
2.4
(iv)
(v)
Mathematical Induction
115
1 is not a successor for any natural number; and
if a property is possessed by 1 and possessed by the successor of every natural
number that possesses it, then the property is possessed by all natural numbers.
These axioms are sufficient to derive all the familiar arithmetic and order properties of N that are listed in the Appendix. The development of all these properties
as consequences of Peano’s axioms is certainly a worthy activity, but it would take
more time than we can devote to the topic here. Instead, we focus our attention on
the inductive property of N given in the fifth axiom. Peano’s fifth axiom can be
restated as a property of sets of natural numbers.
Principle of Mathematical Induction (PMI)
Let S be a subset of N with these two properties:
G)
Gi)
1¢S; and
forallneN,ifne S,thenn+1€S.
Then) S =! N:
A set § of natural numbers is called inductive if it has the property that whenever nmeS, thenn = I 1S: Both (5,6, 7, 8,:.-} and (100, 101) 1020S aa rare
inductive sets. The set {1, 3, 5, 7, 9,...} is not inductive because, for example, 7 1s
a member but 8 is not. Exercise 3 asks you to show that N and © are inductive sets.
Although many sets of natural numbers have the inductive property, only one set is
inductive and contains |. By the Principle of Mathematical Induction, that set is N.
An important use of the PMI is to state definitions. An inductive definition
follows the form of the PMI: We describe a first object, and then the (n + 1)st
object is defined in terms of the nth object. The PMI ensures that the set of all n for
which the corresponding object is defined is N.
It’s easy to define the positive even integers inductively: The first even number is 2, and the (n + 1)st even number is obtained from the nth even number
by adding 2. Calculations involving n terms may also be defined inductively. For
n
example, the sigma notation
sue =A,
terms is defined by
t=]
(Ae
Gi)
1
=
4,4 441 °° 1 x, for the sum O17
ee
n+l
n
is
c=
ForiallmeN, Dox, = Dox; + Xparn
The inductive definition of the product [
|x; = x, - x) -x3-°:+ + x, of nterms is
=i!
@®
1
[]x4=n.
T=
(ii)
n+1
n
j=!
‘i=
ForallneN, []x;= (IIs) tt.
116
CHAPTER 2.
Sets and Induction
Example.
The noninductive definition of the factorial of a natural number n is
n=
Ne (Mee) eta a.
For example, 5! = 5-4. 3-2-1 = 120. The inductive definition of n! is
(i
(ii)
a0
FornEN,(+
1)! =@+
Dn!
To show that the inductive definition defines n! for all natural numbers n, we let
S be the set of n for which n! is defined. First, 1 € S because of part (i). Second, S is
inductive because if n € S, then n! is defined, and, hence, by part (ii), (n + 1)! is also
defined. Thus n + 1 € S. By the PMI, § = N. In other words, the set of numbers for
which the factorial is defined is N, so n! has been defined for all natural numbers.
Because it is useful to extend the definition of the factorial to 0, we define 0! to be 1.
The inductive definition makes clear the relationship between the factorial of a number and the factorial of the next number; if you happen to know that
11! = 39,916,800, then you compute 12! =
12 - 11! = 479,001,600.
The Principle of Mathematical Induction provides a powerful method for proving statements that are true for all natural numbers. For example, we note that the
sum of the first three odd numbers is 1 + 3 + 5 = 9, which happens to be 32: the
sum of the first four odd numbers is 1 + 3 + 5 + 7 = 16, which is 4”; the sum of
the first five odd numbers is 25 = 5”; and the sum of the first 6 odd numbers is 67.
This pattern leads to the conjecture that
foralln, 1-3
+5 ---- -— Qn —
1) = ne’.
We could never verify this statement by checking all possible values for n, but we
can prove it using the PMI.
If we let S be the truth set of the statement “1 + 3+5+---+(2n
—1) =n?,”
then proving the statement is true for all natural numbers amounts to showing that
S = N. The PMI tells us that after we verify that S has properties (i) and (ii) as stated
in the PMI, we can then conclude that S must be equal to N.
Example.
Prove that for every natural number n,
1434+5+4+---+
Qn-—1) =n’.
Proof.
(i)
Because 1 = 17, wehave 1 eS.
(ii)
Let n be a natural number such that n € S. Then
14+34+5+°4+
Qn—- 1) =n’.
(We have assumed that some n is in S. From this assumption, we will show
thatn + 1 € S. This is accomplished by verifying that
143454--4
[27+ 1)-1l=(+4 1).
2.4
Mathematical Induction
117
At this point in the proof, it is essential to compare the statements for n and
for n + 1. Notice that the left-hand sides of the two equations are almost
identical, but the statement about n + 1 has one more term.)
Byadding
2(n + 1) — 1 to both sides
of 1 + 34+5+-:-4+
(Qn —1)=
n* we have
14+34+54--4+
Qn—1)4 [2741-1] =r + [2Qn4+1)- 1]
=n*>+2n+1
= (n+ 1).
This shows that ifn
€ S,thenn + 1 eS.
By the PMI, S = N. That is, 1 + 3 +5 +-:- + (2n — 1) =n’ for every natural
number n.
C
After defining the set S, the first important step in the proof above, called the
basis step, was to verify that the statement is true for n = 1. The second key step,
called the inductive step, was to assume that some natural number n is in S and
verify that n + 1 is in S. The assumption that n € S iscalled the hypothesis of
induction. Notice that we must not assume that n € S for all natural numbers n
because that would be assuming what we want to prove.
The comparison of the statement about n with the statement about n + 1 is the
crucial link in the proof. Every good proof by induction will use the hypothesis of
induction to show that n + | € S. Finding the connection between these statements
is the heart of a proof by induction.
In actual practice, very few induction proofs of statements of the form
(Vn € N)P(n) start by defining the set $. Because “1 € S” is equivalent to “P(1)
is true,” the basis step is a determination that P(1) is true. Since “n € S implies
n+ 1 €S” is equivalent to “P(n) implies P(n + 1),” the inductive step often takes
the form of a direct proof that “for all n € N, P(n) implies P(n + 1).” This gives us
the preferred form for the outline of a proof using the PMI:
PROOF OF (Wn € N)P(n) BY MATHEMATICAL
Proof.
(i)
(ii)
INDUCTION
Basis Step.
Show that P(1) is true.
Inductive Step.
Suppose P(n) is true (assume the hypothesis of induction).
Therefore, P(n + 1) is true.
Therefore, by the PMI, (Vn € N)P(n) is true.
a
Notice in the following examples that it is not enough just to figure out what
the correct statement is for n + 1. To construct a valid inductive step, look for a
118
CHAPTER 2_
Sets and Induction
connection between what we know about some number n and what we want to
know about the next number, n + 1.
Example.
Foralln
@N,n + 3 < 5n’.
Proof.
(i)
1+3
(ii)
Assume that for some
induction.) Then
<5-1?,so the statement is true for 1 .
Gel
eoa=me
née N,n +3
oie
on
Thus, the statement is true for
< 5n’. (This is the hypothesis
+ l=
on
+ l0n + 5 = 57
of
+1)
+ 1.
By the PMI, n + 3 < 5n’* for everyneN.
my
The polynomial x — y divides the polynomials x? — y* and x? — y? because
x2 — y* = (x — y)(@« + y) and x? — y? = (x — y)(x? + xy + y’). The next example extends these divisibility results to x” — y” for every natural number n.
Example.
For every natural number n, the polynomial x — y divides x” — y”.
Proof.
(i)
(ii)
x — ydivides x! —y!=x
y because x — y = 1(x — y). Thus, the statement holds forn = 1.
Assume that x — y divides x” — y”" for some n. To show that x — y divides
x7tl _ yr+l we write
xt l ps ynt!
=
xx?
—
yy”
=
xx?
=
yx" ap yx" =a yy”
Oy
ie
yt
eye).
Now x — y divides the first term because that term contains the factor
(x — y). Also x — y divides the second term because it divides x” — y”
(by the hypothesis of induction). Therefore, x — y divides the sum. That is,
x — y divides x**! — y"t!,
By the PMI, x — y divides x” — y” for every natural number n.
a
n
Our first result proved by induction was that
example involves both factorial and product notation.
.
Example.
.
Prove that for alln € N, [](4i — 2) =
cat
>> (2i — 1) = n. Our next
i=1
(2n)!
n!
Th
2.4
Mathematical Induction
3
(Qe)!
i=]
.
119
Proof.
f
(i)
The statement is true for n =
ae
(ii)
1 because
T[ Gi — 2) i= 2 and
7
= 2
(2n)!
Assume that [](4i — 2) =
; forsomen € N. (We now use the hypothesis
i=!
We
ea!
2
1))!
of induction to prove that I] (47 — 2) _\ at
|
n+l
,
©) .) Then
(n aie 1)!
n
2n)!
i=l
:
[
[Gi -— 2) = nC = 2 |40 +1)=2)= — (an + 2).
i=]
We also compute
“2
(2(n + 1))! ist (2n + 2)! uy 2(n +.1)(2n + 1)(2n)! a
Ga Wh 7 Mecp bS
(a+ In!
(4n + 2)(2n)!
7
n!
Since these expressions are equal, the statement is true for n + 1.
u
2n)!
i=!
n!
By the PMI, |[(4i — 2) = \ : for alln EN.
@
The following examples show some other situations where induction is used.
Example.
Suppose a “map” is formed by drawing straight lines in a plane to rep-
resent boundaries. Figure 2.4.1 shows
10 countries, labeled A through J, formed
by drawing four lines in the plane. The problem is to color the countries so that
adjoining countries (those with a line segment as a common border) have different
colors. This has been done in Figure 2.4.2 using only two colors—blue and white.
We will use induction to prove that every map formed by drawing n straight lines
can be colored using exactly two colors.
Proof.
(i)
If a map is made by drawing one straight line, then there are only two
countries. Thus, every map formed with one line can be colored with two
colors.
Figure 2.4.1
Figure 2.4.2
120
CHAPTER 2
Sets and Induction
(ii)
Assume that for some n, every map formed by drawing n lines can be colored
with exactly two colors. Now consider a map with n + | lines. Before coloring this map, choose any one of the lines and label it L. Now color the map
as though the line L were not there, using exactly two colors. This can be
done, initially, by the hypothesis of induction. (Such a coloring is shown in
Figure 2.4.3, with the line L shown as a dashed line. Of course, only part of
the plane can be shown.) To color the map with line L, proceed as follows.
Call one half-plane determined by L side 1 and the other half-plane side 2.
Leave all colors on side 1 exactly as they were, but change every color on
side 2 to the other color. This gives a coloring to every country in the map
with line L. (See Figure 2.4.4.) It remains to verify that adjacent countries in
this map with n + | lines have different colors.
Suppose we have two adjacent countries. There are two cases to consider:
Case 1.
Figure 2.4.3
Case 2.
Suppose L is the border between the two countries, which means
that one country is on side | and the other on side 2. Initially, the
two countries had the same color because they were parts of the
same country in the map with n lines. When L was added to the
map, the color of the country on side 2 switched to a different
color from the country on side 1.
Figure 2.4.4
Suppose L is not the border between the two countries. Then both
countries are either on side 1 or side 2. If both countries are on
side 1, they were initially colored differently and remain so when
L is added. If both countries are on side 2, their colors were initially different but are now switched and are still different.
In both cases, the two adjoining countries have different colors.
By the PMI, every map can be colored using only two colors.
a
2.4
Mathematical Induction
121
The PMI may be used to prove de Moivre’s’ formula—an important relationship between complex numbers and trigonometric functions.
Theorem 2.4.1
De Moivre’s Formula
Let 6 be a real number. For all n € N, (cos @ + isin 6)” = cos nO + isin né.
Proof.
(Jn this proof, we use addition and multiplication of complex numbers and
the following “sum of angles” formulas from trigonometry:
cos(@ + B) = (cosa)(cosB) — (sina)(sin
B)
sin(a + B) = (sina)(cosB) + (cosa@)(sin
B).)
(i)
Forn = 1, the equation is (cos@ + isin@)! =
cos@ + isin@, which is cer-
tainly true.
(ii) |Assume that (cos@ + i sin@) =
cos k@ + i sink@ for some natural number k.
Then, using the sum of angles formulas,
(cos@ + i sing)kt!
= (cos@ + isin6)(cos@ + i sin@)
(cosk@ + i sink@)(cos@ + isin@)
( by the hypothesis of induction)
cosk@ cosé + isink@ cos@ + icosk@ sind + i? sink@ sind
(cosk@ cos@ —
sink@ sin@) + i(sink@ cos@ + coské sin#@)
cos(kO + 8) + i sin(ké + 6)
II cos(k + 1)@ + isin(k + 1).
By the PMI, (cos@ + i sin)" = cosn6@ + i sinné is true for alln € N.
|
As one more example of the use of the Principle of Mathematical Induction,
we prove a seemingly simple but useful result, known as the Archimedean Principle, about the comparative sizes of natural numbers. Archimedes’ said that given a
fulcrum and a long enough lever, he could move the world. See Figure 2.4.5. This
statement illustrates the principle of physics that relates the forces at the ends of a
lever to their distances from the fulcrum point. Even though it would take a very
large force to move the Earth and a person could exert only a small force, the force
is multiplied when applied to a long lever.
* In his early twenties, Abraham
de Moivre (1667-1754)
fled with the Huguenots from France to
England and eventually became friends with Isaac Newton. De Moivre made significant contributions to
probability and astronomy. He made his living by tutoring and playing chess in London’s coffeehouses.
+ Archimedes (c. 287 B.c.E.-c. 212 B.c.E.) is considered the greatest scientist of his era, having made
fundamental discoveries in mathematics, astronomy, physics, and engineering. Many of his drawings of
proposed machines proved to be very effective devices. His “method of exhaustion” to calculate areas
under curves is similar to the methods of integral calculus used today.
122
CHAPTER 2_
Sets and Induction
Pennsylvania
University
of
Courtesy
Library
aa
Figure 2.4.5
To understand this theorem, think of a and b as any two natural numbers, with
a being much larger than b. The Archimedean Principle says that a can eventually
be surpassed by taking natural number multiples of b. We give a proof by induction.
Theorem 2.4.2
Archimedean Principle for \\
For all natural numbers a and b, there exists a natural number s such that sb > a.
Proof.
(De
Let b be a fixed natural number. The proof proceeds by induction on a.
livi—s
chooser: tolberws Uhenspt——2
Dale sa.
(ii) |Suppose the statement is true when a = n for some natural number n. Then
there is a natural number f¢ such that th > n. Choose s to be ¢ + 1. Then
we have sb = (t+ 1)b = th+b>n +1, so the statement is true when
a=n+1.
By the PMI, the statement is true for all natural numbers a and b.
5
Some statements are not true for all natural numbers but are true for numbers
in some inductive subset of N. To prove such statements in such cases, we need a
slightly generalized form of the PMI, where the basis step starts at some number
other than 1.
Generalized Principle of Mathematical Induction
Let & be a natural number. Let S$ be a subset of N with these two properties:
(i)
(ii)
keS; and
foralln
EeN withn > k,ifne S,thenn+
1 eS.
Then S contains all natural numbers greater than or equal to k.
2.4
Example.
Proof.
(i)
(ii)
Mathematical Induction
123
Prove by induction that n? — n — 20 > 0 foralln > 5.
(We will use the Generalized PMI starting at n = 6.)
Forn = 6, we have 6? — 6 — 20 = 10, which is greater than zero.
Assume for some natural number k > 5 that k2 — k — 20 > 0. Then
(een)
= (= ko wee | =k — 1 — 20
= kek
— 20 4 2k.
Since k* — k — 20 > 0 (by the induction hypothesis) and 2k > O (since kis a
natural number), k* — k — 20 + 2k > 0. (The sum oftwo positive integers is
positive.) Thus (ke + 1)
e+
120
= 0:
By the Generalized PMI, n? — n — 20 > Ois true for all n > 5.
a
We note that an algebraic proof of the last example is possible: Because
n—n—20=(n+4)(n—5)
and
n+42>0
for
all
natural
numbers
n,
n> —n — 20 > 0 for n > 5. Neither proof is “more correct” than the other. We
chose the first proof to demonstrate the Generalized PMI.
Exercises 2.4
1.
Which of these sets are inductive?
(a) A205 21R2 2423a)
(c) > Ale2h4gs, Oicees}
(e) {xe N: x? < 1000}
2.
(Db). 8 { 234, 6,8, 10NRe}
x (d)
{17}
Ar (foul
162-3845 05028)
Suppose a set S is inductive. Which of the following must be true?
(a) Ifn+1e€S,thenneS.
* (b)
Ifne S,thenn+2€eS.
(c) Ifn+1¢€S,thenn
¢ S.
(d)
If6eS,then
11 €S.
(e)
6E€SandllesS.
3.
(a)
(b)
Prove that
N is inductive.
Prove that @ is inductive.
4.
Use the PMI to prove the following for all natural numbers n.
x
(a)
S Gi 2),
(b)
Retiben)
lh
(c)
522! =e
Ph
(d)
te
(e)
BaP
Cn ail):
eee
ay
oe
i
n(n + 2}
tn tae | 5
La,
Do:
124
CHAPTER2 Sets and Induction (f)
5 (2i = 13 =7n?Qn’? — 1).
i=]
Oe
S) i
aeee ie th
a
2!
:
(i)
j
1
:
(CO
n
a
(n+1)! —
3!
=
;
eran
1
ik
oD) H(
—)
a+ 1)!”
n
=
On |
|
Ss
a.
&) iG[1@i-h=—
(1)
hee
iGo)
a ae
n+
1
(2n)!
n!2
(Sum of a finite geometric series)
n—1
;
Yoari
tj
=
i=0
forre Rr
#0, r# Landn
> 1
lar
5.
Use the PMI to prove the following for all natural numbers:
*
(a)
(b)
n> + 5n + 6 is divisible by 3.
4” — 1 is divisible Daan
(c)
(d)
(e)
(f)
n° — nis divisible by 6.
(n° — n)(n + 2) is divisible by 12.
8 divides 57” — 1.
10"*!'+43.4"~! + 5 is divisible by 9.
(g)
(h)
8 divides 9” — 1.
Ifxis areal number and x > 2, thenn < x".
(i)
For every prime p, for every natural number a, if p divides a” then p
divides a.
*
(pars 2a)1 42".
AI(k)NS fer (71 -3):.
d)
4"** = (n + 4).
n
1
(m) )i 552-4
rh
(n)
For every positive real number x, (1 + x)" > 1 + nx.
3
5
ee
(p)
ee
15 18 an integer.
d
—(x") = nx"~!.
Use
d
df
dx
dg
the differentiation
formulas
act ®) ate: we ax.
(q)
Ifaset A has n elements, then P(A) has 2” elements.
<0) = 1 and
2.4
Mathematical Induction
125
Use the Generalized PMI to prove the following. Then show that the equation
or inequality is false for some natural number n.
(a) n°? <n! foralln > 6.
(b)
2” > n? foralln > 4.
(c)
(+1)!
(d)
(e)
2n—8 <n? —8n+ 16foralln
> 6.
n! > 3nforalln > 4.
(f)
ee > 1 2areal number, then (x + 1)” > nx? + 1 for all n > 3.
> 2"*3 forall n > 5.
(g)
min
2n
7
1
for alln > 2.
Velatotallarmece
Leta
i=l
(i)
For alln > 2, the sum of the angle measures of the interior angles of a
convex polygon of n sides is (n — 2) - 180°.
;
reLae
ayforalln > 2.
(j)
Wee face
err
ag Wee + al
Sp
Use the PMI to prove De Morgan’s Laws for an indexed family {A;: i € N}.
You may use De Morgan’s Laws for two sets.
(a)
n
i=l
(b)
Cc
n
(A4,) = VAS for alln EN.
i=l
(U4,) = {)AS for alln EN.
i=l
i=1
Give an inductive definition for each set:
(a)
{nin = 5k forsomekeEN}.
(b) {n:neNandn > 10}.
(c)
(d)
{nin = 2‘ for some k € N}.
{a,a+d,a+
2d,a + 3d,...}, where a and d are real numbers. (The
elements of oe set form an arithmetic progression.)
(e)
{a,ar, ar’, ar*,...}, where a and r are real numbers. (The elements of
the set form a geass progression.)
Py a HSAO S13 7 eee
Let P,, P>,..., P,, be n points in a plane with no three points collinear. Show
that the number of line segments joining all pairs of points is
figure forn = 5.
P
P
Ps
Py,
P
;
— es See the
126
CHAPTER 2_
Sets and Induction
Ww
10.
A puzzle called the Towers of Hanoi consists of a board with three pegs and
several disks of differing diameters that fit over the pegs. In the starting position, all the disks are placed on one peg, with the largest at the bottom, and the
others with smaller and smaller diameters up to the top disk (see the figure). A
move is made by lifting the top disk off a peg and placing it on another peg so
that there is no smaller disk beneath it. The object of the puzzle is to transfer
all the disks from one peg to another.
With a little practice, perhaps using coins of various sizes, you should convince yourself that if there are three disks, the puzzle can be solved in 7 moves.
With four disks, 15 moves are required. Use the PMI to prove that with n disks,
the puzzle can be solved in 2” — | moves. (Hint: In the inductive step, you
must describe the moves with n + | disks and use the hypothesis of induction to
count them.)
11.
In a certain kind of tournament, every player plays every other player exactly
once and either wins or loses. There are no ties. Define a top player to be a
player who, for every other player x, either beats x or beats a player y who
beats x.
(a)
(b)
Proofs to Grade
12.
*
Show that there can be more than one top player.
Use the PMI to show that every n-player tournament has a top player.
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
(a)
Claim.
“Proof.”
All horses have the same color.
We must show that for all n € N, in every set of n horses,
all horses in the set have the same color. Clearly in every set containing
exactly one horse, all horses have the same color.
Now suppose all horses in every set of n horses have the same
color. Consider a set of n + 1 horses. If we remove one horse, the
horses in the remaining set of n horses all have the same color. Now
consider a set of n horses obtained by removing some other horse. All
horses in this set have the same color. Therefore, all horses in the set
of n + | horses have the same color. By the PMI, the statement is true
for everyn EN.
a
2.4
*
(bd) Claim.
Mathematical Induction
127
For all n € N, n> + 44n is divisible by 3.
“Proof.”
(i)
1° + 44(1) = 45, which is divisible by 3, so the statement is true
TOM i—ele
(ii) Assume the statement is true for some n € N. Then n? + 44n is
divisible by 3. Therefore, (n + 1)? + 44(n + 1) is divisible
by 3.
By the PMI, the statement is true for all n € N.
(c)
a
Claim. For all natural numbers n,n >n-+ 1.
“Proof.”
Assume the statement is true for some k € N. Thenk>k+ 1.
Adding | to both sides of this inequality, we have k + 1 > (k+1)+ 1.
Therefore, the statement is true for k + 1. By the PMI, the statement is
true for alln EN.
(d)
&
Claim. For all natural numbers n, 2”~ ! < 2” — 1.
“Proof.”
(i) The statement is true for n = 1 because 2!~!=2°=1<2!—1=1.
(ii) We must show that if n = k, then
n = k + 1. Suppose that
OED)
eT ead. That 18202 “ane
Lneneahave
2*<2*+1_ 9. Dividing by 2, we have 2*—! < 2* — 1.
By the PMI, the statement is true for all natural numbers n.
(e)
Claim. For every natural number n, n* + n is odd.
“Proof.”
The number n = 1 is odd. Suppose n€N
a
and n* + n is
odd. Then
(nt
A)eee im + lantern
= (n’ + n) + (2n 4+ 2)
is the sum of an odd and an even number. Therefore, (n + 1)? + (n + 1)
x
(f)
is odd. By the PMI, the property that n* + n is odd is true for all natural
numbers n.
a
Claim.
Every natural number greater than | has a prime factor.
“Proof.”
(i) Letn = 2. Then n is prime.
(ii) Suppose & has a prime factor x. Then k = xy for some y. Thus,
k+1=xy+
1=(+4
10 4 1), whichis
By the PMI, the theorem is proved.
(g)
aprime factorization.
a
Claim.
For all natural numbers n > 4, 2” < n!
“Proof.”
2* = 16 and 4! = 24, so the statement is true for n = 4.
Assume that 2” < n! for some n EN. Then 2”+! = 2(2”) <2(n!) <
(n + 1)(n!) = (n + 1)!,80 2"*! < (n + 1)!. By the PMI, the statement
is true for alln > 4.
r
128
2.5
CHAPTER2
Sets and Induction
Equivalent Forms of Induction
This section discusses two other versions of mathematical induction. The value of
these new forms of induction is that they may be used in situations where it would
be difficult to apply the PMI. Both new forms are equivalent to the PMI, which
means that either of them could replace the PMI in the list of axioms for N.
To prove that a statement is true for all natural numbers using the PMI, the
inductive step assumes that a statement is true for an arbitrary natural number 7 and
then shows that the statement is true for n + 1. When there might be no apparent
connection between the statement for n and the statement for n + 1, there may be
a connection between the statement for the n + 1 case and the statement for some
value or values less than n. There is a variation of the PMI to handle this situation.
A much stronger assumption is made in this alternate form of induction, called
complete (or strong) induction.”
Principle of Complete Induction (PCI)
Suppose S is a subset of N with this property:
For all natural numbers 7, if {1, 2,3,...,2
—
1} CS, thenneS.
Then Sv N"
The hypothesis of induction for this form of induction assumes that the statement is true for every natural number less than n. Thus, we are allowed to assume
the statement is true for each of k = 1, 2, 3,..., n — 1. Every proof using the PCI
assumes this hypothesis of induction and then proceeds by showing that the statement is also true for n.
Notice that the statement of the PCI does not require a basis step in which we
show that 1 € S. Nevertheless, for n = 1, the PCI property has the form
Qa
S =e
S.
which is equivalent to 1 € S. Practically speaking then, it is best to begin a PCI
proof by verifying that | is in S. Sometimes special consideration is also needed for
n = 2 or larger integers. We saw in Section 2.4 that this caution may be necessary
for the PMI as well (see Exercise 12(a) of Section 2.4).
For our first example of a proof using the PCI, we consider a version of the
game of strategy called Nim. In this game, two players (1 and 2) begin with two
piles of n coins and take turns removing coins from either pile. Player | goes first.
In each turn, a player removes any number of coins, but at least one, all from the
same pile. The winner is the player who takes the last coin. We say a player has a
winning strategy if that player has a way to win every game, no matter what moves
the other player makes.
* Because this form of induction employs such a strong assumption, the Principle of Mathematical
Induction is sometimes referred to as “weak induction.”
2.5
Example.
Equivalent Forms of Induction
129
Prove that Player 2 has a winning strategy for this game of Nim.
Proof. Let S be the set of all natural numbers n such that Player 2 has a winning strategy when there are n coins in each pile. (Now assume the hypothesis of
induction.) Suppose m is a natural number and {1, 2, 3,...,m—
1} CS. In the
special case that m = 1, each pile has only one coin. Player 2 necessarily wins
because Player | takes the only coin from one pile and Player 2 takes the last (only)
coin from the other pile.
For m > 1, if Player | takes an entire pile, then Player 2 wins by taking everything in the other pile. Otherwise, Player 1 takes j coins from one pile, where
1 <j < m. If Player 2 takes j coins from the other pile, the result is two piles with
m — j coins each. Because 1 < m — j < m — 1, by the hypothesis of induction
Player 2 has a strategy to win a game with m — j coins. To win the game with
m coins, Player 2 takes j coins and then follows the winning strategy for a game
with m — j coins.Therefore m € S.
By the PCI, S = N. Therefore Player 2 always has a winning strategy.
rT
The proof above clearly relies on the strong hypothesis of induction because
Player 2 must have a strategy to counter every move Player | makes. The PCI is
applicable because it lets us assume there is a strategy for every k < m.
As with the PMI, proofs using the PCI are often written without referring to
the set we called S. Also like the PMI, the PCI has variations where the induc-
tion starts at some number greater than one. To prove, for example, that some
property holds for all numbers greater than 6, we would verify that for all natural
numbers n > 6, if the property holds for every number from 7 up to n — 1, then
the property holds for n. Our next example is a proof that every natural number
greater than | is prime or is a product of primes. This fact was used without proof
in Chapter 1 because we did not yet have induction available as a method of
proof. It is another good example of a proof where the PCI is much more natural
to use than the PMI.
Lemma 2.5.1
Every natural number greater than | is prime or is a product of primes.
Proof. Let m be a natural number greater than 1. We note that 2 is a prime, so the
statement is true when m is 2. Now assume that k is prime or is a product of primes
for all k such that 1 < k < m. If m has no factors other than | and itself, then m is
prime. Otherwise, m = st for some natural numbers s and ¢ that are greater than
1 and less than m. By the complete induction hypothesis, each of s and f either is
prime or is a product of primes. Thus, m = st is a product of primes, so the statement is true for m.
By the PCI, we conclude that every natural number greater than | is prime or
is a product of primes.
5
Like the PMI, the PCI can be used to create inductive definitions, one of which
is the definition of the sequence 1, 1, 2, 3,5, 8, 13,... examined by Leonardo
130
CHAPTER 2
Sets and Induction
Fibonacci’ in the 13th century (see Exercise 5). These numbers have played important roles in applications as diverse as population growth, flower petal patterns,
and highly efficient file-sorting algorithms in computer science. For each natural
number n, the nth Fibonacci number, f,, is defined inductively by
ae
fi —
—
1, and
tna
=
n+l
+ f,
forn
Ze
it
We see that 75 = 2, Ja.= o55
= Safe
= 9, and
so.On.
Inductive proofs of properties of the Fibonacci numbers Fouls involve the
PCI because we need to “reach back” to both f,_, and f,_» to prove the property
for f,. Here is a typical example.
Example.
Let ¢ be the positive solution to the equation x7 = x + 1. (The value of
(I+ V5)
od is Thges Sees approximately 1.618.) Prove that f, < @”~' forall n > 1.
Proof.
Inthe special cases of m = 1 and m = 2, the inequality f,, < ¢”~' is true
because f, = 1 < ¢° = 1 and ff, =1 < ¢! = (1 + V5)/2. Let m be a natural
number, m > 3, and assume that f, < *~' for all k € {1, 2, 3,...,m — 1}. For
m > 3, we have
ips =fin—1
Hl
Nee,
< pf"? + $"~? (by the induction hypothesis form — \ andm — 2)
=
Ce
=p"
=
a
367)
1)
(because ¢ is a solution tox*
=x +1,¢+1=¢7)
Gis ad
heretotc g(a.
By the PCI, we conclude that f, < ¢”~! for alln > 1.
Ei
A third property that characterizes the set N is the Well-Ordering Principle
(WOP). Although it is quite simple to state, the WOP turns out to be a powerful tool
for constructing proofs. The WOP, like the PCI, may be derived from the Peano
Axioms and is equivalent to the PMI. (See Theorems 2.5.5—2.5.7.)
Well-Ordering Principle (WOP)*
Every nonempty subset of N has a smallest element.
* Leonardo of Pisa, also called Leonardo Fibonacci (c. 1170-c. 1250), was a prominent mathematician in
the Middle Ages. His text, Liber Abaci (Book of Calculation), was influential in the European adoption
of Hindu-Arabic numerals. He did not invent the sequence named for him but used it as an example.
* Some mathematicians refer to the Well-Ordering Principle as the Well-Ordering Property. They
reserve the use of the term Well-Ordering Principle for the statement that for every set there is an order
that makes the set a well-ordered set. Orderings are discussed in Section 3.5.
2.5
Equivalent Forms of Induction
131
Proofs using the WOP frequently take the form of assuming that some desired
property does not hold for all natural numbers. This produces a nonempty set of
natural numbers that do not have the property. By working with the smallest such
number, one can often find a contradiction.
As an example of how the WOP is used, we give an alternate proof of
Lemma 2.5.1. Compare this proof with the earlier proof that used the PCI.
Lemma 2.5.1
Every natural number greater than | is prime or is a product of primes.
Second proof. Suppose there is at least one natural number that is not a prime
and cannot be written as a product of primes. Then the set T of such numbers is
nonempty, so by the WOP there is a smallest such number, m. From the fact that m
is not prime, we have that m = st for some natural numbers s and ¢ that are greater
than | and less than m. Both s and t are less than the smallest element of T, so they
are not in 7. Therefore, each is a prime number or is the product of primes, which
makes m a product of primes. This is a contradiction. Therefore, every natural number greater than | is prime or is a product of primes.
a
The WOP is the tool we need to prove the next three theorems, which were
stated without proof earlier in the text. The Division Algorithm was the primary
result that we used in Section 1.8 to develop interesting results about the greatest
common divisor of two integers and linear combinations of the integers.
Theorem 2.5.2
The Division Algorithm
For all integers a and b, with a # 0, there exist unique integers g and r such that
=aq+rand0<r< |a|.
Proof. Let a and b be integers and a #0. Assume that a > 0. (The proof in the
case a < 0 is similar and is left as an exercise.) We must first show the existence
of g and r.
Let S =
{b — ak: kis an integer and b — ak > 0}. IfOisinS, then
a divides b,
and we may take q to be the integer 7 and r = 0. Now assume that 0 ¢ S.
It follows from the assumption 0 ¢ S that b # 0. (Otherwise, b — a0 = 0 € S.)
Now if b > 0, then b — a0 € S, and if b < 0, then b — a(2b) = b(1 — 2a) € S. In
either case, the set $ is nonempty.
By the WOP,
the set S has a smallest element, which we will call r. Then
r = b — aq for some integer g, so b = aq + r and r = 0. We must also show
‘iat ee a
SUppese a. lhe ag
4~l)—D—
aq
—a=—7
—a=(),
so b— a(q+ 1) ES. But b — ag + 1) < b — aq, and b — aq is the smallest
member of S. This is a contradiction. We conclude that r < |a|.
To complete the proof, we must show that g and r are unique. Suppose there
exist integers g, q’, r, and r’ such that
b=aq+r
withO
<r < |a| and
b=aq' +r'withO
<r’ < fa].
132
CHAPTER 2
Sets and Induction
Without loss of generality, we may assume that r’ > r. (Otherwise, we could
relabel r and r’.) Then aq + r = aq’ + r’, which implies a(q — gq!) =r’ — r.
Then @idivides7°= rf ButO = rir =r’ =a) =a Therefore, r° —'7
0,so0
=
r =r, Because a(ga tq
=O g aac
|
Section 1.8 also utilized the following result about linear combinations of
integers.
Theorem 2.5.3
Let a and b be nonzero integers. Then there is a smallest positive linear combination of a and b.
Proof.
Let a and b be nonzero integers, and let S be the set of all positive linear
combinations of a and b. Then S is nonempty because either a(1) + b(0) > 0 or
a(—1) + b(0) > 0, depending on whether a > 0 or a < 0. By the WOP, this non-
empty subset of N has a smallest element, which is the smallest positive linear
combination of a and b.
a
We now prove the Fundamental Theorem of Arithmetic. We have already
established the existence part of this theorem, in Lemma 2.5.1. What remains is
to show that if we disregard the order in which the prime factors appear, then the
prime factorization of a composite natural number is unique. The proof uses the
WOP and also Euclid’s Lemma (Lemma 1.8.3).
Theorem 2.5.4
The Fundamental Theorem of Arithmetic
Every natural number greater than | is prime or can be expressed uniquely as a
product of primes.
Proof. Suppose there exist natural numbers that can be expressed in two or more
different ways as the product of primes, and let n be the smallest such number. Then
N =P) Pr P3.-.pyandn = q, qo 43---G, for some k, m in N, where eachp;and q;is prime.
Then p, divides n, so p, divides n = q, q> q3...q;. By Euclid’s Lemma, p; = q; for some j,
1 <j <™m. Then n/p, = n/q;is a natural number smaller than n that has two different
prime factorizations. This is a contradiction. We conclude that every natural number
greater than | that is not prime can be uniquely expressed as a product of primes.
gE
The final theorems of this section show that the WOP may be proved from
the PMI, the PCI may be proved assuming the WOP, and the PMI may be proved
assuming the PCI. The theorems together show that all three forms of induction are
equivalent.
Theorem 2.5.5
The Principle of Mathematical Induction implies the Well-Ordering Principle.
Proof.
Assume that the PMI holds for N. (To show that the WOP is true, we show
that every nonempty subset of N\ has a smallest element.) Suppose T is a nonempty
subset of N. Let § = N — T. (We will suppose T has no smallest element and use
the PMI to reach a contradiction.) Suppose that T has no smallest element.
2.5
(i)
Equivalent Forms of Induction
133
Since 1 is the smallest element of N and T has no smallest element, 1 ¢ T.
Therefore, 1 € S.
(ii)
Suppose that n € S. No number less than n belongs to T because, if any of
the numbers 1, 2, 3,..., — 1 were in T, then one of those numbers would
be the smallest element of T. We know n ¢ T because n € S. Therefore,
n + 1 cannot be in T, or else it would be the smallest element of 7. Thus,
n+1eS.
By the PMI, §S= N. Then T = ©, contradicting the assumption that T is nonempty.
Therefore, T has a smallest element. We conclude that the WOP holds in N.
|
Theorem 2.5.6
The Well-Ordering Principle implies the Principle of Complete Induction.
Proof.
Assume that the WOP holds for N. Suppose that S is a subset of N with this
property: For all natural numbers m, if {1, 2, 3,...,m — 1}C S, thenm eS.
Suppose that § # N. Then the complement of S is nonempty, and by the WOP,
there is a smallest natural number k that is not in S. Then {1, 2, 3,...,k —1} CS,
so by the property we assumed for S, k € S. This is a contradiction. We conclude
that the PCI holds in N.
a
Theorem 2.5.7
The Principle of Complete Induction implies the Principle of Mathematical Induction.
Proof. Assume that the PCI holds for N. Suppose S is a subset of N' with two
properties:
Gi)
(ii)
1¢€S; and
For all natural numbers n, ifn € S, thenn + 1 eS.
(To prove that S = N, we show that S satisfies the hypothesis for the PCI: namely,
that for allmeN,
if {1, 2, 3,...,m — 1} CS, then me S.) Let m be a natural
number such that {1, 2, 3,...,m — 1} CS. There are two cases:
Casel
If m=1,
then 1 €S
{1, 2, 3,...,m —
Case2
by the first property of S. Thus, the statement
1} C S impliesm € Sis true when m =
Ifm > 1, then from {1,2, 3,...,m —
1.
1} CS, we have m —
1 €S. But
then by the second property for S, we have m € S. In this case, too, we
have {1, 2, 3,:..,m —
1} CS impliesm eS.
We conclude that the statement { 1, 2, 3,...,™ — 1} CS implies m € S is true for
all natural numbers m. Therefore, by the PCI, S = N. We conclude that the PMI
holds in N.
a
Exercises 2.5
1.
Use the PCI to prove that
(a) every natural number greater than or equal to 11 can be written in the
form 2s + 5t, for some natural numbers s and t.
134
CHAPTER 2_
Sets and Induction
(b)
every natural number greater than 22 can be written in the form 3s + 41,
(c)
every natural number greater than 33 can be written in the form 4s + 5t,
where s and ¢ are integers, s > 3 andt > 2.
where s and ¢ are integers, s > 3 and t > 2.
In an alternate form of the game of Nim, as in the example given in this section, there are two piles with n coins in each pile. As before, each player in
turn must remove at least one coin from just one of the two piles, but in this
version, the last player to remove a coin loses. Prove that for all n > 2, Player
2 has a winning strategy for this game.
Letaa, = 2 a = 4, eand d..,—.9¢,,,, — 0a, tor alln = 1 Provemhat
d, = 2" for all natural numbers n.
In this exercise, you are to prove some well-known facts about numbers
as a way to demonstrate the use of the WOP. Use the WOP to prove the
following:
(a)
Ifnis anatural number, thenn + 1=1-+n.
(b)
(c)
Ifa > 0, then for every natural number n, a” > 0.
For all positive integers a and b, b#a-+ b. (Hint: Suppose that for
some a there is b such that
b = a + b. To make use of the WOP, do not
use the property that b —
b = 0. You may use the successor properties
of N.)
(d)
V2 is irrational.
In 1202, Leonardo Fibonacci posed the following problem: Suppose a particular breed of rabbit breeds one new pair of rabbits each month, except that a
1-month-old pair is too young to breed. Suppose further that no rabbit breeds
with any other except its paired mate and that rabbits live forever. At | month,
we have our original pair of rabbits. At 2 months, we still have the single pair.
At 3 months, we have two pairs (the original and their one pair of offspring).
At 4 months, we have three pairs (the original pair, one older pair of offspring,
and one new pair of offspring).
(a) Show that at n months, there are f, pairs of rabbits.
(b) Calculate the first 10 Fibonacci numbers f,, f, fy,---, io.
(c) Finda formula for f,.3; — f,,+1.
Prove each of the following statements about Fibonacci numbers. Use the
PMI for parts (a), (b), and (d). Use part (b) to prove part (c).
(a)
fs, 1s even and both f;,,, and f;,, , are odd for all natural numbers n.
(b)
gcd(f,,f,4
(c)
d@
gcd(f,,f.42) = | for all natural numbers n.
fithit+At+:: + f= fizz — 1 for all natural numbers n.
1) = | for all natural numbers n.
Use the PCI to prove the following properties of Fibonacci numbers:
(a) f,, is anatural number for all natural numbers n.
(b) fise =4f,43 +f, for all natural numbers n.
(c)
For every natural number a, ff, + fosifn+1 = fo+n+1 for all natural
numbers n.
2.5
(d)
Equivalent Forms of Induction
135
(Binet’s formula) Let ¢ be the positive solution and p the negative
1
solution to the equation x = x + 1. (The values are @ = Je
and
Ve vs Show for all natural numbers n that f, = ae
n= pr .
p= ot
ee.
Prove that for all n € N,
(a) f,, is a multiple of 3.
(b) fs, is a multiple of 5.
(Ca
lat
Jon Jane
Complete the proof of the Division Algorithm (Theorem 2.5.2) for the case
a <Q. That is, show that for all integers a and b, with a < 0, there exist
unique integers g and r such that b = aq + rand0 <r < |a| = —a.
10.
Let Z~ be the set of negative integers. Prove that every nonempty subset of
Z~ has a largest element.
11.
In the tournament described in Exercise 11 of Section 2.4, a top player is
defined to be one who either beats every other player or beats someone who
beats the other player. Use the WOP to show that in every such tournament
with n players (n € N), there is at least one top player.
12.
Let the “Fibonacci-2” numbers g,, be defined as follows:
Sle
13.
14.
2, and Sn+2
or §n+18n
for all n Zz Ne
(a)
Calculate the first five “Fibonacci-2” numbers.
(b)
Show that that for alln
EN, g, = 2”.
Prove the converses of Theorems 2.5.5, 2.5.6, and 2.5.7. That is, prove that
(a)
(b)
(c)
Proofs to Grade
2, Diz
the WOP implies the PMI.
the PCI implies the WOP.
the PMI implies the PCI.
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
(a)
Claim.
“Proof.”
For all natural numbers n, 5 divides 8” — 3”.
Suppose there is n € N such that 5 does not divide 8” — 3”.
Then, by the WOP, there is a smallest such natural number ¢. Now t # 1
because 5 does divide 8! — 3!. Therefore, tf — 1 is a natural number
smaller than t, so 5 divides 8’~! — 3'~!. But then5 divides 8(8’—! — 3‘~!)
and 5 divides 5(3'~!), so 5 divides their sum, which is 8’ — 3’. This is a
contradiction. Therefore, 5 divides 8” — 3” forall ne N.
a
(b)
Claim.
For every natural number n, 3 divides n> + 2n + 1.
“Proof.”
Suppose there is a natural number n such that 3 does not
divide n> + 2n + 1. By the WOP, there is a smallest such number.
Call this number m. Then m —
(m — 1
| is a natural number, and 3 does divide
+ 2m — 1) 4+ 1 =m? — 33m? + 3m
=m
— 14+ 2m—241
— 3m -— 5m — 2.
136
CHAPTER 2_
Sets and Induction
But 3 also divides
3m? — 3m + 3, so 3 divides
the sum
of these
two expressions, which is m? + 2m + 1. This contradicts what we know
about m. Therefore, the set {n € N: 3 does not divide n+ 2n+ 1}
must be empty. Therefore, 3 divides n>? + 2n + 1 for every natural
(c)
a
number n.
Claim.
For every natural number n, the nth Fibonacci number is even.
Let m be anatural number, and assume that the nth Fibonacci
“Proof.”
is even for all k such that 1 < k < m — 1. That is, assume that for k = 1,
2,3,...,m—1,f,is even. Then f,,=f,,-2+/,,-1- By the hypothesis of
induction, both f,,_5 and f,,_ ; are even, so f,, is even. By the PCI, f, is
even for every natural number n.
(d)
Claim.
“Proof.”
z
For every natural number n, n* + 31n is even.
Let § = {n € N: n? + 31n is even}. Suppose m is a natural
numberand {15 2,3,:..,m—
1} CS. Then Gn — 2)? + 31(m — 2) and
(m — 1)* + 31(m — 1) are even. Also, 2m + 30 = 2(m + 15) is even,
and the sum of two even numbers is even, so (m — 1)* + 31(m— 1) +
2m + 30 = m? + 31mis even. Therefore m € S. By the PCI,
S=N, so
a
n> + 31n is even for every natural number n.
(e)
For every natural number n, 5” = 5.
Claim.
The statement is true form = 1 because 5! = 5. Assume that
“Proof.”
5*= 5 forall ksuch that 1
<k<m-—1.Then5”"=(5"~!)(5"-1)/5"-2 =
(5-5)/5 = 5, so the statement is true for m. By the PCI, the statement is
|
true for all natural numbers.
(f)
Every natural number n > 22 can be expressed as a natural
Claim.
number multiple of 3 plus a natural number multiple of 7.
Let
T= {n € N: n = 3a + 7b for some natural numbers
“Proof.”
a and b}. Let m be a natural number greater than or equal to 22, and
assume that {22, 23, 24,...,m— 1} CT. We must show that m € T.
Notice that 22, 23, and 24 are in T because 22 = 3(5) + 7(1); 23 =
3(3) + 7(2); and 24 = 3(1) + 7(3), so we may assume that m > 25. Then
m—3>22,som-— 3 isinT. Therefore, for some natural numbers a and
b, we have m — 3 = 3a + 7b. Thus, m = 3a + 7b +3 =3(a+1)4+ 7b,
(g)
(h)
&
so m € T. By the PCI, every natural number n > 22 is in T.
The PCI implies the WOP.
Claim.
Assume the PCI. Let T be a nonempty subset of N. Then
“Proof.”
T has some element x. Then {1, 2,...,x — 1}
CN — T. By the PCI,
xe N = T.
xe T and
x€N —T. This is a contradiction because
Therefore T has a smallest element.
a
The WOP implies the PCI.
Claim.
“Proof.”
Assume the WOP. To prove the PCI, let § be a subset of N
such that for all natural numbers m, {1, 2, 3,...,m — 1} CS. LetkeEN.
Then k + 1 is an integer, so {1,2,...,(k + 1) — 1} CS. However,
(k + 1) — 1 =k,sok €S. Thus, every natural number is in S, so S = N.
|
2.6
2.6
Principles of Counting
137
Principles of Counting
This section describes some of the fundamental techniques for counting the
number of elements in a finite set. We will for now rely on an informal definition of
finite: A set is finite if it is empty or if it has n elements for some natural number n.
A precise definition is given in Chapter 5.
We write A = 7 to indicate that the set A has n elements. For example, if
A = {p, q} and B = {3, 2, 1,5, 9}, then A = 2 and B = S. For the empty set, we
write
Theorem 2.6.1
@ = 0.
The Sum Rule (for two sets)
Let A and B be finite sets with m and n elements, respectively. If A and B are
disjoint, then
AUB =m
+ n.
Although a formal proof of Theorem 2.6.1 must be postponed until we see
the precise definition of finiteness in Chapter 5, what the basic sum rule for two
sets says is so familiar that we rarely stop think about it. If a password consists of
6 letters and 5 digits, then the password has 11 symbols. If set A contains 12 oil
paintings and set B has 7 watercolors, then A U B has 19 works of art. The next
theorem gives the more general result that the size of the union of n pairwise disjoint finite sets is the sum of their sizes. Watch for where its proof uses the hypothesis of induction and the sum rule for two sets.
Theorem 2.6.2
The Sum Rule
For all n € N and for every family
wise disjoint sets, if A, = 4,106 ly
4 = {A;:i = 1, 2,3,...,n} of n distinct paira7, then
Proof.
@
1
ri
1
Ifn=1,then UA; = A; =a, = da;
t=
(ii)
HI
Suppose for some n € N thatA,= a, fori = 1, 2,...,n and UA; = Se
for every family 4 = {A,:i = 1, 2, 3,...,n} of n distinct pairwise disjoint
sets. Let 4 = {Aj, Ap, Az,...,A,4,} be a family of n + 1| distinct pairwise
disjoint sets with A; = a; fori = 1, 2,....n + 1. Then
n+l
UA;
i=1
n
=
(Ua,)Uar.
i=1
=
=
UA;
i=1
ee ir
+ Anas
n
as dea; + any
Lal
n+1
—
a.
ial
138
CHAPTER 2 _ Sets and Induction
By the PMI, the sum rule is true for every family of n distinct pairwise disjoint sets,
5
for alln Ee N.
The sum rule is useful in situations where it would be practically impossible
for any individual to make an acceptably accurate count. For example, a good estimate of the total population of a country on a fixed date (a census) may be accomplished by summing the results of the combined work of thousands of individuals,
each of whom does a count for a designated small geographic area.
If sets A and B are not disjoint (see Figure 2.6.1), then determining the number
of elements in A UB by simply adding A and B overcounts A U B by counting
twice each element of A M B. Theorem 2.6.3 corrects this error by subtracting A M B
from A + B.
Figure 2.6.1
Theorem 2.6.3
For finite sets AandB, AUB=A+B—ANB.
Proof.
A
LetA and B be finite sets. The set A is the union of the disjoint sets A — Band
B, so by the sum rule A = A —
B + AB.
Therefore, A —
B =A
— ANB.
Applying the same reasoning to the set B, we have B — A = B — AMB. Because
the family {A — B, B — A,
AM B} is a family of distinct pairwise disjoint sets
whose union is A U B, by the sum rule
AND I— Al
be
be
Ar
AB
=A-ADB+B—-AMB+-ANB
Ab
Avene.
a
Example.
Suppose a certain natural number a has 12 divisors and another natural
number b has 48 divisors. All together there are 56 numbers that divide either a or
b. How many numbers are common divisors of a and b?
Letting A and B be the sets of divisors of a and b, respectively, the solution is
ANB=A+B—AUB==
12+ 48 —56=4.
Applying the sum rule to the disjoint sets A — B and A Q B, we could also find
thatA —B=A —ANB=12—4=8, soa has 8 divisors that are not divisors of
b. Similarly, we could determine that there are 44 divisors of b that are not divisors
of a. See Figure 2.6.2.
Oo
2.6
Principles of Counting
139
(
Figure 2.6.2
Example.
French and German are two of the four national languages of Switzerland. Suppose 80% of Swiss residents speak German fluently, 66% speak French
fluently, and 52% are fluent in both languages. What percentage of Swiss residents
are not fluent in either French or German?
To solve this problem, we first find the percentage of residents fluent in at least
one of the two languages (80 + 66 — 52 = 94) and subtract this result from 100%.
Based on the given estimates, 6% of residents are not fluent in either language.
Theorem 2.6.3 can be extended to three or more sets by the Principle of
Inclusion and Exclusion. The idea is that, in counting the number of elements
in the union of several sets by counting the number of elements in each set, we
have included too many elements more than once, so some need to be excluded, or
subtracted, from the total. When more than two sets are involved, this first attempt
at exclusion will subtract too many elements, so some need to be added back, or
included again, and so forth. For three sets A, B, C, the Principle of Inclusion and
Exclusion states that
AU BULG =A 4B + C=
ADR
AMCo
BGG)
Ang bine
The inclusion and exclusion formulas for more than three sets are lengthier.
(See Exercise 5.) The Principle of Inclusion and Exclusion is often applied to determine the number of elements not in any of several sets, as in Exercise 6.
The next theorem counts the number of pairs in the cross product of two sets.
Like Theorem 2.6.1, its formal proof is found in Section 5.1. Theorem 2.6.5 extends
this result to the product of n sets.
Theorem 2.6.4
The Product Rule (for two sets)
If A and B are finite sets with m and n elements, respectively, then A x B = mn.
The product rule may be applied to counting the total number of ways to perform two independent tasks (jobs or activities). By independent, we mean that the
number of ways the second task can be done is not affected by the result of the first
task. For example, if the first task is rolling a six-sided die and the second task is
tossing a coin, then rolling a 1, 2, 3, 4, 5, or 6 will have no effect on whether the
coin toss results in heads or tails. The set {1, 2, 3, 4,5, 6} x {H, T} has6-2 = 12
elements, so there are 12 possible results of performing the two tasks.
The product rule can be extended to more than two sets.
140
CHAPTER 2
Theorem 2.6.5
Sets and Induction
The Product Rule
—
For all n € N and for every family @ = {A;: i = 1, 2,3,...,n} of nsets, if A; = a;
for 1 <i <n, then
Ai
As x
AS
ex
1
Proof.
See Exercise 12.
&
The product rule is interpreted as follows: If there are n consecutive tasks to
be done, the tasks are independent, and there are a; possible ways to do the ith task,
then there are exactly a, a, ----- a,, ways to complete the sequence of n tasks. This
interpretation is useful in many diverse situations.
Example.
To find the number of three-digit positive integers, we must perform
three tasks: Choose each of the three digits. There are 10 possibilities for the units
digit and 10 for the tens digit, but only 9 possibilities for the hundreds digit because
it can’t be zero. By the product rule, there are 10 - 10 -9 = 900 ways to form a
three-digit positive integer. We can check this result easily: Of the 999 numbers
from | to 999, the first 99 have only one or two digits, so 900 have three digits.
Example. To find the number of three-digit positive integers with no repeated
digits, we might begin by observing that there are 10 possibilities for the units digit
and 9 remaining possibilities for the tens digit. At this point, we see that the task
of choosing the hundreds digit is not independent of the other tasks: The number
of possibilities depends on whether 0 is chosen for either the units or the tens digit.
To use the product rule, we must think of a different sequence of tasks, or perhaps
of carrying out these tasks in a different order. Beginning with the hundreds digit,
there are 9 possibilities (everything but 0), then 9 possibilities for the tens digit,
and 8 for the units digit. Thus, there are 9 -9 - 8 = 648 three-digit positive integers
with no repeated digits.
Example.
To find the number of odd three-digit positive integers with no repeated
digits, one method is to choose the units digit (there are 5 possibilities), then the
hundreds digit (8 possibilities, to avoid O and the chosen units digit), and finally
the tens digit (again, 8 possibilities). Thus, there are 320 odd three-digit positive
integers with no repeated digits.
It is instructive to use another method for this problem. A situation similar
to that of the previous example arises if we begin with the units digit, then the
tens digit, and finally the hundreds digit. To resolve this problem, we consider two
disjoint sets of three-digit integers: those with 0 as the tens digit and those with a
nonzero tens digit. For the first set, there are 5 possible units digits, only one possible tens digit, and 8 possible hundreds digits. There are 40 such integers. For the
second set, there are 5 possible units digits, 8 possible tens digits, and 7 possible
hundreds digits, making 280 integers in this set. By the sum rule, there are a total of
320 odd three-digit positive integers with no repeated digits.
o
2.6
Principles of Counting
141
If the set A has n elements, then forming a subset of A amounts to carrying out n
independent tasks, where each task is to decide whether to place an element in the subset.
Because each task has two outcomes, there are 2” ways this process can be carried out,
so 9P(A) has 2” elements. This argument is a restatement of the proof of Theorem 2.1.4.
DEFINITION
A permutation of a set with n elements is an arrangement
of the elements of the set in a specific order.
Example.
To find all permutations of the set
A=
{a, b, c, d}, we might begin by
writing down all the arrangements of elements of A that begin with the element a.
These are
abcd
abdc
acbd
acdb
~adbc
adcb
The other permutations of A, beginning with b, then c, and then d, are:
bacd
cabd
dabc
badc
cadb
dacb
bcad
chad
dbac
bcda
chda
dbca
bdac
cdab
dcab
bdca
cdba
dcba
There are 24 permutations of the four-element set A.
Recall that n! (n-factorial) is the product n(n —
and that 0! =
1.
Theorem 2.6.6
1)----- 3-2-1 forallneN,
The number of permutations of a set of n elements is n!.
Proof.
See Exercise 12(b).
r
Example. A shuffle of the playlist on a portable music device is simply a permutation of the set of song titles. For a playlist of 10 songs, there are 10! = 3,628,800
possible different playlists.
Example.
Find the number of possible user passwords with seven characters that
consist of digits or letters of the alphabet, without repetition.
Ignoring the case of the letters, we can think of the problem as having to select
seven different symbols without repetition from a set of 36 (10 digits and 26 letters),
and then arranging them in some order. For the first symbol, there are 36 choices;
for the second symbol, there are 35 choices; and so on. The number of arrange36!
ments is 36 - 35 - 34-33 -32-31-30= 791° according to the next theorem.
Theorem 2.6.7
The Permutation Rule
If n is a natural number and r is an integer such that 0 < r < n, then the number of
permutations of any r distinct objects from a set of n objects is
n!
(n — ry!
142
CHAPTER 2 _ Sets and Induction
Proof.
See Exercise 13.
=
Example. A club with nine members plans to elect 4 officers by electing first
a president and then a vice president, secretary, and treasurer. In this situation,
the order is important, so the number of possible lists of elected officers 1s
9!
9!
=13,024:
(C= 14) eee!
If the club was selecting 4 members to bring refreshments to the next meeting,
the order of selection would not be important. In this case, the selection is called a
combination of 9 people taken 4 at a time.
DEFINITIONS _ For a natural number n and an integer r with 0 <r <n,
a combination of n elements taken r at a time is a subset with r elements
from a set with n elements.
The number of combinations of n elements taken r at a time is called the
binomial coefficient ()read “n choose r’ or “n binomial r.
e
e
e
n
37°
i
~
.
”
if
Example.
Choosing three people from a set of five people is the same as forming
;
5
a 3-element subset. There are (3)different possible combinations. If we identify
the five people as R, C, W, H, and P, the 3-element subsets are {R, C, W}, {R, C,
PU
Cea
ttt,
Wot
jC, Wold CWP
|G. Halak,
5
and {W, H, P}. Thus, (3)= il),
For any set with n elements, there is only one way to select a subset of n
elements. Therefore,
n
()el etomallereee Ne
n
Also, there is only one way to construct a subset with zero elements. Therefore,
(")ale hOrealle7aeeNe
In particular, there is only one zero-element subset of @—the empty set itself.
Thus
2.6
Principles of Counting
143
Example.
ThesetA = {a, b, c, d} has four subsets with one element and four subsets with three elements. A has six subsets with two elements. They are {a,b}, {a,c},
{a, d}, {b,c}, {b, d}, and {c, d}. Thus, (*)ae
fe = 6, and (5)= 4)
2
The next theorem develops a simple calculation for binomial coefficients. The
proof is a good example of a technique called two-way counting, in which expressions for a given quantity are determined using two different counting approaches,
thereby creating an equality.
Theorem 2.6.8
The Combination Rule
Let n be a positive integer and r be an integer such that 0 < r < n. Then
(A des
rr)
n!
rnn—
nt
Proof.
(The quantity we count in two different ways is the number of ways to
arrange the n objects in an n-element set.) Let A be a set with n elements. By
Theorem 2.6.6, the number of permutations of all n objects in A is n!.
The n elements of the set A may also be arranged as follows: Select r objects,
order them, and then order the remaining n — r objects. Selecting r objects can be
:
n
done in
:
:
F
;
":)ways; ordering the r objects can be done in r! ways; and ordering
'
the remaining n — r objects can be done in (n — r)! ways. Thus, the number of
:
:
¢
a
permutations of all n objects in A is
ff
rl a(
pare
if
Comparing the two methods for counting the number of permutations of the
n!
elements of A, we have g -r!-(n — r)! =n!. Therefore (")=
r
a
r\(n — r)!
Example.
Inacompany with 15 employees, suppose 5 are selected for bonus pay.
The number of ways the 5 employees could be selected is
(is)
Stee
15!! _
ae 1514-13-12
Ave
Sa l ae
te
te |
11 _ gos,
For this calculation, we assumed that all 5 employees will receive the same bonus
amount, so that there is no need to think of the 5 employees as being selected in any
order. They may be selected simultaneously as a subset of the 15.
If the 5 selected employees are to get different bonus amounts, we need to
arrange these employees in order. There are 5! ways to order 5 employees. Thus, the
number of ways to give five different bonuses is the number of combinations times
the number of permutations within each combination, or 3,003(5!) = 360,360.
144
CHAPTER 2
Sets and Induction
If, in the previous example, we know from the start that there will be five differ-
ent bonus amounts, we use the permutation rule to conclude that the five employees
can be selected, in order, in 15 - 14-13 - 12-11
= 360,360 ways.
Example.
Let A = {2, 3, 6, 18, 38, 81, 442, 469, 574, 608}. In how many ways
can four elements of A be selected so that their sum is (a) less than 400? (b) odd?
(c) even and less than 400?
(a)
For the sum to be less than 400, we can choose any four of the six elements
6
of A that are less than 100. This can be done in (‘)= 15 ways.
(b)
There are three odd numbers in A, so for the sum to be odd we must select
3}
either all three of them or exactly one. There is only 0) = 1 way to
i
choose all three of them and then (7)= 7 ways to choose the fourth
summand from the seven even numbers. By the product rule, there are
1-7 = 7 combinations using all three odd numbers. To form an odd sum
with only one odd summand,
number and
xe
35 ways to choose three even numbers from A. By the
product rule, there are 3 - 35 =
(c)
3
there are (;)= 3 ways to choose the odd
105 combinations involving one odd num-
ber. Thus, there are 7 + 105 = 112 combinations of four elements of A
whose sum is odd.
A contains two odd numbers less than 400; for an even sum, we must use both
2\/4
of them or neither. There are (3)(6) = 1-6 = 6 ways to choose two odd
2\(4
and two even numbers and (O)( )= 1-1 =1
4
way to choose four even
numbers less than 400. Thus, there are seven combinations whose sum is
even and less than 400. As an alternative, we could compute that an odd sum
less than 400 requires one of the two odd elements of A and three of the four
4
2,
even elements. There are (*). ()= 2-4 = 8 such combinations, which
3
leaves 7 of the 15 sums that are even.
Theorem 2.6.9
The Chairperson Identity
Let n be a positive integer and r be an integer such that 0 < r < n. Then
OE)
Proof. The name of this theorem comes from thinking of forming a committee
with r people from a set of n people and then selecting one of the r people to be the
2.6
9
‘
:
:
Principles of Counting
145
n
committee chair. This can be done in ( )ways. On the other hand, we could first
if
select the committee chair and then select n — 1 other people to be on the committee.
—1
This can be done in n(" 1 ways.
&
r
_
The next theorem describes some additional relationships among binomial
coefficients. First, part (a) explains why (")is called a binomial coefficient: The
coefficient of a’ b"~" in the expansion of the binomial (a + b)’ is
)For example,
(a + bp = & + 5a*b + 10a°b? + 10a2b> + Sab* + b>. Thus, ne coefficient of
a°b? is (3)= 10, and the coefficient of a'b* is (*)= 5,
Part (b) tells us that there are as many ways to choose r objects out of a set
with n elements as there are ways to choose n — r objects from the set. This must
be true because choosing r elements to take out is the same as choosing n — r
objects to leave behind. We will consider an interpretation of part (c) at the end of
this section.
Relationships among binomial coefficients can often be proved either
algebraically or combinatorially. Combinatorial proofs, like the proofs of Theorems
2.6.8 and 2.6.9, are based on the meaning of the binomial coefficients. Algebraic
proofs apply the formulas derived in these and other results. In Theorem 2.6.10, we
give a combinatorial proof for part (a) and an algebraic proof for part (b).
Theorem 2.6.10
Let n be a positive integer and r be an integer such thatO
(a)
<r <n.
The Binomial Theorem For a, b € R, (a + b)" = yy ("arr
r=0
(c)
(")= ("
(a)
Since (a+ b)" = (a+ b\(a+b)...(a +b), each term of the expansion
r
r
')+ @
\T
r—1
')for re Ih
of (a + b)" contains one term from each of the n factors (a + b). Thus,
each term of (a + b)” contains a total of n a’s and b’s, and, therefore, each
term includes a factor of the form a’b”~" for some 0 < r < n. For a given
r, the coefficient of a’b"~" is the number of times a’b”"~" is obtained in the
expansion of (a + b)”. Since the term with a’b"~" is obtained by choosing
a from exactly r of the factors (a + b), the coefficient for a"b”~" is
n
at
146
CHAPTER 2
Sets and Induction
b
(")
(b)
PP
(c)
Nine
)
-( n
n!
_
n!
aan)
mM Gere
|
See Exercise 20.
Part (a) of Theorem 2.6.10 provides another way to count the number of subsets
:
ni
of a finite set. Ifa set Ahas n elements, we start with be the number of zero-element
subsets of A, plus (")the number of one-element subsets of A, and so on, up through
nA
(”),the number of n-element subsets of A. The sum, >> (") is the number of
r=0
subsets ofA. Byspart. (a), 27 == (1. 1)
has 2” subsets.
=>) @ Therefore, A
=O
To explain the relationship among coefficients in part (c) of Theorem 2.6.10, we
refer to Pascal’s’ triangle, shown in Figure 2.6.3. The triangle provides a simple means
for computing binomial coefficients. For example, we read on row n = 4 (rows are
labeled on the left) the coefficients for a” b”~' for r increasing from 0 to 4. Thus,
(a+b)
= la"
Se Gis
nO \P
4a°b + 6a*b* = tab”
W,
S
‘
n=0
l
n=3
9)
1
n=4
1
|
3
4
6
|
6
15
.
|
4
.
a
Ss
1
15
ne
7
<a
; y
1
5
10
20
.
ne
3
10
5
1
n=5
‘ 7
1
1
ie 2
n=6
NC
i
n=1
1b*.
6
]
Figure 2.6.3
The triangle is constructed by beginning with the first two rows
Le
and constructing the next row by putting | on the far left and far right. All other
entries in a row are found by adding the two entries immediately to the left and right
in the preceding row. Thus, the first 10 in the fifth row is the sum of 4 and 6 from
* Blaise Pascal (1623-1662) was a French mathematician, physicist, and philosopher. He made profound contributions to projective geometry. He used the triangle, which was known centuries earlier by
Chinese, Indian, and Arabian mathematicians, to advance the study of probability.
2.6
the fourth row. Part (c) of Theorem 2.6.10
Principles of Counting
147
tells precisely how each entry in one row
of the triangle is formed from the two entries in the row above.
Exercises 2.6
Find A, where A is
(a)
(ce)
{weZin?
= 41}.
{reRx?
== 1)
(Dee
(dq)
2.6; 2..6,.2):
{nEN:n+
1 = 4n — 10}
Suppose A = 24, B= 21, AUB
= 37, ANC=11,
C — B = 12. Find
(a) ANB.
+. (Dandies,B.
(ce) Bien
(d) BUC.
(e) C.
(Hae
B— C= 10, and
How many natural numbers less than or equal to | million are either squares
or cubes of natural numbers?
Of the four teams in a softball league, one team has four pitchers, and the
other teams have three each. Determine each of the following, and state the
counting rules that apply.
(a) The number of possible selections of pitchers for an all-star team if
exactly four pitchers are to be chosen.
(b)
(c)
(d)
The number of possible selections if one pitcher is to be chosen from
each team.
The number of possible selections of four pitchers if exactly two of the
five left-handed pitchers in the league must be selected.
The number of possible orders in which the four pitchers, once they are
selected, can appear (one at a time) in the all-star game.
State the Principle of Inclusion and Exclusion for four sets, A, B, C, and D.
Of the 32 teams that qualified for the 2010 World Cup in South Africa, 24 were
among the top teams in pretournament rankings. There were 13 European
teams, and of the countries represented there, 28 had flags with colors other
than red or white. Eleven top-ranked teams were from Europe, and 22 topranked teams had flags with colors other than red or white. There were 10
European countries whose flag had a color other than red or white, and 9 of
those were among the top-ranked teams. How many non-European countries
with a red and white flag were not among the top-ranked teams?
(a)
Ifsomeone has 10 left shoes and 9 right shoes and does not care whether
(b)
they match, how many “pairs” of shoes can this person select?
A standard deck of cards consists of 52 cards with four suits (spades,
hearts, clubs, and diamonds), and each suit has | card of each of 13
(c)
ranks (2, 3, 4,..., 10, J, Q, K, A). How many ways are there to select 4
cards, with one card of each suit?
How many ways are there to select 13 cards, with 1 card of each rank?
148
CHAPTER 2 _ Sets and Induction
Calculate the number of even three-digit positive integers with no repeated
digits by finding the number of such integers that have (a) units digit 0 and
(b) nonzero units digit. Verify your answer by comparing the number of odd
three-digit positive integers with no repeated digits with the total number of
three-digit positive integers with no repeated digits.
(a)
(b)
Find the number of four-digit positive integers with no repeated digits.
Find the number of odd four-digit positive integers with no repeated
digits.
(c)
Without using your results from parts (a) and (b), find the number of
even four-digit positive integers with no repeated digits.
10.
A square is bisected vertically and horizontally into 4 smaller squares, and
each of the 4 smaller squares is to be painted so that adjacent squares have
different colors. If there are 20 different paint colors available, in how many
ways can the 4 smaller squares be painted?
11.
Prove
that
MOBS
12.
if A and B are disjoint and
C=
BC
ANGE Bin €:
C is any
other
set,
then
(a)
Use Theorem 2.6.4 to prove Theorem 2.6.5 by induction on the number
(b)
of sets.
Prove Theorem 2.6.6 by induction on the number of elements.
13.
Prove Theorem 2.6.7
(a) by using the product rule.
(b) by induction on n.
14.
Find the number of passwords that use each of the digits 3, 4, 5, 6, 7, 8, and 9
exactly once.
15:
In how many of the passwords of Exercise 14
(a)
(b)
(c)
(d)
16.
are the first three digits even?
are the three even digits consecutive?
are the four odd digits consecutive?
are no two odd digits consecutive?
The number of four-digit numbers that can be formed using exactly the digits
1, 3, 3, and 7 is less than 4! because the two 3’s are indistinguishable. Prove
that the number of permutations of n objects, m of which are alike, is n!/m!.
17-19.
17:
*
A hand is a set of five cards chosen from a standard deck of cards (see
Exercise 7(b)).
Determine the number of
(a)
(b)
(c)
(d)
(e)
possible hands.
hands in which exactly one card is a heart.
hands in which exactly one card is a heart and exactly one card is a diamond.
hands in which exactly one card is a heart and at least one card is a diamond.
hands in which exactly one card is a 7.
2.6
Principles of Counting
149
18.
Determine the number of hands in which
(a) two cards are in the same suit and one card is in each of the other suits.
(b) two cards are hearts and one card is in each of the other suits.
(c) three cards are hearts and two cards are diamonds.
19.
Determine the number of hands in which
(a) all five cards are in the same suit (a flush).
(b) two cards have the same rank and the other cards have different ranks
(a pair).
(c) two cards have the same rank, two other cards have the same rank differ-
ent from the first pair, and the remaining card has another different rank
(two pairs).
(d)
(e)
three cards have the same rank and the other two cards have different
ranks (three of a kind).
all cards are in consecutive ranks (a straight). The A is considered the
lowest rank of 5, 4, 3, 2, A and the highest rank of A, K, Q, J, 10. Include
in your count hands that are also flushes.
20.
PAN
22.
Prove these parts of Theorem 2.6.10 as follows:
(a)
(b)
Prove part (b) using a combinatorial argument.
Prove part (c) using a combinatorial argument.
(c)
Prove part (c) algebraically.
(d)
Use part (c) to prove part (a) of Theorem 2.6.10 by induction.
Find
(a)
(b)
(a+ b)°.
(a+ 2b)*.
(c)
(d)
the coefficient of a°b!°3p10 in the expansion of (a + b)!.
the coefficient of a7b!° in the expansion of (a + 2b)".
(a)
Give acombinatorial proof that if n is an odd integer, then the number of
ways to select an even number of objects from a set of n objects is equal
to the number of ways to select an odd number of objects.
Give a combinatorial proof of Vandermonde’s identity: For positive
(b)
integers m and n and an integer r such thatO
<r<n+™,
CE eG
Ge Ea HONGea,
ONG)
(c)
Prove that
Parl
n
23.
2n
ALi
ene
ise
ne |
n
,
:
Give a combinatorial argument that n? = of
i)+ n. Hint: Arrange dots in
an n by n square array, and count the dots in two ways.
150
CHAPTER 2
Sets and Induction
24.
The nth pyramid number, p,, is the
number of balls of equal diameter that
can be stacked in a pyramid whose base
is ann by n square. The first few pyramid
numbers are p, = 1, p, =5, p3 = 14, and
ps = 30. A pyramid of ps = 55 balls is in
the figure at the right. Show that
siponainmeat
(a)
sso
(b)
p,=("*
2
el)
SLi
,
for every natural number n.
)torn = 2
) for all natural numbers n.
(d) pan("?
25.
|
Tel
Oia =S
Proofs to Grade
pne
1
l
hey
)form 22
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
a}
(a)
Claim.
“Proof.”
For alln > 0,
2)
n- -- 71
ni —n
——
an
Consider a set of n + 1 elements, and let one of these elen+
ments be x. There are (
n—1
‘)rs nz+tn
ways tochoosen
n
elements. Of these, there are (
; =n
n—
—
| of these
ways to choose the n —
:
n
n—n
elements without choosing x and ( y)=,
1
waysito choose
We
2)
n — | elements including x. Therefore,
D
n~-+n
n~—n
=n
-+
a2
(b)
Claim.
Forn > 1,
Dipl
Pig)ee Ong
“Proof.”
0=(-1+1"=)>) (")ede”
k=0
k
=()-Q+
+e)
G)
+-+eo(
--)) «
2.6
(c)
Claim.
Principles of Counting
151
For n > 1, the number of ways to select an even number of
objects from n is equal to the number of ways to select an odd number.
“Proof.”
From part (b) of this exercise (The claim made there is
(+ G)+~=()+G) +>
The left side of this equality gives the number of ways to select an even
number of objects from n, and the right side is the number of ways to
select an odd number.
a
7
wh
éelef?
veer)
Lamy jeldy
fi
*hresi")”
wud)
ist
CHAP
TER
3
Relations and Partitions
We often think of objects, whether they are people, numbers, or something else,
as related to other objects. For example, among the set of all people, we could say
two people are related if they have the same blood type. In mathematics, we could
say two integers a and b are related if a is a divisor of b. In this chapter, we study
the idea of “is related to” by making precise the notion of a relation. We then focus
on certain relations called equivalence relations and look at an alternate way to
describe them. In the last two sections, we describe a form of arithmetic that oper-
ates on sets of related numbers and examine relations that organize sets of objects
by putting them in order.
3.1
Relations
When we speak of a relation on a set, we identify the notion of “a is related to b”
with the ordered pair (a, b). Thus, a relation may be defined simply as a set of
ordered pairs.
DEFINITIONS
Let A and B be sets. R is a relation from A to B if Risa
subset of A x B. A relation from A to A 1s called a relation on A.
If (a, b) € R, we say a is R-related (or simply related) to b and write
aR b. If (a, b) € R, we write a R b.
Let R be the relation “is older than” on the set of all people. If Rick’s age is
32 and IIsa’s is 25, then (Rick, Ilsa) € R and Rick R Ilsa. Because (Ilsa, Rick) ¢ R,
we write Ilsa & Rick. If Victor’s age is 45, then Rick
Victor, Victor R Ilsa, and
Victor R Rick.
153
154
CHAPTER 3
Relations and Partitions
Let A = {—1, 2,3, 4} and B= {1, 2, 4, 5, 6}. The set of ordered pairs
Example.
R=
ear
a), Q; 4), U2, 1): (4, 2)}
R5,2R4,2R 1,
is a relation from A to B. We can also describe R by writing —1
relationship
the
indicating
by
format;
and 4 R 2; by listing the pairs in a two-column
in an arrow diagram; or by drawing a graph. See Figure 3.1.1.
Ae
(a) Table for R
(b) Arrow diagram for R
+
$4
=|
1234
(c) Graph of R
Figure 3.1.1
An equation, inequality, expression, or graph is often used to describe a relation, especially when listing all pairs is impractical or impossible. For example, the
“Jess than” relation on R is the set {(x, y) € R x R: x < y}. This example explains
why it is natural to use the notation x R y when (x, y) is in a relation R; it is the same
as writing x < y when x is less than y.
Consider the relation S on the set N x N given by (m,n) S(k,j) if
m+n=k
+ j. Then (3, 17) S (12, 8), but (5, 4) is not S-related to (6, 15). Notice
that S is a relation from N x N to N x N and consists of ordered pairs whose
entries are themselves ordered pairs. Here the notation (3, 17) S (12, 8) is preferred
over the cumbersome expression ((3, 17), (12, 8)) € S.
The empty set @ and the set A x B are relations from A to B. In fact, every
subset of A x B is a relation from A to B. In Exercise 14, you are asked to prove
that if A has m elements and B has n elements, then there are 2’”” different relations
from A to B.
For any set A, a simple but important relation on A that will arise again and
again in mathematics is the identity relation.
DEFINITION
For any set A, the identity relation on A is the set
I, = {(a,a):a € A}.
For A = {1, 2, a, b}, the identity relation on A is J, = {(1, 1), (2, 2), (a, a),
(b, b)}. The graph of the identity relation on [—2, oo) is given in Figure 3.1.2.
3.1
Relations
155
Figure 3.1.2
DEFINITIONS
The domain of the relation R from A to B is the set
Dom (R) =
{x € A: there exists y € B such that x R y}.
The range of the relation R is the set
Rng (R) =
{y € B: there exists x € A such that x R y}.
Thus, the domain of R is the set of all first coordinates of ordered pairs in R, and
the range of R is the set of all second coordinates. By definition, Dom (R) C A and
Rng (R) C B.
For the relation R = {(—1, 5), @, 4), @, 1), 4, 2)}, Dom (RX) = {—1, 2,4}
and Rng (R) =
{1,2,4,5}.
For the “less than” relation on R, the domain
and
range are the set IR. For every set A, both the domain and the range of J, are equal
to A because (a, a) € I, for every a EA.
Every set of ordered pairs is a relation. If M is any set of ordered pairs, then M
is a relation from A to B, where A and B are any sets for which Dom
(M) CA
and Rng (M) CB.
For a relation R from set A to set B, to prove that a given x € A is contained
in Dom (R), we must show that there exists an element y in B such that x R y. To
show that y € B is in Rng (R), we must prove that there exists x € A such that x R y.
Example.
Let T= {(x, y)
€R x R: x* + 4y’ < 16}. The graph of Tis the shaded
area in Figure 3.1.3. Prove that
(i)
Dom (T) =[-—4, 4] and
Gi)
Rie C2]:
Proof.
(i)
Suppose that x € Dom (7). Then for some y, we have x T y, so Xe e4y
16:
Then x* = 16, sox € [—4, 4].
Now suppose that x € [—4, 4]. Then x? < 16. (We must find a real
number y such that x T y. We choose y = 0.) Then x? + 0? < 16, so x T 0.
Therefore, x € Dom (T).
156
CHAPTER 3
Relations and Partitions
(ii)
Suppose that y € Rng (7). Then for some x, we have x T’y, so x + 4y* = 16.
Then y’ < 4, so ye [—2, 2].
Now suppose that y € [—2, 2]. Then y? < 4. (In this part, we must find
a real number x such that x T y. We choose x = 0.) Then 0? + 4y* < 16, so
fad
O07 y. Theretore ye Rng (7).
Rng (T) = [-2, 2]
, Dom (T) = [-4, 4]
;
Figure 3.1.3
We can use a directed graph or digraph to represent a relation R on a small
finite setA.We think of the objects in A as being points (called vertices) and the relation R as telling us which vertices are connected by ares. Arcs are drawn as arrows:
There is an arc from vertex a to vertex bif (a, b) € R. Anarc from a vertex to itself is
called a loop. For example, let A = {2,5, 6,12} and R = {(6, 12), (2, 6), (2, 12),
(6, 6), (12, 2)}. The digraph for R is given in Figure 3.1.4.
e5
5
12
Figure 3.1.4
The remainder of this section is devoted to methods of constructing new relations from given relations. These ideas are important in the study of relations and
will be used again when we study functions.
Because relations from set A to set B are subsets of A x B, the union and inter-
section of two relations from A to B are again relations from A to B.
Example. Let X = [2, 4] and Y = (1, 3) U {4}. Let S be the relation on R defined
by x S yifx © X, and let Tbe the relation on R defined by x Ty if y € Y. The graphs
of S and T are given in Figures 3.1.5 (a) and (b). Figures 3.1.5 (c) and (d) show
the graphs of SM T and S U T, respectively. Note that S = X x R, T=R
x Y, and
SOW = Xoay,
o
3.1.
;
aaa
a
oe
Ho
Ste
ee
x
=i
=e
Pape
ey
Relations
ie)
x
Ai
—_-.
le
_
oy ee
ails
(b) T
157
we
—
23455
—
1
CSO
3
|
E
>X
(d) SUT
Figure 3.1.5
DEFINITION
relation
If R is a relation from A to B, then the inverse of R is the
R™' = {(,x): @, y) © R}.
Because inversion is a matter of switching the order of each pair in a relation, if
R is arelation from A to B, then R™ is a relation from B to A.
Examples.
R— =
The inverse of the relation R = {(1, b), (1, c), (2, c)} is the relation
{(b, 1), (ec, 1), (c, 2)}. For any set
A, the inverse of J, is J, itself. The inverse
of the “less than” relation on R is the “greater than” relation on R.
Oo
When R is a relation on a set A, the digraph of R~! is obtained from the digraph
of R by reversing the direction of each arc. The digraphs of the relation “divides” on
the set {3, 6, 9, 12} and its inverse relation, “is divisible by,” are given in Figure 3.1.6.
3
6
3
6
9
12
9
12
(a) divides
(b) is divisible by
Figure 3.1.6
Example. Let T= {(x, y) € R x R: y < 4x? — 7}. To find the inverse of T, write
T—! as aset of pairs (x, y) ER x R.
Gay) Gta att
(ax) Gl.
iffx< 4y?—7
iffx+7 <4y?
iff 4y? > x+7
iffy> watt ory < Sgt
Thus T7! = {(@7, y)E Rx Riy> Vett ory < ~My,
158
CHAPTER 3
Relations and Partitions
Example.
Let EXP be the relation on R given by x EXP y if y = e*. The inverse
of EXP is given by x EXP™! y iff x = e. We know thatx = e? iff y = In x, where
In is the natural logarithm. Thus, the inverse of EXP is the relation In. The familiar
graphs of EXP and In are given in Figure 3.1.7.
Ore)
—2
|
|
-l
ie ee ters)
(a) x EXP y: y = e*
(b) In = EXP™!:
y=In x
Figure 3.1.7
In the previous example, Dom (EXP) = R and Rng (EXP) = (0, 00), while
Dom (In) = (0, co) and Rng (In) = R. The next theorem says that this switch of
the domain and range of a relation to the range and domain of the inverse relation
always happens.
Theorem 3.1.1
Let R be a relation from A to B.
(a)
(b)
Dom (R7!) = Rng (R).
Rng (R~!) = Dom (R).
Proof.
(a)
be Dom(R7~') iff there exists a € A such that (b, a) € R7! iff there exists
a €A
(b)
such that (a, b) € R iff b € Rng (R).
See Exercise 1 1(a).
|
Given a relation from A to B and another from B to C, composition is a method
of constructing a relation from A to C.
DEFINITION
Let R be a relation from A to B, and let S be a relation
from B to C. The composite of R and S is
SoR
=
{(a,c): there exists b € B such that (a, b) € R and (b, c) € S}.
3.1
Relations
159
The relation S © Ris arelation from A to C because
S° RCA x C. Itis always
true that Dom (So R) C Dom (R), but it is not always true that Dom (S° R) =
Dom (R). See Exercise 11, parts (b) and (c).
Our definition of composition uses the same right-to-left notation that is commonly used in calculus and other analysis courses. To determine S$ © R, you need to
remember that R is the relation from the first set to the second and S is the relation
from the second set to the third. Thus, to determine S © R, we apply the relation R
first and then S.
EXaniple
salet Ag
Nita 3 45
8 alg)
vast) sand CO =i in ve cane let
R be the relation from A to B:
R=
{1 p), d,@)2rq). G7), G4is)}
and § the relation from B to C:
n=
(Des
Ge x) (Gay
Usa)
fe
Figure 3.1.8
These relations are illustrated in Figure 3.1.8 by arrows from one set to another.
An element a from A is related to an element c from C under $ © R if there is
at least one “intermediate” element b of B such that (a, b) € R and (b,c) € S.
For example, because (1, p) € R and (p, x) € S, then (1, x) € SO R. By following
all possible paths along the arrows from A to B and B to C in Figure 3.1.8, we have
SOR = {(1, x), (1,y), 2, x), 2, y), 4, 2}.
If R is a relation from A to B and S is a relation from B to A, then R° S and
So R are both defined, but you should not expect that
Ro S = So R. Even when
R and S are relations on the same set, it may happen that
ROS# SOR.
Example.
LettR={@, yeERxR:y=x4
lj andS={(%,
yERx
Then
ROS = {(x, y): (x, z) € S and (z, y) € R for some z € R}
= {(x,y):z = x?andy =z + 1|forsome
ze R}
Sci
Se
ees
Riy= x}.
160
CHAPTER 3.
Relations and Partitions
SOR = {(x, y): (x, z) € Rand (z, y) € S for some z € R}
= {(x,y):z =x+4+ landy = 2 for some ze R}
ha
Clearly
So
R#R0©S
Sea
because x” + | is not equal to (x + ies
The last theorem of this section presents several results about inversion, composition, and the identity relation. We prove part (b) and the first part of (c), leaving
the rest as Exercise 12.
Theorem 3.1.2
Suppose that A, B, C, and D are sets. Let R be a relation from A to B, S be a relation
from B to C, and T be a relation from C to D.
Oy
ee
(b)
(c)
(dye
T°(SOR) =(T°S)OR, so composition is associative.
Jg°0R=RandROl,=R.
(SOR) Rao S-
ewe
Proof.
(b)
The pair (x, w) € T° (S° R) for some x € A and w € D
iff (dz € C)[G, z) € S° Rand (z, w) € T]
iff (dz € C)[(Ay € B)((X, y) € R and (y, z) € S) and (z, w) € T]
iff (dz € C)(Ay € B)[(, y) € R and (y, z) € S and (z, w) € T]
iff (dy € B)\Gz € C)[G, y) € R and (y, z) € S and (z, w) € T]
iff (dy € B)[Q, y) € R and (Sz € C)((y, z) € S and (z, w) € T)]
iff (Sy € B)[(@, y) € Rand (y, w) € TO S]
iff (x, w) E (TOS)OR.
Wherctored
1S OR) = (7-0 S) OR,
(c)
(We first show that Iz°
RC R.) Suppose that (x, y) € Jp ° R. Then there
exists z € B such that (x, z) € R and (z, y) € Ip. Because (z, y) € Ip, Z = y.
Thus, (x, y) € R (because (x, y) = (x, z) € R).
Conversely suppose that (p, g) € R. Then (q, q) € Jp, and thus (p, q) € Ip © R.
Thus J,°R = R.
=
1.
The phone faceplate pictured at the right may be used to
Exercises 3.1
define a relation from the set
A = {0, 1, 2,..., 9, *, #}
of digits and symbols to the set [ = {A, B, C,..., Z}
of 26 letters. The relation R from A to I is defined by
“appears on the same phone button.”
(a) Which of these are true?
@) 3RJ
(ii) (8,S)ER
(iii) ORN
(iv) JR5
3.1
(v)
(b)
4RG
(vi)
Relations
161
9RZ
(vii) MR-!6
(viii) TR
Find the domain and range of R.
12
Feu ipber the relation {(G) 1), (2, 3)-G, 5), (252), (1,6), ©, 6), (1, 2)}. Find
(a)
Dom (T).
(Cc)
(b)
Rng (T).
(ds
(re
Find the domain and range for the relation W on R given by x W y if
(a)
y=2x+1.
(Cea
el
(b)
y=x?
+ 3)
(d)
yp:
1
(e)
y <x.
(f)
|x| <2 andy =3.
(2)
ied] <2 208 yu,
(h)
yx.
LetR={@jy)
eR x Roy =2x 4+ 1j,S={@y)
eR x Riy=V4
— 2x},
T= {(x,y)
ER x R:y—2=
x7}, and W= {(x,
y) € R x R: y < 5}. Prove that
(a)
(c)
(e)
(g)
RcDom(R).
1 ¢Rng (7).
4¢€ Rng (R).
4e¢Rng(W).
(b)
(d)
(f)
(h)
3 ¢ Dom (S).
RcDom(T).
Oc Rng (S).
RcDom(W).
Prove that the sets given are the domain and range for each relation.
(a) R={@,y) eR x R: y = 6x}, domain = R, range = R.
(b)
R={(x, y) ER x R: y => x’}, domain = R, range = [0, 00).
(c)
(d)
R={(,y) ER x R: 9x? + 4y’ < 36}, domain = (—2, 2), range =(—3, 3):
R={(@, y) eR x R: x=|y]}, domain = (0, oo), range = R.
Find the inverse of each relation. Express the inverse as the set of all pairs (x, y)
subject to some condition. In parts (1), (j), (kK), and (1), P is the set of all people.
(a)
R= {Q@yERx
Ry =x}
(b)
(Cc)
(d)
(e)
()
(g)
R, = {@ y) ER x R:y = —5x 4+ 2}
Re = {0 y) eR > Riy = 7x — 10}
R,= {@, y)ER x R:y=x* + 2}
Rs = {@,y)ER
x Riy = —4x° + 5}
R={@
yeRx Ry <x+ 1}
R,={@y
ER x Ry > 3x — 4}
(h) R= {oe Rx Ry = x au— 2.}
(i) Ryo={(@, y) € P x P: yis a younger sister of x}
Clio = La) SP Se Psy is theiather or 4}
(k) Ry, = {@
y) eP x P: yisasibling
of x}
(tee — 1s) Se Pera
lovesx}
Let R = {(1, 5), (2, 2), G, 4), 6, 2}. 8 = ((2, 4), 3, 4), 3, D, (5, 5)}, and
f=
(a)
f(1y4); Gy), G4) Dts Find
x (b)
ROS.
(d)
(g)
ROR.
KensieD):
ROT.
(e) SOR.
(h) (ROS)OT.
()
Tos.
erat
162
CHAPTER 3.
Relations and Partitions
Find these composites for the relations defined in Exercise 6.
+
(a)
(d)
R,OR,
ROR;
(b) R, OR,
(e) R,OR,
(c) R, oR,
(f) ROR,
(g)
R,oORs
(h)
Roo R,
(i)
ReoR,
(j)
(m)
Reo Re
R,° Rg
(k)
(n)
R,°R;
R,°R,
(i)
(0)
R5°Rs
Rg° R;
(p)
Ro° Rio
*
(q) Roo Ry
(r) Ryo ° Rio
Give the digraphs for these relations on the set {1, 2, 3}.
(a) =
(b) S = {(1, 3), 2, 1}
(Cli
(dom Wwhclen= 1153). 2L))
(Cre
(Teese Sewheres= 1, 3), (2, 1))
10.
Let A = {a, b,c, d}. Give an example of relations R, S, and T on A such that
(a)
(Oy
(d)
11.
ROSFSOR:
(b) (SOR) 'A#S oR,
sok = Pek, bus #7.
Rand S are nonempty, and R © S and S © R are empty.
Let R be a relation from A to B and S be a relation from B to C.
(a)
Prove that Rng (R~') = Dom (R).
(b)
(c)
(d)
Prove that Dom (S$ ° R) C Dom (R).
Show by example that Dom (S ° R) = Dom (R) may be false.
Prove that one of these statements is true, and give an example to show
that the other is false.
Rng(S) CRng(SoR)
12.
Rng(S°o R) C Rng(S)
Complete the proof of Theorem 3.1.2 by proving that if R is a relation from A
to B and Sis a relation from B to C, then
(yee (Ra) ses
(bj, Kel, = R.
(c)
Ger)
'=R
oS.
13.
Let A, B, and C be nonempty sets. Prove or give a counterexample: If T is a
relation from A to C, then there exist relations R from A to B and S from B to C
such that f= SoR.
14.
Prove that if A has m elements
different relations from A to B.
15:
(a)
and B has n elements,
then there are 2!”
Let R be arelation from 4 to B. For a € A, define the vertical section of
R ata to be V, = {y € B: (a, y) € R}. Prove that (J V, = Rng (®).
acA
(b)
Let R be arelation from A to B. For b € B, define fie horizontal section
ofR at b to be H, = {x € A: (x, b) © R}. Prove that J H, = Dom (R).
beB
16.
We may define ordered triples in terms of ordered pairs by saying that
(a, b, c) = ((a, b), c). Use this definition to prove that (a, b, c) = (x, ye) at
and only iia =x.and
Proofs to Grade
Lie
b = yand¢ = z,
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
3.2.
(a)
Claim.
Let R and S be relations from A to B and from B to C, respec-
tively. Then
“Proof.”
(b)
163
Equivalence Relations
SoR = (RO S)},
The pair (x, y) € S © R if and only if (y, x)
EROS if and only
if (x, y) €(R° S)~!. Therefore, So R = (RO S)7!.
Claim.
Let R be a relation from A to B. Then J, C R710 R,
“Proof.”
a
Suppose that (x, x) € J,. Choose any y € B such that (x, y) ER.
Then (y, x) € R7!. Thus, (x, x) © R7! © R. Therefore, J, C R7'O R.
(c)
Thus, (x, y) = (x, x) and x € A, so @, y) € ,.
(d) Claim.
VSO,
a
IfR={(x,y)ER x R:(x—2)?+ (vy — 2) <4}, then Dom (R) C
[0, 4].
*“Proof.”’
Suppose that x € Dom (R). Then for some y, we have x R
i= 12)4-e OO)
4 ew
vhend(a = 2) <34,.50,—
2,=
—)
Thus, 0 < x < 4, sox € [0, 4].
(e)
|
Claim.
Suppose that R is a relation from A to B. Then R-' 0 RC ite
“Proof.” Let (x,y) €R-!OR. Then for some zEB, (x,z)ER and
(z, y) © R7!. Thus, (y, z) € R. Because (x, z) € R and (y, z) ER, x — y.
Claim.
Dom (k).
“Proof.”
IfR=({(@,y)
=e
|_|
eR x R: @—2)?+ (Wy — 2) < 4}, then [0, Alc
Suppose thatx € [0, 4]. Then 0 < x < 4, so —2 < (x — 2) —e
Thus, .(¢—.2)*.< 4° so. @ = 2)4-F (2. = 2)" = 45 Therefore
he
ace
x € Dom (R).
(f)
Claim.
If S = {(x, y)& N x N: y < 2x}, then every natural number n
is in Rng (S).
“Proof.”
(g)
Letn be a natural number. Then n < 2n, son Sn. Therefore,
|
n & Rng (S).
Claim.
IfS= {(x, y) eR x R: y < 2x}, then every real number z is in
Rng (S).
“Proof.”
Let z be a real number. Then z < 2z, so z S z. Therefore,
z & Rng (S).
3.2
Equivalence Relations
The goal of this section is to describe a way to equate objects in a set according to
some value, property, or meaning. The three properties we define next, when taken
together, comprise what we mean by objects being equivalent.
—~
DEFINITIONS
LetA be a set and R be a relation on A.
R is reflexive on A if for all x € A, x Rx.
R is symmetric if for all x and y € A, if x Ry, then y Rx.
R is transitive if for all x, y, and z € A, if x R y and y R z, then x R z.
164
CHAPTER 3
Relations and Partitions
The relation R, defined as “had the same final exam score,” on the set C of all
students in a given class has all three of these properties. R is symmetric because
if student x had the same score as student y, then student y must have had the same
score as student x. R is transitive because if student x had the same score as student y and
student y had the same score as student z, then x had the same score as Z. Finally, for
every student x in C, x must have had the same score as x. Thus R is reflexive on C.
To prove that a relation R is symmetric or transitive, we usually give a direct proof
because these properties are defined by conditional sentences. A proof that R is reflexive on A is different. What we must do is show that for all x € A, x is R-related to x.
Example.
Prove that the relation S = {(x, y) € R x R: x* = y’} is reflexive on R,
symmetric, and transitive.
Proof.
(i)
Let x be a real number. Then x? = x*. Therefore x S x. Thus, S is reflexive
(ii)
Assume that x and y are real numbers and x S y. Then x* = y*. Therefore,
y* = x’. Thus, y S x. We conclude that S is symmetric.
on R.
(iii) |Assume that x, y, and z are real numbers and that both x S y and y S z. Then
x = y- and y* =z. Therefore x* = z*. Thus, x S z. This proves that S is
transitive.
|
To prove that a relation R on a set A is not reflexive on A, we must show that
there exists some x € A such that x R x. Because the denial of “If x R y, then y R x”
is “x Ry and not y Rx,” a relation R is not symmetric if there are elements x and y
in A such that x R y and y R x. Likewise, R is not transitive if there exist elements x,
y, and
zin A such that x R y and y R z but x R z.
EXAMpleSsmr ome
«(2 )2017). let.S = {(2, 5)... 6), (2.6), 7, 7)} and f=
{(2, 6), (5, 6)}. Because 6 $6 and 2 72, neither S nor T is reflexive on B. The
relation S is not symmetric because 2 S 5 but 5 § 2. Likewise, T is not symmetric
because 5 T 6 but 6 75.
Both S and T are transitive relations. To verify that S$ is transitive, we check
all pairs (x, y) in S with all pairs of the form (y, z). We have (2, 5) and (5, 6) in S,
so we must have (2, 6); we have (7, 7) and (7, 7) in S, so we must have (7, 7). The
relation T is transitive for a different reason: There do not exist x, y, and z in B
such that x Ty and y T z. Because its antecedent is false, the conditional sentence
“If x Tey andy Tz, then
Theorem 3.2.1
Tz Ais true:
Let A be a set. For the power set P(A), the relation “is a subset of” is reflexive on
(A) and transitive but not symmetric.
Proof.
(i)
For every set B € P(A), we have B C B. Therefore, the relation is reflexive
on P(A).
3.2
Equivalence Relations
165
(ii)
By Theorem 2.1.1(c), the subset relation is transitive.
(iii)
If A has at least two elements x and y, then the subset relation is not symmetric
on P(A) because {x} is a subset of {x, y}, but {x, y} is not a subset of {x}.
m
The properties of reflexivity ona set, symmetry, and transitivity can be described
using the identity relation, and the operations of inversion and composition. See
Exercise 4. The properties can also be characterized by traits possessed by the
digraph. For a relation R ona set A:
¢
¢
Ris reflexive on A if and only if every vertex of the digraph has a loop.
Ris symmetric if and only if between any two vertices there are either no ares or
an arc in each direction.
* R is transitive if and only if whenever there is an arc from vertex x to y and an
arc from vertex y to z, there is an arc (a direct route) from x to z.
Examples.
Let A = {2, 3, 6}. The relation “divides” is reflexive on A and transitive but not symmetric. These properties are evident in the digraph shown in
Figure 3.2.1(a). There is a loop at each vertex, exhibiting reflexivity. There is an
arc from 2 to 6 but not in the reverse direction, so the digraph does not exhibit symmetry. Note that every pair of arcs to be checked for transitivity involves a loop.
e
3
2
3
2
6
(a) divides
6
6
(b) >
(@) i
Figure 3.2.1
Figure 3.2.1(b) is the digraph of the relation > on A. It has no loops (x > x is
never true) and has no pair of arcs in opposite directions (is not symmetric). The
relation is transitive.
The digraph of another relation 7 on A is given in Figure 3.2.1(c). We see that
T is not reflexive because the vertices 2 and 3 have no loops. For every arc, there
is another arc in the opposite direction (the loop at 6 is its own opposite), so T is
symmetric. The relation T
is not transitive because there is an edge from 2 to 6 and
another from 6 to 3 but no edge from 2 to 3.
Equivalence relations, which we define next, are a means for relating objects
according to whether they are, if not identical, at least alike in the sense that they
share some common property. For example, if J is the set of all triangles, we might
say two triangles are related when they are congruent. This relation is reflexive on
J (every triangle is congruent to itself), symmetric (if triangle X is congruent to Y,
then Y is congruent to X), and transitive (if triangle X is congruent to Y and Y is
166
CHAPTER 3.
Relations and Partitions
congruent to Z, then X is congruent to Z). These three properties are all we need to
define an equivalence relation.
DEFINITION | A relation R on a set A is an equivalence relation on A if
R is reflexive on A, symmetric, and transitive.
If R is an equivalence relation on a set A and a, b, and c are distinct elements
of A with a R b and b Rc, then seven other pairs must also be related. By transitivity, aR c. By symmetry, b Ra, c Rb, and c Ra. And by reflexivity on A, we have
aRa,bRb,andc Re.
For every set A, the identity relation /, is an equivalence relation on A because
it is reflexive on A, symmetric, and transitive. Equivalence in this case means “is
exactly the same as.”
Example. Prove that the relation R = {(x, y) € Z x Z:x + y is even} is an equivalence relation.
Proof.
(i)
Letx eZ. Thenx + x = 2x is even, so x R x. Therefore, R is reflexive on Z.
(ii)
Suppose that x R y. Then x + y is even, so y + x is even, and, thus, y R x.
Therefore R is symmetric.
Suppose that x R y and y R z. Then x + y is even, and y + z is even. Thus,
(iii)
x + 2y + zis even, and because 2y is even, x + zis even. Thus x R z. There-
fore R is transitive.
a
In the example above, what is the common property shared by related integers?
All odd integers are related to one another because the sum of two odd integers is
even, and all even integers are related to each other for the same reason. No odd
integer is related to any even integer. Thus, integers are related by R if and only if
they have the same parity.
We always make an equivalence relation when we think of objects as being
related by having the same property. For example, on the set P of all people, let L
be the relation on P given by x L y if x and y have the same family name. We have
Lucy Brown L Charlie Brown, James Madison L Dolley Madison, and so on. If
we make the assumption that everyone has exactly one family name, then L is an
equivalence relation on P.
An equivalence relation on a set naturally divides the set into subsets of related
elements. For example, the subset of P consisting of all people who are L-related
to Charlie Brown is the set of all people whose family name is Brown. This set
contains Charlie Brown, of course, but it also contains Sally Brown, James Brown,
Buster Brown, Leroy Brown, and all other people who are like Charlie Brown
in the sense that they have Brown as a family name. The same is true for the
Madisons: The set of people L-related to Dolley Madison is the set of all people
with the family name Madison.
3.2.
Equivalence Relations
167
—
DEFINITIONS
Let R be an equivalence relation on a set A. For x € A, the
equivalence class of x modulo R (or simply x mod R) is the set
¥—{yeA:cky}.
Each element of x is called a representative of this class. The set
A/R= {x:x EA}
of all equivalence classes is called A modulo R.
Other common notations for the equivalence class of x include [x] and x/R. The
latter is especially useful when it is important to indicate the relation.
Charlie Brown, the equivalence class of Charlie Brown modulo L, is the set of
all people whose family name is Brown. Furthermore, Buster Brown is the same
set as Charlie Brown. Lucy Brown, Leroy Brown, and every other person named
Brown are also representatives of this class.
Example.
The relation H = {(1, 1), (2, Dae, SUE) Ox 1)} is an equivalence relation on the set A = {1,253} ,Here l= 2-= {4,2} andi3 33 eis
AIH = {1,3} = {{1, 2}, {3}}.
Example.
Oo
We proved above that S = {(x, y) € R x R: x= y’} is reflexive on R,
symmetric, and transitive. For this equivalence relation, we have Dia yp IN,
7 = {x, —7}, etc. Also, 0 = {0}. For every x € R, the equivalence class of x
and the equivalence class of —x are the same. The set of equivalence classes is
R/S
= {{x, —x}:
x € R}.
o
Note that in both examples above any two equivalence classes are either equal
or disjoint. The next theorem tells that us this is always true and, furthermore, that
equivalence classes are never empty; objects are related if and only if their equivalence classes are identical; and objects are unrelated if and only if their equivalence
classes are disjoint.
Theorem 3.2.2
Let R be an equivalence relation on a nonempty set A. For all x and y in A,
(a)
xeExandxcaA.
(b)
xR yif and only if x= y.
(c)
xRyifand only ifxn y=.
Proof.
(a)
(b)
Every class x is a subset of A by definition. Because R is reflexive on A, x R x.
Theretore, x Sx.
(i)
Suppose that x R y. (We first show that y C x.) Letz € y. Then y R z.
Therefore, by transitivity, x R z, so z € x. Similarly, using transitivity
and symmetry, we have that x Cc y. Therefore x = y.
(ii) | Now suppose that x = y. Because y € y, we have y € x. Thus, x Ry.
168
CHAPTER 3
Relations and Partitions
(c)
(i)
Ifxn y=,
(ii)
(We give a proof by contraposition that x R y implies x N y = ©.)
then y ¢ x because y € y. Therefore, x Ry.
Suppose that x n y # ©. Then there exists z € A such that z € x and
z€ y. Then x R zand y R z. By symmetry, we have x R z and z R y and,
5
by transitivity, x R y.
As a consequence of parts (b) and (c) of Theorem 3.2.2, whenever two equivalence classes x and y have any element or elements in common, then the classes are
equal, and the representatives x and y are related. See Exercise 11.
Congruence relations on Z, defined next, were first introduced by Carl Frie-
drich Gauss.” We will show that congruence mod m is an equivalence relation and
describe its equivalence classes. “Arithmetic” for this relation is mentioned at the
end of this section and covered in much more detail in Section 3.4.
DEFINITIONS
Let m be a fixed positive integer. For x, y € Z, we say
x is congruent to y modulo m and write x = y (mod m) if m divides (x — y).
The number m is called the modulus of the congruence.
Example.
Using 4 as the modulus, we have
3 = 3 (mod 4) because 4 divides 3 — 3 = 0
9 = 5 (mod 4) because 4 divides 9 -5 = 4
—27 = 1 (mod 4) because 4 divides —27 — 1 = —28
20 = 8 (mod 4) because 4 divides 20 — 8 = 12
100 = 0 (mod 4) because 4 divides 100 — 0 = 100.
It is easy to see that 0 is congruent to 0, 4, —4, 8, and —8 and, in fact, is congruent
modulo 4 to every multiple of 4. Conversely, every integer congruent to 0 modulo
4 must be a multiple of 4.
Oo
Theorem 3.2.3
For every fixed positive integer m, congruence modulo m is an equivalence relation
on Z.
Proof.
(i)
(ii)
To show reflexivity on Z, let.x be an integer. We show that x = x (mod m).
Because m- 0 = 0 = x — x, m divides x — x. Thus, congruence modulo m
is reflexive on Z.
Forsymmetry, suppose that x = y (mod m). Then m divides x — y. Thus, there
is an integer
k such that x— y = km. But this means that —(x — y) = —(km),
or that y — x = (—k)m. Therefore, m divides y — x, so y = x (mod m).
* Carl Friedrich Gauss (1777-1855), one of the greatest mathematicians of all time, also made major
contributions to astronomy and physics. Congruence and modular arithmetic (and much more) appeared
in his masterwork, Disquisitiones Arithmeticae, which he completed at the age of 21. He proved the
Fundamental Theorem of Algebra and the Prime Number Theorem, among many other notable results.
3.2
(iii)
Equivalence Relations
169
Suppose that x= y (mod m) and y = z(mod m). Thus, m divides both x — y
and y — z. Thus, m divides the sum x — z, so
x = z (mod m). Therefore,
congruence modulo m is transitive.
a
What do the equivalence classes for congruences look like? For congruence
modulo 4, we have already seen that the integers congruent to 0 are exactly the
multiples of 4. Therefore,
C= eee 16n 10h
8
048
12916,
To form the equivalence class of 1 (mod 4), we first note that | contains 1 by
Theorem 3.2.1(a). Beginning with 1, we can add or subtract any multiple of 4 to
produce a number that is also congruent to 1 (mod 4). For example, 25 = 1 + 4 - 6,
so 4 divides 25 — 1 and 25 = | (mod 4). This observation tells us the other integers
ay Ike
Tos
PAS
1 et
185, 913817
In the same way, we form
=
(eee 14e 106)
9
{13,9
2: 246) 10) 148
Sa
5, 1, 3.7 A 5 aloes
We can stop now because choosing any other representative integer—say, 5—
will not produce a new equivalence class. Adding a multiple of 4 to 5 is the same as
adding a multiple of 4 to 1 because 5 + 4k = 1 + 4(k + 1). Thus 5 and 1 are equal.
The set of equivalence classes modulo 4 is {0, 1, 2, 3}. This set will be denoted Z,.
DEFINITION
The set of equivalence classes for the relation congruence
modulo m is denoted Z,,.
You should verify that x € Z is congruent to 0 (mod 6) if and only if x is a
multiple of 6. Thus, in Z, the congruence class of 0 is 0 = {..., —18, —12, —6, 0,
6, 12, 18, . . .}. The other congruence classes modulo 6 are
II te lal
pero,
1415.19. x.)
II {...,—16, —10, —4, 2, 8, 14, 20, .. .}
II ee, 10, 9
5,5,.9, 1), 21, -..}
...,—14,
—8,
—2,
4,
10, 16, 22, ...}
{
lei Le LS reget:
Dl! {Oe Se
BIW)
A)
There are six different congruence classes modulo 6, so Z, = {0, 1, 2, 3, 4, 5}.
It is tempting to conclude that Z, = {0, 1, 2, 3} is a subset of Z, = {0, 1, 2,
3, 4, 5}, but this is not the case. Although both sets contain, for example, the
170
CHAPTER 3
Relations and Partitions
equivalence class 3, the equivalence class of 3 modulo 4 is not the same set of
integers as the class of 3 modulo 6.
Our next theorem provides an explicit description of Z,,,.
Theorem 3.2.4
Let m be a fixed positive integer. Then
(a)
for integers x and y, x = y (mod m) if and only if the remainder when x is
divided by m equals the remainder when y is divided by m.
(b)
Z,, consists of m distinct equivalence classes: Z,, = {0, 1, 2,...,m — 1}.
Proof.
(a)
Let x and y be integers. By the Division Algorithm, there exist integers gq,
r, t, and s such that
x = mq + r, with
0 < r < m, and
y= mt + s, with
0 <5 < m. (We must show that x = y (mod m) iff r = s.) Then
x = y(mod m) iff-m divides x — y
iff m divides (mg + r) — (mt + s)
iff m divides m(q — ft) + (r — S)
iff m divides r — s
lth 7 = 5,41 nisis because
Or r= mand
(Dyn
We
just. show
0 <=k<m
that. 2 =
10) 1)2....,m-=1}.)
Ror
0s
each
-< 7m.)
k, where
—1, the set k is an equivalence’class, so {0, 1, 2,..:, m —
1}
is a subset of Z,,. Now suppose that x € Z,, for some integer x. By the
Division Algorithm, there exist integers g and r such that x = mg + r, with
0 <r<m. Then x —
r= mag, so m divides x — r. Thus x = r (mod m).
Bye lheoremt(3.2.1(b) x= 7: Because 0 <7 < max en, 2, se ie
diheretore ZG .0;,.152,...5m—1}.
Finally we will know that Z,, has exactly m elements when we show
that the equivalence classes 0, 1, 2,..., 7m — 1 are all distinct. Suppose that
k = 7, where 0 < r<k < m — 1. (We have assumed, without loss of generality, that r < k.) Then k = r (mod m), and, thus, m divides k — r. But
ORS
soe
ar — Os [hen k = 7. hereiore, the mequivalence classes are distinct.
i
It helps to remember that 0, 1>2, ..., and m — 1 are exactly all the possible
remainders when integers are divided by m. For this reason, the elements of Z,, are
sometimes called the residue (or remainder) classes modulo m. When you think
of equivalence classes modulo m, think remainders. In Z,, for example, 109 = 4
because the remainder is 4 when 109 is divided by 7. Mathematicians usually omit
the equivalence class notation and simply write Z,, = {0, 1, 2, ...,m— 1}.
The 12 numbers on the face of an analog clock correspond to the 12 classes in
Z 1, = {0, 1, 2, 3,..., 11}, where we note that 0 = 12 corresponds to 12 o’clock
because 0) = 12 (mod 12). Rather than talking about anything beyond 12 o’clock,
we start over again with | o’clock instead of 13 o’clock because 13 = 1 (mod 12),
3.2.
Equivalence Relations
171
2 o'clock instead of 14 o’clock, and so on. We are so accustomed to working with
these equivalence classes modulo 12 that we routinely do arithmetic with them.
For example, 9 hours after 8 o’clock is 5 o’clock because 8 + 9 = 17 and 17 = 5
(mod 12). We know a train that is scheduled to arrive at 10 o’clock but is 3 hours
late will arrive at 1 o’clock. In these examples, we are saying that in Z,), 8S 925
and 10 + 3 = 1. In fact, dispensing with the equivalence class notation, we say
$9
—-5 and Oe 3hn an Zap.
We shall see that for each natural number m, it is possible to do arithmetic
inside the set Z,,, using rules for addition and multiplication that depend upon the
modulus m. The details of how this modular arithmetic is done and a comparison
of the properties of modular arithmetic and the familiar arithmetic for Z appear in
Section 3.4.
Finally our discussions about sets and equivalence relations provide a glimpse
of how it is possible to build a foundation for mathematics based entirely on the
theory of sets.” To construct zero and the natural numbers, we use the idea of successor x’of a set x, which is the set x U {x}. We define 0 to be the empty set and
inductively define 1 = 0’ = {0}, 2 = 1’ = {0, 1}, and so on. All the familiar properties of the natural numbers can be proved for the system thus created.
We then define the integers as equivalence classes of the “difference” relation
d on (N U {0}) x (N U {0}) given by (m, n) d (7, s) if
m+ 5 =r+ q. The
integer “—5” is the equivalence class that contains (5, 10), (2, 7), etc. Similarly, the
rational numbers are defined as the equivalence classes of the “quotient” relation
gon Z x (Z — {0}) given by (m, n) g (r, s) if ms = rn. The rational number :is the
equivalence class containing (2, 3), (10, 15), etc. Section 7.5 contains brief descrip-
tions of two different methods for constructing the reals from the rationals. One of
the methods is based on equivalence classes of sequences.
Exercises 3.2
1.
x
x
Indicate which of the following relations on the given sets are reflexive on the
given set, which are symmetric, and which are transitive.
(a)
(12)
nom (142)
(b)
sonN
(ce
on
(d)
<onN
(e)
>onN
(f)
#onN
(g) “divides” on N
(h)
{G@.y)eZ x Zx+y = 10}
Wipe )eOal) aC ion thetset Av 1 23,455)
(j) =
{(,m): land mare lines and / is perpendicular to m} on the set of
all lines in a plane
(k) S, where x S y iff 3 divides x + y, on the set N
(1) RR, where (x, y) R (z, w) iff x +
z< y + w, onthe setR x R
“Details may be found in Robert Stoll, Set Theory and Logic, Dover, 1979.
172
CHAPTER 3
Relations and Partitions
(m)
S, where x S y iff x is a sibling of y, on the set P of all people
(n)
T, where (x, y) T (z, w) iff x + y < z+ w, onthe set R x R
Let A =
{1, 2,3}. List the ordered pairs, and draw the digraph of a relation
on A with the given properties.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
not reflexive, not symmetric, and not transitive
reflexive, not symmetric, and not transitive
not reflexive, symmetric, and not transitive
reflexive, symmetric, and not transitive
not reflexive, not symmetric, and transitive
reflexive, not symmetric, and transitive
not reflexive, symmetric, and transitive
reflexive, symmetric, and transitive
For each part of Exercise 2, give an example of a relation on R with the
desired properties.
The properties of reflexivity, symmetry, and transitivity are related to the
identity relation and the operations of inversion and composition. Prove that
(a)
Ris areflexive relation on A if and only if J, C R.
(b)
Ris symmetric if and only if R = RoI.
(c)
Ris transitive if and only if
RORCR.
Let R be a relation on a set A. Prove that
(a) if A is nonempty, the empty relation © is not reflexive on A.
(b) the empty relation © is symmetric and transitive for every set A.
For each of the following, prove that the relation is an equivalence relation.
Then give information about the equivalence classes as specified.
(a) The relation S on R given by x S y iff x — y € Q. Give the equivalence
class of 0; of 1; of V2.
(b)
The relation R on N given by m Rn iff m and n have the same digit in
the tens places. Find an element of 106 that is less than 50; between 150
(c)
and 300; greater than 1,000. Find three such elements in the equivalence
class 635.
The relation V on R given by x Vy iff x = y or xy = 1. Give the
equivalence class of 3; of ae of 0.
(d)
OnN — {1}, the relation R given by a R b iff the prime factorizations
of a and b have the same number of 2’s. For example, 48 R 80 because
48 = Doo and 80 = 2+.5. Name three elements in each of these
classes: 7, 10,727
(e)
Therelation Ton R x R given by (x, y) T (a, b) iff x? + y? = a? + Bb’.
Describe the equivalence class of (1, 2); of (4, 0).
(f)
For the set X = {m,n, p,q, r,s}, let R be the relation on P(X) given
by AR B iff A and B have the same number of elements. List all the
elements in {m}; in {m, n, p, g, r}. How many elements are in X? How
many elements are in 9P(X)/R?
3.2.
vy
(g)
The
relation
P on
R x R defined
Equivalence Relations
173
by (x, y) P(z, w) iff |x — y| =
|z — w|. Name at least one ordered pair in each quadrant that is related
to (3, 0). Describe all ordered pairs in the equivalence class of (0, 0); in
the class of (1, 0).
(h)
Let R be the relation on the set of all differentiable functions defined by
fR g iff
f and g have the same first derivative; that is, f’ = g’. Name
three elements in each of these classes: x7, 4x? + 10x. Describe x3
and 7.
(i)
The relation T on R given by x T y iff sin x = sin y. Describe the equivalence class of 0; of 2/2; of 1/4.
Let R be the relation on © defined Dye,
£ Ne~ iff pt = gs. Show that R i an
equivalence relation. Describe the brent ofthe equivalence class of =5
Which of these digraphs represent relations that are (i) reflexive on the aa
set? (i1) symmetric? (ili) transitive?
(a)
|
4
(b)
EN
Determine the equivalence classes for the relation of
(a)
(c)
congruence modulo 2.
congruence modulo 1.
(b)
(d)
congruence modulo 8.
congruence modulo 7.
10.
Name a positive integer and a negative integer that are
(a) congruent to 0 (mod 5) and not congruent to 0 (mod 6).
(b) congruent to 0 (mod 5) and congruent to 0 (mod 6).
(c) congruent to 2 (mod 4) and congruent to 8 (mod 6).
(d) congruent to 3 (mod 4) and congruent to 3 (mod 5).
(e) congruent to 1 (mod 3) and congruent to | (mod 7).
11.
Let R be an equivalence relation on a set A, and let x and y be elements of A.
Without reference to Theorem 3.2.2, prove that if x and y have an element in
common, then x R y and x = y.
174
CHAPTER 3.
Relations and Partitions
12.
Using the fact that congruence modulo m is an equivalence relation on Z and
without reference to Theorems 3.2.2 and 3.2.4, prove that for all x and y in Z
(a) xex.
(b) x47.
(c)
(e)
13.
ifx=y(modm),thenx = y.
ifxNy#, then
x = y.
(d)
(Hie
ifx = y, then x =y (mod m).
ih piv @ then
x= »:
Suppose that R is an equivalence relation on a set A with elements x, y, and
z, and suppose that z is in both the equivalence class of x and the equivalence
class of y. Use Theorem 3.2.2 to explain why
14.
(a)
(b)
x and yare related.
the classes of x, y, and z are all equal.
(a)
Find the equivalence classes for Z; and for Zo.
(b)
What is the relationship between the five equivalence classes of Z; and
the 10 classes of Z,9?
(c)
Let R and S be equivalence relations on a nonempty set A such that
R CS. Prove that every equivalence class modulo R is a subset of the
corresponding equivalence class modulo S; that is, prove that for all
G@eA. a/R Ee als:
15.
(a)
Suppose that R and S are equivalence relations on a set A. Prove that R
Sis an equivalence relation on A.
(b)
Let R be the relation of congruence modulo 3 and S be the relation of
congruence modulo 4. Describe the equivalence relation R 4 S on Z.
16.
Determine which of the following statements are true. Prove each true statement, and give a counterexample for each false statement.
(a)
(b)
(c))
(d)
(e)
(f)
17.
Ifx=1 (mod 5), then x = | (mod 10).
Ifx = 0 (mod 6) or x = 3 (mod 6), then x = 0 (mod 3).
lt a= lt Gnod 10) or x= 6 (mod 10), then x = 1 (mod 5).
Ifx=1 (mod 5), then x = 1 (mod 10) or x = 6 (mod 10).
Forallm,neN, ifx=0(mod m) and x = 0 (mod n), then x = 0 (mod mn).
ForallmeN, ifx=0 (mod 2m) or x= m (mod 2m), then x = 0 (mod m).
Prove that if R is a symmetric, transitive relation on A and the domain of R is
A, then R is reflexive on A.
Let R be a relation on the set A.
(a) Prove that
RU R™! is symmetric. (RU R~! is the symmetric closure
of R.)
(b)
Proofs to Grade
19.
Prove that if S is a symmetric relation on A and R C S, then R7! C S.
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
:
(a) Claim.
Ifthe relation R is symmetric and transitive, it is also reflexive.
“Proof.”
Because R is symmetric, if (x, y) € R, then (y, x) € R. Thus,
(x, y) € R and (y, x) € R, and because R is transitive, (x, x) © R. There-
(b)
fore, R is reflexive.
a
Claim.
The relation T on R x R given by (a, y) T(r, s) iff x + y =
r + s is symmetric.
3.3
“Proof.”
175
Suppose that (x, y)éR x R. Then (x, y) T(y, x) because
x+y=y-+
(c)
Partitions
x. Therefore, Tis symmetric.
@
Claim. The relation Won R x R given by (x, y) W(r, s) iff x — r=
y — s is symmetric.
“Proof.”
Suppose that (x, y) and (r, s) arein R x Rand (x, y) W(r, s).
Then x — r= y — s. Therefore, r — x = s — y,so(r, s) W(x, y). Thus,
(d)
W is symmetric.
=
Claim.
Ifthe relations R and S are symmetric, then RM S is symmetric.
“Proof.”
Suppose that (x, y) € ROS. Then (x, y) € R and (x, y) ES.
Because R and S$ are symmetric, (y,x)e@R and (y,x) € S. Therefore,
(a8)
(e)
*
GIR AS:
B
Claim.
Let R be an equivalence relation on A, and x, y, z € A. If x € y,
and y € z, then x € z.
“Proof.”
Suppose that x € y and y € z. Then by definition of an equivalence class, x R y and y R z. By symmetry, we have z R y and y R x, so
by transitivity z R x. Therefore x € z.
a
(f)
Claim.
Let R be an equivalence relation on A, and x, y, z € A. If
WIM z tien ZC
we
(g)
“Proof.”
Assume that x € yM z. Then x € y and x € z. Then y R x and
z R x. Now suppose that w € z. Then z R w. Therefore, z R w and x R z,
so x R w. Thus w € x.
=
Claim.
Let R be an equivalence relation on A, and x, y, z € A.
li x ey
iz, them z < x.
“Proof.”
Assume thatx € MZ. Then x € Z, soz Rx. By Theorem 3.2.2,
we have z = x. Therefore z € x.
3.3
|
Partitions
Partitioning is frequently used to organize the world around us. The landmass of
the United States, for example, is partitioned in several ways—by postal zip codes,
state boundaries, time zones, etc. In each case, nonempty
subsets of the United
States are defined that do not overlap and that together comprise the entire country.
This section introduces this concept of partitioning of a set and describes the close
relationship between partitions and equivalence relations.
DEFINITION
Let A be a nonempty set. Y is a partition of A if P is
a set of subsets of A such that
(Gi)
if XeP,
then X 4S.
(ii) |)1k
Gi)
e@Piand Ye, then X = Yorxn Y= @
LUX=A.
XEfP
a
176
CHAPTER 3
Relations and Partitions
A partition of a set A is a pairwise disjoint collection of nonempty subsets of
A whose union is A.
The set W of all employees in a large work area can be partitioned into work
groups by putting up physical partitions (walls) to form cubicles. If we are careful
so that (i) every cubicle contains at least one worker, (ii) no worker is assigned
to two different cubicles, and (iii) every worker must be in some cubicle, then we
have formed a partition of W. Notice that the workers are not elements of the partition; each element of the partition is a set of workers within a common cubicle. In
Figure 3.3.1, Wis a set of 6 workers, and the partition of W consists of four sets—
two sets each with two workers and two sets each with a single worker.
Figure 3.3.1
Example.
Prove that the family G = {[n,n + 1):n € Z} is a partition of R.
Proof.
(i)
(ii)
For every n € Z, the half open interval [n, n + 1) contains n. Therefore, for alln € Z, [n,n+ 1)4# ©.
Let mandn
be integers. (We show that either [m, m+ 1)=[n,n+ 1)
or({m, m+ 1) [n,n+ 1) =.) Suppose that [m,m + 1)#[n,n+ 1).
Then m ¥ n, and, without loss of generality, we may assume that
m <n. Then, because.m and n are integers, m + 1 <n. Thus, for any
real numberx in [m, m+
1), we have x < m+
1, sox <n. Therefore,
[m,m+1)O[nn+1)=2.
(iii)
Because
every
half
open
interval
is a
subset
of
the
reals,
LU [n,n + 1) C R. Now letx € R. Choose mto be the greatest integer
neZ
less than or equal tox.Then
x € [m,m-+ 1). Therefore,
x € U [n,n + 1).
We conclude that 7 [Inon+1)=R.
a
neZ
Therefore
is a partition of R.
=
3.3
Examples.
Partitions
177
Each of the following is a partition of Z:
P = {E, D}, where E is the set of even integers and D is the set of odd integers.
# = {N, {0}, Z~}, where Z~ is the set of negative integers.
4 = {A; k € Z}, where A, = (3k, 3k + 1, 3k + 2}.
Oo
Two other partitions
of Z are {..., {—3}, {—2}, {—1}, {0}, {1}, {2}, {3}, ...}
and {Z}. In fact, for any nonempty set A, the families { {x}:x @A} and {A} are
partitions of A.
Example.
For the set A =
Cr
{a, b,c, d, e}, the family C =
(0.e\7 CO, —
|
G,c.d) and C81
{C,, C,, C3}, where
Dae).
is a partition of A even though the sets C, and C; are not disjoint. The family
{C,, C,, C3} is the same as the family {C,, C3}.
Let W be a set of six people and C = {blue, green, red, white}. For each
Ce Onlet
B. = {x € W: xis wearing clothing with color c},
and let
B= {Bytes Berens Breas Bwnite}- The family % may not be a partition of
W because any of the three parts of the definition might be violated. If no one
is wearing red, then B,., is empty, so condition (i) fails. If someone is wearing
green only and a second person 1s wearing green and blue, then the different sets
Boye ANd Byreen Overlap, in violation of condition (ii). If someone is wearing only
yellow clothing, then that person does not belong to any set in %, in violation of
condition (iii).
Theorems 3.3.1 and 3.3.2 describe the connection between partitions and
equivalence relations.
Theorem 3.3.1
If R is an equivalence relation on a nonempty set A, then A/R is a partition of A.
Proof. By Theorem 3.2.1, every equivalence class x is a subset of A and is nonempty because it contains x, and any two equivalence classes are either equal or
disjoint. All that remains is to show that the union over A/R is equal to A.
First, Lx C A because each x C A. To prove A © rs suppose that y € A.
xEA
Because y € y, we have y € |) x. Thus, A = Ux.
xEA
Example.
Let
A =
xEA
=
xEA
{4,5, 6, 7} and T be the equivalence relation
{(4, 4), (S, 5), (6, 6), (7, 7), (5, 7), (7, 5), (7, 6), (6, 7), (5, 6), (6, 5)}The equivalence classes of T are 4 = {4} and5 = 6=7 = (5, 6,7}. By Theorem 3.3.1,
the partition produced by Tis A/T= {{4}, {5, 6, 7}}.
178
CHAPTER 3
Relations and Partitions
The Five Boroughs of New York City
B,: Manhattan
B,: Brooklyn
B;: Queens
B,: The Bronx
B,: Staten Island
Figure 3.3.2
New York City is divided into five boroughs (counties). The boroughs are
labeled B, through B, in Figure 3.3.2. If A is the set of all residents of New York
City, then A is partitioned into five subsets: the set of residents living in B,, the resi-
dents living in B,, and so on. How can we use this fact to define an equivalence relation on A? We say that two residents of New York City are equivalent if and only
if they are in the same partition element; that is, they reside in the same borough.
The method we will use to produce an equivalence relation from a partition is
based on this idea that two objects will be said to be related if and only if they belong
to the same member of the partition. The next theorem states that this method for
defining a relation always produces an equivalence relation and, furthermore, that
the set of equivalence classes of the relation is the partition.
Theorem 3.3.2
Let P be a partition of the nonempty setA. For x and y € A, define x Q y if and only
if there exists C € P such that x € C and y € C. Then
(a)
(b)
Qis an equivalence relation on A.
A/O=P.
Proof.
(a)
We prove that Q is transitive and leave the proofs of symmetry and reflexivity on A for Exercise 10. Let x, y, z € A. Assume that x Q y and y Q z. Then
there are sets C and D in
such that x, y € C and y, z € D. Because # is a
partition of A, the sets C and D are either identical or disjoint, but because
y is an element of both sets, they cannot be disjoint. Hence, there is a set
(b)
C (= D) that contains both x and z, so x Q z. Therefore, Q is transitive.
We first show A/Q C PP. Let x € A/Q. Then choose B € # such that x € B.
We claim x = B. If y € x, then x Q y. Then there is some C € # such that
3.3
xe Cand y€C.
Because
x €
CN B,
Partitions
179
C = B, so y€ B. On the other hand,
if y € B, then x Q y, so y € X. Therefore, x = B.
To show % C A/Q, suppose B € Y. As an element of a partition,
B#©. Choose any y & B; then we claim B = y. If s © B, then y Qs, so
s © y. On the other hand, if s € y, then y Q's, so s and y are elements of the
same member of %?, which must be B.
B
Example.
Let A = {1,2, 3,4}, and let ® = { {1}, {2,3}, {4}} be a partition
ofA with three sets. The equivalence relation Q associated with # is {(1, 1), (2, 2),
(3, 3), (4, 4), (2, 3), (3, 2)}. The three equivalence classes for Q are T= /1)72=
3 = {2,3}, and 4 = {4}. The set of all equivalence classes is precisely P.
o
Example.
The set of = {Ap, Aj, A>, A3} is a partition of Z, where
fo [4k:k EZ}.
= {4kK+1:keEZ}.
= {4kK+2:keEZ}.
= {4k+3:keEZ}.
Then integers x and y are in the same set A; iff x = 4n + i and y = 4m + i for
some integers n and m or, in other words, iff x — y is a multiple of 4. Thus, the
equivalence relation associated with the partition
is the relation of congruence
modulo 4, and each 4A; is the residue class of imodulo 4 for i = 0, 1, 2, 3.
o
We have seen that every equivalence relation on a set determines a partition
for the set and every partition of a set determines a corresponding equivalence relation on that set. Furthermore, if we start with an equivalence relation, the partition
we make is the set of equivalence classes, and if we use that partition to form an
equivalence relation, the relation formed is the relation we started with. Thus, each
concept may be used to describe the other. This is to our advantage, for we may use
partitions and equivalence relations interchangeably, choosing the one that lends
itself more readily to the situation at hand.
Exercises 3.3
1.
2.
Describe four different partitions of the set of all students enrolled at a
university.
For the given setA, determine whether
is a partition of A.
ko
(a)
(Dy)
(ce),
(d)
itn L255 ale eet
fe) e123} 3.404
al $2534 3G, TA
a
lie 2S ts ahi}
Atl
2 34
Or ats yf Oh ef 2n4h (77)
tieNin S25}
32455)
2
Age NS
(Clee
(f)
co,
IU [1 Ty
cs)
A=R,% = {S,: ye Randy > 0}, where S, = {xe
Rix < y}
180
CHAPTER 3
Relations and Partitions
3.
Prove that each of these collections is a partition of the given set.
(a) {{—x,x}:x
EN U {0}} is a partition of Z.
(b)
{0}}, where each A, = {2‘n: n € N and n is odd}, is a
{Az ke NU
partition of N.
(c)
{A,: r € R}, where each A, =
(d)
Rx R.
{Et € [0, 1)}, where each E, = {z+1t:z€ Z}, isa partition of R.
{(x, y): x+y
= r}, iS a partition of
Describe the partition for each of the following equivalence relations:
(a) Forx,yER,xRyifx—yeZ.
(b)
Forn,meZ,nRmifn
and m have the same tens digit.
(c)
Forx,yE€R,xRyif sinx = siny.
(d)
(e)
()
(g)
(h)
Forx,yeR xRyifx? = y’.
For (x, y) and (u, v) ER x R, (x, y) S (u, v) if xy = uv = Oorxyuv > 0.
For,
y) and (u,v)inR x R, Gy) Rw, vifx+v=yt+u.
Forx,ye Q — {x},xSyif(—x(a—y)>0.
For x, y € N, x S.y if x and y have the same prime divisors.
Let C = {i, —1, —i, 1}, where i? = —1. The relation R on C
given by x Ry
if xy = +1 is an equivalence relation on C. Give the partition of C associated
with R.
Let C be as in Exercise 5. The relation S on C x C given by (x, y) S (u, v) if
xy = uv is an equivalence relation. Give the partition of C x C associated
with S.
Describe the equivalence relation on each of the following sets with the given
partition.
(Ayes
ets Jone
Ot TON Ie 99t {100 LTOT A. 999}. ...+
(DZ
ee =), EA UMC
(CPR (00y'0)){ OF) (0 00)}
CO ee 2
2)
(Erez A 2) es) y eta}
(e)aeZ,
(DeNe
(g)
lhl ahNeha 5 Soe a
1.
0), (03, (0, D4 1:
{ASB where A= {xe 72x
3) and B= Z—A
Onl 2.3) (4,.5,6,71218,9,
10 111...)
R, {{@,
y) eR x R: xy =0}, {(%, y) EeR x R: xy > 0},
{(x,
y) E R x R: xy < OF}
For eacha € R, let A, = {Qpy)
ER x Riy =a — x’}.
(a)
Sketch a graph of the set A, for a = —2, —1, 0, 1, and 2.
(b)
(c)
Prove that {A,: a € R} is a partition of R x R.
Describe the equivalence relation associated with this partition.
List the ordered
pairs in the equivalence
relation on A =
{1, 2, 3, 4, 5}
associated with these partitions:
(a)
(C)
le}
273A
nao
Odiatel |}
(b)
(d)
itt} {2}, {3,4},
{5}}
til si 12,4),
13))
3.3
10.
181
Complete the proof of Theorem 3.3.2: Suppose that Y is a partition of A and
suppose that x Q y if there exists C € P such that x € C and y € C. Prove that
(a)
(b)
11.
Partitions
Qis symmetric.
Qis reflexive on A.
Let R be a relation on a set A that is reflexive and symmetric but not transitive. Let R(x) = {ye A:xRy}. (Note that R(x) is the same as X except
that R is not an equivalence relation in this exercise.) Does the set 4 =
{R(x): x € A} always form a partition of A? Prove that your answer is correct.
Repeat Exercise 11, assuming R is reflexive and transitive but not symmetric.
13.
Repeat Exercise 11, assuming R is symmetric and transitive but not reflexive.
14.
Let A be a set with at least three elements.
(a) IfP = {B,, B,} isa partition of Awith B, # B,, is {BS, BS} a partition
of A? Explain. What if B, = B,?
(b)
IfP =
{B,, B,, B3} is a partition of A, is {BS, BS, BS} a partition of A?
Explain. Consider the possibility that two or more of the elements of P
may be equal.
(c)
Proofs to Grade
IfP = {B,, B,} isa partition of A, €, is a partition of B,, €, is a partition of B,, and B, # B,, prove that €@, U “, is a partition of A.
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
(a)
Claim.
Let R be an equivalence relation on the set A, and let x, y, and
z be elements ofA. Ifx € y andz ¢ x, thenz € y.
“Proof.”
Assume that x € y and z € x. Then y R x and x R z. By transitivity, y R z, so z © y. Therefore, if x€ y and z ¢ x, then z ¢ y.
a
(b)
Claim.
Let R be an equivalence relation on the set A, and let x, y, and
z be elements of A. If x € y and z ¢ x, then z € y.
“Proof.”
Assume that x € y, and assume that z € y. Then y R x and
y Rz. By symmetry, x R y, and by transitivity, x R z. Therefore, z € x.
We conclude that if x € y and z ¢ x, then z € y.
(c)
XeA
*
(d)
|
Claim.
If & isa partition of a set A and &% is a partition of a set B, then
A U Bis a partition of A U B.
“Proof.”
(i) IfXe AUS, then X € A, or X € R. In either case, X # ©.
(ii) If Xe AURBand YEAHUS,
then XE A
andYeA,orXeH
and Y € R,orX € Rand Y € A,orX € Rand
Y € BR. Because both
A and & are partitions, in each case either X = YorXN Y= ©.
(iii) Because |) X = A and L)xX=B,
U) x’ =AUB.
=
xEB
XEAUB
Claim.
If % isa partition of A and if x Q y iff there exists C € % such
that x € C and y € C, then the relation Q is symmetric.
“Proof.”
First, x QO y iff there exists C € % such that x e Candy EC.
Also, y Q x iff there exists C € % such that y € C and x € C. Therefore,
0 yitiey OX
a
182
3.4
CHAPTER 3.
Relations and Partitions
Modular Arithmetic
This section continues the study of congruence relations on Z and develops the
arithmetic of congruence classes. Recall from Section 3.2 that for a fixed natural
number m, we may define the relation of congruence modulo m on Z as follows:
a = b (mod m) if m divides b — a.
We showed that congruence modulo m is an equivalence relation on Z and proved
that a = b (mod m) if and only if a and b have the same remainders when divided
by m. Because the only possible remainders when an integer is divided by m are
0, 1,2,..., and m —
1, there are exactly m distinct congruence classes (mod m),
which we represent as {0, 1, 2,...,m — 1}. We call this set Z,,, the set of integers
modulo m.
As mentioned in Section 3.2, we routinely make computations doing what
might be called “clock arithmetic,” with the 12 hours that appear on the face
of an analog clock. We know that 7 hours after 10 o’clock is 5 o’clock because
10+ 7= 17 and 17 =5 (mod 12). What we’re doing when we compute in this way
is addition with the elements of Z,, because
10 + 7 = 5 (mod 12). The only dif-
ference is that we usually say 12 o’clock instead of 0 o’clock, but they’re the same
thing in Z,, because 12 = 0 (mod 12).
For each natural number m, it is possible to do arithmetic with numbers in the
set Z,,,, using rules for addition and multiplication that depend upon the modulus m.
The same idea of addition that works for a 12-hour clock works for any modulus.
Thus, with a 6-hour “clock” as in Figure 3.4.1, we could say that 3 hours after 5
o’clock is 2 o’clock because 5 + 3 = 8 and 8 = 2 (mod 6).
Figure 3.4.1
If we think of the numbers 5 and 3 as representatives of the equivalence classes
5 and 3 in Z,, then the sum of those classes, written 5 + 3, is the class containing
the integer 5 + 3. That class is 2 in Z,.
3.4
Modular Arithmetic
183
DEFINITION _ For each natural number m, the sum of the classes x and
y in Z,, is defined to be the class containing the integer x + y. In symbols,
we write
J
Because a = b (mod m) if and only if a and b have the same remainders when
divided by m, the sum x + y may be represented by the remainder when x = y is
divided by m. Thus for Z¢
2 ae 3 = 2 because 2 is the remainder when 5 + 3 is divided by 6
it2= 3 because 3 is the remainder when | + 2 is divided by 6
4 + 2 = 0 because 0 is the remainder when 4 + 2 is divided by 6.
When we do arithmetic inside Z, or another Z,,, for a fixed integer m, it is more
convenient to omit writing the bars above 0, 1, 2,3
3,4, and 5 and simply say Z, =
{0, 1, 2, 3, 4, 5}. Just remember that the element we are calling 2 in Z, is really ine
congruence class 2 = {..., —16, —10, —4, 2, 8, 14, 20,
. .}. We will continue to
use the bar notation where it is important to emphasize that the object is an equivalence class.
We can now compute all possible sums in Z,, as shown in the addition table
in Figure 3.4.2(a).
OGL
tie
eto
Manes
ON
Ll
UP
tl
ib
eae
ee ey
SEA
a
ee
Del
2
es
aes
Om
o)
|Parruiae
igh er
SNP Set
Fa
Ae
BO,
OEY
O|1o®
Loh Om
Foc
ere
©
©
©
©
Vleet
PO
MAY
oe
aE
2
er
sr
a
a
OS
We lheet Fp
ee Be
a
ed
AN
mat
a
ete
28
a)
ees
a2
a8
@
aes
2!
Figure 3.4.2
By reading the Z, addition table in Figure 3.4.2(a), we see that 3 + 4 = 1. This
may seem bizarre to someone not familiar with congruences, but remember that the
objects we’re calling 3, 4, and 1 in Z, are really infinite sets (congruence classes).
Each class is represented by the remainder after division by 6.
The rule for multiplication in Z,, mirrors the rule for addition.
DEFINITION
For each natural number m, the product of the classes x
and y in Z,, is defined to be the class containing the integer x - y. In symbols,
we write
184
CHAPTER 3.
Relations and Partitions
In Z,, we see that 4-5 = 2 because 4-5 =4-5=20
= 2 and2-3=0
because 2 - 3 = 2-3 = 6 = 0. The complete multiplication table for Z, is given in
=o
Figure 3.4.2(b).
The rule we use for addition in Z,, looks simple: x + y = x + y. But the objects
we are adding are not really numbers; they are infinite sets. And each infinite set can
be represented by any of its elements, so how can we be sure that we will get the
same answer if we use different representatives? To be more specific, consider this
computation in Z,: To find 4 + 3, we find 4 + 3 =7= 1 (mod 6). But 4 = —14 and
3 = 33, so when we use the addition rule, we find 4 + 3 = —144+ 33 =—-14 + 33=
19 = 1 (mod 6). In this case, the answer turned out to be the same when we used dif-
ferent representatives, but we must prove that this is always the case.
Theorem 3.4.1
Let m be a positive integer and a, b, c, and d be integers. If
(mod m), thena +
b=c +d (mod m).
a= c (mod m) andb=d
Proof. Suppose that a = c (mod m) and b = d (mod m). Then m divides both
a — cand b — d, so m divides their sum, (a — c) + (b — d) = (a+ b) — (c+ d).
Therefore,
a+
b=c +d (mod m).
Restating the theorem above in Z,,,ifa =
|
candb=d,thena+b=c+d.
This is what is meant when we say that addition in Z,, is well defined. A similar
result holds for multiplication.
Theorem 3.4.2
Let m be a positive integer and a, b, c, and d be integers. If a = c (mod m) and b =d
(mod m), thena-b=c-d(modm).
Proof.
Exercise 3.
a
With a little practice, you can perform arithmetic for any modulus m. Be sure
to take advantage of the fact that we can, at any time, replace one representative
of a congruence class with another. Note carefully in the next example that the
arithmetic is performed in Z,. The equivalence classes in this example are not the
same as those in Ze.
Example.
To compute (5 + 7) - (6 +5) in Z, we write
(56+7):-(6+5=12-13=4-5=20=4
because 12 = 4 (mod 8), 13 = 5 (mod 8), and 20 = 4 (mod 8).
Modular arithmetic simplifies the work of finding remainders. Suppose that we
want to find the remainder when 7 divides 215 - 699. We could do the multiplication by hand or calculator (and get 150,285) and then divide by 7. The easy way to
find the answer, by hand or otherwise, is to do the computation in Z,.
Example: To compute the remainder when 7 divides 215 - 699 we first find smaller
representatives for 215 and 699 in Z,: 215 = 5 (mod 7) and 699 = 6 (mod 7). Then
215 - 699 = 5 - 6 = 30 = 2 (mod 7). Therefore, the remainder is 2.
oO
.
3.4
Example:
Modular Arithmetic
185
To find 3° in Z,, calculate the first few powers of 3 modulo 7:
31=3
37=9=2
33=3.37=3-2=6
34 = 3.33 =3-.6=18=4
3 =3.34=3-4=12=5
3° =3.3=3-5=15=1.
Because 3° = 1, we have 3° = (3°)* = 1‘ = 1 (mod 7) for all natural numbers k.
Therefore, 3° = 3 . 33 =1 .33=1-6=6(mod 7).
Oo
Because computer computations can be done very rapidly with modular
arithmetic, problems can be solved that would otherwise be too time-consuming,
even on powerful computers. As an example of how modular arithmetic can be
applied to a less complex problem, we give an alternate proof that k° + 5k is always
divisible by 6. Note that in the proof the number 5 in the equation k’ + 5k =
0 represents the usual integer 5, not an element of Z,. This does not affect the
computation.
Example:
Proof.
Prove that for every integer k, 6 divides k? + 5k.
Let k be an integer. To show that 6 divides k° + 5k, we show that
k> + 5k = 0 (mod 6). There are six cases because k may be congruent to 0, 1, 2, 3,
4, or 5 (mod 6).
Case 1.
Case2.
k=0(mod 6). Then k = 0 and 5k =0, sok + 5k =0+0=0
(mod 6).
k= 1(mod6). Then k? = 1 and 5k=5,sok+5k=1+5=0
(mod 6).
Case3. k = 2 (mod 6). Then k? =
2+4=0
(mod 6).
8 =
2 and 5k =
10 = 4, sok
+ 5k =
Case 4.
k = 3 (mod 6). Then k = 27 = 3 and 5k = 15 = 3, sok + 5k =
Case 5.
k = 4 (mod 6). Then k? = 64 = 4 and 5k = 20 = 2, sok’? + 5k =
Case 6.
4+ 2=0 (mod 6).
k = 5 (mod 6). Then k? =
5+ 1 =0 (mod 6).
3 +3 =0 (mod
6).
125 = 5 and 5k = 25 =
In every case, k> + 5k = 0 (mod 6), so 6 divides k? + 5k.
1, sok? 4+ 5k =
te
Most of the properties for addition and multiplication listed in the Appendix to
describe the integers also hold for addition and multiplication in Z,, for every positive integer m. Because addition is commutative in Z (a+ b=b + a for all integers a
and b), the corresponding addition operation in Z,, inherits commutativity:
GO
a4
b Sb
a=
bP a.
Exercise 6 asks you to prove some other properties for Z,, that follow directly from
properties of the integers. For example, in Z,, the equivalence class | serves as the
186
CHAPTER 3
Relations and Partitions
multiplicative identity: For every xin Z,,x - 1 =x. The element 3 is the multiplicative inverse of the element 5 because 5 - 3 = | (mod 7).
There are other properties of the integers that do not transfer to Z,,. We are
used to the idea that if a product of integers is 0, then one of the factors must be 0.
However, in Z,, we saw that neither 2 nor 3 is 0, yet 2 - 3 = 0. Similarly in Z,),
6-8 = 48 = 0, but neither 6 nor 8 is 0. We prove next that this phenomenon always
occurs when the modulus m is composite. The proof uses equivalence class notation to distinguish classes from their integer representatives.
Theorem 3.4.3
Let m be a positive composite integer. Then there exist nonzero equivalence classes x
and y in Z,, such that x y = 0.
Proof. Because m is a composite, m = xy for integers x and y, with 0 < x < mand
0 < y < m. Then neither x nor y is 0, but xy = m = 0 (mod m). That is, x y= 0. @
One consequence of Theorem 3.4.3 is that cancellation does not work in Z,,
when m is composite. In Z,,, both 3 - 8 =0 and 6- 8 = 0. This means 3 - 8 =6:
8,
but 3 does not equal 6 in Z,,. The situation is much better when m is prime.
Theorem 3.4.4
Letp be a prime. Whenever x y = 0 in i then either x = 0 or
Proof.
y= 0.
Let x and y be elements of Z,, and suppose that x y = 0. Then xy =
(mod p). Therefore, p divides xy. By Euclid’s Lemma (Theorem 1.8.3), we conclude that either p divides x or p divides y. Thus, either x = 0 (mod p) or y = 0
(uod)p). Theretore, either x = 0 or y = 0.
E
The cancellation law for multiplication in Z,, follows directly from Theorem 3.4.4.
Theorem 3.4.5
Cancellation Law for Z,
Letp be a prime. If xy = xz in Z, and x #0, then y = z.
Proof.
See Exercise 9(a).
a
More about the algebraic properties of modular arithmetic can be found in
Chapter 6, where we look to Z,, as a guiding example for arithmetic operations and
their properties for other algebraic systems.
Exercises 3.4
1.
Perform each of these calculations in Z,.
(a) 6+6
(b) 5-4
xo
(C) ae Oke Tope
(C) PESela
as
(dq) 2-44+3-5
(Hi 20S 304
3.4
(g)
3°+4
eC)ieezes
(k) 4%
« 2,
Modular Arithmetic
(h)
(4+ 5)-(5+6)
(I)
226
uease:
Repeat Exercise | with calculations in Zo.
3.
Prove Theorem 3.4.2: If a, b, c, and d are integers,
(mod m), thena-b=c-d(modm).
4.
(a)
For x, yin Z,,, give a definition of subtraction: x — y.
(b)
(c)
Prove that subtraction is well defined.
Find the results of
(i) 8 — 6in Zp.
(iii) 3 — Sin Z,.
5.
187
a= c (mod m), andb=d
(ii) 5 — 8 in Zo.
(iv) 1 —4inZ,.
Suppose we try to define < on Z,, by saying that for x, yin Z,,, x < yifx <y.
What is wrong with this definition?
6.
Let a, b, and ¢ be classes in Z,, for some fixed integer m. Explain why each of
the following properties holds in Z,,.
(a)
(b)
(c)
Multiplication is commutative: a - b=b-a.
Addition is associative: a + (b + €) = (4 +b) +6.
Multiplication distributes over addition: ab+t)=
(d) (is the additive identity: a + 0 = a.
(e)
7.
*
*
8.
9,
*
1 is the multiplicative identity: a- 1 = a.
Find the remainder when
(a) 238+ 496 — 44 is divided by9.
(c) 421 + 366 — 14 isdivided by 11.
(e) 152- 156 is divided by 7.
(g) 182. 144 is divided by 13.
(b)
(d)
(f)
(h)
238 +496 — 44 is divided by 7.
317 - 403 is divided by 9.
280-318 is divided by 11.
288 - 301 is divided by 13.
Use the methods of this section to prove that
(a)
for every integer k, 5 divides k> — 5k? + 4k.
(b)
for every integer k, 5 divides k*(k*+ — 1).
(c)
for every integer k, 5(k® — k°) = 0 or 3 (mod 7).
(a)
(b)
Prove the Cancellation Law for Ls (Theorem 3.4.5).
Solve the equation 2x = 5 in Z,.
(c)
Solve the equation 2x = 5 in Z,).
(d)
Solve the equation 2x = 9 in Z,3.
10.
Prove that for every natural number m, if x = y (mod m), then x* = y* (mod m)
for all natural numbers k.
11.
Between 1970 and 2007, commercial books were published with an ISBN-10
(International Standard Book Number) consisting of 10 digits in the format
a-bcd-efghi-j. The last digit j is a check digit, meaning that is takes one of the
values 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, X (where X is a single letter symbol for 10)
and 10a + 9b + 8c + 7d + 6e + 5f+ 4g + 3h + 21+ j =0 (mod 11).
In this
way, the check digit makes sure the ISBN number is valid.
*Since 2007, an ISBN-13 standard has been used.
188
CHAPTER 3.
Relations and Partitions
Which of these are valid ISBN-10 numbers?
(a)
(c)
Proofs to Grade
12.
0-495-56203-5
2-015-39351-X
(b)
(d)
4-851-60772-4
0-618-51471-3
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
(a) Claim. Let m be a positive integer and a, b, c, and d be integers. If
a =c (mod m) and b = d (mod m), then a + b = c + d (mod m).
Let a=c (mod m) and b =d (mod m). From a = c (mod m),
“Proof.”
a+b=c+b(modm). Substituting b = d (mod m), we have
we have
a+b=c+d(modm).
z
(b)
Claim.
For all integers x and y, x + 3y = 0 (mod 8) if and only if
2x + 3y = 0 (mod 8).
“Proof.”
Let x and y be integers. Then
x + 3y = 0 (mod 8) iff 2(x + 3y) = 0 (mod 8)
(c)
Claim.
iff 2x + 6y = 0 (mod 8)
iff 10x + 6y = 0 (mod 8)
iff 5(2x + 3y) = 0 (mod 8)
iff 2x + 3y = 0 (mod 8).
=
For all integers x and y, 3x + 2y = 0 (mod 11) if and only if
5x + 7y = 0 (mod 11).
Let x and y be integers. By Theorem 3.4.4, a product in Z,,
“Proof.”
is 0 if and only if one of the factors is 0. Therefore,
3x + 2y = 0 (mod 11) iff 4x + 2y) = 0 (mod 11)
iff 12x + 8y = 0 (mod 11)
x + 8y = 0 (mod 11)
iff
iff S(x + 8y) = 0 (mod 11)
iff 5x + 40y = 0 (mod 11)
(d)
(e)
iff 5x + 7y = 0 (mod 11).
Claim. Let p be a prime. If xy= xz in Z, and x ¥ 0, then y = z.
Let x, y, and z be in Z,, and suppose that xy = xz and x
“Proof.”
=
0.
Thengyi—xz = 0; Therefore, x(y — z) = 0. Thus, x(y — z) =x - 0. By
cancellation, y — z= 0, so y = z.
=
Claim.
Let m be an integer greater than 1. If there exists an integer
n such that n* + 1 = 0 (mod m) and (n + 1)? + 1 = 0 (mod m), then
Mhs== By,
“Proof.” Suppose that n? + 1 = 0 (mod m) and (n + 1)? + 1 = 0
(mod m). Then n? + 2n +2 = 0 (mod m) and n? + 2n +2 — (n? +1) =
0 (mod m), so 2n + 1 = 0 (mod m). (Call this result 1.)
Then (2n + 1)? = 0? = 0 (mod m), so 4n? + 4n + 1 = 0 (mod m).
(Call this result 2.)
From n? + 2n + 2 = 0 (mod m), we have A(n? + 2n + 2) = 4(0) =
0 (mod m). (Call this result 3.)
3.5
Ordering Relations
189
Subtracting the second result from the third, we have 4(n? + 2n + 2) —
(4n? + 4n+ 1)=0-—0=0 (mod m), so 4n + 7 = 0 (mod m). (Call this
result 4.)
Now subtracting 2 times the first result from the fourth, we
4n
+ 7 — 2(2n+ 1) =0 — 2(0) = 0 (mod
m), so 5 = 0 (mod
m).
have
Therefore, m divides 5. Butm > 1,som=S5.
3.5
|
Ordering Relations
This section describes properties of relations like “less than” and “less than or equal
to” on the number systems N, Z, and R—relations that impose an order on the
elements of a set. For example, in a large department store, some employees can
give directives to other employees. The transitivity property is essential to maintain order because if the store president can give directives to the manager and the
manager can give directives to the clerk, then the president can give directives to
the clerk. On the other hand, the store ordering relation must not have the symmetry
property because it would be chaotic if two different people could give directives
to each other.
Ordering relations are not equivalence relations and must be characterized by
a different set of properties.
DEFINITIONS
Let R be a relation on a set A. Then
R is antisymmetric if for all x, ye A, if x R y and y Rx, thenx = y.
R is irreflexive on A if x R x for all x € A.
R has the comparability property if for all x, y € A, eitherx
Ry ory Rx.
The antisymmetry property is an opposite of the symmetry property. When
R is a symmetric relation, whenever x is related to y, then y is related to x. With
antisymmetry, we never have both x related to y and y related to x unless x and y
are the same object. Another way to describe antisymmetry is to say that for all
x,y €A,ifxRyandx#y, then
y Rx.
The irreflexivity property is a complete opposite of reflexivity. In a reflexive
relation on a set, every element is related to itself, but with an irreflexive relation,
no element is related to itself.
Comparabilty requires that distinct objects must be related in one way or the
other. It is a mistake to assume comparability always holds: In a multinational
corporation, workers in one branch office usually don’t have authority over those
in another branch.
190
CHAPTER 3
Relations and Partitions
Examples.
The relation < on R is reflexive on R, transitive, and antisymmetric
and has the comparability property. The relation < on R is transitive, antisymmetric, and irreflexive on R and has the comparability property. The reason < is
antisymmetric is that for x, y € R, it never happens that x < y and y < x. Because
the antecedent is false, the implication “if x < y and y < x, then x = y” is true for
allx, ye R.
Example.
Oo
The relation“‘divides” js reflexive on N and transitive. Also, if a divides
b and b divides a, then a = b, so “divides” is antisymmetric. However, “divides”
does not have the comparability property because there are natural numbers a and b
(e.g., 12 and 21) such that both “a divides b” and “b divides a” are false. “Divides”
is also not irreflexive because at least one (in fact every) natural number divides
itself.
0
Although, as noted above, the “divides” relation on N is antisymmetric,
“divides” is not an antisymmetric relation on the set Z. This is because, for example, 6 divides —6 and —6 divides 6, but 6
—6.
Examples.
Let X be a set. The set inclusion relation C on the power set of X is
reflexive on P(X) and transitive. Also, set inclusion is antisymmetric because if A
and B are subsets of X with A C B and B CA, then A = B. Unless X is empty or
has only one element, this relation does not have the comparability property. For
example, if X = {1,2,3,4}, then {1,3} and {1,4} are elements of 9(X), but
bouts Si 4) and {ly 4} cc {1,3} are false.
The strict subset relation C (subset, but not equal to) on P(X) is irreflexive on
P(X), transitive, and antisymmetric.
oO
Example. Let Y be the relation “is the same age in years or younger than” on
a fixed set P of people. Then Y is reflexive on P and transitive. This relation also
has the comparability property. However, the relation Y is not antisymmetric. If a
and b are two different people in P and both a and b are 20 years old, then a Y b and
bYa,buta#b.
Oo
A relation may be antisymmetric and not symmetric,
antisymmetric, both, or neither. See Exercise 2.
symmetric
and not
DEFINITIONS
Let R be a relation on a set A. Then
R is a partial order (or partial ordering) for A if R is reflexive on A,
antisymmetric, and transitive.
R is a strict partial order (or strict partial ordering) for A if R is irreflexive
on A, antisymmetric, and transitive.
A set with a partial order is called a partially ordered set, or poset.
3.5
Ordering Relations
191
The relations < on R and € on (A) for a fixed setA are partial orders, whereas
the relations < on R and C on (A) are strict partial orders. The only difference
between a strict partial order and a partial order is the difference between “less
than” and ‘less than or equal to.” In a strict partial order, no element is related to
itself, but in a partial order, every element is related to itself. Exercise 7 asks you
to show that every partial order has an associated strict partial order and vice versa.
For the remainder of this section, we will consider only partial orders.
Notice that the comparability property is not required for a partial ordering. Comparability holds for the partial order < on R but fails for the partial order
“divides” on N. The partial order C on P(X) does not have the comparability property when X has two or more elements.
Example.
Let W be the relation on N given by x Wy if x + y is even and x < y.
For example, 2W4,4W6,6W68,...,
and 1 W3,3 W5,5 W7,..., but we never
have mWn where m and n have opposite parity. We verify that W is a partial order.
Proof.
(i)
(ii)
(iii)
(Show Wis reflexive on N.) Let x € N. Then x + x = 2xis even, and x < x,
sox Wx.
(Show W is antisymmetric.) Suppose that x Wy and y Wx. Then x + y is
even, x < y, and y < x. By antisymmetry of < on N, x = y.
(Show W is transitive.) Suppose that x Wy and y Wz. Then x < y, x + y
is even, y < z, and y + z is even. By transitivity of < on N, x < z. Also,
x + z is even because x + z = (x + y) + (y + z) + (—2y) is the sum of
three even numbers. Therefore, x W z.
i]
Suppose that R is a partial order on the set A and a, b, and c are three distinct elements of A. Further suppose that a R b, b Rc, and cc R a. A portion of the digraph of
R is shown in Figure 3.5.1. The chain of relationships a Rb, b Rc, c Rais called
a closed path in the digraph. The path is closed because as we move from vertex to
vertex along the path, we can start and end at the same vertex. From a R b and b Rc,
by transitivity we must have a Rc. (The arc from a to c is not shown in the portion of the digraph in Figure 3.5.1.) But c R a is also true, and this contradicts the
antisymmetry property of R. Using this reasoning, we can prove by induction that
the digraph of a partial order can never contain a closed path except for loops at
individual vertices. See Exercise 6.
ne
a
Figure 3.5.1
192
CHAPTER 3
Relations and Partitions
DEFINITION
Let R be a partial ordering on a set A, and let a, beEA
with a ~ b. Then a is an immediate predecessor of b if a R b and there does
not exist
c € A such thata#c,b#c,aRc,andcRb.
In other words, a is an immediate predecessor of b when aRb and-no other element
lies “between” a and b.
Example.
For A = {1,2,3,4,5}, P(A) is partially ordered by the set inclusion relation C. For the set {2, 3,5}, there are three immediate predecessors in
P(A): {2,3}, {2,5}, and {3,5}. The empty set has no immediate predecessor.
Also, @ is the only immediate predecessor for {3}. We have {4} C {2, 4, 5},
but {4} is not an immediate predecessor of {2, 4, 5} because {4} C {4,5}, and
igie Sk aly
Let M =
{1, 2, 3,5, 6, 10, 15, 30} be the set of all positive divisors of 30. The
relation “divides” is a partial order for M whose digraph is given in Figure 3.5.2(a).
We can simplify the digraph significantly. First, because we know that every vertex
must have a loop, we need not include them in the digraph. Also, because there are
no closed paths, we can orient the digraph so that all edges point upward; thus we
may eliminate the arrowheads, assuming that each edge has the arrowhead on the
upper end. We can also remove edges that can be recovered by transitivity. For
example, because there is an edge from 2 to 10 and another from 10 to 30, we do
not need to include the edge from 2 to 30. In other words, we need include only
those edges that relate immediate predecessors. The resulting simplified digraph,
Figure 3.5.2(b), is called a Hasse” diagram of the partial order “divides.”
(a) Digraph of “divides.”
(b) Hasse diagram for “divides.”
Figure 3.5.2
“Helmut Hasse (1898-1979) was a German mathematician who did not invent these diagrams but made
use of them in his work in algebraic number theory and field theory.
3.5
Ordering Relations
193
Example.
Let A = {1, 2,3}. The Hasse diagram for (A) partially ordered by
C is given in Figure 3.5.3. It bears a striking resemblance to Figure 3.5.2(b) for
good reason. Except for the naming of the elements in the sets, the orderings are the
same. In fact, it can be shown that every partial order is “the same” as the set inclusion relation on subsets of some set. Although we need the concepts of Chapter 4 to
make precise what we mean by “same,” Exercise 21 outlines how one might start
to show this.
(aaa)
PLP
a
Hasse diagram for C.
Figure 3.5.3
DEFINITIONS
Let R be a partial order for a set A. Let B be any subset of
A anda € A. Then
ais an upper bound for B if b Ra for every b € B.
a is a lower bound for B if a R b for every b € B.
For the subset [1, 4) of RR with the usual < order, some upper bounds are 646,
1/50, 4.0001, and 4. Some lower bounds for [1, 4) are —3, —12, 0.99, and 1.
Examples.
ForA = {1,2,3,4,5,6,7,8,9, 10}, letB = {{1,4, 5,7}, {1,4, 7, 8},
{2, 4, 7} }. The set B is a three-element subset of P(A). Using the partial order C
for P(A), we see that {1, 2, 3, 4, 5, 6, 7, 8} is an upper bound for B because
{RIADTh eCetil,
LR ARTES
Wendy
25.3, tends Os TOs
vce (bs Say, 0,17208)}
pand
G4 123.4, 5, 0,7, 3}.
Another upper bound for Bis {2, 4, 5, 7, 8, 9, 10}. Elements of P(X) that are lower
bounds forB are @, {4}, {7}, and {4, 7}.
194
CHAPTER 3
Relations and Partitions
Let R be a partial order for a set A. Let B be a subset of A
DEFINITIONS
and a € A. Then
ais a least upper bound for B (or supremum of B) if
(i) ais an upper bound for B and
(ii) a R x for every upper bound x for B.
ais a greatest lower bound for B (or infimum of B) if
(i) ais a lower bound for B and
(ii) x R a for every lower bound x for B.
We write sup(B) to denote the supremum of B and inf(B) for the infimum of B.
We shall soon see (Theorem 3.4.2) that there is at most one supremum and one
infimum for a set.
The definitions do not require that sup(B) and inf(B) be elements of the set B.
For example, with the usual < order on R, each of the three sets {1, 4}, (1, 4), and
[1, 4] has least upper bound 4 and greatest lower bound |. Some other sups and infs
for subsets of IR are
for A= {5}, sup(A) = 5 and inf(A) = 5.
forB = {1, 6, 3, 9; 12,—4, 10}, sup’) =
12 and inf(’) = —4.
for C = {2*:k © N}, sup(C)does not exist and inf(C) = 2.
for D = {2-*:keN}, sup(D) = 5 and inf(D) = 0.
Example. Figure 3.5.4 is the Hasse diagram for a partially ordered set A =
{a, b, c, d, e, f}. Both a and b are upper bounds for {a, f}, and a, b, c, d, and e are
all upper bounds for {e,
f}. For the set {a}, the lower bounds are a, e, andf, and the
set {a, b, c} has e and fas its lower bounds. The only lower bound for A in A is f.
There is no upper bound for A or for the set {b, d}. Furthermore,
sup({a,
f}) =a
sup({e, f}) =e
inf(A) = f
inf({a, d}) =e
SupCianG) \==.b
sup(A) does not exist
inf({b,d})
=e
imib,
=e
ae
NI
(2
yp
Figure 3.5.4
3.5
Ordering Relations
195
For the earlier example with the set inclusion order on (A), where the set A =
{1,2,..., 10} and B= { {1,4,5,7}, {1, 4,7, 8}, {2, 4, 7} }, the least upper bound
for Bis sup(B) = {1, 2,4, 5, 7, 8}, and the greatest lower bound for B is inf(B) =
{4, 7}. Notice that in this example sup(B) is the union of the sets in B and inf(B) is
the intersection of the sets in B. This is true in general: For any nonempty setA with
P(A) partially ordered by C, if B is a set of subsets of A, then sup(B) = (J X and
inf(B) = [ ) X. See Exercise 16.
aoe
XEB
Theorem 3.4.2
Let R be a partial order for a set A and B C A. Then if sup(B) exists, it is Pees
Also, if inf(B) exists, it is unique.
Proof. Suppose that x and y are both least upper bounds for B. (We prove that
xX = y.) Because x and y are least upper bounds, then x and y are upper bounds.
Because x is an upper bound and y is a least upper bound, we must have y R x.
Likewise, because y is an upper bound and x is a least upper bound, we must have
x Ry. From x Ry and y Rx, we conclude that x = y by antisymmetry. Thus, if it
exists, sup(B) is unique.
The proof for inf(B) is left as an exercise.
=
We have seen examples of sets B where, when they exist, the least upper and
greatest lower bounds for B are in B and other examples where they are not in B.
DEFINITION
Let R be a partial order for a set A. Let B C A. If the greatest lower bound for B exists and is an element of B, it is called the smallest
element (or least element) of B. If the least upper bound for B is in B, it is
called the largest element (or greatest element) of B.
If a partial ordering also has the comparability property, the partial ordering is
called linear.
DEFINITION A partial ordering R on A is called a linear order (or total
order) on A if for any two distinct elements x and y of A, either x R yor y Rx.
Examples. Each of N, Z, and R with the ordering < is linearly ordered. P(A) with
set inclusion, where A = {1, 2, 3}, is not a linearly ordered set because the two
elements {1,2} and {1,3} cannot be compared. Likewise, the relation “divides”
is not a linear order for N because 3 and 5 are not related (neither divides the other).
If R is a linear order on A, then by antisymmetry, if x and y are distinct elements
of A, then either x R y or y R x, but not both. The Hasse diagram for a linear ordered
set is a set of points on a vertical line.
196
CHAPTER 3
Relations and Partitions
For a given linear order on a set, it is not always true that every subset has a
smallest or largest element. The set of integers with < is linearly ordered, but the
set B = {1,3,5, 7, ...} has neither upper bounds nor a least upper bound. Likewise, {—2, —4, —8, —16, —32,...} has no greatest lower bound (and hence no
smallest element).
DEFINITION _ Let L bea linear ordering on a set A. L is a well ordering
on A if every nonempty subset B of A contains a smallest element.
In Chapter 2, we proved the Well-Ordering Principle from the Principle of
Mathematical Induction. Using the terminology of this section, the Well-Ordering
Principle says that the natural numbers are well ordered by <. The integers, Z, on
the other hand, are not well ordered by < because as we have seen { —2, —4, —8,
—16, —32,...} is anonempty subset that has no smallest element.
Finally, we state without proof a remarkable result.
Theorem 3.4.3
Well-Ordering Theorem
Every set can be well ordered.
The Well-Ordering Theorem should not be confused with the Well-Ordering Principle of Section 2.5, which is a property of the natural numbers. The theorem says that
for every nonempty set A there is always a way to define a linear ordering on the set
so that every nonempty subset of A has a least element. Even the set of real numbers,
which we know is not well ordered by the usual linear order <, has some other linear
ordering so that R is well ordered by that ordering. The proof of the Well-Ordering
Theorem requires a new property of sets, the Axiom of Choice. (See Section 5.5.)
Exercises 3.5
1.
ae
Which of these relations on the given set are antisymmetric? Which relations
have the comparability property?
(A) MAb fle 405) ian CL, SCID) (254), G, 2), (5, 4), (4 2)}.
(b)iAr St
L273 e43 Re — 11, 4), 182) (253) (3; 4), (5, 2), (4, 2), C1) 3)}:
A
(Cc)
Zo
x
(d)
(ec)
(f)
R,xRyifx <2.
IRoGR,
x Syat y=ax = 1:
A= {1,2,3,4}, Ras given in the digraph:
Roy ity
ead
©
aay)
ae
3.5
(g)
Ordering Relations
197
A= {1, 2,3, 4}, Ras given in the digraph:
| ______+ 2
4
5
Let A = {a, b, c}. Give an example of a relation on A that is
(a)
(b)
(c)
antisymmetric and symmetric.
antisymmetric, reflexive on A, and not symmetric.
antisymmetric, not reflexive on A, and not symmetric.
(d)
(e)
symmetric and not antisymmetric.
not symmetric and not antisymmetric.
(f)
irreflexive on A and not symmetric.
(g)
irreflexive on A and not antisymmetric.
(h)
(i)
antisymmetric, not reflexive on A, and not irreflexive on A.
transitive, antisymmetric, and irreflexive on A.
Let R be an antisymmetric relation on the nonempty set A. Prove that if R is
symmetric and Dom (R) = A, then R = I.
Let R be the relation on N given by a R b if there exists an integer k > 0 such
that b = 2*a. Show that R is a partial ordering. Does this relation have the
comparability property?
Define the relation R on C by (a + bi) R(c + di) if a +b?
<c? + d?.IsR
a partial order for C? Justify your answer. Does this relation have the comparability property?
Prove that except for loops at vertices, the digraph of a partial order has no
paths that begin and end at the same vertex. That is, prove that if R is a partial
order foraisetAland if ZK x, sx no
RX, Ee). x, Kx, and x, Kz, then
es Oe
This exercise demonstrates that partial orders and strict partial orders are
related by adding or removing the equality condition as a defining property of
the relation.
(a) Suppose that R is a strict partial order on a set A. Let R'’ = RU Jy. (In
other words, for all a, b € A, a R’b if and only if a R b or a = b.) Prove
that R’ is a partial order for A.
(b) Suppose that R is a partial order on a set A. Let R* = R — J,. (In other
words, for all a, b € A, a R* b if and only if aR b and a ¥D.) Prove that
R* is a strict partial order for A.
Prove that if R is a partial order on a set A, then R~' is also a partial order on A.
Let R be a partial order for the set A and S be a partial order for the set B.
(a)
Let Tbe the relation on A x B defined as follows: For all a, b € A and
x, y € B, we have (a, x) T(b, y) ifa Rb and x S y. (Tis called the product
order of R and S.) Prove that Tis a partial order on A x B.
198
CHAPTER 3
Relations and Partitions
(b)
Let R* be the companion strict partial order for R (see Exercise 7(b)).
Let L be the relation on A x B given by (a, x) L (b, y) if (i) a R* b or
(ii) a= band x S y. (Lis called the lexicographical order on A x B, as
in a dictionary.) Prove that L is a partial order on A x B.
10.
Draw the Hasse diagram for
(a)
(b)
(c)
the relation of set inclusion on the power set of {a, b, c, d}.
the relation “divides” on the set {1, 2, 6, 7, 9, 24, 27, 42, 54}.
the relation “is a subset of” on the set {{0}, {1},.{3}, {5}, {1, 6}, {3,5},
(d)
the relation “has fewer prime factors” on the set {2, 7, 21, 30, 33, 210,
{57} AG
etl Gel
IN
330, 390}.
Lt
For each Hasse diagram, list all pairs of elements in the relation on the indicated set.
(a) A = asbea}
(b)
a
b
SS
G
A =
{a,b,c,d}
©
(Cc)
A = 4a.0.G.0}
b
ve GKAL BX.
a
a
Gc
12.
Let A be a set partially ordered by the relation R and C C B CA. Prove one of
the following, and show by example that the other is false.
(a) Every upper bound for C is an upper bound for B.
(b) Every upper bound for B is an upper bound for C.
13.
For the partially ordered set A = {a, b, c, d, e, f g, h} with the Hasse diagram
below, find
(a) allupper bounds for the sets {g,h}, {e,h}, {a,h}, {c.f g}, and {b,f, g,h}.
(b) all lower bounds for the sets {g, h}, {fh}, {b, c, d}, {c,f,g}, and {a, d}.
(c) the supremum, if it exists, for the sets {g, }, {g, c}, {e, g, h}, {b.f,g},
and {c, e}.
(d) the infimum, if it exists, for the sets {b, e,f}, {e, g}, {b,c, d}, and {b,f, g}.
(e) the smallest element, if it exists, for the sets {b, c, f}, {e, g}, {b,c d},
and {b, f, g}.
wy
*
Osean,
va
14.
A
Let A be a nonempty set, and let (A) be partially ordered by set inclusion.
Show that
(a) if Be P(A) and x € B, then B — {x} is an immediate predecessor of B.
(b) if B © P(A) and x ¢ B, then B is an immediate predecessor of BU {x}.
3.5
15.
16.
199
Let R be the rectangle shown here, including the edges.
Let H be the set of all rectangles whose sides have positive
length, are parallel to the sides of R, and lie within R. H is
partially ordered by set inclusion.
(a) Does every subset of H have an upper bound? a least upper bound?
(b)
Does every subset of H have a largest element?
(c)
(d)
Does every subset of H have a lower bound?
Does every subset of H have a smallest element?
Let A be a set and C be the ordering for P(A).
(a) Let Cand D be subsets ofA. Prove that the least upper bound of {C,D}
is CU D and the greatest lower bound of {C, D} is CN D.
(b)
Let % be a family of subsets of A. Prove that the least upper bound of B
is U X and the greatest lower bound of B is { ) X.
XER
17.
Ordering Relations
XER
Which are linear orders on N’? Prove your answers.
(a)
(b)
(c)
(d)
T,wheremTnifm < 2n
V, where m Vn if
mis odd and nis even, or
m and n are even and m < n, or
mand n are odd and m <n
S= {(m,n):m,neN,m < nandm#5} U {(m,5):meEeN}
W=
{(m,n):m,neéN,m
< nandn #5} U {(5,m):meEN}
18.
Prove that the relation V in Exercise 17(b) is a well ordering.
19.
To prove that a given relation is a well ordering, we may show that it has the
three properties of a partial order and the two additional properties that make
it a well ordering. There are other ways to describe a well ordering.
(a) Prove that a partial order R on a set A is a well ordering if and only if
every nonempty subset of A has a smallest element.
(b) Prove that a relation R on a set A is a well ordering if and only if every
nonempty subset B of A contains a unique element that is R-related to
every element of B.
20.
Prove that every subset of a well-ordered set is well ordered.
21.
This exercise provides the steps necessary to prove that every partial ordering is in a sense the same as the set inclusion relation on a collection of
subsets ofa set. Let A be a set with a partial order R. For each a € A, let
S,= {xeA:xRa}.Let#
= {S,:a eA}. Then ¥ is a subset of P(A) and
thus may be partially ordered by C.
(a)
Show thatif aR b, then S, C S,.
(b)
(c)
Show that if S, C S,, then a R b.
Show that for every b € A, an immediate predecessor of b in A corresponds to an immediate predecessor of S,, in #.
Show that if B CA and x is the least upper bound for B, then S, is the
least upper bound for {S,:b € B}.
(d)
200
CHAPTER 3
Proofs to Grade
Relations and Partitions
22.
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
(a) Claim. Let S be the relation on R x R defined by (x, y) S (a, b) if
either x < a or y < b. The relation S is antisymmetric.
Suppose that (x, y) S (a, b) and (a, b) S (x, y). Then x < a or
“Proof.”
y < band a < x orb < y. Therefore, x = a and y= J, so (x, y) = (a, b).
(b)
a
Thus S$ is antisymmetric.
Let A be a set with partial order R, and let B and C be subsets
Claim.
of A. If B has an upper bound and C has an upper bound, then B U C has
an upper bound.
“Proof.”
Suppose that r € A is an upper bound for B and s € C is an
upper bound for C. Let t be the larger of r and s. Then x R t for every
element of B, and x R t for every element of C, so x R t for every element
rT]
of B U C. Thus fis an upper bound for B U C.
(c)
Claim.
Let < be the usual order relation on R, and let B and C be sub-
sets of IR. If B has an upper bound and C has an upper bound, then B U C
has an upper bound.
“Proof.” Suppose that r € R is an upper bound for B and s € R is an
upper bound for C. Let t be the larger of r and s. Then x < r for every element of B, and x < t for every element of C, so x < t for every element
(d)
cf
of B UC. Thus fis an upper bound for B U C.
Let A be a nonempty subset of R, s = sup(A), and B = {—x:
Claim.
x € A}. Then B is bounded below, and —s = inf(B).
“Proof.”
(i) Suppose that x € B. Then —x € A. Because s = sup(A),
s 18 an upper bound for A, so —x < s. Then —s < x. Therefore, —s is a
lower bound for B, and B is bounded below. (ii) Suppose that ¢ is a lower
bound for B. Then for every w € B, t < w. For every x in A, —x is in B,
so t < —x. Thus, x < —t. This means that —f is an upper bound for A.
By definition of supremum, s is less than or equal to every upper bound
for A, so s < —t. Therefore,
t < —s. We have seen that for every lower
bound ¢ for B, —s is greater than or equal to t. Thus —s = inf(B).
x
x
(e)
(f)
Let A be a set with partial order <. If
Claim.
exists, then sup(B) exists.
r
CC B C A and sup(C)
“Proof.”
Suppose that sup(C) exists. If sup(C) € B, then sup(C) =
sup(B). Otherwise, there must be an element b of B such that sup(C) < b.
B
Then b = sup(B).
Let A be a set with a partial order R. If B C A, u is an upper
Claim.
bound for B, and u € B, then sup(B) exists, and u = sup(B).
“Proof.”
(g)
Because u € B, u R sup(B). Because u is an upper bound,
sup(B) Ru. Thus, u = sup(B).
B
Claim.
Let A be a set with a partial order R. If B C A, u is an upper
bound for B, and u € B, then sup(B) exists, and u = sup(B).
“Proof.” We are given that u is an upper bound for B. Suppose that t is
an upper bound for B. Then u R t because u is in B. Thus, u satisfies both
3.5
Ordering Relations
201
conditions to be the supremum of B, so u = sup(B). Therefore, sup(B)
*
(h)
exists.
Claim.
LetA bea set with partial order <. If
and C is nonempty, then sup(C) exists.
a
CC B CA, sup(B) exists,
“Proof.”
Suppose that sup(B) exists. Let b = sup(B), and let c € C.
Then b is an upper bound for B and c € B, soc < b. If there is noc’ in C
such that c < c’, then c = sup(C). Otherwise, for some natural number
k there must be elements c,, cy, ¢3, ... , and c,in C such that c, < c, <
C3 <...< (<b and no ¢,,, in C such that c, < c,,;. Then c,,, =
sup(C).
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7
CHAPTER 4
Functions
Functions are the main objects of study in most branches of mathematics. Although
the concept of a function is very old, the word function was not explicitly used until
the end of the 17th century by G. W. Leibnitz.”
In this chapter, our goals are to develop a deeper understanding of methods
of constructing functions and properties of functions such as one-to-one and onto,
while developing skills with proofs involving functions.
4.1
Functions as Relations
In many of your previous courses, and in our Appendix, a function f is described as
a rule of correspondence where, for example, we might simply say “the function
fix) = x’.” Given an object x, a number, the value of the function associated with x
is another number called f(x), which in this case is x*. No ambiguity is allowed—to
each x must be assigned only one f(x), making the function “singled valued.”
The rule “f(x) = x?” drives a process for forming a collection of ordered pairs
that contains (—2, 4), (10, 100), ..., because f(—2) = 4, f110) = 100, etc. We can
turn things around and say that the set of ordered pairs describes the function: The
pairs (3, 9) and (—1, 1) in the set imply the correspondence includes f(3) = 9 and
f(—1) = 1. A collection of ordered pairs is a relation, so we can say a function is a
relation from one set to another.
However, not every relation can serve as a function. The relation S$ = {(1, 6),
(2, 7), (1, 8)} cannot describe a function because there is more than one object
paired with 1. A function must be a relation with additional special properties. It
has become standard practice to treat a function as we define it below.
* Gottfried Wilhelm Leibnitz (1646-1716) was a German scholar, lawyer, and diplomat who made
major contributions to mathematics, philosophy, logic, technology, and physics. Working independently, both he and Isaac Newton developed calculus. Leibnitz devised the now standard a and | f(x)dx
notations. His development of the binary number system is the basis of modern computing devices.
203
204
CHAPTER 4
Functions
A function (or mapping) from A to B is a relation f
DEFINITIONS
from A to B such that
(i)
(ii)
the domain of fis A and
if (@, y) € f and (x, z) € f, then y = z.
We write f: A — B, and this is read “f is a function from A to B,” or
“f maps A to B.” The set B is called the codomain of f. In the case where
B =A, we say fis a function on A.
When (x, y) € f,we write y = f(x). We say that y is the image of f at x (or
value of f at x) and that x is a pre-image of y.
Condition (i) of the definition ensures that every element of A is a first coordinate of some pair. It is condition (11) that makes fa single-valued correspondence.
This condition enables us to refer to the image of f at x, or the value of fat x, and to
unambiguously denote this unique object as f(x).
Examples.
Let A =
{1, 2,3} and B =
{4,5, 6}. All ofthe sets
aI (1, 4), (2, 5), (3, 6), (2, 6)}
R={ C1, 4), (2, 6), (3, 5)}
II (1, 5), 2, 5), G, 4)}
Sa
Wa (1, 4), (3, 6)}
are relations from A to B. Because (2, 5) and (2, 6) are distinct ordered pairs with
the same first coordinate, the relation P is not a function from A to B. Relations R
and S satisfy conditions (i) and (11) and are functions from A to B. Thus, we can say,
for example, that R(2) = 6 or, equivalently, 6 is the image of 2. For S, the numbers
1 and 2 are pre-images of 5 because S(1) = 5 and S(2) = 5. The domain of T is the
set {1, 3}, which is not equal to A, so the relation T is not a function from A to B.
However, T is a function from {1, 3} to B.
The codomain B for a function f: A > B is the set of all objects available for
use as images (second coordinates), As with any relation, the range of f is
Rng(f) =
{b © B: there is a € A such that (a, b) € f},
which is the set of objects that actually appear as images.
In the examples above, Rng(R) = {4, 5, 6} = B, but Rng(S) = {4, 5} #B.
Although we say that S is a function from A to B, we could also say S$ is a function
from A to {4, 5}, or from A to {V3, z, 4, 5, 8}, or from A to any set that contains
both 4 and 5. A function has only one domain and one range but many possible
codomains.
4.1
Example.
Functions as Relations
205
Prove that g = {(x, y): x € N and x + y = 5} isa function from N to Z.
Proof.
(i)
- (ii)
Suppose that x is a natural number. Then 5 — x is an integer, and (x, 5 — x) € g,
so g is arelation from N to Z with domain N.
Suppose that (x, y) and (x, z) € g. Thenx + y=5=x+z,soy=z.
From (i) and (ii), we conclude that g is a function from N to Z.
a
Example.
Any set of ordered pairs that satisfies condition (ii) will be a function if
you make judicious choices for the sets A and B. For example, g = {(a, 6), (b, 4),
(c, 6), (d, 8)} is a function from A to B if we specify A to be {a, b, c, d} and let B be
any set that includes Rng(g) = {4, 6, 8}.
Functions whose domains and codomains are subsets of R are often referred
to as_real functions. The domain of a real function is usually understood to be the
largest possible subset of R. Thus, the domain of f(x) = x’ is R, while the domain
of g(x) = Vx — 2 is [2, co). The statement “f is defined on the interval I’ means
that 7 C Dom (f).
To prove that a given relation r from A to B is not a function from A to B, we
may either (i) show that some element of A is not a first coordinate (that is, some
element of A is not in Dom(r)) or (ii) find some element x of A that is a first coordinate with two different second coordinates.
The relation
H = {(x, y)<€R x R:x* + y? = 25} with domain [—5, 5] is
not a function from [—5, 5] to R because, for example, (3, 4) € H and (3, —4) € H.
See the graph in Figure 4.1.1. Because we have the graph to view, an easy way to
tell that H is not a function is to apply the Vertical Line Test for a relation r on R:
r is a function if and only if no vertical line intersects its graph more than once.
Adding the vertical line x = 3 to the graph helps us discover that H is not a
function because (3, 4) and (3, —4) are both in H.
> X
Figure 4.1.1
206
CHAPTER 4
Functions
If we apply the Vertical Line Test to the graph shown in Figure 4.1.2, the
relation appears to be a function. However, this observation does not constitute a
proof because only a portion of the graph is shown and our representation might not
reveal small vertical segments of the graph.
>~X
Figure 4.1.2
Arrow diagrams for relations with small finite domains may be used to determine whether the relation is a function. The diagram in Figure 4.1.3(a) represents a
function with domain {a, b, c}, but the diagram in Figure 4.1.3(b) does not.
(a)
(b)
Figure 4.1.3
Let h: {—2, 3} {4, 9} by given by h(x) = x’, and let k: {—2,3} > R be given
by k(x) =x + 6. Although h and k have different rules and different codomains, they
have the same domain {—2, 3}, and both are equal to {(—2, 4), (3, 9)}. Because h
and k are the same sets of ordered-pairs, we say that / and k are equal functions.
You should be aware that there are some situations in mathematics where two functions are equal as sets but are thought of as different functions because they have
different codomains. We have adopted the view that a function may have many
different codomains.
Because functionsf and g are equal if and only if they are equal as sets, we
may prove that f= g by showing f C g and g C f. Another very natural and useful
way to express the idea that two functions are equal is to assert that they have the
same domain (so they act on the same objects) and that for each object in the common domain the function images are the same.
4.1
Theorem 4.1.1
Functions as Relations
207
Two functionsfand g are equal if and only if
(i)
(ii)
Dom(f) = Dom(g) and
for allx € Dom(f), f(x) = g(x).
Proof.
(We prove that conditions (1) and (ii) hold when f = g. The converse is
left as Exercise 7.) Assume that f = g.
(i)
Suppose that
xe Dom(f). Then (x,y) ef for some y, and because
f =g, we have (x,y)€g. Therefore
x€Dom(g). This shows that
Dom(f)
(ii)
€ Dom(g).
Similar
reasoning
shows
that
Dom(g)
C Dom(f).
Therefore Dom(g) = Dom(f).
Suppose that x © Dom(f). Then for some y, we have (x, y) € f. Because
j= 2, &y) €.2, Therefore, | (x) = v= 2a).
i”
Examples.
Suppose that f, g, and / are real functions given by f(x) = : Eo
es
landi(x) = at Then f# g because they have different domains: The number 0
is in Dom(g) but not in Dom(f). The functions
f and h are different because they
have different function values. For instance, f(—2) =
1 and h(—2) = —1.
oO
The remainder of this section describes some functions that find uses in many
branches of mathematics.
Let A be any set. The identity relation /, is the identity function /,: A > A
given by /,(x) = x. IfA is asubset of B, we define the inclusion function 7: A > B
by i(x) = x for all x € A. Because both have domain A and J,(x) = i(x) = x for
all x € A, 7, = i by Theorem 4.1.1. There is no difference between these functions, but this is a case where
equal functions
with different codomains
are
called by different names.
For the next function, assume that a universe U has been specified. For A C U,
the characteristic function of A is the function X,: U >
{0, 1} given by
| ifxeA
AOS i ifxeU—A
For example,
if A =
[1,4),
with
the universe
being
the real numbers,
X,(x) = 1 if and only if 1 < x < 4. Figure 4.1.4 is a graph of X;,_4).
y
A
2
X14)
os
Ji
=)
Se
ee
= 2 Sil
ee
Me
—]|
—
Figure 4.1.4
ee
Sy
eee
TAS
then
208
CHAPTER 4
Functions
The greatest integer function is an example of a function with domain
R and range Z. It assigns to each real number x the integer part of x, by which
we mean the largest integer n such that n < x. On graphing calculators, this function is usually denoted as “int.” For instance, int(5.9) = 5, int(V2) = 1, and
int(—z) = —4. Some authors refer to this function as the “floor function” and use
the notation |x| for the greatest integer of x.
DEFINITION
A function x with domain N is called an infinite
sequence, or simply a sequence. The image of n is usually written as x,
instead of x(n) and is called the nth term of the sequence.
5
l
For sequence x given
by x, = Tees
the 63rd term is x63 = me The range of
xis {Libis
a
k
tS
aie
}
Ye
The terms of a sequence need not be distinct. The first 10 terms of the
Sequence x, given by x, =((— Ly" a 1)n*. are, 0,8} .0,.32..0, 72,.0,
128, 0, and
200.
If R is an equivalence relation on the set X, then the function from X to X/R that
maps each a € X to the equivalence class of a is called the canonical map for the
relation R. If fis the canonical map for the relation of congruence modulo 5 on Z,
the images of —3 and 9 are
=2=...,—8,—3, 2,7, 12,...} and
fOr
0 = 4S (es =6) 1740 142.
The canonical map is a natural function to consider, and it plays an essential role in
the development of many mathematical structures.
Rules of correspondence between sets of equivalence classes have interesting
properties. Consider, for example, the classes 0, 1, 2, and 3 of Z, and the rule that
x in Z, corresponds to the equivalence class [2x] in Z,,. (For clarity, we use here
the bar notation for the equivalence classes of Z, and the bracket notation for the
equivalence classes of Z;,9.) Under this rule,
fO) = [0], f@) = [2], f@ = [4], FB) = [6],
However, 0 = 4, but f(0) # f(0). Thus fis not a function.
When a rule of correspondence assigns more than one value to an object in the
domain, we say “the function is not well defined,” meaning that it is not really a
function. A proof that a function is well defined is nothing more than a proof that
the relation defined by a given rule is single valued.
4.1
Example.
Functions as Relations
209
Consider the functionf:Z, = Z, given by f(x) = [x + 2]. For exam-
ple,
f sends 0 to [2], 1 to [3], 2 to [0], 3 to [5] = [1], etc. To prove that
f is well
defined, we must show that if x = z, then [x + 2] = [z+ 2].
Proof. Suppose that x = zin Z,. Then x = z (mod 8), so 8 divides x — z. Then 4
divides x — z, so 4 divides (x + 2) — (z + 2). Then x + 2 =z+2 (mod 4). Thereforest f@)u= ilxi2) = [zr] =f @).an Za
a
Exercises 4.1
1.
Which of the following relations are functions? For those relations that are
functions, give the domain and two sets that could be a codomain.
*
(a)
{(0,
ES)s (A,
LI), (ay
UE
CAs
U),
(Oy
0)}
(D) TCs 2) F305 Gi 4105),
CL Gt
(Cyl 25)
(d)
{(@,y)ER x R:x=siny}
(e—)
{@yeENxNix<y}
(f): of @, y) SZ ecZay?
= x}
(g) {@,y)eN
x N:x*+y? <5}
(h) {{(,y)eNxN:2*=4}
(i)
(j)
See
ee SS
2.
il
Give two reasons why the “rule” f(x) = + Vx does not define a function from
R to R.
3.
Identify the domain, range, and another possible codomain for each of the
following mappings. Assume that the domain is the largest possible subset of R.
*
(a)
{ae
(b)
{~yeERx
(c)
{@ y)ER
(dd)
{@yeER x Ry = Xy@}
(e)
{Or eR x Ry
*
RxR
1
y=}
Ry=x4+5}
x R: y=
() 4@»eZx Zy=
tanx}
te
2
x— 4
210
CHAPTER 4
Functions
4.
Assuming that the domain of each of the following functions is the largest
possible subset of IR, find the domain and range of
eae
al
|
(a)
f@)=
MTs
(b)
f=
rar
()
f@M=V5—x4+Ve—-3.
@
f=
Vx4+24+ V-2—x.
(a)
(b)
Prove that the empty set @ is a function with domain ©.
Prove that if f: A > B and any one of f,A, or Rng(f) is empty, then all
three are empty.
(a)
Let A be the set {1, 2, 3, 4} and let R be the relation on A given by
(b)
function with domain A.
Let A be the set {1, 2,3}, and let R be the relation on A given by
{(x, y): 3x + y 1s prime}. Prove that R is a function with domain A.
{(x, y) € A x A: 2x + y is prime and not equal to 5}. Prove that R is a
(c)
LetR={(%, y) eZ x Z:x° + y= 2}. Prove that R is a function with
domain Z.
(d)
Let R= {(x,y)
domain N.
EN x N: 2x*— y = 1}. Prove that R is a function with
Complete the proof of Theorem 4.1.1. That is, prove that if (1) Dom(f)
=
Dom (g) and (11) for all x € Dom(f), f(x) = g(a), then f = g.
4
Explain why the functions f(x) =
Xe
3
and g(x) = 3 — x are not equal.
Let the universe be R, and A = [1, 3). Find
(b)
(d)
(a) X,(1).
(c) X4(z).
10.
ee WE,
(x)
acne W2X (==
=}
21
(bye 0: Xa) =0}
(d)
Maeouex (x) = 1}
Prove that the given functions are equal. In parts (c) and (d), A and B are subsets of a set S.
(a) f(x) =In@ — x) —Inx
2x) =n : = =
,
12.
X02):
Let U be the universe. Suppose A C U with A # ©, and A ¥ U. Let X, be the
characteristic function of A. Find
CA)
(C)
11.
xe)
$2.0)
0
ifx EZ
b fix) == int(x)
int(x) +
int(—x
(b)
+ int(—x)
x)= fe
g(x)
(Ce)
2) avn)
(d)
g(x) = X4(x) + Xp@) — X40) - X p(X)
fix) =X,yup®
ifr eZ
ee)
Find the Ist, 5th, and 10th terms of each sequence.
(a) x,=2+2
(b) x,=n+(-1y
(Cae n = sin
(d)
to n Sen
4.1
13.
14.
Functions as Relations
211
For the canonical map f: Z > Z,, find
(a)
(3).
p
(b)
the image of6.
(c)
a pre-image of 3.
(d)
all pre-images of 1.
By naming an equivalence class in the domain that is assigned at least two
different values, prove that the following are not well-defined functions.
15:
16.
(a)
f:Z,— Z, given by f(x) = [x]
(b)
f:Z,— Z, given by f(x) = [2x + 1]
(c)
f:Z,— Z, given by f(x) = [x]
(d)
f:Z,— Z, given by f(x) = [x + 4]
(ee
(52,
@)i= (v2)
Prove that the following functions are well defined:
(a) the functionf:Z, > Z,. given by f@&) = [3x]
(b)
the functionfiZ,,5 > Z,9 given by f(x) = [7x]
(c)
the functionf Z,, > Z, given by f(@%) = [x —
(d)
the functionf:Z,, > Z, given by f(x) = [2x + 1]
(e)
the functionfiZ, > Z, given by f(x) = x7 + 3
1]
Let S be a relation from A to B. We define two projection functions
mw,: S—> A and m,: S > B as follows: For all (a, b) in S, 1,(a, b) = a and
,(a, b) = b. In terms of S, find
(a)
(b)
17.
Ze siven, by
Rng(z)).
Rng(z,).
Suppose that set A has m elements and set B has n elements. We have seen
that A x B has mn elements and that there are 2’””” relations from A to B. Find
the number of relations from A to B that are
(a) functions from A to B.
(b) functions with one element in the domain.
(c) functions with two elements in the domain.
(d) functions whose domain is a subset ofA.
18.
(a)
(b)
(c)
Let
f be a function from A to B. Define the relation T on A by x T y
iff f(x) = f(y). Prove that T is an equivalence relation on A.
Inthe case when f: R > R is given by f(x) = x’, describe the equivalence class of 0; of 2; of 4.
Inthe case when f: R — R is the cosine function, describe the equiva-
lence class of 0; of 2/2; of 1/4.
Proofs to Grade
197
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
(a) Claim.
The functions fand g are equal, where f and g are given by
f= cs and g(x) =
‘ il
inks
S@
J
eGiiceny
212
CHAPTER 4 Functions
“Proof.”
Let x be a real number. If x is positive, then fe =
.
;
Om)
ae
x
x
so f(x) = g(a). If xis negative, then - >= — =
Invevery case, f
(b)
Claim.
x
1, so f(x) = gQ).
=e Ge), SOf S12:
The functions f(x) =
il,
gi
1
1 + . and g(x) =
:
are equal.
“Proof.”
The domain of each function is assumed. to be the largest
possible subset of R. Thus, Dom(f) = Dom(g) = R — {0}. For every
x
€R — {0}, we have
1
in
(c)
BY
ges mae; 1
]
x+
aw
1
ht)
Therefore, by Theorem 4.1.1, f = g.
Claim.
The relation x? = y° defines a function from R to R.
“Proof.”
The graph of the relation x7 = y° is given here:
a
Because no vertical line crosses the graph more than once, x? = y*
defines a function.
H
(d)
Claim.
The rule that assigns to each equivalence class x in Z, the class
[x + 1] in Z, is well-defined.
“Proof.”
Suppose that (x, [y]) and (x, [z]) are two ordered pairs
in the relation determined by the rule. We must show that [y] = [z].
According to the rule, [y] = [x, + 1] and [z] = [x, + 1] for some
X,,Xz in the class x. Because x, and x, are in the same equivalence
class (mod 4), x, — x, = 4k for some integer k. Then (x, + 1) —
(X_ + 1) = 4k = 2(2k), so x; + 1 and x, + 1 are in the same equiva-
lence class (mod 2). Therefore, [y] = [z].
4.2
=
Constructions of Functions
This section discusses several methods for creating new functions from given ones.
One familiar approach for real functions is to form the sum, difference, product, or
quotient of two functions. Because every function is a relation, another approach
4.2
Constructions of Functions
213
is to perform the operations of composition and inversion on functions, just as we
did in Chapter 3.
For functions f: A > B and g: B — C, the inverse of f is the relation from
BtoA:
fT ={@y): 0,» Ef},
and the composite of f and g is the relation from A to C:
g ° f= {(, z): there exists y € B such that (x, y) € fand (y, z) € g}.
We are careful to say these are relations because nothing in their definitions guarantees that either is a function.
Examples. Let fand g be real functions given by f(x) = 2x + 1 and g(x) = x’. The
domains and ranges of f and g are subsets of R. The inverse of fis
f= (6,9): 0,9 ef) = (Gy): x= 2 +
= {yy = Sh.
The inverse of fis a function, and we write f~!(x) =
The inverse of g is
go
(Gy)
Oss) © oe}
Gy)
x —
5
1
ae
In this case, g~! is a relation from R to R but not a function because it contains pairs
such as (4, 2) and (4, —2).
The composite of f and g is
g © f= {(, z): dy © B such that (, y) € fand (y, z) € g}
= {(x, z): dy € B such that y = 2x + 1 andz=y?}
= {(x, z): y= 2x+ 1 andz=(2x
+ 1)*}.
Here g © fis a function.
The examples show that sometimes the inverse of a function is a function.
Conditions on f to ensure that f ~! is a function are given in Section 4.4. The next
theorem shows that the composite of two functions is always a function.
Theorem 4.2.1
Let A, B, and C be sets, and let f, A > B and g:
B— C. Then g °f is a function
fromA to C, and Dom(g of) = A.
Proof. In Section 3.1, we proved that g°/f is a relation from A to C. To show
that g °f is a function from A to C, suppose that (x, y) € g Of and (x%,z) Ee g of.
(We must show that y = z.) Because (x,y) € g °f, there exists u € B such that
(x,u) €f and (u, y) € g. Likewise, there exists v € B such that (x,v) ef and
(v, y) € g. Because f is a function and (x, u) and (x, v) are in f, we have u = v.
Because g is a function, (u, y) and (v, z) are in g, and u = v, we have y = z.
214
CHAPTER 4 Functions
(We next show that Dom(g°f) =A.) In Section 3.1, we proved that
Dom(g © f) C Dom(f) = A. Suppose that a € A. Because A = Dom(/f), there is
b € Bsuch that (a, b) € f. Because B = Dom(zg), there is c € C such that (b, c) € g.
Then (a, c) € g Of. Therefore, a = Dom(g © f).
a
We can take advantage of the fact that each element of the domain of a function has a unique image to simplify the notation for composition. Forf: A > B and
g: B = C, instead of writing (x, y) € fand (y, z) € g, we write y = f(x) and z = g(y).
Then the composite is z = g( f(x)). In other words, for all x € A,
(Corea)
eee)
Notice that the first function applied in composition is the function on the right,
which is closer to the variable x. As shown in Figure 4.2.1, fis function applied first
to evaluate (g © f) (x).
nee
B
8
OD
Figure 4.2.1
Examples.
Letf,g, and A be real functions given by f(x) = sin x, g(x) = x + 6,
andiCoi—e™
Then
(g ° f) @) =8(f Od) = g(sin x) = (sin x) + 6 = sin’ x + 6,
(f ° g)@)=f(g (x) =f? + 6) = sin (? + 6),
(g © h) (x) = g(h (x) = g(e*) = (e*) + 6 = e* + 6, and
(h © g) (x) =h(g (x) =A? + 6) = er,
a
Composition of functions is not commutative. The examples above show that
g ° fandf © gare not always equal.
and ZO)
L112 iit
Example. Leth: Z, > Z, and k: Z, > Z; be given by h(0) = [0], h(1) = [1], A(2)
= [2], AGB) = [0], AA) = [1], hS) = [2]; and k(LO}) = [0], (1) = [2], A((2]) = C1).
Then
(k° h)(O)0 = kh)0 = K([0]) = [0]
(ko h)(2) = kh) = K((2]) = [1]
(ko hy(4) = kh) = K((1]) = [2]
(ko hy(1) = KA) = KEN) = [2]
(ko hY(3) = k(h@)) = K({O}) = [0]
(ko hy(5) = k(A(5)) = K(E21) = [1]
O
4.2
Constructions of Functions
215
Care must also be taken when determining the domain of g°f. We saw in
Chapter 3 that Dom (g © f) is a subset of Dom (f ). If it happens that Rng(f) is not
a subset of Dom (g), then the composite g o f(x) is undefined whenever f(x) is not
in Dom(g). For example, for the real functions f(x) = x* and g(x) = ot , we have
2 € Dom (f), but f(2) = 4 ¢ Dom (g). In this example, Dom (g ° f) = R — {2} is
not the same as Dom (f) = R.
In Chapter 3, we proved that composition of relations is associative. As a
result, composition of functions is associative. We also proved that forming the
composite of a function f with the appropriate identity function yields the same
function f. These properties are restated for functions with proofs that take advantage of functional notation.
Theorem 4.2.2
Let A, B, C, and D be sets, and let f: A > B, g: B—> C, and h: C—
D. Then
(ho gle f=hateos),
Proof.
By Theorem 4.2.1, the domain of each function is A. Now let x € A. Then
((h° g)° f(x) = (hO g) (f(x) = A(g(f@)) = Ag © f(x) = (h° (g © FQ).
By Theorem 4.1.1, the functions are equal.
a
The relationship (h © g)° f= h°(go0 f) in Theorem 4.2.2 is represented in
the diagram in Figure 4.2.2. For any x € A, by following the diagram from A to
D along the upper route, the image of x is ((h ° g) ° f)(x). Along the lower route,
the image of x is (h ° (g ° f ))(x). Theorem 4.2.2 says that these images are always
the same, and consequently the figure is called a commutative diagram. This theorem enables us to avoid the use of parentheses for composition and to simply say
ho g° f isa function from A to D and the image of x is (h © g ° f(x).
Figure 4.2.2
Theorem 4.2.3
ge
Let f:A > B. Then fol, =f andl,p° f=f.
Proof.
Dom(f°J,) = Dom (J,) = A = Dom (f). If x € A, then (f° 1,)(x) =
fU,@))
= f(). Therefore, ae
The proof that J, °f = f is left as Exercise6.
a
216
CHAPTER 4
Theorem 4.2.4
Functions
f
Let f:A— B with Rng(f) =C. If f7! is a function, then f~!°
oy
=I, and
ee
Proof. Suppose that f:A — B and f~! is a function. Then DomGs.0 7) =
Dom(f) (by Theorem 4.2.1). Thus, Dom(f~!of) = A = Dom(I,). Suppose
that x € A. From the fact that (x, f(x))
(fof
0)
fee 1)
The proof that
eee]
ef, we have (f(x), x) ¢ f~!. Therefore,
(ep ts Proves tat Meo. eal
fo f~! = Ic is left as Exercise 7.
ie
Selecting just some of the ordered pairs from a function f: A — B is another
way to create a function. If g C f, we say g is a restriction of f. The relation g will
be a function because f is single valued. We define the restriction by specifying the
new, smaller domain.
DEFINITIONS
is the function
Let f: A > B, and let D C A. The restriction of f to D
flp = {0% y): y = fe) and x € D}.
If g and h are functions and g is a restriction of h, we say h is an
extension of g.
Examples.
Let
A=
{1,2,3,4},
B=
{a,b,c,d},
and
g be
the
mapping
(C1, a), (2,4), 3, d), (4,c)}. Then g|,14) = {C1, 4), 4, 0)}, 813) = (G,d)},
and g|4 = &.
Let f: R — R be given by f(x) = 2x + 1. For comparison, the graph of f is
shown with the graphs of f|,; >, and f|;_>, _1,0, 1,2) in Figure 4.2.3.
y
y
5st
4-
5
/
3h
f|
2
BNE
(1,2]
2
Ib
SF
ee
Se
S|
Fea, -150,1,2)
1
Le
It
e
4
es x
23
—2b
=a
Figure 4.2.3
4
ae ieee
Poses
—2
Ores
|)
ee
th Gene She zh
ESS
4.2
Constructions of Functions
217
Because functions are sets (of ordered pairs), it is appropriate to ask about
unions and intersections of functions. For the functions
H = {(1, 2), (2, 6), (3, —9), (5, 7)} and
G = {(1, 8), (2, 6), (4, 8), 5, 7), (8, 3)},
HM Gis {(2, 6), (5, 7)}, which is a function. Notice that Dom(H MG) = {2,5}
is a proper subset of Dom(H) MNDom(G) = {1, 2,5}. This is because 1 € Dom(H)
and 1 © Dom(G), but H(1) # G(1).
It turns out that if / and g are functions, then / g is always a function (see
Exercise 10), but hM g can just as easily be expressed as a restriction of either
hor g.
The situation regarding h U g is much more interesting and useful. First, in general, h U g need not be a function. In the case of the functions H and G above, the union
is not a function because (1, 2) € H U Gand (1, 8) € H U G. However, if the domains
of h and g are disjoint sets, then / U g is a function. The next theorem states that we can
put together functions with disjoint domains to define a function “piecewise.”
Theorem 4.2.5
Let h and g be functions with Dom (h) = A and Dom (g) = B. If
hU gisa
Cee)
Proof.
AN B = ©, then
function with domain A U B. Furthermore,
h(x)
ifxeA
1
ifxeB
See Exercise 11.
a
Example.
Let h(x) = x’ and g(x) = 6 — x. The restrictions h|(_,,,9 and 2,2, 60)
have disjoint domains. Their union f is an extension of each (but not an extension
of h or g). See Figure 4.2.4. The function
f may be described in two pieces:
m={*
ae
it
6
= 22
Fifa S 2)
f= h\—«, 21 U gle, «)
Figure 4.2.4
218
CHAPTER 4
Functions
The absolute value function and the characteristic functions discussed in
Section 4.1 are examples of piecewise defined functions.
Functions can be constructed piecewise from three or more
functions, and
Theorem 4.2.5 may also be extended to the case where domains are not disjoint, provided that the functions agree on the intersection of the domains. See Exercise 13.
We conclude this section with examples of proofs about increasing and decreasing properties for real functions.
DEFINITIONS
Let fbe a real function defined on an interval /. Then
fis increasing on IJ if x < y implies f(x) < f(y) for all x, y in . And, fis
decreasing on J if x < y implies f(x) > f(y) for all x, y in /.
The graph of the function fin Figure 4.2.4 is increasing on the interval [0, 2]
and decreasing on the intervals (—oo, 0] and [2, oo). This seems clear from the
graph, but remember that looking at its graph does not constitute a proof because
we cannot see the entire graph or all the details of its behavior. We give two examples of proofs of these properties.
Example.
Proof.
Prove that the function f of Figure 4.2.4 is increasing on the interval [0, 2].
Let x, y € [0, 2], and suppose that x < y. Then g(x) = x? and g(y) = y’.
3
Because x < y and x > 0, x” < xy. Because x < y and y > 0, xy < y’. Therefore,
x? < y*. Thus, g(x) < g(y), so g is increasing on [0, 2].
a
Example.
Proof.
Let f(x) = 2 +
1
Prove that fis decreasing on the interval (0, 00).
]
I
Suppose that 0 < x < y. Then f(x) = 2 + 3 and f(y) = 2+ ae Because
x and y are positive and x < y, we have rae ae Thus..2.4-> = 2 +- =; that ds,
x
x
y
F(x) > f(y). Therefore,
fis decreasing on (0, 00).
a
Exercises 4.2
1.
For which of the following functions
f is the relation f~! a function? When
f—' is a function, write an explicit expression for f'(x). Use the understood
domain for each function.
* (a) f@)=Se+2
©) f@=22+4+1
* ©
fas
I
XxX
(d)
f(@)=sinx
1
@ f®=7—
* (e)
f(x) =e*t3
(fuer
ety
@
a
@ f@=-x4+3
‘
ye ae %
a
fo=—
4.2
Constructions of Functions
219
Find fo g and g © f for each pair of real functions fand g. Use the understood
domains for fand g.
(a)in
(b)
(C)
CN
f@) = 2x
5,
9G) = 67x
f@ =x? +2x,
g@)=2x+1
(Desires
$e)
sd
(eS shies
Cyan
(CET)
Ns
Ce Un 2 (x)= 1UK.S), (F5), (Ss, k)}
Cr
Or OE 32, GG, o)4, 2).(0,.2)15
g(x)= (C1, 5), @, 3), G, 7), 4, 3), 6, 4}
@ fa= > smart
(h)
f(x)= 3x4 2, g(x) = |x|
(i)
f@=
:
if
es
¢
see 3
() foy= 1S
x
5s
ee
g(x) = ee
iy ae
x ifx
> —-l
Gries
e 3)
Tile
ifx > 3’ HOE
a
ns
eee
«ah x S22,
ifx >2
Find the domain and range of each composite in Exercise 2.
Give two different examples of
(a)
a pair of functions
f and g such that (f ° g)(x) = (3x + 7).
(b)
(c)
apair of functions
fand g such that (f ° g)(x) = V 2x? — 5.
apair of functions fand g such that (f ° te = sin |2x + 4|.
Let Z,= {0, 1,2,3,4,5,6,7}
fi Ze > 2g
FR) =
Lg
BP)
and Z, = {[0], [1], [2], [3]}. Define
2 hele
Sj 2a he
and
ae
é2 —+Z,
aan
k(t
as
follows:
|e
ey,
comparing images, verify the following equalities.
(a) (koOk)(x) = [2] forall xin Z,.
(b) (g°f)(x) = h(x) for all x in Ze.
(c) (f° x2)\@) =%&G@) forallx in,Z,.
(d)
(hoO(hOA))\(x) = 4 for all xin Z..
Prove the remaining part of Theorem 4.2.3: If f: A — B, then Ip ° f =f.
Prove the remaining part of Theorem 4.2.4: If f: A > B with Rng (f) =
and it f
‘isa function, then fof” =I.
Let f(x) = 4 — 3x with domain R and A =
{1, 2, 3, 4}. Sketch or describe
the graphs of the functions f|4, fl;—1,3) Fl, 4}, and S146}.
Describe two extensions with domain R for the function f
(a)
f= {@y EN
x Niy=2'}.
(b)
f= {@ y)EN x Ny
= 3}.
(C)
faa) el ane
Sie Aly
— x
(d)
f=.
10.
Prove that iffand g are functions, then fM g is a function by showing that
le 18 la, Were Av. 1.0 20(X) = 27
(x)\.
11.
Prove Theorem 4.2.5.
220
CHAPTER 4
Functions
12.
Let f be a function with domain D, and let g be an extension of f with
domain A. Then by definition, f = g|p and D CA. Let 7 be the inclusion
mapping from D to A given by i(x) = x for all x € D. Prove that f= g ° 1.
13.
Let h: A > B and g: C > D, and suppose that E =
A/C. Prove that h U g
is a function from A U C to B U Dif and only if h|; = g\p.
14.
For each pair of functions h and g, determine whether h U g is a function. In
each case, sketch a graph of h U g.
(a) h:(—o,0] > R, A(x) = 3x + 4
g:(0,00) > R, g@) =4
(b) h:[-1,c0)>R,
(c)
(d)
¢: (2,00)
(e)
15.
h@) =x? +1
g:(—o0o, -I] ~ R, ga) =x+3
h:(—o, 1] ~ R, h@) = |x|
g: [0, co) > R, g@) =3 — |x—-—3|
h:(—oo,2] > R, hi) = cos x
= R, eG) =x
h:(—00,3) —~ R, hw) =3-x
g: (0,00) > R, ga) =x+1
Let f:
A— B and g: C—
D. Define f x g = {(a,c), (b, d): (a, b) € fand
(c,d) € g}.
16.
Proofs to Grade
(a)
Prove that f x gisamapping from A x Cto B x D.
(b)
For (a,c)e€A
x C, write (f x g)(a, c) in terms of f(a) and g(c).
Prove each of these statements.
(a) fis increasing on R, where f(x) = 3x — 7.
(b) gis decreasing on R, where g(x) = 2 — 5x.
(c)
his increasing on [0, 00), where h(x) = x°.
(d)
fis increasing on (—3, 00), where f(x) = jae
x+3
17.
Prove or give a counterexample:
(a) Iffis a linear function with positive slope, fis increasing on R.
(b) Iffand g are decreasing functions on an interval / and f © g is defined on
I, then f° g is decreasing on J.
(c) Iffand g are decreasing functions on an interval J and f° g is defined on
I, then f° g is increasing on J.
(d) IfDom(f) = R and fis increasing on the intervals [—2, —1] and [1, 2],
then fis increasing on [—2, 2].
(e) If f is decreasing on (—oo,0) and decreasing on [0, 00), then f is
decreasing on R.
18.
Let fand g be real functions. Prove that if fand g are increasing on an interval
I, then the pointwise sum f+ g , given by (f+ g)(x) = f(x) + g(x), is increasing on J.
19.
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
4.2
*
(a)
Constructions of Functions
221
Claim. Let f: A > B. If f~' is a function, then f-! 0 f= Ij.
“Proof.” Suppose that (x,y) ¢ f~!of, Then there is z such that
(x,z)ef and (z,y)ef—!. But this means that (z,x)ef—! and
(z, y) € f |. Because f—! is a function, x = y. Hence, (x, y) € f~! of
implies (x, y) € I,; that is, f~!o f © I,. Now suppose that (x, y) € Jy.
Because
A = Dom(f),
there is w © B such that (x, w) € f. Hence,
(w, x) € f—!. But (x, y) eI, implies x = y, so (w, y) € f~!. But from
(x,w)ef and (w,y)ef-', we have (x,y)ef~!of,
I, < f7' of. Therefore, 1, = f-!9f.
(b)
Claim.
If fand f—! are functions on A and fof = f, then f = J.
“Proof.”
Suppose that f: A > A and f~!: A > A. Because f = f of,
f
lof = f—!o(fof). Byassociativity,
wehave
This gives I, = 1,0f. Because
La,
(c)
f-!of= (f~!of) of.
I,of =f, by cancellation
we have
|
Claim.
If f,g, and f~! are functions on A, then g = f~!o(gof).
“Proof.”
Using associativity and Theorems 4.2.3 and 4.2.4,
fTHO(C Of) = fmno( foe) = (fu oO} op
(d)
This shows
a
Claim.
necro
Sy
If f’(x) > 0 on an open interval (a, b), then fis increasing on
(a, b).
“Proof.”
Assume that f’(x) > 0 on the interval (a, b). Suppose that
x, and x, are in (a, b) and x; < x. We must show that f(x,) < f(x»).
We know from calculus that because fis differentiable on (a, b), it is
continuous on [x,, x,] C (a, b) and differentiable on (x,, x). By the
Mean Value Theorem, there exists c in (x, x) such that
iD)
2
(e)
= fc).
Therefore, f(x.) — f(x,) =f'(c)(%, — x,). By hypothesis, f’(c) > 0
and'x, =x, = 0 because, "=< x5.- Therefore, f(x,) — f(%,) = 0. We
conclude that f(x,) < f(x).
|
Claim.
If h: A — Bandg:C > D,thenhUg:AUC—>
BUD.
“Proof.”
Suppose (x, y)&hUg and (x,z)ehUg. Then % y) Eh
or (x, y) € g, and (x, z) Gh or (x, z) Eg. If (, y) EA and (@, z) Eh,
then
(f)
i oa)
= Si
y =z.
Otherwise,
(x, y)e@g
and
(,z)€g;
so
again
y = z.
Therefore,
Ug is a function. Thus, we have that Dom(hU g) =
Dom(h) U Dom(g) =A U CysohU gs AUC
BUD.
a
Claim.
If fand g are real functions that are increasing on an interval /,
then f © g is increasing on /.
“Proof.” Assume that f and g are increasing on /, and suppose that
x and y are in J and x < y. Then f(x) < fi) and g(x) < g(y). Then
(f 2 9) @) =f)
° gx) <fO) °
f © gis increasing on /.
80) =(F © 8) 0). Therefore,
=
222
4.3
CHAPTER 4
Functions
Functions That Are Onto; One-to-One Functions
The definition of a function f: A ~ B is stated in terms of conditions on the first
coordinates of f, Two important properties of functions are defined by requiring
additional conditions on the second coordinates of f. We begin with the condition
that every element of B must be used at least once as a second coordinate—that 1s,
that every element of the codomain is in fact an image under the function.
DEFINITION
A function f: A > B is onto B (or is a surjection) if
onto
Rng(f) = B. When fis a surjection, we write f: A —> B.
From our discussion in Section 4.1, we know that the functions
f:N—
R, where f(n) = 2n and
g: N > E*, where g(n) = 2n
are equal because they are the same sets of ordered pairs. The range of this function is the set E+ of even natural numbers. This range is the same as the given
codoniain for g, so we say g maps onto E*. It would be incorrect to say f: N meas
Properly speaking, when we say a function
fis onto, we should finish the sentence by saying what set it is that
f maps onto. However, when the codomain of a
surjection f is clear from the context, it is common practice, even if not perfectly
correct, to say simply that “fis onto.” If we are given a function f and we want to
say that fis onto, what we must do first is determine the range of f.We can then say
fis onto that range.
Because Rng (f) C B is always true, f: A > B is a surjection if and only if
B € Rng(f)—that is, if and only if every b € B has a pre-image. Therefore, to
prove thatf: A—B is onto B, it is sufficient to show that for every b € B there
exists a € A such that f(a) = b. The scratch work for such a proof usually proceeds
by working backward from the desired image to find a pre-image. To prove that f
is not onto B, it suffices to find at least one element of B that is not an image under
the function.
Example.
Prove thatf: R > R, where f(x) = x + 2, is onto R.
Proof. Let we R. We choose t = w — 2. (Because we want w = f(t) = t + 2,
the pre-image we need is
t=w— 2.) Then f(t)=t+2=w. Therefore,
f: R = Ris onto R.
a
Example.
Let g: R > R be defined by g(x) = x? + 1. Then g is not onto R. To
show this, we find an element y in the codomain R that has no pre-image in the
4.3
Functions That Are Onto; One-to-One Functions
223
domain R. Let y be —2. Because x? + 1 > 1 for every real number x, there is no
x € R such that g(x) = —2. Hence, g is not onto R.
When fis a real function, it is often helpful to look at the graph of f and apply
this Horizontal Line Test for onto functions:
f maps onto B if and only if for every b € B, the horizontal
line y = b intersects the graph of fA visual check is not the same as a proof, but it can help us decide whether to attempt
a proof that fis onto or else to identify an element of B that is not in the range.
Examples.
Figure 4.3.1 shows the graphs of two functions, h and k, from [1, 3] to
[1, 4]. The function h can be shown to be onto [1, 4] because every horizontal line
with y-intercept b, where | < b < 4, intersects the graph. The function k is not onto
[1, 4]. The line y = 1.5 does not intersect the graph.
)
y
A
Aw
y = h(x)
4
5
2 |=
ei
i
DSacesita
a
wlea a
Lp
ees
5
y = K(x)
[Le et
a5
ors
(a)
(b)
Figure 4.3.1
Example.
onto R.
Let p: K — R be a polynomial function of odd degree. Prove that p is
Proof. Let
we R. The equation p(x) — w = 0 is a polynomial equation of odd
degree with real coefficients. By the Complex Root Theorem, the nonreal roots of
p(x) — w = 0 occur in conjugate pairs, so the equation has at least one real root
a € R. Thus p(a) — w = 0. Therefore p(a) = w. Hencep is onto R.
a
Example.
The mapping M: Z x Z— Z, where M(x, y) = xy, is a surjection. For
any
zEZ, (z,1)€Z x Z and Mz, 1) =z-1 =z. Even though some integers
have many pre-images (for example, 24 = M(3, 8) = M(12, 2) = M(4, 6)), to
prove that M maps onto Z, we need to show only that there is at least one pre-image
for each z € Z.
224
CHAPTER 4
Functions
When you prove that a given function f maps A onto B by showing that every
element of B has a pre-image, be sure to verify that each pre-image is in the
domain A. This very important step is the statement “Then x € (—oo, 0]” in the
following example.
Example.
Let s: (—00, 0] + [—4, 00) be defined by s(x) = x? — 4. Prove that
sis onto [—4, oo).
Proof.
Let we [—4, oo). Thenw > —4,sow + 4 > 0. Choose x= —Vw + 4.
(Note that we do not choose x = Vw + 4.) Then x € (—ox, 0]. It follows that
s(x) = (-Vw
+ 4)? —-4=(W+4)-4=w.
Therefore the function f maps onto [—4, oo).
a
The next two theorems relate composition and the property of being a surjection.
Theorem 4.3.1 says the composite of surjective functions is a surjection. Theorem
4.3.2 says that if a composite is a surjection, then the second function applied in the
composition is also a surjection.
Theorem 4.3.1
Iff:AB
Proof.
Theorem 4.3.2
is onto B and g: BC
is onto C, then g © f is onto C.
See Exercise 5.
Letf:A>B and g: BC.
w
If g °f is onto C, then g is onto C.
Proof.
Suppose that c € C. Because g ° f maps onto C, there is a € A such that
(g OF \@) =e Letb = f(a), whichis inB.Then (g ° f)(a) = g(f(a)) = g(b) =e.
Thus, there is b € B such that g(b) = c, and g maps onto C.
a
For a mapping f: A — B, we said that fis onto B if every element of B is used
at least once as a second coordinate. If fhas the property that every element of B is
used at most once as a second coordinate, we say f is one-to-one.
DEFINITION _ A function f: A > B is one-to-one (or is an injection)
if whenever f(x) = f(y), then x = y. When f is an injection, we write
f:A BETS Hy
A direct proof that f: A > B is one-to-one begins with the assumption that x
and y are elements of A and that f(x) = f(y); the rest of the proof shows that x = y.
A proof by contraposition assumes that x # y and shows that f(x) 4 f(y). To show
that fis not one-to-one, it suffices to exhibit two different elements of A with the
same image.
4.3
Example.
Functions That Are Onto; One-to-One Functions
225
Show that the function f: R — R defined by f(x) = 2x + 1 is one-
to-one.
Proof.
Suppose that x and z are real numbers and f(x) = f(z).
2x + 1 = 2z + 1. Therefore, 2x = 2z, so x = z. Therefore
fisone-to-one.
Example.
Let g: R — R be given by g(x) = wy)a
injection by assuming that g(x) = g(y). Then
l
Then
B
We attempt to show g is an
1
P+l y+)
Therefore, x7 + 1 = y? + 1, so x? = y’. It does not follow from this that x = y.
This unsuccessful proof suggests a way to find distinct real numbers with equal
images. We note that 2? = (—2)? and then compute g(2) = g(—2) = . Thus g is
not an injection.
Given the graph of a function f: A — B where the sets A and B are subsets of
IR, we can apply this Horizontal Line Test for one-to-one functions:
fis one-to-one if and only if every horizontal line intersects
the graph of f at most once.
From Figure 4.3.1(a), we see that the line y = 3 meets the graph of h twice.
Therefore, 3 has two pre-images, so / is not one-to-one. The graph in Figure
4.3.1(b) suggests that the function k is one-to-one.
The next example shows that we need to consider the domain when we determine whether a function is one-to-one.
Example.
Let f: [0, 00) — [0, 00) be given by f(x) = x”. Show that
f is one-
to-one.
Proof.
Suppose that x, y € [0, oo) and f(x) = f(y). Then x” = y’, so either
x = yor x = —y. Because x > 0 and y > 0, we conclude that x = y. Thereforef
is One-to-one.
=
The next two theorems are similar to Theorems 4.3.1 and 4.3.2. The composite
of injections is an injection, and if a composite is an injection, then the first function
applied must also be an injection.
Theorem 4.3.3
If f: A>B is one-to-one and g: B—C is one-to-one, then g ° f is one-to-one.
\\
Proof.
Assume that (g © f)(x) = (g ° f)(z). Thus g(f(x)) = g(f(z)). Because g is
one-to-one, we have f(x) = f(z). Because fis one-to-one, we have x = z. Therefore,
rT]
g Of is one-to-one.
226
CHAPTER 4
Theorem 4.3.4
Functions
Letf:A>B and g: B>C. If g 0 f is one-to-one, then fis one-to-one.
Proof.
See Exercise 6.
Ls
Mappings that are constructed by means of restrictions or unions may share
injective or surjective properties. These results will be used in the study of cardinality in Chapter 5.
Theorem 4.3.5
If f: A>B is one-to-one, then every restriction of fis one-to-one.
Proof.
Theorem 4.3.6
See Exercise 7.
Let h: A>C and g: BD
a
be functions.
(a)
IfA and B are disjoint sets, h is onto C, and g is onto D, then
(b)
hUg:AUB—> CUD rs onto CUD.
IfA and B are disjoint sets, h is one-to-one, g is one-to-one, and C and D
are disjoint, then h Ug: AUB -— CU Dis one-to-one.
Proof.
(a)
(b)
See Exercise 8.
Suppose that h: A—> C, g: B —> D,
ANB = ©, and CND = @. Then
by Theorem 4.2.5, h U g is a function with domain A U B.
Suppose that x, y ¢ A U B. Assume that (h U g)(x) = (hU g)(y).
(i)
(ii)
(iii)
If x, y € A, then h(x) = (AU g)(x) = (AU g)(y) = hy). Because h is
one-to-one, x = y.
If x, y € B, then by a similar argument g(x) = g(y), and g is one-toOne sor cy:
Suppose that one of x, y is in A and the other is in B. Without loss of
generality, we may assume that x € A and y € B. Then A(x) = g(y),
h(x) € C, and g(y) € D. But
CN D = ©. This case is impossible.
In every possible case, x = y. Therefore, h U g is one-to-one.
z
Examples.
Let h be the function {(r, 1), (s, 2), (¢, 3), (u, 4)}. Then h is oneto-one and onto {1, 2, 3, 4}. The function g = {(x, 4), (y, 5)} is one-to-one and
onto {4, 5}. Because the domains are disjoint, / Ug is a function. By Theorem
4.3.6(a), h Ug is onto {1, 2,3,4} U {4,5} = {1, 2, 3, 4, 5}. Notice that we can-
not apply part (b) of Theorem 4.3.6, because the ranges of h and g are not disjoint.
See Figure 4.3.2(a).
If we let k =
{(w, 5), (z, 6)}, then the domains of h and k are disjoint, and
the ranges of A and k are disjoint. In this case,
hUk is a function that maps
{r, 5, t, u, w, Z} One-to-one and onto {1, 2, 3, 4, 5,6}. See Figure 4.3.2(b).
4.3
Functions That Are Onto; One-to-One Functions
227
Cpe
K
(ajhUg
(b) hUk
Figure 4.3.2
Exercises 4.3
Which of the following functions map onto their indicated codomains? Prove
each of your answers.
(a)
(b)
(c)
f: RR,
f® = aN
f: Z — Z, f(x)
= —x + 1000
iN
N x N, f@) = (x,x)
d)
(e)
(R= R, fe
f: RR, f@)
= Vx? 45
(f)
(g)
f2:R—R, f@)=2"
f:R— R, f@ = sinx
(h)
f:Rx
R-R,f@y)=x-y
Gi)” fol
[—ill, feces
(j)) f:R-> [1, 0), f@M=x 41
yee
(Kk) f: 12,3) > [0, 00), fle) = =—
x
(I)
ae
aL
20).
\1)
ae
(m) f: Zig > Ziq, (fx) = 2x + 1
Which of the functions in Exercise | are one-to-one? Prove each of your answers.
\
For each function, determine whether the function maps onto the given
codomain and whether it is one-to-one. Prove your answers.
(a) The identity function /, from A to A
(b) The canonical map f from Z to Z,
(c) The greatest integer function int: R > Z
(d) The sequence a: N > N whose nth term is a, = 2”
228
CHAPTER 4
Functions
4.
Let A = {1, 2, 3,4}. Describe a codomain B and a function f: A > B such
that fis
(a) onto B but not one-to-one.
(b)
(c)
(d)
one-to-one but not onto B.
both one-to-one and onto B.
neither one-to-one nor onto B.
Prove that if f: A ““S Band g: B “$ C, then g0 f: A ~$ C (Theorem 4.3.1).
onto
Prove that if f:A—- B, g:B—C,
(Theorem 4.3.4).
and gof:A —
C, then f:A 5
Prove Theorem 4.3.5.
Prove part (a) of Theorem 4.3.6.
Find sets A, B, and C and functions f: A > B and g: B > C such that
teen
(a) fis onto B, but g ° f is not onto C.
(b)
(c)
(d)
(e)
(f)
10.
gis onto C, but g ° f is not onto C.
g°f is onto C, but fis not onto B.
fis one-to-one, but g ° f is not one-to-one.
gis one-to-one, but g ° f is not one-to-one.
g°f is one-to-one, but g is not one-to-one.
Prove that
(Ayer
(Db)
*
f@) =
(Chef
iS
Y= 39
|
ili ze < Il
=
x
hee Se
x+4
ifx<-—2
4—x
if —2 <x
jp = Gh
alise Se
11.
<2 __ is onto R but not one-to-one.
ee
|
a ee
Halil
1
(G)
is one-to-one but not onto R.
|x|
x—3
is one-to-one and onto R.
ifx = —4
ie
;
ere
ifx>2
:
is neither one-to-one nor onto R.
Prove that if the real-valued function fis increasing (or decreasing) on R, then
fis one-to-one.
12.* (a)
Let f: Z, > Zz be given by f(x) = [2x], for each x € Z,. Prove that
fis one-to-one but not onto Ze.
(b)
Let f: Z,— Z, be given by f(x) = [3x], for each x € Z,. Prove that
(c)
fis onto Z, but not one-to-one.
Let f: Z, > Z, be given by f(x) = x + 1, for each x € Z,. Prove that
fis one-to-one and onto Z,.
4.3
(d)
13.
229
Let f: Z,— Z, be given by f(x) = 2x. Prove that f is not one-to-one
and not onto Z,.
Give two examples of a sequence x of natural numbers such that
(a)
(b)
(c)
(d)
14.
Functions That Are Onto; One-to-One Functions
xis neither one-to-one nor onto N.
xis one-to-one and onto N.
x is one-to-one and not onto N.
xis onto N and not one-to-one.
Suppose that the set A has m elements and the set B has n elements. By
Exercise 17 in Section 4.1, there are 2’”” relations fromA to B and n” functions
from A to B.
(a)
(b)
(c)
Ifm <n, find the number of one-to-one functions from A to B.
Ifm =n, find the number of one-to-one functions from A to B.
Ifm > n, find the number of one-to-one functions from A to B.
(d)
(e)
(f)
Ifm <n, find the number of functions from A onto B.
Ifm =n, find the number of functions from A onto B.
Ifm =n + 1, find the number of functions from A onto B.
(g)
If m=n,
find the number of one-to-one
correspondences
from A
onto B.
Proofs to Grade
15.
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
(a),
Claim.
The function f: R x R— R given by f(x, y) = 2x — 3yisa
surjection.
“Proof.”
Suppose that (x, y)é R x R. Then x € R, so 2x € R. Also,
y ER, so 3y ER. Therefore 2x — 3y € R. Thus, f(x, y) € R, so fis a
surjection.
g
(b)
Claim.
The function f: [1, co) — (0, oo) defined by f(x) = !maps
onto (0, oo).
‘
“Proof.”
T=:
(c)
(d)
Suppose
that w € (0, 00). Choose
x = *. Then
f(x) =
Therefore the function fis onto (0, 00).
w
Claim.
onto
a
onto
If f: A —> B and g: B —~ C, then g°f: A — C maps onto
C (Theorem 4.3.1).
“Proof.”
Suppose that ae A. Then f(a) € B. Because f(a) € B,
e(f(a)) € C. Therefore, (g °f)(a) = g(f(a))
EC, so g Of is onto C.
=
Claim.
The function f: R — R given by f(x) = 2x + 7 is onto R.
“Proof.”
Suppose that f is not onto R. Then there exists
b € Rng(f).
a
sb — 7) is a real number, and f(a) = b. This is a contradiction.
Thus fis onto R.
Thus,
for all real numbers
x, b#2x
+7.
be R
with
But
rT]
230
CHAPTER 4
Functions
(e)
Claim.
Let / be the interval (0, 1). The function f: 7 x [— I given
by f(x, y) = x is a surjection.
“Proof.”
Let tel. Then 0<t<1,s0o0<P<t<1,sorel.
Choose
x=?
Therefore,
and
y=
: el,
Then
{Gy
=x
= (7) =
/ C Rng(f), so the function fis onto /.
(f)
(g)
(Theorem 4.3.3).
“Proof.”
We must show that if (x, y) and (z, y) are elements of g ° f,
then x = z. If (x, y) € g Of, then there is u € B such that (x, u) € f and
(u, y) € g. If (z, y)€ g Of, then there is v € B such that (z, v) ef and
(v, y) € g. However, (u, y)€& g and (v,y)€g imply u = v because
g is one-to-one. Then (x,u) ef, (z,v) ef, and u =v; therefore,
x = z, because
f is one-to-one. Hence, (x, y) and (z, y) in g Of imply
x = z. Therefore g ° f is one-to-one.
a
Claim.
The function f: R — R given by f(x) = 2x + 7 is one-to-one.
“Proof.”
eM
(h)
If f-At-}B
and
g:B->C,
then
a
Claim.
gof-A->C
Suppose that x, and x, are real numbers with f(x,) # f (x9).
d, 7 2X.
ie ANC, UU ae2 kyr 2X5. Lene, 44. 45, WICH
shows that fis one-to-one.
a
Claim.
The function
fin part (e) is an injection.
“Proof.”
Suppose that (x, y) and (x, z) arein] x Jand f(x, y) = f(x, z).
Then x” = x. Dividing by x*, we have x”~* = x° = 1. Because x ¥ 1
(i)
and\2°~* = 1, y='z’ must be’ 0! Therefore y =.z. This shows that
(x, y) = G, z), so fis an injection.
a
Claim.
The function
fin part (e) is not an injection.
“Proof.”
roo
jr
Lr
Both
=, =*)and (;
= ;)are in/
1
oO |
(;
x J.‘ But
Apa)=(G=(G))=G)e3)
1
(j)
|
l
Claim. The functionf:Z\y > Z,9 given by f(x) = 3x + 1 is onto Zyo
“Proof.” Let w € Z,). Then w € Z and 3 divides exactly one of w,
Wier 1 Or Ww. 2.
a3: Then ft) = ae . = L)=
Case 1. If3divides w— 1, choose
x = ~—_—
w—
|
als
Case 2.
=
fi
+ 3. Then fiz) = f(“ + 3)
(3(¢+3)) +1=w
+10=w.
wif Sidivides! wi 2 «choose x — “= ey Then
re
If 3 divides w, choose x = us
Oo
Case3.
eS
ve Cee
In every case, w is in Rng(f), sof maps onto Zo,
l=w+20=w.
a
4.4
4.4
Inverse Functions
231
Inverse Functions
In this section, we consider functions that have the desirable properties of both being
one-to-one and mapping onto their codomains. The key role played by these functions in succeeding chapters suggests their importance in advanced mathematics.
DEFINITION
A function f: A > B is a one-to-one correspondence
(or a bijection) if fis one-to-one and onto B.
For every set A, the identity function /, is a one-to-one correspondence from
A to A. There is a natural one-to-one correspondence from the set of letters of the
English alphabet to the set of numbers from | to 26:
Example.
a
b
Cc
d
y
+
c
u
’
y
|
2
3
4
Ds)
26
The mappingf: (0, 0o)—(0, 1) given by f(x) =
Z
x
a
is a bijection.
Proof.
(i)
(We first show that f is one-to-one.) Let x and y € (0, co), and suppose that
fix) = fy). Then
is one-to-one.
=
»80 x
FOESN 9 Berl
1 =
y+ 1 andx
= y. Therefore,f
1
(We next show thatf is onto (0, 1).) Let r € (0, 1) . Then — > 1, and, hence,
(ii)
]
r
==
1
if
| ss07Theas—
1
r
1 = (0,coyand (+—
]
I
=
= pete.
+35)
1)ee
fore, every element of (0, 1) is an image, so fis onto (0, 1).
a
The importance of the next example will become clear in Section 5.2.
Example. Let F:N x
NWN
be defined by F(m,n) = 2”~'(2n — 1). For
example, Hl, 3) (2:3 0) =>, FG.) = 2°01 = 1). 4, and HG,5) =
2*(2.-5 —
1) =
144. The function F is a one-to-one correspondence.
Proof.
(i)
To show that F is onto N, let s € N. We must show that s = F(m, n) for some
(m,n) inN x N. Ifsis even, then s may be written as 2't, where k > 1 andt
is odd. Because tis odd, t = 2n — 1 for somen € N. Choosing m = k + 1,
we have F(m, n) = 2"—'(2n — 1) = 2*t = s. If s is odd, then s = 2n — 1
232
CHAPTER 4
Functions
for some n € N. For this n and m = 1, we find F(m, n) = 2°(2n — 1) =.
(ii)
Therefore, F is onto N.
To show that F is one-to-one, suppose that (m, n) and (r, s) arein N x N and
F(m,n) = F(r, s). We first prove that m = r. Without loss of generality, we
may assume that m > r. (Ifm < r, we could relabel the arguments.) From
F(m,n) = F(r, s), we have 2”—!(2n — 1) = 2’~!(2s — 1), which implies
2™-"(2n — 1) = 2s — 1. Because the right side of the equality is odd, the
left side is odd. Thus, 2’”"-” =
thate7ie—we
Dividing
both
1. Therefore, m — r = 0, and we conclude
sides of the equation
2”~'(2n —
1) = pile
ty
Rye— |W)
by 2™-!(2m—-1 = 97-1) we have 2m — 1 = 2s — 1, which implies n =s.
Thus, m = r andn = s. Therefore (m, n) = (r,s). Hence, the function F is
one-to-one.
Example.
a
We can prove by cases that the function k: RR
given by
Dene ew
CARS ee ifx¢Q
is a one-to-one correspondence from R to R. However, we can also writek =hU g,
where h: Q— @ and g: Q°— O° are given by h(x) = 2x and g(x) = 3x. Simple proofs
show that / and g are bijections. Because @ and ©*° are disjoint and R = QU O°,
we know by Theorem 4.3.6 that h U g: R > Ris a bijection. The technique of writing a function in piecewise form as union of one-to-one correspondences can be
useful as a way to show that the function is a bijection.
Combining Theorems 4.3.1 and 4.3.3, we see that the composite of one-to-one
correspondences is a one-to-one correspondence:
Theorem 4.4.1
Iff: A > B is a bijection and g: B > C is a bijection, then
bijection.
Example.
go f: A >
Cisa
The function f given by h(x) = x + 1 is a one-to-one correspondence
from (0, 00) to (1, 00), and the function g given by g(x) = 2 iS a One-to-one corre-
spondence from (1, co) to (0, 1). By Theorem 4.4.1, the composite g © h is a one-toone correspondence from (0, oo) to (0, 1). Because (g ° h)(x) = g(h(x)) = g(x + 1) =
cs for all x € (0, oo), the composite g ° h is the function fin our first example.
o
We know that the inverse of a function is not always a function. To understand
why this happens, we must understand the connection between inverses and oneto-one mappings.
Theorem 4.4.2
Let fbe a function from set A to set B.
(a)
fis a function from Rng(f)
(b)
If f~! is a function, then f~! is one-to-one.
to A if and only if fis one-to-one.
4.4
Inverse Functions
233
Proof.
(a)
Assume
that f~' is a function
from
Rng(f)
to A. Suppose
that
f(x) = f(y) = z. Then (x, z) ef and (y, z) € f. Therefore, (z, x) € f~! and
(z, y) € f—'. Because f—! is a function, x = y. Therefore fis one-to-one.
Now assume that f is one-to-one. Suppose that (x,y) <¢f~! and
(x, z) € f—!. Then (y, x) ef and (z, x) € f. Because f is one-to-one, y = z.
Therefore f~! is a function.
(b)
See Exercise 4.
w
When an inverse is needed for a function that is not one-to-one, we restrict the
domain so that the resulting function is one-to-one. With trigonometric functions,
we refer to this restriction as the principal branch of the function.
Example.
;
When the domain of sin: R—R
,
,
u 0
is restricted to the interval | ZI
the principal branch of the sine function, sometimes written as Sin (with capital S),
is one-to-one and onto [—1, 1]. See Figure 4.4.1. By Theorem 4.2.2, Sin has an
inverse function. Sin~!:[—1, 1] >
= . . See Figure 4.4.1.
(a) Sin = Sule
oe
(b) Sin7!
Figure 4.4.1
We must be careful not to assume that if f: AB
is one-to-one, then function
f~' maps B to A. We know by part (a) of Theorem 4.4.2 that f~! is a function from
Rng (f) to A and by part (b) that f~' is one-to-one. If falso happens to be onto B—
that is, if fis a bijection—then f ~! is a bijection from B to A.
Corollary 4.4.3
The inverse of a one-to-one correspondence is a one-to-one correspondence.
The next result gives a simple, practical method using composition to determine whether a given function is the inverse of a function.
234
CHAPTER 4
Theorem 4.4.4
Functions
If fis a function from A to B and g is a function from B to A, then
(a)
¢ =f ‘ifand only ifgof=J,
and fog = Jp.
(b)
If fis a one-to-one correspondence, then g = f—! if and only if g ° f = I, or
fog = Ip.
Proof.
(a)
If g=f—', then
gof=J, and fog =Jy, by Theorem 4.2.3. (We use
the fact that Rng(f) = Dom(f~!) = B.)
Assume now that
g°f=J, and f° g = Jp. Then f is one-to-one by
Theorem 4.3.4, and
f maps onto B by Theorem 4.3.2. Thus, f~! is a function
on Bandf-!=f-'ol, =f-!o(fog) =(f-!of)og=hog =z.
(b)
See Exercise 5.
|
Theorem 4.4.4(a) captures the essential idea of an inverse—whatever f does to
a domain element x, applying the inverse to f(x) takes you right back to x. For the
natural logarithm and exponential functions, part (a) gives the familiar properties:
e'™ — x for allx > 0
inves — ro tonmallegesrine
Example.
Letf:R — {1} ~
R — {4} and g: R — {4} —
R — {1} be given by
JX) = tet and g(x) = . = 2.Rather than writing out proofs that both fand g are
one-to-one and onto their codomains, we may instead compute the composites to
verify that g 0 f = Ip_,,, and f° g = /p_ 4). It follows from Theorem 4.4.4(a) that
both functions are one-to-one (because their inverses are functions) and onto (using
Theorem 4.3.2).
Oo
Theorem 4.4.4(b) provides a useful shortcut for verifying a function g is the
inverse of a one-to-one correspondence f. It suffices to test only one of the equations-g
of = 1, and fo g = Tp.
We conclude this section with the special case of functions that are bijections
from a set to itself. These functions will appear again in Chapter 6 because they are
essential to the understanding of the algebraic structures called groups.
DEFINITION
Let A be a nonempty set. A permutation of A is one-toone correspondence from A onto A.
A permutation of a set A has the effect of arranging (or permuting) the elements
ofA:For the'set A=)1)2;3 }; thefunction's = 4(172), (2,3)/ G, 1) asa permutation
ofA.The arrangement produced by s is obtained by listing the images of 1, 2, and 3
in order: 2 3 1. The arrangement resulting from the permutation t = {(1, 2), (2, ip)
(353) tigre ls)
4.4
Inverse Functions
235
For every set A, the identity function J, is a permutation of A, called the identity
permutation on A.
Looking at the result of a permutation on a finite set provides the basis for a
simplified notation. The result of the permutation t above is that 2 is in the first
position, so 1 was mapped to 2.The second position is occupied by 1, so the image
of 2 is I, and finally the image of 3 is 3. We list the images in order, and write the
permutation ¢ as [2 | 3].
Examples.
The function h = [426531] is a permutation of the set B=
{1, 2, 3, 4,5, 6}. It maps 1 to 4, 2 to 2, 3 to 6, 4 to 5, 5 to 3, and 6 to 1. The identity
permutation on B is [1 2345 6].
Previous results about one-to-one correspondences and inverses can be restated
in the language of permutations.
Theorem 4.4.5
Let A be a nonempty set. Then
(a)
(b)
(c)
(d)
the identity mapping /, is a permutation of A.
the composite of permutations of A is a permutation of A.
the inverse of a permutation of A is a permutation of A.
iffis a permutation of A, then fol, = 1,of=f.
(e)
(f)
if fis a permutation of A, then fo f-! = f-'o f = I.
if fand g are permutations of A, then (g 0 f)~! = f-!o gl.
Proof.
(a)
(b)
See Exercise 3 of Section 4.3.
See Theorem 4.4.1.
(c)
See Corollary 4.4.3.
(d)
(e)
(f)
See Theorem 4.2.3.
See Theorem 4.2.4.
See Theorem 3.1.2(d).
a
Composites and inverses of permutations can be easily computed using the
notation described above. Remember to evaluate a composite from right to left
because the function on the right is applied first.
Examples.
For A =
{1, 2, 3,4}
and the permutations
t= [2431]
[2 3 1 4], the composite permutation t © 5 is [243 1] 0 [2314]
=
and s =
[4321]. The
thought process begins with [2 3 | 4], where “1 goes to 2.” Then we see that in
[2 4 1 3], “2 goes to 4.” Thus, in the composite, “1 goes to 4.” We find the other images
similarly. For example, s sends 2 to 3 and f sends 3 to 3, so the composite sends 2 to 3.
To find the inverse of the permutation tf = [2 43 1], we think of reversing its
action. The permutation ¢ sends | to 2, so the inverse of t sends 2 to 1. Because ¢
sends 2 to 4, t~! sends 4 to 2, etc. In this way, we find that r~! = [4 1 3 2]. We can
verify that this is correct by computing [2 43 1] o [413 2] = [1 23 4], which is
the identity permutation.
236
CHAPTER 4
Functions
Exercises 4.4
Show that each of these functions is a one-to-one correspondence.
(a) f: R— Rgiven by f(x) = mx + b,m#0
—Xx
(b) f: (2, 00) > (—oo, —1) given by f(x) = eas)
(c)
(d)
g:(—oo, —4) > (—co, 0) given by g(x) = —|x + 4| h:8N— 10N, where 8N = {8k:k EN}, 1ON = {10k: ke N}, and
Lh) ale
(e)
G:N x N—8N
given by G(m, n) = 2"*7(2n — 1)
l
(f)
f: [1, cc) —[2, oo) given
by f(x) Seale
(g)
f:Z 3—> Z, given by f(x) = 5x — 1
Find a one-to-one correspondence between each of these pairs of sets. Prove
that your function is one-to-one and onto the given codomain.
(a)
{a, b,c, d, e, f} and {2, 4, 8, 16, 32, 64}
(b)
NandN
(c)
(d)
(3, co) and (5, co)
(—oo, 1) and (—1, o&)
(e)
(f)
12N and 20N, where 12N = {12k:k E N} and 20N = {20k:k EN}
x Rand the set C of complex numbers
— {1}
For each one-to-one-correspondence, find the inverse function. Verify your
answer by computing the composite of the function and its inverse.
(a)
f: (0, 00) > (0, o&) given by f(x) = .
(b) g:(—2, 00) > (00, 4) given by g(x) = an
(c)
(d)
h: R= (0, 00) given by h(x) = e**?
x
5
—— sd)
G: (3, 00) — (, oo) given by G(x) = sitio
(e) f:Z,— Z, given by f(x) =x
+2
Prove part (b) of Theorem 4.4.2: If f: A > B and f~—' is a function, then
fis
one-to-one.
(a)
Assume that f: A — Band g: B > A. Prove that g = f—'iff
gof=I,
or f° g = I, (Theorem 4.4.4(b)).
(b)
Give an example of sets A and B and functions f and g such that
f:A> Bg: BoA, g°f=I,,andg#f-!.
Let f:A — B and g: B — A. Prove that if go f= I, and f° g = Jp, then
f:A—>
Band g:B A.
onto
onto
Use the one-to-one correspondences In: (0, 00)
> Rand f: (2, 00) — (0, 00),
where f(x) = x — 2, to describe a one-to-one correspondence
(a) from (0, co) onto (2, 00).
(b) from (2, oo) onto R.
(c) from R onto (2, 0).
(d) from R onto (0, oo).
4.5
8.
Set Images
237
Prove that if f: A > B, g: B > C, and h: C > A are one-to-one correspond-
ences, then f~! 0 g~! 0 A~! is a permutation ofA.
9.
Use the notation of this section to write these permutations of the set
Ce 4 3.6, 1)e
(a) Ic
(D) Beh) 5(650)
442. LG, 2); (2,6). (753).)
RCV)
(2a) Gray Om
no) On7), (43) }
(d) w= {(1, 2), Q, 1), G, 4), G, 3), G, 5), ©, 7), (7, 6)}
x
Proofs to Grade
10.
(ce)
(uoy
(ft) ay ou
(g)
wow
(h)
uovow
(i)
vt
(KROv)
2) Rye cus
(mM)
tus oye
*
(i)
uw!
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
(a) Claim.
If f is a one-to-one correspondence from A to B and g is a
one-to-one correspondence from B to A, then g = f~!.
“Proof.”
Suppose that fis a one-to-one correspondence from A to B
and g is a one-to-one correspondence from B to A. Then by Theorem
4.4.1, g © f isaone-to-one correspondence fromA to A. Likewise, f° gis
a one-to-one correspondence from A to A. Then
f~! = f~!°(g°f) =
fri o(fog)=(f-lof)og=hog =z.
a
(b)
Claim.
If fis a permutation of A, then J, ° f= fol,.
“Proof.” Because J, is the identity, J, o f = f, Also, J, is the identity,
Sony Od 7 —o/ slherefore
ey — fel
a
(Cc) eeClaim-ns
eter —. [21453 5) gandess—s|4e2 sale i|ebe Perna Ons of
{
1,2, 3,4,5}. Then the inverse of no sis (3.24 [5]
“Proof”? Let r= [12435] ands = [423 15]. Thenr
= = mand
s~! = 5s. Therefore, (ro s)“! = r7!os-! = ros = [32415].
|
(d)
Claim.
“Proof.”
If fand g are permutations of A, then (f° g)"!' = g
Weknow by Theorem 4.4.5 that f° gis
lof !.
a permutation of A, so
(f° g)~! is a permutation ofA.Also, by Theorem 4.4.5, g~! and f~! are
permutations of A, so g~! o f~!is a permutation of A. By Theorem 4.4.4,
we can check whether g~! o f~' is the inverse of f © g by computing their
composite.
We find that (f° g)° (g7!of7!)= fo(ge (7 eye a) =
fo(gog of! = folof'=(fol)of-! = fof = Iy. There-
fore, (fog) =g
(e)
56x — 1) = 3=257—
4.5
lof.
a
Claim.
The two permutations fand g of Z, given by f(x) = 5x — | and
g(x) = 5x — 3, respectively, are inverses.
“Rroofsmelet, bein. Then (gf Xx) = gif(x)) = sG6x — 1) =
8 =x
r
Set Images
We have considered images and pre-images of individual elements under a mapping. The next step is to ask about collections of elements in the domain and codomain of a function, and what corresponds to them in the other set.
238
CHAPTER 4
Functions
DEFINITIONS
Let f: A > B, and let X C A and Y C B. The image of
X or image set of X is
Xx) = (ye By =f@) fomsome x = X7,
and the inverse image of Y is
fo Wy = (xe As)
ey):
Example. Let A = {—3, —2, —1, 0, 1, 2, 3}, B = {0, 1, 2, 4, 6, 9}, and
f be the
function from A to B given by f(x) = x’ and displayed in Figure 4.5.1.
The set X = {—1, 3} is a subset ofA. The image of X is f(X) = {1, 9} because
J
= Mand 3) = O We also have —3, 3}) = (9) (0, 1, 2}) = {Os
ay,
and f(A) = {0, 1, 4, 9}.
The set Y = {4, 6} is a subset of B. To find the inverse image of Y, we look for
the pre-images of 4 and 6. We find f~'(Y) = {2, —2} because both —2 and 2 are pre-
images of 4, and 6 has no pre-image in A. Also, f~'({9}) = {—3, 3},
f '({2}) =©,
and fia (b) =A.
Note that f—! is not a function from B to A, so it would not make sense to consider f~'(1). However, f—'({ 1}) is meaningful and equal to {1, —1}.
Figure 4.5.1
Examples. Let f: N — N be given by f(x) = x + 4. Then we have f(1) = 5
6 Alsoy FC 10-11, 12) = 14 SiG)
and (2) ="O,-so™ 7 (1) 2)}) =a
and f({xeN: x > 20}) = {xe N:x > 24}. The image set of N is f(N) =
{5,6,7,...}, which is the range off. There is no x in N such that f(x) = 2
or f(x) = 3, so f~'({2,3}) = ©. Also, f—'({5,6}) = {1,2} is the same as
f ‘C1, 2, 3, 4, 5, 6}). The inverse image of N is Dom(f)
same as f-'({x E N: x > 5}).
= N, which is the
4.5
Examples.
Set Images
239
Let f: R > Rbe given by f(x) = x. Then f({ —2, 2}) = {4} because
both f(2) = 4and f(—2) = 4. From Figure 4.5.2(a),
we see that f({1, 2]) = [1, 4].
Also, f({—1,0]) = [0, 1].
F([-1, 2]) = [0,4]
(b)
Figure 4.5.2
In this example, it is tempting to guess that f({—1,
2]) = [(—1)’, 27] =
[1,4], but this is incorrect. By definition, f({[—1, 2]) is the set of all images of
allthe elements of [—1, 2]. Because at , 0, and 0.7 are in [—1, 2], their images i 0,
and 0.49 must be in f({—1, 2]). Figure 4.5.2(b) shows that f({[—1, 2]) =
[0, 4].
Wealsohave f—'({16}) = {—4, 4}, f-'({1, 4, 9}) = {—3, —2, -1, 1, 2, 3},
and f—'({—4,-—3])
=@ because all images of f are nonnegative. Even
though /f((0,2])= [0, 4], f((0, 41) = [—2, 2]. Figure 4.5.3 illustrates why
f~'({1, 41) =[—2, -1] U [1, 2].
Oo
f(A
eater,
Figure 4.5.3
2)
240
CHAPTER 4
Functions
Proofs involving set images take some special care because of the interplay
among points, sets, and images of sets. Let f: A > B,
DC A,
ECB, andaeéa.
Here are some facts about images of sets that follow from the definitions:
(a)
IfaeD, then f(a) € f(D).
(b)
Ifaef '(E), then f(a) € E.
(c)
If f(a) EE, thenae f—(E).
(d)
If f(a) € f(D) and fis one-to-one, then a € D.
For part (d), we note that it is not correct to say that f(a) € f(D) always implies
@e D. For the function fG)='2- and’D'= [1,2], we see that f(—1) e/(),
but —1 ¢ D.
Examples. Let f: R > R be the function given by f(x) = x’. Let A = [—3, 2]
and ©C =s| too), Then (AC) ((—3, o)):— [0,23] “and Ay y(cy =
[OpoT
25 = 0,25), so 7A UC) =f Aya
C):
On the other hand. fA) Cy=7(1, 2)]) = [1,4] and fA f(C) = [1 97,
so f(ANC) C f(A) N F(C), but it is not true that f(A N C) = f(A) N f(C).
Theorem 4.5.1
Oo
Let f: A > B, C and D be subsets of A, and E and F be subsets of B. Then
(a) f(CND) CS F(C)N f(D).
(b) f(CUD)=f(C) Uf).
(Cee
Ei yf (Bing().
(Die
OF) = fs (2) Us).
Proof.
(a)
(Cc)
Suppose that b € f(CN D). Then b = f(a) for some ae CN D. Because
aéC
and b= f(a), be f(C). Also,
ae D and b = f(a), so be f(D).
Thus, b € f(C) M f(D).
a ef (ENF)
iff fae ENF
iff f(a) € Eand f(a)E F
iff a ef ‘\(E) anda eéf—\(F)
iff ae f—\(E) Nf -\(F).
The proofs of parts (b) and (d) are left as Exercise 9.
|
Finally we note that if f: A > B, then every subset X of A has a corresponding image set f(X) that is a subset of B. This correspondence is a function from
P(A) to P(B) called the induced function, f: (A) + P(B). Note that we use the
same name, f, for the function and the induced function. The appropriate interpretation is usually clear from the context.
Likewise, every subset Y C B has a unique inverse image f—!(Y) that is a
subset of A. This correspondence is another induced function, f~!: P(B) > P(A).
4.5
Set Images
241
Exercises 4.5
Pet
Az—s| le, 3
(a)
(b)
Find the image under h of each of the 8 subsets of A.
Find the inverse image of each of the 8 subsets of B.
Let (Qe
(a)
(©)
Oy
07
abo 4.5. 6} xand fi 4(1.4), (2,4), (3, 5)}.
1 Find
f({1, 3)).
ie ih):
1 ee
Cb)
tO se 2341):
(de (2.3):
(f) Br1, 51 117, 26):
Let f@) = 12x, Find
(a) f(A), where A = {—1,0,1,2,3}.
(b)
(c)
(ejay
(cd) oy(12551):
) FG (F341):
f—'(R).
(l4)
Let f: R —
(a)
CP
{0}
f(N).
> R be given by f(x) = x + - Find
f((0, 2)).
Ge
AAD) ey (| Ee):
(yay (LO),
(e) f(f-"®)).
(f) ¢(¢7'([=5 10.1])).
Let f: N x N > N be given by f(m, n) = 23". Find
(a)
f(A Sc B), where
(b)
f—'({5, 6, 7, 8, 9, 10}).
A= {152,,3),
and
B=.{3, 4}.
Let f: Z;,> Zj, be given by f%) = 3x + 2. Find
(a)
fU3, 4)).
9}).
yA}:
(Dy Ca):
(d) f({2,
6,10}).
©) Ma OL):
Dye
b
iit xe Mh
I ee shige <A
. Give an example of
(a)
(b)
aéRand DCR
such that f(a) € f(D) anda ¢ D.
subsets A and C of R such that f(A N C) # f(A) N f(C).
(c)
(d)
asubset D of R such that D # f—'(f(D)).
asubset E of R such that E 4 f(f—!(E)).
Let f: A > B,C,
© C, GA, and D, © D, SB.
(a)
Which must be true, f(C,) S f(C,) or f(C,) SAC)? Prove that your
answer is correct, and give a counterexample to show that the other set
inclusion may be false.
(b)
Which must be true, f~'(D,)
©f~'(D,) or f-'(D,)
S f~'(D,)? Prove
that your answer is correct, and give a counterexample to show that the
other set inclusion may be false.
Prove parts (b) and (d) of Theorem 4.5.1.
10.
Let f: A > B, and let {D,: a € A} and { Eg: BET}
of A and B, respectively. Prove that
be families of subsets
242
CHAPTER 4
Functions
* (a) f(P.)¢ AhiCy
een.) f( UP.) = Uae:
acA
(c)
11.
acA
= LIFE).
A)fey). @) f(UEs)
= pel
(Es)
pel
pel
pel
Give an example of a function f: R > R and a family {D,: a € A} of sub-
sets of [IR such that f(‘al D,)ol
aecA
12.
Let f: A > B,
(a)
(b)
(c)
(d)
(e)
(f)
(g)
33
acA
DCA, and E C B. Prove that
V7 (eer.
A —f-\(E) Cf-\(B - EB).
f-\(B -— E)CA —f-(E).
E = f(f—'(£)) if and only if E C Rng (f).
DSF *GiD)):
D = f—'(f(D)) if and only if f(A — D) C B — f(D).
(G- (Byeas MRas):
Let f: A > B, and let X,
(a)
(b)
(c)
Ds):
acA
Y
CA and U, V CB. Prove that
f(X) € U if and only if X C f-(U).
oleae)
G(X = 1).
ON (Vif
(UV).
14.
Let f: A — B. Prove that
(a) if fis one-to-one, then f(X) ONf(Y) = f(X M Y) for all X, YEA.
(b) if f(XX) OfF(Y) = f(X NM Y) for all X, Y CA, then fis one-to-one.
(c) if X C A and fis one-to-one, then f(A — X) = f(A) — f(X).
(dm ti XGA; Y
CB, and f is abijéction, then f(X) = Y if and only if
i 08t
ee
15.
Let f: A = B.
(a) Whatcondition onfwill ensure that the induced function f: P(A) > P(B)
is One-to-one?
(b) What condition onf will ensure that the induced function f: P(A) > P(B)
is onto P(B)?
16.
Let f: A — B. Let T be the relation on A defined by x Ty if and only if
f(x) = f(y). By Exercise 18(a) of Section 4.1, Tis an equivalence relation on
A. Describe the partition of A associated with T.
Proofs to Grade
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
(a) Claim. If f: A > Band X CA, then f~!( f(X)) CX.
“Proof.”
Assume that f: A > Band X C A. Suppose that x € f~'( f(X)).
Then
by definition
fim
iG60. O)S).6
(b)
of f~', f(x) € f(X). Therefore,
xX.
Thus,
mi
Claim. If f: A — Band X CA, then X C f-!(f(X)).
“Proof.”
Assume that f: A > Band X C A. Suppose that z € X. Then
f(z) € f(X). Therefore, we conclude that z € f~'(f(X)), which proves
the set inclusion.
a
4.6
(c)
Claim.
Sequences
243
If f: A > Band {D,: a € A} isa family of subsets of A, then
Nf) < f(a D,
acA
“Proof.”
acl
Suppose that y € [{) (D>). Then y= 7D") tor alla, Uaus,
aeA
there exists x € D, such that f(x) = y, for all w. Then xe
[) D,, and
acA
f(x) =y,soye f(() D, Therefore, {) f(D,) €
f(() Da
acd
(d)
Claim.
acA
Let f: A > B bea
=
ae
function such that for all subsets X and Y of
A, it f(x) = fl), then X = Y. Thenis one-to-one.
“Proof.”
Suppose that fsatisfies the condition. Suppose that x, y € A,
and f(x) =f). Then f({x}) = {fix} = (f(y)}. By the condition, we
have (x)= 4}. SOx =.
|
4.6
Sequences
Sequences play a central role in the representation of functions using infinite series.
Sequences are also important because of their usefulness in characterizing a number
of important properties of the real numbers. This section is devoted to sequences of
real numbers and the fundamentals of convergent and divergent sequences.
As we saw in Section 4.1, a sequence
is a function x with domain
N. For
n &N, the image of n is called the nth term of the sequence and is written as x,.
All sequences we consider will be sequences of real numbers.
Examples.
The sequence x of odd positive integers has nth term x, = 2n — 1.
The first féewsterms
of x are
153, 5) (32250) ne
sequence y, where y, = (—1)”, has range
nately —1 and 1.
SUih tenets 79 —
500 whe
{—1, 1} because ifs terms are altero
n
The sequence x whose nth term is x, =
Ht
n
property of sequences. The first few terms of x are
:
é
5
1
illustrates the convergence
(pills
Det OY!
Rent pean
Bees)”,
.., aS Shown
(5°
2
1
in Figure 4.6.1. The 99th term is x99 = —7p5; and the 1000th term is x;99 = Fo97-
i
2
=i
ee
Lg
4
ape
8
ec
I
a
Lo
_|_ee—e
6
fe
nN
aL
5
(=F
n aie 1
Figure 4.6.1
3
f
>
244
CHAPTER 4
Functions
The figure shows that x, is near 0 when n is large, and the further out we go in the
sequence, the closer the nth term is to 0. We say that the limit of this sequence is 0
and make this notion precise in the following definition.
——
DEFINITIONS | Fora sequence x of real numbers and a real number L,
we say x has limit L (orxconverges to L) if for every ¢ > 0 there exists
a natural number N such that ifn > N, then |x, — L| < e.
When x converges to the real number L, we write lim x, = Lor x, > L.
n>
©
If no such number L exists we say x diverges or lim x,, does not exist.
n—>
co
In the definition of x, — L, we usually think of ¢ as being a small positive
number, so the expression |x,, — L| < € means that the distance between x, and L
is small. What the definition guarantees is that no matter how small ¢ may be, all
terms beyond a certain point in the sequence are within ¢ distance of L. The point
where we can be sure this happens is the Nth term. When we work with a particular
sequence, we need to be aware that if we were to make ¢ smaller, we would prob-
ably need to go further out in the sequence (choose a larger number for NV) to be
sure the terms are close enough to L.
In symbols, we may write the definition of lim x, = L as
n>
(Ve > O)(ANEN)\(Vn
Thus, a proof of the statement
co
EN) > NS |x, — L| < ©).
lim x,, = L will usually have the following structure:
n>
oo
EPSILON-DELTA PROOF THAT
lim x, = L
Proof.
sei
Let ¢ be a real number greater than 0.
Choose
N= ____ (Specify some value for N, typically in terms of e.)
Let n € N, and suppose that n > N.
Therefore, |x, — L| < ¢.
We conclude that lim x, = L
n>
a
©}
Z|
The intermediate steps to verify that |x, — L| < e is true are often discovered
through some preliminary scratch work, working backward from |x, — L| < ¢ and
continuing until we find a relationship between n and ¢ that will suggest a choice
for N.
Example.
Earlier we claimed that the sequence x with nth term be
verges and
lim
n>coo
i
(-1)" = 0. Before proving
a this result, we first use thealinequality
n+
1
4.6
Sequences
245
|x, — 0| < e€ to derive a relationship between x, and ¢. For any ¢ > 0, because
(1),
siaey
|p —O lr ie
p> cies p We need
1
n+ 1
mre,
1
==,
pal
é
n>—-l.
8
This last inequality tells us how large n must be (and hence tells us a value for NV)
to ensure that |x, — 0| < e. Here is the proof:
Proof.
Let € be a positive real number. Let N be any integer greater than - Il.
Suppose
that
n > N. Then
==) 4
a=
OS
n > 1 —l,son+1>
‘and
|
\
"=
n+
1
Because
=
i
m
< ¢.
n+
0)
ea hereiore
ae i
mali
NW
CO
‘au
n=
— 0
|
The next theorem says that once we know that a sequence converges to a limit
L, we know it cannot converge to any other number. The proof by contradiction
supposes that there are two different limits L and M and selects ¢ so small that for
sufficiently large n the terms x,, cannot simultaneously be within e of L and within
€ of M. See Figure 4.6.2.
ee
—_(_e-#
eee
e
I
—
$j
—e —
|L — M|
3
~
e
~
+
as x
>|
e=5|L—M|
Figure 4.6.2
Theorem 4.6.1
If a sequence x converges, then its limit is unique.
Proof. Suppose that x, — L, x, > M, and
L#M. Lete = sIL — M|. Because
x, — Land x, — M, there are natural numbers N, and N, such thatn > N, implies
|x, — L| < e and n > N, implies |x, — M| < e. Let N be the larger of N,, N5.
Suppose that n € N andn > N. Thenn > N, andn > N,, so
Oh
CE
ea
6)
<|L — x,| + |x, — M| (by the Triangle Inequality)
=
Xn re L\|+ [Xn =
ZOE
A
M|
246
CHAPTER 4
Functions
Thus, the assumption L # M leads to |L — M| < s|L — M|. Because |L — M| > 0,
this is impossible. We conclude that the limit of x is unique.
a
With practice, you will be able to identify limits of many sequences, when they
exist. Sometimes it helps to calculate several terms to see the trend. For example,
a.
[fll
sin (-)
fee
for the sequence x, where x, =
will be correct_in guessing that
n
lim x, = 1 by calculating several values, such as x; © 0.993346, x9) © 0.995833,
n>
©
and x, * 0.999954. Be careful when guessing limits this way because the exact
:
ae
value of the limit might not be obvious. For instance, the sequence a, = ( :
)
has terms 419 = 2.59374, dsq = 2.65159, and aj99 = 2.70481. The limit of a is the
irrational number e.
All terms of a constant sequence c given by c,, = K for some real number K
are very close, in fact equal, to the number K. The proof that c converges to K is
Exercise 4(a).
Limits of sequences of rational expressions can usually be evaluated quickly.
12n* + 5n + 1
When computing the terms of a sequence such as x, =
for large
72n* — 18n
values of n, the value of the numerator is essentially determined by 12n? and the
value of the denominator by 72n>. Thus, for large values of n, the sequence x
12728
el
behaves much like the constant sequence w,, = On
6 We claim (correctly)
thats lin
n>
==
©
For the next example of a proof that a sequence converges, we need to first deter-
mine the limit and then find a relationship between n and ¢ before giving the proof.
3n?
converges.
9)
n+ |
We make a guess that x converges to 3. Next we must show that for every
Example.
Prove
that
the
sequence
x
:
given
by
x, =
~—Tt
€ > 0, there is a natural number N such that n > N implies
|— 5 ‘a 3
3n?
os
3
;
Because
= 3=
|=
|
- , we require an integer N such
et]
Neal
Cpt
that n > N implies
rare <€
ie
or, equivalently, n* + 1 > 2, ae know
that
n+ 1>n,s80 by selecting N to be any natural number greater than >= we see that
n > N implies n? +1>N> = The formal proof now follows.
Proof.
Then
A
Let e > 0. Let N be a natural number greater than =:
> Suppose that n > N.
n > ,, and because
1
oy
LhUs
2)(ke
ii =F 1
n*?+1>n,
=o)
nr+1>
< €. Therefore,
2 Therefore,
lim
3n?
n— oo n- +
a
]
if n > N, then
a
4.6
Sequences
247
We use a denial of the definition of convergence to prove that a sequence x
does not converge to a number L:
(de > OWWNEN)(FrENn>NA
ee
Tell) es 2),
Such a proof starts with a judicious choice for ¢, which is usually found by examining the pattern of the terms of the sequence x. Here are two examples.
Example.
Siar aa) e
Prove that the sequence y given by y, = pire diverges.
You would expect that the sequence y diverges because its terms are the alternating sequence 1, 2, 1, 2, 1, 2, ... . If the limit L existed, then both 1 and 2 would
have to eventually be within distance e of L. By choosing ¢ = l we will guarantee
»
that this cannot happen.
Proof.
Suppose that the limit for y exists and is L. Let e = 5.LetNEN JEL > 3,
let n be any odd integer greater than N. Then y, = 1 and |y, — L| > u =e, Onithe
a
other hand, if L < 2. let n be any even integer greater than N. Then y, = 2 and
Yue Ll 2 : = &. In both cases, there exists a natural number n such that n > N
and |y, — L| = e. Therefore, y does not converge to L. This is a contradiction.
Thus, y diverges.
a
Example.
Prove that the sequence x given by x,, = 2n diverges.
The terms 2, 4, 6, 8, 10, ... of x, are increasing in such a way that they even-
tually surpass any number that might be suggested as a limit. Any positive real
number is an appropriate choice for ¢.
Proof.
Assume that
lim x, = L for some positive real number L. Let ¢ = 1.
n>
Then for some NEN,
©
|x, — L| < 1 for all n > N. Suppose that n > N and
n> ZL +1. Then 2n > L +1, so |x, — L| = |2n—L| > 1 =e. This is a
contradiction. We conclude that x diverges.
®
The next theorem is useful for determining and verifying limits without directly
using the definition. Sometimes referred to as the Sandwich or Squeeze Theorem,
it states that if a sequence b has its nth term “sandwiched” below by a, and above
by c, for all n € N and if both a and c converge to a number L, then b must also
converge to L.
Theorem 4.6.2
\
Suppose that a, b, and c are sequences of real numbers such that a, < b, < c,, for
alln e N. Ifa, — Land c, — L, then b, — L.
Proof. Suppose a, — L and c, — L. Let ¢ > 0. There are natural numbers N,
and N, such that n > N, implies |a, — L| < ¢ andn > N, implies |c, — L| < ¢.
Met beahelarvernolNV,N, Becaise d=0) oC
4a, — L=b,—- Lac, — L.
Therelote, tor = N,-6€
—a,—L=b,— L=<c —L < e. Thus, |b, —L| <«
forall > N, sob, — L.
|
248
CHAPTER 4
Functions
To illustrate Theorem 4.6.2, consider the sequence whose nth term is
Example.
sin
1
i
saaig Because sine is a function with range [—1, 1], =a <
1
< 55for all
1
1
sin 11
=> OL
n € N. Because both ah ak O and Fie 0, we conclude that
oO
Exercises 4.6
List the first five terms of each sequence.
=
n+
1
b)
OE eeonl erana
: (>)
b, = sin|—
bait
2
=n
1
=
(c)
Ch
(e)
ef= =
n!
d,, =
(f)
7 =
he
(it)
ee)
ee
2
a
1
(d)
2
n!
2
l
(oye. =
ieee
Vn
1
lee) n
Decide whether each sequence in Exercise | converges. If the sequence converges, identify or estimate the limit.
For each sequence x, estimate
lim x, or determine that it does not exist.
Ti—>
(a)
x, = 10n
oy
=
:
CO)
(oy
An® + 7
iO
2
se
10
Fa
3
(jee
eee
8n° + An 4-1
2
eee
10n+ 7n + 5
6n
OL a= ae
line —-n+5
+ 5n* + 3n + 8
(f) x, = 10n? + Tn?: + 5n — 8
1 \2n
(2)
1
(1— z
(Dither = 0:9)2
1\
(kK) x, =(-1" + (1 4 7)
n!
(m) x, = 7
(a)
(b)
\ae?
Daa
(1+ 7)
(Dares
11)"
OY
gn
eas
=
32
1
(nO
ena) a
Let KER, and let c be the constant sequence given by c,, = K. Prove
that c converges.
Describe all possible sequences x of natural numbers that converge with
limit 2.
Determine whether each sequence converges, and prove your answer.
1
(2) Be
a
(D)
a
55
4.6
(c) x,=2"
(d)
exe
(Ci
(fee
Sequences
249
ib
Salli
2n+
1
cos n
(3)
apo
5
(i)
a
*
x,=Vn+l-—
id Sai)
x,
—
a
a
no +2
(k) x,=
5,000
(ny
—
n
0)
(h)
Vn
6
oa
1
Wo
(Tye
n!
n!
=
e
—
Uae
sin |—
n
(i)
1
x, = iT
(p)
(5)
Dy
x, =(Vn+n—n)
n
Prove that if x, — L, y, > M, andre
(a)
Rot
R, then
Vn ae eva
(C)
(e)
x,y, — LM. (Assume L # 0.)
CO
ee ile PSE
Prove that if x, — L, y, —~ M,
(a)
(b)
(b)
Be DS Pipe ofbe
(CN
eee aes aby
L#0, M #0, and x, #0
(Ub
for all n, then
there is a number N such that if n > N, then |x,,| > ee
y,,
M
ae
(a)
Given convergent sequences x and y, prove that if x, = y,, for all natural
numbers n, then lim x, = lim y,.
(b)
Use part (a) to prove that if a sequence x converges to a real number L
and x, = 0 for all natural numbers n, then L = 0.
Prove or find a counterexample: Given convergent sequences x and y, if
x, <y, for all natural numbers n, then lim x, < lim y,.
(c)
=a (2,2)
n—
co
n— oo
n—> oo
A sequencex diverges to infinity, written lim x, = oo,orx, — ov, iffor
n>
wo
all natural numbers / there exists a natural number n such that if n > N,
10.
then x,, > M. Prove that each of these sequences diverges to infinity.
1
(a)
tan
(b)
(c)
x, = Vn
eee
(a)
Give a precise definition of the notion that a sequence x diverges to
negative infinity, written lim x, = —oo, or x, > —O0.
(b)
(c)
Prove that the sequence x, = 10 — 3n diverges to —oo.
n>
©
Prove that the sequence x,, = —n’ diverges to —oo.
250
CHAPTER 4
Functions
11.
A sequence y, is a subsequence of x, if and only if there is an increasing function f: N — N such that y, = x/q). For example, y, = Xz, is the
sequence whose terms are just the even-numbered terms of the sequence x,,.
n
(a)
Letxy = sere
Describe the subsequences x5, and x,,,_ |.
(b)
Prove that if a sequence x converges to L, then for every real e > 0,
(c)
there exists a subsequence y of x such that |y, — L| < ¢ foralln Ee N.
Prove that if x, converges to L and y, is a subsequence of x, then y
converges to L.
(d)
12.
Prove that if x contains two convergent subsequences y and z, y, > M,
Z, > L, and M # L, then x diverges.
In some cases a sequence x may be defined inductively by specifying a value
for the first term and then specifying x, for n > 1 in terms of earlier values in
the sequence.
(aoe
(D))
(c)
(d)
Let x,
310 and tor
= —] x27. Find the first six terms of 4,
w= 18
and determine whether x converges.
Leta, = 1 and, forn >
x, = 1— x, 2) Find the tirst.six terms or x,
and determine whether x converges.
Find the first ten terms of the Fibonacci sequence f, where f; = 1,
j= Leand, torn = 27) = 7
pad, 2 Does
7 COlVverse?
If it exists, estimate the limit of the sequence x, where
3
x,
=
al
5
a
ifn IV 2
Xn
(e)
If it exists, estimate the limit of the sequence x, where
See
i
Proofs to Grade
13.
{
1
nS
iN aol
ie
2
;
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
(a) Claim.
If two sequences x and y both diverge, then x + y diverges.
“Proof.”
Suppose that lim (x, + y,,) = L. Because x diverges, there
n>
Co
exists €, > 0 such that, for all NE N, there exists
ts)
ib
8)
all
NEN,
n > N such that
= &. Because y diverges, there exists ¢, > 0 such that, for
there
exists
n> N_ such _ that
-¥
Zab
SS ,min{ €,, €)}. Then for all N € N, there exists n > N such that
(X, a Yn) =
L| >
lV
LL
4.7
Limits and Continuity of Real Functions
251
= €} =P E>
= ne ar re
Pe
Therefore,
lim (x, + y,) #L.
n+
x
(b)
5
©
Claim.
If the sequence x converges and the sequence y diverges, then
x + y diverges.
“Proof.” Suppose that x, + y,, — K for some real number K. Because
X,—> L for some number L, (x, + y,) — x, K —L; that is,
Y, — K — L. This is a contradiction. Thus, x, + y,, diverges.
a
(c)
Claim.
Sn — 0 7
lim
n>oo
“Proof.”
n> + |
Let ¢ > 0 be given. We may assume that ¢ < 11. Then
i
wai
D> 0; Chooses
n>,
iLih
/——
€
r+
iil
err. 1, and suppose that n > N. Then
=
i
1
ne > ——
€
I, so
ly thus.
;
|S
11
€
somwenhave
5n*—6
—11
11
< €. Therefore), |x, —5|=|—>
—5/=
=
Kn
|
n= 1
| aa
ee
We conclude that
4.7
=i.5)5
lim
n>oo
5n2
— 6
.
ces
n +
;
1
a
Limits and Continuity of Real Functions
This section examines the concepts of limits and continuity of real functions. Our
emphasis is on proof strategies involving limits rather than the computational aspects
of limits usually found in a beginning calculus course. We will connect these ideas
with the results about sequences of real numbers from the previous section.
DEFINITION
Let a be a real number, / be an open interval that contains
a, and
fbe a real function such that fis defined on /, except possibly at a. We
say that the limit of fas x approaches a is L, and write
lime x)
a (Ona
L asx >a)
X— > a
if for every number ¢€ > 0, there exists a number 6 > 0 such that
if 0 < |x —a| <6, then |f(x) —L| <.e.
252
CHAPTER 4
Functions
This definition says that we can make the distance between f(x) and the limit L
as small as we want (smaller than ¢, no matter how small ¢ may be) by choosing x
to be any number that is close enough to a (that is, x is at a distance less than 6 from
a) but different from a. The value of f at a plays no part in this definition, so fmay
be undefined at a or f(a) may be equal to L or to some other number.
As was the case with sequences, to prove that a limit exists and is equal to L,
we must begin with an arbitrary ¢. Before proceeding with the proof, we generally
work backward from the conclusion, |f(x) — L| < ¢, to find a relationship between
x and a that we use to choose a 6 that will satisfy the definition.
The choice of 6 usually depends on f, a, and ¢, and for a smaller ¢, we must
find a smaller 5. The greatest integer function int(x) is an exception. Suppose that
we must prove that _jim int(x) = 4, (See Figure 4.7.1.) Given ¢ = 3, we could
choose 6 = 2 because when 0 < |x — 4.8| < 2, int(x) is one of 2, 3, 4, 5, or 6, so
|int(x) — 4| < 3. However, the choice of 5 really doesn’t depend on ¢ because for
every positive ¢, we may choose 6 = 0.2. This is because if |x — 4.8| < 0.2, then
int(x) is 4 exactly and thus |int(x) — 4| =0 <e.
(
6
e—6
5
Ah
o——
4
=
3
e—o
2
1
|
|
|
|
eo)
|
|
e—o
|
'a=4.8
rl ae a te
tae
LOM
ile
us NE
TH We STUNG:
ony
Figure 4.7.1
For a linear function f, the choice of 5 depends on ¢ and on the slope of f but
not on the value of a. To show that the linear function f(x) = mx + b (with m 4 0)
has limit f(a) as x approaches a, we work backward from the inequality we want:
|f(x) — fla)| < ¢. Rewriting this as |mx + b — (ma + b)| = |m| |x — a| < €, we
see that we need |x — a| < iar Thus, we choose 6 to be —. With this 5, we can
construct the proof.
7
4.7
Theorem 4.7.1
Limits and Continuity of Real Functions
253
Let fbe a linear function f(x) = mx + b. Then for every real number a, the limit of
fas x approaches a is f(a).
Proof.
Let ¢ > 0 be given. There are two cases:
Casel.
If m=O, then fis the constant function f(x) = b. Let 6 be any positive
real number. Then |f(x) — f(a)| =|b—b| =0 <e.
Case2.
If m#0, choose 6 = =tes) and suppose that 0 < |x — a| < 6. Then
|m|?
|) — fla) | = |mx + b — (ma + b)| =|m| |x-—a| <|m|d=e.
Therefore lim f(x) = fla).
|
wt
Sometimes |f(x) — L| < € can be rewritten because f(x) — L may be expressed
as a product of two or more factors. To complete a proof in this situation, we need
a two-step approach to choosing 6. Because we eventually intend to find a small 6,
we begin with a choice for 6 that serves to keep at least one of those factors small.
We can then choose, if necessary, an even smaller 6 that will keep the product from
getting large.
Example.
Let fbe given by fix) = x* + x — 6. Prove that lim Ax) = 6.
x3
To discover a proof, we work backward:
(Ax) —L) =|?
+x —6—6|
=|?
+2
—12|S |x 4+4| |x -—3] He:
Obviously |x + 4| will be small when x is close to —4, so we need worry only
about how large |x — 3| might be in comparison to ¢. For instance, if |x + 4] < 1 (we
chose 1 arbitrarily), then we would have |x — 3| =|x+4—7|<|x+4]|+|-—7| <8.
This suggests that we want 6 to be no larger than | and also no larger than e
Proof.
Let ¢ > 0 be given. Choose 6 = min {1, a Suppose that |x + 4| < 6.
Then |x+ 4| < 1, so |x — 3| < 8. Thus, |x
+ 4| |x — 3| < 86 < e. Therefore,
lim | fx) ==) 6
®
iL —
We remarked above that the value of the function fat a is not relevant when we
find the limit as x approaches a. If two functions have the same values near a but
are not necessarily equal at a, then their limits as x approaches a are the same. This
useful fact is stated more precisely in the following theorem.
Theorem 4.7.2
Suppose that / is an open interval containing the real number, a; f and g are two
functions defined on J, except possibly at a; and f(x) = g(x) for all x € J — {a}. If
lim f(x) exists, then lim g(x) exists, and lim g(x) = lim f(x).
ae at04
Proof.
Xa
Xa
X=
a
Assume that f and g are as specified, and let lim f(x) = L. Suppose ¢ > 0.
Choose 6 so that if 0 < |x — a| < 4, then |f(x) — i) = €. If necessary, choose a
254
CHAPTER 4
Functions
smaller 5 so that |x — a| < d implies x € I. (This is where we use the fact that I is
an open interval.) For that 5, suppose that 0 < |x — a| < 6. Then |f(x) — L| <e.
a
Because f(x) = g@(x) for all x € J — {a}, we conclude that |g(x) — L| < €.
Theorem 4.7.2 enables us to find the limit of a function by replacing that function with a different function whose limit is easier to determine.
Example.
Let g be the function given by g(x)= x + =,—
=
iim1 g(x), we begin by simplifying: g(x) =
ae) Fy Seiil
G+ M4) y
If we let
f be given by
ae=x +5, then fand g agree on the “punctured” interval (1, 3) — {2}. Because
lim f(x) = 7, it follows that lim g(x) =.
=
x=
RZ
The limits of real functions and the limits of real sequences are closely
connected. We saw above that lim (x + x — 6) =6. Now think of any sequence
=
s of real numbers within an open interval containing —4 that converges to —4.
For instance, let’s suppose that s the sequence with terms in the interval (—8, 0) is
given by
Then the corresponding sequence of images under
f is f(s,;) = 14, f(sy) = 6.71,
f(s3) = 6.0701, fts,).= 6.007001,..., which converges to 6. The same is true for
every sequence that converges to —4: The corresponding sequence of images
always converges to 6. This property provides a necessary and sufficient condition
for a function to have a limit. The proof involves two kinds of limits: the limit of a
function and limits of sequences.
Theorem 4.7.3
Sequential Characterization of Limits
Let a be a real number, / an open interval that contains a, and fa function defined on
I, except possibly ata.Then im f(x)= Lif and only if tim: J%,) = Lior every
Xi—> A
sequence x, with terms in/ — {a} such that lim x,= a.
n>
Proof.
@®
Assume that lim f(x)= L, and suppose that
xa
lim x, = a for a sequence
n>
x whose terms are in J — {a}. (We must show that the pee
limit jim1 fXy)
exists and equals L.) Let ¢ > 0. For this ¢, there exists 6 > 0 such that |- —ae <6
implies |f(x) — L| < e. Let N € N be such that n > N implies |x, — a| < 6. (We
know such an N exists because 5 > 0 and we are given that the sequence x, converges to a.) But |x, — a| < 6 for alln > N implies |f(x,) — L| < e for alln > N.
Therefore, the sequence f(x,,) converges to L.
Suppose that f does not have limit L as x approaches a. (Rather than proving
the converse directly, we prove its contrapositive.) Then there exists some ¢* > 0
such that for all 6 > 0 there exists x such that 0 < |x — a| < 6 and |f(x) — L| > e*.
4.7
Limits and Continuity of Real Functions
255
(We next buildasequencey,,that converges to a.) For a néEN, letd=- Then
for each 4,,, there is a point y, in / su that 0 < |y, —a| <>“and oa
L|>
2a
for eachn € N. From0 < |y, — a| <_, we see thaty, 4 a andalso that |y, — a| 1s
sandwiched between the constant moquenee 0 and the sequence , both of which converge to 0). Thus, by the Squeeze Theorem (Theorem 4.6.2), the sequence |y, — a|
converges to 0, which implies that y, converges to a. (Now show the corresponding
sequence of images f(y,) does not converge to L.) However, |f(y,) — L| = e* for
each n € N, which means f(y,,) does not converge to L. Thus, iffdoes not have limit
L as x approaches a, then there is a sequence y in J — {a} that converges to a such
that the corresponding sequence of images does not converge to L. Therefore, if all
sequences y in / — {a} that converge to a also have their corresponding sequences
fy,,) converge to L, then the limit as x approaches a of f exists and is L.
=
Theorem 4.7.3 provides a rationale for the way you may have first learned to
guess the limit of a function. Authors of beginning calculus texts usually ask the
reader to get a feeling for limits such as caeoe 7 by giving a sequence of values
De
approaching 3, such as 2.8, 2.9, 2. aay} sinh fh asking the reader to calculate
the corresponding images of f(x)=
We detect the pattern for the images oe mi the correct guess: lim,
or
24
5
Ve;
= 0),
Theorem 4.7.3 explains that this guessing process is a sound ne
but also
explains why the process is not infallible. We have not checked all the infinitely
many sequences that converge to 3. This leads to a second observation: The precise
é-6 definition of the limit of a function may seem challenging at first, but it provides
a way to prove limits that is generally easier than verifying the sequential property
for every possible sequence.
Theorem 4.7.3 also provides two ways to show that a function does not have
a limit at a. We can either look for a sequence x, that converges to a such that the
corresponding sequence f(x,) diverges or look for two sequences, both converging
to a, but such that their images under f converge to different points.
Example.
Let g be given by g(x)= sin > ston
n € N. Then s, converges to 0, but g(s,,) = sin Cn
7 On Letts, 2 apse for all
In is | when n is odd and
2,
—1 when n is even and thus does not converge. Therefore, lim18) does not exist.
x0
The alternate approach to showing that the function g does not have a limit at
7
Ois to lets, = and
t=
. Then s, and ¢, both converge to 0, but
(4n — 3)
n
(4n —
1)x
the corresponding sequences g(s,,) and f(s,,) are constant sequences that converge to
1 and to —1, respectively. Again, lim,g(x) does not exist.
Oo
xX?
There are other important consequences of Theorem 4.7.3. For example, we
can use the results about the arithmetic of convergent sequences from Exercises 6
256
CHAPTER 4
Functions
and 7 of Section 4.6 to establish corresponding results for limits of functions. Under
certain conditions, the limit of the sum, difference, product, or quotient of two func-
tions is the result of the corresponding operation on the limits of the functions. See
Exercise 3:
We have seen that the limit of a function f at a point a can sometimes be found
by simply calculating the value of f at a. In such cases, we say fis continuous at a.
Intuitively, saying that a function fis continuous at the point a means that its graph
is unbroken at (a, f(a)). In other words, lim f(x) exists (there is no abrupt change
in the graph for x near a), f(a) exists (there is no missing point on the graph corresponding to a), and the two numbers
lim f(x) and f(a) are equal (the value f(a)
does not create a “hole” in the graph). Inshort, we write
lim f(x) = fia).
Examples.
The function fgiven by f(x) = 1/xis not continuous at 0 because lim 1/x
=>
does not exist and f(0) is not defined. The function g, where g(x) = (sin x)/x, does
have a limit of | at x = 0, but g is not continuous at 0, because g(0) does not exist.
The function h given by h(x) = 2x + 3 is continuous at 0 because lim,(2x? + 3)=3
3
and h(O) = 3. The graphs of f,g, and h are shown in Figure 4.7.2.
y
10 =
A
10,4
5in
=10
ae eee
—5
A
5
een Se
=e
5
10
ae
-10
—5
=5
|
3%
=10-
10-3
g(x) = im
h(x)
= 2x? + 3
ree
Figure 4.7.2
The definition of limit and the above description of continuity may be combined to give a precise definition of continuity in terms of ¢ and 6.
DEFINITION
Letfbe areal function defined on an interval D containing a.
We say that fis continuous at a if for every ¢ > 0 there exists 5 > 0 such that
|fix) — fla)| < ¢ whenever |x — a| < dandxeD.
4.7
Limits and Continuity of Real Functions
257
A proof that a function fis continuous at a given point a is precisely the same
€-d argument used earlier to verify limits, only now the limiting value is f(a). Earlier
in this section, for the function f given by f(x) = x* + x — 6, we gave an e-6 proof
that dm fe) = 6. Because f(—4) = 6, that same proof shows that fis continuous
Aah
By Theorem 4.7.1, every linear function is continuous at every real number a.
It is also true that every polynomial function is continuous at every real number.
See Exercise 5. Other familiar functions such as the rational functions, root functions, and trigonometric functions are continuous at every point where they are
defined.
.
We finish this section with a proof that the composite of continuous functions
is continuous. To that end, it helps to first have a lemma about limits of composites.
Lemma 4.7.4
For functions f and g, suppose that lim f(x) = b and g is continuous at b. Then
lim g( fi) = g(@).
a
Proof. Let ¢ > 0 be given. (We must choose 5 > 0 such that if0 < |x —a| <6,
then |g(f(x)) — g(b)| < &.) Because g is continuous at b, for e > 0, there exists
o > O such that
if 0< |y
— b| <a, then |g(y)
— g(d)| < «.
And because lim f(x) = b, for ao > 0, there exists 6 > 0 such that
eal
if
0 < |x —a| <4, then |f(x) —b| <o.
Combining these statements, we see that 0 < |x — a| < 6 implies |f(x) — b| <o,
which in turn implies |g(f(x)) — g(b)| < e. Therefore, lim g(f(x)) = g(d).
a
Bee(8!
Theorem 4.7.5
For functions f and g, suppose that fis continuous at a and g is continuous at f(a).
Then g °f is continuous at a.
Proof.
Because
f is continuous
at a,
lim f(x) = f(a).
X=
By
Lemma
4.7.4,
a
lim (g © f)(x) = lim g(f(x)) = g(f(a)) = (g 0 f (a). Therefore, g © f is continuous
3 Sa!
Nae &
at a.
B
Exercises 4.7
1.
Find the limit and prove your answer using the e-6 definition.
:
ee
lim (0.05x + 1)
(d)
Cee ea ag
(c)
fe al
:
x—4
lim |x|
x0
258
CHAPTER 4
Functions
ee
MP Ele
|
e
tae
tim? x
Uy
wyBN
2S
(i)
lim(x? + 8x + 4)
(j)
lim @? — 8x + 3)
Mera
* (k) lim(? — 5)
()
(m) limx?
(n)
x>4
(a) f="
On =
Oe
tim Vx
lim!
x>3%*
(b) fix) =cos~I
(@d) fx)=3se
Nee sal0)
3x + 2 ifx <0
be Teilieif ia Vlgred
sivas Lat fs 43 ifx>0
Letabe areal number; / an open interval that contains a; and fand g ee:
defined on J, except possibly at a. Prove that if lim f(x)= L, iim Bt) ==
and r is a real number, then
*
ice
Prove that lim f(x) does not exist.
x0
3.
x2
(3)
re te Il
2.
lim
eee
oe
(a)
jim f+ g(x) = L+M.
(b)
lim(—
g)\x%) =L-M.
()
lim(/2)@)
= LM.
(@) limrfix)=
(e) If g(x) #0 forall xe J, L #0, and M £0, then lim (£je =4
xX a
*
(f)
forallneN,
limx” =a".
Jara!
4.
Use the results of Exercise 3 to show that
* (a) ime)
=15
x>5
b
L
:
(c)
x2
+
5x
=
6
——_—.
:
Bock (eee al \apeat
tim ( x +1
ae.
5.
Let abe areal number. Prove that every polynomial p(x) = b,x" + b,_x"~! +
- + bp is continuous at a.
6.
(a)
A continuous function f with domain R has the property that for x # 1,
fo) = eae What is f(1)?
(b)
Ifpossible, provide a definition for f(1) so that fis continuous at 1, where
bs
Of) = f8aL)
hige Se ||
1
ace
4.7
Limits and Continuity of Real Functions
259
ye == By lta Ss |
ii
=
USA) = " —x
ifx<0O
sin =
iiioe SI
2)
(iti) f(x) =
a
cos 53 +x)
ifx<l
7.
Prove that if fand g are continuous at a, then
f+ g is continuous at a.
8.
Using the e-6 definition, prove that each of these functions is continuous at the
given point a.
9.
(a)
f{o=]=x —3ata=2
(b)
fa) =3x 41ata—1
(c))
xy)
=/x=
Oo)
= |x
(d)
Die ts
f(x) = ticer
3
—
hata)
ie SS 2
bas
es Alt G3 = 2!
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
(a) Claim.
If lim f(x)= L, then im1 LA) | === eh
Xa
“Proof.”
Assume that lim oe= L. Lete = 0 be given: There exists
xa
5 > 0 such that|f(x) — L| < ¢ whenever 0 < |x — a| < 6. Suppose that
0 < |x — a| < 6. Then, using Exercise 6(g) of Section 1.4, we have
ILA) |— |£I|< |fx) — L| < «. Therefore, iim1 LA =U)
a
(b)
Claim.
Let f be a function defined on Aegal
Tanda
eI. If
lim f(x) exists, it is unique.
xa
“Proof.”
Suppose
that iim1 fx) = EL and lim1 A) = M. For each
neN, lets —a. then the sequence ic converges to a, and the sequence
(c)
fis,,) converges to f(a). By Theorem 4.7.3, if fhas a limit as x approaches
a, then for every sequence that converges to a, the corresponding
sequence of images under f converges to the same number, which is the
limit of the function. Thus, M = f(a) and L = f(a), soM = L.
a
Claim.
The function f given by f(x) = |x| is continuous at a = 0.
“Proof.” First, f(0) = |0| =0. Second, let e > 0. Because ||x| — 0| =
esl
hoe) = lee 0] aie, letd— e, Thenst |x| <:6, |x |<. 6. Therefore.
ae = (0. Thus, fis continuous at 0.
a
ie}
eT
*%
x
%
——,
»4
\a
a
j
7
-
al
a
7
{
|
4
ery,
o
“p
AE ED nieti*i!
zs
S Wie fon
_
ey
Li)
Pk
Pe
roy| dy
ee ha al a6 Wh
car Fwy the May
‘
fear
pvt?
ee
Liipwuil)
eru* H| Gaapiptendl he
cbirbapat|
An) (i) geo (D
Le
Wy
very (ett {s @ bist 4 © Hh
=
p>
4H
(fi
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Rinne
da
7
D0
We
ca
ti
‘
:
;
CHAPTER 5
Cardinality
How many elements are in the following set?
A=
{x29a.
1
5° —3, x, X, ab
After a short pause, you said “eight.” You probably looked at z and thought “1,”
then looked at 28 and thought “2,” and so on up through @ and thought “8.” What
you have done is match up the set A and the known set of eight elements {1, 2, 3, 4,
5, 6, 7, 8} to conclude that A has eight elements. Counting the number of elements
in a Set is essentially setting up a one-to-one correspondence between the set and a
known (standard) set of elements. Here is another example.
A shepherd has many dozens of sheep in his flock, but he cannot count beyond
10. Each day he takes all his sheep out to graze, and each night he brings them back
into the fold. How can he be sure all his sheep have returned? The answer is that
he can count them with a one-to-one correspondence.
He needs two containers,
one empty and one containing many pebbles, one pebble for each sheep. When the
sheep return in the evening, he transfers pebbles from one container to the other,
one at a time for each returning sheep. Whenever there are pebbles left over, he
knows that there are lost sheep. The solution to the shepherd’s problem illustrates
the point that even though we may not have counted the sheep, we know that the
set of missing sheep and the set of leftover pebbles have the same number of
elements—because there is a one-to-one correspondence between them.
In this chapter, we will make precise the idea of the number of elements in a
set. We will discuss finite and infinite sets and discover that there are different sizes
of infinite sets.
5.1
Equivalent Sets; Finite Sets
To determine whether two sets have the same number of elements, we see whether
it is possible to match the elements of the sets in a one-to-one fashion. This idea
261
262
CHAPTER 5
Cardinality
may be conveniently described in terms of a one-to-one correspondence (a bijection) from one set to another.
DEFINITION
Two sets A and B are equivalent if there exists a oneto-one function from A onto B. The sets are also said to be in one-to-one
correspondence, and we write A ~ B.
If A and B are not equivalent, we write
A # B.
The sets A = {5, 8, 7} and B = {r, p, m} each have three elements and thus are
equivalent. If we wanted a formal proof that A ~ B, we could describe any of the
six possible one-to-one correspondences from A to B—for example, the function
16,7); (8, p)s Gr m)}.
The sets C =
{x, y, z} and D =
{p, q} are not equivalent. There are eight
different functions from C:to D, but none of them is one-to-one. (We will see later
in this section why this must be so.)
Example.
integers.
Prove that the set E of even integers is equivalent to the set D of odd
Proof. Let f: E— D be given by f(x) = x + 1. Then f is one-to-one because
f@) =f)
implies x + 1=y-+1, which yields
x=y. Also, f is onto
DM becausew at Zz “isany odd integer) then w=%z—1
is even, and
fw)
=w+1=(<-1)+
1 =z. Therefore
E © D.
|
Example.
For a, b,c, d € R, with a < b and c < d, the open intervals (a, b) and
(c, d) are equivalent.
Proof.
(There are many bijections from (a, b) to (c,
d). We choose the simplest:
a linear function.) Let f: (a, b) > (c, d) be given by
fa) =
ad=c
b—a
(x —a) +c.
See Figure 5.1.1. Exercise 4 asks you to prove that fis a bijection.
y
A
dt
y= sf)
lL
|
[sememnes
a
b
Figure 5.1.1
a
5.1
Equivalent Sets; Finite Sets
263
This last example shows that any two open intervals are equivalent, even when
the intervals have different lengths. In particular, the example says the open interval
(1, 3) is equivalent to (0, 20), even though the interval (0, 20) includes (1, 3) as a
proper subset and is 10 times as long as (1, 3). Your intuition might tell you that
(0, 20) has to have “more elements” than (1, 3), but that would be incorrect.
Example.
Let % be the set of all binary sequences—i.e., the set of all functions
from N to {0, 1}. Prove that
# ~ P(N), the power set of N.
Proof.
Every function in & is a characteristic function with domain N. We associate each function with the appropriate subset of N by defining h: # + P(N) as
follows:
torge@o,
(eg) = {re N-e@) = 1),
Thus, under the function h, every function in ¥ has an image in P(N).
To show h is one-to-one,
let g,,g,€
# and g, #g,.
Because
g, and g,
are different functions with the same domain N, there exists m€N such that
g,(m) ¥ g,(m). Without loss of generality, assume that g,(m) = 1 and g,(m) = 0.
Then me {xEN:¢,%) = 1} =h(g,) and mé€ {xEN: g(x) = 1} = he).
Thus, h(g,) # h(g>).
To show that h is onto P(N), let A e P(N). Then A CN, and the characteristic function X,:
N— {0,1} is an element of &. Furthermore, h(X,) =
{x
Theorem 5.1.¥
EN: Xx) = 1} = A. Therefore h is onto P(N).
Because h is a bijection, # ~ P(N).
a
Equivalence of sets is an equivalence relation on the class of all sets.
Proof.
See Exercise 1.
P|
The next lemma will be particularly useful for showing equivalences of sets.
Lemma 5.1.2
Suppose that A, B, C, and D are sets with A © C and B © D.
(a)
(D)i
IfA and B are disjoint and C and D are disjoint, then
Ae Be Gap:
AU BY
CU
D.
Proof. Assume that A ~ C and B ~ D. Then there exist one-to-one correspondences h:A > C and g:B => D.
(a)
By Theorem 4.3.6, h Ug: A
Therefore,
AU B > CUD.
UB — CU Dis a one-to-one correspondence.
(b)
Let f:A x B>CxD
be given by f(a, b) = (h(a), g(b)). We leave
it as Exercise 5 to show that f is a one-to-one correspondence. Therefore
AxBrCxD.
a
264
CHAPTER 5
Cardinality
Examples. The set {a, b} is equivalent to {1, 2}, and the set {3, 4, x} is equivalent
to {5, 6, x}. To apply Lemma 5.1.2(a), we note that {a, b} and {3, 4, x} are disjoint
and that {1,2} and {5, 6, x} are disjoint. Therefore, {a, b, 3,4,x} ~ {1, 2,5, 6, x}.
By Lemma 5.1.2(b), the cross product of {a, b} and {3, 4, x} is equivalent to
the cross product of {1, 2} and {5, 6, x}:
{(a, 3), (a, 4), (a, x), (b, 3), (b, 4), (b,x) } = {C1,5),C1,6), (1,¥), (2,5), (2, 6), (2, *) J.
DEFINITIONS
For each natural number k, let N, =
A set S is finite
if S=
{1, 2, 3,..., k}.
© or S > N, for some
k E N.
A set S is infinite if S is not a finite set.
You should think of the set N, as the standard finite set with k elements against
which the sizes of other sets may be compared. For example, N, = {1, 2, 3, 4}
is the standard set with four elements. The set S = {f, ae 99} is finite because
SN,
Sets such as N, R, and {M 4 i : wih are examples of infinite sets. These and
other infinite sets will be discussed in the next section.
DEFINITIONS © Let S be a finite set.
_ifS= ©, then S has cardinal number 0 (or cardinality 0). We write
Ss) 0;
If S ~ N, for some natural number k, then S has cardinal number k
(or cardinality k), and we write S = k.
The set A = {z, 28, V2, 5,—3, x, X, a} has cardinality 8 because A ~ Ne.
Thesetb = {8, 7,3, 7, 2, 7, 8} 1s finite, and B = 4because B =
(8, 7,3,2)
SNe
The identity function Jy:N;, — N, is a one-to-one correspondence, so N, ~ N,
and N, = k.
_
Our definition of A for a finite set A agrees with our intuitive notion that Ais
the number of elements in A. We use the same symbol A for the cardinality of a
finite set A as we used in Section 2:6 for the number of elements in A.
Because the definition of finite has two parts, proofs that a set is finite usually
have two cases—the empty set case and the case in which the set is equivalent to
N, for some k € N.
Theorem 5.1.3
If A is finite and B © A, then B is finite.
Proof.
Suppose that A is finite and B ~ A. If A = @, then B = © (see Exercise
6). If A © N, for some natural number k, then B © N, by transitivity of ~. In either
case, B is finite.
|
5.1
Equivalent Sets; Finite Sets
265
Our next goal is to show that every subset of a finite set is finite. The proof uses
the results of the following two lemmas.
Lemma 5.1.4
If Sis a finite set with cardinality k and x is any object not in S, then § U {x} is finite
and has cardinality k + 1.
Proof. If
S= ©, then S has cardinality 0. Then SU {x} = {x} is finite
because it is equivalent to N,. In this case,
SU {x} has cardinality 0 + 1 = 1.
If SA, then S + N, for some natural number k. Also, {x} © {k + I}.
Therefore, by Theorem 5.1.2(a),$ U {x} ~ N,U {k + 1} = N,.4. Thus, SU {x}
is finite and has cardinality k + 1.
a
Lemma 5.1.5
For every k € N, every subset of N, is finite.
Proof. Let k be a natural number. (We prove by induction that every subset of N,
is finite.)
@)
Ifk=1andACN,,
(ii)
Suppose, forsomek € N, that every subset of N, is finite. Let AC N,, ;. Then
A — {k+ 1} is a subset of N, and, by the induction hypothesis, is finite.
then A =
O orA =N,. In both cases, A is finite.
If A=A — {k + 1}, thenA is finite. Otherwise,
A = (A — {k + 1})U
{k + 1}, which is finite by Lemma 5.1.4. In both cases, A is finite.
By the PMI, every subset of N, is finite for every k € N.
Theorem 5.1.6
|
Every subset of a finite set is finite.
Proof.
Assume that S is finite and TC S. If T = ©, then T is finite. Thus, we
may assume that T# © and hence S # ©. Then S © N, for some k € N and there
is a one-to-one function f from S onto N,. Then the restriction of f|, of
fto Tis a
one-to-one function from T onto Rng (f|7). Therefore, T is equivalent to Rng (f|7)
(see Figure 5.1.2). But Rng (f|,) is a subset of the finite set N, and is finite by
Lemma 5.1.5. Therefore, because T is equivalent to a finite set, T is finite.
rT]
1-1, onto
Figure 5.1.2
266
CHAPTER 5
Cardinality
Although the theorems of this section may seem nothing more than obvious facts, the real value of these results lies in the reasoning and in the use of
functions to establish facts about cardinalities of finite sets. This experience
will be helpful when we deal with infinite sets because there our intuition may
fail us.
The next main result is that the union of a finite number of finite sets is finite.
We first prove the case for two disjoint sets. The proof is a rigorous development
of the sum rule for two sets, which says that if A has m elements, B has n elements,
andA and B are disjoint, then
A U B has m + n elements. Part (b) of the theorem is
a restatement of Theorem 2.6.3.
Theorem 5.1.7
(a)
(Theorem 2.6.1) IfA and B are disjoint finite sets, then A U B 1s finite, and
(b)
(Theorem
AUB=A+B.
2.6.3) If A and B are finite sets, then AUB
is finite, and
AUB=A+B—ANB.
(c)
If A,,A>,...,A, are finite sets, then ) A, is finite.
i=]
Proof.
(a)
Suppose that A and B are finite sets and
AUB = B; if B = ©, then
AN B=.
If
A=,
then
AUB = A. In either case, A U B is finite, and
because
@ = 0, AUB=A+B.
Now
suppose that
A#@ and B¥M.
Let
AXN,, and
BYN,
and suppose that f: A> N,, and g: B— N,, are one-to-one correspondences. Let H = {m+ 1,m+2,...,.m+n}. Then h:N, — Z given by
h(x) =m +x is a one-to-one correspondence, and, thus, N, ~ H. See
Figure 5.1.3. Therefore,
AUBSN,
B ~ H by transitivity. Finally, by Lemma
UH=N,,,,- Therefore A U B is finite and
1
2)
NG
3
Nin
:
m
Naor
|
Z
2
N,
h
n
Figure 5.1.3
H
ear
Il
igh
9 2
m
ae 3
maint
5.1.2,
AUB = m + n.
5.1
(b)
Assume that A and B are finite sets. Because
Therefore,
by part (a),
AUB=A+B—AN
(c)
Equivalent Sets; Finite Sets
AUB=AU(B-—
267
B — A C B, B — A is finite.
A) is finite. The
proof that
Bis Exercise
8(a).
The proof of this part uses mathematical induction. See Exercise 8(b).
a
Lemma 5.1.4 shows that adding one element to a finite set increases its cardinality by one. It is also true that removing one element from a nonempty finite set
reduces the cardinality by one. The proof of Lemma 5.1.8 is left as Exercise 13.
Lemma 5.1.8
Letre N withr > 1. Forallxe N,N, — {x} © N,_).
There is a property of finite sets popularly known as the Pigeonhole Principle. In its informal version, it says: “If a flock of n pigeons comes to roost in a
house with r pigeonholes and n > r, then at least one hole contains more than one
pigeon.” If we think of the set of pigeons as N,, and the set of pigeonholes as N,,
then the Pigeonhole Principle says any assignment of pigeons to pigeonholes (function from N,, to N,) is not one-to-one.
Theorem 5.1.9
The Pigeonhole Principle
Letn,
re Nand f: N, > N,. Ifn > r, then fis not one-to-one.
Proof. The proof proceeds by induction on the number n. Because n > r and r is
a natural number, we begin with n = 2.
(i)
Ifn = 2, thenr = 1. In this case, fis a constant function with f(1) =
1. Thus fis not one-to-one.
1 and
f(2) =
(ii) |Suppose that the Pigeonhole Principle holds for some integer n; that is, sup-
pose that for all r < n, there is no one-to-one function from N,, to N,.. Let
r <n-+
1. (The proof now proceeds by contradiction.) Suppose that there
is a one-to-one function h: N,,,; > N,. The restriction hj, of h to N,, is
one-to-one. The range of this function OR not contain h(n ‘ 1). We rie
assume that r > | because otherwise h would be a constant function, which
is not one-to-one. Now by Lemma 5.1.8, there is a one-to-one function
eg. N. — {htm + 1)} > N,_). Let f= g° (Aly ). Then f: N, — N, is oneto-one because the composite of one-to-one functions is one-to-one. This is
a contradiction to the hypothesis of induction.
By the PMI, for every n EN if r <n, there is no one-to-one function from N,,
to N,.
a
The Pigeonhole Principle is surprisingly powerful. See the discussion and
references in Martin Gardner’s The Last Recreations (Springer-Verlag, New York,
1997) and the examples in Exercise 21. It also provides the following useful results
about finite sets.
Corollary 5.1.10
If A =n,
B =r, andr < n, then there is no one-to-one function from A to B.
Proof:
See Exercise 14.
r
268
CHAPTER 5
Corollary 5.1.11
Cardinality
If A is finite, then A is not equivalent to any of its proper subsets.
Proof. We will show that N, is not equivalent to any of its proper subsets and
leave the general case as Exercise 15.
If k = 1, the only proper subset of N, = {1} is ©, and {1} is not equivalent
to @. Thus, we may assume that k > 1. Suppose that A is a proper subset of N, and
A ~ N,. Then there is a one-to-one function f: N, — A that is onto A.
Case 1.
Case 2.
Suppose that k ¢ A. Then A C N,_,. In this case, the function
f maps N,
to N,_,, and fis one-to-one. This contradicts the Pigeonhole Principle.
Suppose that k € A. Because A is a proper subset of N,, the set N, — A is
nonempty. Choose an element y € N, — A. Let A’ = (A — {k})U {y}.
Then A ~ A’ because the function J,_;,; U {(k,y)} is a one-to-one
correspondence. Thus, A’ + N,, A‘ is a proper subset of N,, and k ¢ A’. This
is the situation of Case 1 with N, and A’ and again yields a contradiction. m
The Pigeonhole Principle tells us that our definition of cardinality for a finite
set A corresponds to our informal understanding that A is the number of elements
in A: The cardinality of a finite set is unique. That is, if A+ N, and A © N,,,m? then
fe — mt. see Exercise 20.
Exercises 5.1
1.
Prove Theorem 5.1.1. That is, show that the relation
ric, and transitive on the class of all sets.
2.
Find the cardinality k of each of these finite sets and a one-to-one correspondence from the set to N,.
(A) lez 4 8, 16, 32) 6471287 256, 512}
3.
*
(bye
(Cy
ire
Co x= 1}
ee 1
(d)
{@ y)EN
~
is reflexive, symmet-
x Nix+y
<6}
Use the results of this section to explain why each set is finite.
(a) {1,2,3,...,
479, 480}
(b) [3,5] — (3,5)
(CP Sao ers Wl}
(Qo
eo.
(e) {1, Co 2,3 Se
oan |
Le 2A Ot
8 A830)
(g)
(h)
{1,2,3..em 480) Uae 83-2 3 e
{4,5,479, 480, 3-20, 3-2!. ead a
(i)
J A,, where for each i, 1 < i < 9, A, = {10i, i+ 103}
b]
3ee en
3
%)
peal
9
(j)
LU) A;, where for each i, 1
<i < 9,A;=N,
i=1
4.
Complete the proof that any two open intervals (a, b) and (c, d) are equivalent
:
d-—c
by showing that f(x) = (<= a — a) + c maps one-to-one and onto (c, d).
5.1
Equivalent Sets; Finite Sets
269
Complete the proof of Lemma 5.1.2(b) by showing that if h: A > C and
g: B + D are one-to-one correspondences, then f:A x B—
by f(a, b) = (h(a), g(b)) is a one-to-one correspondence.
Show that if A ~ @, then
C x D given
A = ©. (See also Exercise 5, Section 4.1.)
Let A and B be sets. Prove that
(a) if A is finite, then A / B is finite.
(b) if A is infinite and A C B, then B is infinite.
Complete the proof of Theorem 5.1.7 by proving that
(a)
if A and B are finite sets, then
AUB=A+B—ANB.
(D) sdAisAge ae os A,, are finite sets, then UJ A, is finite.
pill
(a)
(b)
Show thatA ~ A x {x}, for every set A and every object x.
Use part (a) and Theorem 5.1.7(c) to prove that if A and B are finite,
A x Bis finite.
10.
Define B4 to be the set of all functions from A to B. Show that if A and B are
finite, then B“ is finite.
11.
If possible, give an example of each of the following:
(a) an infinite subset of a finite set.
(b) acollection {A;: i € N} of finite sets whose union is finite.
(c) a finite collection of finite sets whose union 1s infinite.
(d) finite sets A and B such that A UB#A + B.
Prove that if A is finite and B is infinite, then B — A is infinite.
Prove that ifr > 1 and x EN,, then N, — {x} ~ N,_, (Lemma 5.1.8).
Prove Corollary 5.1.10.
Complete the proof of Corollary 5.1.11 by showing that if A is finite and B is
a proper subset of A, then B # A.
Prove or disprove:
(a)
(b)
If Cis an infinite set and C = A UB, then at least one of the sets A or B
is infinite.
Suppose that A is a set and p is an object not in A. If A © A U {p}, then
A is infinite.
iy:
Prove by induction on n that if r < n and f:N,— N,, then fis not onto
No
18.
LetA and B be finite sets with A © B. Suppose that f: A > B.
(a) If fis one-to-one, prove that fis onto B.
(b) If fis onto B, prove that fis one-to-one.
19.
Prove that if the domain of a function is finite, then the range is finite.
20.
Let A be a finite set. Prove that if A + N, and A ~ N,,, then n = m.
21.
Give a proof using the Pigeonhole Principle:
(a) The Italian village of Solomeo has a population of 400. Prove that there
are at least two village residents with the same birthday.
270
CHAPTER 5
Cardinality
(b)
Let § C Nog such that S contains exactly 10 elements. Prove that S has
two disjoint subsets with identical sums. For example, if S contains
4, 12, 18, 27, 36, 50, 61, 62, 70, and 98, then the elements of the sets
(c)
{4, 12, 27, 36} and {18, 61} both add up to 79.
Suppose that n + 1 different numbers have been selected from the set
{1,2,...,2n}. Prove that there are two selected numbers whose sum is
(d)
(e)
Dine tale
Let S be a set of n + | integers. Prove that there exist distinct integers a
and b in § such that a — b is a multiple of n.
The English market town of Newton-le-Willows has a population of
21,307. Assuming that every resident has a first, middle, and last name,
(f)
(g)
Proofs to Grade
22.
prove that there are two residents with identical three-letter initials.
There are 10 hotel rooms numbered 5, 10, 15, ... , 50. Prove that if 6
of these rooms are occupied by hotel guests, then at least two occupied
rooms have numbers that differ by 5.
There are 50 hotel rooms numbered 1, 2, 3,... 50, and 26 of these rooms
are occupied by guests. Prove that at least two occupied rooms have
numbers that differ by 5.
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
(a)
Claim.
If A and B are finite, then A U B is finite.
“Proof.”
If A and B are finite, then there exist
that
A~N,, andB~N,.
onto
Claim.
“Proof.”
such
Let f: A aay N,, and h: B ae N,,- Then
i
A OB — Nintn
A U Bis finite.
(b)
m,n eWN
Which
shows
onto
that
AUB*N,,_,.
Thus
=
If S is a finite, nonempty set, then § U {x} is finite.
Suppose that S is finite and nonempty. Then S ~ N, for
some integer k.
Case 1.
x€S. Then SU {x}
finite.
S, so
SU {x} has k elements and is
Case 2. x ¢S. Then SU {x} Re© N,U {x}
© N,UN, © Ni,|.Thus
SU {x} is finite.
5
(c)
Claim.
If A x B is finite, then A is finite.
“Proof.”
‘Choose
any
b*eB.
Then
A#A x {b*}. ~ But
A x {b*} = {(a,b*):aeA} CA x B. Because A x B is finite,
A x {b*} is finite. Because A is equivalent to a finite set, A is finite. m=
(d)
Claim. The set N is finite.
“Proof.” For every n in N, the set N,, is finite because N,, ~ N,. By
(ee)
[o,@)
n=1
n=
Theorem 5.1.7, we know UN, is finite. Because (J N., = N, we see
that N is finite.
|
5.2
5.2
Infinite Sets
271
Infinite Sets
In this section, we will verify the not-at-all-surprising result that some familiar sets,
such as the sets of natural numbers, integers, and real numbers, are infinite. The
result that many people find surprising is that there are different sizes of infinite
sets. We will describe two infinite cardinal numbers and find that we can use them
to “count” all of the elements of certain infinite sets.
Recall that an infinite set is defined as a nonempty set that cannot be put into
a one-to-one correspondence with any of the sets N,. To prove that a set is infinite
using this definition, we assume that the set is finite and that such a correspondence
exists for some natural number k. We then find a contradiction. Another approach to
proving that a set A is infinite is to make use of the contrapositive of Corollary 5.1.11:
If A is equivalent to one of its proper subsets, then A is infinite.
We can interpret this statement as a test for whether a set could be finite. To use
this test, we look for a suitable proper subset of A and a one-to-one correspondence
between A and the subset. If we find such a set and correspondence, we conclude
that A is not finite.
To demonstrate the use of these methods, we give two different proofs that N
is infinite. Notice that the first proof resembles Euclid’s proof that there are infinitely many primes (see Section 1.5).
Theorem 5.2.1
The set N of natural numbers is infinite.
First Proof. Suppose that N is finite. Because N # ©, there exists a natural number k such that N ~ N,. Therefore, there exists a one-to-one function f from N;, onto
N. (We will show that f is not onto N by constructing a number that is not an image.)
Let n = max{ f(1), f(2),..., f(m)} + 1. Then n ¥ f(i) for any i € N,. Therefore,
fis not onto N. This is a contradiction. Hence N is an infinite set.
a
Second Proof. Let E* be the set of even positive integers. The function f: N > Et
defined by f(x) = 2x is a one-to-one correspondence from N onto E*. Thus,
N ~ Et. Because E* is a proper subset of N, we conclude that N is infinite.
|
The set of natural numbers is our first example of a set with infinite cardinality.
The standard symbol for the cardinality of N uses the letter 8, aleph, which is the
first letter of the Hebrew alphabet.
DEFINITIONS
The set S is denumerable if S~N. For a denumerable set S, we say S has cardinal number X, (or cardinality
No) and write S = No.
272
CHAPTER 5
Cardinality
Because N is equivalent to itself, N is denumerable, and the cardinality of
N is No. The subscript 0, read “naught” or “null,” indicates that No is the smallest
infinite cardinal number, just as the integer 0 is the smallest finite cardinal number.
The set N is our “standard” set for the cardinal number No.
We showed earlier that the set E* of even positive integers is equivalent to N.
Therefore, E* is denumerable. Even though E* is a proper subset of N, E~ has
the same number (No) of elements as N. Thus, although our intuition might tell us
that only half of the natural numbers are even, it would be misleading to say that
N has twice as many elements as E*, or even to say that N has more elements
than E*.
Results like this may be surprising if you rely only on your knowledge of
finite cardinal numbers to guide your insight into infinite cardinals. We have seen that
if A and B are finite disjoint sets, where A has m elements and B has n elements, then
A U B has m + n elements. The situation is more complicated when either A or B is
infinite. We must rely on one-to-one correspondences to determine cardinality.
In Section 5.5, we will see that every infinite set is equivalent to one of its
proper subsets. Together with Corollary 5.1.11, this will characterize infinite sets:
A set is infinite if and only if it is equivalent to one of
its proper subsets.
The next theorem will show that the set of all integers is denumerable. Our
proof constructs a bijection between N and Z.
Theorem 5.2.2
The set Z is denumerable.
Proof.
Define the function f: N > Z by
Wieascerthatsy
Ch s=nOay(2)
"hey (3) =e)
o,f (5) = 2, f (6) =
f(/) = —3, and so on.
To show that fis one-to-one, assume that f(x) = f(y) for some x, ye N. We
first observe that if one of x or y is even and the other is odd, then only one of f(x)
or f(y) is positive, so f(x) * f(y). Therefore, x and y must have the same parity. If
x and y are both even, f(x) = f(y) implies . = » and, therefore, x = y. If x and
y are both odd, then
l—;
5 ae
~~
5 Z and, again, x = y.
To show that
f maps onto Z, suppose that w € Z. If w > 0, then 2w is even,
2w
and f(2w) = ban w. If w <0, then
fd
—2w) =
onto Z.
1—(1—2w)
y
2w
=
1 — 2w is an odd natural number,
and
= w. In both cases, we Rng (f). Thus, f maps
Therefore, Z is equivalent to N.
|
5.2
Example.
Infinite Sets
273
Let P be the set of reciprocals of positive integer powers of 2. By writ-
ing the setP as
1
a
tarke
2
nt
we see that there is a natural one-to-one correspondence between N and P. Because
the function h: N — P given by h(n) = an is a bijection, P is denumerable.
Oo
Example. To prove that the set K = {p, g, r}
U{n © N:n #5} is denumerable,
we construct a bijection from N to K. We prefer a function that can easily be seen
to be one-to-one and onto K. By arranging the elements of N and K as shown in
Figure 5.2.1, we can see how to construct this function. Let g: N > K be given by
P
ocak
qd
itn =
g(n)= Vr
if 3
n—-3
if4<n<7
n—-2
ifn>8
A proof by cases shows that g is both one-to-one and onto K. Therefore, N ~ K, so
K is denumerable.
Theorem 5.2.3
(a)
(b)
Theset N x N is denumerable.
IfA and B are denumerable sets, then
A x Bis denumerable.
Proof.
(a)
The function
F in Section
4.4 given by F(m, n) = 2”~'(2n — 1) is a
bijection from N x N to N. Therefore N x N is denumerable.
(b) | Suppose A and B are denumerable sets. Because A ~ N and B ~ N, we have
A x
B= N x N by Theorem 5.1.2(a). By part (a), N x
NN. Therefore,
Ax BN. Hence, A x Bis denumerable.
zB
DEFINITIONS A set S is countable if S is finite or denumerable. We say
S is uncountable if S is not countable.
Sets that are finite or denumerable are called countable because their elements
can be “counted” using some or all of the natural numbers. “Counting” elements in a
nonempty countable set S means setting up a one-to-one correspondence between S
and N, (when Sis finite) or between S and the entire set N (when Sis denumerable).
274
CHAPTER 5
Cardinality
Figure 5.2.2 shows the relationships among finite, infinite, denumerable, counta-
ble, and uncountable sets. We see that every finite set is countable and every uncountable set is infinite. Because denumerable sets are those sets that are both infinite and
countable, denumerable sets are sometimes referred to as countably infinite sets.
Denumerable
Uncountable
Sets
Sets
Countable Sets
Infinite Sets
Figure 5.2.2
Examples.
Some sets that are both infinite and countable (that is, denumerable)
are N, Z, and the set of even positive integers. Some countable finite sets include
N, ©, {11, 7,77, 3, 15,79}, and {xe R:x° + 12x? — 21x? + 3x + 11 = O}.
There are infinite sets that are uncountable, such as [R and (0, 1), as we will see next.
Before showing that the open interval (0, 1) is uncountable, we need to review
decimal expressions for real numbers. Every real number can be written in normalized
form, which simply means that the decimal expansion does not end with all 9’s
from some point on- For example, we write 0.4999... in normalized form as 0.50.
The importance of normalizing decimals is that two decimal numbers in normalized
form are equal if and only if they have identical digits in each decimal position.
Theorem 5.2.4
The open interval (0, 1) is uncountable.
Proof.
(We must show that (0, 1) is neither finite nor denumerable.) The interval
(0, 1) includes the subset taeke nt.which is infinite. Thus, by Theorem 5.1.6,
(0, 1) is infinite.
:
(We now assume that (0, 1) is denumerable and reach a contradiction.) Suppose
that (0, 1) is denumerable. Then there is a function f: N — (0, 1) that is one-to-one
and onto (0, 1). (The contradiction arises when we construct a number in (O, 1) that
is not in Rng (f).). Write the images of f, for each n € N, in normalized form:
FO)
= 0.4) 14424)
34)4415.
Ff)
—— 0.04 19747307405.
+:
FB) = 0.43)437033434035.-FA) = 0.04 Ayr y3Qg4dgs.
fm)
=i 0.414243
Ings
st
5.2
Infinite Sets
275
Now let b be the number b = 0.b,b,b3b,b5..., where
one { fag eS
a
itat=15
(The choices of 3 and 5 are arbitrary.)
Then b € (0, 1) because of the way it has been constructed. However, for each
natural number n, b differs from f(n) in the nth decimal place. Thus, b # f(n) for
any n € N, which means b ¢ Rng (f). Thus, fis not onto (0, 1). This contradicts
our assumption that fis onto (0, 1). Therefore, (0, 1) is not denumerable.
=
The interval (0, 1) is our first example of an uncountable set and is assigned
cardinal number c.
DEFINITION
A
set S has cardinality c (or cardinal number c) if S is
equivalent to (0, 1). We write
S = c, which stands for continuum.
The cardinal number c is the only infinite cardinal other than NX, that will
be identified by name. Implicit in this statement is that there are other infinite
cardinals—an issue that will be addressed in Section 5.4. The remainder of this
section will describe other sets with cardinality ce.
Theorem 5.2.5
(a) | Every open interval (a, b) is uncountable and has cardinality c.
(b)
The set R of all real numbers is uncountable and has cardinality c.
Proof.
(a)
Letaand b be real numbers with a < b. In Section 5.1, we showed that any
two open intervals are equivalent. In particular, (a, b) is equivalent to (0, 1).
Therefore, (a, b) has cardinality c.
(b)
Define f: (0, 1) ~ Rby f(x) = tan(mx — 5 . See Figure 5.2.3. The function
fis a contraction and translation of one branch of the tangent function and is
one-to-one and onto R. Thus, (0, 1) + R.
y
A
Figure 5.2.3
Exercise 10 asks you to use a different function to show that R ~ (0, 1).
|
276
CHAPTER 5
Cardinality
Let C be the circle of radius ,with center (0, ) and the point (0, 1)
Example.
ed by
removed, as shown in Figure 5.2.4. For any point p in C, the line determin
function
a
define
We
point.
one
exactly
in
x-axis
the
(0, 1) and p will intersect
of
f: C = Ras follows: For each p in C, let f(p) be the x-coordinate of the point
and
p,
points
different
that
see
We
and:p.
1)
(0,
by
ed
determin
intersection of the line
P> will generate nonparallel lines and hence different values for f(p,) and f(D).
Thus fis one-to-one. Any point m € R along the x-axis will determine a line through
(0, 1) that will intersect C in exactly one point q. For this point q, f(q) = m. Therefore, fis onto R. Hence, fis a bijection, and the set C is equivalent to R.
(0, 1)
NS
FP)
Figure 5.2.4
Example.
The set A = (0, 2) U [5, 6) also has cardinal number c. The function
f: O, 1) > A, given by
l
4x
Ole
5
IKED)
=
il
2x + 4
a
>
4
iS a One-to-one correspondence between (0, 1) and A. See Figure 5.2.5. We note
that (0, 1) is a proper subset of A and A is a proper subset of RX but that all three sets
have the same infinite cardinality.
ay
A
ae
Me
5
=
Ans
3
2
l
aa
ae
ilih
2
Figure 5.2.5
5.2
Infinite Sets
277
Exercises 5.2
—_
Prove that if A is an infinite set and A ~ B, then B is an infinite set.
Prove that each of these sets is infinite.
(a)
Ae
(c)
(0, 0.001)
(e)
(a,b) M Q, for rational numbers a, b with a < b
lana aaa,
(b)
N—N,s
(d)
(0, co)
Prove that the following sets are denumerable.
(a) D*, the odd positive integers
(b) 3N, the positive integer multiples of 3
(c) 3Z, the integer multiples of 3
(d)
{n:nEN andn > 6}
(e)
{x:xeZandx < —12}
(f) N — {5,6}
(2) poLoow
ie Nis Rey1}
(h) {xe Z:x = l(mod
5)}
Prove that the following sets have cardinality c.
(a) (1, oo)
(b) (a, co), for a real number a
(c) (—co, b), for a real number b
x/(d)
[1,2)U6, 6)
(e) PA, 6)
1020)
(FE) BOs L123
as)
(g)
RU {i}
(h)
R—
{0}
State whether each of the following is true or false.
(a)
(b)
(c)
Ifaset A is countable, then A is infinite.
Ifaset A is denumerable, then A is countable.
IfasetA
is finite, then A is denumerable.
(d)
(e)
(f)
IfasetA is uncountable, then A is not denumerable.
IfasetA is uncountable, then A is not finite.
Ifaset A is not denumerable, then A is uncountable.
6. *(a)
Give an example of a bijection g from N to the set
integers such that g(1) = 20.
(b)
E* of positive even
Give an example of a bijection 4 from N to E* such that h(1) =
HO)
16,
andei(6) 2s
Which sets have cardinal number X,)? ¢?
(a)
R— [0, 1)
(c) {1:neN}
tee
)
(ff)
()
(b)
(5, oo)
(@) {2:xeN}
{~ geEeRxRpt+q=1)}
{(p,q eR x Riq= V1 —p’andg > 0}
{@yeZx Zy=Cl1Y} @) {(5+7:reR}
Give an example of denumerable sets A and B, neither of which is a subset of
the other, such that
(a) AM B is denumerable.
(c) A — Bis denumerable.
(b)
(d)
ANB is finite.
A — Bis finite and nonempty.
It can be shown that the sets [0, 1] and (0, 1) U N have cardinality c. Use these
facts and other sets to show that there are sets A and B such that A =
B=c
and A
B is
(a)
(c)
empty.
denumerable.
(b)
(d)
finite and nonempty.
uncountable.
278
CHAPTER 5
Cardinality
10.
Give another proof of Theorem 5.2.5 by showing that f(x) = =
is a
one-to-one correspondence from (0, 1) onto R. (You may apply methods from
calculus.)
11.
Proofs to Grade
12.
It can be shown that R x R has cardinality c. Use this fact to prove that the
set C of complex numbers has cardinality c.
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
(a) Claim. Let Wbe the set of all natural numbers with tens digit 3, and let
D* be the set of odd natural numbers. Then W and D* are equivalent.
“Proof.”
W contains 30, 130, 230, 330,..., and many other natural
numbers, so W is an infinite subset of N. Therefore W is denumerable.
D* is also denumerable, by Exercise 3(a). Therefore,
(b)
Pome:
oN
(c)
(d)
rT
eey
GCS.Ly)at eel
is one-to-one and onto A U {x}. Thus, N +AU
{x}, so
AU {x} is
infinite.
a
Claim.
If A U B is infinite, then A and B are infinite.
“Proof.”
Assume that A and B are finite. Then by Theorem 5.1.7,
A U Bis finite. Therefore, if A U B is infinite, A and B are infinite.
|
Claim.
If a set A is infinite, then A is equivalent to a proper subset of A.
eR LOOL. am Leal
(e)
W ~ D~.
Claim. If A is infinite and x ¢ A, then A U {xx} is infinite.
“Proof.”
Let A be infinite. Then A ~ N. Let f: N > A be a one-toone correspondence. Then g: N > A U {x}, defined by
4
is, 2.) Choose. B=
1X5, x5,.--}. Then Bb isia
proper subset of A. The function f: A + B defined by f(x,) = x,4, is
clearly one-to-one and onto B. Thus, A ~ B.
a
Claim.
The set
T= {n € Z:n = 2(mod 6)} is denumerable.
“Proof.”
DefineafunctionF on theintegers by setting F(z) = 6z + 20.
F is one-to-one because if F(u) = F(v), then 6u + 20 = 6v + 20, so
u = v. Every element t of T has the form 6k + 2 for some integer k, and
t = F(k — 3), so F maps onto 7. Therefore, T is equivalent to Z, which
(f)
(g)
(h)
is denumerable.
=
Claim. The set N is infinite.
“Proof.”
The function given by f(n) = n + 1 is a one-to-one correspondence between N and N — {1}, so N is equivalent to a proper
subset of N. Therefore, by Corollary 5.1.11, N is infinite.
cei
Claim.
The set SN = {5n:n €N} is infinite.
“Proof.” =5N is a subset of N, and N is infinite. Then 5N is infinite
because every subset of an infinite set is infinite.
=
Claim.
If A and B are denumerable, then A U B is denumerable.
“Proof.”
Assume that A and B are denumerable. Then there
exist bijections h and g such that h:A —N and g:B—N. Then
hUg: (AUB) > Nisa bijection, so A U B is denumerable.
a
5.3
5.3
Countable Sets
279
Countable Sets
The first two sections of this chapter have presented several examples of countable
sets—all of the finite sets in Section 5.1 and several denumerable sets such as N, Z,
and N x N in Section 5.2. This section presents the essential facts needed by anyone who works with countable sets. Our first task is to determine the cardinality of
the set of rational numbers.
Because N C @ C R, and the cardinalities of N and R are X, and c, respec-
tively, it is reasonable to believe that Q is either N, or ¢, or possibly some infinite
cardinal number between them. There are infinitely many rationals between any
two rational numbers (see Exercise 2(e) of Section 5.2), so you might also suspect
that
is not denumerable. But Georg Cantor* showed that Q* (the positive rationals) is denumerable through a clever rearrangement of Q*.
Every element in @* may be expressed as 2for some p, g € N. Thus, the elements of this set can be presented as in Figure 5.3.1, where the nth row contains all
the positive fractions with denominator n.
Figure 5.3.1
To show that Q@* is denumerable, Cantor listed the elements of @*
in the
order indicated by the arrows in the figure. First are all fractions in which the sum of
:
i
:
2
the numerator and denominator is 2 (only i):then those whose sum is 3 (;and 5).
then those whose sum is 4 (3,S and 5);and so on. Some rational numbers appear
* Georg Cantor (1845-1918) created set theory, primarily in papers that appeared in 1895 and 1897.
This work can be seen as a revolution in mathematics because he made it possible to think of actual
infinite quantities, rather than the infinite as unattainable. He was the first to use one-to-one correspondences to describe set size, the first to show the rational numbers are countable, and the first to show the
reals are not countable. Several of his contemporaries did not accept some parts of his work.
280
CHAPTER 5
Cardinality
oes
PALE’
multiple times: For example, 2 and 3are repetitions of the fraction T Disregard all
fractions that are not in lowest terms: > i >4,e. .. The remaining fractions have
no repetitions and are circled in Figure 5.3.1.
Starting at the top left corner of the array; traverse the array from upper
right to lower left along each diagonal, assigning the natural numbers to the circled fractions, starting with n then the diagonal a * and so forth. This pattern
defines a one-to-one correspondence f from N to @* shown in Figure 5.3.2, where
fl) =4,F@ =2/@ =5, fA =3 fO) = 7 ete.
Nestle
Qt
Sea
2.0
54m oO, 6. Fess
9, LOL,
tessa)
P2heLeszLsres
se
ee oe
Sg
ao Pe ee
oT ee
eee cies
Figure 5.3.2
This correspondence can be used to establish the following theorem. We omit the
details of the proof.
Theorem 5.3.1
The set @* of positive rational numbers is denumerable.
The principal results of this section are that (1) every subset of a countable set
is countable and (2) the union of countably many countable sets is countable. These
theorems are two of the most useful facts about cardinalities.
Theorem 5.3.2
Every subset of a countable set is countable.
Proof.
Let A be a countable set. Assume that B C A. If B is finite, then B is countable. Otherwise, B is infinite, so A is infinite. Because A is infinite and countable, A is
denumerable. Let f be a bijection from A to N. The restriction of a one-to-one function is one-to-one, so f|g is a bijection from B to C = Rng (f| a): Therefore, B ~ C.
We now define a function g: N > C by induction. (We also make use of the
Well-Ordering Principle.) Let g(1) be the smallest integer in C. The set C — {g(1)}
is nonempty because C is infinite. For each n > 1, define g(n + 1) to be the smallest element in the nonempty set C — {g(1), g(2),..., g(n)}.
If r,s © N and r < 5s, then g(r) is an element of {g(1), g(2),...,g(s — D},
but g(s) is not. Therefore, g(r) # g(s). Thus, g is an injection. Also, if
te C and
there are k natural numbers less than t in C, then g(k + 1) = t. Therefore, g is onto
C. Thus, g is a bijection from N to C.
Therefore,
N ~ C © B, so B is denumerable.
|
* Many different one-to-one correspondences are possible. For another interesting example, see
N. Clakin and H. S. Wilf’s article “Recounting the Rationals” in the American Mathematical Monthly,
April 2000, pp. 360-363.
5.3
Countable Sets
281
We have seen that the set @+ is denumerable and therefore countable. Thus,
the subsets {* ne N}, @ 1 (0, 1), and {2,g,3| are countable sets.
Corollary 5.3.3
A set A is countable if and only if A is equivalent to some subset of N.
Proof.
If A is countable, then A is either finite or denumerable. Thus,
AN,
of N.
for some ke N, or A © N. In each case, A is equivalent to some subset
A = ©, or
If A is equivalent to some subset of N, then A is equivalent to a countable set
because all subsets of (countable) N are countable. Therefore, A is countable.
-|
We have seen (Theorem 5.1.7) that adding one or any finite number of elements
to a finite set yields a finite set with larger cardinality. In the next three theorems,
we consider adding elements to a denumerable set and find an important distinction
between finite and denumerable sets: Adding finitely many or even denumerably
many elements does not change the cardinality of a denumerable set.
Theorem 5.3.4
If A is denumerable, then A U {x} is denumerable.
Proof.
If x €A, then A U {x} =A, which is denumerable. Suppose that x ¢ A.
Because N ~ A, there is a one-to-one function f: N — A that is onto A. Define
g3.N>AU
{x} by
x
i ie = Ml
a(n) = oe Uppcla = wl.
It is straightforward to verify that g is a one-to-one correspondence between N and
A U {x}, which proves that A U {x} is denumerable.
w
Theorem 5.3.4 may be loosely restated as Ny + 1 = Np. Its proof is illustrated
by the story of the Infinite Hotel,” attributed to the mathematician David Hilbert.
The Infinite Hotel has Xp rooms numbered
1, 2, 3, 4,...and is full to capacity with
one person in each room. You approach the desk clerk and ask for a room. When
the clerk explains that each room is already occupied, you say, “There is room for
me! For each n, let the person in room n move to room n + 1. Then I will move
into room 1, and everyone will have a room as before.” There are Ny + 1 people,
and they fit exactly into the Xp) rooms. See Figure 5.3.3.
* The Infinite Hotel is one of the topics discussed in Aha! Gotcha: Paradoxes to Puzzle and Delight by
Martin Gardner (Freeman, New York, 1981).
+ David Hilbert
(1862-1943) spent most of his career at the University of Gottingen. He is considered
the most influential and creative mathematician of his time and was a staunch supporter of Cantor
and his set theory. At the International Congress of Mathematicians in Paris in 1900, he proposed 23
open problems (the first one being the continuum hypothesis—see Section 5.5), which set the stage
for much research in the 20th century. Some of the 23 problems remain unsolved.
282
CHAPTER 5
Cardinality
Figure 5.3.3
Rooms in the Infinite Hotel can also be found for any finite number k of additional people by asking each guest to move to room n + k (Theorem 5.3.5). In the
event of a fire alarm at the Grand Infinity Hotel across the street, the Infinite Hotel
could even accommodate denumerably many additional guests by sending the current guest in room n to room 2n and assigning new guests to the odd-numbered
rooms (Theorem 5.3.6). Later we shall see that the clerk could find rooms if a
denumerable number of additional people arrive a finite number of times (Corollary
5.3.9(c)) or even a denumerable number of times (Corollary 5.3.9(d)).
Theorem 5.3.5
If A is denumerable and B is finite, then A U B is denumerable.
Proof.
Theorem 5.3.6
See Exercise 3.
=
If A and B are disjoint denumerable sets, then A U B is denumerable.
Proof.
Let f: N —. A andg:N a B. Define h: N > AUB via
n+ |
(3) enon
) if n is odd
NO
B\5
ifn 1S even
5.3
Countable Sets
283
(The effect of h is to map the odd natural numbers to elements of A and the even
natural numbers to B.) It is left as Exercise 4 to show that h is a one-to-one
a
correspondence from N onto A U B. Therefore, N ~ A U B.
We may apply the previous theorems to produce many new examples of
denumerable sets. We know from earlier examples and exercises that the sets E +
(the positive even integers) and 3N (the integer multiples of 3) are denumerable.
Therefore, the sets Et U {5}, 3N U {1, 2, 4,5}, and E+ U 3N are denumerable.
Theorem 5.3.7
The set © of all rational numbers is denumerable.
Proof.
The set @* is denumerable, and the mapping that sends each positive
rational number to its negative is a bijection, so the set Q~ of negative rationals
is denumerable. By Theorem 5.3.5, @* U {0} is denumerable, and by Theorem
5.3.6, (QT U {0}) U @ is denumerable.
a
The other major theorem of this section is presented here because of its importance in dealing with countable sets. Because the proof requires the use of a new
property of sets (the Axiom of Choice) that will be introduced in Section 5.5, the
proof will appear in that section.
Theorem 5.3.8
Let f be a countable collection of countable sets. Then
(J A is countable.
Aes
:
We have already seen some theorems that are in fact special cases of Theorem
5.3.8: The union of finitely many finite sets is finite, and the union of a denumerable
set with a finite set or of two disjoint denumerable sets is denumerable. Four other
special cases of Theorem 5.3.8 are stated in the following corollary. Exercises 6, 7,
and 8 ask you to prove the first three of these by methods we have already used
(without relying on the Axiom of Choice.) The fourth statement, part (d), cannot be
proved without using the Axiom of Choice.
Corollary 5.3.9
(a)
If A is a finite pairwise disjoint family of denumerable sets, then
(b)
(c)
Aes
countable.
IfA and B are countable sets, then A U B is countable.
If Misa finite collection of countable sets, then ly, A is countable.
LJ A is
(d)
If
AEA
is adenumerable family of countable sets, then
) A 1s countable.
AEA
Example.
Theorem
5.3.8 and its corollaries provide another method that we
can
to prove
that @
use
LJ A, = {t:meN}.
neN
For
is countable.
The
set @*
be written
as
each
the
set A, is denumerable.
By
neN,
may
B
Corollary 5.3.9, this union is countable. After this step, we can observe
that Q~ is countable because it is equivalent to 2+. Then @ = Q* U {0} UO is
countable because it is a union of countably many countable sets.
284
CHAPTER 5
Cardinality
Our final example of a denumerable set is significant for anyone with an interest in the theory of computation. A computer program is written in a given programming language and consists of a finite sequence of symbols. These symbols
are selected from a finite set called an “alphabet” (typically all 26 upper- and lowercase letters, the 10 digits, a blank space, certain punctuation marks, arithmetic
operations, etc.).
For each n EN, let P,, be the set of all programs with precisely n symbols.
Because there is only a finite number of symbols in our alphabet, there can be
only a finite number of programs of length n. Therefore, P,, is finite (and therefore
countable) for all n € N. Also, because any computer program is finite in length,
every program is an element of P,, for some n. Thus, the set of all possible programs
is LU P,.
neN
By Theorem 5.3.8, this countable union of countable sets is countable. Hence,
only a countable number of programs could ever be written in a given language.
However, we saw in Section 5.1 that the set of all functions from N to {0, 1} is
equivalent to 9 (N), and we shall see in the next section that 9? (N) is uncountable.
For a given programming language and its finite alphabet, this means that there are
many functions from N to {0, 1} for which there can be no computer programs to
compute them. Put a different way, there are not enough solutions (programs) for
all the possible problems to be solved (functions).
Exercises 5.3
1.
What is the 28th term in the sequence of positive rationals produced by the
counting process described in the discussion of Theorem 5.3.1?
2.
Use a counting process similar to that described in the discussion of Theorem
XxX
5.3.1 to show that
By" x,y
© N?
is denumerable.
3.
Prove Theorem 5.3.5 by induction on the number of elements in the finite set B.
4.
Complete the proof of Theorem 5.3.6 by showing that the function h as
defined is one-to-one and onto A U B.
5.
The Infinite Hotel is undergoing some remodeling, and consequently some
of the rooms will be taken out of service. Show that, in a sense, this does not
matter as long as only a few rooms are removed.
(a) Prove that if A is denumerable and x € A, then A — {x} is denumerable.
(b) Prove that if A is denumerable and B is a finite subset of A, then A — B
is denumerable.
(c) Explain how it is possible to have denumerably many rooms being remodeled at the same time but still have denumerably many rooms in service.
(d) To accommodate more guests, the owners of the Infinite Hotel are considering building an “Infinite Strip” of a denumerable number of Infinite
Hotels. Why are the owners wasting their money?
5.3
Prove part (a) of Corollary 5.3.9: If # =
Countable Sets
285
{A;: i = 1, 2, 3,..., n} is a finite
pairwise disjoint family of denumerable sets, then (J A is countable. Do not
refer to Theorem 5.3.8.
fs
Prove part (b) of Corollary 5.3.9: If A and B are countable sets, then A U B is
countable. Do not refer to Theorem 5.3.8.
Proves palra(C)sOle Corollary
.).0-O7lingd (Aa!
=
1,2,3,9,..
71} 4s a
finite collection of countable sets, then
|) A is countable. Do not refer to
Theorem 5.3.8.
ai
Use the theorems of this section to prove that
(a)
an infinite subset of a denumerable set is denumerable.
(b)
if A is a countable
uncountable.
(c)
ON (1, 2) is denumerable.
20
J(QA(™m,n + 1)) is denumerable.
(d)
subset of an uncountable
set B, then B — A is
(eal
(ec)
W(Qn (n,n + 1)) is denumerable.
neN
n
(fy)
oneke utis denumerable.
neN
10.
11.
Prove or disprove:
(a)
(b)
(c)
If A C Band Bis denumerable, then A is denumerable.
If A C Band A is denumerable, then B is denumerable.
J UK
is denumerable, where J is the set of all linear functions with
(d)
slope 1 and rational y-intercept and K is the set of all linear functions
with slope 2 and integer y-intercept.
© — Z is denumerable.
(e)
IfA and B are denumerable, then A — B is denumerable.
Prove that if {B;:
ie N)} is
adenumerable family of pairwise disjoint distinct
finite sets, then J B, is denumerable.
ieN
12.
Give an example, if possible, of a family A,, A>, A3,... of sets such that
[o,2)
(a)
each set A; is finite and |} A, is denumerable.
n=1
(b) each set A, is finite and ) A, is finite.
n=
(c)
each set A; is finite, the family {A;: i e N)} is pairwise disjoint, A; # A;
whenever i ¥ j, and UJ A, is finite.
n=1
13.
Prove that each set is uncountable.
(a)
the set of irrationals
(c)
{@, y)ER x Rix? 4+ y <1}
x
(b)
(1,2)
(d)
RxN
U{2 nEN}
286
CHAPTER 5
Cardinality
14.
(a)
(b)
Let S be the set of all sequences of 0’s and 1’s. Forexample, 1010101...,
1001101001..., and 011111... are in S. Using a proof similar to that for
Theorem 5.2.4, show that S is uncountable.
For each n € N, let T,, be the set of all sequences in S with exactly n 1’s.
Prove that T7,, is denumerable for all n € N.
(c)
15.
Lett T=
U T,. Use a counting process similar to that described in the
k=1
discussion of Theorem 5.3.1 to show that 7 is denumerable.
Let A be a denumerable set. Prove that
(a)
(b)
(c)
(d)
the set {B:B CA and B = 1} of all one-element subsets of A is
denumerable.
the set {B:
BCA and B= 2} of all two-element subsets of A is
denumerable.
a
for every
ke N, {B:
BCA and B = k} is denumerable.
the set {B: B C A and B is finite} of all finite subsets of A is denumerable. (Hint: Use Theorem 5.3.8.)
Proofs to Grade
16.
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
(a)
Claim.
IfA,B, and C are sets,
C C A U B, B is denumerable, and C is
uncountable, then A is uncountable.
“Proof.”
Because C C A U B, we have C = (Af) C)
U(BN C). The
set B is countable, and subsets of countable sets are countable, so
BM C
is countable. The union of two countable sets is countable, so if
AMC
were countable, the union (A % C)
U(B MC) would be countable. But
the union is C, and C is uncountable, so A M C is uncountable. Subsets
of countable sets are countable, and A M C is an uncountable subset of
A, so A must be uncountable.
a
x
(b)
Claim.
If A is denumerable, then A — {x} is denumerable.
“Proof.”
Assume that A is denumerable.
Case 1. If x¢@A, then A — {x} =A, which is denumerable
by
hypothesis.
Case 2.
Assume
that
x € A. Because A is denumerable,
TN —> A. Define
gi N nee
(A — {x}), so N ©A
onto
is denumerable.
(c)
there exists
g by setting g(n) = f(n + 1). Then
—
{x}. Therefore,
A—
{x}
5
Claim.
If A and B are denumerable, then A x B is denumerable.
“Proof.”
Assume that A and B are denumerable but that A x B is
not .denumerable. Then A x B is finite. Because A and B are denumerable, they are not empty, so we can choose a € A and b € B. Then
AXA x {b} and B® {a} x B. Because A x B is finite, the subsets
A x {b} and {a} x B are finite. Therefore, A and B are finite. This
contradicts the statement that A and B are denumerable. We conclude
that A x Bis denumerable.
|
5.4
(d)
Claim.
“Proof.”
The Ordering of Cardinal Numbers
287
The set Q* of positive rationals is denumerable.
Consider the positive rationals in the array in Figure 5.3.1.
Order this set by listing all the rationals in the first row, then the second
row, and so forth. Omitting fractions that are not in lowest terms, we
have an ordering of @* in which every positive rational appears. There-
fore, Q* is denumerable.
(e)
rs
Claim.
If A and B are infinite, then A ~ B.
“Proof.”
Suppose thatA and Bare infinite sets. LetA = {a,, a5, d3,...}
and B =
{b,, by, b3,...}. Define f: A > B as shown:
{ A), Gy, Az, d4,...}
an
Dis
an
O55 DasDyacie ah:
Then, because we never run out of elements in either set, fis one-to-one
and onto B,so A © B.
|
Claim.
R — Q is uncountable.
“Proof.”
is uncountable, and R — Q is a subset of R. Every subset
of an uncountable set is uncountable, so R — Q is uncountable.
5.4
|
The Ordering of Cardinal Numbers
When Georg Cantor developed set theory, he described a cardinal number of a set
M as “the general concept which, with the aid of our intelligence, results from M
when we abstract from the nature of its various elements and from the order of their
being given.” This definition was criticized as being less precise and more mystical than a definition in mathematics ought to be. Other definitions for the number
of elements or “size” of the set were given, and eventually the concept of cardinal
number was made precise. The double overbar notation S used today for the cardinality of a set S is suggestive of the double abstraction referred to by Cantor.
We have chosen to define cardinal numbers informally by picking one standard
representative set for each cardinal number we introduced:
Cardinal Number
0
k, for every
Xo
c
ke N
Standard Set
©
N,
N
(0, 1)
The essential point is that every set equivalent to one of the standard sets has the
same size (cardinality) as that of the standard set and distinct standard sets are not
equivalent.
The following definitions enable us to compare the sizes of two sets.
288
CHAPTER 5
Cardinality
DEFINITIONS
LetA and B be sets. Then
A = B if A © B; otherwise, en
A < B if there exists a one-to-one function f: A > B.
A < BifA<BandA#B.
We read A < B as “the cardinality of A is strictly less than the cardinality of
B,” and < is read “less than or equal to.” In addition, we use
A < B and A ¥ B to
denote the denials of A < Band A < B, respectively.
Example:
For sets:A"="{a, b, c}; B=
{a, 8) }, and
C —"{6, 7, 3; 9};
A = B because f= {(a, 1), (b, a), (c, B)} is a one-to-one correspondence
from A onto B.
A < CC because g = {(a, 6), (b, 7), (c, 9)} is a one-to-one function fromA to B.
>|
|| << G because A < C and there is no one-to-one
correspondence from A to C. 5
Because 0, 1, 2, 3, ... are cardinal numbers, the set N U {0} may be viewed as
a subset of the collection of all cardinal numbers. In this sense, the properties of <
and < that we will prove for cardinal numbers in the next theorem may be viewed
as extensions of those same properties of < and < that hold for N U {0}.
Theorem 5.4.1
For sets A, B, and C, ©
(a)
A < A.
(Reflexivity)
(b)
ifA=BandB=C,thenA=C.
(Transitivity of =)
(c)
if A < Band B < C, then A << G
(Transitivity of <)
(d)
A <BifandonlyifA
(ec)
ifACB,thenA <B.
(f)
A <Bifand only if there is a subset W of B such that W — A.
< BorA =B.
Proof.
(c)
Suppose that A < B and B < C. Then there exist functions f: A BE FS and
g:B += C. Because the composite g © f: A — C is one-to-one, we conclude
AC
(e)
Let A CB. We note that the inclusion map i: A —> B, given by i(a) = a, is
one-to-one, and therefore A < B.
Proof of parts (a), (b), (d), and (f) are left as Exercise 2.
Theorem 5.4.2
(a)
(b)
(c)
Form,neNU
{0}, ifm < nas integers, then m < nas finite cardinals.
Every finite cardinal is less than No.
The cardinal Np is less than the cardinal ec.
|
5.4
The Ordering of Cardinal Numbers
289
Proof.
(a)
Let msne NU {0} and m < n. If m = 0, the empty map from © to N,,
is one-to-one. Otherwise, the inclusion map i: N,,,> N,, is one-to-one,
som =N,, < N, =n. By the Pigeonhole Principle, m = NAN, =2.
(b)
Therefore, m < nas cardinal numbers.
Let
ke NU {0}. We have @ CN when
k = 0 and N, CN
when
k is a
natural number. Thus, by Theorem 5.4.1(e), k < No. Because
© and N, are
finite, neither set is equivalent to N. Therefore k < No.
(c)
See Exercise 8.
a
Summarizing the results of Theorem 5.4.2, we have
Os
2 Say
1
Before Cantor’s work, everyone assumed that there was only one infinity. Now
that we have two different infinite cardinals, we can ask: Are there infinite cardinal
numbers greater than ¢? And if so, how many are there? The theorem named for
Cantor leads to answers for these questions.
Theorem 5.4.3
Cantor's Theorem
For every setA, A < P(A).
Proof.
Part
To show A < (A), we must show that (i) A < P(A) and (ii) A # P(A).
(i) follows
from
the fact that f: A — P(A)
one-to-one.
defined
by f(x) = {x}
is
-
To prove (ii), suppose that
A = YP(A); that is, assume that
A ~ 9(A). Then
there exists g: A —, P(A). LetB = {y EA: y € g(y)}. Because
B CA, B € P(A),
and because g is onto (A),
B = g(z) for some z € A. Now either z € Bor z ¢ B. If
z& B, thenz € g(z) = B. This is acontradiction. Similarly, z ¢ B implies z ¢ ¢(z),
which implies z € B. This is also a contradiction. We conclude that A is not equivalent to P(A) and hence A < P(A).
r]
Cantor’s Theorem tells us that there are infinitely many infinite cardinal
numbers. One of them is Ny) = N. Another infinite cardinal is P(N) because
N < P(N). Yet another is P(P(N)) because P(N) is a set and P(N) < P(A(N)).
In this fashion, we may generate a denumerable set of cardinal numbers, each
greater than its predecessor:
Ny < PIN) < PAWN) < PPM)
<
Cantor’s Theorem not only tells us that all these infinite cardinal numbers are distinct
but also guarantees that there can be no largest cardinal number. See Exercise 5.
290
CHAPTER 5
Cardinality
least as many elements
For sets A and B, it appears to be obvious that if B has at
as A and A has at least as many elements as B, then A = B. The proof, however,
is not obvious. The situation may be represented in Figure 5.4.1. From A < B and
B <A, there are one-to-one functions f: A > B and g: B > A. The problem is to
construct h: A — B such that h is both one-to-one and onto B. The figure suggests
that h will be some combination of f and g~!, but finding the right combination
turns out to be challenging.
A
is
D=Rng(f)
C = Rng(g)
¢
——<————
Figure 5.4.1
Cantor solved this problem in 1895, but his result was
accepted because his proof used the Axiom
not immediately
of Choice (Section 5.5). Proofs not
depending on the Axiom of Choice were given by Ernst Schréder’ in 1896 and by
Felix Bernstein’ two years later.
Theorem 5.4.4
Cantor-Schroder-Bernstein Theorem
If A < Band B < A, then A = B.
Proof. We may assume that A and B are disjoint because otherwise we could
replace A and B with the equivalent disjoint sets A x {0} and B x {1}, respectively. Let f: A 4+ Bwith D = Rng (f), and let g: B EA with
C= Rng (g). If
B = D, we already have A © B, so assume that
B—
Define a string to be a sequence s: N — A UB
D# ©.
such that
s(1) EB — D,
if s(n) € B, then s(n + 1) = g(s(n)),
if s(n) € A, then s(n + 1) = f(s(n)).
The first term of a string s is in B — D, but thereafter the terms are alternately in
C and in D. If n is even, s(n) € C, and if n is odd and n > 1, then s(n) € D. See
Figure 5.4.2. Every element of B — D is first term s(1) of some string.
* Ernst Schréder (1841-1902) was known mostly for his work in logic and its applications to other
areas of mathematics. He advanced the methodical use of quantifiers. The design of Schréder’s proof of
Theorem 5.4.3 was correct, but his proof contained an error.
* Felix Bernstein (1878-1956), while he was still a student under Cantor, corrected the error. Bernstein
made contributions in many fields, including applied mathematics, statistics, and especially genetics.
5.4
The Ordering of Cardinal Numbers
291
D= Rng (f)
sine Je KD, KOE SO, 6G, S@)h K@)oon
Figure 5.4.2
Let W = {x € A: xis aterm of some string}. We note that W C Rng (g) and
that x € W if and only if x = s(2n) for some string
f and natural number n. Let
h: A — B be given by
f(x)
AG) =
{ by
PX)
ifx ¢ W
,
kh ee
‘
We will show that / is a one-to-one correspondence from A onto B.
Suppose that x, ye A and h(x) = h(y). We will first show that x and y must
both be in W or both be in A — W. For a proof by contradiction, assume that this
is not the case. Without loss of generality, we may assume that x © A — W and
y € W. Then from h(x) = h(y), we have f(x) = g7!(y), so y = g(f(x)). Because
y € W, y = s(2n) for some string s and some natural number n. Therefore,
s(2n — 1) = f(x) (because g(sQn — 1)) =s(2n) = y = g(f(X)) and g is oneto-one). If n = 1, then s(1) = f(x), which implies s (1) € Rng (f), a contradiction to the definition of string s. Thus, n > 2. But then s(2n — 2) = x because
1) = f(x). This implies that x is a term in the string s, a contradiction to
x € A — W. Therefore, we know that x and y are either both in W or both in A — W.
s(2n —
If x and y are both in W, then h(x) = h(y) implies that g~'(x) = g7!(y). Therefore, x= y because g~! is one-to-one. If xand
y are bothin A — W, then f(x) = f(y)
and x = y because f is one-to-one. In either case, we conclude x = y and h is
one-to-one.
Next we show that / is onto B. Let
some x € A. There are two cases:
b € B. We must show that b = h(x) for
Case 1. If g(b) € W, let x = g(b). Then A(x) = h(g(b)) = g'(g(b)) = b.
Case 2.
If g(b) ¢W, then be Rng(f). Uf b¢Rng(f), then be B-— D.
Therefore, b is the first element of some string, and g(b) is the second
element of that string, so g(b) € W. This is a contradiction.) Because
b € Rng (f), there exists x € A such that f(x) = b. Furthermore, x ¢ W.
292
CHAPTER 5
Cardinality
(If x € W, then x is a term in some string, and, therefore, f(x) and g(f(x))
are the next two terms of the same string. But this is a contradiction
because g(f(x)) = g(b) and we have assumed that g(b) is not on any
string.) From x € A — W, we conclude h(x) = f(x) = b.
In both cases, h(x) = b, so his onto B.
r
The Cantor—Schréder—Bernstein Theorem may be used to prove equivalence
between sets in cases where it would be difficult to explicitly exhibit a one-to-one
correspondence.
Example.
Proof.
Prove that (0, 1) ~ [0, 1].
First, note that (0,1)
[0,1], so (0, 1) < [0,1]. Likewise, because
[0,1]
<(—1, 2), we have [0,1]
C(—1,2). But we know (0,1) © (—1, 2)
and thus (0, 1) = (—1, 2). Therefore, we may write [0,1] < (0, 1). We conclude (0, 1) = Tosi) by the Cantor—Schréder—Bernstein Theorem and thus
(0,
1) © [0, 1].
a
We can use the Cantor—Schréder—Bernstein Theorem to show thatR x RR.
(See Exercise 13.) This means that there are just as many points on the real line as
there are in the entire Cartesian plane. (In a letter to a friend, Cantor said of this,
“T see it, but I don’t believe it.’’)
The cardinal number c did not appear in the increasing sequence
N < DIN) < PAN) < PPM)
< ---
SO we may wonder whether c is one of these numbers and if so, which one. The next
theorem shows that cardinal number c is the second cardinal in the list.
Theorem 5.4.5
(N=;
Proof.
First, recall that any real number in the interval (0, 1) may be expressed
in a base 2 (binary) expansion 0.b,bb3b,..., where each b; is either 0 or 1. If we
exclude sequences that terminate with infinitely many 1’s, such as 0.0101111111
... (which has the same value as 0.01100000...), then the representation is unique.
Thus, we may define a function f: (0, 1) + 2 (N) such that for each x € (0, 1),
f(x) = {n EN: b, = 1 in the binary representation of x}.
The uniqueness of binary representations ensures that the function is defined and is
one-to-one. Because fis one-to-one, (0, 1) < P(N).
Next define g: P(N) > (0, 1). For A € P(N), let g(A) = 0.d,d,a3d4..., where
———
y
t ifne A
2
Ss itn eA
5.4
The Ordering of Cardinal Numbers
293
For any set A € P(N), g(A) is a real number in (0, 1) with decimal expansion con-
sisting of 2’s and 5’s. (Any pair of
digits not including 9 will do.) The function g is
one-to-one but certainly not onto (0, 1). Therefore, P(N) <€ (O, 1)
=
the
Cantor-Schréder-Bernstein
Theorem,
#(N)= (0, 1). Therefore,
(N) =
a
Additional ordering properties that hold for all cardinals are listed in the next
theorem. Each one is obviously true for finite cardinals (0 and the natural numbers),
but each requires the Cantor—Schréder—Bernstein Theorem for its proof in general.
(See Exercise 11.)
Theorem 5.4.6
For sets A, B, and C,
(a)
ifA=<B,then BZA.
(b) ett Agendus
G@ntnene see Oe
(c)
ipAu—wBandupeaGeathenpAmnGe
(d)
if A <
BandB
< C,thenA
< C.
It is tempting to extend our results even further to include the converse of
Theorem 5.4.6(a): “If B < A, then A < B.” (As far as we know now, for two given
sets A and B, both A < Band B < A may be false.) The Cantor—Schroder—Bernstein
Theorem turned out to be more difficult to prove than one would have guessed from
its simple statement, but the situation regarding the converse of Theorem 5.4.6(a)
is even more remarkable. This is discussed in Section 5.5, where “If B <A, then
A=B87 is rephrased as “Either A < B, or A = B, or B < A.”
Exercises 5.4
1.
For each set, name a proper subset that has the same cardinality.
(a) NU {0}
(b) Z
(c)
(d)
R
NxN
2.
Prove the remaining parts of Theorem 5.4.1.
3.
Provethat
(a)
(b)
4.
won
_
r
a
if A < Band B = C, then A <
if A < Band A = GrtheniGr= B
State whether each of the following is true or false. For each false statement,
give a counterexample.
(a)
A <= B implies that A C B.
(D)
ANB = B.
(c)
A < B implies P(A) & P(B)
(d)
(e)
A= B implies A= B.
If B — A is nonempty, then A <AUB.
(f) PN) < PR).
294
CHAPTER 5
Cardinality
Prove that there is no largest cardinal number.
Arrange the following cardinal numbers in order:
(a)
(0,1), [0,1], {0,1}, {0}, ACR), Q, 0,
(b),
{055}, 105). 1053; SPR
()
R—N
OUitz) R=
R— N, PPCR)), R
i438), PA 057). ((0;5)), 0/5) = 131;
txt, PCO),
(0721, ONco),. 2, RZ)
Suppose that there exist three one-to-one functions f: A >-B, g: B > C, and
h: C — A.The functions may or may not map onto: their codomains. Prove
that Amare ese
Complete the proof of Theorem 5.4.2 by proving that Np < ¢.
<S
If possible, give an example of
(a) aone-to-one function f: R > N.
(b) aone-to-one function f: P(N) > N.
(c) aone-to-one function f: [4,5] > Z.
(d) one-to-one functions f: [5,7] — [3,6] and g: [3,6] — [5,7] such
that neither fnor g maps onto its codomain.
(e) one-to-one functions f: N > Q and g: @ —> N such that neither fnor g
maps onto its codomain.
10.
Prove that if there is a function f: A — N that is one-to-one, thenA is countable.
11.
Prove Theorem 5.4:6: For sets A, B, and C,
(a) ifA < B,thenB <A.
(b) Ries =ebeandes ea1GaihenvAn <a @
(c) A < Band B < C,thenA
< C.
(d) if A < BandB < C,thenA
< C.
12.
Use a cardinality argument to prove that there is no universal set of all sets.
132
Use the Cantor—Schréder—Bernstein Theorem to prove the following.
(a)
The set of all integers whose digits are 6, 7, or 8 is denumerable.
(b)
Rx
(c)
If A C R and there exists an open interval (a, b) such that (a, b) C A,
R&R.
then A = c.
14.
In Section 5.1, we proved that the set ¥ of all functions from N to {0, 1}
is equivalent to P(N). Use the Cantor-Schréder—Bernstein Theorem to give
another proof that # ~ P(N).
15.
Apply the proof of the Cantor-Schréder—Bernstein Theorem to this situation:
=
Gee
zp fall lle ll
aceee : ee
and g:: B — A, where g(x) Tiga
aef:A—B, where
|
f(x)me= Geet
cod
1
ab Oe that = and | are in B — Rng (f).
Let s be the string that begins at s and let ¢ be the string that begins at i
(a)
Find s(1), s(2), s(3), s(4).
(b)
(c)
Find (1), (2), 1(3), ¢(4).
Define h as in the proof of the Cantor—Schréder—Bernstein Theorem,
and find (2), h(8), h(13), and h(20).
5.5
Comparability and the Axiom of Choice
295
Ab, Consider the family #€ = {f: fis a function from [0, 1] to [0, 1]}.
*
(a) Prove that there is no bijection from [0, 1] to #.
Proofs to Grade
17.
(b)
Show that 9 is uncountable by showing that 9 has a subset equivalent
(c)
to [0, 1].
What is the relationship between 10, 1] and #€?
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
(a)
Claim.
“Proof.”
If A <
BandA = C, thenC < B.
Assume that A < B and A = C. Then there exists a function f
such that f: A 1-5 B. Because A= CafnG eam ;) Therefore,
x (b) Claim.
C < B. m
If BC C and B = C, then B= C.
“Proof.”
Suppose that B # C. Then B is a proper subset of C. Thus,
C— Bs @e Ths implies"C— pe 0. But C= BU(C
8) and
because B and C — B are disjoint, C = B + (C — B). By hypothesis,
B = C. Thus, (C — B) = O. This is a contradiction.
(c)
Claim.
If A < Band B < C,thenA
|
< C.
“Proof.”
Assume that A < Band B < C.
Then there exists f: A — B that is one-to-one and not onto B, and
there exists g:B — C that is one-to-one and not onto C. iin
g of: A — C is one-to-one
and not onto C, soA < C.
*
(d)
Claim.
If A #@ and A < B, then there exists a function f: B aes a
“Proof.”
Assume that A < B. Then there exists a function giA ed)
Because g is one-to-one, every b in B has exactly one pre-image in A.
Thus, the set f= {(b, y): y is the pre-image of b under g} is a function. This function is onto A because for each a in A, g(a) € B, and so
f(g(a))= a. Thus, f: B “3 A.
5.5
=
Comparability and the Axiom of Choice
One of the most useful properties of the natural numbers is the trichotomy
property: [fm and n are any two natural numbers, thenm > n,m =n,orm < Nn.
The analog for cardinal numbers is stated in the Comparability Theorem.
Theorem 5.5.1
The Comparability Theorem
_
IfA and
B are any two sets, then A < B,A
=B,orB <A.
Surprisingly, it is impossible to prove the Comparability Theorem from the
axioms and other theorems of Zermelo—Fraenkel set theory (see Section 2.1). Ina
formal study of set theory, one can build up, starting with a few axioms specifying that certain collections are sets, to the study of the natural, rational, real, and
complex numbers; polynomial, transcendental, and differentiable functions; and all
the rest of mathematics. Still, comparability cannot be proved. On the other hand,
it is impossible to prove in Zermelo—Fraenkel set theory that comparability is false.
296
CHAPTER 5
Cardinality
Theorem 5.5.1 is undecidable in our set theory; no proof of it and no proof of its
negation could ever be constructed in our theory.
At this point, we could choose either to assume the truth of Theorem 5.5.1 (or
assume the truth of some other statement from which comparability can be proved)
or else to assume the truth of some statement from which we can show comparability is false. Of course, we have revealed the fact that we want comparability to be
true by labeling the statement as a theorem. It has become standard practice by most
mathematicians to assume that the Comparability Theorem is true by assuming the
truth of the following statement.
The Axiom of Choice
If 4 is any collection of nonempty sets, then there exists a function F (called a choice
function or choice rule) from «& to U A such that for every A € A, F(A) € A.
AEA
The Axiom of Choice at first appears to have little significance: From a collection of nonempty sets, we can choose an element from each set. If the collection is
finite, then this axiom is not needed to prove the existence of a choice function. It
is only for infinite collections of sets that the result is not obvious and for which the
Axiom of Choice is independent of other axioms of set theory.
Many examples and uses of the Axiom of Choice require more advanced
knowledge of mathematics. The first example we present is not mathematical in
content, but it has become part of mathematical folklore.
A shoe store’s stockroom has an infinite number of pairs of shoes and an infinite number of pairs of socks. A customer asks to see one shoe from each pair.
When the clerk has an explicit rule for making a choice, he does not need to invoke
the Axiom of Choice to know there is a choice function. His rule may be to choose
the left shoe from each pair. If the socks in each of the infinitely many pairs are
indistinguishable and a customer asks to see one sock from each pair, then the clerk
has no rule for making a choice. Without the Axiom of Choice, we can’t say there
is a function that chooses one sock from each pair.
Example.
Let 4 = {A:A C RandA # ©}. If we are to select one element from
each setA in 4, then we will need to use the Axiom of Choice. However, if we let
BR = {A:ACR,A# ©, andA is finite}, then, even though % is infinite, we do
not need the Axiom of Choice to select one element from each set in %. Our choice
rule might be this: For each B € 9%, choose the greatest element in B. Because B is
finite, such an element exists for each B € BR.
Proofs of the Comparability Theorem may be found in undergraduate textbooks that include a thorough development of axiomatic set theory.* In order to
round out our results on countable sets, we will use the Axiom of Choice to give a
proof of Theorem 5.3.8, which we postponed in Section 5.3. The proof uses the fact
* See, for example, Elliott Mendelson, Introduction to Mathematical Logic, 5th ed. (Chapman & Hall/
CRC Press, Boca Raton, FL, 2009).
5.5
Comparability and the Axiom of Choice
297
that every denumerable family of sets can be replaced by another denumerable family of pairwise disjoint sets whose union is the same as that of the first collection.
Lemma 5.5.2
Let
{A;:ie@N} be a denumerable family of sets. For each ie N, let B,=
i-1
An (a) Ai)Then {B;: i e N} is a denumerable family of pairwise disjoint sets
k=1
such that LJ A; = LU B..
ieN
ieN
Proof. Let x <A andj and k be natural numbers. Ifj < k and x € B,, then x € By.
Thus B; and B, are disjoint. Therefore {B;: i € N} is pairwise disjoint.
By definition of the B, UB,
neN
¢ UA, If x € A; for some i € N, then there
neN
is a smallest natural number k such that x € A, so x € B,. Thus
Menges,
neN
Therefore
Theorem 5.3.8
LJA, =
U Be
neN
neN
Let & be a countable
neN
i]
collection of countable
sets. Then
LJA is countable.
Ac
(restated)
Proof.
Let & be a countable collection of countable sets. We may assume that
A = {A,, A,,...A,,,--.} is denumerable because if there were only k sets in A, we
could extend 4 by defining A,,,; = A,,. = --- = ©. By Lemma 5.5.2, we may
also assume that the sets in & are pairwise disjoint. From the fact that each set A,,
is countable, we know that each A,, is equivalent to a subset of N: to N,, for some
néN, to ©, or to N itself. Thus, for each A,,, there is a (possibly empty) bijection
f,, from A,, to a subset of N.
We now define a function g from
LU) A to N. Let
xe LU A. Because the
Aced
AEA
collection # is pairwise disjoint, x € A,, for exactly one natural number m. Let p,,
be the mth prime number. Define g(x) = (p,,)/". (See the example below.) We
claim that g is one-to-one. Suppose that g(a) = g(b) for some a, b € |) A. Then
Aecd
a € A,and b € A, for some i, j € N. Then (py = (pi, so by the Fundamental
Theorem of Arithmetic, i = j and f,(a) = f,(b). Because f; is one-to-one, a = b.
Because g is a one-to-one function from U,.4A to N, J A is equivalent to a
AeEdd
subset of N. And because every subset of N is countable,
) A is countable.
a
Aes
As an example of how the function g in the proof above works, suppose that
As = {r,s,t}. Then A; © Ns, sof; is a bijection from A; to N,. Let’s suppose that
f(r) = 2, f5(s) = 3, and f5(t) = 1. Because the fifth prime is ps; = 11, the images
of g for the elements of A are g(r) = 11? = 121, es) = 11° = 1,331, and e@) =
11' = 11. Every element of the next set, Ag, is mapped to distinct powers of the
sixth prime, 13. If Ag happens to be denumerable, then every power of 13 will be
in Rng (g).
298
CHAPTER 5
Cardinality
Where was the Axiom of Choice used in the proof of Theorem 5.3.8? For
each m EN, A,, is countable, and there are generally many bijections from A,,, to
a subset of N. In our proof, we select one such bijection and call it f,,. We do this
infinitely many times, once for each m € N. Our collection consisted of sets of
bijections, and we needed one bijection from each set of bijections. There is no way
to select these functions without the Axiom of Choice.
Many important theorems, in many areas of mathematics, cannot be proved
without the use of the Axiom of Choice. In fact, several crucial résults are equiva-
lent to it. Some of the consequences of the axiom are not as natural as the Comparability Theorem, however, and some of them are extremely difficult to believe.
One of these is that the real numbers can be rearranged in such a way that every
nonempty subset of R has a smallest element—in other words, that the reals can
be well ordered. Another, called the Banach—Tarski paradox, states that a ball can
be cut into a finite number of pieces that can be rearranged to form two balls the
same size as the original ball. Actually, this “paradox” is hardly more surprising
than the result in Section 5.4 that R x R ~ R, although that result can be proved
without the Axiom of Choice.
The Axiom of Choice has been objected to because of such consequences and
also because of a lack of precision in the statement of the axiom, which does not
provide any hint of a rule for constructing the choice function F. Because of these
objections, it is common practice to call attention to the fact that the Axiom of
Choice has been used in a proof so that anyone who is interested can attempt to find
an alternate proof that does not use the axiom.
The proofs of the next three results about cardinality are typical of proofs that
rely on the Axiom of Choice. The first theorem says that if there is a function f from
a set A onto a set B, then A must have at least as many elements as B. The proof
uses the Axiom of Choice to choose, for every b € B, ana € A such that f(a) = b.
Theorem 5.5.3
If there exists a function from a set A onto a set B, then B < A.
Proof.
If
B= ©, then
BCA.
Therefore,
in this case,
B < A. Suppose
that
B # ©, and suppose that f: A — B is onto B. Let b € B. Because
f is onto B, the
set C, = {a € A: f(a) = b} of pre-images of b is a nonempty subset of A. See
Figure 5.5.1. Therefore, 4 = {C,: b € B} is a nonempty collection of nonempty
sets. Note that the union |_) C, = A because the domain of the function {sam
beB
By the Axiom of Choice, there is a function g: 4 >
(J C, such that g(C,) € C,
beB
for every b € B. (The choice function g chooses one element g(C ») from each nonempty set C,,.) Because Ly C, = A, the mapping g is a function from «4 to A.
beB
Define h: B > A by h(b) = g(C,) for every b € B. We will show that h is
one-to-one. Let b and d be elements of B, and suppose that h(b) = h(d). Then
* Paul Howard and Jean E. Rubin, Consequences of the Axiom of Choice (Mathematical Surveys and
Monographs, v. 59) (American Mathematical Society, Providence, RI, 1998).
5.5
Comparability and the Axiom of Choice
299
Figure 5.5.1
g(C,) = g(C,). Call this object x. Then x € A, and by definition of g, we have x € C,
and x € C,. Thus, f(x) = b and f(x) = d, so b = d. Therefore, h is one-to-one from
A to B. We conclude that B < A.
Theorem 5.5.4
|
Every infinite set A has a denumerable subset.
Proof.
Suppose that A is infinite. We inductively define a denumerable subset of A. First, because A is infinite,
A#@. Choose a, € A. Then A — {aj}
is infinite, hence nonempty.
a, € A. Continuing
Choose
a,¢A — {a,}. Note that a,#a,
in this fashion, suppose
and
that a,,..., a, have been defined.
‘ThenvAr et are a,} # ©, so select any a,,, from this set. By the Axiom of
Choice, a,, is defined for all n € N. The a, have been constructed so that each
a, ©A and a;#a, for i#j. Thus, B = {a,:n eN} is a subset of A, and the
function
f given by f(n) = a, is a one-to-one correspondence from N to B. Thus
B is denumerable.
rT]
Theorem 5.5.4 can be used to prove that every infinite set is equivalent to one
of its proper subsets. (See Exercise 8.) This result characterizes infinite sets because,
as we saw in Section 5.1, no finite set is equivalent to any of its proper subsets.
Theorem 5.5.4 also confirms that Np, is the smallest infinite cardinal number.
For any set A with infinite cardinality, there is a denumerable subset B of A. There-
forepNg =p =":
Corollary 5.5.5
A nonempty set A is countable if and only if there exists a function from N onto A.
Proof.
Exercise 9.
a
We have seen that 8) = N < ce= P(N) < P(P(N)) < PPAIN))) < ....
The fact that Np) and ¢ are the first two cardinal numbers in this sequence does
not necessarily mean that ¢ is the next largest cardinal number after No. Cantor
conjectured that this is so: That is, no set X exists such that Xy < Xe This conjecture, called the continuum hypothesis, is one of the most famous problems in
300
CHAPTER 5
Cardinality
modern mathematics. The combined work of Kurt Gédel” in the 1930s and Paul
Cohen‘ in 1963 shows that the continuum hypothesis can be neither proved nor disproved in Zermelo—Fraenkel set theory. Like the Axiom of Choice, the continuum
hypothesis is undecidable.
Exercises 5.5
Indicate whether the Axiom of Choice must be employed to select one element from each set in the following collections.
(a)
an infinite collection of sets, each set containing one odd and one even
integer
(b) a finite collection of sets such that each set is uncountable
(c) aninfinite collection of sets, each containing exactly four natural numbers
(d) adenumerable collection of uncountable sets
(ec)
{A:nNCACR
(f)
(g)
(h)
{A:A CN
{A:A CR
{A:A CR
— Q}where zN = {2n:neEN}
(a)
Prove this partial converse of Theorem 5.5.3 without using the Axiom
and bothA and N — A are infinite}
and bothA and R — A are infinite}
and A is denumerable}
of Choice: Let A and B be sets with
B # ©. If B < A, then there exists
g: A — B that is onto B.
(b)
Use the Axiom of Choice toprove that if there exists f: A oS B, then
there exists a function g: B a
A
Let A and B be any two nonempty sets. Prove that there exists f: A > B that
has at least one of these properties:
(i)
fis one-to-one.
(ii) fis onto B.
Prove that
*
(a) if f: A — B, then Rng (f) <A.
(b) for every sequence x, the range of x is a countable set.
Suppose thatA is a denumerable set and B is an infinite subset of A. Prove that
BOSS
Suppose that B < C and B # A. Prove that A < C.
A
Let {A;: i ¢ N} be a collection of distinct pairwise disjoint nonempty sets.
That is, if andj are in N andi ¥ j, then A; # A; and A;
A; = ©. Prove that
(e) A, includes a denumerable subset.
ieN
* Kurt Gédel (1906-1978) was a logician best known for his Incompleteness Theorem, which says
(roughly) that in any logical system rich enough to include the theory of the natural numbers, there will
always be true statements that are unprovable.
+ Paul Cohen (1934-2007), created a method of proof that he used to show that neither the Axiom of
Choice nor the continuum hypothesis can be proved in the set theory based on the Zermelo—Fraenkel
axioms.
5.5
Comparability and the Axiom of Choice
301
Let A be an infinite set. Prove that A is equivalent to a proper subset of A.
Prove Corollary 5.5.5: A nonempty set A is countable if and only if there is a
function f: N— A that is onto A.
Proofs to Grade
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
(a) Claim.
There is a denumerable set B of irrational numbers such that
any two elements of the set differ by an integer.
“Proof.”
Define the sequences by setting the nthtermtobes, = a + n,
and let B be the range of s. Then s is a one-to-one correspondence between
N and B, so B is denumerable. The difference between any two elements
ma + mand za + nof Bis m — n, which is an integer.
rT]
(b)
Claim.
Every infinite set A has a denumerable subset.
“Proof.”
Suppose
that A is infinite and that no subset of A is
denumerable. Then all subsets of A must be finite. In particular,
Thus, A is finite, contradicting the assumption.
(c)
A C A.
x
Claim.
Every infinite set A has a denumerable subset B.
“Proof.”
If A is denumerable, let B = A, and we are done. Other-
wise, A is uncountable. Choose x, € A. If A — {x,} is denumerable, let
B=A = {x,}. Otherwise; choose x, <=A — "(xj|e it Ars ire ae
denumerable, let B = A — {x,, x,}. Continuing in this manner, using
the Axiom
(d)
“Proof.”
(e)
of Choice, we obtain a subset C =
{x,,x,...}
(g)
a
Let A be an infinite set. By Theorem 5.5.4, A has a denu-
merable subset B. Then A — B is infinite because A is infinite and is
disjoint from B. By Theorem 5.5.4, A — B has a denumerable subset C.
Then B and C are disjoint denumerable subsets of A.
=
Claim.
If A#©
and A & B, then there exists a function f: B me A.
“Proof.”
Assume that A < B. Then there exists a function gi A oH
Then g™! is a function that maps Rng (g) onto A. Let a* be some fixed
element er and define f = g—! U {(b, a*): b € B — Rng (g)}. Then
fe fBes
onto
(f)
such that
B =A — Cis denumerable.
Claim.
Every infinite set has two disjoint denumerable subsets.
P|
Claim. fib infinite set has two disjoint denumerable subsets.
“Proof.”
Let A be an infinite set. By Theorem 5.5.4, A has a denumer:
}
(=i
able subset B. Because B is denumerable, there is a function f: N a B.
Let
C= {fQn):neEN}
and
D={fQn—1):neEN}.
Then
Cra hi (2), (4) of ©)scr-} and D.= {7 (1), 7G),
7 O),.--}.are disjoint
denumerable subsets of A.
|
Claim.
Every subset of a countable set is countable.
“Proof.”
Let A be a countable set, and let B C A. If B is finite, then B
is countable by definition. If B is infinite, because B C A, A is infinite.
Thus, A is denumerable. By Theorem 5.5.4, B has a denumerable subset
C. Thus,
CCBCA,
which
implies
8) = C and
C<
B<A=RX,.
Therefore A= B = Xp. Thus, B is denumerable and hence countable. =
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Concepts of Algebra
The broad meaning of algebra refers to systems of computation and the study of
properties of such systems. In this chapter, we make precise the idea of an algebraic
system and introduce several different types of systems. The goal is to make available additional opportunities to sharpen your proof-writing skills, while providing
a first experience with some of the topics in this important field.
6.1
Algebraic Structures
A system of computation (or an algebra) consists of a set and an operation on the set.
The set of natural numbers with the addition operation is a system of computation.
In this example, the operation acts on pairs of elements of N, such as the pair (5, 3),
to produce an element of N (in this case, 8). In other words, addition in N is a
function from N x N to N.
DEFINITION
Let A be a nonempty set. A binary operation « on A is a
function from A x A to A. The notation for the image of (x, y) € A x A is
x * y under the operation *.
“Binary” refers to the fact that the operation acts on two elements of A to
produce a third. There are operations other than binary operations. For example, a
ternary operation maps A x A x A to A. In this text, when we say “operation,” we
mean a binary operation.
The most familiar operations include addition and multiplication on sets of
numbers. Other operations include the union and intersection on the power set of a
set and composition for a collection of functions whose domain is R.
303
304
CHAPTER 6
Concepts of Algebra
We
will usually denote an operation by one of the symbols-,©°,
or *.
Sometimes, as is the case for multiplication of numbers, the operation symbol is
omitted, and we write xy for the image of (x, y).
The images x - y, x ° y, x * y, and xy are called products, regardless of whether
the operations have anything to do with multiplication. Similarly, x + y is referred to
as the sum of x and y, even when the operation + does not involve adding numbers.
DEFINITION
An algebraic system (or algebraic structure) is a nonempty set A with a collection of one or more operations on A and a (possibly
empty) collection of relations on A.
The notation (IR, +, -, <) refers to the system of real numbers with the operations of addition and multiplication and the relation “less than.” For all but the last
section of this chapter, “algebraic system” will refer to a system with one binary
operation and no relations. We write, for example, (Q, -) for the rationals with
the usual multiplication and (P(A), U) for the power set of a set A with the union
operation.
DEFINITION
Let (A, *) be an algebraic system. Let B be a subset of A.
We say B is closed under the operation « if x « y € B forall x, y € B.
For an algebraic system (A, *) the set A is, of course, closed under * because
the operation * 1s a function that maps to A. The three statements “‘A is closed under
*,”
99
66
“x 1g an operation on A,” and “(A, *) is an algebraic system” are all equivalent.
Examples. The algebraic system (IR, - ) has many subsets that are closed under
multiplication. The set @ of rational numbers is closed under multiplication
because the product of any two rational numbers is rational. Other subsets of R that
are closed under multiplication include N, Z, the set of even integers, the intervals
(0, 1) and [0,1], and the sets {—1,1} and {—1, 0, 1}.
Within the system (IR, +), the set of even integers is closed under addition, but
the set of odd integers, the irrationals, and the set (0, 1) are not.
For a system (A, *) with n elements, the operation may be displayed by means
of ann x narray called a Cayley’ table or operation table. The result of the operation x * y is the entry found in row x and column y.
* Arthur Cayley (1821-1895) was the leader of the British school of pure mathematics in the 19th
century. To earn a living, Cayley was a lawyer the first 14 years of his adult life, specializing in property
law. During that time, he wrote nearly 300 mathematical papers. His work included many contributions
to the algebra of matrices, non-Euclidean geometry, and n-dimensional geometry.
6.1
Example.
Algebraic Structures
305
The table in Figure 6.1.1 defines an operation * on the set A= {1, 2, 3}.
We see, for example, from row 2 and column | that 2 * 1 = 3. A sample computation in this system is (3 * 2) *(1 *3)=3*1=2.
Oo
*
|
2
3
3
3)
2
2
3)
2}
1
3)
1
g)
3)
Figure 6.1.1
Cayley tables are impractical when n is large and impossible to construct for
infinite algebraic systems. In these cases, the operation must be described by a rule
or algorithm.
In Section 3.4, we defined addition and multiplication for the equivalence
classes in Z,,:x + y =x + yandx.y =x-y. Theorem 3.4.1 shows that + and are well defined, which means that they are operations on Z,,. Thus, (Z,,, +) and
(Z,,. *) are algebraic systems with m elements.
It helps to become proficient in modular arithmetic because these operations
have many applications. We follow the usual practice of omitting the bar notation
for equivalence classes, referring to the elements of Z,, simply as 0, 1, 2, 3,...,
m— 1. In Z,, we have 2 + 4 = 1 because 2 + 4=6 = 1 (mod 5) and4.3
=2
because 4 - 3 = 12 = 2 (mod 5). The Cayley tables for addition and multiplication
for Z, are given in Figures 6.1.2(a) and (b), respectively.
a
0
Z
3
4
0
1
2
3
4
0
0
0
0
0
0
1
2
3
4
0
2
4
]
3
0
3
1
4
2
0
3
2)
1
+
0
1
2
3
=
0
1
2
3
4
1
2
3
0
2
S
4
0
]
3
4
0
]
2
4
0
1
2
3
0
1
2
3
+
Figure 6.1.2
DEFINITIONS
Let (A, «) be an algebraic system. Then
* is commutative on A if for allx, ye A,x* y = y*x.
* is associative onA if for all x, y, z € A, (x * y) *Z =x * (y * Z).
anelemente
of A is anidentity element for*ifforallx € A,x*e =e*xx =x.
if A has an identity element e and if a and b are in A, then b is an inverse of a if
a*b=bx«a =e. Inthis case, a would also be an inverse of b.
306
CHAPTER 6
Concepts of Algebra
The system of integers with addition has all four of the properties listed. Addition is commutative and associative in (Z, +), the integer 0 is the identity element,
and every element has its inverse (its negative). Multiplication is commutative and
associative in (Z, -), and | is the identity element, but no integer other than | and
—1 has a multiplicative inverse.
Some very important operations are not commutative. For example, subtraction and division of real numbers and composition of functions are not commutative operations. Neither is multiplication of square matrices. (See Exercise 5.)
When an operation is displayed with a Cayley table, it is easy to see whether
the operation is commutative—the table is symmetric about its main diagonal from
the upper left to the lower right. The system will have an identity element e if
and only if the row labeled e is the same as header labels on the columns and the
column labeled e is the same as the header labels on the rows. Elements x and y will
be inverses if and only if e is the entry in both row x, column y and row y, column x.
Examples.
Figure 6.1.3 displays the Cayley tables for three algebraic systems
(Byio) A(Cxsand (Dice), where.B == (a,b, c}, C= {(pyger}, and D = {x, y; z}.
Operation © is not commutative because its table in Figure 6.1.3(a) is not symmetric about the main diagonal. No column duplicates the labels for the rows, so
the system does not have an identity. Without an identity, the issue of inverses does
not arise for (B, o).
The table for (C, -) in Figure 6.1.3(b) is symmetric about the main diagonal,
and the entries in the row and column labeled qg duplicate the headers. Therefore,
(C, -) has a commutative operation with identity element g. Every element of C
has an inverse because
p-r = r- p = q and q- q = q. The elements p and r are
inverses, and gq is its own inverse.
The table in Figure 6.1.3(c) for (D, +) is not symmetric, so the operation +
is not commutative. The element z is an identity for +. The element x is its own
inverse, as is the element z. However, the identity z does not appear in the row for
y, So y does not have an inverse.
oO
b
he
0G
y
Z
b
b
x
)
Zz
2
z
x
x
a y
b
is
x
)
Z
(a) (B, 0)
(b) (C, -)
(c) (D, +)
Figure 6.1.3
It is not easy to tell by looking at a table whether an operation is associative. For a system of order n, verification of associativity may require checking
n° products of three elements, each grouped two ways. For the system (A, *) in
Figure 6.1.1, the operation is not associative because (1 * 1) #241 x (1 * 2).
The operations for systems (B, 0) and (C, -) in Figure 6.1.3 are associative, but
6.1
Algebraic Structures
307
the + operation for (D, +) is not. You should find elements i, j, k € D (not necessarily distinct) for which (i + j) +kK#Ai+ G+h).
The associative property is a great convenience in computing products. First,
it means that so long as factors appear in the same order, we need no parentheses.
For both x(yz) and (xy)z, we can write xyz. This can be extended inductively to
products of four or more factors: (xy)(zw) = (x(yz))w = (xy(zw)), and so forth.
Second, for an associative operation, we can define powers. Without associativity,
(xx)x might be different from x(xx), but with associativity, they are equal, and both
can be denoted by x°.
Theorem 6.1.1
Let (A, *) be an algebraic structure.
(a)
(A, *) has at most one identity element.
(b)
Suppose that * is associative with identity e. If a € A has an inverse, then a
has only one inverse.
Proof.
(a)
(We need to show that if e andf are both identities for *, then e = f.)
Suppose that e and f are both identities for *. Then because e is an identity,
the product e « f = f. And because f is an identity, e « f = e. Therefore,
Ht
(D)eScelexcicisem:
Theorem 6.1.2
|
For every natural number m,
(a)
(Z,,, +,,) is an algebraic system that is associative and commutative with
(b)
identity element 0. Every element has an inverse.
(Z,,, :,,) is an algebraic system that is associative and commutative. Ifm > 1,
the system has identity element 1.
Proof.
See Exercise 12.
Po
Multiplication in Z,, does not have all the properties of multiplication of the
integers. When m is a composite, there are elements a and b in Z,, such that a # 0,
b # 0, and ab = 0 (Theorem 3.4.3).
e
DEFINITIONS
Let a be a nonzero element of (Z,,, -).
If ab = 0 for some b # 0, then we say a (and also b) is a divisor of zero.
If a has a multiplicative inverse in Z,,,, then a is called a unit in Z,,.
The set of all units in Z,, is denoted U,,,.
Example. To find the zero divisors in the system (Z,,, -), we look for nonzero
elements a and b for which ab = 0 in Z,,. For example, 9 and 10 are zero divisors
308
CHAPTER 6
Concepts of Algebra
because 9 - 4 = 36 = 0 (mod 12) and 10 - 6 = 60 = 0 (mod 12). The set of zero
divisors for Z,, is {2, 3, 4, 6, 8, 9, 10}.
To find the units in Z,5, we look for the elements that have inverses. These are 1,
5,7, and 11 because 1 -1=1,5-5=25=1,7-7=49=1,and11-11=121=1.
Therefore ly
allo
lia
Theorem 3.4.3 says that whenever m is composite, (Z,,, -) has at least one zero
divisor. If m > 2, Z,,, must have at least two distinct units. (See Exercise 13.)
Theorem 6.1.3
Let m and a be natural numbers with a < m. Then
(a)
(b)
aisaunit in (Z,,, -) if and only if a and mare relatively prime.
aisa divisor of zero in (Z,,,m -) if and only if a and m are not relatively prime.
Proof.
(a)
Suppose that a and m are relatively prime. Then there exist integers r and
s such that ar + ms = 1. Then ar — 1 = —ms, so m divides ar — 1. Thus,
ar — 1 =0 (mod m) and ar = 1 in Z,,. Therefore, a and r are inverses, which
(b)
implies that a is a unit in (Z,,, -).
The proof of the converse is left as Exercise 14.
Suppose that a and m are not relatively prime. Let d be the greatest common
divisor of a and m. Then d > 1, and there exist positive integers a, < a and
m, <msuch that a= a,d and m=mz,d. But then am, = (a,d)m, = a,(dm,) =
a,m = 0 (mod m). Therefore, am, = 0 in Z,,,, so a is a divisor of zero.
The proof of the converse is also left as Exercise 14.
a
It follows from Theorem 6.1.3 that every nonzero element of Z,, is either a
unit or a zero divisor but never both. Furthermore, when p is prime, every element
except 0 in Z, is a unit because every element in Z, except 0 is relatively prime to p.
Thus, whenp is prime, U, =si(
Theorem 6.1.4
1, 2 Sha
piel he
For every natural number m > 1, the system (U,,, -) is an algebraic system, has
identity element 1, is associative, and is commutative. Furthermore, évery element
in U,, has a multiplicative inverse.
Proof.
(We first show that (U,,, -) is closed under multiplication. We do so by
proving that if a and b are units in Z,,, then ab is a unit in Z,,.) Suppose that a
and b are units in Z,,. Then ac = | and bd = 1 for nonzero elements b and d. Then
(ab)(cd) = (ac)(bd) = 1 - 1 = 1. Thus, ab is a unit in Z,,,.
The proofs that | is the identity in U,, and that multiplication is associative and
commutative in U,, are left as Exercise 15. Finally, if a € U,,, then by definition, a
has an inverse in Z,,, and a is the inverse of that element. Thus, the inverse of a is
a unit and is an element of U,,.
isd
6.1
Example.
Algebraic Structures
309
The Cayley table for multiplication on U, = {1, 2, 4, 5, 7, 8} is given
below. Note that 1 is the inverse of 1 and 8 is the inverse of 8. Also, 2 and 5 are
inverses, and 4 and 7 are inverses.
1
2
4
5)
7
8
1
2)
+
>)
qe
1
2)
4
5
I
8
D
4
8
|
S)
ii
4
8
7
2,
1
5
3)
qo
Dh yp a
1
5
8
4
4
2
2
i
4
7
8
+
(Us, »)
Figure 6.1.4
Exercises 6.1
1.
Determine which of the following are algebraic structures—i.e., which sets
are closed under the given operation. For those systems that are algebraic
structures, is the operation commutative? associative?
x
(a)
(d)
(sy
GQ
2.
(Z, +)
TEN)
(R—- Q,-)
G.Orlh
()
Gk, =)
(Hy
(OF)
(P(A), N)
({0, 1}, +)
Consider the set A = {a, b, c, d} with operation o given by the Cayley table
at the right.
lo)
a
b
@
d
(a) Name the identity element of this system.
(b) Is the operation o associative on A?
a
a
b
Cc
d
(c) Is the operation o commutative on A?
b
b
a
d
Cc
(d)
3.
(Zo)
(b)
(R, =)
ra A
(Ors)
(h)
CA
EAS?~ ee He
For each element of A that has an inverse,
Cc
Cc
d
a
b
name the inverse.
d
d
Cc
b
a
*
a
b
Cc
d
a
b
Cc
d
G
d
a
b
d
a
b
€
a
b
Cc
d
b
Cc
d
a
(e)
Is B, =
(f)
(g)
(h)
Is B, = {a,c} closed under o?
Name all subsets of A that are closed under o.
True or False? For all x, ye A, xox =yoy.
{a,b,c} closed under 0?
Repeat Exercise 2 with the operation * given by the
table on the right.
310
CHAPTER 6
Concepts of Algebra
4.
The Cayley tables for operations 0, *, +, and x are listed below.
oO | @
Gi \| @
fo) || fo
IB
al
Sil@w
bb
€
tN
eD
gi | ©
iy | @
C|
pb
Ge
Ip
€
—€
€
fb
oO | @
iy \\
a
oi
Sell
a
2
G
(a)
(b)
Which of the operations are commutative?
Which of the operations are associative?
(c)
Which systems have an identity? What is the identity element?
(d)
For those systems that have an identity, which elements have inverses?
Let m,n € N and Wl = {A: A is anm x n matrix with real number entries}.
(a) Let - be matrix multiplication. Under what conditions on m and n is
(Mt, -) an algebraic system? Is the operation commutative? Explain.
(b) Let + be matrix addition. Under what conditions on m and n is (At, +)
an algebraic system? Is the operation commutative? Explain.
Let - be an associative operation on nonempty setA with identity e. Suppose
that a, b, c, and d are elements of A, b is the inverse of a, and d is the inverse
of c. Prove that db is the inverse of ac.
Let (A, o) be an algebraic structure, a € A, and e be the identity for o.
(a) Prove that if o is associative and if x and y are inverses of a, then x = y.
(b)
Give an example of a nonassociative structure in which inverses are not
unique.
Suppose that (A, *) is an algebraic system and * is associative on A.
(a)
Prove that if a,, a5, a3, and a, are in A, then
(ay * Ay) * (Az * Ay) = A, * (Ay * Az) * 4).
(b)
Use complete induction to prove that any product of n elements
Gj, Ay, 43,...,d, in that order is equal to the left-associated product
(...((@; * dy) * a3)...) * a,. Thus, the product of n elements is always
the same, no matter how they are grouped by parentheses, as long as the
order of the factors is not changed.
/
Let (A, o) be an algebra structure. An element / € A is a left identity for o if
loa=aforeveryaeéA.
(a) Give an example of a 3-element structure with exactly two left identities.
(b)
(c)
Define a right identity for (A, 0).
Prove that if (A, ©) has a right identity r and a left identity /, then r = J,
and that r = / is an identity for o.
10.
11.
Construct the operation table for each of the following:
(a)
(Ze, +); (Ze, ), and
(c)
(Zi;
1)s
(Zio;
(U,, )
-), and
(Uo,
)
(b)
(Zs, +); (Z., -), and
(d)
(Zi,
+);
(Za.
-), and
Find all divisors of zero in
(a)
Zy4.
(b)
Zs.
(C)
Zi.
(d)
(U,, +)
Zo.
(U;,;
)
6.1
12.
(a)
Algebraic Structures
311
Prove that (Z,,,, +) is associative and commutative and has an identity
and that every element has an additive inverse (Theorem 6.1.2(a)).
(b)
Prove that (Z,,, -) is associative and commutative and has an identity
when m > | (Theorem 6.1.2(b)).
Proofs to Grade
1S:
Suppose m € N and m > 2. Prove that 1 and m— | are distinct units in (Z,,, -).
14.
Let m and a be natural numbers with a < m. Complete the proof of Theorem
6.1.3 by proving that
(a) ifaisaunit in (Z,,, -), then a and mare relatively prime.
(b) if ais a divisor of zero in (Z,,, -), then a and mare not relatively prime.
15:
Complete the proof of Theorem 6.1.4. First, show that | is an element of
Une).
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
(a) Claim.
Let (A, o) be an algebraic structure. If e is an identity for o and
if x and y are both inverses of a, then x = y.
“Proof.”
Because x and y are inverses of a,x°a = eandy°a=e.
Thus, x ° a = ya. By cancellation, x = y.
(b)
Claim. If every element of a structure (A, 0) has an inverse, then
commutative.
a
© is
“Proof.”
Let x and y be in A. The element y has an inverse, which
we will call y’. Then y © y’ = e, so y is the inverse of y’. Now x = x,
and multiplying both sides of the equation by the inverse of y’, we have
yOx =x0y. Therefore, © is commutative.
a
(c)
(d)
Claim.
If a and b are zero divisors in (Z,,,-), then ab is a zero
divisor.
“Proof.”
If a and bd are zero divisors, then ab = 0. Thus, (ab)(ab) =
0-0 = 0, and ab is a zero divisor.
|
Claim.
If a and b are zero divisors in (Z,,,m? -) and ab # 0, then ab is a
zero divisor.
“Proof.”
(e)
Because a is a zero divisor, ax = 0 for some x ¥ 0 in Z,,.
Likewise, by =0 for some y#O in Z,. Therefore, (ab)(xy) =
(ax)(by) = 0-0 = 0. Thus, ab is a zero divisor.
a
Claim.
Let m be a composite natural number. Then there exist zero
divisors a and b in (Z,,,, -) such that ab is not a divisor of zero.
“Proof.”
Suppose that m = kj, where k and j are natural numbers
strictly between | and m. Let a = k and b =/j. Then a and b are divisors
of 0 in (Z,,, :), but ab is not a divisor of 0 because ab = Oin (Z,,,-).
(f)
™
Claim. Let m be a natural number. If a and b are units in (Z,,,, -), then ab
is a unit.
“Proof.”
Suppose that ab is not a unit in (Z,,, -). Then m is not relatively prime to ab, so some prime p divides both m and ab. Then by
Euclid’s
Lemma
(Lemma
1.8.3), either p divides a or p divides b.
Therefore, at least one of a or b is not relatively prime to m, so one of a
or b is not a unit in (Z,,, -).
a
312
CHAPTER 6
6.2
Concepts of Algebra
Groups
In this section, we focus on one particularly important algebraic structure, the group.
It was the work of Evariste Galois’ on polynomial equations that led to the study
of groups as an aid to solving equations. The concept of a group has influenced and
enriched many other areas of mathematics. Group theory has applications outside
of mathematics, too, in fields such as nuclear physics and crystallography.
The properties of associativity, the existence of an identity, and the existence
of an inverse for each element are just the properties needed to define a group. Our
approach to defining a group is axiomatic, in the sense that we shall list the desired
properties (axioms) of a structure and any system satisfying these properties is
called a group.
DEFINITION
(G, o) is a group if (G, ©) is an algebraic system such that
(i) the operation o is associative on G.
(ii) there is an identity element e in G for o.
(iii) every x € G has an inverse x! in G.
If G is a finite set, the order of the group is the number of elements in G.
When G is infinite, the group has infinite order.
The systems (R, +), (Q, +), and (Z, +) are all groups with identity 0. The
algebraic system (R, -) is not a group because 0 has no multiplicative inverse. The
system (Rt, -), where Rt denotes the positive real numbers, is a group with identity 1. The system ({0}, +) is the smallest group.
The system (N, +) is not a group because it fails to satisfy group axioms (ii)
and (iii). There is no identity, and therefore it makes no sense to discuss inverses.
The algebraic structure (Z — {0}, -) is not a group because although multiplication is associative and the number | is an identity, only the elements | and —1 have
multiplicative inverses in Z.
/
¥
Theorem 6.2.1
Let m be a natural number. Then
(a)
(b)
Proof.
(a)
(b)
(Z,,, +) is a group of order m with identity 0.
ifm>1,(U,,, -) is a group with identity 1.
These results are immediate consequences of
Theorem 6.1.2(a).
Theorem 6.1.4.
|
* Evariste Galois (1811-1832) was a French mathematician who discovered elegant necessary and sufficient
conditions for a polynomial equation to be solvable by radicals. He introduced the concept of a finite field
and was the first to use the word group in reference to a group of permutations. He died at age 20 as the
result of a duel, but his work led to the development of an area of algebra that is known as Galois Theory.
6.2
Groups
313
When we refer to “the group G” without specifying the operation, the result
of the operation on x and y is called the product of x and y, written simply xy, even
when the operation is not multiplication.
Example.
Let G = {e,a,b, c,d} with the operation given by the Cayley table
shown below. Then G is a group of order 5.
e
a
b
G
d
e
a
b
G
d
e
a
b
c
d
a
b
Cc
d
e
b
G
d
e
a
c
d
e
a
b
d
e
a
b
c
Proof. To show that G is a group of order 5, we first note that G has five elements
and is therefore nonempty. The table defines an operation on G because the product
of every pair of elements of G is specified by the table.
The element e is the identity for G because ex = e (examine the first row of the
table) and xe = x (examine the first column) for all x € G.
The verification that G satisfies (xy)z = x(yz) may be done by considering all
53 = 125 possible assignments of values to x, y, and z. This work can be shortened
considerably by noting that the equation is clearly true when any of x, y, or z is the
identity e. As for the remaining 64 cases, we see, for example, that (bd)a = b(da)
because both expressions have value b and that (ca)b = c(ab) because both expres-
sions have value a.
Finally, every element of G has an inverse. The inverse of a is d because ad = e
and da = e. These equations also prove that a is the inverse of d. The elements
b and c are inverses because bc = cb = e, and e is its own inverse because ee = e.
Therefore G is a group.
a
You may have noticed that the commutative property is not included among
the group axioms. All of the groups considered so far in this section, including
(R, +), (Q, +), (R — {0}, -), and (Z,,, +), have operations that are commutative,
but not all groups have commutative operations. Commutative groups are called
abelian groups and are so named in honor of Niels Abel.”
DEFINITION _ A group Gis abelian if the group operation is commutative.
The abelian property is independent of the group axioms; that is, it cannot be
proved from those axioms. It could have been considered as another axiom for
* The Norwegian Niels Abel (1802-1829) was a pioneer in the development of group theory. He made
fundamental contributions to the theory of functions and proved that no general solution involving
radicals exists for fifth degree polynomial equations. Tuberculosis ended his brilliant career at age 26.
314
CHAPTER 6
Concepts of Algebra
defining a group. Because there are many important algebraic structures that are
groups but do not satisfy the commutative property, mathematicians choose not to
include commutativity in the definition of a group.
Our work with one-to-one correspondences revealed algebraic properties that
we can now use to form groups whose elements are functions. In this way, we
encounter our first examples of nonabelian groups.
A permutation on a nonempty set A was defined in Section 4.4 as a function
f: A = A that is both one-to-one and onto A. If the elements of A are listed in order,
the effect of the permutation is rearranging (or permuting) the elements of A.
The simplified notation of Section 4.4 is useful for working with permutations.
For example, the permutation ¢ = {(1;3), @, 1), G, 4), G, 5). ©, 2)} on the set
A= 11, 263,74, 5) is wiitten'as ¢ = [34 4/5 2]" [he permutation 2 — [54312]
is the function given by h(1) = 5, h(2) = 4, h(3) = 3, A(4) = 1, and A(5) = 2.
From Theorem 4.4.5(b), we know that the composite of two permutations on a
set A is again a permutation of A. Therefore, the set of all permutations on A, with
composition as the operation, is an algebraic structure. We know by Theorem 4.2.2
that composition is associative. Theorem 4.4.5 goes on to say that /, is an identity
(parts (a) and (d) of the theorem) and that every permutation has an inverse (parts
(c) and (e)). Combining these results, we have established the following.
Theorem 6.2.2
Let A be a nonempty set. The set of all permutations on A with the operation of
function composition is a group, called the group of permutations of A.
The group of permutations on the set N, = {1, 2, 3,...,} is given a special
name.
DEFINITION
Let n be a natural number. The group of all permutations
of N,, is called the symmetric group on n symbols and is designated by S,,.
There are exactly n! arrangements of the elements of N,, s6 the order of S,, isn!
Example.
For the set N, =
{1, 2,3}, there are 3! = 6 permutations in $3: J, =
eo ies eo ee ey,
=
(nl 3), ae 2S
and A= St OT.
Peta that we compute products as function composites. For example,
[3e2eL L132) 7 fOr
sh
301 2) because
ei)
=p GO)
=
y= 3
(2° FQ) = AGH) =C),= 1, and
(fo ° FG) = f(AG3)) =f) = 2.
Likewise, [3.0 2)(2)3 11 = hoo = (12 si
for $3 is
[,. The complete Cayley table
6.2
Groups
kes
ey
ety
whey
23 1]
ee
|
fl
33
[ies]
[92 Wy
2s
Zeal
aie]
ale
eS)
(23)
(23
3 12)
(B2A
[21 3
nat
(eal)
[Sle]
[1 2 3]
[2S
[2 1 3]
Ps zy
aia
Aw Sy
231)
exe |
(ee)
Se]
(32 Wh
[23 1]
2
iy
[2 1 3]
Se
[32 i
enle2]
2
[312]
Bale
(2M
(2a)
eS.
[1 2 3]
[23 i
315
Sy
Note that the two products (2 1°31(3 12) and [3 1 2]/2 03)" are different:
Because’ [2 *1"3 [391 23°21
"and"
3.2) 2713) = | 13.2)" the sroupss; 1s
not abelian.
o
The next results are consequences of the group axioms and facilitate calculations involving elements of a group. Notice that proving a statement like x = y7!
is not like proving, say, a trigonometric identity. The statement x = y! is read “x
is the inverse of y” and is proved by showing that x plays the role of an inverse for
y—.e., that the product of x and y is the identity.
Theorem 6.2.3
Let G be a group with identity e. For all a, b, and c in G,
(ajmee
(b)
(c)
(d)
(ae) —G(ab)! =b"'a"!.
if ac = bc, then a = b (Right Cancellation Law).
if ca = cb, then a = b (Left Cancellation Law).
Proof.
(a)
Because a7! is the inverse of a, we have a~'a1 = aa™! 1 = e. Therefore, a acts
(b)
We know (ab)~! is the unique element x of G such that (ab)x = x(ab) = e.
We see that b~'a~! meets this criterion by computing
as the (unique) inverse of a~', so (a~!)"! =a.
(ab)(ba-) = abba
(c)
= ale)a“! = aa
=e.
Similarly (b~!a~')(ab) = e, so ba! is the inverse of ab.
Suppose that ac = be in the group G. Then c™! is in G, and (ac)c7! =
(bc)c—!. Using the associative, inverse, and identity properties, we see that
(ac)c-! = a(cc~'!) = ae =a
and
(bc)c"! = b(c c~!) = be = b.
(d)
Therefiorera’=1d:
See Exercise 8.
Py
316
CHAPTER 6
Theorem 6.2.4
Concepts of Algebra
Let G be a group with identity e. For every a € G,
the function A,: G > G, where A,(x) = ax for each x € G, is a permutation
of G.
the function p,: G > G, where p,(x) = xa for each x € G, is a permutation
of G.
(a)
(b)
Proof.
(a)
To show that A, is one-to-one, let x, y € G, and suppose that A, (x) = A,().
Then ax = ay, so by the Left Cancellation Law, x = y.
To show that X,, is onto G, let b € G. (We need to findx in G so that A,(x) = b.)
Choosex = a~'b.ThenA,(x) = A,(a~'b) = a(a~'b) = (aa™')b = eb = b.
(b)
The proof of part (b) is similar to part (a) and is Exercise 9.
|
Lambda (A,) is the function that performs left multiplication by a, and rho
(p,) is the function that performs right multiplication by a. For a finite group G,
Theorem 6.2.4(a) says that for any row (specified by a € G) in the Cayley table, the
row is a permutation of the list of elements in G in the order presented by the table’s
row heading. The same is true for every column in the table. Therefore,
if G is a finite group, then every element of G occurs exactly once in every row
and exactly once in every column of the Cayley table.
The converse of this statement is false. It is possible to have each element
occur exactly once in every row and once in every column of an operation table for
a structure that is not a group. See Exercise 4.
We use the standard notation for powers of elements in a group. If G is a group
with identity e and a € G, then
a® =e.
a”! — q"a for alln EN.
if m is a negative integer, a” = (a~!)—™.
For all integers m and n, these familiar laws of exponents hol in a group:
aq!
=
qntn
(a’"y"
_
ghee
and
(a)! = a™ forn > 0.
When the group operation is +, we use different words for these concepts. For
gd, b GG),
the operation is called addition, and a + b is the sum of a and b.
the additive inverse of a is —a, called the negative of a.
the difference of a and b is a — b, which is an abbreviation for a + (—b).
for n € N, the nth multiple of a, denoted na, is a + a + --- + a (n times).
forn € N, (—n)a = —(na).
6.2
Groups
317
Thus, for example, 4a is a ++ a + a + a and —2b is —(2b), the negative of 2b. The
expression 4a — 2b is an abbreviation fora +a+a+a+(—b)
+ (—b).
For every natural number ¢, the set ¢Z of all integer multiples of t is an additive
group. (See Exercise 1(e).) For example, the group (5Z, +) contains the elements
(—4)5 = —20 and (7)5 = 35. Notice that —4 and 7 are natural numbers and not elements of 5Z. Their sum is (—4)5 + (7)5 = (—4 + 7)5 = (3)5 =15, and the negative
of 30 = (6)5 is (—6)5 = —30.
We close this section with some important observations about the axiomatic
approach used to define a group. First, a small set of axioms is advantageous,
although challenging to produce, because a small set means that fewer properties
need to be checked to be sure a given structure satisfies the axioms. The definition
of a group uses just three axioms.
Second, it may be best to leave a desired property out of the axioms if it can
be deduced from the remaining axioms. Theorem 6.1.1 tells us that the identity and
inverses of elements in a group are unique. In the definition of a group, we could
have said “there is a unique identity element e in G for © .” Verifying this property
requires showing the existence and uniqueness of the identity rather than only the
existence. We stated the definition of a group as we did to make it easier to verify
that a structure is a group.
Finally, the fact that axioms may be altered by adding or deleting specific axioms does not mean that the axioms are chosen at random or that all of the axioms
are equally worthy of study. The group axioms are chosen because the structures
they describe are so important to mathematics and its applications. Treating an
additional property, such as commutativity, as a property that holds for many but
not all groups enables us to keep the basic axioms for a group minimal.
Exercises 6.2
1.
Show that each of the following algebraic structures is a group. Which groups
are abelian?
(a) ({1,—1},-), where - is integer multiplication.
(b)
*
(ce)
(d)
Hai wn sme nge ecrdae
ah Ege ne, and - is complex
pos Me
2
2
number multiplication.
({1, -1,i, —i}, -), where - is complex number multiplication.
(P(X), A), where X is a nonempty set and A is the symmetric difference operation
(e)
(f)
2.
A A B = (A — B) U(B — A).
(tZ, +), where ¢ is a natural number.
The set of 2 x 2 real matrices with determinant 1, where the operation
is matrix multiplication.
Given that G =
{e, u, v, w} is a group of order 4 with identity e, u? = v, and
y? = e, construct the operation table for G.
3.
Given
that G =
{e,u,v,w}
is a group
of order 4 with identity e and
u2 = y* = w? = e, construct the operation table for G.
318
CHAPTER 6
Concepts of Algebra
4.
Give an example of an algebraic system (G, 0) that is not a group such that in
the operation table for o, every element of G appears exactly once in every row
and once in every column. This can be done with as few as three elements in G.
Construct the operation table for S. Is S, abelian?
(a)
What is the order of S,, the symmetric group on four elements?
(b)
Compute these products in S,: [124 3][42 13], [432 1][432
1],
and [2 1 4 3][1 3 2 4].
Compute these products in S,: [3 12 4][32 14], [432 1][3 124],
and [1 43 2][1 43 2].
Find the inverses of [1 3 4 2], [4 1 2 3], and [2 1 4 3].
Show that S, is not abelian.
(c)
(d)
(e)
Let G be a group and a; € G foralln Ee N.
(b)
State and prove a result similar to part (a) for n elements of G, for all n € N.
Prove part (d) of Theorem 6.2.3. That is, prove that if G is a group, a, b, and
c are elements of G, and ca = cb, thena = b.
Prove part (b) of Theorem 6.2.4.
Let G be a group. Prove that if a* = e for all a € G, then G is abelian.
Give an example of an algebraic structure of order 4 that has both right and
left cancellation but that is not a group.
12.
Let G be a group. Prove that
(a)
(b)
13°
Gis abelian if and only if ab? = (aby for all a,b € G.
€ N anda, beG.
Gis abelian if and only if a"b” = (ab)" for alln
Show that the structure (R — {1}, ©), with operation ° defined by ao b =
a + b — ab, is an abelian group. You should first show that (R — {1}, ©)
is an algebraic structure.
(a)
In the group G of Exercise 2, find x such that v © x = e; x such that
(b)
Ona — we SUCH Nath
lOpu—ay aang oa Such tiation
Let (G, *) be a group and a, b € G. Show that there exist unique ele-
ments
x and y in G such that a * x = bandy*«a = b.
Show that (Z, #), with operation # defined by
Find x such that 50 # x = 100.
a#b = a + b + 1, is a group.
Let m be a prime natural number and a € (U m? -). Prove that a = a7! if and
only ifa=lora=m-—l.
Galois discovered a connection between certain groups and the solutions to
polynomial equations. Refer to the finite sets in Exercises I(a), (b), and (c).
Each of these sets forms a group (although these groups do not represent the
general case of Galois’ work). Find a polynomial equation with integer coefficients such that
(a)
(b)
(c)
the equation has degree 2 and {1, —1} is the solution set.
the equation has degree 3 and { 1, a, B} is the solution set.
the equation has degree 4 and {1, —1, i, —i} is the solution set.
6.2
Proofs to Grade
18.
Groups
319
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
(a) Claim. If G is a group with identity e, then G is abelian.
“Proof.”
Let a and b be elements of G. Then
ab = aeb
= a(ab)(ab)~'b
a(ab)(b~!a7'!)b
=1(aa)(Dbe
jaa D
= (aa)a'b
= (aa)(b"'a)"!
= a(a(b~
a) *)
= a((b~!a)“a)
= (((a-D)a)
= (aa~')(ba)
= e(ba)
=f 0
x
(b)
Therefore, ab = ba and Gis abelian.
5
Claim.
If G is a group with elements x, y, and z and if xz = yz, then
ay
“Proof.” If Z = é€, then xz = yz implies thatwe —veusOn— sy ale
ze,
then the inverse of z exists, and xz = yz implies & =
x = y. Hence, in all cases, if xz = yz, then x = y.
.
= and
“og
Anke)
Claim.
The set Q* of positive rationals with the operation of multiplication is a group.
‘Proof.’
The product of two positive rationals is a positive rational,
so Q* is closed under multiplication. Because 1-r=r=r-1 for
every r © Q, | is the identity. The inverse of the positive rational ; iS
the positive rational 2.The rationals are associative under multiplication
because the reals are associative under multiplication.
i=]
(d)
Claim.
If m is prime, then (Z,, — {0}, -) has no divisors of zero.
“Proof.”
Suppose that a is a divisor of zero in (Z,, — {O},-).
Then a #0, and there exists b #0 in Z,, such that ab = 0. Then
ab = m(mod m), so ab = m. This contradicts the assumption that m is
prime.
(e)
Claim.
B
If m is prime, then (Z,, —
{0}, -) has no divisors of zero.
“Proof.” Suppose that a is a divisor of zero in (Z,, — {0},-).
Then a #0, and there exists b #0 in Z,, such that ab = 0. Then
ab = 0 (mod m), so m divides ab. Because m is prime, m divides a, or
m divides b. But because a and b are elements of Z,,, both are less than
m. This is impossible.
a
320
CHAPTER 6
Concepts of Algebra
(f)
Claim.
For every natural number m, (Z,, — {0}, -) is a group.
We know that (Z, -) is associative with identity element 1.
“Proof.’
Therefore,
(Z,, — {0}, -)is associative with identity element 1. Itremains
to show every element has an inverse. For x € Z,, — {0}, x #0. Therefore, 1/x € Z,, — {0} and x - 1/x = x(1/x) = 1. Therefore, every ele-
ment of Z,, — {0} has an inverse.
(g)
Claim.
If (Z,, —
|
{0}, -) is a group, then m is prime.
“Proof.”
Assume that (Z,, — {0}, -) is a group. Suppose that m is
not prime. Let m = rs, where r and s are integers greater than | and
less than m. Then r- s = m = O (mod m). Because r has an inverse f in
Z-= (Ores), hens = 1s
= (ft 7) os
2 (7 os) =
Oe 0:
That is, s = 0 (mod m). This is impossible because 1 < s < m.
|
6.3
Subgroups
A subgroup of a group is a subset that forms a group when the operation is restricted
to the subset. To fully understand a group, we must know its subgroups and their
properties. In this section, we give a simple test for checking whether a subset
forms a subgroup and look at examples of subgroups, including the special (cyclic)
subgroups that arise from a single element of the group.
DEFINITION
Let (G, o) be a group and Ha
subset of G. Then (H, o) is a
subgroup of G if (H, o) is a group.
Examples.
The operation of integer addition, when restricted to the set E of even
integers, forms an algebraic system because E is closed under addition. (The sum
of two even integers is an even integer.) The system (E, +b) is associative because
associativity holds for all integers; the integer 0 in E works as the identity for (E, +)
because it acts as the identity for (Z, +); and every element x of E has its additive
inverse in E because —x is also even. We have verified that (FE, +) is a group, so
(E, +) is a subgroup of (Z, +).
We note that E is the set of multiples of 2. In general, for every integer t, the
set tZ of all multiples of t is a subgroup of (Z, +). However, the set of odd integers
is not closed under addition and therefore cannot be a subgroup of (Z, +).
o
Example.
The group (G, o) with G = {e, a, b, c} and the Cayley table shown
below has identity e. That portion of the Cayley table corresponding to the subset
J = {e, b} is displayed at the right. J is closed under 0 because ee = e, eb = be = b,
and bb = e. The system (J, 0) is associative and has an identity e, and every element
of J is its own inverse. Therefore, (J, 0) is a subgroup of G of order 2.
6.3
€
a
b
Cc
Subgroups
| e
b
e
le
a
b
Cc
e
e
b
a
a
e
Cc
b
b
b
e
b
b
Cc
€
a
Cc
1G
b
a
e
321
Similar reasoning shows that the set K = {e, a} is another subgroup of order 2.
If e is the identity for a group (G, 0), then ({e}, 0), is a subgroup called the
identity (or trivial) subgroup of G. Also, (G, ©) is a subgroup of itself. All subgroups of G other than G are called proper subgroups.
So far, every subgroup we have discussed has had the same identity as the
main group, and every element in the subgroup has had the same inverse it had in
the main group. This must always be true.
Theorem 6.3.1
Let H be a subgroup of G. Then
(a) _ the identity of H is the identity e of G.
(b)
if x EH, the inverse of x in H is its inverse in G.
Proof.
(a)
_Ifiis the identity element of H, then ii = i. But in G, ie = i, so ii = ie and,
by cancellation, i = e.
(b)
See Exercise 3.
|
When we use the definition of subgroup to check whether a subset of a group is
a subgroup, we need to check whether the subset is closed under the operation and
whether it satisfies all the group axioms. The next theorem makes it easier to prove
that a subset of a group is a subgroup. Although the theorem is a biconditional statement, the important result is that only two properties must be checked to show that
H is a subgroup of G. The first is that H is nonempty. This is usually done by showing that the identity e of G is in H. The other is to show that ab~' € H whenever a
and b are in H. This is usually less work than showing both that H is closed under
the group operation and that b € H implies b~! € H.
Theorem 6.3.2
Let G be a group. A subset H of G is a group if and only if H is nonempty and for
alla, b € H, ab= € H.
Proof.
First, suppose that H is a subgroup of G. Then H is a group, so by
Theorem 6.3.1, H contains the identity e. Therefore,
H # ©. Also, if a and b are in
H, then b~! € H (by the inverse property) and ab~! € H (by the closure property).
Now suppose that H is nonempty and for all a, b € H, ab! € H. (We will show
H satisfies the group axioms and is closed. It is best to proceed in the order that
follows.)
322
CHAPTER 6
Concepts of Algebra
The associative property holds for all elements of G and therefore for all
elements of H.
(ii) Because H is nonempty, there is some a € H. Then aa“! = e € H.
(iii) |Letxbe in H. Then by (ii), e € H. Therefore, by our hypothesis, x! = ex EH.
(iv) To show that H is closed, suppose that.x, y € H. Then by part (iii). y~! € H.
(i)
Then x and y~! are in H, so by our hypothesis, x(y~!)~! = xy € H.
Example.
r
The symmetric group S; with six elements has six subgroups, two of
which are the trivial subgroup {[1 2 3]} and S, itself. Let
J = {[1 2 3], [2 1 3]}.
Then J contains the identity element [1 2 3] of S,. By computing
12 Sie We
2
es [23 |
tesla
etal? Sie Ls 12.1.3]
Prilps \deeO lnc (Ales leet
ales |
Dee Wel leer. ales I Dl eS|| L235]
ee
we see that ab~! € J for all a, b € J. Therefore, J is a subgroup of S;. Similar computations show that K = {[1 2 3], [3 2 1]} and Z = {[1 2 3], [1 3 2]} are also subgroups of $3. The only subgroup of S, of order 3 is M = {[1 2 3], [2 3 1], [3 1 2]}.
Oo
Example. The group (Z,5, +) has five subgroups: {0}, Z,,, H = {0, 6}, J =
{0, 3, 6, 9}, and K = {0, 2, 4, 6, 8, 10}. Remember that when the operation is
addition, inverses are negatives, so to use Theorem 6.3.2, we need to calculate sums
of the form a + (—b) rather than products of the form ab~!. To show that K is a
subgroup of Z,5, we could calculate each of 67 = 36 sums, including these three:
4+(-6)=4+6=10
eye
0 ee
6+(—4)=64+8=14=2.
A quicker method is to note that all the elements of K are multiples of 2, so all their
inverses and all sums of the form a + (—b) are multiples of Z. By either method, we
find that a + (—b) € K for every pair a, b in K. Therefore, K is a subgroup of Z,5.
Oo
If a is an element of the group G, then all powers of a are in G by the closure
property. Furthermore, {a":n € Z}, the set of all powers of a, forms a subgroup of G.
Theorem 6.3.3
If Gis a group and a € G, then {a": n € Z}, is an abelian subgroup of G.
Proof. Because a! = a is a power of a, {a":n € Z}, is nonempty. Suppose that
x =a
and y = a" for some integers mand n. Then xy~! = a’"(a")"! = a"q~" = q"—"
is a power of a. Therefore, by Theorem 6.3.2, {a": n € Z} is a subgroup of G. The
subgroup is abelian because aa" = a™ = q”” = q"q".
a
6.3
Subgroups
DEFINITIONS
Let G be a group and a € G. Then (a) = {a":n€Z}
called the cyclic subgroup generated by a.
323
is
The order of the element a is the order of the subgroup (a). If (a) is an
infinite set, we say a has infinite order.
Examples.
The group (U,, -) has six elements: 1, 2, 3, 4, 5, 6. Computing powers
of 2 modulo 7, we have 2° =1, 2! = 2,2? =4, and 23 =8 =1 again. It follows that
all positive and negative powers of 2 in U, are equal to 1, 2, or 4, so (2) = {1, 2, 4}
and the order of the element 2 is 3.
The nonnegative powers of 3 are 3° =1, 3' = 3, 3? = 9 = 2, 3? = 313? =
3(2) = 6, 34 = 3°37 = 2(2) = 4, and 3° = 3°32 = 6(2) = 12 = 5. Because 3° = 1,
the powers of 3 are, in order, 1, 3, 2, 6, 4, 5. Every element of U, is in the cyclic
subgroup generated by 3. The order of 3 is 6 because (3) has six elements.
The order of | is 1 because the cyclic subgroup generated by 1 is {1}. Also, 4
has order 4, 5 has order 6, and 6 has order 2.
oO
Examples.
In the additive system (Z,,, +), we use multiples instead of powers, so
to find the order of the element 3, we compute 1(3) = 3, 2(3)
3)
+3 = 9,and 43) = 3 +) 3455 4
=3 +3 = 6, 33) =
= 12 —.07 It tollows thatallumultiplesies
3 are 0, 3, 6, or 9, so the order of 3 is 4. The multiples of 5 (listed in order) are 5,
10)3;,5,.1, 6,110,497 2. 7, and 0, so. (5) = Z,, and the order Of suse)
Examples.
subgroup
For the group (R —
(2) =
{2"7sneEZ}
=
42
{0}, -), the element 2 generates the infinite cyclic
(hem zedyala eek S24.
ect Thus
2 has
infinite order in this group. We note that (5) is the same subgroup as (2).
The element —1 generates the cyclic subgroup {—1, 1}. Thus, the order of —1
in the subgroup { —1, 1} 1s 2.
Oo
DEFINITIONS
Let G be a group. If there is an element a € G such that
(a) = G, then we say G is a cyclic group. Any element a of G such that
(a) = Gis called a generator for G.
For the group U,, we found that (3) = U, and that (5) = U,. Thus the elements
3 and 5 are generators for U,. These are the only generators because no other ele-
ment of U, has order 6.
For m > 1, every group (Z,,, +) is cyclic with generators | and m — I. See
Exercise 14. The cyclic group (Z,, +) has only two generators, 1 and 3. This is
because the multiples of 1 and 3 are
Nea
20)=1+1=2
and
ies
20) ssa
oo —
324
CHAPTER 6
Concepts of Algebra
See
ACG)
alee
3G) Se
She eid)
43)=1+3=0.
The element 2 does not generate Z,; it has order 2 and generates the subgroup
1 One
Example.
The group S; is not cyclic because none of its elements generates the
entire group. For example, the cyclic subgroup generated by [3 1 2] is { [3 1 2],
[2 3 1], [1 2 3] }. Similar calculations show that none of the other elements of S,
generates S,. See Exercise 9(a). A simple way to conclude that $3 is not cyclic is to
note that every cyclic group is abelian and 5; is not abelian.
Theorem 6.3.4
Let G be a group and a be an element of G with order r. Then r is the smallest posi-
tive integer such that a” = e, the identity, and (a) = {e, a,a’,...,a"~'}.
Proof.
Because the order of a is finite, the powers of a are not all distinct. Suppose
a” = a” with 0 < m < n. Then a”
= e with n — m > O. Therefore, the set of
positive integers j such that a/ = e is nonempty. Let k be the smallest such integer.
(This k exists by the Well-Ordering Principle.) We prove that k = r by showing
that the elements of (a) are exactly a® = e, a', a’,..., a&~!.
First, we show that the elements e, a!, a*,..., a! are distinct. If a’ = a’ with
O<s<t<k,thena’*=eand0 <t-—=s
< k, contradicting the definition of k.
Second, we show that every element of (a) is one of e, a‘, a?,..., a‘—!. Consider a’ for t € Z. By the Division Algorithm, t = mk + s with 0 < s < k. Thus,
a = a™ts —
gms = (a)"a' = ea’ = ea’ = a‘, sothata’ = a withO <5 < k.
We have shown that the elements a* for 0 < s < k are all distinct and that
every power of a is equal to one of these. Because (a) has exactly r elements, r = k
An diaa—rex
P
If a € Ghas infinite order, then all the powers of a are distinct and
Exercises 6.3
1.
Find all subgroups of
x
(a)
(c)
(Zoe),
(Zs, +).
(€)s
Ua*) pwith St
a; bse d,enf}
and the table for « shown at the
right.
(b)
(d)
(Uj), -).
(U,,-).
*
Re
b
C
d
e
f
a
Dae
D
c
G
died
e
e
a Al ae
b
a
e
i
Cc
d
e
if
a
e
d
b
d
e
d
b
G
i
a
ia
a
e
if
a
b
G
6
d
b
a
e
6.3
Subgroups
325
In the group S,,
(a) find two different subgroups that have three elements.
(b) find two different subgroups that have four elements.
(c)
[2314][3 124] = [1234]. Is there a subgroup of S, that contains
[2 3 1 4] but not [3 1 2 4]? Explain.
(d)
(e)
find the smallest subgroup that contains [4 2 1 3] and [3 2 4 1].
find the smallest subgroup that contains [2 3 1 4] and [3 4 2 1].
Prove that if G is a group and H is a subgroup of G, then the inverse of an
element x € H is the same as its inverse in G (Theorem 6.3.1(b)).
(a)
(b)
Prove that if H and K are subgroups of a group G, then H 1 K is a subgroup of G.
Prove that if {H,: a € A} is a family of subgroups of a group G, then
0) H, is a subgroup of G.
acd
(c)
Give an example of a group G and subgroups H and K of G such that
H U K is not a subgroup of G.
Let G be a group and H be a subgroup of G. If H is abelian, must G be abelian?
Explain.
Prove or disprove: Every abelian group is cyclic.
Let G be a group. If H is a subgroup of G and K is a subgroup of H, prove that
K is a subgroup of G.
Find the order of the element 3 in each group.
(a)
(e)
(Zy,+)—
(db) (Zs, +)
(Zone) ee (Lea)
(©)
(g)
(Ze)
(;,°)
(Za)
(hy) (Uj), °)
Find the order of each element of the group
(A)
953.
(b)
(Z,, +).
(c)
(Ze, aa):
(d)
(U;;, -).
10.
List all generators of each cyclic group in Exercise 9.
11.
Let G be a group with identity e.
(a) Yet a] GG. Prove that the set C7 — {xe G: xa — ax}, called: the
centralizer of a in G, is a subgroup of G.
(b) Let C = {x &€G-: forall y € G, xy = yx}. Prove that C, the center of G,
is a subgroup of G.
(c)
12.
Let aeéG. Prove that the center of G is a subgroup of the centralizer of
ainG.
Let G be a group, and let H be a subgroup of G. Let a be a fixed element of G.
Prove that K = {a~'ha: h € H} is a subgroup of G.
13:
Let (C — {0}, -) be the group of nonzero complex numbers with complex
number multiplication. Leta =
(a)
(b)
1+iv3
5
Find (a).
Find a generator of (a) other than a.
326
CHAPTER 6
Concepts of Algebra
14.
Prove that for every natural number m greater than |, the group (Z,,. +) is
cyclic with generators | and m — 1.
15.
Prove that every subgroup of a cyclic group is cyclic.
16.
Let G = (a) be acyclic group of order 30.
*
Proofs to Grade
17.
*
(a)
(c)
What is the order of a°?
List all elements of order 3.
(b)
(d)
List all elements of order 2.
List all elements of order 10.
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
(a)
Claim.
If H and K are subgroups of a group G, then HK
is a
subgroup of G.
“Proof.”
Let a,b & HK. Then a,b €H and a,b eK. Because H
and K are subgroups, ab~! € H and ab™! € K. Therefore, ab“!
(b)
Claim.
If H is a subgroup of a group G and x €H,
{xh: h € H} is a subgroup of G.
“Proof.”
First; the identity
ee H. Thus,
x = xe
© xH.
© HO K.
=
then xH =
Therefore,
xH # ©. Second, let a,b © xH. Then a = xh and b = xk for some
h,ke&H.
Then
we
have
ab-! = (xh)(xk)~! = Gh)k-YGW) =
x(hk~! x7!) © xH. Therefore, xH is a subgroup of G.
=
6.4
Operation Preserving Maps
One of the most important concepts in algebra involves functions that relate the
calculations in one system (A, ©) to the calculations in another system (B, *). For
example, the logarithm function log: (0, 00) — R has the property that log(x - y) =
log x + log y. Through this equation the logarithm function creates a correspondence between the multiplication operation in the system ((0, 00), -) and the addition
operation in (IR, +). The log function also matches up the identity elements: The
identity in ((0, 00), -) is 1, and log 1 = 0, the identity in (IR, +). The log function
matches up inverses, too, because log (x~!) = —log x.
When a function f: A > B aligns the algebraic structure of B with that of A, we
say that f“preserves” the structure of A.
DEFINITION
Let (A, o) and (B, *) be algebraic systems and
f be a function from A to B. Then fis operation preserving (OP) if for all x, y € A,
f(xy) =f) * f(y).
The equation f(x © y) = f(x) * f(y) is a statement about elements in B. Calculating each of f(x ° y) and f(x) * f(y) requires two steps: performing an operation
and applying the function. If f is an OP map, the equation f(x ° y) = f(x) * f(y)
6.4
Operation Preserving Maps
327
means that the result is the same whether the operation x © y in A is performed first
or the images f(x) and f(y) in B are determined first. Another way to say a mapping
is Operation preserving is “the image of the product is the product of the images.”
Example.
Let
fbe the function from R to R given by f(x) = x”. Then fis an OP
map from (R, -) to (R, -) because f(x - y) = (x- y)? = x*- y* = f(x) - fy) for all
x, y € R. The function
fis not an OP map from (R, +) to (R, +) because (x + y)* =
x* + y* is not true for all real numbers.
Examples.
Let (%, +) be the set of all polynomial functions with real coef-
ficients, and let + be
staan s polynomial addition. Let D be the o
ferentiation mapping D: (#, +) > (#, +), where for each fe ¥, D(f)=
the first derivative of the Nee eb eat The function D is an OP map because Ge
know that
d
Df
+ 2) hee fas
2)
dx
df
dg
Shs lars, «= DU)
+ DG).
dx
dx
The differentiation map D: (#, -) > (&, -) is not operation preserving because
the derivative of a product is not always equal to the product of the derivatives. oO
Example.
Let the operation o on R x R be defined by setting (a, b) ° (c,d)
=
(a+ c,b +d), and let - be the usual multiplication on R. Then the function
f:(R x R, 0) > (R, -) given by f(a, b) = 243° is operation preserving.
Proof.
Let (x, y) and (u, v) be elements of R x R. Then
FG, y) © (U, v)) = fe + uy y + v) = 297% 397" = (2%3")(23") = fix, y) - flu, v).
Therefore, fis operation preserving.
a
The next theorem explains in more detail what we mean by saying that an OP
map f: (A, °) > (B, *) preserves the structure of (A, ©).
Theorem 6.4.1
Let
f be an OP map from (A, co) to (B, *).
(a)
(b)
(c)
(d)
(e)
(Rng (f), *) is an algebraic system.
If fis onto B and o is associative on A, then * is associative on B.
If fis onto B and o is commutative on A, then « is commutative on B.
Proof.
We prove parts (a) and (c) and leave the rest as Exercise 8. Let
If fis onto B and e is the identity for A, then f(e) is the identity for B.
If x has an inverse in A, then f(x) has an inverse in B and Go) 9= fx).
fbe an OP
map from (A, o) to (B, *)
(a)
First, note that because A is nonempty, Rng (f) is nonempty. Assume that
u, v € Rng (f). Then there exist elements x and y of A such that f(x) = u
328
CHAPTER 6
Concepts of Algebra
and f(y) = v. Then ux v = f(x) * f(y) = f(x° y), so u * v is the image of
x0 y, which is in A. Therefore, u * v € Rng (f). Thus Rng (f) is closed
under the operation *.
(c)
Assume that fis OP, fis onto B, and
© is commutative on A. Let u and v be
elements of B. Then there are x and y in A such that u = f(x) and v = f(y).
Then
uxv = f(x) *f(y) =f@°y) =fly°x) = f(y) *f@) = v * u.
Therefore,
u*v=v*
wu.
|
Example. For each natural number m, the canonical map is an important operation preserving map from (Z, +) onto (Z,,, +). We define H: Z — Z,, by setting
H(x) = x, the equivalence class determined by x. The canonical map is operation
preserving because, as we have seen,
foralla,be Z,H(a+b)=a+b=a+b=H(a)+
Hb).
Oo
We note that operation preserving mappings need not be limited to an algebraic structure with a single operation. Because
H(ab) = ab=a-b=H(a)
- Hb),
the canonical map H also preserves multiplication. Thus, H: (Z, +, -) > (Z,,, +, :)
is OP for both multiplication and addition.
Special terminology is used for operation preserving mappings where the algebraic structures involved are groups.
DEFINITIONS
Let
(G,0)
and
(H,*)
be groups.
An
OP
mapping
h: (G, o) + (H, *) is called a homomorphism from (G, 0) to (H, *). The
range of h is called the homomorphic image of (G, ©) under h.
j
The function f: (Z, +) — (SZ, +) where 5Z is the set of all integer multiples
of 5 and f(x) = 5x, is an example of a homomorphism. To verify this, we note
that for all x,y eZ, fx + y) = 5 + y) = 5x + S5y = f(x) + f(y). The homomorphism f maps onto 5Z because for each w € SZ, w = 5k for some integer k,
and therefore f(k) = 5k = w. Thus, (SZ, +) is the homomorphic image of (Z, +)
under f.
Example.
Define g: (Z,, +) > (Z3, +) by g(x) = [2x]. For all x, ye LZ65
g(x + y) = [2% + y)] = [2x] + [2y] = go) +26).
Therefore, g is a homomorphism from (Z,, +) to (Z3, +).
o
6.4
Theorem 6.4.2
Operation Preserving Maps
329
Let (G, -) and (K, *) be groups. If f: (G, -) > (K, *) is a homomorphism, then
(Rng (f), *) is a subgroup of (K, *). Furthermore, if the operation - is commuta-
tive, then * is a commutative operation on Rng (f). In other words,
(a)
(b)
the homomorphic image of a group is a group, and
the homomorphic image of an abelian group is an abelian group.
Proof. This theorem isa restatement of previous results using our new terminology.
The image of a group under a homomorphism is an algebraic system by Theorem
6.4.1(a). The homomorphic image is associative, has an identity element,-and every
element has an inverse by Theorem 6.4.1(b), (d), and (e). Finally, if a group is
abelian, then its image is abelian (Theorem 6.4.1(c)).
a
Example. Let (G, o) be a group with identity e. The mapping f: G > G given
by f(x) =e for all x€G is a homomorphism, called the trivial homomorphism. We can verify this by observing that for x, ye G, both f(x ° y) = e and
S(x) ° f(y) = e°e =e. Therefore, f(x ° y) = f(x) © f(y). In this case, the homomorphic image of (G, ©) is ({e}, ©).
Oo
Example. The function h: (R, +) ~ (R — {0}, -) given by h(x) = 2* is a homomorphism because
hee y) = 222 = 2D)
enya
By Theorem 6.4.2, (Rng (), -) must be a subgroup of (R — {0}, -). We verify this by noting that the range of h is (0, c0), which is indeed a subgroup of
(R — {0}, -).
DEFINITIONS
Let (G,o) and (K,*) be groups. A homomorphism
h: (G, 0) — (K, *) that is one-to-one and onto K is called an isomorphism.
If h is an isomorphism, we say (G, o) and (K, *) are isomorphic.
The word isomorphic comes from the Greek words isos (equal) and morphe
(form), literally meaning “equal form” because two isomorphic groups will differ only
in the names or nature of their elements. All their algebraic properties are identical.
Theorem 6.4.3
The relation of being isomorphic is an equivalence relation on the class of all
groups.
Proof.
See Exercise 18.
=
The three groups (Z5, +), ({1,—1},-), and ({W,A}, A) of order 2 are
isomorphic, where A is a nonempty set and A is the symmetric difference
330
CHAPTER 6
Concepts of Algebra
operation defined by XA Y = (X — Y) U(Y — X). The Cayley tables for the three
groups
are
Te
Mee
a
i
i
i
sieeO
;
aki
VANE
were
1
=e)
jee
al
|
@|@
ie)
fe
A
In fact, any two groups of order 2 are isomorphic (see Exercise 9(a)).
Example. The Cayley tables for the groups (Z,, +) and (Us, -) are shown in Figure
6.4.1. The elements of Us are listed in the order 1, 2, 4, 3 to make it easier to see that
the two groups have the same structure. The function h: (Z4, +) > (Us, -), given by
h(O) = 1, h(1) = 2, h(2) = 4, and h(3) = 3, is an isomorphism.
FeAl,
Oma]
BeOS
0
I
2
0
|
2
2
3
2)
3
0
Sele
oe
Oe
lene
3
0
live
De
Ae
Ss
2
a
it
2
-
2
4
3
4
3
3
1
2
Sa
hs a
(Z4, ol)
|
(U,, -)
Figure 6.4.1
We have seen examples of groups whose elements are permutations of a set.
Groups whose elements are some (but not necessarily all) permutations of a set
are called permutation groups. The importance of permutation groups is that for
every group of elements of any kind (numbers, sets, functions,...), there is a permutation group with the same structure. This result is due to Arthur Cayley.
Theorem 6.4.4
Cayley’s Theorem
Every group G is isomorphic to a permutation group.
Proof.
|
We choose the set G itself to be the set of objects to be permuted. By
Theorem 6.2.4(a), for every a in G, the function A,: G >
G, where X,(x) = ax
for each x € G, is a permutation of G. Let #€ = {A,: a € G}. By Theorem 6.2.2,
the set
of all permutations of the elements of G is a group with the operation
of function composition. We claim that 9€ is the image of a one-to-one homomorphism from G to G and conclude that G is isomorphic to the permutation
group 4.
Let f: G > Gbe given by f(a) = A,. Let a, b € G. Let x € G. Then
f(ab)(x) = Xa,(x) = (ab)x = a(bx) and
(fla) 2 f{B)VX) = (Ago Ap(X) = AQ(Az (x) = a(bx).
6.4
Operation Preserving Maps
331
Therefore, f(ab)(x) = (fla) © ftb))(x) for all x € G. Thus, f(ab) = f(a) © f(b) for
all a, b € G. Therefore,
fis a homomorphism.
To show that fis one-to-one, suppose that f(a) = f(b). Then X,, = A,. Therefore,
ax = bx for every x € G. In particular, if e is the identity for G, ae = be, soa = b.
By definition, every permutation in 9 is A, for some a € G. Therefore, fmaps
onto df.
We have shown that d€ is the homomorphic image of G, so by Theorem 6.4.2,
d€ is a group. Because
f isone-to-one,
fis an isomorphism. Therefore, G and 9€ are
isomorphic groups.
a
Example. Let (G, -) be the group {1, —1, i, —i} of four complex numbers, where is the usual multiplication of complex numbers. The corresponding group 9 of
permutations, as described above, consists of the four left multiplications by the
elements of G. For example, A; is the mapping that multiplies each element of G
on the left by 7:
A) =i-1 =i,
Aj)
=
NA
Ke)
hie eta
= Tea
(= 1)
7,
Thus, A; = [i —i —1 1]. The other three permutations are 4, = [1 —1 7 —i],
A_, =[-1 1 -i i], and A_; = [—i i 1 —1). The tables for G and 3€ show that
they have the same structure.
(G2
1
=
(4, ©)
i
all
1
==
al
1
=
I
aaa
th
I
need
1
i
=
l
oat
°
r,
A_|
A;
A_;
i
ei
Ay
AL,
A;
r_;
Ad,
NSE
Al
A
N=
Ay
ee
A_;
A_|
A,
r_;
A;
1
1
=
A_;
d;
Ay
Ny
Finally, we note that G (and therefore 9) is isomorphic to (Zy, +). (See
Exercise 16.) This fact suggests the possibility that all groups of order 4 are
isomorphic to (Z4, +). See Exercise 9(c) for a counterexample to this conjecture.
Exercises 6.4
1.
Define f R+ > R* by fix) = Vx where R°* is the set of all positive real
numbers.
(a)
(b)
Is f: (R*, +) > (R*, +) operation preserving?
Is f: (R*, -) > (R*, -) operation preserving?
332
CHAPTER 6
Concepts of Algebra
2.
Definef: R > R by fi) = x°.
(a)
(b)
Isf: (R, +) > (R, +) operation preserving?
Is/f: (R, -) > (R, -) operation preserving?
Define @ on R x R by setting (a, b) ® (c, d) = (ac — bd, ad + be).
(a) Show that (R x R, ®) is an algebraic system.
(b) Show that the function h from the system (C, -) to (R x R, @) given
by h(a + bi) = (a, b) is a one-to-one function from the set of complex
numbers that is onto IR x R and is operation preserving.
Let # be the set of all real-valued integrable functions defined on the interval
[a, b]. Then (#, +) is an algebraic structure, where + is the addition of
functions. Define J: (#, +) > (R, +) by (f) =
2 fdx. Use your knowledge of calculus to verify that J is an OP map.
Letf: (A, -) > (B, *) and g: (B, *) > (C, X) be OP maps.
(a) Prove that g ° f is an OP map.
(b)
Prove that if f~! is a function, then f—! is an OP map.
Let WW be the set of all 2 x 2 matrices with real entries. Define Det:
by Det ¢
ce
b
@
4 > R
|
= ad — be.
(a)
Prove that Det: (Wl, -) > (IR, -) is operation preserving, where (J, -)
(b)
Prove that Det: (Wl, +) — (IR, +) is not operation preserving, where
(M, +) denotes Wl with matrix addition.
denotes MU with matrix multiplication.
Let Conj: C — C be the conjugate mapping for complex numbers given by
Conj (a + bi) =a — bi.
(a) Prove that Conj: (C, +) — (C, +) is operation preserving, where
(C, +) denotes the complex numbers with addition.
(b) Prove that Conj: (C, -) > (C, -) is operation preserving, where (C, -)
denotes the complex numbers with multiplication.
Prove the remaining parts of Theorem 6.4.1.
x (a)
10.
11.
(b)
Show that any two groups of order 2 are isomorphic.
Show that any two groups of order 3 are isomorphic.
/
(c)
Prove that there exist two groups of order 4 that are not isomorphic.
Let 3Z and 6Z be the sets of integer multiples of 3 and 6, respectively. Let f
be the function from (3Z, +) to (6Z, +) given by f(x) = 4x.
(a)
(b)
Prove that fis a homomorphism.
Prove that fis one-to-one.
(c)
What group is the homomorphic image of (3Z, +) under f?
Let (3Z, +) and (6Z, +) be the groups in Exercise
10, and let g be the
function from 3Z to 6Z given by g(x) = x + 3. Is g a homomorphism?
Explain.
6.4
12.
Operation Preserving Maps
333
Let ({a, b, c}, 0) be the group with the operation table shown here.
°
a
G
|
a
b
c
a
b
G
G
G
a
a
b
Verify
that
the
mapping
g:(Z,,+)— ({a,b,c},o)
defined
by
g(0) = 9(3) = a, g(1) = g(4) = b, and g(2) = g(5) = cisahomomorphism.
13. * (a)
* (b)
14.
15.
Prove that the function f: Z,, > Z5, given by f(x) = 4x is well defined
and is a homomorphism from (Z,., +) to (Z54, +).
Find Rng (f), and give the operation table for the subgroup Rng (f) of Z,.
Define f: Z,; > Z,, by fx) = 4x.
(a)
Prove that f is a well-defined function and a homomorphism
(b)
Find Rng (/), and give the operation table for this subgroup of Z,.
from
Let
(G,o) and (H,*) be groups, i be the identity element for H, and
h: (G, 0) — (H, *) be a homomorphism. The kernel of f is ker(f) =
{x € G: f(x) = i}. Thus, ker(/) is all the elements of G that map to the identity in H. Show that ker(f) is a subgroup of G.
16.
Show that (Z,, +) and ({1, —1, i, —i}, -) are isomorphic.
17.
Is S; isomorphic to (Z,, +)? Explain.
18.
Prove that the relation of isomorphism is an equivalence relation. That is,
prove that
(a) if (G, -) is a group, then (G, -) is isomorphic to (G, -).
(b) if (G, -) is isomorphic to (H, *), then (H, *) is isomorphic to (G, -).
(c) if (G, -) is isomorphic to (H, *) and (H, *) is isomorphic to (K, ®), then
(G, -) is isomorphic to (K, ®).
19.
Proofs to Grade
20.
Use the method of proof of Cayley’s Theorem to find a group of permutations
isomorphic to
(a) (Z3; -F).
(b)
(Zs, +).
(c)
SCR):
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
(a) Claim. Let © be the operation on
R x R defined
by setting
(a, b)° (c,d) = (a + c, b + d), and let — be the usual subtraction on
RR. Then the function f given by f(a, b) = a — 3b is an OP map from
(R x R, o) to (R, —).
SProot, gun ao pallid. (gunk) pareninghte
aie Then f((4; 2) > (3, 1) =
i} Gao)
Ware
2, Whereas, (4,2) — (3, 1) = —2 — 0 = —2,
so f is operation preserving.
B
334
CHAPTER 6
Concepts of Algebra
(b)
Claim.
Let f:(G, *) > (H,-) and g: (H,-) > (K, ®) be OP maps.
Then the composite g © f: (G, *) — (K, ®) is an OP map.
“Proof.” g © flab) = g(f(ab)) = g(f@fb)) = s(fas(f)) =
(g 0 flag ° f(b).
2
6.5
Rings and Fields
Thus far we have considered algebraic structures with exactly one binary operation,
and in this setting, we have explored the derivation of structural properties (such
as uniqueness of the identity element and cancellation) and the concepts of substructure and operation preservation. We have considered systems such as (N, +)
and (N, -) as distinct algebraic systems, ignoring any interaction between the two
operations. In this section, we extend our study of algebraic structures by investigating systems with two binary operations (addition and multiplication). We will
identify those algebraic properties of Z, Q, and R that distinguish them from the
natural numbers and from one another.
The two common properties of all these systems that describe the interaction
between addition and multiplication are the distributive laws. Thus, we begin by
making the distributive laws part of the definition for the algebraic structure called
a ring.
DEFINITION
A
ring (R,+,-)
is a set R together
with
two
binary
addition.
For all
operations + and - that satisfy the following axioms:
(1)
(R, +) is an abelian group. For all a, b,c € R,
(a) there is an identity element 0 € R.
(b)
for every ae R, there is an additive inverse —a € R.
(c) the operation + is associative.
(d)
the operation + is commutative.
/
(2) The operation - is associative.
(3) The multiplication
operation
is distributive
over
a,b,c ER,
a-(b+c)=
(a+ b)-c =
(a-b)+(a-c)
(a-c)+(b~-c).
and
The definition of a ring says a great deal about addition but requires only the
associative property for multiplication. As we did for groups, we often write ab
instead of a - b, and we write a — b for a + (—D).
6.5
Exampies.
Rings and Fields
335
The real number system (R, +, -) with addition and multiplication is
a ring. The systems (Z, +, -) and (Q, +, -) of integers and rational numbers also
form rings. If E is the set of even integers, then (E, +, -) is a ring.
The number system (N U {0}, +, -) is not a ring because (NU {0}, +) is not
a group—only the element 0 has an additive inverse.
Oo
Example.
Let .t be the set of all 2 x 2 matrices with real number entries. Then
(Wt, +, -) is a ring because matrix addition is associative and commutative, matrix
multiplication is associative, and multiplication distributes over addition. The additees
biaire:
,
| @
ite
Die
tive identity is the zero matrix F a and the additive inverse of k ‘|is the
G
Gand
matrix — wile
Example.
lis
=o
le
Fe
e
ee
:
al" Multiplication in this ring is not commutative.
—
—
Gq
Let & be the set of all functions from R to R. Then (%, +, -) is a ring.
(See:exercisem)
Theorem 6.5.1
For every
Oo
m EN, (Z,,, +, -) 1s a ring.
Proof.
We know by Theorem 6.1.2(a) that for every natural number m, (Z,,,, +)
is an abelian group and by Theorem 6.1.2(b) that the operation - is associative on
Z,,. We need only verify the distributive axioms to show that (Z,,, +, -) is a ring.
Let a, b, and c be integers. Then
a-(b+¢)=a
-(b+¢c)
(definition of addition in Z,,)
= a(b +c)
(definition of multiplication in Z,,)
= (ab
+ ac)
(distributivity of + and - in Z)
= ab + ac
(definition of addition in Z,,)
=a-b+a-c
(definition of multiplication in Z,,).
The proof of the other distributive axiom is Exercise 3. Therefore, (Z,,, +, -) is a
ring.
a
We next consider properties that are shared by every ring (R, +, -). As a first
step, we note that because (R, +) is an abelian group, properties that hold for every
abelian group hold for (R, +).
Theorem 6.5.2
Let (R, +, -) be aring, and a, b, c € R. Then
(a)
(b)
(c)
the additive identity (zero) of R is unique.
additive inverses (negatives) of elements of R are unique.
left and right cancellation hold in R.
(d) ee(=G) — Gand —(a. 4b) — (=a) +.(—)).
336
CHAPTER 6
Concepts of Algebra
(e)
forall integers m and n, m(a + b) = ma + mb, (m + n)a = ma + na, and
m(na) = (mn)a.
Proof. All the above are restatements of properties of abelian groups developed in
a
Section 6.2.
The operations in property (e) of Theorem 6.5.2 must be interpreted carefully. Addition in the expression m(a + b) = ma + mb involves the ring addition
operation + for both sides of the equation. The terms ma, mb, and m(a + b) do not
represent multiplication of ring elements but instead are expressions for multiples
of a, b, and a + b. The expression (m + n)a = ma + na, however, involves two
different addition operations. The term m + n adds two integers, but the + sign in
the term ma + mb means the (ring) addition of two ring elements. Likewise, in the
equation m(na) = (mn)a, mn refers to multiplication of integers, whereas m(na) and
(mn)a represent the ring sums na + na +---+ na (m times) anda +a-+---+a
(mn times), respectively.
The distributive axioms allow us to derive properties that relate multiplication
to the zero and negatives in a ring.
Theorem 6.5.3
Let (R, +, -) be a ring with zero element 0. Then, for all a, b,c € R,
(@)
(b)
(c)
(dq)
OR a—yae
0 =.10!
a-(—b)
= (—a)-b= —(a-b).
(—a)-(—d) =a-b.
a-(b—c)=a-b—a-cand(a—b)-c=a-c—b-c.
Proof.
(a)
(b)
(c)
(d)
LetaeR. (We use the fact that 0 is the additive identity in two different ways.)
Then (0-a) +0=0-a=(0+0)-a=(0-a)+(0-a). By the left cancellation property, 0 = 0 - a. The proof that a - 0 = 0 is Exercise 7(a).
Let a,beER. (To show that a-(—b) = —(a-b), we must show that
a-(—b) plays the role of the inverse of a-b.) Then a; (—b) + (a:b) =
a-(—b + b)=a-0=0. Because a-(—b) is a negative of (a-b) and
negatives are unique, a - (—b) = —(a- b).
The proof that (—a) - b = —(a- b) is Exercise 7(b).
Let a,beR. Then —a and b are in R, so by part (b), (—a)- (—b) =
—[(—a) - b] = —[-—(a- b)]. By Theorem 6.5.2(d), —[—(a- b)] = a-b, so
(—a)-(—b) =a-b.
Let a,b,cER. Then a-(b—c)=a-[b+(-c)] =a:b+a-(-—c)=
a-b+[-(a:-c)] =a-b—-a-c.
The proof that (a — b)-c =a-c —b-c
is Exercise 7(c).
a
In Section 6.4, we discussed operation preserving maps and defined a group
homomorphism as a function that preserves the operation. A ring homomorphism
must be operation preserving for both addition and multiplication.
6.5
Rings and Fields
337
DEFINITIONS
Let (R, + , -) and (S, 6, ®) be rings. A function h: R > S
is a ring homomorphism if for all a, b, € R,
|
h(a + b) = h(a) @ h(b) and
h(a- b) = h(a) @ h(b).
If h is one-to-one and onto S, then h is a ring isomorphism.
=
Example.
Let
h:(Z,+,-)— (Z,,,+,-)
be
the canonical
function
given
by
h(x) = x. We have seen in Section 6.4 that h:(Z, +) > (Z,,, +) and h:(Z, -) + (Z,,,°)
are both operation preserving. Therefore, h is a ring homomorphism.
Example.
Proof.
The function g: Z, > Z, defined by g(x) = 3xis aring homomorphism.
Let
x,yeZ,.
Then
g(x + y) = 3(x + y) = 3x + 3y = g(x) + @(y).
eo)
Alsop Gy) = 3xy = Oxy = 3x 2 3yr
a
2G):
Ring multiplication does not have to be commutative, need not have an iden-
tity element, and allows the possibility of zero divisors (nonzero elements a and b
such that ab = (0). Rings that have various combinations of these properties are
defined next.
DEFINITIONS
Let (R, +, -) be aring.
(R, +,-) is a ring with unity if there is an element | € R such that for all
Ge RAG las lie,
(R, +, -) is a commutative ring if for alla, b,e R,a-b=b-a.
(R, +, -) is an integral domain if R is a commutative ring with unity element
1, 1
#0, and R has no divisors of zero.
The integers (Z, +,-) are the prototypical example of an integral domain.
Multiplication is commutative. The integer | is the unity element (multiplicative
identity), and (Z, +, -) has no zero divisors.
Examples.
Both of the systems (Q,+,-) and (IR, +,-) are integral domains,
although we will see later that they have even more properties of multiplication.
The ring of even integers (E, +, -) is a commutative ring with no zero divisors.
It is not an integral domain because it does not have a unity element.
The ring (4, +, -) of 2 x 2 matrices is a ring with unity: The multiplicative
;
ae
j
:
;
identity is the identity matrix J =
ik
0
@
:
een
Net
il Matrix multiplication is not commuta-
tive, so this ring is not an integral domain. The product of the nonzero matrices
Sale|. lets a}
338
CHAPTER 6
Concepts of Algebra
The ring (%, +, -) of functions from R to R with pointwise addition and multiplication is commutative because multiplication of real numbers has this property.
The constant function C, € ¥, where C,(x) = | for all x € R, is the multiplicative identity. However, ¥ has zero divisors: The product X(_..,9)* Xj0,0) Of the
nonzero characteristic functions X/_ ,,. 9) and Xo, 0) is the zero function Co, where
Co) = 0 forall v=] R:
Theorem 6.5.4
(a)
For every natural number m > 1, the system (Z,,,, +, -) is acommutative ring
with unity.
The ring (Z,,, +, -) is an integral domain if and only if m is prime.
(b)
Proof.
Part (a) follows immediately from Theorem 6.5.1 and Theorem 6.1.2(b).
Part (b) follows from Theorem 6.1.2(c).
|
In an integral domain, we can apply cancellation laws to simplify products
with a nonzero common factor.
Theorem 6.5.5
Let (R, +, -) be an integral domain and a, b,c
bi=—ewandil
bia = ce-.a, then b= ce
Proof.
Assume
that a#0
€
and a-b=a-c.
Rwitha #0. Ifa-b =a-c, then
Then
a-b—a-c=0O,
so by
Theorem 6.5.3(d), a- (b — c) = O. Because there are no divisors of zero in R and
a#0,b —c = 0. Therefore, b = c.
The proof that b -a = c- aimplies b = c is Exercise 13.
&
Let (R, +, -) be an integral domain. One might hope that (R, -) would be an
abelian group because - is associative and commutative and | is the multiplicative
identity. However, this is impossible, because the element 0 cannot have a multiplicative inverse. (See Exercise 12.) It is possible, however, that the nonzero elements
of R all have inverses. An integral domain with this property is called a field.
*
DEFINITION
The ring (R, +, -) isa field if (R, +, -) is an integral domain
and (R — {0}, -) is an abelian group.
A field can also be described as an algebraic structure (R, +, -) such that
(i)
(ii)
(R, +) is an abelian group with identity 0.
(R — {0}, -) is an abelian group with identity 1.
(iii)
(iv)
foralla,b,ceR,a-(b+c)=(a-b)4+(a-c).
OF 1.
The proof of this fact is Exercise 14.
6.5
Rings and Fields
339
We now have an important distinction between the integers and the rationals
and reals. The rational numbers and real numbers are fields. The system (Z, +, -)
is not a field because no elements other than 1 and —1 have multiplicative inverses.
Finally, not all fields are infinite. In the proof of the following theorem, remember that when m is a prime, Z,, — {0) = U,,,, the group of units of Z,,.
Theorem 6.5.6
The ring (Z,,, +, -) is a field if and only if m is prime.
Proof.
By Theorems 6.5.4, and 6.2.1(c), we see that (Z,, +, -) is an integral
domain if and only if m is prime and that (Z,, — {0), -) is a group when m is prime.
Finite fields like (Zz +, -), where p is prime, are important in coding theory
and cryptography and are intimately related to the study of polynomials and their
roots.
Exercises 6.5
1.
*
Which of the following is a ring with the usual operations of addition and
multiplication? For each structure that is not a ring, list the ring axioms that
are not satisfied.
(a) N
(b) the closed interval [—1, 1]
(c)
(d)
2.
{a+ bi: a,b eZ}, where i? = —1
{bi:b eZ}, where i? = —1
Let Z[ V2] be the set {a + bV2: a,b € Z}. Define addition and multiplication on Z [V2] in the usual way. That is,
(a + bV2) + (@+4V2)—G+or
mo av?
(a+ bV2) =~ (€
fdV2) =ac
and
fadvV2 + beV2 + bd(v2)
= ac + 2hd + (ad + bc) V2.
Prove that (Z [V2], +, -) is aring.
3.
4.
Complete the proof that for every m € N, (Z,,, +, -) is aring (Theorem 6.5.1)
by showing that (b + c)a = ba + ca for all a, b, and c in Z,,.
Define addition © and multiplication @ on the set Z x Z as follows.
For a,b,c,d€Z, (a,b)
@ (c,d) =(a+c,b+d)
(ac, bd). Prove that (Z x Z, ®, ®) is a ring.
5.
and
(a,b) @(c,d) =
Let # be the set of all functions from R to RR, and define addition and
multiplication operations on % as follows. For fge# and xeER,
(f + g(x) =f@) + g(x) and (f- g)(x) = f(x) - g(x). Prove that (¥, +, -) is
a ring.
340
CHAPTER 6
Concepts of Algebra
6.
Let (R, +, -) be a ring and a, b € R. Prove that b + (—a) is the unique solution to the equation x + a = b.
Prove the remaining parts of Theorem 6.5.3: For all a, b, c © R,
(a)
a= 0 = 0.
(b)
(—a)-b=
(ce)
(a—b)-c=(a-c)—(-C).
— (a:b).
We define a subring of a ring in the same way we defined a subgroup of a
group: (S, +, -) is a subring of the ring (R, +, -) if-S'CR, and (S, +, -) is
a ring with the same operations. For example, the ring of even integers is
a subring of the ring of integers, and both are subrings of the ring of rational
numbers.
(a)
Prove that the ring ({0}, +, -) is a subring of any ring (R, +, -) (called
the trivial subring).
(b)
(Subring Test) Prove that if (R, +, -) is aring, Tis anonempty subset of
R, and T is closed under subtraction and multiplication, then (7, +, -) is
a subring.
Let 3Z = {3k:k € Z}. Apply the Subring Test (Exercise 8(b)) to show that
(3Z, +, -) is a subring of (Z, +, -).
10.
(a)
(b)
(c)
11.
Show that the function h: Z — Z defined by h(x) = 3x is not a ring
homomorphism.
Show that the function h: Z, > Z, defined by h(x) = 2x is not a ring
homomorphism.
Let Z[x] be the set of all polynomials p(x) in the variable x with integer
coefficients. Show that the function g: Z[x] — Z defined by g(p(x)) =
p(O) is a ring homomorphism.
Let R be the equivalence relation on Z x (Z — {0}) given by (x, y) R (u, v) if
xv = yu. Let P be the set of equivalence classes of Z x (Z — {0}) modulo R.
For (a, b) and (c, d) in P, define the operations ® and @ by
(a, b) @ (c,d) = (ad + be, bd)
(a)
(b)
and
(a, b) ® (c,d) = (ac, bd).
Find the additive identity in this ring, the unity element, and the additive
and multiplicative inverses of (2, 5). Hint: For each answer, you must
give a representative of the equivalence class in P.
Suppose that f: (Q, +, -)—> (P, @, ®) is given by f(p/q) = (p,q).
Prove that fis a ring homomorphism.
12.
Let (R, +, -) be an integral domain. Prove that 0 has no multiplicative inverse.
13.
Complete the proof of Theorem 6.5.5. That is, prove that if (R, +, -) is an
integral domain, a, b,c € R, and a #0, then
14.
b- a = c-aimplies b = c.
Let (R, +, -) be an algebraic structure such that
(i) (R, +) is an abelian group with identity 0.
(ii) (R — {0}, -) is an abelian group with identity 1.
(iii) Foralla,b,c
€
R,a-(b +c) =(a-b)4+(a-c).
(iv)
OF 1.
6.5
Rings and Fields
341
Prove that (R, +, -) is a field by showing that
b)-c=(a-c)+(b-0c).
(a) for alla, b, cE R, (a+
(b)
Proofs to Grade
15.
R has no divisors of zero.
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
(a)
(b)
Claim.
If (R,+,-) is a ring, a,b
© R, anda #0, then the equation
ax = b has a unique solution.
“Proof.”
Suppose that p and q are two solutions to ax = b. Then
ap ='b and ag = b. Thetefore, ap = aq. Therefore, p = q.
=
Claim.
If (R, +, -) is a finite integral domain, then (R, +, -) is a field.
“Proof.”
Suppose that R has n elements. Let x € R. Then the n + 1
POWEIS
Ol
ue —= 1X4
eee
ware noteall distinct, Wherefore,
x' = x" for integers t and r, where we may assume that ¢ < r. Then
x tx! = xx’, and therefore e = x’~'. Thus, e = x- x’~'~!. Therefore,
x has an inverse. Hence R is a field.
a
(c)
Claim. Let (R, +, -) be an integral domain with a, b, c € R and unity
element 1. If ab = ac and a = 0, than b = c.
“Proof.” Suppose that ab = ac and a # 0. Then b = 1b =(a“!a)b =
a'(ab) = a~'(ac) =(a~!a)c =1c = c. Therefore b = c.
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7
Concepts of Analysis
Analysis is that major branch of mathematics that deals with functions and associated concepts, including limits, differentiation, integration, infinite series, and a
myriad of applications within and outside mathematics. Although some aspects of
analysis can be traced back to ancient Greece, most of today’s concepts and terminology were developed in the 1800s. This chapter will discuss the most common
setting for doing analysis —the real numbers. We will lay the foundation for further
study by considering the real number system as a field (a set of objects with addition
and multiplication operations) that is ordered (so that all the real numbers may be
thought of as forming a line) and complete (so that all the numbers that “should be”
on the line are there). The rational number system is also an ordered field, but the
rationals lack the property that is the subject of this chapter—completeness.
Field axioms and properties of fields in general are discussed in Chapter 6, but
it is not necessary to have studied that material before reading this chapter. It is
sufficient to think of R as the set of all decimal numbers along the number line and
Q as the subset of R consisting of the repeating or terminating decimals.
We begin by explaining what we mean by completeness and by assuming
that the reals have this property. We then develop a sequence of three famous
theorems, each proved from the previous and each illustrating completeness from
a new perspective. The third result leads back to a proof that R is complete. This
reasoning is circular and therefore does not in fact prove the completeness of R.
However, along the way, we will have seen different ways of understanding completeness and proved that completeness is equivalent to each of the theorems. A
separate proof that the reals are complete requires a more careful definition of the
real numbers, as described in the last section.
343
344
CHAPTER 7
7.1
Concepts of Analysis
The Completeness Property
The main goal of this section is to draw an important distinction between R
and @. Both fields are ordered by the relation “less than,” which enables us to think
of both sets as numbers along a line. Furthermore, the rationals are like the reals in
this respect: There is never a “gap” between two numbers. The difference is that Q
is “missing” some numbers that are needed to work with limits and other topics in
analysis. Our first task is to make precise these ideas of gaps and missing numbers.
When we say that RR and @ have no gaps, we mean that there are never any
empty spaces between their elements. An example of an ordered number system
with gaps is Z—there are, for example, no integers between 3 and 4 or between 0
and 1. In the set of rational numbers, there are rationals between 3 and 4, such as
Tand 3.920941,
We can prove that this property of never having gaps is shared by every
ordered field. The following theorem is stated in general for “the field F,” but you
may think of F as being either the reals or the rationals. The proof uses only properties common to every ordered field, such as the distributive property, the existence
of a multiplicative identity 1, and the existence of a multiplicative inverse for every
nonzero element.
Theorem 7.1.1
If a < b in an ordered field F, then there exists c € F such thata <<c <b.
Proof.
Suppose that a < b. Then
ata<a+b<b+4+b
d+la<a+b<(1+1)b
Yeh
OG FEI
2 NG
a<}(a+b)<b.
Here we have used 2 for the element 1 + 1 and ;as the symbol for the multiplicative inverse of 2. We conclude that c = ;(a + b) is an element of the field that is
between the elements a and b.
&
The reasoning used to conclude that between any two distinct elements a and
b there is always a third element c (call it c,) can be repeated to produce another
element c, between a and c,, and so on. In this way, we can by induction produce
infinitely many elements between a and b.
Another way to think of this property is that nowhere on the number line is
there a “next rational number” or a “next real number.” For example, there is a
next integer after 0 (namely, 1), but there is no next real number after 0. Between 0
and any positive real number, there are always infinitely many real numbers.
Although the rational number system has no empty spaces between numbers, it
is missing some numbers that in some sense ought to be there. For example, every
7.1
The Completeness Property
345
rational number in the set B = {x € Q: x? < 2} is less than V2. There are many
rational numbers that are larger than any number in B: The rational number 1.5
is larger than every element of B; the rational number 1.42 is larger than every
element of B; and so are 1.415 and 1.4143 and 1.41422. We could continue this list
by finding smaller and smaller rational numbers, each of which is larger than every
element of B. Ideally, we'd like to find the smallest rational number that is larger
than every element of B. We will soon prove that there is no such rational number.
Because the field of rational numbers is missing numbers like these, we say that Q
is not complete.
In order to describe the completeness property, we need the idea of bounds for
a set. Upper and lower bounds, suprema, and infima are discussed in Section 3.5,
but it is not necessary for you to have studied that section.
These terms are defined here in the context of an ordered field F, and again you
may think of F as being either Q or R.
DEFINITION
Let A be a subset of an ordered field F’. Then
u € F is an upper bound for A iff a < u for every aE A. If A has an
upper bound, we say A is bounded above.
1 € F is a lower bound for A iff / < a for every a € A. If A has a lower
bound, we say A is bounded below.
If A has an upper bound and a lower bound, we say A is bounded.
In the ordered field R, the half-open interval [0, 3) has 3 as an upper bound.
In fact, zr, 18, and 206 are also upper bounds for [0, 3). Both —0.5 and 0 are lower
bounds for [0, 3). We note that a bound for a set may be, but does not have to be,
an element of the set.
Any finite nonempty subset A = {x,,
X), x3,...,x,} of R is bounded both
above and below:
u = max {x;:x; € A} is an upper bound for A, and
] = min {x;:x, € A} is a lower bound for A.
In R, the subset N is bounded below by 0. Later in this section, we will prove
that N is not bounded above in R. The sets @ and Z are neither bounded above nor
bounded below.
In the ordered field Q, the set {a
es
at of negative integer powers
of 2 is bounded above by 5 and below by 0. The set
A = {x € Q: x° < 4} has
many upper bounds: 8, 1.6, 1.59, and so on. However, A has no lower bounds. The
set B = {x € Q: x? < 2} is bounded above by 3 and below by —3.
The best (smallest) possible upper bound for a set A is called the supremum
of A.
346
CHAPTER 7
Concepts of Analysis
Let A bea subset of an ordered field F. Then s € F is a least
DEFINITION
upper bound for A (or supremum for A) if
(i)
(ii)
sis an upper bound for A and
s <x for every upper bound x for A.
We write sup(A) to denote a supremum of A.
i € F is a greatest lower bound for A (or infimum for-A) if
(i)
(ii)
iis alower bound for A and
x <i for every lower bound x for A.
An infimum of A is denoted inf(A).
Although a set Amay have many upper bounds, when sup(A) exists it is unique.
Likewise, inf(A) is unique if it exists. See Exercise 8.
Examples.
In the ordered field R,
inf({0, 3)) = 0 and sup([0, 3)) = 3.
inf(N) = 1, but sup(N) does not exist because N is not bounded above.
For A= {2~*: ke N}, inf(A) = 0 and sup(A) = 5.
For B = {x € Q:x? < 2}, inf(B) =
— V2 and sup(B) = V2.
In the ordered field Q,
For A = {27*:k © N}, inf(A) = 0 and sup(A) =
Ni]
For B = {x € Q: x* < 2}, Bis bounded, but as we shall see, inf(B) and sup(B)
do not exist in the field Q.
The following theorem provides a characterization of the supremum of a set. It
tells us first that every number larger than sup(A) is an uppef bound for A and every
number smaller than sup(A) is not an upper bound for A. Second, it says that if a
number has these two properties, then it must be the supremum for A.
Theorem 7.1.2
Let A be a subset of an ordered field F. Then s = sup(A) if and only if
@)
(ii)
forall e > 0, if x € A} then x < 5 + e«.
forall ¢ > 0, there exists y € A such that y > 5 — e.
Proof. First, suppose that s = sup(A). Let ¢ > 0 be given. Thenx <5 <s+e
for all x € A, which establishes property (i).
To verify property (ii), suppose that « > 0 and there is no x € A such that
x > s — €. Then s — € is an upper bound for A less than the least upper bound of
A, which is a contradiction.
7.1
The Completeness Property
347
Suppose now that s is anumber that satisfies conditions (i) and (ii). To show that
S = sup(A), we must first show that s is an upper bound for A. Suppose that there
y—s
is
yeA such that y > s. See Figure 7.1.1(a). If we let ¢ =
,theny>s+
6,
which violates condition (i). (The idea here is that ¢ is only half the distance from s
to the larger number y, so s + € must still be less than y.) We conclude that y < s
for all y € A, so s is an upper bound for A.
To show s is the least of all upper bounds, suppose that there is another upper
bound f such that t < s. (We will show that t cannot be an upper bound.) If we
let € = s — ¢, then by condition (ii), there is a number ze A such that z > s — e.
See Figure 7.1.1(b). Thus, z > s —¢=s —(s — ft) =t. This contradicts the
assumption that ¢ is an upper bound for A. Therefore, s is indeed the least upper
bound for A.
a
—__—_—_——:s—
OOO Oe
A
S
s+6é
y
(a) Could y be an element of A?
(b) Could ¢ be an upper bound for A?
Figure 7.1.1
In the field R, the supremum of the set (2, 4) is 4. Even if we take e to a
very small positive number, say ¢ = 0.0001, every element of (2, 4) is less than
4 + « = 4.0001. Furthermore, 4 — ¢ = 3.9999 is not quite big enough to be the
supremum because 3.99995 is an element of (2, 4) and is greater than 4 — e.
We define next a property that can be used to distinguish the real number system from the rationals.
DEFINITION
An ordered field F is complete iff every nonempty subset
of F that has an upper bound in F has a supremum that is an element of F.
Theorem 7.1.4
The field (R, +, -) is a complete ordered field.
A proof of the completeness property requires considerable preliminary study
of the real number system. Section 7.5 includes a brief description of how the real
numbers may be rigorously defined. The details of the construction of the real
numbers and of the proof of Theorem 7.1.3 are beyond the goals of this text.
One important application of the completeness property is the Archimedean
Principle for R. It tells us that the set N is not bounded above by any real number.
348
CHAPTER 7
Theorem 7.1.3
Concepts of Analysis
The Archimedean Principle for R
For every real number x, there is a natural number n such that n > x.
Proof. Let xc R. If x < 1, we use m = 1, and then n > x. For x = 1, let
A = {ne N:n <x}. The set A is nonempty (because it contains 1) and is A
bounded
above by x. Therefore, by the completeness
property of R, sup(A)
exists. Let s = sup(A). Because s — | is less than sup(A), there exists m € A such that
s—1<m. Thus,s<m+1,andsom + 1 ¢ A. Therefore, m+ 1 is a natural number, andm-+
1 > x.
B
The proof of the unboundedness of N needs to be more than simply claiming
that wherever r is on the number line, there must be a natural number farther to the
right of r. The completeness property is the key to this proof of the Archimedean
Principle. Formal proofs of completeness, proper rules for the manipulation of infinite expansions, and careful definitions of limits, continuity, and other concepts are
all part of the development of modern real analysis.
We can now see why we say that @ is missing some numbers—numbers that
would have to be there to make @ a complete ordered field. We give an example
of a bounded subset of © that has no supremum in @. Exercise 19 shows that there
are, in fact, many points missing from Q.
Example. The set B = {x € Q: x* < 2} is bounded above in the field Q but has
no supremum in @.Proof.
There are many rational numbers, such as 2and 1.44, that are upper bounds
for B, so B is bounded above in Q.
os
Now suppose that there is a rational number s such that s = sup(B). We will
first show that both s < V2 and s > V2 are false.
If s < V2, then V2 — s is positive. By the Archimedean Principle for R,
;
1
there is a natural number K such that K > eae
‘
Then s +
1
Kx
V2ands
1
+ K
is rational (because both s and é are rational). Thus, s + i is an element of B,
which contradicts s being an upper bound for B.
lfs = V2
then. s — V2 is positive. By the Archimedean
is a natural number
M such that M > : Bays.
:
. Then
Principle, there
s — 7,> V2 and s — +is
M
rational. Then for all x in B, x < V2 <5 — 4 < 5, SO 5 — i is an upper bound
for B that is less than s. This contradicts the assumption that s is the least upper
bound for B in Q.
Because both s < V2 ands > V2 are false, we conclude that s = V2. But
this is impossible because s is a rational number. Therefore, sup(B) does not exist
in the ordered field Q.
5
7.1.
The Completeness Property
349
Although the set B in the example has no supremum in Q, when B is considered as a subset of R, it does have a supremum in R: sup(B) = V2. The fact that
sup(B) in R exists is a consequence of the completeness of R.
Finally, for everything that has been said about upper bounds and suprema,
there is a corresponding statement about lower bounds and infima. In particular,
the definition of completeness could have been stated in terms of lower bounds and
infima. That is, an ordered field F is complete iff every nonempty subset of F that
has a lower bound in F has an infimum in F. See Exercise 18.
Exercises 7.1
1.
Find two upper bounds (if any exist) for each of the following sets.
{x eR:x? < 10}
(c)
jreRix+tes|
(d}oe
(ate Rox
(e)
(g)
(Caelkeee =. 2 Ser
(),
(xe Rix =0 andx? = 30)
{27:x ER}
(h)
{xeER:x— 10 < logx}
1
* (b)
1
(a)
tare Nt
+ Ox — 30}
Find a lower bound in R (if one exists) for each of the sets in Exercise 1.
3.
Find the supremum and infimum, if they exist, of each of the following sets.
*
(a)
{hnen}
n
(b)
x
(c)
{2.xEZ}
(d)
{ar(1 +h)nen}
x
“(e)
teegnen}
(f)
{xeQ:x?
< 10}
x
(g)
Sete
CD) ie eet
(i)
x
farxyent
isa
GU
(2th nen}
n
sei
:
tle 0)}
2}
4.
LetA and B be subsets of IR. Prove that
(a) if A is bounded above and B C A, then B is bounded above.
(b) if A is bounded below and B C A, then B is bounded below.
(c) if A and B are bounded above, then A U B is bounded above.
(d) if A and B are bounded below, then A U B is bounded below.
5.
Let x be an upper bound for A C R. Prove that
(a)
(b)
if x < y, then y is an upper bound for A.
if x€ A, then x = sup(A).
350
CHAPTER 7
Concepts of Analysis
6.
Let A C R. Prove that
(a) if A is bounded above, then A‘ is not bounded above.
(b) if A is bounded below, then A° is not bounded below.
Give an example of a set A C R for which both A and A‘ are unbounded above
and below.
Let A C R. Prove that
(a) if sup(A) exists, then it is unique. That is, ifxand y are both least upper
bounds for A, then x = y.
(b) if inf(A) exists, then it is unique.
Let
(a)
(b)
A < B CR. Prove that
if sup(A) and sup(B) both exist, then sup(A) < sup(B).
if inf(A) and inf(B) both exist, then inf(A) > inf(B).
10.
Formulate and prove a characterization of greatest lower bounds similar to
that in Theorem 7.1.2 for least upper bounds.
11.
If possible, give an example of
(a) aset A CR
such that sup(A) = 4 and4 € A.
(b) aset A C Q such that sup(A) = 4 and 4 ¢ A.
(c) aset A CN
such that sup(A) = 4 and 4 € A.
(d) aset
ACN
such that sup(A) > 4 and 4 ¢ A.
Let A C R. Prove that
(a) if sup(A) exists, then sup(A) = inf {u: uw is an upper bound of A}.
(b) if inf(A) exists, then inf(A) = sup {/: / is a lower bound of A}.
Let A and B be subsets of R.
(a) Prove that if sup(A) and sup(B) exist, then sup(A U B) exists, and
sup(A U B) = max {sup(A), sup(B)}.
(b) State and prove a similar result for inf(A U B).
14.
(a)
(b)
15.
(a)
Give an example of sets A and B of real numbers such that
AN B 4 ©,
sup(A ™ B) < sup(A), and sup(A 9 B) < sup(B).
For sets A and B such that
AM B # ©, state and prove a relationship
among sup(A), sup(B), and sup(A M B).
Give an example of sets A and B of real numbers such that
AN B 4 ©,
inf(A 1 B) > inf(A), and inf(A M B) > inf(B).
(b)
For sets A and B such that A
B # ©, state and prove a relationship
among inf(A), inf(B), and inf(A M B).
16.
An alternate version of the Archimedean Principle for the reals has the effect
of saying that there are no infinitesimal (infinitely small) real numbers. It says
(Ve > O)(dne ni( =a °)
Prove that the two versions are equivalent.
Le
Give an example of a set of rational numbers that has a rational lower bound
but no rational greatest lower bound.
7.2.
The Heine-Borel Theorem
351
18.
Prove that an ordered field F is complete iff every nonempty subset of F that
has a lower bound in F has an infimum in F.
19.
Prove that every irrational number is “missing” from Q. Begin with an irrational number x and find a subset A of @ such that A is bounded above in
@ and sup(A) does not exist in Q, but when A is considered a subset of R,
Sup(A) =x.
Proofs to Grade
20.
x
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
(a) Claim. Let A CR. If i= inf(A) and e > 0, then there is y € A such
thaty <i+e.
.
“Proof.”
(b)
Let y=i+ sfThen i < y, so y EA. By construction of y,
Vie
es
a
Claim.
Let A C R. If A is bounded above, then A‘ is bounded below.
“Proof.”
If A is bounded above, then sup(A) exists (because R is
complete). Because sup(A) = inf(A°) (see the figure), inf(A‘°) exists.
Thus, A‘° is bounded below.
a
A
A Cc
at
sup(A) = inf(A°) oes
(c)
Claim.
If
ACBCR,
and if sup(A)
and sup(B) both exist, then
sup(A) < sup(B).
“Proof.”
Assume that A C B and sup(A) > sup(B). We choose ¢ =
5(sup(A) — sup(B)). Then ¢ > 0 and sup(B) < sup(A) — € < sup(A).
By part (ii) of Theorem 7.1.1, there is y € A such that y > sup(A) — e.
Then
yeB
and y~> sup(B). This is impossible.
Therefore,
sup(A) < sup(B).
a
(d)
Claim.
If f; R — R and A is a bounded subset of R, then the image
set f(A) is bounded.
“Proof.”
Let m be an upper bound for A. Then a < m for alla EA.
Therefore, f(a) < f(m) for all a € A. Thus, f(m) is an upper bound for
JIA).
(e)
7.2
a
Claim. The set N is not bounded above.
“Proof.”
Suppose that N is bounded above by the natural number m.
Then m + 1 is a natural number larger than m. This is a contradiction. m
The Heine-Borel Theorem
This section begins by describing certain subsets of R called “open” sets that have
properties similar to those of open intervals. We use open sets to define another
type of set (the “compact” set) that, even when the set is infinite, has some proper-
ties of a finite set. Finally we use the completeness of [IR to prove a necessary and
352
CHAPTER 7
Concepts of Analysis
sufficient condition for a set to be compact—the Heine—Borel Theorem. Before we
get to these results, we describe the “interior” points of a set.
DEFINITION
Let a and 6 be real numbers with 6 > 0. The
6-neighborhood of a is the set
N(a,
6) = {x E R: |x —a| < 5}.
The 5-neighborhood N(a,5) of a real number a is the open interval
(a — 5,a + 5). For example, the 1-neighborhood of 2 is N(2, 1) = (1, 3), the
0.3-neighborhood of —1 is N(—1, 0.3) = (—1.3, —0.7), and the 0.1-neighborhood
of 0.5 is N(0.5, 0.1) = (0.4, 0.6). These neighborhoods are shown in Figure 7.2.1.
N(—1, 0.3)
4
as)
Oe
=2
=I
N(0.5, 0.1)
IN, MY
2) Sh ssa
a
0
0.5
1
2
ara)
3
Figure 7.2.1
Neighborhoods have the advantage of using sets instead of distances to define
“closeness” and therefore enable us to define analysis concepts in terms of sets. For
example, rather than using the expression |x — a| < 6 to say x is within distance 4
of a, we write x € N(a, 5). Thus, the definition of lim f(x) = L,
ye.
Oso.
O)(x — ia)
0 = FG) — Ly =< e)
may be written as
(Ve > 0)(46 > 0) € N(a, 6) > f(x) € N(L, &))
This definition can be used for systems other than the real numbers.
DEFINITION
Fora set A C R, a point x is an interior point of A if there
exists 6 > O such that N(x, 6) C A.
If x is an interior point of A, then not only is x contained in A, but also all ele-
ments of some neighborhood around x are contained in A.
For the interval [2, 5), 3 is an interior point because N(3, 0.5) C [2, 5). Also,
4.9998 is an interior point because N(4.9998, 0.0001) € [2, 5). See Figure 7.2.2.
In fact, every point in (2, 5) is an interior point of [2, 5). The point 2 is not interior
7.2
The Heine-Borel Theorem
353
to [2, 5) because every 5-neighborhood of 2 contains points that are less than 2 and
hence not in [2, 5).
N(2, 8)
N(3, 0.5)
ee a ee
4.9997 4.9998
4.9999
5
Figure 7.2.2
DEFINITION
The set A C R is open if every point of A is an interior
point of A. The set A is closed if its complement A‘ is open.
The interval (2, 5) is open because every point in (2, 5) is an interior point of
(2, 5). Thus its complement, (— oo, 2] U [5, 00), is a closed set.
In ordinary conversation, with references to objects like doors and eyes,
the concepts of open and closed are opposites, but the interval [2, 5) is an
example of a subset of R that is neither open nor closed. The interval is not
open because 2 is an element, but is not an interior point, of [2, 5). On the other
hand, [2, 5).1s not.closed because its complement, |2, 5)° = (—co, 2] U5, co);
is not open. The complement
contains 5, but 5 is not an interior point of
(—oo, 2] U[S, oo).
Examples. The set R is open because for every x € R, N(x, 1) C R. The empty
set © is open because the antecedent is false in the statement “for all x, if x € U,
x is an interior point of ©.” The sets RK and © are complements, so both are also
closed sets. It can be shown that there are no other subsets of the reals that are both
open and closed.
A set is open if about each element in the set there is a 6-neighborhood that
lies entirely within the set. This means no point of the set can be on the “boundary”
or outer edges of the set (see Exercise
14). The next results describe open and
closed sets and explain why “open intervals” and “closed intervals” are appropriately named.
354
CHAPTER 7
Theorem 7.2.1
Concepts of Analysis
Every open interval of real numbers is an open set.
Proof.
Let (a, b) be an open interval, and let x € (a,b). Choose 6 = min
{x — a,b —x}. (This minimum is the largest possible 5 we can use. See
Figure 7.2.3.) Then & > 0. To show that N(x, 5) C (a, b), let y € N(x, 6). Then
|
ye (a,odb).
a=x—(x—a)<x-—8<y<x+65<x+(b—x)=b,ands
N(x, 6)
¢
—L
K—
x-a
)
—>|
b—-x
—____ ¢- _____|
—}
a
—> x
b
xX
N(x, 6) in the case
6 =x —a
Figure 7.2.3
Neighborhoods are open intervals, so we know from Theorem
N(x, 5) is an open set for all x € R and every positive 6.
Theorem 7.2.2
7.2.1 that
Let & be a nonempty collection of open subsets of RR. Then
(a)
J A is an open set.
Aen
(b)
.
if A is finite, ()A is an open set.
Aces
Proof.
(a)
Let & be a nonempty collection of open sets.
Suppose
that
xe
(JA.
Then
there
exists
Be
such
that
xe B.
AcA
Because B is in the collection , the set B is open. Thus, x is an interior point
of B.
BC
(b)
Therefore,
there
exists
56 > 0 such
that
N(x,5)
CB.
Because
U A, N(x, 6) € iy A. Therefore, x is an interior point of U A.
AEA
Aen
AEA
Suppose that sf is finite and xe
()A. Then x€A
for all
Ae XA. For
AEA
each open set A € H, there corresponds d, > 0 such that N(x, 6,4) CA.
Let 6 = min {d4:A € A}. (Note that the minimum of a finite set of positive numbers must be positive.) Then 5 > 0 and N(x, 8) C 1(x, 64)
CA
for all
AG A. Thus
N(x, 5) C (\A. Therefore,
x is an interior point
AEA
of ()A.
B
AEA
For the proof of part (b), we chose 5 = min {6,: A € 4} and relied on the
fact that it is always possible to find the minimum of a finite set of real numbers.
However, if the set {54: A € } is infinite, we cannot be sure of finding a minimum.
7.2
The Heine-Borel Theorem
355
The set {= Me N}, for example, does not have a minimum element. The statement
of Theorem 7.2.2(b) is false if we omit the word “finite.”
Theorems 7.2.1 and 7.2.2 can be used to produce many examples of open subsets of R.
Examples.
For every a € R, the open rays (a, 00) and (—on, a) are open sets. (See
Exercise 8.) The set (1, 2) U (5, 7) U (15, 00) is open because it is the union of
three open sets, and R — {2} is open because R — {2} = (—oo, 2) U(2, 00). oO
Corresponding to Theorem 7.2.1, it can be shown that every closed interval of
real numbers is a closed set. See Exercise 9. The analog of Theorem 7.2.2 for closed
sets is Exercise 10.
Examples. For every a € R, the rays [a, 00) and (—oo, a] are closed sets because
their complements are open rays.
hesetA
ja54
Seal
2 ot
0
be) (tttne nt = ce isa g,...
n
AES)
the complement of A is A° = (—00, 1)
isctosed because
ee
D
i
U(2, 00) U LU (it cae )which is a
n
a
union of two open rays and infinitely many open intervals.
Oo
Every finite subset of R is closed because the complement of a finite set is the
union of two open rays and finitely many open intervals. Finite sets are good examples of sets that are both closed and bounded. We shall see that infinite closed and
bounded sets have other properties in common with finite sets. For example, if A is
a finite set, then the infimum and supremum of A exist and are elements of A. The
next theorem shows this is true for all closed and bounded sets.
Theorem 7.2.3
If A is
anonempty closed and bounded subset of IR, then sup(A) € A and inf(A) € A.
Proof. Suppose that A is a nonempty closed and bounded set. Let s = sup(A),
and suppose that s ¢ A. Then s € A‘, which is open because A is closed. Thus,
N(s, 5) C A° for some positive 5. This implies s — 6 is an upper bound for A
because the interval (s — 6, s + 5) is a subset of A‘. (No element of A is greater
than s, or equal to s, or between s — 6 and s.) This contradicts Theorem 7.1.2.
Therefore, s € A.
The proof that inf(A) € A is similar.
o
Examples.
The set X = [2,5] U[7, 10] U {12} is closed and bounded. In
accordance with Theorem 7.2.3, inf(A) = 2 and sup(A) = 12 are elements of X.
n+ 1
Earlier we showed that A = {1} U {
ine ntis closed. Because A is
Nn
also bounded, inf(A) = 1 and sup(A) = 2 must be elements of A, and they are.
O
356
CHAPTER 7
Concepts of Analysis
Both properties of being closed and bounded are necessary for Theorem 7.2.3.
For example, [0, 1) is bounded but not closed, and it does not contain its supremum.
The set [2, 00) is closed but not bounded, and it has no supremum.
To understand how the completeness property of R is related to properties of
closed and bounded sets, we need one more concept.
DEFINITIONS
Let A be a set of real numbers. A collection€ of open
subsets of R is a cover forA if A € U G&:
Cee
If B C € and ZF is also a cover for A, we say & is a subcover of “.
A cover for a set A is a collection of open sets whose union includes A.
Examples.
Let A = {2, 4, 7, 12} and‘@ = {(—3,3), (1,5), (3, 10), (5, 9), (8, 00)}.
The collection 6 is a cover for Abecause each member of © is an open set and A is
a subset of the union of the five sets in €.
The five sets in @ are not all needed to form a cover for A. The collection
H, = {(—3, 3), (3, 10), (8, o0)} is also a cover of A. Because %, & @, the collection 2), is a subcover of @. The collection #, = {(1, 5), (5, 9), (8, o0)} is another
subcover of 6.
Oo
Example.
1
1
For each n EN, let K, be the open interval (>ae n+ 1),and let
J = {K,:n © N}. For each natural number n, the only set in # that contains n is
K,,. Therefore, J is a cover for N with no subcovers other than 7 itself.
Oo
CO
Example.
Because \s) (—oo, n) = R, the collection # =
{(—co,n):n EN} isa
i
cover for R. The collection # = {(—oco, n):n € 2N}, where 2N is the set of even
natural numbers, is a subcover of d€ for R. There are many subcovers of # for R,
but there is no finite subset of #€ that is a cover for IR. (See Exercise 15.)
Oo
A cover
of a set A may be imagined by thinking of the covering sets of €
as providing shade for the set A. In Figure 7.2.4, we have the sun directly over the
set A. The covering sets C,, Cg, Cs,... are drawn slightly above A. For this analogy, think of the sun’s rays as parallel beams of light. As the sun shines straight
down, “ is a cover for A iff A is within the region shaded by the sets in €—that
is, tft AcSule) C.
Ce%
Let A = {X), x, X3,...x,} be a nonempty finite set of real numbers, and let
% = {O,: a € A} be a cover for A. We may not need all the sets in 6 to make a
7.2
The Heine-Borel Theorem
357
cover for A, so we look for a subcover. For each i = 1, 2, 3,..., 7, there is a, in A
such that x; € O,,. Then the collection 8 = {O,:i = 1, 2, 3,...n} is a finite subset of @ whose union includes A. In other words, given any cover for the finite set
A, we can always construct a finite subcover for A. Sets of real numbers that have
this property are called compact sets.
Sun
Sun’s rays
G
&
a
C5
cae
.
ps expel
|
|
:
A
es
$f
—}—_______-x
Figure 7.2.4
DEFINITION A subset A of R is compact if for every cover @ for A,
there is a finite subcover of © for A.
By the discussion above, every nonempty finite set is compact. The next
example is an infinite set that is compact.
Example.
Proof.
1
The set A = {1} U tt- ne ntis compact.
Let {O,:a¢€ A} be any cover for A. One of the covering sets—call
it O,»—contains
the element
1. Because
O,, 18 open, there is a 6-neighborhood
358
CHAPTER 7
Concepts of Analysis
on
On
cE
SS
Ow,
Seca
Oz,
PERCE TERETE ILS
a 6
a
_6—_o—
)
ce
@———_
a
3
—=+
-@
3
2
2
>~X
Figure 7.2.5
N(,
6) € O,«. See Figure 7.2.5. (We will show that all but a finite number of elements
of A are in N(1, 6) and therefore in O,x.) Choose N such that N > * Ifn > N, then
i
n, so 1 < né. This implies n + 1 < n + né, which means
to)
n+1
< ||=o.
+ |
n+ 1
Therefore, ifn > N, then bi a
© N(1, 6) and
=
eC.
N=
N
Now choose O,,, Og,; Quy +++» Og,
Such that 2 € O,,, 3a0), ;€ O,,,+--, and
a
1
ne
e
€ O,,. Then A € Oyx U Og, U Og, U Og, U +**U Og, (The first N elements
ofA are in OF OEE OF and all other elements are in O,«.) We have succeeded in
finding a finite subcover of N + 1 sets for the cover {O,: a € A}. Therefore, A
is compact.
&
We have seen that # = {(—oo, n):n € N} is acover for R with no finite subcover. Thus R is not compact. Neither is the open interval (0, 1) compact because
the collection
@ = {ie 1)axe CL 00)} is a cover for (0, 1) that has no finite
subcover.
It is no coincidence that all the examples of sets that are compact have been
closed and bounded and all the examples of sets that are not compact are either not
closed or not bounded. The next theorem is a beautiful characterization of compact
sets based on the work of Edward Heine* and Emile Borel.+ The proof uses the following lemma. Watch for the places where the proof of the Heine—Borel Theorem
depends on the completeness of R.
Lemma 7.2.4
Let A be a closed set and x € R. If
Proof.
Exercise 16.
AN N(x, 6) # © for all 6 > 0, then x € A.
r
* Edward Heine (1821-1881) was a mathematician at the University of Halle (the same university
where Georg Cantor spent his entire career) who made several contributions to analysis, especially to
the descriptions and solutions of equations involving infinite series. As a senior professor, Heine gave
the young Cantor a problem in analysis whose solution and generalization demonstrated the need to
define the term set precisely.
+ Emile Borel (1871-1956) was a mathematician and politician who contributed substantially to probability and game theory and the creation of the branch of mathematics called measure theory. He stated
and proved the Heine—Borel Theorem for countable sets. He served many years in French politics and
was a member of the French resistance in World War II.
7.2.
Theorem 7.2.5
The Heine-Borel Theorem
359
The Heine-Borel Theorem
A subset A of R is compact if and only if A is closed and bounded.
Proof.
(i)
Suppose that A is compact. We first show that A is bounded. We note that
A EIR =
a (—n, n). Therefore, #€ =
{(—n, n):n € N} is acover forA.By
neN
compactness, d€ has a finite subcover
{(—n, n): a Spiro
aes
we
then
(SLA
choose
N = max
{n,,M,...,n,},
AC
Ue i)
Mie
i—!
Therefore, A is bounded above by N and below by —N.
We show that A is closed by proving A‘ is open. Suppose that y € A°.
(We must show y is an interior point of A°.) For each x € A, x # y, and thus
One é |x — y| is a positive number. The collection {.N(x, 6,): x € A} is a
family of open sets that covers A. By the compactness of A,
A CNX, 5,.) UN (x, 6.)
Us
UN, 5.)
for some x), X5,...,x, A. Leté = min {6,, 6,,...,6,,}. See aaa 2.0:
(We will show ‘ist NGS ) ONG; 5) Eh por ail it 2
Suppose
that
ze WN(y, 6) M N(x, by). Thetetore
ede
miles= <6
and
|z — x,|< 6,.Then
Ki
SP
ete — y < 5, + OS 20, = 2; — YI.
This is acontradiction. Because A C N(x), 5.) U Np, 6,,)
U-* U Ny, 5.)
we have N(y, 5) © A‘. Therefore, A‘ is open, so A is closed.
¢
(
¢
ca
|
|
|»
x4
ssoy,
wa
ee)
A
sa
cl
|
)
’
____» ____» ___» ___»5 __|______»
x
X5
x3
Xo
—___>x
year
A
Figure 7.2.6
(ii)
Suppose that A is a closed and bounded set and
For each x € R, let A, = {a € A: a < x}. Also, let
“ is a cover for A.
D = {x €R: A, is included in a union of finitely many sets from ¢
}.
360
CHAPTER 7
Concepts of Analysis
We claim that (a) D is nonempty and (b) D has no upper bound.
(a) The set A is bounded, so inf(A) exists by the completeness of IR. For every
z < inf(A), the set A, is empty, and, therefore, z € D. Thus, D is nonempty.
Suppose that D is bounded above. Then x) = sup(D) exists, by the
completeness of R. By Theorem 7.1.1, for every 5 > 0, there exists
(b)
t € D such that x,» - 6 <t Sx. If
AM N(x, 6) = ©, it follows from
t € D that x) + 6 € D, contradicting the fact that x) = sup(D). Therefore, for all 6 > 0, we have
AN N(x, 6) # GW. By Lemma 7.2.4, xo is
in the closed set A.
Let C* be an element of @ such that x9 € C*. Because C* is open,
there exists ¢ > 0 such that N(%, €) C C*. Choose x, € D such that
Xp — G6 ay = xy. because, ©), there are,open sets; Cy, Cys. 2G
in @ such that A, © C; UC, U-:- UC, Now let x) = x9 + 2 Then
Ke Cr randrA nC il Cc) Ce C*. Thus, x, ¢€. Ihis is a
contradiction, because X, > Xp and x) = sup(D). We conclude that D
has no upper bound.
Because A is bounded and R is complete, sup(A) exists. The set D has
no upper bound, so there exists y € D such that y > sup(A). Thus A, = A.
Therefore, A is a subset of a union of finitely many sets from @. We conclude
that A is compact.
a
Example.
n+1
Earlierinthis section,weshowed that A = {1} U {
ine I isa
closed and bounded subset of RR. By the Heine—Borel Theorem, A is compact.
Exercises 7.2
1.
Find x and 6 € R such that
(2) oN), 7,15).
(C) a Nis.0) =31(6.023, 6,024).
2.
AS (Deeds
(d)
(CO) =a (5-03 568):
N(x, 6) = (40, 49).
For real numbers x, 5,, 55,... 6, describe
(a) () NG, 4).
(b)
i=1
UNG, 3).
i=l
3.
State the definition of continuity of the function f at the number a in terms of
neighborhoods.
4.
Find the set of interior points for each of these subsets of R.
(a)
aly)
(Db)
(c)
Q
(7d)
1
* (@) {gkeN}
(g) R-N
() {a:keN}
u {0}
(hy) R-{J:keN}
—
Ri
ee Ge 0a
neN
at]
R-Q
1 e
en0.2)
(j=
eZ
7.2
The Heine-Borel Theorem
361
Suppose that x is an interior point of a set A. Prove that
(a)
if ACB, then x is an interior point of B.
(b)
(c)
if x is an interior point of B, then x is an interior point of A U B.
if x is an interior point of B, then x is an interior point of AQ B.
Show by example that the intersection of infinitely many open sets may not
be open.
Classify each of the following subsets of R as open, closed, or neither.
(b)
WN(a, 6) — {a} forae Randd > 0
(on. 8 0) 1Oh
(a)
coy 3)
x
(d)
Q
(ec)
R-—N
CONG
(C2)
()
Ee eles
Sf [a
a
ie S|
(h)
{x: |x —5| #7}
Ci) eer 0
es
7}
Let a € R. Prove that the rays (a, oo) and (— 00, a) are open sets, using
(a)
the definition of open sets.
(b)
Theorem 7.2.2.
Let a, b € R. Prove that every closed interval [a, b] is a closed set.
10.
Let A be a nonempty collection of closed subsets of R.
(a)
Prove that i A is a closed set.
(b)
Prove that if 4 is a finite collection, then
AEA
(c)
L) A is aclosed set.
Acs
Show by example that part (b) is false if we do not assume that 4 is
finite.
11.
If possible, give an example of each:
(a) an open set with no interior points.
(b) anonempty closed set with no interior points.
(c) an open set with one interior point.
(d) aclosed set with one interior point.
12.
Let A and B be subsets of R and x € R. Prove that
(a) if Ais open, then A — {x} is open.
(b)
(c)
if A is open and B is closed, then A — B is open.
if A is open and B is closed, then B — A is closed.
13.
Let A be a subset of R. Prove that the set of all interior points of A is an
open set.
14.
A point x is a boundary point of the set A if and only if for all 6 > 0,
Nx, d)NA #O and N(x, 6) N A 4S.
(a)
Find all boundary points of (2, 5], (0, 1), [3, 5] U {6}, and Q.
(b)
Prove that x is a boundary point of A if and only if x is not an interior
point of A and not an interior point of A°.
Prove that A is open if and only if A contains none of its boundary points.
Prove that A is closed if and only if A contains all of its boundary points.
(c)
(d)
15.
Show that the collection # = {(—0o, n):n € N} is a cover for R but #€ has
no finite subcover for R.
362
CHAPTER 7
Concepts of Analysis
16.
Prove Lemma 7.2.4.
17.
Which of the following subsets of R are compact?
*
+
18.
(a)
(c)
Z
[, V10]
(e)
(192439409
9123 Is)
(f)
(OU {amen
20.
[0, 10] U [20, 30]
RR —A, where A is finite set
(g)
© 3)5]
(h)
[0,1]
NO
Give an example of
(a)
(b)
(c)
19.
(b)
* (d)
abounded subset of R and a cover of that set that has no finite subcover.
aclosed subset of R and a cover of that set that has no finite subcover.
sets A, B, C, and D of real numbers such that A C B C C C D, Ais open,
B is closed, C is neither open nor closed, and D is compact.
Let A and B be compact subsets of R.
(a) Use the definition of compact to prove that A U B is compact.
(b) Apply the Heine—Borel Theorem to prove that A U B is compact.
(c) Apply the Heine—Borel Theorem to prove that A M B is compact.
2
Lets = (0,1,and
ter€ = {(2Qn22,2!) nent.
z
21.
(a)
(b)
(c)
Prove that & is a cover for S.
Is there a finite subcover of @ for S?
What does the Heine—Borel Theorem say about S$?
(a)
Use the Heine—Borel Theorem to prove thatif {A,: a € A} isanonempty
collection of compact sets, then () A, 18 compact.
(b)
Give an example of a collection {A,: a € A} of compact sets such that
acA
J A, is not compact.
acA
Proofs to Grade
22. Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
(a)
Claim.
Let a eR. The open ray (a, oo) is an open set.
[o@)
“Proof.”
Write the set (a, 00) as (a, 00) =
U (atneatn+l).
n=0
Each set (a + n, a +n + 1) is an open interval (each of length 1) and
(b)
*
(c)
therefore an open set. By Theorem 7.2.2, (a, oo) is an open set.
S
Claim.
If A and B are compact, then A U B is compact.
“Proof.” If A and B are compact, then for any cover {O,:a€ A}
for A, there exists a finite subcover O, , O,,..., O, , and for any open
cover {U,: B € V} forB, there exists a inte: Roce Uy Ug,.-+,
Ug .
HUSA
2 Ope OT
eaCO,Peat
ke “UUs. Thee
fore, AUBCO, UO, ng non YU Ug, U Ug, U --U Ug, a union of
a finite number af open sets. Thus, A U B is compact.
a
Claim.
If A is compact, B C A, and B is closed, then B is compact.
“Proof.” Let {O,:a€ A} be a cover for B. If {O,:a€ A} is a
cover for A, then there is a finite subcover of {O,: a © A} that covers
7.3
(d)
363
A and hence covers B. If {O,: a € A} is not a cover for A, add one
more open set O* = R — B to the collection to obtain a cover for A.
This cover for A has a finite subcover of A that is a cover for B. In
either case, B is covered by a finite number of open sets. Therefore, B
is compact.
=
Claim.
If A is compact, B C A, and B is closed, then B is compact.
“Proof.”
B is closed by assumption. Because A is compact, A is
bounded.
(e)
The Bolzano-Weierstrass Theorem
Because
B CA, B is also bounded.
Thus, B is closed and
bounded. Therefore, B is compact.
al
Claim.
The set (5, oo) is compact.
“Proof.”
The set
@ = {(4, 12), (10, c0)} is a cover for (5, 00). Then
is a finite subcover of € for (5, 00), so (5, 00) is compact.
7.3
The Bolzano-Weierstrass Theorem
In this section, we use the Heine—Borel Theorem to prove another classical result
of analysis, the Bolzano—Weierstrass* Theorem. Let’s begin with a closed interval
[a, b] and imagine that we must build a subset A of [a, b] by selecting elements for
A, one at a time, from [a, b]. Because A C [a, b], A will necessarily be a bounded
set. If A is finite, we could choose elements that are spread out across the interval.
In other words, if A is finite, there need not be any point in the interval where the
elements of A pile up or “accumulate.” What the Bolzano—Weierstrass Theorem
says 1s that if a set A is infinite and bounded, there must be at least one point in
[a, b] around which an infinite number of elements of A will be congregated. Before
we get to that result, we give an example and define carefully what it means for
elements of a set to accumulate around a point.
Example.
For the bounded infinite set
A = {1,5. > — ; be of each element
ofA is contained in the interval [—1, 1]. Figure 7.3.1 suggests that the number 0
is a point around which the elements of A gather. We say that 0 is an accumulation
point for the set A.
@
-1
el
2
So
@@
en iyesalee
A678
wee
oe
V0)
@
@
1Nek ae
aS
3
@
[
l
x
Figure 7.3.1
* Bernard Bolzano (1781-1848) was a mathematician, philosopher, and logician. He pioneered several
modern mathematics concepts (such as the rigorous definition of limit), but because of his strong antimilitary beliefs, most of his work appeared in obscure publications. The Bolzano—Weierstrass Theorem
was for several years called the Weierstrass Theorem until Bolzano’s independent proof was discovered
many years after his death.
Karl Weierstrass (1815—1897) was the foremost leader in the 1800s in developing highly rigorous
definitions and theorems that characterize much of modern mathematics. He was the first to develop
rigorous proofs for the Intermediate Value Theorem, the Heine—Borel Theorem, and several theorems
whose titles now bear his name.
364
CHAPTER7
Concepts of Analysis
Let A be a set of real numbers. The number x is an accuDEFINITION
A if for all 6 > 0, N(x, 6) contains at least one point of
for
point
mulation
A distinct from x.
The set of accumulation points for A is called the derived set of A and is
denoted A’.
To prove that a point x is an accumulation point for a set A, we need to show
that for every 5 > 0, the “punctured” neighborhood N(x, 6) —
{x} contains a point
of A.
Example.
Proof.
‘
;
:
it
Prove that 0 is an accumulation point for the set A = {I— a i <r) et Let 5 > 0. By the Archimedean Principle, there exists an integer k such
that k > 4.Let m = 2k + 1. Then mis odd, +€ A, and m > {.But then >< 5 and
J.# 0, so i e€ N (0, 5) —
{0}. Because every neighborhood of 0 contains a point
of A distinct from 0, we conclude 0 is an accumulation point for A.
Example.
tu
Find the derived set for [3, 7).
Suppose that x is a real number. We consider several cases for x.
Casel.
3 <x <7. Let 5 > 0. Because x € (3, 7) and (3, 7) is open, there exists
fB > Osuch that N(x, 8) C (3, 7). Let y be the smaller of 6 and 8. Then
N(x, y) € (3, 7), and x + 4 is a point in N(x, y) that is in A and distinct
Case 2.
Case 3.
Case 4.
Case 5.
from x. Thus, if x is in (3, 7), x is an accumulation point for A.
x = 3. Let 6 > O. If 6 > 4, then 5 is a point of N(x, 5) that is in [3, 7)
and distinct from 3. If 6 < 4, then 3 +
is a point of N(x, 6) that is in
[3, 7) and distinct from 3. Thus, 3 is an accumulation point for A.
x = 7. By an argument very similar to that in part (ii), 7 is also an accumulation point for A. (See Exercise 1.)
x < 3. Let 6 =3 — x. Then N(, 5) and A are disjoint, so x is not an
accumulation point for A.
x > 7. Using reasoning similar to that in part (iv), x is not an accumula-
tion point for A.
The derivediset tor| 3.47) 1s 33 7].
is
Examples.
Let A = (2,6) U {9}. The number 9 is not an accumulation point for
A because, for example, NV (9, 0.5) contains no points of A other than 9. The derived
Set of Anisiv AN ==8[266)).
If B= {1.4, 1.41, 1.414, 1.4142, 1.41421, ...} is a set of successive decimal
approximations to V2, then B’ = { V2}. Other examples of derived sets of subsets of R are ((2, 6) U (6, 8))’ = (2, 8) and N’ = @.
Oo
7.3
The Bolzano-Weierstrass Theorem
365
The next theorem shows that every real number is an accumulation point of
@. The proof uses both the Archimedean Principle for R and the Well-Ordering
Principle.
Theorem 7.3.1
The derived set of the rationals is Q’ = R.
Proof.
Letrbe areal number and 5 > 0. We must show that N(r, 5) = (r — 5, r+ 8)
contains at least one rational number different from r. More generally, it can be
shown that every interval (x, y) contains rationals. It suffices to consider the case
¢
when x and y are both positive. (See Exercise 13.)
Let x and y be positive real numbers with x < y. Then y — x > 0, and by the
Archimedean Principle, there exists a natural number n such that n(y — x) > 1.
Therefore,
n > 0 and nx + 1 < ny. Also because N is not bounded above in R,
there exists a natural number k such that k > nx. Thus, the set A =
{te N:t > nx}
is nonempty.
By the Well-Ordering Principle, the set A has a smallest element m. Then
m > nx, and because m is the smallest element in A, m —
1 < nx,som
< nx + 1.
Therefore, nx < m < ny, which implies that x < ™ < y. Because m and n are integers andn > 0, = is a rational number between x and y.
By this reasoning, there is a rational number f in (7, r + 5), so N(r, 5) contains
at least one rational number different from r.
a
We see from our examples that an accumulation point for a set is not necessarily an element of the set and conversely that an element of a set is not necessarily
an accumulation point for the set.
We also see that for x to be an accumulation point for the set A, much more is
required than just that there is an element of A distinct from x that is within 6 of x. The next
theorem shows that there must be infinitely many points of A that are within 6 of x.
Theorem 7.3.2
A number x is an accumulation point for a set A if and only if for all 6 > 0, N(x, 5)
contains an infinite number of points of A.
Proof. If every neighborhood of x contains an infinite number of points of A, then
each neighborhood certainly contains at least one point of A distinct from x. Therefore, x is an accumulation point.
Now suppose that x is an accumulation point for A. Suppose that N(x, 6)
is finite for some
6 > 0. Let
uw= min{|x — y|:yE N(x, 6)N
A, x Fy}.
A
(Our
choice of x is so small that N(x, 4) will have no points ofA other than perhaps x
itself.) Then N(x, 4) A = {x}, which contradicts the initial assumption that x is
an accumulation point for A. Therefore, every neighborhood of x must contain an
a
infinite number of points of A.
From Theorem
7.3.2, it follows that no finite set can have any accumulation
points. The following theorem relates derived sets and closed sets.
366
CHAPTER7
Theorem 7.3.3
Concepts of Analysis
A set A is closed if and only if A’ C A.
Proof.
Suppose that A is closed and x € A’. If x ¢ A, then x € A‘, an open set.
Thus, N(x, 6) C A° for some positive 6. But then N(x, 6) can contain no points ofA.
Thus, x is not an accumulation point of A, so x ¢ A’, which is a contradiction. We
conclude that x € A. Therefore A’ C A.
Now suppose that A’ C A. To show thatA is closed, we show that A° is open. If A°
is not open, there is at least one x € A° that is not an interior point of A’. Therefore, no
6-neighborhood of x is a subset of A‘; that is, each 6-neighborhood of x contains
a point of A. This point must be different from x because x € A®. Thus x € A’. But
A’
CA, so x €A. This is a contradiction. We conclude that A 1s closed.
|
At the beginning of this section, we suggested that every bounded infinite set
sas , must have at least one accuthe Heine—Borel Theorem.
Theorem 7.3.4
The Bolzano-Weierstrass Theorem
Every bounded infinite set of real numbers has an accumulation point in R.
Proof.
Suppose that the set A is bounded and infinite but has no accumulation
points. Then
A’ = ©. Because
A’ CA,
A is closed.
Then
by the Heine—Borel
Theorem, A is compact.
Because A has no accumulation points, for each x € A there exists 6, > O such
that N(x, 6,)
NA =
{x}. Thus, ify € A andx # y, then y € N(x, 6,). But this means
the family {.N(x, 6,): x € A} is an infinite collection of open sets that covers A and
has no subcover other than itself. Hence, A has no finite subcover. This contradicts
the fact that A is compact. Therefore, A must have an accumulation point.
B
Exercises 7.3
1.
Prove that
(a) 7 is an accumulation point for [3, 7).
(b)
5 is an accumulation point for (4, 5) U (5, 6).
(c)
Ois an accumulation point for jest ine vt
*
(d)
eis an accumulation point for {(1+ 1)ie
n
2.
Find an example of an infinite subset of R that has
(a) no accumulation points.
(b) exactly one accumulation point.
}
;
(c)
1
J
‘
exactly two accumulation points.
=
sive
i]
n
ut
7.3
The Bolzano-Weierstrass Theorem
367
Find the derived set of each of the following sets.
(a) et nent
(b) (2% neN}
(c)
(d)
{6n:nEN}
iainen}
(e) O,1]
(NG, 2G, 8}
(2) {i+ olmen}
(hy @NO,1)
(i)
jdt
act as
th (j)
{sinx: xe (GFSh)=
ae
SS
(k)
{et iknen}
(1) {A:xvez}
Pets =O; ty Pmd Soa 6S).
Prove that if A C R, z = sup(A), and z € A, then z is an accumulation point
of A.
(a) Prove thatif
AC B CR, then A C B’.
(b)
Is the converse of part (a) true? Explain.
Let A and B be subsets of R.
(a) Prove that (A U BY = A’ UB’.
(b) Prove that (ANB)
CA NB’.
(c) Find a counterexample for (A 9 B)’ = A’/N B’.
Let A and B be sets of real numbers. Prove that
(a) if Bis closed and A C B, then A’ C B.
(b) A UA’ is closed.
(a) Prove that if x is an interior point of the setA, then x is an accumulation
(b)
(c)
(d)
10.
point for A.
Is the converse of part (a) true? Explain.
Prove that if $SC R is open, then every point of S is an accumulation
point of S.
Is the converse of part (c) true? Explain.
Which of the following must have at least one accumulation point?
(a)
(c)
an infinite subset of N
an infinite subset of [0, 100]
(b)
(d)
an infinite subset of (—10, 10)
an infinite subset of Q
(e)
ta ke nt
(f)
an infinite subset of QO [0, 1]
Let A be a set of real numbers. Prove that (A’)° € (A‘)’.
12.
Let A and F be sets of real numbers, and let F be finite. Prove that if x is an
accumulation point of A, then x is an accumulation point of A — F.
133
In the proof of Theorem 7.3.1 that Q' = R, it is asserted that to prove that
every interval (x, y) contains rationals, it suffices to consider the case when x
and y are both positive. Explain.
368
CHAPTER7
Proofs to Grade
Concepts of Analysis
14.
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
(a) Claim.
For A,B CR, (AUB) =A UB’.
(b)
*
(c)
“Proof.”
(i) Because
AC AUB, A’C (AUB) by Exercise 6(a). Likewise,
B’ Cc (A UBY. Therefore, A’ U B’ C (A UBY.
(ii) To show that (A UB)’ CA’ UB’, let x € (A UB)’. Then for all
5 > 0, N(x, 6) contains a point of A U B distinct from x. Restating
this, we say, for all 6 > O, that N(x, 6) contains a point of A distinct from x or a point of B distinct from x. Thus, for all 6 > 0,
N(x, 6) contains a point of A distinct from x, or for all 6 > 0,
N(x, 5) contains a point of B distinct from x. But this means x € A’
or x € B’. Therefore, x € A’ U B’.
@
Claim.
For A CR, (A°) = (A).
“Proof.”
x € (A’)° iff x EA
Claim.
“Proof.”
(d)
(e)
iff x is not an accumulation point for A
iff x 1s an accumulation point for A‘
iff x € (Ac).
ForA, BCR, (A — By) CA — B’.
r
=(ANB*)’
(definition of A — B)
CAM (BS) (Exercise 7(b))
GAT VU
ebecause(B°) GB )°)
= A — B’.
a
Claim.
[fA is closed, then A’ C A.
“Proof.”
Suppose that A is closed. Then A‘ is open. Let x € A°. Then x
is an interior point of A°. Therefore, there exists 6 > 0, so N(x, 6) C AS.
Hence, N(x, 6) A = ©. Thus, x is not an accumulation point for A.
Because x € A° implies x ¢ A’, we conclude A’ C A.
=
Claim.
(A — By
If A is a set with an accumulation point, B C A, and B is infi-
nite, then B has an accumulation point.
“Proof.”
First, A is infinite because B C A and B is infinite. Because
A has an accumulation point, by the Bolzano—Weierstrass Theorem A
must be bounded. Because B C A, this means B is bounded. Hence, by
the Bolzano—Weierstrass Theorem again, B has an accumulation point.
|
7.4
The Bounded Monotone Sequence Theorem
We showed in Section 4.6 that when a sequence of real numbers converges (has a
limit), then the limit is unique. In this section, we prove that the Bolzano—Weierstrass
Theorem implies that certain kinds of sequences of real numbers (bounded monotone sequences) must converge.
7.4
The Bounded Monotone Sequence Theorem
369
DEFINITIONS
Let x be a sequence of real numbers.
If there exists a real number M such that x, < M for all n € N, we say x is
bounded above (by /).
If there exists a real number M such that M < x, for all n € N, we say x is
bounded below (by M).
If x is bounded above and bounded below, we say x is bounded.
The sequence z given by z, = 1 + sinn for all n € N is bounded above by 2 and
bounded below by 0. We may, of course, use other bounds and say, for instance,
that z is bounded above by 13 and below by —2.
Remembering that a sequence x is a function, an equivalent way to say x is
bounded is to say its range {x,: n € N} is a bounded subset of R. For the sequence
z above, Rng (z) = {1 + sinn: n € N} has an upper bound of 2 and a lower bound
of 0. Because —2 is also a lower bound for z, we have that |z,,| < 2 for alln EN.
This is a convenient way to say that the sequence is bounded.
Theorem 7.4.1
A sequence x of real numbers is bounded if and only if there exists a real number B
such that |x,| < Bforallne N.
Proof.
See Exercise 5.
a
The sequence y given by y, = 2” is unbounded. Intuitively it seems that y must
diverge because its terms never approach any possible limit L. Our next theorem
confirms that every unbounded sequence diverges.
Theorem 7.4.2
If a sequence of real numbers converges, then it is bounded.
Proof. Suppose that x is a sequence convergent to the real number L. For e = 1,
there is a natural number N such that if n > N, then |x, — L| < 1. Because
I|x,,| — ILI < |x, — LI, we have for all n > N, |x,| — |L| < 1. Thus, for alln > N,
|x, |< |L| + 1. (All but the firstN terms are bounded by |L| + 1. We now find an
upper bound for those terms
as well.) Let B= max {|x,|, |X |,--., |Xy|,|L] + 1}.
Then |x,,| < B for all n € N. Therefore, by Theorem 7.4.1, x is bounded.
a
The proof shows that after the first few terms (that is, after N terms), the
remaining terms of the sequence must be close to the limit.
Example.
Let x be the sequence
(—2)”
y=
1<n<
ilbya
n+
1
n > 1000
1000
370
CHAPTER7
Concepts of Analysis
This sequence converges to 15 and therefore must be bounded. In this case, the
first “few” (1000) terms jump around before the terms settle in close to 15. The
sequence is bounded because for all n, |x,| < 210°.
Because a sequence is a function, a sequence may be increasing, decreasing, or
neither. (See Section 4.2.) The sequence y given by y, = 2” is increasing because
n < m implies 2” < 2”. The sequence whose terms are z, = e-” is decreasing
because e~” > e~" ifn < m. The alternating sequence 1, —1,.1,-—1, 1, ... 1s neither
increasing nor decreasing.
A proof that a sequence x is increasing (or decreasing) is similar to the proof
that a real valued function is increasing (or decreasing) on an interval /.
Example. The first few terms of the sequence x given by x, =
1234
;
ai.
—,——)—,..-. Prove that this sequence 1s increasing:
i) Ai Noes VS
Proof.
n
i
are
I
Suppose that m and n are natural numbers and n < m. Then
Mi
=
ne
1
n(m + 1) < m(n + 1), and
n
n+l
<
m
m+1
Ad
(because n + | andm + | are positive).
Therefore, x,, < x,,. Hence the sequence x is increasing.
E
A monotone sequence is one that is either non-decreasing or non-increasing.
Thus the sequence x is monotone if either x, S x,, whenever n < m, or x, = X,,
whenever n < m. All increasing and decreasing sequences are monotone. A constant sequence k, for which every term k, = c, where c is some real number c, is
monotone. The sequence 1, 1, 2, 2, 3, 3, 4, 4, ... is monotone but not increasing.
The sequence x above is not only increasing but also bounded above by 1.
Intuitively we think of the terms of the sequence as getting closer and closer to | or
possibly to some number less than |. (See Exercise 3.) In other words, we expect
the sequence x to converge. It is true that every bounded monotone sequence has a
limit, and this fact can be proved using the Bolzano—Weierstrass Theorem.
Theorem 7.4.3
The Bounded Monotone Sequence Theorem
Every bounded monotone sequence converges.
Proof. Assume that x is a bounded and non-decreasing sequence. (The proof in
the case where x is non-increasing is similar.)
If the range of x, {x,:n © N}, is finite, then let L = max {x,:n € N}. For
some NEN, x, = L, and because x is non-decreasing,
Therefore, x converges to L.
x, = L for all n > N.
7.4
The Bounded Monotone Sequence Theorem
371
Suppose that {x,,: 2 € N} is infinite. Then by the Bolzano—Weierstrass Theorem,
the bounded infinite set {x,,: n € N} must have at least one accumulation point.
Let L be an accumulation point for {x,: 1 € N}. (We will show first thatx, = L for
alln € N and second that x converges to L.)
Suppose that x, > L for some n € N. Because {x,,: 1 € N)} is an infinite set
and L is an accumulation point for {x,: 1 € N}, the number L is also an accumulation point for {x,:n > N}. (See Exercise 12 of Section 7.3.) However, if we
choose 6 = xy — L, then 6 > 0 and N(x, 4) contains no points of {x,:n > N}.
This is a contradiction. Therefore, x, < L for alln € N.
To show x converges to L, let ¢ > 0. Because L is an accumulation point
for {x,,:n € N}, there exists M € N such that x,, € N(L, €). But then x, € N(L, €)
for all n > M because L — ¢ < x,, < L < L + «. In other words, if n > M then
aL)
re, Theretores linn. —
a
n>
cw
One application of the Bounded Monotone Sequence Theorem is that it can be
used to establish the existence of some significant real numbers.
Examples.
One way to define the constant e is e =
lim (1 + . For this to
n>
©}
be a valid definition, we must show that the sequence x given by x,, = (1 si ai is
convergent.
Every term of x is positive, so x is bounded below by 0. The sequence x is
bounded above by 3 because, by the Binomial Theorem (Theorem 2.6.10(a)), for
any n,
Et
xn =
Lee 1
ia Il
n(n —
7
nae gee
=
or
1 n(n —
i
A
=
DN
n
1) 1
n(n —
ee
1)(n — 2) 1
bag
31
1)
1)(n — 2)
1 nin —
nu 3
ne
1
ee
1 n!
~
n! vale
iu
I
b+l+—o4+54¢--4+—
2
Oh
n!
HarteQn]
‘ 1 HorasGh
=e
2" — |
gn-l
<1+2
=i:
The sequence x is increasing. (See Exercise 8.) Therefore, by the Bounded
Monotone Sequence Theorem, the sequence x converges.
372
CHAPTER7
Concepts of Analysis
Exercises 7.4
For each sequence x, determine whether x is bounded, bounded above, or
bounded below.
10
(ip
$3,
0s
(b)
x, = --
(c)
ea *Oe
(di
ee
(e)
x, = oe
(Le
Ge
(g)
x, =ntan ey
(h)
x, =(-l)"n
10
(ya
0.9).
Cee
teh
Gen
a
(k)
n
slog),8
(m) x,=n+(-D'n
(I)
cosn
7
ey
xX, =
(eke
ay
(n) x,=n sin
L(=1)"
ar 1)
n
Give an example of
(a) a bounded sequence that is not convergent.
(b) an increasing sequence that is not convergent.
(c) aconvergent sequence that is not monotone.
(d) a divergent sequence x such that the sequence whose nth term is |x,,|
converges.
(e) an increasing sequence bounded above by 2 that does not converge to 2.
(f) an increasing sequence that converges to oe
Prove that if x is a convergent sequence with x, = B for all ne N, then
lim x, <B.
n—
CO
Prove Theorem 7.4.1.
Prove that if x, — 0 and y is a bounded sequence, then x,,y,, > 0.
Determine whether each sequence is monotone. For each sequence that is
monotone, prove your answer.
(a) x,= 7
2
(ce) x, =(n— 2)(n — 5)?
(cia
n+2
—
n!
Ons
(ete
SS
|(
i )
poe
=
(b) y,=27
@
»=4
(Dae
ae
2n —5
—
ae
n!
es
Vn Al
Give a proof of the Bounded Monotone Sequence Theorem for the case in
which the sequence x is bounded and non-increasing.
7.4
8.
Complete
The Bounded Monotone Sequence Theorem
the proof that x, = (1 + =) is increasing
by showing
373
that
Xy < X,4,forallne N.
9.
Let x be a bounded non-decreasing sequence. Use the completeness property
of the reals and the properties of suprema and limits to prove directly (without reference to the Bolzano—Weierstrass Theorem) that x converges. (Hint:
Consider the supremum of the set of terms of x.)
10.
Recall from Exercise 11 of Section 4.6 that the sequence y, is a subsequence of x,, if and only if there is an increasing function f: N — N such that
Yn = Xp). Prove that
11.
(a)
if x is bounded, then every subsequence of x is bounded.
(b)
if xis monotone, then every subsequence of x is monotone.
A sequence x of real numbers is a Cauchy* sequence if for every ¢ > 0,
there exists an integer M such that if m,n > M, then |x, — x,,| < ¢. That is,
terms in the sequence are arbitrarily close together if the terms are chosen far
enough along the sequence.
(a) Prove that if x is a Cauchy sequence, then x is bounded.
(b)
Prove that if x is a convergent sequence, then x is a Cauchy sequence.
(It can also be shown that every Cauchy sequence converges.)
12.
Let x and y be positive real numbers
b, =
with x > y. Let a, =
x
ca. and
Vxy. The numbers a, and b, are called the arithmetic and geometric
;
pats
means, respectively. For n > 1, let a,,; = ——— and b,,,; = .Va,b,.
(a) Use induction to show that for alln € N, a, > a,4, > Dp) > Dy
(b)
Let a and b be the sequences whose terms are a, and b,, respectively.
(c)
Show that both sequences a and b converge.
Show that lim a, = lim b,. This number is called the arithmeticn>
ow
nC
geometric mean.
Proofs to Grade
13.
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
(a) Claim.
Every bounded non-increasing sequence converges.
“Proof.” Let x be a bounded non-increasing sequence. Then y, = —x,,
defines a bounded non-decreasing sequence. By the proof of Theorem
7A.3, lim y, = Lfor some L. Thus, lim x, = —L.
a
noo
(b)
Claim.
nC
The sequence x, where x, = mn converges.
* Augustin Louis Cauchy (1789-1857) was a creative mathematician and pioneer in the efforts to bring
rigor to the infinitesimal calculus. He was the first to define complex numbers as a pair of real numbers.
Cauchy’s name is associated with concepts and results in many fields of mathematics, including geometry, analysis, and mathematical physics.
374
CHAPTER7
Concepts of Analysis
“Proof.”’
Because
Inn
Inn < n for all natural numbers n, apr < |. There-
eae
: a bounded sequence. The derivative of
fore, x is
Inn,
7
;
1 —Inn
is
a)=——, which
is less than 0 for every natural number n greater than e. Therefore, except
for the first two terms, x is a decreasing sequence. Because x is bounded
and decreasing, x converges.
7.5
:
=
Equivalents of Completeness
We began the chapter by stating without proof that the real numbers are a complete
ordered field. In subsequent sections, we (1) used completeness to prove that a set
of real numbers is compact if and only if it is closed and bounded, then (2) used that
property to prove that every bounded infinite set of reals has an accumulation point,
and then (3) used that property to prove that every bounded monotone sequence of
real numbers converges. In this section, we use the bounded monotone sequence
property to prove that R is complete. This result completes a cycle of implications
about the real numbers. See Figure 7.5.1.
When we finish this section, we will not have proved that the real numbers
have any of the four properties we have studied. Rather, we will have shown that
the completeness of R is equivalent to each of the other properties, so that we have
a deeper understanding of the meaning and importance of completeness.
Before we get to the main theorem, we need two lemmas about the convergence
of sequences. Their proofs are Exercises | and 2.
Completeness
Bounded Monotone
Sequence Theorem
NC
Heine-Borel
Theorem
Bolzano—Weierstrass
ae
Theorem
Figure 7.5.1
Lemma 7.5.1
If x and y are two
hens
lie
as
sequences
such that
lim Y, = 8 and
noo
lim (x, — y,) =0
n—> oo
<
nyo
Lemma 7.5.2
If x is a sequence with jim x, = s and ¢ is a real number such that ft < s, then
there exists N € N such that x, > ¢ for alln > N.
2
7.5
Theorem 7.5.3
Equivalents of Completeness
Suppose that R has the property that every bounded monotone
converge. Then R is complete.
Proof.
375
sequence must
Let A be a nonempty subset of R that is bounded above by a real number b.
To prove completeness, we must show sup(A) exists and is a real number. Because
A # {, we may choose a € A. If a is an upper bound for A, then a = sup(A) (see
Exercise 5(b) of Section 7.1), and we are done. Assume that a is not an upper bound
for A. If (a + b)/2 is an upper bound for A, let Siu ovandy,— (a = b)/2: if not,
let x; = (a + b)/2 and y,= D. In either case, y, — x, = (b — a)/2, x, is not an
upper bound, and y, is an upper bound for A.
Now if (x; + y,)/2 is an upper bound forA, let x, = x, and y,= (x, + y,)/2;
otherwise, let x. = (x, + y,)/2 and y,= y,. In either case, the result is that
Vy — X_ = (b — a)/4, xy > x,, and x, is not an upper bound forA, whereas y, < y,
and y, is an upper bound for A.
Continuing in this manner, we inductively define an a non-decreasing sequence
x such that no x,, is an upper bound for A and a non-increasing sequence y such that
every y, iS an upper bound for A. (If y,, = sup(A) for some m, the sequence y
is constant from that point on. See Exercise 3.) By the hypothesis, because y is
bounded below, y converges toa point s € R. In addition, y,,— x, = (b — a)/2",so
limy(y,—)x;,)= jimn ( = .a)i2"=.) Therefore, by. Lemma 7-5.
noo
ulimia as.
n>
We claim eas s is an upper bound for A. If z > s for some z € A, then z > yy
for some N (because lim | y, = 8). This contradicts the fact that yy is an upper
bound for A.
-
Finally, if fis a real number and ¢ < s, then t < x, for some N € N by Lemma
7.5.2. Because x, is not an upper bound for A, f is not an upper bound.
Thus, s is a real number that is an upper bound for A, and no number less than
s is an upper bound; that is, s = sup(A). Therefore, R is complete.
a
We saw in Section 7.1 that (Q, +, -) is an ordered field that is not complete.
Thus, all the properties described by the main theorems of this chapter must fail for
the rational numbers. We give an example for each property.
Example. The set A = {x € Q: x” < 2} is a bounded subset of @ that is closed
in Q. (The set of boundary points of A in @ is empty, so A contains all its boundary
points.) A is not compact because {(—x, x): x € A and x # 0} is a cover for A with
no finite subcover. This example shows that the Heine—Borel Theorem fails for Q.
Example.
The
set B=
{1, 1.4, 1.41, 1.414, 1.4142,...}
of (rational)
decimal
approximations of V2 is a bounded and infinite subset of @ with no accumulation
point in Q. Thus, the Bolzano—Weierstrass Theorem fails in Q.
oO
Example. The Bounded Monotone Sequence Theorem fails in Q. Let x be the
sequence whose terms are the elements of the set B in the previous example.
376
CHAPTER7
Concepts of Analysis
Thus, x, ='1, x, = 1:4, 23 = 1.41, and'so forth. Then x is a sequence of rational
numbers that is increasing and bounded above (by 3) but does not converge to any
Oo
rational number.
Why is completeness such a crucial property of the real number system? In
:
(Lg
Section 7.4, we saw that the sequence whose nth term is x,, = (1 af | is bounded
and increasing, so by the completeness of IR (via the Bounded Monotone Sequence
Theorem), the terms x,, must approach a unique real number, which is the number e.
The fact that important constants such as e must exist in R is a consequence of
completeness.
Not only must the limit of a bounded monotone sequence of rational numbers
be in R, but also the same is true for a bounded monotone sequence of irrational
numbers, or of rationals and irrationals. The completeness property and its equivalents assure us that every number that is a limit of a sequence of reals is in fact a
real number.
A proof that R is a complete ordered field requires a much more rigorous
definition of a real number than simply saying “all the decimal numbers on the real
line.” A rigorous definition requires construction of the reals from the rationals in
such a fashion that the completeness property holds. What this means is that we
must identify some set of objects based on the rational numbers; tell how to add,
multiply, and order these objects; and then show that all the properties of a complete ordered field hold for these objects.
One approach considers Cauchy sequences of rational numbers (see
Exercise 11 of Section 7.4). Two Cauchy sequences
{x,} and {y,} are equiva-
lent if the sequence |x, — y,| converges to zero. In this approach, R is the set
of equivalence classes of Cauchy sequences. For example, the real number
\/2 is represented by the equivalence class containing the rational sequence
1, 1.4, 1.41, 1.414, 1.4142,1.41421, 1.414213, 1.4142135, ... and all other Cauchy
sequences equivalent to this one. After carefully crafting the definitions of addition, multiplication, and the order properties for equivalence classes of Cauchy
sequences, one can show that the resulting system is a complete ordered field. See
Charles Chapman Pugh’s Real Mathematical Analysis (Springer, New York, 2002)
for an explanation of how to define addition and multiplication and why this system
forms a complete ordered field.
A different approach to constructing the reals from the rationals sets up twoelement partitions of ©, called Dedekind* cuts as a method for defining irrational
numbers. For example, the pair {A,,A,}, where A, = {x € Q: x < Oorx? < 2}
and A, = {x € Q: x > O and x* > 2}, is a cut that partitions Q into two sets: A, is
* Richard Dedekind (1831-1916) was Gauss’ last graduate student at the University of Gottingen. A
strong supporter of Cantor, he was noted for his work with infinite sets and axiomatic definitions of
number systems. He was the first to show that a set is infinite iff it is equivalent to one of its proper
subsets and among the first to point out the importance of groups in algebra.
7.5
Equivalents of Completeness
377
all rational numbers less than V2, and A, is all rational numbers greater than V2. The
cut {A,, A,} represents the real number V2. Again, one must carefully define addition
and multiplication of cuts and the ordering of the set of all cuts. It can be shown that the
set of Dedekind cuts forms a complete ordered field. See Walter Rudin’s Principles of
Mathematical Analysis, 3rd ed. (McGraw-Hill, New York, 1976).
With much work, we could show that R is essentially the only complete
ordered field. In Chapter 6, we discussed the concept of isomorphisms of algebraic
structures: one-to-one correspondences that preserve the algebraic structure. We
could apply this concept to ordered fields, and the end result would be that every
complete ordered field is isomorphic to the field of real numbers.
Equivalence classes of Cauchy sequences and Dedekind cuts give us vastly
different mental images of the real numbers, but the study of these approaches is the
means to explain why the real numbers possess the powerful properties described
in this chapter.
Exercises 7.5
1.
Prove Lemma 7.5.1.
Prove Lemma 7.5.2.
3.
For the sequence y defined in the proof of Theorem 7.5.3, explain why the
sequence is eventually constant if y,, = sup(A) for some m.
4.
Give an example of
(a) aclosed subset A of @ such that A C [7, 8] and A is not compact.
(b) a bounded infinite subset of @ M [7, 8] that has no accumulation point
in Q.
(c)
5.
a bounded increasing sequence x of rational
{x,:n € N}C [7, 8] and
x has no limit in Q.
numbers
such
that
For the set R — Q of irrational numbers, give an example of
(a) aclosed subset A of R — @ such that A C [3, 4] and A is not compact.
(b) abounded infinite subset of (R — @)/N
[3, 4] that has no accumulation
point inR — Q.
(c)
a bounded
increasing
sequence
x of irrational
numbers
such that
{x,:2 © N} C [3, 4] and
x has no limit in R — Q.
6.
Let J be a sequence of intervals. Then for each n € N, /, is an interval on the
real line. If /,,, CJ, for all n e N, the sequence / is called a sequence of
nested intervals.
(a)
Give an example of a sequence / of nested open intervals such that
Lae.
neN
(b)
Give an example of a sequence / of nested open intervals such that
() 1,=, 21.
neN
378
CHAPTER7
Concepts of Analysis
(c)
(The Nested Interval Theorem) Use the results of this chapter to prove
that if J is a sequence of nested closed intervals, then (a) [,, is nonempty.
(d)
Give an example of a sequence / of nested closed intervals such that
neN
() 7,=01, 21.
neN
(e)
Give an example of a sequence / of nested closed intervals such that
() 2, = {1}.
neN
Proofs to Grade
7.
Assign a grade of A (correct), C (partially correct), or F (failure) to each.
Justify assignments of grades other than A.
(a) Claim. If every bounded monotone sequence in the reals is convergent, then the reals are complete.
“Proof.”
Suppose that the reals are not complete. Then there is a
bounded infinite subset A of RR such that A has no supremum in R. Let
x, € A. Then x, is not an upper bound for A, or else x, would be the
least upper bound (because x, € A). Thus, there is an x, € A such that
X, < Xy. Likewise, x, € A, and x, is not an upper bound, so there exists
x3 € A with x, < x. Continuing in this fashion, we build an increasing
sequence X,, X>, %3,.... This sequence is bounded because it is a subset
ofA. Therefore, L = lim x, exists. Because L > x, for alln EN, Lis
n>
wo
the supremum of A. Therefore, sup(A) exists, which is a contradiction.
* (b)
Thus, R is complete.
m
Claim.
The Bolzano—Weierstrass Theorem implies the completeness
of R.
“Proof.”
Suppose that every bounded infinite subset of the reals has
an accumulation point. Let A be an infinite subset of R with an upper
bound ay. Then B = [0, a )] NA is a bounded set. If B is finite, then
B has a least upper bound, which is a least upper bound for A. If B is
infinite, then by the Bolzano—Weierstrass Theorem, B has an accumulation point a;, which, by construction, is the least upper bound of A. =
APPENDIX
Sets, Number Systems,
and Functions
As noted in the Preface to the Student, this Appendix includes important definitions
and notations that will be used throughout the text and reviews some basic concepts. You are not expected to have complete mastery of these topics at this time.
In fact, a main goal of the text is for you to gain a richer understanding of the topics
in each section of this Appendix.
I. Sets
A set is a collection of objects called the elements or members of the set. When
the object x is an element of the set A, we write x € A; if x is not a member of A,
we write x € A.
We can present a set by listing all its elements or by naming the property that
determines membership in the set. We can also list enough elements to establish
the membership criterion. For example,
{15, 16, 17,...} is the set of all integers
greater than 14.
Example.
The set K = {6, 7, 8, 9} has four elements. We see that 7 € K but 3 ¢ K.
We may also write the set K as
K = {x: x is an integer greater than 5 and less than 10},
which we read as “K is the set of x such that x is....”
Oo
The empty set © is a set with no elements. Note that © is not the same as
{@}. There is one object in {@}—that element is @. By the same reasoning, the
set whose only element is 5 is not the same as the number 5; that is, {5} #5.
A set is finite if it is empty or if it has n elements for some natural number n.
Otherwise, the set is infinite. The sets K = {6, 7, 8,9}, L = {—10, —9, —8,..., 18,
19, 20}, and M = {x: xis areal number and x? = —1)} are finite. K has 4 elements,
379
380
Appendix
Sets, Number Systems, and Functions
L has 31 elements, and M is the empty set. The set of all integers and the set of all
real numbers are infinite.
We say that A is a subset of B, and write A C B, if every element of A is an
element of B.
Examples. The set {7, 9} is a subset of K = {6, 7, 8, 9}. K has in fact a total
of 16 subsets, including {6}, {8, 9}, {6, 7, 8}, ©, and K itself. Some subsets of
G = (=10)-29) =8) 1.2, 18.919,.20 Watesl2 43,45 55S tou 10, 2041055, 107135}.
and {—9, —7, —5, ..., 15, 17, 19}.
Oo
If sets A and B have exactly the same elements, we say they are equal and
write A = B.
Examples. The set {x: x is a solution to x* = 25} is equal to {5, —5}. All three sets
\a,Di G, ad}, (c,d, bya}, and. (a, b,c b,d\ are equal.
Exercises I
Use these exercises to check your understanding. Answers appear at the end of the
Answers to Selected Exercises.
1.
Write each set in two ways: by listing its elements and by stating the property
that determines membership in the set.
(a) The set of integers between 6 and 12
(b) The set of integers whose square is less than 17
(c)
The set of solutions to x — 81 =0
(d)
(e)
The set of integer powers of 2
The set of ingredients in a peanut butter and jelly sandwich
2.
What is the set of all real numbers x such that (a) x7 = 4?
3.
Which of these sets are finite?
(a) The set of grains of sand on Earth
(b) The set of integer powers of 2
(c) The set of three-letter words in English
(d)
4.
(b) x? = —4?
The set of solutions to the equation 3x!7 — 5x? — 437 = 0
(e)
The set of real numbers between 0 and 1
(a)
(b)
List all eight subsets of the set A = {3, 5, 7}.
LetA = {j, m, h}. Explain why {A} is not a subset ofA.
ll. Number Systems
We use these notations for the number systems in this text:
N = {1, 2, 3,...} is the set of natural numbers.
Z={...—3, —2, —1,0, 1, 2,...} is the set of integers.
Il. Number Systems
381
Q is the set of all rational numbers.
R is the set of all real numbers.
C is the set of all complex numbers.
The Natural Numbers
The properties below describe the basic arithmetical and ordering structure of the
natural numbers.
1.
Successor properties for N\.
1 is a natural number.
Every natural number x has a unique successor x + 1.
1 is not the successor of any natural number.
2.
N is closed under addition and under multiplication.
The sum of two natural numbers is a natural number.
The product of two natural numbers is a natural number.
3.
Addition and multiplication are associative.
For allx,y,z
EN x+Q+z2=(+4+y)
+z.
Forallx,
y,z EN, x-Q-D=(-y)-z.
4.
Addition and multiplication are commutative.
For allx,y EN, x+y=y+x.
For all x, y E N, xy = yx.
5.
Distributivity.
For all x, y,z
For allx, y,z
6.
EN, xy + 2) = xy 4+ xz.
EN, W+2x=yx4+u.
Cancellation.
For all x, y,z
EN, if x+z=y+z, thenx=y.
For all x,y,z, © IN) if xz = yz then sy;
7.
The number 1 is the multiplicative identity for N.
For allxeN, 1-x=x.
For natural numbers a and b, we say a divides b if there is a natural number k
such that b = ak. When a divides b, we also say a is a divisor of b, or a is a factor
of b, or b is a multiple of a. Thus, 4 is a divisor of 36, and 36 is a multiple of 4. The
natural numbers 1, 2, 3, 4, 6, 9, 12, 18, and 36 are all divisors of 36.
A natural number p is prime if p is greater than 1 and the only natural numbers
that divide p are 1 and p. The set of prime numbers is infinite. There are 21 primes
less than 75:
eo
eal
elo Ml Onc one
oS
oi ea
Ld, Ae d.9,.01,,0/5,/1, and 73.
A composite is a natural number that is neither 1 nor prime. Therefore, to be
composite, a number must have at least one divisor other than | and itself. We saw
above that the composite 36 has several divisors other than | and itself. Using these
382
Appendix
Sets, Number Systems, and Functions
divisors, we can write 36 as a product in various ways, such as 36 = 4 - 9, 36 =
2-18, or 36 = 2-3-6. However, there is only one way to write 36 as a product of
its prime divisors: 36 = 2-2-3-3 = 27-37. The same is true for every composite
number.
The Fundamental Theorem of Arithmetic
Every natural number larger than | is prime or can be expressed uniquely as a
product of primes.
Example.
2,200 = 23- 57- 11. Although the prime divisors could be written in a
different order, there will always be three 2’s, two 5’s, one 11, and no other prime
divisors of 2,200.
Oo
The Integers
The set of integers,
Z = {...—3, —2, —1, 0, 1, 2,3,...}, shares properties 2 through
7 listed above for the natural numbers, except that when z = 0, xz = yz does not
imply x = y. We say x is positive when x > 0 and x is negative when x < 0.
Other important properties for Z are as follows:
For all x in Z, x < 0, x = 0, or x > 0. (Trichotomy property)
For allxinZ,x+0=x,
x+(—x)=0,andx-0=0.
For all x in Z, there is y € Z such that x + y = 0. (Every integer has an additive
inverse.)
For all x, y, zin Z, if x< y and zis positive, xz < yz
For all x, y, zin Z, if x < y and zis negative, xz > yz
For all x, y in Z, if x and y are both positive or both negative, then xy is positive.
For all x, y in Z, if one of x or y is positive and the other negative, then xy is
negative.
The definition of “divides” for natural numbers extends to the integers: For
integers a and b with a # 0, we say a divides b if there is an integer k such that
b = ak. (When we say a divides J, it is understood that a ~ 0.) Thus, 4 divides —36
because —36 = 4 - (—9), and 12 divides 0 because 0 = 12 - 0.
Real and Rational Numbers
We think of the real numbers as being all the numbers along the number line,
stretching infinitely far both left and right. Each real number is a point on the line
and can be represented as an integer together with a finite or infinite decimal part.
The real numbers include terminating decimals such as 7/16 = 0.4375, repeating
decimals such as —13/11 = —1.1818181..., and nonrepeating decimals such as
€ = 2.718281828459045....
For real numbers a and b with a < b:
(a, b) = {x: x © IR and a < x < b} is the open interval from a to b.
la, b] = {x: x € R and a < x < Db} is the closed interval from a to b.
Il. Number Systems
383
(a, 00) = {x: x € R anda < x} and (—oo, b) = {x:
x € Rand x < b} are
open rays.
[a, 00) = {x: x € R and a < x} and (—o, b] = {x: x € R andx < b} are
closed rays.
Note that the infinity symbol “oo” is simply a notational convenience and does
not represent any real number. Do not confuse (1, 6) with either {1,6} or {2, 3, 4, 5}.
The interval (1, 6) is the set of all real numbers between | and 6 and contains, for
example, 2, 7, V13, and ae
A real number x is rational if there are integers p and g, with g # 0, such that
x = p/q. The set © of rational numbers contains exactly those numbers along the
number line that have terminating or repeating decimal expressions. All other real
numbers are irrational. In Chapter 1, we prove that V2 is irrational.
The number systems R and @ share many of the arithmetic and ordering properties of the naturals and integers, along with the property that every nonzero real
number has a multiplicative inverse:
For all x in R except 0, there is y € R such that that xy = 1.
Complex Numbers
A complex number has the form a + bi, where a and b are real numbers and
i = V—1. Because every real number x can be written in the form x + Oi, R is a
subset of C. Complex numbers do not share the ordering properties of the reals.
Arithmetic operations are performed the same way that two binomial expressions are combined, remembering that i? = —1. The conjugate of a + bi is a — bi.
When a complex number is multiplied by its conjugate, the result is a real number:
(a + bi) - (a — bi) = a? + Bb.
Complex numbers provide solutions for equations that cannot be solved in the
real number system.
The Fundamental Theorem of Algebra
Every polynomial in one variable with complex coefficients and degree n > 0 has
exactly n zeros in the complex number system.
Example.
(twice),
The four solutions of the equation x4 + 4x7 + 13x? = 0 are x = 0
+= —2 + 3i.and x4= —2, — 33.
o
Exercises II
1.
2.
Write each number as a prime or as a product of primes. If the number is
composite, list at least eight divisors.
(a) 672
(b) 673
(c) 675
Use the definition of “divides” to explain (a) why 5 divides 65 and (b) why 7
does not divide 23.
384
Appendix
Sets, Number Systems, and Functions
3.
True or false. When the answer is false, explain your reasoning.
(b) (5+ 2i)(5 — 21) is a real number.
(a) Ois anatural number.
(d) QOCZ.
(cia:
(e) [2,5] ={2,3,4, 31
(£)eeal32, 4.7] € (3.2, 4-7).
(2) (6;,9] C16; 10).
(h)
(G:2.4.7) © [3.2,74.7].(i) (7, 10GB:
Cima eer OSs. 95 10},
(kK) 105.050:
d)
(-—8,5) Cc (-2, 6).
(m) There are exactly three prime numbers between 40 and°50.
4.
Name the properties of number systems that enable us to write
5.
(a)
6x+ 3x = 9x.
(b)
7(6x + x) = 7x + 42x.
(c)
(3x)(2y)=6xy.
(d)_
forall real numbers x, either x* = 0 or x? > 0.
Which of the following equations have exactly eight solutions, counting multiplicities, in the complex numbers?
(a) 1=0
(b) x®-1=0
(c))
7
+74 4+-7=0
(d).
G2+1)@+1) =0
lll. Functions
A function is a rule of correspondence that associates to each element in a set A
a unique element in a second set B. The features that make fa function from A to
B are that every element of A must have a corresponding element in B and, most
importantly, that no element of A has more than one corresponding element in B. It
is this single-valued property that makes functions so useful and enables us to write
J(@ = b whenever a € A and the corresponding element of B is b.
The set A is the domain of f. The elements of A are called the arguments or
inputs of the function. The set B is the codomain of f. If f(a) = b, we say that b is
the image of a or that b is the value of the function f at a. We also say that a is a
pre-image of b.
The set Rng (f) = {ftx): x € A} of all images under the function f is called
the range of f. Because all images under f are in B, the range of fis a subset of the
codomain B.
Example. Let g be the function from the reals to the reals given by g(x) = x7 + 1.
The function g assigns to each real number x the number that is one more than the
square of x. Because g(3) = 10, the value of g at 3 is 10. The pre-images of 10 are
—3 and 3. The domain of g is R. The range of g is [1, 00).
Oo
In some settings, a function is described by giving only the domain and a
rule. For functions whose domains and codomains are subsets of R, the domain
is sometimes left unspecified and assumed to be the largest possible subset of R
for which image values may be obtained. With this assumption, the domain of
h(x) = Vx + lis [—1, oo) because this is the largest set of real numbers for which
Vx + 1 may be calculated.
Ill. Functions
385
Exercises Ill
1.
A rule of correspondence assigns to each person in Nashville, Tennessee, that
person’s 10-digit telephone number. Explain why this correspondence is not
a function.
For each function, find the value of fat 3 and the image of 5. If possible, find
pre-images of 0 and —2. Then give the domain and range of the function.
(a) f(x) =2x+6
(b) fx) =2x* -2
(ce) fx) =Va-4)
CH)
Seyes:
LetA be the set {1, 2, 3,4} and B = {0, 1, 2,3}. Give a rule of correspondence
from A to B that is
(a)
nota
(b)
(c)
a function with range B.
a function with range {0, 1}.
function.
17
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ep uli papebae
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(Stig ad
Pou
Gone
al
oA which
Answers to Selected Exercises
Exercises 1.1
Ts(c)
(f)
2.
(b)
not a proposition; the symbol x acts as a variable
a true proposition
P A Qis false. P V Q is true.
PH QOis false. P Vv Os false:
Para
(c)
te
T
F
P
F
4
Dir
F
F
SON
SEIN AD
a
F
ie
F
ar
ai
F
F
F
F
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F
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F
(e)
PO
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(i)
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[oe ae
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387
388
Answers to Selected Exercises
4.
5.
6.
7.
(a)
(h)
false
(c) true
(f) false
(g) true
The truth tables for ~ (P A Q) and ~P Vv ~Q are identical, as shown in
the fourth and seventh columns of this truth table.
I
QO
PAO
~(P A Q)
~P
~Q
~PV
li
F
Tr
F
T
I
F
F
TT
F
F
F
F
L
it
1
F
ae
F
i)
F
F
I
ly
F
1
a
if
~Q
Thus, ~( P A Q) and ~P Vv ~@Q are equivalent propositional forms.
(a) not equivalent
(c) not equivalent
(c)
8.
(a)
9.
(a)
(c)
(PV Q)AR, where P is “Caesar was born in 1492,” Q is “Caesar was
born in 1493,” and P is “Caesar died in 1776.” The statement is false.
Because P is equivalent to Q, P has the same truth table as Q. Therefore,
Q has the same truth table as P, so Q is equivalent to P.
The forms are not necessarily equivalent.
The forms are equivalent. P \ S has value T exactly when P and S both
have T. Q has the same truth table as P, and R has the same truth table
as S, so this happens exactly when Q and R both have the value T, which
happens exactly when Q A R has value T. Therefore, the truth tables for
10.
(c)
P A Sand Q A R are the same.
tautology
P
Q
Tie.
F
iy
F
PAQ
~Pv~Q-
(PAQ)V(~PvV
~Q)
| |T
F
a
F
F
F
Tr
1
lg
it
i
T
T
F
F
11.
(a)
xis not positive. (The statement “‘x is negative” is incorrect.)
IS
(Cc)
(e)
(a)
3 = 3.
Roses are not red or violets are not blue.
PO.
a OO
i
ae
In
ae
wv
8
Ee
F
T
Tr
F
Exercises 1.2
1.
(a)
(d)
(f)
Antecedent: squares have three sides.
Consequent: triangles have four sides.
Antecedent: fis differentiable.
Consequent: fis continuous.
Antecedent: fis integrable.
Consequent: fis bounded.
Answers to Selected Exercises
(i)
(a)
(d)
(f)
(i)
(a)
cdl (a)
(b)
(c)
10.
11.
(a)
(c)
(e)
(b)
389
Antecedent: An athlete qualifies for the Olympic team.
Consequent: The athlete has a time of 3 minutes, 48 seconds or less.
Converse: If triangles have four sides, then squares have three sides.
Contrapositive: If triangles do not have four sides, then squares do not
have three sides.
Converse: If fis continuous, then fis differentiable.
Contrapositive: If fis not continuous, then fis not differentiable.
Converse: If fis bounded, then fis integrable.
Contrapositive: If fis not bounded, then fis not integrable.
Converse: A time of 3 minutes, 48 seconds or less is sufficient to qualify
for the Olympic team.
|
Contrapositive: If an athlete records a time that is not 3 minutes and 48
seconds or less, then that athlete does not qualify for the Olympic team.
true
true
P
(c)
true
Q
~P
ea)
One
a
F
i
F
ay
Tt
F
FP
F
Tr
F
T
a
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Ag
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PQ
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AP
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T
i
T
T
it
F
I
f has arelative minimum at xo A f is differentiable at x9 > f'(xo) = 0.
aI
ees
Xo is a critical point for f <= f’(xo) = 0 V f'(xo) does not exist.
There are three nonequivalent ways to translate the sentence, using the
symbols D “The Dolphins make the playoffs” and B “The Bears lose all
the rest of their games.”
DiS BreOt( Gb haa)
B= Dor (~D) = (—B)
Di= Bor (B) S&S (~D)
The conditional meaning of unless (the first translation) is preferred, but
the speaker may have intended any of the three.
12.
(b)
Pi
OmeR
tl
il
tah
sl
af
Teal
Taa|
as)
teal)
(Slt)
as]
sl eal
ta ag]
al
PS)
las]
fa}
a9)
fl
‘sal
Po OmeP INO
— Rapa hieoO,
Pee R
(PN
oR)
0
Fl)
Sal
ssl
Sh
il
12s)
ls]
I
'25|
‘sal
‘sl
Jeol!
il
fs['tzs}
yl
el
sl
Sl
as)
tl
Pal
Is)
Is}
aa)
‘ssf
JS]
sa|
(aa)
ag|
Sh
fash)
bts)
sf
Sh
Ia
asl
aol
aml
SS)
iS)
sl
al
390
Answers to Selected Exercises
13.
(a)
Because the fifth and ninth columns are the same, the propositions
PA Q=> Rand (P A ~R) => ~Q are equivalent.
If6 is an even integer, then 7 is an odd integer.
16.
(c)
(a)
not possible
tautology
1.
(a)
(b)
Gi)
(k)
(m)
~(Vx)(x is precious => x is beautiful).
Or (Sx)(x is precious and x is not beautiful).
Hint: This exercise is not the same as I(a).
(Wea xe ZSx>—4V x < 6)or(VxeE Z)a > —4 V x < 6).
~(Ax)(Vy)(x => y) or (¥x)(4y)@ < y).
(4x)(x is a positive integer and x is smaller than all other positive
integers). Or (4x)(x is a positive integer and (Vy)(y is a positive
Integer ——w = V)))
(n)
(Wx)(~(Vy)(& loves y)). Or ~(4x)(Vy)(x loves y).
(a)
(ij)
(Vx)(x is precious => x is beautiful). All precious stones are beautiful.
(dx eZ) < —4 A x < 6). There is an integer that is both less than or
equal to —4 and greater than or equal to 6.
(dx)(Vy)@ => y). There is an integer that is greater than or equal to every
(d)
neither
Exercises 1.3
2.
(k)
integer.
(Vx)(x is a positive integer > (Jy)(y is a positive integer A x > y)). For
every positive integer there is a smaller positive integer.
Or ~(4x)(xis a positive integer A (Vy)(yis
apositive integer > x < y)).
There is no smallest positive integer.
(n) (4x)(Vy)~@ loves y). Someone loves everyone.
@) CAVxe 6A SxS 73).
The first interpretation may be translated as
(m)
a
5.
(Vx)[x is a person => (Vy)(y is a tax => xdislikes y)].
6.
(a)
T,U,V, and W
7.
(b)
Hint: Every sentence of the form P(x) is equivalent to
~~ P(x). Use this
fact to rewrite (Vx)(~A(x)) and then simplify by using part (a).
8.
9.
(b)
(b)
(d)
true
(e) false
(h) true
Only one real number is both nonnegative and nonpositive.
10.
(a)
true
11.
(a)
Hint: Begin by supposing that U is any universe and A(x) is an open
(b)
Hint: You must name a specific universe and a specific open sentence
such that the converse is false.
(V¥x)(~AQ)) Vv (Ay)(Sz)(AQ) A A(z) A y #2)
Not always true. Two polynomials are unequal if and only if they differ
in at least one coefficient. That coefficient might not be the coefficient
LOE
There is exactly one realnumber whose natural logarithm is 1.
(d)
false
(f)
false
(i)
false
sentence.
(e)
12.
(a)
Answers to Selected Exercises
13.
391
(d)
14.
This statement is not a denial. It implies the negation of (4!x) P(x), but
if (Vx) ~ P(x), then both the statement and (4! x) P(x) are false.
For every backwards E, there exists an upside down A!
1.
(a)
Exercises 1.4
Suppose that (G, *) is a cyclic group.
Thus, (G, *) is abelian.
Therefore, if (G, *) is a cyclic group, then (G, *) is abelian.
4.
(a)
5.
(h)
The crime took place in the library, not the kitchen. By fact (i), if the
crime did not take place in the kitchen, then Professor Plum is guilty.
Therefore, Professor Plum is guilty.
Proof. Suppose that x is even and y is odd. Then x = 2k for
some integer k, and
y=2j+ 1 for some integer j. Therefore,
x+y
6.
=2k
+ Qj + 1) = 24 +7) + 1, which
is odd.
(We
use the
(i)
fact that k + 7 is an integer.)
Hint: Use the words without loss of generality. You may use earlier
parts of this exercise.
(j)
(d)
Hint: Consider cases. You may use earlier parts of this exercise.
Hint: The four cases to consider are case 1, in which a => 0 and b > 0;
case 2,in whicha < Qandb < 0; case3,in whicha > Oandb < 0; and
case 4, in which a < Oandb = O. Incase 3, it is worthwhile to consider
two subcases: In subcase (i), a + b => 0, so that |a + b| = a+ bd; in
subcase (ii), a + b < 0, so that |a + b| = —(a + b). Now, in subcase
(i), we have |a+ b| = a+b <a (from b < 0) and a < a+ (—b)
(from 0 < —b). Thus, |a + b| < a + (—b) = |a| + |b|. Subcase (11)
is similar. Case 4 is the same as case 3 except for the names of the vari-
7.
(g)
ables a and b.
Hint: In the case when a < 0 and b > 0, rewrite |a — b| by replacing
a and b with expressions involving absolute values: |a| = —|a| and
(b)
b = |b|. Then use the Triangle Inequality (Exercise 6(d)).
Proof. Let a be an integer. Assume that a is even. Then a = 2k for
some integer k. Therefore, a + 1 = 2k + 1, soa + 1 is odd.
(d)
(g)
Proof. Hint: Let abe an integer. Use the fact that a is either even or odd
to give a proof by cases. It is acceptable, but not necessary, to use the
definitions of even and odd in proving these cases: Previous exercises
have laid the foundations we need.
Proof. Suppose that a and b are positive integers and a divides b.
Then for some integer k, b = ka. (We must show that a < b, which is
the same as a < ka. To show a < ka, we could multiply both sides of
1 < k by a, using the fact that a is positive. To do that, we must first
show 1 <k.) Because b and a are positive, k must also be positive.
Because k is also an integer,
soa <b.
| < k. Therefore,a=a-l1<a-k=b,
392
Answers to Selected Exercises
(i)
(f)
(a)
Proof. Suppose that a and b are positive integers and ab = 1. Then a
divides 1 and b divides 1. By part (g), a < 1 and b < 1. But a and b are
positive integers, soa = | andb = 1.
Hint: Begin working backward by factoring the inequality.
Proof. Suppose that A > C > B > 0. Multiplying by the positive
numbers C and B, we have AC > C? > BC and BC > B?, so AC > B?.
ACis positive, so 4AC > AC. Therefore, 4AC > B2,s0 B? — 4AC < 0.
11.
(c)
Thus, the graph must be an ellipse.
F. This proof, although it appears to have the essence of the correct
reasoning, has too many gaps. The first “sentence” is incomplete, and
the steps are not justified. The steps could be justified either by using the
definitions or by referring to previous examples and exercises.
C. The order in which the steps are written makes it look as if the author
i
of this “proof” assumed that x + ae 2. The proof could be fixed by
beginning with the (true) statement that (x —
1
the conclusion that x + - >.
1)? > 0 and ending with
(d) F, This is not a proof of the statement. It is a proof of the converse of the
statement.
Exercises 1.5
(a)
Suppose that (G, *) is not abelian.
Thus (G, *) is not a cyclic group.
Therefore, if (G, *) is a cyclic group, then (G, *) is abelian.
(c)
Suppose that the set of natural numbers is finite.
Therefore statement Q.
(e)
Therefore statement ~Q.
This is a contradiction.
Therefore, the set of natural numbers is not finite.
(i) |Suppose that the inverse of the function f from A to B is a function
from B to A.
Therefore f is one-to-one.
Therefore fis onto B.
Therefore fis one-to-one and fis onto B.
(ii) Suppose that fis one-to-one and fis onto B.
Therefore, the inverse of the function f from A to B is a function from
BtoA.
Answers to Selected Exercises
(a)
393
Proof. Suppose that the integer x + 1 is not odd. Then x + 1 is an
even integer. Thus, there exists an integer k such that x + 1 = 2k. Then
x = 2k — 1 = 2(k — 1) + 1, so x is not even. We have shown that if
x + 1 is not odd, then x is not even. Therefore, if x is even, then x + 1
(e)
(b)
is odd.
Hint: The contrapositive statement for “if x + y is even, then either x
and y are odd or x and y are even” is “‘if it is not the case that either x and
y are odd or x and y are even, then x + y is not even.” This is equivalent
to “if either x is even and y is odd or x is odd and y is even, then x + y
is odd.”
Proof. Suppose that it is not true that 2 < x < 3. Then either x < 2 or
x > 3.Ifx < 2,thenx — 2 < Oandx — 3 < O(becausex — 3 < x — 2).
Because the product of two nonpositive numbers is nonnegative,
(x — 2)(x — 3) = x* — 5x + 6 => O. In the other case, if x > 3, then
x — 3 => Oandx — 2 > 0. Therefore, (x — 2)(x —3) = x* — 5x +620.
In either case, x7 — 5x + 6> 0. We have shown that if x <2 or
x = 3, then x? — 5x 45.61
OTherefore: if x7— 5x + 6 = 0: then
(b)
Deh
Proof. Suppose that a and b are positive integers. Suppose that ab is
odd, and suppose that a and b are not both odd. Then either a is even, or
b is even. If a is even, then ab is even by Exercise 5(b) of Section 1.4.
(b)
If b is even, then ab is even by the same exercise. Either case leads to
a contradiction to the hypothesis that ab is odd. Therefore, if ab is odd,
then both a and b are odd.
Hint: We can prove “P if and only if Q” by proving “~P if and only
it ey”
A.
(a)
(c)
Choose m = —3 and z = 1, Then. 2m -- 7n = 1.
Suppose that m and n are integers and 2m + 4n = 7. Then 2 divides
(d)
12.
Exercises 1.6
2m, and 2 divides 4n, so 2 divides their sum 2m + 4n. But 2 does not
(f)
(h)
(a)
(h)
(i)
divide 7, so this is impossible.
Hint: See the statement of part (d). Can you prove that m and n are both
negative whenever the antecedent is true?
Hint: Use the fact that k(k + 1) is an even integer for every integer k.
Proof. Assume that a divides b — 1 and c — 1. (We must show that
be — 1 is a linear combination of b — 1 and c — 1.) Then a divides the
product
(b — 1)(c — 1) = be —b—c + Land the sum (bec —-b-—c
+1)
+ (b — 1) = be — ¢, Therefore, a divides the sum (bc — c) + (¢ — 1) =
be — 1.
Hint: For a counterexample, choose x = 1. Explain.
Hint: For a proof, choose y = x. For a different proof, choose y = 1.
394
Answers to Selected Exercises
(c)
Proof.
Let n be a natural number. Then both 2n and 2n + | are natural
numbers. Let M = 2n + 1. Then M is a natural number greater than 2n.
Hint: Let (0, 6) be the open interval of times between 9 a.m and 3 p.m. Suppose that two people walk the trail simultaneously: one going up the trail at
the exact same pace as the hiker on Monday and the second going down the
trail at exactly the pace set by the hiker on Tuesday.
(a) The implication says that if some object has property P, then all objects
have property P. If the universe is all integers and P(x) is the sentence
“x is odd,” then P(5) is true, and P(8) is false. Thus, (Ax)P(x) is true, and
(Vx)P(x) is false, so the implication fails.
(c)
The implication says that if every object has property P implies that
every object has property Q, then every object that has property P must
also have property Q. Let the universe be the integers, and let P(x) be “x
is odd” and O(x) be “x is even.” Because (Vx)P(x) is false, (Vx)P(x) >
(Vx)QO(x) is true. However, (Vx)[P(x) > QO(x)] is false.
(a)
(b)
(d)
(i)
F. The false statement referred to is not the opposite (denial) of the claim.
C. The “proof” shows that there is a polynomial with the required properties, but it must also show that there is no other polynomial with these
properties.
A.
Hint: The grade should be C. What error must be corrected?
Exercises 1.7
(a)
Proof. Suppose
that the negative
real number
a is a solution
to
x’ — x — 6 = 0. Then by the definition of solution, a2 — a — 6 =
(a — 3)(a + 2) = 0. Thus, a = 3 or a = —2, and because a is negative,
a = —2. Then a’ + 2a’ +.a+2=(—2)' + 2(-2)? + (-2)
+2 =0, so
ais a solution to x? + 2x7 +x+2=0.
(b)
Proof.
Suppose that x is a real number such that 0 < x < 3. Thus, x > 0
and ~= 3,so0 x(x — 3)= 0. Thenx° — 3x = 0, sox” —2x +1 <x +1.
Therefore, x + 1 > (x — 1)”.
(c)
Proof. Assume thataand bare positive integers such that a # 0. Suppose
that there is a positive integer x such that ax* + bx + b — a = 0. Then
eae: me
OES
es MAG = 20) =
i OS a)
BN;
a
1 or
2b
2
a
:
:
‘
ere
2a
* “ Using again the fact thatxis a positive integer, we find
2
that x = 1 — 2and a divides b. Therefore, if a does not divide b, then
there is no positive integer x such that ax* + bx + b —a=0.
(d)
Proof. Suppose that x is a real number and x > 1.
Case 1. x — 2 < 0. Because x > 1, we have 2 < 4x. Then |x — 2| =
EBs 9h
ox, so (using the fact that x is positive) list Nee
x
Answers to Selected Exercises
Case 2.
395
x—2>
0. Because xis positive, we have Fx <2,so0 |x —2|=
x—-2<
a so (using the fact that x is positive) sna 4 al
In every case, oa — 21
l<4,
(e)
Proof. Suppose that there is point (a, b) inside the circle (x — 3)? +
y* = 6 that is on the line y = x + 1. Then
b = a + 1 and (a — 3)?+
(a+ 1) <6, soa? —6a+9
+ a?+ 2a+ 1 < 6. From this inequal-
ity, we have a? — 2a + 2 = (a — 1)? +1
< O. This is a contradiction,
because (a — 1) is never negative. Thus, no point inside the circle is on
the line.
(f)
Proof. Suppose that x7 = 2x + 15 and x > 2. Then (x — 5)(x + 3)=
0. Because x > 2, x must be 5. Then x — 4 and x — 3 are positive, so
(x — 4)/@ — 3) > 0.
(g)
Proof. Let x be a real number. Then —x? + 7x — 10 < 0
if and only if x7— 7x +10 > 0
if and only if (x — 5)@ — 2) > 0
if and only if x < 5 andx <20rx>S5andx>2
if and only if x = 2 orx > 5.
(h) Proof. Let L, and L, be nonvertical lines. Suppose that L, and L, are
perpendicular. (Recall that the slope of a nonvertical line is tan(a@),
where a is the angle of inclination of the line.) Let a, and a be the
angles of inclinations of L, and L,, respectively. See the figure. Without
loss of generality, we may assume that a, > a. Because L, and L, are
perpendicular, a, = a) + a Therefore,
tan (@,) = tan (@, + =) = ley)
= =
ne
3
Thus, tan (@,) - tan (@,) = —1, so product of the slopes is —1.
y
A
>X
(i)
Proof.
(j)
12.9664991 < 13.
Proof.
(i)
Let (x, y) be (—0.001, 3.9901). Then (x — Bye
Letx=2andy=1.Then0
<x <2,0<y<1,and
3(2)? + 2(1) = 14> 14.
GQ = 2° =
3x4 2y*=
396
Answers to Selected Exercises
(ii)
3.
5.
(d)
(c)
9.
(a)
Suppose that 0 < x < 2,0 < y < 1, and 3x° + 2y’ > 14. Then
x* < 4andy’ < 1, so 3x7 + 2y’ < 14. If eitherx < 2 ory <1, then
3x? + 2y? < 14. Therefore x = 2 and y= 1.
Hint: Write n°? — nas the product of three consecutive integers.
Hint: Is it possible that for all irrational numbers x and y, x + y 1s irra-
tional? Or could x + y = 0?
Proof.
Suppose that (x, y) is inside the circle. Then from the distance
formula,
Wied
Seid ee
(x — 3)? + (y — 2)* <4. Therefore, |x —3|* <4 and
4 Ietollows that |x 3) <2) and) |y— 2) < 27550
ee)
AN
ee
Ve)
ed,
LMS, el =e vend
0 < y <4. Therefore, x2 < 25 and y* < 16,sox? + y’ < 41.
10.
11.
(b)
Aint: If ¢ is rational, then d = Ffor some natural numbers a and b. You
may assume that 5is a fraction in lowest terms. Then use the formula in
part (a).
(b)
A.
(h)
Hint: Look for a situation where the claim is false without the condition
that a # 0. Then determine how that condition should have been used in
the “proof.”
(d)
(c)
—36 = (—8)5 + 4. The quotient is —8, and the remainder is 4.
Hint: Assume that d = gcd(a, b) = | and a divides bc. Write
Exercises 1.8
1.
7.
1 as a
linear combination of a and b, and then multiply by c.
9. (a) Proof. Suppose that p is prime and a is any natural number. The only
divisors of p are | and p, and gcd(p, a) divides p, so gcd(p, a) = 1 or p.
(i) Assume that gcd(p, a) = p. Then p the divides a by the definition of
gcd. (ii) Suppose that p divides a. Then p is a common divisor of p and
a. Because p is the largest divisor of p, it is the largest common divisor
of p and a, so gcd(p, a) = p.
11. Proof. Let gcd(a, bc) = d, and suppose that a and be are not coprime. Then
d > 1, sod has a prime factor p. Then p divides a and bc. By Euclid’s Lemma,
p divides b, or p divides c. Because p is either a divisor of a and b or a divisor of a and c, we see that either a and b are not coprime or a and ¢ are not
coprime. Therefore, if a and b are coprime and a and c are coprime, then a and
be are coprime.
16. 42
17. (a) Hint: Use a two-part proof. For the part of the proof that assumes
(c)
a divides b, show both conditions (i) and (ii) for the Icm are satisfied
by b.
Hint: Assume that gcd(a, b) = 1. Because b divides m = Icm(a, b),
m = kb for some integer k. Use part (b) to show that k < a. Then use
part (c) of Exercise 7 to show that a divides k. Conclude that a = k, so
in — kb = ab,
Answers to Selected Exercises
(f)
397
Hint: By Exercise 7(d), ged(S, “)= |. Use part (c) to find an expres-
sion for Iem(S, 2).Then use part (e) to obtain another expression for
Iem(S, 2).Equate the two expressions and simplify.
19.
(a)
84=22.3!.7!and30=2! . 3! . 5! Therefore, gcd(84, 30)
=2! - 3'=6.
ik,
(@))
tl@, tl
(c)
(2)
ee!
Gor)
(a)
Suppose that X is a set. If X € X, then X is not an ordinary set, so X € X.
Exercises 2.1
3.
Not possible, 0
(e)
1, not possible
On the other hand, if X ¢ X, then X is an ordinary set, so X € X. Both
X © X and X € X lead to a contradiction. We conclude that the collec-
tion of ordinary sets is not a set.
4.
Sy
(a)
(@))
(g)
false
tiame
false
6.) (a)
Aj
8.
iG
(c)
(c)
(i)
(2p
false
true
false
ee
(e)
(e)
(k)
aC
(Cc) An Fe
Bee
aye
AC B and
Hint: To prove that if
true
false
true
et
=
eyo}
ACC, begin by assuming
BCC, then
that A C C means
recall
we
that A C B and BCC. To show that A CC,
x be any object.
Let
follows:
as
are
(WV x)(x€A=>x€C). The first steps
Suppose that x € A. Now use the fact that A C Band
BCC.
by
Hint: To prove that A = B, use the hypothesis A C B, and show B CA
using Theorem 2.1.1(c).
(b) Proof. LetX = {x ER: |x —4| =2|x| —2}.
(i) Substituting x = —6 and x= 2, we have, respectively, |—6 —4| =
2 are
— 4| = 2 = 2|2| — 2. Thus, —6and
10 = 2| —6| —2 and |2
elements of X.
(ii) Suppose that x€ X. Then |x — 4| = 2|x| — 2.
|x| =x.
x>4.Thenx —4>0,so|x—4|=x—4and
Case1:
Therefore, x — 4 = 2x — 2,sox = —2. However, x = —2
is impossible when x > 4. Thus, X has no elements that
are greater than 4.
0 <x < 4. Then |x — 4| = —(@ — 4) and |x| = x.
Case2:
Therefore, — (x — 4) = 2x —2, sox= 2.
Case3: x <0. Then |x —4| = —(x — 4) and |x| = —x. Thereforey— (7 — 4) = —2x — 2, sox = —6.
In every case, if x € X, then x el= 602), Uheretore, Xi
(e)
—6, 2}:
By G)and\(ii),X = {—6, 2}.
X = {x ER: |x + 3| < —4x — 2} and Y=(—oo, —1]. To show
Hint: Let
that YC X, assume thatx€ Y, and consider cases. In case 1,-3<x<-l.
Use the fact that |x + 3| =x + 3 to show that |x + 3| < —4x — 2. In case
398
Answers to Selected Exercises
2,x < —3. Use the fact that |x + 3| = —x — 3 to show that |x + 3| <
—4x — 2. This is a good place to use the method of working backward
from the desired result to find a proof.
14.
(a)
(c)
{{0}, {A}, {O}, (0, A},{0,
O}, {A, O},x,O}
{{O}, fa}}, (by, (La, b}, {D, (a)}, {D, {db}, {D, (a, by}, {La}, {bo},
{{a}, {a, b}}, {bd}, (a, bi}, (©, {a}, {b}}, (©, {a}, (a, dj},
{O, {b}, {a, b}}, {fa}, {b}, (a, bj}, X, O}.
15.
(a)
false
16.
(a)
no proper subsets
(e)
false
(g)
true
true
true
(e)
(k)
true
true
(CMO le
17.
(a)
(g)
true
true
19.
(c)
C. The “proof” asserts that x € C but fails to justify this assertion with
a definite statement that x € A or that x € B. This problem could be
corrected by inserting a second sentence “Suppose that x € A” and a
fourth sentence “Then x € B.”
(e)
(c)
(i)
F. The error repeatedly committed in this proof is to say
A C B means
x €A and x € B. The correct meaning of A C B is that for every x, if
x € A, then x € B.
(f)
F. The claim is false. In part (ii), the proof fails to consider a third case,
when —4 < x < 0. It can be shown that X = {4, —4/3}.
(i)
C. The proof could be considered correct, but it lacks a statement of
the hypothesis, helpful explanations, and connecting words. How much
explanation you include depends on the presumed level of the reader’s
knowledge. We prefer the use of words, not just symbols.
(j)”
F: The:claim is false“(For example, let A = {1,2},B = {1,2, 4% and
Cx (112) 5, On 7). Lhe statement “Since x
2B, x © A...” would be
correct if we knew that B C A.
(ayer
On) nee
Exercises 2.2
Le
nOn ia
Ol
{3,9}
(1,8)
3.
(a)
{0, —2, —4, —6, —8, —10,...}
()
(0, 274,70, '8,...}
4.
(2) ari, 01
(Cc) 1274)
Serle) eat
(e)
2. (a)
A and B are disjoint.
(a) A Venn diagram is helpful.
325,79}
eth) Si leio7)}
(g) (—oo, 3) U[8, co)
(d)
ZU
{0}
Answers to Selected Exercises
399
Because C C A UB, every element of C is in either A or B. In the diagram, the shaded area must be empty. Because A ™ B is not a subset of
C, there is some element x that is in A M B but not in C. To ensure that C
is nonempty, we must place an element y in any one of the three available regions of C. Our solution is A = {x},
B= {x,y},
C= {y}. For
other correct examples, we could place other elements anywhere in the
diagram (except in the shaded region).
(a)
Hint: To prove that A C B implies
(c)
and show that x € A — B is false for every object x. To prove the converse, assume that A — B = ©, and suppose that x is an object in A.
Proof.
(i)
Assume that
CC ANB.
A — B = ©, assume
Suppose that x € C. Then x
that A C B,
©
AM B, so
x € A and x € B. Because x € C implies x € A, C CA. Similarly,
Ce
(ii)
10.
(c)
Assume that
CC A and C CB. Suppose that x € C. Then from
C CA, we have x € A, and from C C B, we have x € B. Therefore,
x € ANB. We conclude that
CC ANB.
Proof.
Suppose that
CC A and D C B. Assume that C and D are not
disjoint. Then there is an object x € CM D. But then x € C and x € D.
Because
CC A and D C B,x € Aandx € B. Therefore, x
11.
(a)
(c)
(e)
and B are not disjoint.
Asal 102) Bh It FC
ae |
Ava
Os Bes alee Venue
Alea) Wee ey es)eee
12.
(a)
Proof.
13.
(b)
A x B=
BEGA =
€AMB,soA
S <= P(ANB)
it HSAs
iff (by Exercise 9(c))
SC Aand$CB
iff S e@P(A) and S € P(B)
iff S = P(A) NM PCB).
2), Crh),
{Cage (i) )
Ce AAC AGEs
OBIGE 4 Feu)
Ga DG 2.1), 2); Gt, 2),
(GC
(Gel
GV he ez) ats 4 V2)
iS:
(a)
Proof. (a,b) <&A x (BNC)
iffaecAandbe
BNC
iff ae Aandbe BandbeC
iffaeAandbe BandacAandbeC
iff (2. Db) eA x Banda, byeA x C
iff (a,b) € (A x B)N(A x C).
17.
(a)
Hint: Let x be the smaller of 5aand 5(a + b). Show that x € (a, b) and
(b)
a6 G2 (al, Ao).
Hint: First, by considering the quantity (b — a) + (d — c), show by
contradiction that c < b. Then determine where the number c is located.
(a)
F. One serious error is the assertion that x ¢ C, which has no justification.
The author of this “proof” was misled by supposing that x € A, whichis an
20.
400
Answers to Selected Exercises
(b)
(d)
acceptable step but not useful in proving A — C C B — C. After assuming that A C B, the natural first step for proving that A CC B — Cis
to suppose that x € A — C.
Hint: The second sentence of this proof says “Suppose A — C,” which
doesn’t make sense. (What should we suppose about the set A — C?)
Consider that the author of this proof may have meant to say “Suppose
xEA-—C.”
C. The proof that
AM B = A is incomplete.
Exercises 2.3
mae)
A=, 2,324,
5,6, 78
AEew
(c)
1nl)| Ait 4,5}
AcA
UA, = {5,6, 10, 11, 12, 15, 16, 17,18} U {neEN:n > 20}; (\A,=@
neN
neN
(lane A =i (a) An 1 10)
AEA
Aces
(gy) neN
UA, = 0,0; neN
1A, = @
Giuls)4, = 10,00); {
|An= 2
reR
reR
(p)
2.
The union is the triangular region bounded by y = 0, x = 1, and y = x.
The intersection consists of two line segments: from the origin to (1, 0)
and from (1, 0) to (1, 1).
The family in Exercise (a) is not pairwise disjoint. The family in 1(b) is
pairwise disjoint.
7. (a) Proof.
xe BN UA, iff xe Bandxe UA,
aecA
acA
iff x € Band x € A, for some a € A
iff xe BNA, for some a € A
iff xe.) (BO A,).
acA
8.
(a)
Using the result of Exercise 7(a) twice, we have
(Ua)o((ys)= Ul(Ua)om) = U( Yoon
acA
Bel
Bel \ \aeA
9.
(c)
Hint: The statement is correct.
10.
(a)
Proof. Suppose that
BeT
\aeA
xe JA,. Then x € A, for some a € I’. Because
ae.
TCA,
11. (2)
X=
we A. Thus xe A, for some we A. Therefore xe
[)A.
AEA
|) A,.
acd
Answers to Selected Exercises
16.
(c)
Proof.
Let x € JA,. Then there exists j @N such that
i=k
[e,2)
and x € A. Because j € N and x € Aj, x € An
i=1
Alternate Proof. The set [ =
401
k<j<m
{k,k + 1,k + 2,...,m} is a subset of
2X 294 lle 3,4. a A herefore, by Exercise 10(a), UA, Se LUA.
i=k
i=1
18.
19.
Let Ap = [8 1+
r)foreachk EN.
(a)
C. This proof omits an explanation of why there is some f in A such that
Ag & {A,: a © A}. The explanation is that by definition of an indexed
family, A # ©. If we allowed A = ©, the claim would be false.
(b)
C. No connection is made between the first and second sentences. The
connection that needs to be made is that if
some a € A, so x € B because A, C B.
xe (J A,, then x € A, for
es
Exercises 2.4
(b)
Slee (b)
(a)
not inductive
true
Proof.
(d)
(e)
not inductive
false
(f)
not inductive
1
(i)
The statement is true for n = 1 because
Yi
FQ) —
ee
=I.
— 2)=1
and
(ii) Assume that for some n€ N, >) Gi — 2) = 5Gn — 1). We must
i=l
n+1
n ae 1
show > Gi —2)=
;
+ 1) — 1). The equation’s left-hand
age
i=!
side is
5 Gi — 2)+ [3+
i=l
32+ 3n+1,
and
the
1)-2])= 5 Gn —1)+Gn+1)=
equation’s
right-hand
side _is
Hn + 1)(3n + 2), which also simplifies to sn ae >n +f.
Thus, the statement is true forn + 1.
(b)
By the PMI, the statement is true for every n € N.
Proof.
(i) Forn = 1,4! — 1 = 3, which is divisible by 3.
(ii) Suppose for some k € N that 4* — | is divisible by 3. Then
A4k+l _ 1 = 4(45- 1
= 44 -1)-1+4
= 4(4* — 1) + 3.
Both 3 and 4* — 1 are divisible by 3, so 44+! — 1 is divisible by 3.
By the PMI, 4” —
1 is divisible by 3 for alln EN.
402
Answers to Selected Exercises
(i)
(k)
Hint: Use Euclid’s Lemma (Lemma 1.8.4).
Proof.
(i)
(ii)
33+! = 34 = 81> 64 = (1 4+ 3), so the statement is true forn = 1.
Assume that 3”+3 > (n + 3)? for some n € N. Then
3in+1)+3
=
an+4
=
3. an+3
> 3(n + 3)° = 3(n? + On? + 27n + 27)
= 3n? + 27n* + 81n + 81
> n+
12n? 4+ 48n4+ 64=(n+ 4% =(n4+1)+ Be
That is93@7 +? = (n 4 1) #3).
By the PMI, 3+? > (n + 3) for everyn EN.
(a)
Proof.
(i) 6° = 216 < 720 = 6!, so the statement is true for n = 6.
(ii) Assume that n* < n! for some n > 6. Then
(n+
132 =n? + 3n* + 3n4+ 1
<n
<p
an
2(8
+ 3n? + 3n+n=n? + 3n? + 4n
+3 +n =n + 4n?
fr =2n
= we
<(n+
(c)
(i)
l)n! = (n+
I)!.
By the PMI, n° < n! for all n > 6.
Hint:
"= 2)(@ + 1)!] = @ + 2)27*3'= 277°),
Hint: For a convex polygon, any line segment drawn from an edge
point to another edge point lies inside the polygon. For the inductive
step, when you consider a polygon of n + | sides, draw a line as shown
between two vertices. (Only part of the polygon is shown.)
The new line segment separates the upper left triangle from a convex polygon that has exactly n sides, so we can apply the hypothesis of
induction to the n-sided polygon. Then use the result to compute the sum
of the interior angles for the polygon of n + 1 sides.
(c)
10.
A = {n:n = 2‘ for somek € N} may be defined as
(i)
2EA.
(ii)
if x€ A, then 2x € A.
Hint: Use n = 3 for the basis step. For the inductive step, assume that the
statement is correct for any collection of n points with no three points collinear. Consider a collection of n + 1 points, but apply the hypothesis of
induction to only n of those points. Then calculate the total number of line
segments determined by all n + 1 points.
Hint: For the induction step, visualize the starting position with n + 1
disks. When you think about the moves you would make to transfer all
Answers to Selected Exercises
12.
403
n + | disks from one peg to another, try to break down the task into three
separate tasks so you can use the assumption about how many moves are
required to move n disks. The first task is to move the top n disks from the
stack to another peg.
(a) F. The claim is obviously false, but this example of incorrect reasoning
is well known because it’s fun and the flaw is not easy to spot.
Let S = {neEN: all horses in every set of n horses have the
same color}. It is true that 1 € S. It is also true, for n > 2, that ifn e S,
thenn + | © S. The “proof” fails in the case when n = | because 1 € S
but 2 ¢ S. In this case, when either horse is removed from the set, the
(b)
(d)
remaining horse has the same color (as itself) [because there is only one
horse left], but the two horses may have different colors. We conclude
that the set S$ is not inductive, and in fact S$= {1}.
F. The basis step and the assumption that the statement is true for some
n are correct. Perhaps the author hopes that just saying the statement
is true for n + | is good enough. For a correct proof, one must use the
statement about n to prove the statement about n + 1.
Hint: One problem with this alleged proof is the sentence that says we
must “show that if nm = k, then
(f)
n = k + 1.” This does not make sense.
The sentence could be corrected to say “We must show that if the statement is true for some natural number k, then it is true for k + 1.”
Before assigning a grade, determine what the hypothesis of induction
should be and whether each step in the inductive part of the proof follows from previous steps.
.
F. The factorization of xy + 1 is wrong, and there is no reason to believe
x + lor y+ 1 1s prime.
Exercises 2.5
(b)
Hint: Let
S=
{ne N:n > 22 and n = 3s + 4t for some
integers
s > 3 and t > 2}. Show that 23, 24, and 25 are in S. Then let m > 22
be a natural number, and assume that for all k € {23, 24,...m — 1},
k € S. To show that m € S, proceed as follows: If m = 23, 24, or DS).
we already know m is in S. Otherwise, m > 26, so m — 32 235 BY
the hypothesis of induction,
m — 3 € S, som — 3 = 3s + 4t for some
integers s and ft, where s > 3 and t = 2. Then m = 3(s + 1) + 4t.
(d)
(a)
(d)
Hint: Let
A= {beN:
there exists
a@N
such that a* = 2b*},
assume that A is nonempty, and use the WOP to reach a contradiction.
Hint: The induction hypothesis is “Suppose that f,, is even and both
fay, and fy, are odd for some natural number k.” From this, use the
definition of Fibonacci numbers to show that f,,4;) is even and both
Praceryar and fray 1y+2 are odd.
Proof.
(i) In the case of n = 1, the formula is f, = f,,. — 1, which is
1 = 2 — 1. Thus the statement is true forn = 1.
404
Answers to Selected Exercises
(ii)
Suppose for some k that
ff +f+A+°°
+ h=ha2-
=APhPAR
HP tT
1.
Then
fiththt--+tht
ha
Sei
= (fra — D+ Sets
=
(fe+2 Fi fea) =Al
= fir3 —
1.
Therefore, the statement is true for n + 1.
Thus, by the PMI, f, +
(d)
H+ 4+---+ Ff, = fiz — 1 for all natural
numbers n.
Hint: Consider the cases n = | and n = 2 separately. For n > 2, you
will find it useful to multiply the equation ¢? = @ + 1 by $"~* and
p=p+t iby po”.
Hint: Modify the proof given for the case a > 0. Slightly different argu-
13.
ments are needed to show (i) that S is nonempty and (ii) that r < |a| = —a.
(c) Hint: Assume that S is a subset of N with the property that for all natural
numbers m, if {1, 2, 3,...,m— 1} CS, then m € S. Use the PMI to prove
that for every natural number k, the set {1, 2, 3,..., k} C S$. Conclude that
14.
(b)
SsaN
F. The claim is false. The flaw in the “proof” is the incorrect assumption
that m —
1 is anatural number. In fact, | is the smallest natural number n
such that 3 does not divide n? + 2n + 1. There is no contradiction about
a smaller natural number because there is no smaller natural number.
Exercises 2.6
t
(b)
16
Hint: Because 1,000,000 = (10°)? = (107)* = 10°, there are 10° squares less
than or equal to 1,000,000; 10° cubes less than or equal to 1,000,000; and 10
natural numbers that are both squares and cubes (sixth powers) less than or
equal to 1,000,000.
Hint: Complete this formula:
AL peeps
Ane
10.
AR
ee SANE
G +
“ALB OCD.
Hint: The answer is not 20-19-19 - 18; itis 130,340.
Consider the cases
where the bottom right is colored the same as or differently from the upper
left corner, and give two products that yield the correct sum.
17.
(a)
32
(5
18.
(b)
Ay
Ale
WAR
We
(SYED)
19.
(e)
= 52-51-50-49-48/120 = 2,598,960
<4 a3
2 = 171.360
There are 10 possible highest cards in a straight. The highest card can
come from any of the 4 suits, and there are 4 choices for the next highest
Answers to Selected Exercises
405
card, etc. This yields 10-4-4-4-4.-4 =10,240 straights. (This count
includes 40 straight flushes.)
20.
(b)
Proof.
The number
of subsets of A with r elements
is i Choose
ti
one particular element x from a set A of n elements. The collection of
r-element subsets may be divided into two disjoint collections: those
subsets containing x and those subsets not containing x. We count the
number of subsets in each collection and add the results. First, there are
n—1
;
.
(
)
r-element subsets of A that do not contain x because each is a subS
set of A — {x}. Second, there are is
)r-element subsets of A that do
:
r—
4
contain x because each of these corresponds to the (r — 1)-element
subset of A — {x} obtained by removing x from the subset. Thus, the
sum of the number of subsets in the two collections is
a se
r
(d)
i
=
r—1
r
:
Hint: The algebra in the inductive step is
it cs Nrerttnr
70
=
li
prt! of Il" ‘ aronetar a qit!
=
=
prt!
a
me)
+(, is:)
Je
rpitiq-r
—
prtl al alt! a’ptt!- rane pa
4 gntl
Jarontior
q?t!
ofr +E (Meow) ofS," )orwres a)
=A S("ew-) +e
- &("o0-) +
Guan
ee
uf
N
ee
NE
oe
=(a+ eas
22.
(b)
24.
(a)
Hint: Consider two disjoint sets Rola
tsn and m elements.
Hint: First show p, = 1? + 2? + --- +n’ by counting the number of balls
in each layer of the structure of the pyramid. Then show 1? + 2 + ++»+
i
n(n + 1)(2n + 1)
6
by mathematical induction.
406
Answers to Selected Exercises
Exercises 3.1
3.
4.
(a)
(c)
(e)
(a)
(c)
domain R, range R
domain [1, 00), range [0, 00)
domain R, range R
Proof.
Let r be a real number. Then 27 + | is a real number, and
(r, 2r + 1) € R. Thus r € Dom (Rk).
:
Proof.
Suppose that 1 © Rng (7). Then for some xe R, 1 — 2 = ee,
so —1 = x. Because x* > 0, this is impossible.
6.
(ap
Res=a
Coy
ak
ah
— eh
(©) Ry! = {@%y)ERx Ry =1@ + 10)
(e)
fe! =
{se Rx
(g)
he! = {ope Rx
By = 4
a
x4
Biy> 2 \
(i)
Re =
(j)
Hint: Ryo! # {(x, y) © P x P: yisachild of x}
{(x, y) € P x P: yis an older sibling of a female x}
fae
(Dinh Od =. 1(3, 2),(4.5)}
(d) ROR= {(1, 2), @, 2), G, 2)}
Smale ons = +(x, y) x = y) = Ry
(d) R,°OR, = {@, y)ER
(2)eeKee
Re = {@y) eR
x R: y = —35x + 52}
x Rey= 16° = 40n7 + 27)
(j))
x Riy < x4
ReoRo = {@, y)ER
2}
Cine = {we RxRy=
= 10}
a6
—
(p)
{(@,y)eP
(r)
Hint: Ro © Rois not {(x, y): y is a grandfather of x}.
x P: y isa younger sister of x’s father}.
17.
(g)
Hint: The grade should be C.
1.
(a)
(e)
(m)
not reflexive, not symmetric, transitive
reflexive, not symmetric, transitive
not reflexive, symmetric, not transitive
2.
(a)
{(1, 1) 2.2) 2a 3nGe by}
Exercises 3.2
1
(‘cane TS
Answers to Selected Exercises
(d)
407
(C1, 1), 2, 2), G, 3), C, 3), 2, 3), G, D, G, 2)}
a)
<—)
we, 2
(a)
(d)
(b)
{(@, y) Ee Rx
3
Rey = 5x}
= yorxy ="0}
{Cage Rex Rex
Proof.
Assume that R is symmetric. Then (x, y) € R iff (y, x) € R iff
(x, y) € R7!. Thus, R = R~!. Nowsupposethat R= R~!. Then (x, y) € R
implies (x, y) € R~!, which implies (y, x) € R. Thus, R is symmetric.
(d)
(g)
Hint: To show that R is reflexive, let a be a natural number. All prime
factorizations of a have the same number of 2’s. Thus a R a. It must also
be shown that R is symmetric and transitive. Three elements of 10 are
[0 25240" = 2783 9 7eand 2427 Je
lee:
Hint: First show P is reflexive on R x R and symmetric. To show transitivity, begin by supposing that (x, y) P (z, w) and (z, w) P (u, v). Then
|x
— y| = |z — wl and |z — wl = |u— vi.
-
12.
16.
(a)
(a)
(c)
(b)
The equivalence class of (0, 0) is the line y = x.
transitive, but not reflexive on {1, 2, 3, 4} and not symmetric
0 = {...,-6, —4, —2, 0, 2, 4,
6,...}
We S81,
193
5)
Hint: oe that x = y (mod m). Part 1: Suppose that z € x. Then
x = z(modm) By symmetry y = x (mod m), and by transitivity, y = z
(mod m). Thus z € y. This shows x C y. In part 2, we must show y C x.
Proof.
Suppose that x = 0 (mod 6) or x = 3 (mod 6). Then 6 divides x
or 6 divides x — 3. Thus, either x = 6k for some integer
k or x= 6j + 3
for some integer j. Because 3 divides 6k and 3 divides 6j + 3, in either
case 3 divides x. Therefore x = 0 (mod 3).
18.
(a)
Proof. Suppose that (x, y) € RU R™!. Then (x, y) € R or (x, y) € Raz
If (x, y) € R, then (y, x) € R7!. Likewise, if (x, y) € R7!, then (y, x) € R.
(f)
A (or C). All steps are correct. The grade you assign should be based on
whether you think the proof should provide more explanation by mentioning the symmetry and transitivity properties.
(d)
The elements of 4 are natural numbers, not subsets of N, so of is not a
In either case, (y, x) €
19.
RU R-!. Thus, R U R“ is symmetric.
Exercises 3.3
(b)
onN.
of N. Note: {{1, 2, 3,4}, {ne N: n > 5}} is apartitiof
partition
9,
2,...,
1,
0,
contains
0
class
The
classes.
There are 10 Ean
100 101s
109) 200,201.)
209.2...
and. all the negatives
of ree
408
Answers to Selected Exercises
numbers. The class of 10 modulo R contains all integers that have | as
the tens digit, and so forth.
6.
Hint:
There
are
four
sets
in the partition.
One
of them
is the set
{C.,—D, (—1, 1), G), (sn =)}.
7.
(b)
(d)
9. (a)
11.
«Ry
it x ry orbotha = 2 andy > 2.
xRy iff (i) x = yand x € Z or (ii) int(x) = int(y) and x, y ¢ Z.
(Recall that int(x) denotes the greatest integer function.)
{Gls
D2).
(5, 3),.(3, 4), S, 3)J
Not always.
Proof.
Ci3)6 4) (3.5) 4,3), (4,4) 54. 5),
Let R be the relation {(1, 1), (2, 2), (3, 3), (1, 2),
(a3) (2m)
wl) eonethensche Atak 2.5 eal Deny
L 1s
RG) a1 lee sand
RS) a ila od ine setie—" {122 3) ah leet
is not a partition of A.
237
ttt
14.
(a)
Yes. {B‘,B5} is a partition of A when B, # B, because {Bi, Bj} =
PAS
NSO
I
cel Pirie ait ae, 0
{B{, B5} is not a partition ofA.
15.
(d) A (orC). The proof is correct because the ideas are all there and every
statement is true. You may give it a C if you feel the ideas are not well
connected.
Exercises 3.4
Lae
20
(Chae:
-(C), cel
4.
(b)
Aint: Let w, x, y, and z be integers. Prove that if w = y (mod m) and x = z
(mod m), then w — x = y — z(mod™m).
7.
(a)
(GY
238+ 496 —-44=4+4+1-8=6
OS PAS lel tarp
9.
13.
(b) |Hint: Rewrite 2x = 5 as 2x = 12 in Z..
(a) F. The claim is true (Theorem 3.4.1), but there are two errors. The first
error is stating without justification that a = c (mod m) implies a + b =
’
(eee
OY Oe
20.)
Wee
c + b (mod m). The next sentence makes the same error, but calls it
“Substituting.” This “proof” essentially uses what it claims to be true to
“prove” the claim.
Exercises 3.5
1.
(a)
(c)
(f)
Not antisymmetric. Does not have comparability.
Not antisymmetric. Does not have comparability.
Not antisymmetric. Does not have comparability.
11.
(a)
{(a,a), (b, b), (c, ce), (c, a), (c, b)}
13.
(a)
The upper bounds for the set {g, 1} are f, b, and c. The only upper bound for
the set {e, h} is b.
There is no lower bound for the set {g, h}. The only lower bound for the
Seti (if, 2), isu
Answers to Selected Exercises
14.
(a)
Proof.
Suppose that B — {x}
©
409
CCB. To show that B — {x} = Cor
C = B, we consider two cases.
Case 1.
Suppose that x © C. Because B —
{x} © C, every element
of B other than x is in C, and x is also in C, we have
Case 2.
BEC.
Because C © B, we have C= B.
Suppose that x ¢ C. Then from C © B, we see that if y is in C,
then y€ Band y # x. Therefore,
C © B — {x}
© C,soB— {x}
VG
is:
16.
22.
(b)
(c)
(a)
(b)
(e)
(f)
In either case, B — {x} = C or
C= B, so B — {x} is an immediate
predecessor of B.
No. For example, consider a set of two squares where the squares are
side by side within the rectangle.
A set containing two disjoint squares does not have a lower bound.
Hint: Use parts of Theorem 2.2.1 and Exercise 9 of Section 2.2.
Hint: Use parts of Theorems 2.3.1 and 2.3.2 and Exercise 10 of Section 2.3.
F. The statement that if sup(C) € B, then sup(C) = sup(B) is false, as
is the assertion that if sup(C) ¢ B, then there must be an element b of B
such that sup(C) = b and the assertion that if there is an element b of B
such that sup(C) S b, then b = sup(B). Finally, the claim is false, too.
F. This proof does not show that sup(B) exists. All it shows is that if
sup(B)
exists,
then
u = sup(B).
A correct
proof would
show
that
u = sup(B) by showing wu has the two supremum properties (wv is an
upper bound and u R v for all other upper bounds v).
(h)
EF. The claim that if there is no c’in C such that c < c’, thenc == sup(C)
is false because c might not be comparable with other elements of C.
The assertion that otherwise there must be elements ¢;, Cy, C3,---, C, in C
such that c, < Cc, < ¢3°**< cy, < bandnoc,,,inCsuch that c, < cj4)
is false for the same reason and also because there might be no element
of C that is the “last one” in C. Even if there were such a chain of elements of C, the element c, may not be the supremum because it would
not necessarily be comparable with other elements of C. Finally, the
claim is false.
Exercises 4.1
(a)
(a)
(c)
(a)
(a)
This is a function with domain and range {0, A, U1, U, M }. Other possible codomains are {0, A, O, U.N, $}, (0, A,O, UM, +,#,%}, and
VASO
OF 253.4.
Domain = R — {—1}. Range = {y € R: y
#0}. A possible codomain
is
Domain = R — {a + kr:ke Z} . Range = R. A possible codomain
1s Moe
Dom (f) = R — {3}, Rng (f) = R — {—1}.
Proof. a ipsEN
prime and not 5 iff y = 1.
y ="
For
1,253,014; 2 = yis
410
Answers to Selected Exercises
If x = 2, then 2x + y= 4+.
and not 5 aff y= 3:
If x = 3, then 2x + y= 6+.
and Noe Sit ye
If x = 4, then
2x + y=
For
y = 1, 2,3, or 4,4
+ y is prime
For
y = 1, 2, 3, or 4,6
+ y is prime
8 + y. For
y = 1, 2, 3, or 4,8 + y is prime
and not'S iff y —=3:
For each x € A, there is a unique y in A such that (x, y) € R.
(c)
Proof.
Let x be an integer. Then x? — 2 is an integer, and (x, x” — 2) € R.
Therefore Dom (R) = Z. Now suppose that (x, y) € R and (x, z) € R.
10.
(a)
13.
14.
(a)
(c)
Then x7 + y=2 andx*
valued.
A.
+z=2,soy=z=2 — x’. Thus
R is single
f GY
8 S98,
373) 0n Ae
Proof.
In Z, we have 0 = 3 = 6 = 9. The rule assigns to 0 these dif-
16.
(a)
17.
(c)
ferent images: [0], [3], [2], and [1] in Z,.
Dom (S)
m(m — 1)
n2. We choose two elements from A for first coordinates; each
(a)
bya
el eet
f-@M=
2
(c)
AC
pi
may then be assigned any element of B as its image.
Exercises 4.2
ieee
1 — 2x
ewe
(e)
(a)
(c)
(e)
fog ={69,6N,6 DE
(i)
Observe that (f° g)(x) = ea
fo)
23)
inx
(f° g)(x) = 17 = 14x, (g Of (x) = —29 — 14x
(fo g)(x) = sin 2x? + 1), (g Of )(x) = 2 sin’x + 1
gof=O
fn) = ihe
ean
We must consider cases.
Ifx < —1, f(g@)) = f(2x). Because in this case 2x < 0, f(2x) = 2x + 1.
If x> —1, f(g(x)) = f(—x). There are two subcases:
If x = 0, then —x < 0, sof(—x) = —x + 1.
f—l=7= 0, then
[Sex > 0. so f(x) = —2x.
Dyes IW
Therefore, (f 9 g)(x) = 4 —2x
—-x+1
ale gee]
if —L
ifx>0
a
0):
Answers to Selected Exercises
411
Similarly we find g ° f as follows:
4
¥
g(x+1)
ifx<0
(g °D)G) = es
ih vee Oh
If x < 0, then g( f(x)) = g(x + 1). There are two subcases:
If
x < —2, thenx + 1 < —l,sog(x+ 1) = 2x42.
Ifx > —2,thenx + 1> —-—l,sog@+1)=-x- 1.
Thee OthanmeG (c))i=seOx)ebBecause
=108 2xe=-0) souk = — 1.
Thus, g(2x) = —2x.
Qx+2
Theretore, (ee
f)\(@)—4
—x—
ifx<-—2
I
—2%
“if —2"=— > = 0.
ie
Soa0)
3. (a) Dom(fog) = R = Rng (f° g) = Dom (g° f) = Rng (g ° f)
(c)
(e)
Dom(fo g) = R, Rng (f° g) = [—1, 1]
(g0f) = R, Rng (gof)= [1,3]
Dom
Dom (f¢ s)\= {h.75)
Domieoy 2
Rog @ og)
i, 1
— Rngi ees)
(i) “Dom (fo?) = RK, Rug Gog) = (— ce,2)
Dom (g° f) = R, Rng
(g ° f) = (—00, 1)
4.
(a)
9.
(a)
Example 1: f(x) = x’, g(x) = 3x + 7.
Example 2: f(x) = (« + 7)*, g(x) = 3x.
13.
14.
{x,y ERx
Riy=0 if x < 0 andy =x’ if x = 0}
{(x,y)—eR x Riy =x}
AUC
Hint: To prove that h Ug is a function when h|, = gl, write
and
sets),
(as
AU(C— EE). Then show that hUg =hU(glc_g)
Theorem 4.2.5.
(a)
hUg
as
use
isa function.
><
Y tay
at
fel
oma
16.
(a)
(d)
Proof. Let x, y¢R. Suppose that x<y. Then 3x <3y and
3x— | = 3y — fe lheretore, f(x) =f).
Suppose that x andy are in (—3, 00) and x < y. Then 4x < 4y,
Proof.
— Oi ey ot Sy — x — 3 OF
ee ENS) Ve
SOK Ee ey
(x —
1)(y + 3) < (y —
1I)@ + 3). Using the fact that x and y are in
412
Answers to Selected Exercises
(—3, 00), we know x + 3 and y + 3 are positive. Dividing both sides
of the last inequality by x + 3 and y + 3, we conclude that
el
rye et
x+3
y+3.
Thus; (Gs) = (QO):
17.
(d)
The function f given by
go el | ghibeonesal)
ile
ei O < oT
eS
19.
(a)
eit x > 1
is one counterexample. Another is f(x) =(x + 1)(x — ibe
A. On line 1, the author of this proof chose not to mention that x € A
(because A is the domain of f). The reader is expected to observe this to
verify that (x, y) € I,.
Exercises 4.3
1.
(a)
Onto R. Proof. Let w € R. Then for x = 2(w — 6), x € R, and f(x) =
4[2(w — 6)] + 6 = w. Thus w &€ Rng (f). Therefore
fmaps onto R.
(c)
NotontoN
x N. Proof. Because (5, 8) EeN x Nand (5, 8) € Rng (f),
f does not map onto N x N.
(k)
Onto [0, co). Proof.
First, if x€ [2, 3), then x —
2 >
Oand3 —x>
0,
r—2
so f(x) = oe > 0. Therefore Rng (f) € [0, 00). Now let w € [0, 00).
3
Choose x =
2
or . Then w > 0, so 2w +2 <3w+2 <3w43.
Dividing by w + 1, we have 2 < x < 3. Thus, x € [2, 3), and
3w + 2
.
Ae
| 1
>
2.
(a)
3w + 2
| |
Bw +2 —2w+
w+
1 |
1)
a are
Therefore f maps onto [0, 00).
One-to-one. Proof.
Suppose that f(x) = f(y) for some x, y € R. Then
a +-6= y a 6. Then 5x = +, SOM. "ys
(c)
(k)
One-to-one.
Proof.
Cn, m=
1) SO nt an:
fis one-to-one. Proof.
x—2
Suppose
that m,n e€ WN and f(m) = f(n). Then
Suppose thatx, z © [2, 3) and f(x) = f(z). Then
z—2
aay = Fp 80 3t —ae — 6 + 2g = 3g — xe — 6 + 2x, Thus x = ve
4.
(a)
B=
{0,3}; f= (G70), @, 3), G, 0), 4, 0)}
Answers to Selected Exercises
Let f: RR
be given by f(x) = 2x and g:R—R
413
be given by
g(x) = x. Thenf maps onto R, but g ° f is not onto R.
By Theorem 4.3.3, the function fcannotbe one-to-one. LetA = {a, b,c},
Be Teles
ands Colney.7) eletmnaa)((a,72): (br2)n(c, 3)} and
g ={(1, x), (2, y), 3, z)}. Then g is one-to-one, but the composite
g° f={(a,y), ©, y), (c, z)} is not.
10.
(c)
Proof.
We verify that f maps onto R as follows: 1 € Rng (f) because
f(4) = 1. For w ¥ 1, choose x =
4
a ss
Dy)
. Then x ~ —4 and
fea) = |? | 2]= [SPE ya] =
w
l1—w
ow
6
Therefore, f maps onto R.
To show that fis ai
riiy suppose that f(x) = f(z). Then for x # —4
—2
—2
and z + —4, we see pee
as
. Therefore, xz — 2z ++ 4x —8 =
x+4
z2+4
xz — 2x + 4z — 8, so 6x = 6z and x =z.
We
must
also consider
—2
whether f(x) = 1 forsome x# —4. Butif x# —4 and f(x)= ai =1l7
12.
(a)
then x — 2 = x + 4,so —2 = 4. This is impossible. Therefore fis oneto-one.
Proof. f is not a surjection because [1] has no pre-image in Z,. This
is because if f(x) =
[1], then [2x] = [1]. But then 8 divides the odd
number 2x — 1, which is impossible. To show f is an injection, suppose
that f(x) = f@). Then [2x] = [2z], so 2x = 2z (mod 8). Therefore, 8
13;
(b)
divides 2x — 2z, so 4 divides x — z and thus x = z.
Example 1: x, = n for all n € N (the identity function).
2: x,, =
Examplele 2:
14.
(c)
(f)
Sn
{ Zs
we
< se
ifn <50 é
None.
Because m = n + 1, one element of B has two pre-images. This ele:
42 le
>
ment of B can be selected in n ways, and the two pre-images in
ways. We can assign each of the remaining n — 1 elements of B as the
image of exactly one of the n — | remaining elements of A in (n — 1)!
n+ 1
iOsP
ways. Thus, there are (eee) (n—-1)!= n(24P") functions
from A onto B.
15.
(a)
(b)
(d)
F. To show that fis onto R, we must prove that R C Rng (f). The proof
shows only that Rng (f) CR.
Hint: What additional information should be included in this proof?
A. Notice that a direct proof would be easier to follow.
414
Answers to Selected Exercises
Exercises 4.4
3.
(b)
4y
Let h be the inverse of g. Then (x, y) €/A
iff(y, x) € g iff x = ED
iff xy +2x=4y
.
iff y=
Leo
iff ya—4)=—2x
Dex
ae
‘
Therefore
Ds
:
reer: where Dom (h) = Rng (g) = (—ow, 4). To verify that
—Xx
this formula is correct, suppose that x > —2. Then
he g)@) =heG@))
4x
=f
amen
=
Ax
& = 5) ~
Vinge
acs.
x+2
Because the composite is the identity function on the domain of g, we
conclude that h = g7!.
7.
(a)
The function f~! is a one-to-one correspondence from (0, co) onto
Dm
1(c) P2004 641-73)
(2500).
(e)
[6147532]
@ibadad.2.0)
Exercises 4.5
2. (a)
[2,10]
(ce)
{0}
40 (b)) (2,52)
“(c)
3 — V5
34+ V5
E ~ v32=™ |U GY:
Go (aye
MIRED
(cy Mt 1S! 9}
10.
Proof. Suppose thatb ef( ia D,,).Then b = f(a) forsomeaeé
(a)
er
aecA
Thus,
a€D,
for every
that
be f(D,)
for
every
acA
ae A.
ae A.
Because
Thus
b = f(a), we
conclude
be
Therefore
() f(D,).
acA
acA
Hint: There must be at least two sets D, and D, in the family, and b € f(D,)
for all w € A, but no element a in { } D, such that f(a) = b. The functionf
cannot be one-to-one.
12.
{| Dy.
acA
r(a D,)< (1 f,).
11.
+ V3
(a)
aeA
Suppose that be f(f~'(E)). Then there is ae f—'(E) such that
f(a) = b. Because ae f~'(E), f(a)€ E. But f(a) =b, so DEE.
Therefore f(f~'(E)) C E.
(d) First, suppose that E = f(f ~'(E)) and that b € E. Then b € f(f ~(B)).
Thus, there is a€ f~'(E) such that b = f(@, so b € Rng (f). Therefore EF G Rng (7).
Answers to Selected Exercises
Now
ipa
assume
(2)
that EC Rng(f).
We
know
415
by part (a) that
GESconto prove equality, we must show E C TCE. (ED):
Suppose that b € E. Then b € Rng (f), so b = f(a) for some a € A.
Because b = f(a) € E, ae f~'(E). Thus, b = f(a) and ae f~'(E),
so b € f(f ~'(E)). Therefore E C f(f ~'(E)).
13.
(b)
Proof.
Suppose that te f(X) — f(Y). Then t € f(X), so there exists
x € X such that f(x) = t. We must have x ¢ Y because t = f(x) € f(Y).
Thus, x © X — Y, and therefore t = f(x) € f(X — Y).
14.
(b)
Proof.
Assume that f(X) M f(Y) = f(X OM Y) for all X, Y C A. Suppose
thatx,y€ Aandx # y. Then f({x}) N fy) =f} O (yp) =f@
= ©. Therefore, f({x}) #f({y}), so f(x) # f(y). Therefore
f is oneto-one.
15.
(a) _ If f is one-to-one, then the induced function is one-to-one.
17.
(a)
F. The claim is not true. We cannot conclude x € X from f(x) € f(X).
Exercises 4.6
3. (a) does not exist
(g)
() 2
e* (Recall that lim (1= ;) = €,)
1
n— co
5.
(e) 0
n
nN
(i)
O
(a)
Proof. Let ¢ be a positive real number. Let N be any natural number
1
1
,/—, so Ne t Thus
= Suppose that n > N. Thenn >
greater than
5 <é. Because
|x, — 0| =
4-0|=4<e
for
all iN.
Lines)
n>
oo
(c)
(e)
Hint: To show x diverges, suppose the limit is L and let e = 1.
Hint: x diverges; use € = I.
(h)
Hint: x, — 0; for ¢ > 0, use N > (2e)~* and
ari
6.
(a)
Vn+1+ Vn
1—- Vn=(Vn4+1- Vi (
AR
A Vatl+Vn
Proof. Let « > 0. Then ; > 0. Because x, > L, there exists N, e€N
such that if n > N,, then |x, — L| < = Likewise, there exists N, © N
Let N, = max {N,, N5},
such that n > N, implies |y, — M| <
and assume that n > N3. Therefore, we have |(x, + y,) — (EL + M)| =
I, —
+O, — | Sb —
+n — Mi < gt
= &-There-
7.
(ie
(a)
fore
aya
Wales alle bls po
78s
Hint: Because x, > L, |x,| > |L| by Exercise 6(f). Now apply the defi-
13.
(b)
A. The proof uses Exercise 6(b).
nition of |x,| — |L| with e = El
416
Answers to Selected Exercises
Exercises 4.7
1.
(g)
Hint: The limit pee to be ~.3As usual, the real work is finding 6 that
will ae cae
anil -4 <i as: BS Sie
hae with the desired
pe eu ;|<e,
F — ; 26)
and
sry
Xx
to
:
get
ethan
is,
ge
3
:
faa
a
Add 4 and simplify to get : ss
* =
We can always assume that € is small, so we make the assumption that
Gees i.This ensures that 1 — 3¢ > 0, and, therefore, we can rewrite the
:
previous
:
:
inequality
3
as
ise
ape
3
lmemee
or;
3}
meee
3
<a NES
ears
This last inequality tells us how large we can allow x to be, and the
next step is to determine how large we can allow x — 3 to be—that
is, how close x must be to 3. Subtracting 3 and simplifying, we have
22
T=
Tae
Oe
'
= Choosing between eer
Oe
and = —7,;) we see
9e
is es ane so we can get the previous ee
if we
Poe aie
. This means that we can be sure
<x-3<
&
require that
T+ 3
mG t 1
1 + 3¢
— 3] < © when |x — 3) <
9
eon can now write the proof in the proper order. Begin with the
hypothesis that ¢ is a Pals real number
eae ll
lie
(k)
4
3| er Leto
The limit is 3. Proof.
— Tae
i “
Let
such that ¢ < L and that
3
and reverse the steps above.
¢ > 0 be given. Choose 6 = min
{1, Tee
Suppose that |x — 2| < 6. Then |x —2| <1, so |x7+2x4+4/S
x?|+ |2x| + |4| <9+6+4 = 19. Thus, |(x? — 5) — 3] = |? — 8) =
|x — 2| |x? + 2x + 4] < 198 < e. Therefore Jim Ge" = 5)=3:
(a)
Proof.
Assume
that
lim fix) = L and im ga)
X=
(f)
(a)
Me Vetix
besa
4
sequence in J — {a} that converges to a. Then the sequence f(x,,) converges
to L, and the sequence g(x,,) converges to M. By Exercise 6 of Section 4.6,
(f + g\(x,) = f(x,) + g(x,) converges to L + M. This is true for every
sequence in / — {a} that converges to a, sof+ g has limit L + M at a.
Hint: The proof is by induction on the exponent n. Use Theorem 4.7.1.
Proof. Because tim, (x — 5) = 0, we cannot apply the quotient rule
for limits (Exercise 3(e)). However, the functions f(x) = x + 4 and
et Se,
g(x) = *—+— 70arebothdefined on R — {5},
and they agree
on R — {5},
2 —x — 20
so by Theorem 4.7.2, lim (?
x>5
x—5
=
’
lim (@&+ 4) = 9.
See
Exercises 5.1
3:
Ts
(a)
(a)
By definition, the set Ngo is finite.
Proof.
Suppose that A is finite. Because AM B is a subset of A,
is finite.
AN B
Answers to Selected Exercises
9.
10.
11.
13.
(a)
Hint: Define f:A > AU {x} by f(@ = (a), for each
417
ae A. Now
show that f is one-to-one and onto A U {x}.
Hint: First show P(A x B) is finite.
(c)
not possible
Hint: When x = r, N, — {x} =N,_, so the sets are equivalent. If x #r,
define a function on N, — {x} by considering first the images of elements that
are less than x (if any) and then the images of elements that are greater than x.
18.
(a)
Hint: Suppose that f is not onto B, and consider the range of /-
(b)
Hint: Suppose that f is not one-to-one. Then A is not empty, and because
A and B are finite and A ~ B, there is some n € N such that N,, ~ A and
B ~ N,,. Construct a function F from N,, onto N,, that is not one-to-one.
Then for some x, y,
19.
21.
zEN,, f(x)
= f(y) = z. Removing (y, z) from F
produces a function from a proper subset of N,, onto N,,. Now apply
Exercise 17.
Hint: Use induction on the number of elements in the domain.
(b) Hint: The largest possible sum of 10 elements of Nog is
90 + 91 +--- + 99 = 945. Thus, there are no more than 945 possible
sums. However, there are 2!° — 1 = 1,023 nonempty subsets of S.
(f)
(g)
Apply the Pigeonhole Principle, and delete any common elements from
two subsets that have the same sum to form disjoint subsets.
Proof. Suppose that there were no occupied rooms whose numbers
differ by 5. Then at most one of the two rooms in the set {5, 10} could
be occupied. Similarly at most one room in each of the sets {15, 20},
{25, 30}, {35, 40}, and {45, 50} could be occupied. However, there is
no one-to-one function from a set with six elements to the set consisting
of these five sets, so this is impossible.
Hint: You may use the result of part (f). You may also apply the reasoning of part (f) to other sets of room numbers.
22.
(b)
C.InCase 2, it is not correct that N, UN, ~ N,,;. In fact, N,
3.
(a)
Proof.
UN; = Ny.
Exercises 5.2
Let f:
N— Dt be given by f(n) = 2n — 1 for each
nEN.
We show that f is one-to-one and maps onto D*. First, to show f 1s
one-to-one, suppose that f(x) = f(y). Thus, 2x —
1 = 2y —
1, which
implies x = y. Also, fmaps onto D* because if dis an odd positive integer, then d has the form d = 2r — | for some r € N. But then f(r) = d.
4.
6.
(e)
Hint: Consider f(x) = —@ + 12) with domain N.
(a)
Hint: Let f: (0, 1) > (1, 0) be given byes)
(d)
Hint: Find a function f: (0, 1) > [1, 2) U(S, 6) that maps (0, ¥2) to
(5, 6) and [¥y, 1) to [1, 2).
(a)
Define g: N > Et by
Dy
Glau a Ogee
De
~
shige
al
it og 0
titses= Ilse a= 10)
418
Answers to Selected Exercises
The
LED)
12.
(a)
xe
(c) Np
(e) c¢
F. Wiscertainly an infinite subset of N, and D* is denumerable, but this
“proof” claims without justification that every infinite subset of N is
denumerable. This assertion is true but requires a proof.
(c)
F. The claim is false. Also, “A and B are finite” is not a denial of “A and
B are infinite.”
(d)
F. Writing an infinite set A as {x,, x5,...} is the same as assuming A 1s
denumerable.
é
Exercises 5.3
7.
Hint: This theorem has been proved in the cases where A and B are finite
(Theorem 5.1.7(b)), where one set is denumerable and the other is finite
(Theorem 5.3.5), and where A and B are denumerable and disjoint (Theorem
5.3.6). The only remaining case is where A and B are denumerable and not
disjoint. Write A U B as A U(B — A), a union of disjoint sets.
11.
Hint: For each m € N, let B,, = k,,. Then there is a bijection f,,: B,, > N,.
m—1
Define h: LJ B; > N by setting h(x) = (Ss, )jy
ieN
i=1
LOL 2 ee
13.
(b)
Proof.
If the set were countable, then every subset would be countable. However, the subset (1, 2) is uncountable.
14.
(b)
Hint: First prove that each set 7, is infinite. Then for a € T,, define
fil) 2 21s 13 Says SB She Pe Pk’s where p, is the kth prime and a; = 0 for
alli > k. Explain why T,, is equivalent to Rng (f) and why Rng (f) is
countable.
16.
(b)
:
C. The proof is valid only when f(1) = x. In the case when f(1) 4x,
we need a new function g that is almost the same as f except that the
image of | will be x. This involves removing the two ordered pairs
with second coordinates x and f(1) and replacing them with two other
ordered pairs. Let t be the unique element of N such that f(t) = x, and
define g= (f — {(1, f(),
&9})U {(,»),(t f())}.
Exercises 5.4
4.
(b)
true
6."
(a)
-O<.(0)
:
9.
(a)
Not possible
11.
(a)
Proof.
{On
=] O 20)
=
[01]
= RN
= R < A(R) < P(A(R)).
a
Suppose that B << A. Then A # B and B < A. Combining this
with the hypothesis that A < B, we conclude by the Cantor—Schréder—
Bernstein Theorem
BA.
(c)
Hint:
Assume
A = Cis false.
that A = 15), which is a contradiction.
Therefore,
that A < B and B < C. Derive A = CG and show that
Answers to Selected Exercises
16.
(a)
419
Hint: Assume that there is a bijection. Associate each function f in 4
with the corresponding real number a such that 0 < a < 1, and write f
as f,. Define g: [0, 1] —
[0, 1] by
go) = ‘\
pig’
eet (ea
Then g = f, for some b € [0, 1]. Compute g(b) to obtain a contradiction.
(b)
17.
(b)
(d)
[Ref.: J. Robertson,
“A
Student
Exercise
on
Cardinality,”
Mathematics and Computer Education 32 (1998), 17-18.]
Hint: Consider the set of constant functions in #.
F. Theclaim is false. We have not defined or discussed the special properties of operations such as addition for infinite cardinal numbers. The
equation C = B + (C — B) applies only to finite sets.
F. The “proof” assumes that every element of B is in the range of f-
Exercises 5.5
1.
5.
The Axiom of Choice is not necessary because there are only a finite
number of sets in the collection.
(h) The Axiom of Choice is necessary.
Binfinite andA denumerable. Because B C A, B < A.
Let B CA with
Proof.
Because A is denumerable, A = N. Because B is infinite, B has a denumerable subset D by Theorem 5.5.4. Thus, A = N= D= Ba By thes Cantor
(b)
Schréder—Bernstein Theorem, B = A. Thus B © A.
Alternate Proof. The set B is an infinite subset of the countable set A, so B
is countable and denumerable. Because both A and B are equivalent to N, A is
equivalent to B.
8.
10.
Hint:
Let xe A. By Theorem
5.5.4, A —
{x}
has a denumerable
subset
{a,:n © N}. Construct a one-to-one correspondence betweenA and A — {x}.
a A.
(c) EF. The idea of this “proof” is to take out countably many elements,
one at a time, until denumerably many elements are left. But if A is
uncountable and C is countable, then the set B = A — C of leftover elements will always be uncountable. (See Exercise 9(b) of Section 5.3.)
Exercises
6.1
1.
(a)
(e)
2.
(a)
(b)
In this structure, the operation is not commutative and not associative.
Not an algebraic structure.
ais the identity element.
Yes. This is tedious to verify because one must verify that 64 equations of the form (x © y)0z = x ° (y © z) are all true. It helps to observe
z).
that if x —a (the identity), then Woy)Ooz=yoz=xo(yo
Similarly, if y= a or z = a, the equation is easily seen to be true. This
420
Answers to Selected Exercises
leaves only 27 cases to verify when x, y, and z are not a. For example,
(bOc)0ob=dob=c, and bo (cob) = (bod) =¢, so the equation
is true when x = b,
(c)
y= c, and z = BD.
Yes, because the table is symmetric about its main diagonal. To verify
by cases that the equation
x° y = y©x is true for every choice of
x and y, consider first the case that one of x or y is a, then the case that
(d)
(e)
(f)
(g)
(h)
x = y, and finally the other three cases.
The inverses of a, b, c, d, are a, b, c, d, respectively.
No. The product b ° c = d is not in B,, so B, is not closed under ©.
WSR? GOGS EG CSC = 6 COGZ=]64ClOCzEE
(jl OND Val GC) aid al
DG, 1).
True. In fact, for allx EA, xOx =a.
Hint: Compute (ac)(db) and (db)(ac).
eS) (a) Hint: Compute x © (a © y) and (x ° a) Oy.
(b)
Hint: Assume that for some natural number n, every product of ¢ elements dj), dy,..., a, of A is equal to (...((a, * ad) * a3)...) * a, for every
t <n. Now consider a product of n + 1 factors a), dy,..., 4,4, in that
order. This product has the form b, * b,, where b, is a product of some
k factors (k < n) ay, ay,..., a, in that order and b, is a product of the
remaining factors a, ,1,..., 4,4 in that order. First, use the induction
hypothesis to write b, and b, in left-associated form.
Now consider two cases. If k > 1, then there are at least two factors
a, and a, in the product b,. Denote the product a, * ay by c, which is an
element of A. Replace a, * a, by c, and use the hypothesis of induction
to write b, * b, as a left-associated product.
If k = 1 and b, has only one factor, then the product b, * by is a, * a5,
which is already in left-associated form. Otherwise, b, has at least two
factors, and we may denote by d the product of the first two factors
dy and a; of b,. Replace a, * a; by d in the product b, *b,, and apply
the hypothesis of induction. Finally, apply the associative property to
a, * (ay * a3) to write the entire product in left-associated form.
(b)
(d)
F. The claim is false. One may premultiply (multiply on the left) or postmultiply both sides of an equation by equal quantities. Multiplying one
side on the left and the other on the right does not always preserve equality.
F. The proof makes the assumption that xy # 0, which may be false.
Exercises 6.2
(c)
La
1
iP
Seti
i
py
aU
pews
|
=
ee
eea)
=
Pa
ar
Sil
i
Ite
i
]
eal
The set is closed under the operation -,
the element | is the identity, and every
element has an inverse. It can be
checked that the operation is associative.
The operation - is commutative, so the
group is abelian.
(d) Hint: © is the identity.
Answers to Selected Exercises
421
Ms
6.
(b)
10.
Hint: For a, b € G, compute and compare a7b? and (ab)”.
11.
17.
Hint: To have both cancellation properties, every element must occur in every
row and in every column of the table.
(a) v,w,e, and u
(b) Hint: Let a, b € G. For an element x such that a * x = b, try a7! *b.
(a).
= DG Dias 1 0
18.
(b)
F.A minor criticism is that no special case is needed for e. The fatal flaw
(c)
is the use of the undefined division notation.
A. The proof is correct but provides minimal explanation.
14.
[3214], [1 23 4], and [2 4 1 3].
Exercises 6.3
Ie
9.
11.
(a)
Ol 210.4), 10,2, 4.0)
(e)
Hint: There are six subgroups.
The order of 0 is 1. The elements 1, 3, 5, and 7 have order 8. The ele-
(c)
(a)
ments 2 and 6 have order 4. The order of 4 is 2.
e€C,,
Proof. The set C, is not empty because ea =a=ae, so
sides
both
Multiplying
ay.
=
ya
and
ax
=
xa
Then
Let x,y@C,.
of the last equation by y~!, we have y—!(ya)y7! = y"'(ay)y!.
Thus, (y~!y(ay~)) = (y~!a)(yy-), or ay! =y7'a. Therefore,
(xy—)a = x(y7!a) = x(ay7!) = (a)y7! =(axy~! = a(xy™!).
This
shows xy~! € C,,. Therefore, C, is a subgroup of G, by Theorem 632
12.
Proof.
The identity e € H because H is a group, and, thus, a~!ea € K. Thus
Kis not empty. Suppose that b, c € K. Then b = a~'hyaandc = a“ ‘haa for
a\@ 4h;
some h,, h, € H. Thus, be~! = (@'h,a\@'hya) = (Gh
a)—
a'h,(aa~')hy!a = a~'h,hz'a. But H is a group, so h,hz' eH. Thus
bc~! € K. Therefore, K is a subgroup of G.
15.
Hint: Let H be a subgroup of (a). If H is the trivial subgroup, it is clearly
cyclic. Otherwise, show that there is a positive integer ft such that a’ is in H.
Now use the Well-Ordering Principle to find the smallest such 7, and use the
Division Algorithm to show that this power of a is a generator for H.
16,
(©
(a)
17.
a? a
C. The proof omits the step of verifying that H M G is nonempty.
Exercises 6.4
4. Let ge. Thenl(f +g) = [ft dx
perl):
= f?fadx + J? ga)dx =
422
Answers to Selected Exercises
9,
(a)
(c)
13.
(a)
Hint: Suppose that
G = ({e, a}, 0) and H = ({i, b}, *) are two groups
with identity elements e and i. To define an isomorphism from G to H,
first determine the image of e.
Hint: Use Theorem 6.4.1(e) to show that the algebraic system in
Exercise 3 of Section 6.2 is not isomorphic to (Z,, +).
Proof. Suppose that y = x in Z,,. Then 18 divides x — y. Therefore, 6
divides x — y, so 24 divides 4(x — y) = 4x — 4y. Thus, f(x) = 4x =
4y = f(y) in -Zoy, sovf 1s) well defined Now let~x,ye Z. Then
fix + y) = 4@ + y) =f) + fd).
(b)
Rng (f) =
0
4
8
1,
16
20
{0, 4, 8, 12, 16, 20} The table is
0
4
8
12
16
20
0
4
8
12
16
20
4
8
12
16
20
0
8
12
16
20
0
4
12
16
20
0
4
8
16
20
0
4
8
We
20
0
4
8
12
16
Exercises 6.5
1.
(b)
The interval [—1, 1] is not a ring because it is not closed under addition.
(For example, ; te ;in not an element of [—1, 1].)
4.
Hint: First show that Z x Z is closed under © and ® and that the
additive identity is (0, 0). The inverse of (a, b) is (—a, —b) because
(a, b) ® (—a, —b) = (a + (—a), b + (—b)) = (0, 0) and
(—a,—b) @ (a, b) = (—-a + a, —
:
15.
b+ b) = (0, 0).
Explain why the remaining axioms hold for Z x Z.
(a)
Hint: F. The claim is false.
Exercises 7.1
1. (b) 5.2
2.
3.
4.
(b)
(a)
(c)
Oand all negative real numbers are lower bounds.
supremum: 1; infimum: 0
supremum: does not exist; infimum: 0
(e)
supremum:
(g)
(a)
supremum: 5; infimum: —1
Hint: Show that an upper bound for A is an upper bound for B.
|; infimum: 5
Answers to Selected Exercises
423
6.
(a)
Hint: Suppose that b is an upper bound for A. Therefore, for all x,
if x © A, then x < b. This means that for all x, if x > b, then x € A.
Explain why A‘ is not bounded above.
12.
(a)
Proof. Lets = sup(A) and B = {u: wis an upper bound for A}. Then
B is bounded below (by elements of A) so inf(B) exists. Let t = inf(B).
We must show s = t.
(i)
To show t < s, we note that because s = sup(A), s is an upper
bound forA. Thus s € B. Therefore t < s.
(ii)
To show s < t, we will show ¢ is an upper bound for A. If ¢ is
not an upper bound for A, then there exists a € A with a > t. Let
e —
13.
(a)
4—!
ue B
Because t = inf(B) and t < t + ¢, there exists
such that wu< t + ¢. Butt + € < a. Therefore, u < a, whichis a
contradiction because u € Banda € A.
Proof. Suppose that sup(A) and sup(B) exist. Then A U B is bounded
above by m = max {sup(A), sup(B)}
(see Exercise 4(c)). By the com-
pleteness property, sup(A U B) exists. We show that sup(A U B) = m.
(i)
Because
ACAUB,
BCAUB
implies
we
that
m = max {sup(A), sup(B)}
(ii)
have
sup(A) < sup(A U B).
sup(B) < sup(A UB).
It follows
Also,
that
< sup(A U B).
It suffices to show m is an upper bound for A U B. Let x € AU B. If
x € A, then x < sup(A) < m. Ifx € B, then x < sup(B) < m. Thus,
18.
20.
mis an upper bound for A U B. Hence sup(A U B) < m.
Hint: Let F be an ordered field. Assume that F is complete and let A be a
nonempty subset of F that has a lower bound in F. To show that A has an
infimum in F, begin by defining the set AW = {—x: x € A}. Prove that A~ has
a supremum and then find an infimum for A.
(a) F. The claim is true, but y=7 + 5 might not be in A.
Exercises 7.2
(b)
7="93:825;76' = 0.025
4. (e) @
@
Ujm+0.1,n +0.2)
neN
19.
(b)
(a)
(d)
(g)
(a)
22.
(c)_
7.
17.
(i) closed
(e) open
open
not compact (not bounded)
not compact (neither closed nor bounded)
not compact (not closed)
Hint: First, show that a cover for A UB is a cover for A and a cover
for B.
C. With the addition of O* to the cover {O,:a€ A}, we are assured
that there is a finite subcover of {O*} U {O,: a € A} forA, but not nec-
essarily a subcover of {O,: a € A}. However, because Oe
useless in a cover for B, it can be deleted from the subcover.
Aes
424
Answers to Selected Exercises
Exercises 7.3
1\n
1.
(d)
Aint: Use the fact the lim (1oH =) = &
3. (a) {5}
(g)
4.
5.
7.
10.
©) ©
{0,2}
(k)
N
{0,1}
Hint: Leta € A. Then z > a. Show z is an accumulation point of A by using
Theorem 7.1.1.
(b) Hint: Use Exercise 6(a).
(a) The set has no accumulation points.
(c)
The set has at least one accumulation point.
14.
(a) _ F. Hint: Identify the misuse of quantifiers.
(c) F. The claim is false. (B°)’ need not be a subset of (B’)°.
1.
(a)
bounded below by 19, not bounded above
(c)
bounded; bounded above by ir bounded below by 0
Exercises 7.4
(e) bounded; bounded above by 10, bounded below by 0
(g) not bounded above, not bounded below
(i) |bounded; bounded above by 0.81, bounded below by —0.9
5.
Hint: Let y be bounded and B be a number such that |y,,| < B for alln Ee N.
Use the definition of x, > 0 with ri
(ital)!
6.
(g)
Hint: It suffices to show that for n € N, (n =. Dye
8.
Hint: By the Binomial Theorem,
1 n(n —
palin
1
and x41 = (1ole
=1+1
1)
1 n(n —
Dame aA
n!
=
n
1)\(m — 2)
bau
“s
BS eab Persea
n+1
)
1 (+
1)n
2!(@@+1)?
Us
n!
DG
1 (n+ 1)(n\(n —
3!
= ee
Gly
1)
m:
(n + 1)
1
(n
+ 1)!
(n+ 1)! (n 4 1th
Answers to Selected Exercises
425
The steps in the proof that x, < x,,,, for all n € N are:
(i) Show that
7— ee
n
n+
rm
1
tonal1.
1
Bry
(ii) Show that for all k < n, AH i
!
1
1
=
=
ade
ieee
7
5 BHo
n
0
#
ae
(el
Less
=
Ge =
2 is less
9)
2
; ish
(nb 1)
(iii) Show that each term in the expansion of x, is less than the corresponding
term in the expansion of Xp41.
than or equal to — fest
k}
(iv)
Oey
Observe that the binomial expansion of x,,,, has one more positive term
than the binomial expansion of x.
Exercises 7.5
4.
7.
(a)
(a)
Hint: Consider rational numbers in [7, 8] whose square is less than 50:
Hint: Consider a set A that includes [0, 2] and the sequence
or
(b)
Ot ae Li
F. The claim is correct, but there is little that is correct in this proof.
For instance, the upper bound ay for A may be negative, in which case
B would have to be defined differently. The most serious error is that
there is no connection between being an accumulation point and being
an upper bound for a set.
Answers to Exercises in Appendix
(a)
(b)
(c)
(d)
(e)
(a)
{x: x is an integer and 6 < x < 12} = {7, 8,9, 10, 11}
{: x is an integer and x? is less than 17} = {—4, —3, —2, 1, 0, 1, 2,
3, 4}
{x: x is a solution to ie Sie 0\t 9, 9}
{x: x is an integer power of 2} = {ae 6,445 1251; 2, 4, 8, 16, 32, ar
{x: x is a component of a peanut butter and jelly sandwich} = {peanut
butter, jelly, bread}
(—2,2}
(b)
Sets (a), (c), and (d) are finite.
(a)
(b)
(a)
AO EIS
ONE TRE, CHUL ENSUE AN)
{A} is a set with one element. That element, A, is not a member of the
set A, whose only elements are j, m, and h.
672 = 25- 3-7. Divisors of 672 are 1, 2, 3, 4, 6, 7, 8, 12, 14, 16, 21, 24,
28, 32, 42, 48, 56, 84, 96, 112, 168, 224, 336, and 672:
426
Answers to Selected Exercises
(b)
(c)
673 is prime. Its divisors are 1 and 673.
(a)
5 divides 65 because 65 = 5 - 13.
7 does not divide 23 because there is no natural number b such that
23 = 7b. (For b = 1, 2, or 3, 7b is less than 23. For all b > 3, 7b is greater
than 23.)
675 = 3°- 5*, Divisors of 675 are 1, 3, 5, 9, 15, 25, 27, 45, 75, 135, 225,
and 675.
2.
(b)
3.
4.
(a)
False
(b) True
(c) True
(d) False
(e)
False
(f)
False
(g)
False
(h)
True
(i)
(m)
True
(j)
False
(k)
False
(1)
True
(a)
Distributivity
Distributivity and commutativity of addition
Associativity and commutativity of multiplication
By trichotomy, every real number is equal to 0, or positive, or negative.
The product of 0 with itself is 0, and the product of two positives or two
negatives is positive.
(b)
(c)
(d)
True
5.
(a), (b), and (c) because they have degree 8.
1.
Some people in Nashville, Tennessee, have more than one telephone number.
Therefore, the rule of correspondence is not single valued, so it is not a function. Note that the fact that some people do not have a telephone does not prevent this correspondence from being a function: If everyone who has a phone
had only one number, then those without a phone would not be in the domain
of the function. If each person with a phone had only one number and several
people shared the same number, the rule would still be a function.
2
ena)
(b)
(c)
(d)
3.
(3) = 12, f(5) = 16. The pre-images of 0 and —2 are —3 and —4,
respectively. Dom (f) = R and Rng (f)=R.
f(3) = 16, f(5) = 48. The pre-images of0 are + 1. There is no pre-image
of —2. Dom (f) = R and Rng (f) = [—2, oo).
(3) = —1 and f(5) = 1. There is no pre-image of 0. The pre-image of —2
is 7/2. Dom (f) is the set of all reals except 4, and Rng (f) is the set of
all reals except 0.
f3) = 2V 15, f(S) = 10. The pre-image of 0 is 0. There is no pre-image
of 7. Dom (== (0) co) and. Rng (7) = [0N.c0):
(a)
For all x in A, x corresponds to 0 and 3. There are other correct answers.
(b)
For all x in A, x corresponds to x — 1. There are other correct answers.
For all x in A, x corresponds to 0 if x is even and to | if x is odd. There
(c)
are other correct answers.
IN
DE
X
A
Abel, Niels, 313
Abelian groups, 313
Accumulation point, 365
Algebraic proofs, 145
Algebraic structures, 303-309
Algebraic system
definition of, 304
properties of, 304-309
Antecedent, 10
Antisymmetric property of relations, 189
Appel, Kenneth, 38
Archimedean Principle, 122, 347, 364
Arcs, 156
Arithmetic mean, 373
Associative Laws, 5
Associative property, 5, 96, 160, 215, 305, 381
Axiomatic set theory, 86
Axiom of Choice, 295-296
Axioms
consistent systems of, 48
definition of, 28
of natural numbers, 381
B
Banach-Tarski paradox, 298
Bernstein, Felix, 290
Biconditional sentences
definition of, 12
proof of, 46,70
Bijections. See also One-to-one correspondence
construction of, 233
definition of, 231
permutation as, 234
Binary operations, 79, 95, 303
Binomial coefficients, 142
Binomial Theorem, 145-147
Bolzano, Bernard, 363
Bolzano-Weierstrass Theorem, 363-366
Borel, Emile, 358
Bound, upper and lower, 345, 369
Bounded above, 345, 369
Bounded below, 345, 369
Bounded Monotone Sequence Theorem, 368-371
Bounded sequences, 369
Bounded sets, 345
(€
Cancellation Law, 186
Canonical map, 208
Cantor, Georg, 279, 287
Cantor-Schréder-Bernstein Theorem, 290-292
Cantor’s Theorem, 289-290
Cardinality
comparability of cardinal numbers and, 295
countable sets and, 279-284
equivalent sets and, 264
finite sets and, 264
infinite sets and, 271
ordering of cardinal numbers and, 287-293
Cardinal numbers
comparability of, 295
definition of, 264, 271, 275, 287
finite, 264
infinite, 271
ordering of, 287-293
Cartesian product 100
Cauchy, Augustin Louis, 373
Cauchy sequence, 373, 376
Cayley, Arthur, 304
Cayley’s Theorem, 330-331
Cayley tables, 304
Characteristic function, 207
Choice function, 296
Closed set, 352
Closed under an operation, 304
Codomain, 204
Cohen, Paul, 300
427
428
Index
Combination Rule, 143-144
Definitions, 29
Combinatorial proof, 145
Common divisor, 76
Commutative (Abelian) group, 313
DeMorgan, Augustus, 5
De Morgan’s Laws, 5, 99, 125
Commutative property, 5, 96, 159, 214, 305, 381
Denial, 6, 24, 25
Commutative ring, 337
Compact sets, 357
Comparability property, 189
Comparability Theorem, 295-296
Complement of a set 98
Complete (strong) induction, 128
Completeness
Denumerable sets, 271, 274
Derived sets, 364
Descartes, René, 100
of an ordered field, 348-349
equivalents of, 374-377
Complex numbers, 383
Complex Root Theorem, 55-56
Composition
of functions, 213
of relations, 158
Conditionals, 10-15
Conditional sentences
converse and contrapositive of, 11-12
definition of, 10
De Moivre’s formula, 121—122
Difference, set operation of, 95
Digraph 156
Diophantine equation, 80
Direct proof 32, 51, 68
Disjoint sets, 95
Disjunction, 3
Distributive Laws, 5
Dividend, 76
Divides, 33-34
Division Algorithm, 76, 131-132
Divisor, 76. See also Common divisor
Domain
of a function, 204
of a relation, 135
Double Negation Law, 5
direct proof of, 36
Congruence, 168-171, 182-186
Conjunction 2
Connectives, 2
Consequent, 10
Consistent axiom systems, 48
Continuum, 275
Continuum hypothesis, 299-300
Constant function, 253
Constant sequence, 246
Contradiction
E
Element-chasing proof, 91
Empty set, 86
Equivalence relations
definition of, 166
partitions and, 177-179
Equivalence class, 169
Equivalent sentences, 5
Equivalent sets, 262-264
Euclidean axioms, 29, 48
definition of, 4
Euclid of Alexandria, 29
proof by, 43, 56, 59, 69
Euclid’s algorithm, 78
Euclid’s Lemma, 80, 132
Euler, Leonard, 55
Exhaustion, proof by, 37
Existence theorems, 54-56, 58
Existential quantifier, 19
Contraposition, proof by, 42, 68
Contrapositive, of sentence, 11
Convergence, of sequence, 255-256
Converse, of sentence, 11
Coprime, 80
Countable sets, 279-284
Counterexamples, 59
Counting, two-way, 143
Cover, 356
Cross product, 100
Cyclic group, 323
Cyclic subgroup, 323
D
Decimal numbers, normalized form, 274
Dedekind, Richard, 376
Dedekind cuts, 376
Deductive reasoning, |
F
Factorial, 116
Family of sets. See also Nested family
definition of, 104
indexed, 107-110
Fibonacci, Leonardo, 130
Fibonacci numbers, 130
Fields
algebraic properties of, 338-339
complete, 347
ordered, 343-349
Fifth postulate (Euclid), 48
Index
Finite set, 264
Four-Color Theorem, 38
Fraenkel, Abraham, 86
Functions
characteristic, 207
choice, 296
codomain of, 204
composite, 213
decreasing, 218
definition of, 204
domain of, 205
H
Haken, Wolfgang, 38
Hasse diagram, 192, 198
Heine, Edward, 358
Heine-Borel Theorem, 351-360
Hilbert, David, 281
Hippasus, 44
Homomorphic image, 328
Homomorphism
group, 328
ring, 337
equality of, 206
Howard, Paul, 298
extension of, 216
Hypothesis of induction, 117
greatest integer, 208
identity, 207
image of sets and, 238
inclusion, 207
increasing, 218
induced, 240
inverse of, 213, 231
limit of, 251
one-to-one, 224-226
I
Identity element, 305
Identity function, 207
Identity permutation, 235
Identity relation, 154
Images of sets, 238-240
Immediate predecessor, 192
Inclusion function, 207
one-to-one correspondence and inverse, 231-234
onto, 222-224
Index, 107
Indexed families of sets, 107
piecewise defined, 217-218
Indexing set, 107-109
Indirect proofs, 42-45
Induced function, 240
range of, 204
real, 205
restriction of, 216
Fundamental Theorem of Arithmetic, 132,
297, 382
Fundamental Theorem of Algebra, 55
Induction
Generalized Principle of Mathematical, 122
Principle of Complete, 128-130
Principle of Mathematical, 115-118
G
Inductive set, 115
Infimum, 194, 346
Galois, Evariste, 312
Gardner, Martin, 267, 281
Infinite cardinality, 276
Infinite Hotel, 281-282
Gauss, Carl Friedrich, 168
Generalized Principle of Mathematical Induction,
122-123
Infinite order, 323
Infinite sequence, 208
Generator, 323
Geometric mean, 373
Godel, Kurt, 300
Goldbach, Christian, 61
Goldbach Conjecture, 61
Greatest common divisor, 77
Greatest element. See Largest element
Greatest integer function, 208
Greatest lower bound (infimum), 194, 346
Groups. See also Subgroups
abelian, 313
cyclic, 323
definition of, 312
homomorphism, 328
permutation, 314-315
symmetric, 314
Infinite sets, 271
Injection, 224
Integers, 382
Interior point, 352
Integral domain, 337
Intersection
over family of sets, 105
set operation of, 95
Inverse element, 312
Inverse functions, 231-235
Inverse permutation, 235
Inverse relation, 157
Irreflexive relation, 189
Isomorphic
groups, 329
rings 337
Isomorphism, 329
429
430
Index
K
Kernel, 333
L
Largest element (greatest element), 195
Law of Excluded Middle, 4
Least element. See Smallest element
Least upper bound (supremum), 194, 346
Left identity, 310
Leibnitz, G. W., 203
Open sets, 351
Operation preserving (OP), 326
Operation tables, 304
Order
of an element, 323
of a group, 312
Ordered fields, 377
Ordered n-tuples, 99
Ordered pairs, 99
Ordering relations, 189
Leonardo of Pisa (Leonardo Fibonacci), 130
Limit
of a sequence, 244
of a function, 251
Linear combination, 53, 77-78
Linear order, 195
Logical equivalences, 30
Loop, 165
Lower bound, 194, 345
P
Pairwise disjoint family 109
Paradox, 2
Parity, 46
Partially ordered set (poset), 190
Partitions
definition of, 175
equivalence relation and, 177
Pascal, Blaise, 146
M
Mapping (see Function)
Mathematical induction, 114
Mean Value Theorem, 221
Modulus, of congruence, 168
Modular arithmetic, 182
Modus ponens, 30
Monotone sequences, 370
Pascal’s triangle, 146
Peano, Giuseppe, 114
Permutation groups, 314, 350
Permutation Rule, 141
Permutations
composites of, 235
counting number of, 141
definition of, 141, 234
identity, 235
N
Natural numbers, 114
inverses of, 235
Piecewise-defined functions, 217
infinite sets and, 243-248 271
Negations
of proposition, 2
simplified form of, 24
Pigeonhole Principle, 267
Poset, 190
Postulates, 28
Power set, 90
Negative, 316
Pre-image, 204
Nested family, 113
Neighborhood, 352
Principle of Complete Induction (PCI), 128
Principle of Inclusion and Exclusion, 139
Nim (game), 128-129
Principle of Mathematical Induction (PMI)
Non-decreasing sequence, 370
Non-increasing sequence, 370
generalized, 122
proofs using, 117
statement of, 115
Product notation, 115
Product Rule, 91
Products, 304
Normalized form, 274
Null set, 86
Number theory, 76
O
One-to-one correspondence. See also bijections
definition of, 231
number of elements in set and, 262
One-to-one functions,
definition of 224
Horizontal Line Test, 225
inverse of ,233
Open sentence, 18
Proofs, |
algebraic, 145
basic methods I, 28-38
basic methods I, 42-48
of biconditional sentences, 45
combinatorial, 145
by contradiction, 48
by contraposition, 42
direct, 32
Index
by cases (exhaustion), 37
indirect, 42
by induction, 117
involving quantifiers, 51
strategies for writing, 67
writing style, 64
Proper subgroup, 321
Proper subset, 89
Propositions
ambiguity and, 6
biconditional, 12
compound, 3
conditionals, 10
contrapositive, 11
converse, 11
definition of, |
denial of, 6
equivalent, 4
negation of, 2
inverse of, 157
irreflexive property of, 189
ordering, 189
range of, 155
reflexivity property of, 163
symmetric property of, 163
transitive property of, 163
Relatively prime, 80
Remainder, 76
Replacement rule, 30
Residue, 170
Restrictions, of function, 216
Right identity, 310
Ring homomorphism, 337
Ring isomorphism, 337
Rings
commutative, 310, 311
definition of, 334
with unity, 337
Pugh, Charles Chapman, 376
Pythagoras, 29
Pythagorean Theorem, 29
Rubin, Jean E., 298
Q
S
Quantified sentence
definition of, 18
strategies for dealing with, 61
Quantifiers
existential, 19
explanation of, 18—25
hidden, 20-21
incorrect deductions and, 55-56
proofs involving, 48-56
unique existential, 25
universal, 20
Quotient, 76
R
Range of a function, 204
of a relation, 155
Rational functions, 257
Rational numbers, 382
Real functions, 205
Real numbers, 382
Reasoning, deductive, |
Reflexive property of relations, 163
Relations
antisymmetric property of, 189
comparability property of, 189
definition of, 153
domain of, 155
equivalence, 166
functions as, 204
Russell, Bertrand, 2
Russell Paradox, 86
Schroder, Ernst, 290
Sentence
conditional, 10
contrapositive of, 11
converse of, 11
definition of, 1
denial of, 6
negation of, 2
quantified, 19
Sequences
bounded, 369
Cauchy, 373
convergence of, 244
decreasing, 370
definition of, 208
divergence of, 244
increasing, 370
limit of, 244
monotone, 370
nth term of, 208
Sets
bounded, 369
cardinality, 275
closed, 352
compact, 357
countable, 273
denumerable, 271
derived, 364
equality, 88
equivalent, 262
431
432
Index
Sets (continued)
family of, 104
Towers of Hanoi, 126
1, 28
Theorem
finite, 264
images of, 238
Transitive property of relations, 163
Trichotomy property, 295
infimum of, 196, 346
Trigonometric functions, 257
infinite, 264
Trivial subgroup, 321
open, 253
Truth set, 18, 19
partially ordered, 190
partition of, 179
permutations of, 141
supremum of, 196, 346
truth, 18
uncountable, 273
Set theory
Truth table, 3
Two-way counting, 143
U
Unary operations, 98
Uncountable set, 273
Undecidable statements, 48
axiomatic, 90
Undefined terms, 29
basic concepts of, 85
Union
Sigma notation, 115
over family of sets, 104
Single-valued correspondence, 204
Smallest element (least element), 195
Strict partial order (strict partial ordering), 190
Strong (complete) induction, 128
Subcover, 356
Subgroups, 320. See also Groups
set operation of, 95
Unique existential quantifier, 25
Unit, 307
Universal quantifier, 20
Universe of discourse, 18
Upper bound, 193, 345
Subsequence, 250, 373
Subsets
Vv
definition of, 86
Venn, John, 89
proper, 89
Successor, 114
Sum Rule, 137, 138
Supremum, 196, 346
Surjection, 222
‘
Venn diagrams, 89, 95
Vertex, 156
Vertical Line Test, 205
Ww
Symmetric closure, 174
Weierstrass, Karl, 363
Symmetric group on n symbols, 314
Symmetric property of relations, 163
Well ordering, 196
Well-Ordering Principle (WOP), 130-131, 324, 365
Well-Ordering Theorem, 196
T
Tautologies, 4
"
Z
Tautology rule, 30
Ternary operation, 303
Zermelo, Ernst, 86
Zermelo-Fraenkel set theory, 86, 295
Total order. See Linear order
Zero divisor, 307
HA Hay AA
LIST OF SYMBOLS
Appendix
gcd(a, b)
xis anelement of setA
379
Iem(a, b)
xis not anelement of setA
379
set of all objects x such that P(x)
empty set
379
379
{x: P(x)}
set A and set B are equal 380
set of natural numbers
380
2)
set of realnumbers
A= B
381
P(A)
381
set of complex numbers
381
open interval of real numbers between
aand b, excluding aandb 383
closed interval of real numbers between
aand b, includingaandb 383
383
(a, CO), (— 00,a) open rays of realnumbers
closed
rays
of
real
numbers
383
[a, OO), (—oo, a]
fis a function from A toB
384
f:AvzB
image (or value) of x under f 384
f(x)
domain of the function f 384
Dom(f)
range of the function f 384
Rng(f)
Chapter 1
PaO
PNZO
(P= (Q
IP
(),
Piff@Q
P(x)
(Ax) P(x)
negation of P 2
conjunction,
PandQ 2
disjunction,
PorQ 3
conditional sentence, “if P then Q”
10
biconditional sentence, “P if and only
her
WD
open sentence
18
existentially quantified sentence, “there
exists x such that P(x)”
(Vx)P(x)
19
aZ
AUB
ANB
A-B
AS
(a, b)
(Ginna
AxB
AAB
LJA
set of all objects x such that P(x)
empty set 86
AisasubsetofB
86
Ais notasubset of B 86
AcE
intersection of the family of sets 4
105
indexed family of sets, indexed
by set A 107
union of an indexed family of sets
108
intersection of an indexed family of
sets
108
union of a family of sets indexed by
natural numbers
109
intersection of a family of sets indexed
by natural numbers
109
sum Of aj, d,... > an
20)
(ax € A)P(x)
there exists x € A such that P(x)
(Wx € A)P(x)
for allx € A, P(x)
(A!x)P(x)
unique existentially quantified sentence,
“there is a unique x such that
TAGay? 2S)
22
22
82
set A and set B are equal 88
power set of the setA 90
set of integer multiples of a 93
union of sets A and B 95
intersection of sets A andB
95
difference of sets AandB 95
complement of setA
98
ordered pair with coordinates
aandb
99
ordered n-tuple
100
cross product of AandB_
100
symmetric difference of A and B 103
union of the family of sets 4 104
Principle of Mathematical Induction
universally quantified sentence,
Sito alllas JAGay’
ACB
AGB
380
set of rational numbers
~P
Chapter 2
Aisasubset of B 380
set of integers
greatest common divisor of integers a
andb
76
least common multiple of integers
aandb
82
115
115
product of aj, a,... ene
n-factorial
116
Principle of Complete Induction
Well Ordering Principle
130
128
number of elements inthe setA
137
binomial coefficient, n binomialr
142
x,
limit of the sequence xisL
co" n
SS iby ag, 2 Ib,
lim fix) = L limit of fas x approachesa
lim n=
244
251
Seis
|
aa,
We
Chapter 5
Chapter 3
aRb
aRb
Dom(R)
Rng(R)
sup(B)
inf(B)
ais R-relatedtob
153
ais not R-related tob
153
domain of the relation R 155
range of the relation R 155
identity relation on the setA
154
inverse of the relation R 157
composite of relations
Rand S$ 158
x mod R, equivalence class of object x
modulo the relation R 167
A modulo R, the set of equivalence
classes
167
x is congruent to ymodulom
168
set of equivalence classes for congruence modulom
169
sum of equivalence classes x and y
modulom
183
product of equivalence classes x and y
modulom
183
supremum (least upper bound) for the
setB
194
infimum (greatest lower bound) for
the setB
194
Chapter 4
f:A->B
f(x)
Dom(/)
Rng(f)
I,
AB
fiA-SB
[abc]
F(X)
fa)
greatest integer of x 208
nth term of the sequence x 208
inverse of the functionf 213
composite of functions fandg 213
restriction of the function f to the
setD
216
function fis onto the set B 222°
function fis one-to-one
224
permutation of a,b,c 234
image of the set X 238
inverse image of the set Y 238
set A is equivalent to set B. 262
set A is not equivalent to set B. 262
set of natural numbers from | throughk
A
cardinality, or cardinal number of the
setA
264
cardinal number aleph naught, the cardinality
of N 271
Xo
207
264
c
cardinal number ec (for continuum), the
A=B
cardinal numbers A and B are equal
A<B_
cardinal number is A less than or equal to
A <B
cardinal number is A less than cardinal
cardinality of (0,1)
cardinal number
number B
275
288
B 288
288
Chapter 6
(A, *)
* is an operation on set
A, (A, *) is an
Lu
x!
algebraic system 303
set of units of the system Z,, 307
inverse of the element x 312
[abc]
S,,
permutation of a,b,c
314
symmetric group on n symbols (or, of
—a
negative of theelementa
<a>
G
OP
ker()
(R, +, :)
cyclic subgroup generated bya
centralizer of aina group 325
operation preserving (mapping)
kernel of the OP map f 333
ring with two binary operations
permutations of N,)
fisafunction from A toB 204
image (or value) of x underf 204
domain of the function f 204
range of the function f 204
identity function on the setA 207
characteristic function of the set A
AXB
AX®B
Ny
314
316
323
326
334
Chapter 7
sup(B)
supremum (least upper bound) for the set
inf(B)
infimum (greatest lower bound) for the
B
N(a, 8)
A'
346
set B 346
S-neighborhood ofa 352
derived set (of accumulation points) of set
A 364
eater
ela ¢)e110
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