Course Development Team
Head of Programme
: Dr Lau Jing Feng
Course Developer(s)
: Dr Ng E-Jay
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: Gevin Leow, ETP
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How to cite this Study Guide (APA):
Ng, E. J. (2021). MTH105 Fundamentals of mathematics (Study Guide). Singapore: Singapore
University of Social Sciences.
Release V1.4
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Table of Contents
Table of Contents
Course Guide
1. Welcome.................................................................................................................. CG-2
2. Course Description and Aims............................................................................ CG-3
3. Learning Outcomes.............................................................................................. CG-6
4. Learning Material................................................................................................. CG-7
5. Assessment Overview.......................................................................................... CG-8
6. Course Schedule.................................................................................................. CG-10
7. Learning Mode.................................................................................................... CG-11
Study Unit 1: Propositional Logic
Learning Outcomes................................................................................................. SU1-2
Overview................................................................................................................... SU1-3
Chapter 1: Compound Statements, Truth Tables, and Logical
Equivalences............................................................................................................. SU1-4
Chapter 2: Conditional Statements..................................................................... SU1-17
Chapter 3: Constructing Arguments................................................................... SU1-29
Summary................................................................................................................. SU1-44
Formative Assessment.......................................................................................... SU1-45
References............................................................................................................... SU1-68
i
Table of Contents
Study Unit 2: Predicate Logic and Quantified Statements
Learning Outcomes................................................................................................. SU2-2
Overview................................................................................................................... SU2-3
Chapter 1: Predicates and Quantifiers.................................................................. SU2-4
Chapter 2: Properties of Quantified Statements............................................... SU2-18
Chapter 3: Arguments with Quantified Statements......................................... SU2-30
Summary................................................................................................................. SU2-37
Formative Assessment.......................................................................................... SU2-38
References............................................................................................................... SU2-56
Study Unit 3: Methods of Proof
Learning Outcomes................................................................................................. SU3-2
Overview................................................................................................................... SU3-3
Chapter 1: Introduction to Some Terminology and Concepts.......................... SU3-4
Chapter 2: Direct Proofs and Usage of Counter-Examples............................. SU3-11
Chapter 3: More Proof Techniques, Indirect Proofs.......................................... SU3-22
Summary................................................................................................................. SU3-34
Formative Assessment.......................................................................................... SU3-35
References............................................................................................................... SU3-49
Study Unit 4: Sequences, Induction, the Euclidean Algorithm
Learning Outcomes................................................................................................. SU4-2
Overview................................................................................................................... SU4-3
ii
Table of Contents
Chapter 1: Sequences............................................................................................... SU4-4
Chapter 2: The Well-Ordering Property and Induction................................... SU4-13
Chapter 3: Greatest Common Divisor and the Euclidean Algorithm............ SU4-30
Summary................................................................................................................. SU4-52
Formative Assessment.......................................................................................... SU4-53
References............................................................................................................... SU4-77
Study Unit 5: Set Theory
Learning Outcomes................................................................................................. SU5-2
Overview................................................................................................................... SU5-4
Chapter 1: Introduction to the Concept of a Set................................................. SU5-5
Chapter 2: Set Theoretic Operations.................................................................. SU5-15
Chapter 3: Properties of Sets................................................................................ SU5-23
Chapter 4: Counting and Probability................................................................. SU5-32
Chapter 5: The Mathematics of Voting............................................................... SU5-60
Summary................................................................................................................. SU5-78
Formative Assessment.......................................................................................... SU5-79
References............................................................................................................. SU5-100
Study Unit 6: Relations and Functions
Learning Outcomes................................................................................................. SU6-2
Overview................................................................................................................... SU6-3
Chapter 1: Relations and Congruency.................................................................. SU6-4
Chapter 2: Definition and Basic Properties of Functions................................. SU6-33
iii
Table of Contents
Chapter 3: Injective, Surjective, Bijective Functions, and Composition of
Functions................................................................................................................. SU6-46
Summary................................................................................................................. SU6-64
Formative Assessment.......................................................................................... SU6-65
References............................................................................................................... SU6-80
iv
List of Tables
List of Tables
Table 1.1 Truth table for negation.............................................................................. SU1-5
Table 1.2 Truth table for conjunction......................................................................... SU1-6
Table 1.3 Truth table for disjunction.......................................................................... SU1-6
Table 1.4 Truth table for
................................................................................. SU1-8
Table 1.5 Truth table for
........................................................................... SU1-8
Table 1.6 The logical equivalence of
and
Table 1.7 The logical equivalence of
Table 1.8
.............................................. SU1-10
and ............................................ SU1-10
is not logically equivalent to
................................ SU1-11
Table 1.9 Truth table demonstrating that
is a tautology......................... SU1-12
Table 1.10 Truth table demonstrating that
is a contradiction................. SU1-12
Table 1.11 An important collection of logical equivalences................................. SU1-13
Table 1.12 Truth table for
................................................................................ SU1-17
Table 1.13 Truth table for
................................................................ SU1-18
Table 1.14 Logical equivalence of
Table 1.15
and
............................................. SU1-19
is not logically equivalent to
Table 1.16 Logical equivalence of
and
Table 1.17 Logical equivalence of
Table 1.18 Truth table for
............................... SU1-20
...................................... SU1-20
and
................. SU1-22
................................................................................ SU1-24
v
List of Tables
Table 1.19 Logical equivalence of
and
....................... SU1-25
Table 1.20 Truth table determining the validity of an argument......................... SU1-30
Table 1.21 Truth table determining the validity of an argument......................... SU1-31
Table 1.22 A summary of the rules of inference.................................................... SU1-37
Table 2.1 Commonly used sets and the symbols denoting them.......................... SU2-6
Table 2.2 The approach to multiply quantified statements.................................. SU2-25
Table 2.3 The negation of multiply quantified statements................................... SU2-28
Table 2.4 Rules of inference for quantified statements......................................... SU2-30
Table 5.1 An important collection of set identities................................................ SU5-23
Table 5.2 An important collection of logical equivalences................................... SU5-24
Table 5.3 A schedule showing the approvals each candidate has received....... SU5-61
Table 5.4 A preference schedule showing the rankings each candidate has
received......................................................................................................................... SU5-62
Table 5.5 A preference schedule showing the various ranking
combinations................................................................................................................ SU5-63
Table 5.6 A preference schedule showing various ranking combinations......... SU5-65
Table 5.7 Revised preference schedule.................................................................... SU5-65
Table 5.8 A preference schedule showing various ranking combinations......... SU5-67
Table 5.9 A revised preference schedule................................................................. SU5-67
Table 5.10 A revised preference schedule............................................................... SU5-68
Table 5.11 A preference schedule showing various ranking combinations....... SU5-69
vi
List of Tables
Table 5.12 Result of Pairwise Comparison............................................................. SU5-69
Table 5.13 A preference schedule showing various ranking combinations....... SU5-70
Table 5.14 Result of Pairwise Comparison............................................................. SU5-70
Table 5.15 A preference schedule showing various ranking combinations....... SU5-71
Table 5.16 A tabulation of the various fairness criteria satisfied by each method
of determining an election result............................................................................. SU5-73
Table 5.17 A preference schedule showing various ranking combinations....... SU5-73
Table 5.18 Result of Pairwise Comparison............................................................. SU5-74
Table 5.19 A preference schedule showing various ranking combinations....... SU5-75
Table 5.20 Result of Pairwise Comparison............................................................. SU5-75
Table 5.21 A preference schedule showing various ranking combinations....... SU5-76
vii
List of Tables
viii
List of Figures
List of Figures
Figure 5.1 Depiction of
Figure 6.1 Graph of
,
,
, and
using Venn diagrams............ SU5-17
............................................................................................ SU6-7
Figure 6.2 Depicting relation ................................................................................... SU6-9
Figure 6.3 Depicting relation
.............................................................................. SU6-10
Figure 6.4 A non-example of a function................................................................. SU6-33
Figure 6.5 A non-example of a function................................................................. SU6-34
Figure 6.6 A non-example of a function................................................................. SU6-34
Figure 6.7 Graph of
................................................................................... SU6-35
Figure 6.8 Failing the vertical line test.................................................................... SU6-36
Figure 6.9 An example of a function....................................................................... SU6-37
Figure 6.10 Graph of
............................................................................ SU6-40
Figure 6.11 Depiction of a function......................................................................... SU6-42
Figure 6.12 Graph of
,
....................................................................... SU6-47
Figure 6.13 An example of a function that is injective but not surjective........... SU6-50
Figure 6.14 An example of a function that is surjective but not injective........... SU6-50
Figure 6.15 An example of a function that is neither injective nor
surjective....................................................................................................................... SU6-51
Figure 6.16 An example of a function that is both injective and surjective........ SU6-51
ix
List of Figures
Figure 6.17 Graph of
with
....................................................... SU6-55
Figure 6.18 Graph of
with
....................................................... SU6-55
x
Course
Guide
Fundamentals of Mathematics
MTH105
Course Guide
1. Welcome
Welcome to the course MTH105 Fundamentals of Mathematics, a 5 credit unit (CU) course.
This Study Guide will be your personal learning resource to take you through the course
learning journey. The guide is divided into two main sections – the Course Guide and
Study Units.
The Course Guide describes the structure for the entire course and provides you with an
overview of the Study Units. It serves as a roadmap of the different learning components
within the course. This Course Guide contains important information regarding the
course learning outcomes, learning materials and resources, assessment breakdown and
additional course information.
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Course Guide
2. Course Description and Aims
This course provides you with knowledge of the basics and foundations of mathematics,
including Propositional logic, Predicate logic, and the usage of quantified statements.
Other foundational mathematical knowledge that will be taught in this course includes
set theory, relations, functions, and techniques of proof such as Mathematical Induction.
Topics such as divisibility, the Euclidean Algorithm, properties of prime numbers,
congruency and modulo arithmetic will also be covered.
Course Structure
This course is a 5-credit unit course presented over 6 weeks.
There are six Study Units in this course. The following provides an overview of each Study
Unit.
Study Unit 1 – Propositional Logic
The study of mathematical arguments and mathematical logic begins with Propositional
Logic. Propositional Logic deals with statements (also known as propositions), which are
sentences having a definite truth value, that is, they can be assigned "true" or "false", but
never simultaneously true and false. In Propositional Logic, we first examine compound
statements, which are more complex statements built up from simpler statements using
logical connectives. The study of Propositional Logic will also involve examination of
many of the fundamental concepts concerning the construction and flow of mathematical
arguments.
Study Unit 2 – Predicate Logic and Quantified Statements
In this study unit, we introduce predicates and quantified statements using the quantifiers
and . An existential statement is a statement of the form
statement is a statement of the form
while a universal
. We will also learn how to interpret
quantified statements and how to negate quantified statements. Finally, we will learn rules
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Course Guide
of inferences that involve quantified statements, and look at some common logical fallacies
and flawed arguments that must be avoided when dealing with quantified statements.
Study Unit 3 – Methods of Proof
In this Study Unit, we examine how to state definitions and theorems using precise
mathematical language, in particular, using the concepts from propositional logic and
predicate logic that we have learnt so far. We will then examine various methods of
proving statements in mathematics. These include direct proofs, proofs by contraposition,
proofs by contradiction, and proofs by division into cases. We will also examine
how to disprove existential or universal statements that are false. Some fundamental
mathematical concepts that will also be covered include even and odd integers, prime
numbers, rational and irrational numbers, the concept of divisibility, and the absolute
value of a real number.
Study Unit 4 – Sequences, Induction, the Euclidean Algorithm
In this Study Unit, we explore another important technique of proof known as induction,
or mathematical induction. Induction is used to prove that a certain mathematical
statement holds for all natural numbers, or holds for all integers at least as great as a
given certain integer. To provide many examples of the use of induction, we first discuss
sequences, as well as the usage of sigma and product notation. We then discuss the WellOrdering Property and derive the principle of mathematical induction from the WellOrdering Property. We then discuss many examples of induction, particularly in proving
statements about sequences, and statements concerning divisibility of integers. Finally, we
learn about the Division Property of the integers, the notions of greatest common divisor
and lowest common multiple, and learn how to implement the Euclidean Algorithm.
Study Unit 5 – Set Theory
In this Study Unit, we formally introduce the notion of a set and establish the basic
properties of sets. The notion of sets is a fundamental concept in mathematics. In every
branch of mathematics, definitions and theorems are stated using the language of set
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MTH105
Course Guide
theory. In the previous Study Units, we have already used the notions of sets and subsets.
Here, we will make these ideas rigorous and demonstrate various set properties by using
proof techniques such as the element method or by establishing a chain of set identities. We
also introduce two interesting applications of what we have learnt so far – the mathematics
of counting and probability, and the mathematics of voting.
Study Unit 6 – Relations and Functions
In this Study Unit, we studied two more important mathematical concepts – the concept of
a relation, and the concept of a function. Although functions are a special kind of relation,
they are so fundamental to mathematics that most people who encounter mathematics
usually encounter the concept of a function before they encounter the concept of a relation.
We will examine the concept of a relation and pay particular attention to an important class
of relations known as equivalence relations. Congruence classes modulo an integer
are
a key example of equivalence classes. We will discuss the concept of congruency modulo
and use the notion of congruency to prove interesting results in divisibility. Having
discussed relations in details, we discuss functions, which are a special kind of relation. We
will learn about the important properties of functions such as injectivity and surjectivity,
and study how to find the domain and range of functions. In this Study Unit, we will
also examine a very useful application of congruence arithmetic – the RSA Encryption
Algorithm.
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Course Guide
3. Learning Outcomes
Knowledge & Understanding (Theory Component)
By the end of this course, you should be able to:
1.
Show certain mathematical statements by rigorous mathematical arguments.
2.
Give counterexamples to disprove certain mathematical statements.
3.
Use mathematical induction or well-ordering principle to prove mathematical
statements.
Key Skills (Practical Component)
By the end of this course, you should be able to:
1.
Describe equivalence classes of a given equivalence relation.
2.
Employ truth tables to determine whether given arguments are valid.
3.
Determine whether given functions are injective and/or surjective.
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Course Guide
4. Learning Material
The following is a list of the required learning materials to complete this course.
Required Textbook(s)
Epp, S. S. (2019). Discrete mathematics with applications (5th ed.). Boston, MA: Brooks/
Cole Cengage Learning.
If you are enrolled into this course, you will be able to access the eTextbooks here:
To launch eTextbook, you need a VitalSource account which can be created via
Canvas (iBookStore), using your SUSS email address. Access to adopted eTextbook is
restricted by enrolment to this course.
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MTH105
Course Guide
5. Assessment Overview
The overall assessment weighting for this course is as follows:
Assessment
Description
Weight Allocation
Assignment 1
Computer Marked Assignment
10%
Assignment 2
Tutor Marked Assignment
20%
Examination
Open book exam
70%
TOTAL
100%
The following section provides important information regarding Assessments.
Continuous Assessment:
There will be continuous assessment in the form of one Computer Marked Assignment
(CMA) and one Tutor Marked Assignment (TMA). In total, this continuous assessment
will constitute 30 percent of overall student assessment for this course. The two
assignments are compulsory and are non-substitutable. It is imperative that you read
through your Assignment questions and submission instructions before embarking on
your Assignment.
Examination:
The final (2-hour) written exam will constitute the other 70 percent of overall student
assessment. All topics covered in the course outline will be examinable. To prepare for
the exam, you are advised to review Specimen or Past Year Exam Papers available on
Learning Management System.
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Course Guide
Passing Mark:
To successfully pass the course, you must obtain at least a mark of 40 percent for the
combined continual assessments and also at least a mark of 40 percent for the final
exam. For detailed information on the Course grading policy, please refer to The Student
Handbook (‘Award of Grades’ section under Assessment and Examination Regulations).
The Student Handbook is available from the Student Portal.
Non-graded Learning Activities:
Activities for the purpose of self-learning are present in each study unit. These learning
activities are meant to enable you to assess your understanding and achievement of the
learning outcomes. The type of activities can be in the form of Formative Assessment,
Quiz, Review Questions, Application-Based Questions or similar. You are expected to
complete the suggested activities either independently and/or in groups.
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Course Guide
6. Course Schedule
To help monitor your study progress, you should pay special attention to your
Course Schedule. It contains study unit related activities including Assignments, Selfassessments, and Examinations. Please refer to the Course Timetable in the Student Portal
for the updated Course Schedule.
Note: You should always make it a point to check the Student Portal for any
announcements and latest updates.
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Course Guide
7. Learning Mode
The learning process for this course is structured along the following lines of learning:
a.
Self-study guided by the study guide units. Independent study will require at
least 5 hours per week.
b.
Working on assignments, either individually or in groups.
c.
Classroom Seminar sessions (3 hours each session, 6 sessions in total).
iStudyGuide
You may be viewing the iStudyGuide version, which is the mobile version of the
Study Guide. The iStudyGuide is developed to enhance your learning experience with
interactive learning activities and engaging multimedia. Depending on the reader you are
using to view the iStudyGuide, you will be able to personalise your learning with digital
bookmarks, note-taking and highlight sections of the guide.
Interaction with Instructor and Fellow Students
Although flexible learning – learning at your own pace, space and time – is a hallmark
at SUSS, you are encouraged to engage your instructor and fellow students in online
discussion forums. Sharing of ideas through meaningful debates will help broaden your
learning and crystallise your thinking.
Academic Integrity
As a student of SUSS, it is expected that you adhere to the academic standards stipulated
in The Student Handbook, which contains important information regarding academic
policies, academic integrity and course administration. It is necessary that you read and
understand the information stipulated in the Student Handbook, prior to embarking on
the course.
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Course Guide
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Study
Unit
Propositional Logic
1
MTH105
Propositional Logic
Learning Outcomes
By the end of this unit, you should be able to:
1.
Construct a truth table for a given compound statement.
2.
Employ a truth table to determine if two given compound statements are
logically equivalent.
3.
Use a chain of logical equivalences to prove that two compound statements are
logically equivalent to each other.
4.
Employ a truth table to determine whether a given argument is valid or invalid.
5.
Construct an argument using the rules of inferences.
6.
Identity and name common logical fallacies in mathematics.
7.
Determine if a given argument is sound or unsound.
SU1-2
MTH105
Propositional Logic
Overview
The study of mathematical arguments and mathematical logic begins with Propositional
Logic. Propositional Logic deals with statements (also known as propositions), which are
sentences having a definite truth value, that is, they can be assigned "true" or "false", but
never simultaneously true and false. In Propositional Logic, we first examine compound
statements, which are more complex statements built up from simpler statements using
logical connectives. Propositional Logic also examines many of the fundamental concepts
concerning the construction and flow of mathematical arguments. Indeed, it can be said
that Propositional Logic is the most basic branch of logic. It forms the key foundation not
just of the vast field of mathematical logic, but indeed, of all of mathematics itself.
This Study Unit introduces essential concepts in Propositional Logic that will be
needed throughout a typical undergraduate mathematics programme. In Propositional
Logic, when constructing compound statements, we restrict our consideration to logical
connectives such as "and", "or", "implies". The concepts of predicates and quantifiers such
as "for all" or "there exists" are aspects of Predicate Logic, which is the next higher level of
mathematical logic. These concepts will be discussed only in the next Study Unit (Study
Unit 2).
SU1-3
MTH105
Propositional Logic
Chapter 1: Compound Statements, Truth Tables, and
Logical Equivalences
1.1 Statements
Definition
A statement or a proposition is a sentence that is either true, or false, but not both.
For example, within the usual number system, the sentence “
sentence “
” is true, and the
” is false. Both these sentences are statements.
We only accept a sentence as a statement when it is meaningful, unambiguous, and not
subject to opinion. For example, "n is probably an integer" is subjective and ambiguous,
and therefore cannot be considered a statement. "Physics is easy to master once you have a
good teacher" is also not a statement as that is an opinion. The word "easy" and the phrase
"good teacher" are also subjective and not well-defined without further qualification.
Activity 1.1
Classify the following sentences into statements or non-statements, and determine
whether the statements are true or false.
a.
“
b.
“
c.
“Mathematics is a difficult subject”
d.
“This statement is false”
”
”
SU1-4
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Propositional Logic
1.2 Logical Connectives, Compound Statements, Truth Tables
Definition
If
is a statement, the negation of , denoted by
The statement
is true when
The negation of ,
is false, and
, has the opposite truth value from .
is false when
is true.
, is also referred to as “not ”. In simple terms,
is the opposite of
. The truth values for negation are summarised in the following truth table. In a truth
table, T denotes true, and F denotes false. True and false can also be represented by 1 and
0 respectively.
Table 1.1 Truth table for negation
T
F
F
T
Definition
If
and
are statements, the conjunction of
that is true whenever both
and , denoted by
, is the statement
and are true.
The conjunction of and is also referred to as “ and ”. The truth values for conjunction
are summarised in the following truth table.
SU1-5
MTH105
Propositional Logic
Table 1.2 Truth table for conjunction
T
T
T
T
F
F
F
T
F
F
F
F
Definition
If and are statements, the disjunction of and , denoted by
is true whenever either
The disjunction of
and
, is the statement that
or (or both) are true.
is also referred to as “ or ”. The truth values for disjunction
are summarised in the following truth table. Note that when we state “ or ”, we always
mean “ or or both”.
Table 1.3 Truth table for disjunction
T
T
T
T
F
T
F
T
T
F
F
F
SU1-6
MTH105
Propositional Logic
The symbols
,
and
are known as logical connectives. Logical connectives are used
to build up more complicated statements known as compound statements from simpler
statements.
Rules of Precedence for Evaluating the Truth Value of Compound Statements
1.
Brackets take precedence over all logical connectives.
2.
Negation is evaluated before disjunction and conjunction.
3.
Conjunction and disjunction are of equal precedence, and they are evaluated
from left to right.
For example, the truth value of the compound statement
is
evaluated in the following sequence:
1.
The truth value of
2.
The truth value of
3.
The truth value of
4.
The truth value of
5.
The truth value of
6.
The truth value of
is first evaluated.
is evaluated.
is evaluated.
is evaluated.
is evaluated.
is evaluated.
By the rules of precedence, the compound statement
is equivalent to
. Note that in the above example, the truth value of
can
be evaluated at any stage prior to stage 6. The sequence of evaluation of truth values is
not unique and any answer is acceptable as long as the rules of precedence are followed.
The truth table for a compound statement displays the truth values of the statement
corresponding to all possible combinations of truth values of its variables. When
constructing a truth table, the rules of precedence must be followed. For example, the
following is the truth table for the compound statement
SU1-7
.
MTH105
Propositional Logic
Table 1.4 Truth table for
T
T
F
T
T
F
T
T
F
T
F
F
F
F
T
T
The following is the truth table for the compound statement
.
Table 1.5 Truth table for
T
T
T
F
T
T
T
T
F
T
T
T
T
F
T
F
F
F
T
F
F
T
F
T
F
T
T
F
F
F
F
T
F
T
F
T
F
F
T
F
F
F
F
F
F
T
F
T
SU1-8
MTH105
Propositional Logic
Activity 1.2
Construct truth tables for the following compound statements:
a.
b.
1.3 Logical Equivalence, Tautologies, and Contradictions
Definition
Two compound statements
and
are said to be logically equivalent if
and
attain
the same truth value for every possible truth value combination of their component
statements.
In other words, two compound statements
and
are logically equivalent if they have
identical truth tables when all the possible truth value combinations of their component
statements are listed in the same order.
For example,
is logically equivalent to
SU1-9
.
MTH105
Propositional Logic
Table 1.6 The logical equivalence of
Similarly,
and
is logically equivalent to
. These two laws are known as the
Commutative Laws.
is logically equivalent to . This is known as the Double Negation Law.
Table 1.7 The logical equivalence of
and
T
F
T
F
T
F
The statement
is not logically equivalent to
truth table below:
SU1-10
, as demonstrated in the
MTH105
Propositional Logic
Table 1.8
is not logically equivalent to
Activity 1.3
Employ truth tables to demonstrate the following:
a.
is logically equivalent to
.
b.
is logically equivalent to
.
These logical equivalences are known as De Morgan’s Laws.
Definition
A tautology is a compound statement that is always true regardless of the truth values of
its component statements.
A contradiction is a compound statement that is always false regardless of the truth values
of its component statements.
SU1-11
MTH105
Propositional Logic
For example,
is a tautology.
Table 1.9 Truth table demonstrating that
Also,
is a tautology
is a contradiction. These two laws are known as the Negation Laws.
Table 1.10 Truth table demonstrating that
is a contradiction
From now on, if the compound statement
is logically equivalent to the compound
statement
is a tautology, we write
, we denote this by
contradiction, we write
. If
. If
is a
. The letters t and c (in bold) by themselves are also used to
denote a tautological statement and a contradictory statement respectively. An important
collection of logical equivalences is summarised below:
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Propositional Logic
Table 1.11 An important collection of logical equivalences
Commutative Laws
Associative Laws
Distributive Laws
Identity Laws
Negation Laws
Double Negation
Law
Idempotent Laws
Universal Bound
Laws
De Morgan’s Laws
Absorption Laws
Negations of t and c
Tautologies and contradictions must be distinguished from statements whose truth or
falsity depends on the meaning of its symbols or its component statements. For example,
the statement “
” is false in the decimal system, but true when integers are added
modulo 2. It cannot be called a tautology. In Propositional Logic, we regard a statement
as a tautology or as a contradiction only when justified by the rules laid out in Table
1.11 above. In general, we do not call true statements as tautologies if not directly justified
by the rules of Propositional Logic (or by a truth table). The following Activity will clarify
this point.
SU1-13
MTH105
Propositional Logic
Activity 1.4
Classify the following statements into tautologies, contradictions, or neither:
a.
b.
c.
“
”
d.
“
e.
“
or
f.
“
and
g.
“
”
h.
“
”
”
”
”
1.4 Chain of Logical Equivalences
We have employed truth tables to show the logical equivalence of two compound
statements.
We can also prove that two statements
chain of logical equivalences from
and
are logically equivalent by constructing a
to , using the rules in Table 1.11.
For example, let us demonstrate that the compound statement
logically equivalent to the statement
itself.
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is
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Very often, there is more than one possible chain of logical equivalences that allow us
to demonstrate that two compound statements are logically equivalent. In the example
above, we could have used this alternative chain:
In this case, this chain of logical equivalences allows us to prove the equivalence of the
statement
and the statement
acceptable.
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faster. However, both answers are
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Activity 1.5
Use a chain of logical equivalences to demonstrate the logical equivalence of each
pair of compound statements given below. You are allowed to use any of the logical
equivalences listed in Table 1.11 without further justification.
a.
is logically equivalent to
b.
.
is logically equivalent to the
statement
.
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Chapter 2: Conditional Statements
2.1 Definition of Conditional Statements
Let
and be statements. The statement
is defined by the following truth table:
Table 1.12 Truth table for
The symbol
T
T
T
T
F
F
F
T
T
F
F
T
is a logical connective that means "implies". The statement
read as " implies ". We call
the hypothesis and
can be
the conclusion. As an illustration,
consider the following statement:
Such a statement is called a conditional statement because the truth of the conclusion
is conditioned on the truth of the hypothesis .
As can be seen in the above truth table, the statement
both true, or when
is true whenever
and
are
is false. A conditional statement that is true by virtue of the fact that
its hypothesis is false is said to be vacuously true. For example, in the real number system,
the statement
"If
, then
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"
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is vacuously true because its hypothesis, the statement "
", is false.
Rules of Precedence for
The logical connective
logical connective
then
has a lower precedence than
is performed first, followed by
or
and
. In other words, the
(from left to right), and
is performed.
For example, let us construct a truth table for the compound statement
.
Table 1.13 Truth table for
T
T
F
F
T
F
T
F
F
T
T
F
F
T
T
F
F
T
F
F
T
T
T
T
As can be seen in the above truth table,
is in fact logically equivalent to
.
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2.2 Logical Equivalences Involving
(a) Let
and be statements. Then
Table 1.14 Logical equivalence of
is logically equivalent to
.
and
T
T
F
T
T
T
F
F
F
F
F
T
T
T
T
F
F
T
T
T
The logical equivalence of
example, let
and
is used widely in everyday speech. For
be the statement "You do not return the library book on time", and let
the statement "You will pay a fine". Then
be
is the statement "Either you return the
library book on time, or you will pay a fine". This is logically equivalent to
, which is
the statement "If you do not return the library book on time, then you will pay a fine".
(b) Let
and be statements. Then
is logically equivalent to
.
Rather than using a truth table to demonstrate this equivalence, we can use the following
chain of logical equivalences:
by De Morgan's Laws
by Double Negation Law
Care must be taken when using Propositional Logic to analyse everyday speech, due to
nuances and peculiarities of our language.
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For example, let
be the statement "You will be given a job promotion this month", and
let be the statement "You will be given a pay raise next month".
Then
is the statement "It is false that if you get a job promotion this month,
then you will be given a pay raise next month". The statement does not assert that "if
you get a job promotion this month, then you will not get a pay raise next month". The
, which is not logically equivalent to
latter statement is actually the statement
.
Table 1.15
(c) Let
is not logically equivalent to
and be statements. Then
Table 1.16 Logical equivalence of
is logically equivalent to
and
T
T
F
F
T
T
T
F
F
T
F
F
F
T
T
F
T
T
F
F
T
T
T
T
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the contrapositive of
We call the statement
.
This is useful when we wish to prove statements or theorems in mathematics. If we wish
, a direct proof would involve assuming that is true, and then using
to demonstrate
our assumption to prove that must also be true.
A proof using the contrapositive would involve assuming that
showing that as a consequence,
demonstrated
is false, and then
has to be false as well. In other words, we would have
. For example, let the letter
denote a specific (fixed) integer,
like 0, 4, -7, and so on. In order to show that
we could provide a proof that
Sometimes, it may be the case that the contrapositive statement is easier to prove than
the original statement. In later chapters, we will call this "proving the contrapositive" or
"proof by contraposition".
Activity 1.6
Employ a truth table to show that the statement
is not logically equivalent to
. This serves to caution the reader that we cannot prove that statement
statement by demonstrating that
(d) Let
,
, and
implies
be statements. Then
.
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implies
.
is logically equivalent to
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Table 1.17 Logical equivalence of
and
T
T
T
T
T
T
T
T
T
T
F
T
F
F
F
F
T
F
T
T
T
T
T
T
T
F
F
T
F
F
T
F
F
T
T
T
T
T
T
T
F
T
F
T
F
T
F
F
F
F
T
F
T
T
T
T
F
F
F
F
T
T
T
T
This logical equivalence is called division into cases. It tells us that if we wish to prove
that either
or
leads to the conclusion , we can demonstrate that
also implies . For example, let the letters
and
implies , and that
denote specific integers. In order to
show that
we could show that
Of course, the reader will realise that by symmetry (that is, using a symmetrical argument),
in order to show that
that
even implies
is even whenever either term is even, it suffices to demonstrate
even. The symmetrical argument can be exploited because
multiplication of integers (of real numbers in general) is a commutative operation. But
it must also be understood that the symmetrical argument is a unique feature of this
particular question (because it involves numbers, and not, for instance, matrices, where
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multiplication need not be commutative). It is not a result of Propositional Logic and
cannot be universally applied in all situations.
2.3 The Converse, the Inverse, and the Biconditional
Definition and Facts
The converse of the statement
The inverse of the statement
is the statement
is the statement
Neither the converse nor the inverse of
.
.
is logically equivalent to
. This can be
verified using truth tables.
However, the converse and the inverse are logically equivalent to each other, that is,
is logically equivalent to
. This can be verified either by using truth tables, or by
using a chain of logical equivalences.
Activity 1.7
Employ both a truth table as well as a chain of logical equivalences to demonstrate
that
is logically equivalent to
.
For example, the converse of the statement "If I can have a month-long vacation, then I
can find time to practise golfing" is the statement "If I can find time to practise golfing,
then I can have a month-long vacation". Clearly, the converse of a conditional statement
is not logically equivalent to the original statement.
As another example, the inverse of the statement "If today is New Year's eve, then
tomorrow is a public holiday" is the statement "If today is not New Year's eve, then
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tomorrow is not a public holiday". Clearly again, the inverse of a conditional statement is
not logically equivalent to the original statement.
Definition
Given statements and , the biconditional of and , denoted by
that is true if and only if both
In other words,
and are true, or when both
, is the statement
and are false.
is true when statements and have the same truth value, and false
when statements
and have different truth values.
Table 1.18 Truth table for
The symbol
T
T
T
T
F
F
F
T
F
F
F
T
is a logical connective that means "implies and is implied by". The
statement
can be read as " implies and is implied by ", or as " if and only if ".
For example, let denote a specific positive integer. If
denotes the statement
" is a prime number",
and denotes the statement
"
then
is an integer with exactly three distinct positive divisors",
is the statement
" is a prime number if and only if
is an integer
with exactly three distinct positive divisors".
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Rules of Precedence for
The logical connective
connective
and
has the same precedence as
is performed first, followed by
and
. In other words, the logical
(from left to right), and then
are performed (from left to right).
The statement
is logically equivalent to
Table 1.19 Logical equivalence of
.
and
T
T
T
T
T
T
T
F
F
T
F
F
F
T
T
F
F
F
F
F
T
T
T
T
This logical equivalence tells us that if we wish to prove the statement
so by proving both
and
.
For example, let be a specific integer. Suppose we wish to prove that
is an even integer if and only if
is an even integer.
We can demonstrate this by proving both the statements
is an even integer implies
is an even integer
and
is an even integer implies is an even integer.
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, we can do
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Activity 1.8
Employ a truth table to demonstrate that
is logically equivalent to
.
2.4 Necessary and Sufficient Conditions
Definition
Let
and be statements.
a.
The assertion " is a sufficient condition for ", or more simply, " is sufficient
for ", means the statement
b.
, or equivalently,
The assertion " is a necessary condition for ", or more simply, " is necessary
for ", means the statement
c.
.
, or equivalently,
.
The assertion " is a necessary and sufficient condition for " , or more simply,
" is necessary and sufficient for ", means the statement
, or equivalently,
.
The assertion that
is sufficient for means that the truth of
is enough to guarantee the
truth of .
It may be the case that can be true without
For example, and
are positive integers is a sufficient condition for
integer. But we can have
to be a positive
being a positive integer without having both and
positive integers. For example,
The assertion that
being true.
to be
.
is necessary for means that in order for to be true, we need
true.
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to be
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It is possible that even though
addition to
is true,
is still false. This may be due to the fact that in
being true, other conditions need to be satisfied in order to make true.
For example,
is divisible by 5 is a necessary condition for
is divisible by 10. But it can
be true that is divisible by 5, and yet is not divisible by 10.
The assertion that
is necessary and sufficient for
is true whenever
is
is divisible by both 2 and 5 is a necessary and sufficient condition for
is
true, and is true whenever
For example,
means that
is true.
divisible by 10.
A list of equivalent ways of expressing the notion that
of expressing the notion that
, and a list of equivalent ways
, are presented below. The reader should be aware of
these lists as these expressions are widely used throughout mathematics.
Equivalent ways of expressing
The following list consists of all equivalent ways of expressing the assertion that
a.
implies
b.
is implied by
c.
If
then
d.
if
e.
only if
f.
is sufficient for
g.
is necessary for
h.
i.
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Equivalent ways of expressing
The following list consists of all equivalent ways of expressing the assertion that
a.
implies and is implied by
b.
if and only if
c.
is necessary and sufficient for
d.
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Chapter 3: Constructing Arguments
3.1 Valid and Invalid Arguments
An argument is a series of statements leading to a conclusion. For an argument to be valid,
the conclusion must be a logical consequence of the preceding statements. The premises
of an argument refer to the statements preceding the conclusion that are assumed to be
true, and which are not deduced from previous statements or assumptions.
Consider for instance the following argument:
If he grows crops, he will have food.
He grows crops.
He will have food
In the above argument, the symbol
stands for "therefore", or "as a result". The premises
are "If he grows crops, he will have food", and "He grows crops". These are the statements
that are assumed to be true. The conclusion is "He will have food". The conclusion is a
logical consequence of the premises.
Let
and be statements. The above argument has the abstract form
If
then .
.
.
This is clearly a valid argument, because if the conditional statement
the hypothesis
is true, then the conclusion
is true, and
must also be true. In daily life, we very
often employ this line of reasoning at an intuitive level without considering the logic
behind our reasoning. In mathematics, the underlying logic must be clearly expressed and
understood.
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In order to test a mathematical argument for validity, we first identify the premises and
the conclusion of the argument. Then we construct a truth table showing the truth values
of all the premises as well as the conclusion. A row of the truth table in which all the
premises are true (that is, have the truth value T) is called a critical row. The argument is
valid if and only if the conclusion is true in all the critical rows. If there exists some critical
row in which the conclusion is false, then the argument is said to be invalid.
For instance, consider again the following argument:
If
then .
.
.
In order to test the argument for validity, we employ the following truth table:
Table 1.20 Truth table determining the validity of an argument
The first row is the only critical row. In the critical row, the conclusion is also true. Hence,
this is a valid argument.
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As another example, consider the following argument:
.
.
.
We use the following truth table to determine whether or not this argument is valid:
Table 1.21 Truth table determining the validity of an argument
In the above truth table, we can see that the conclusion
rows. We therefore conclude the argument is invalid.
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is false in one of the critical
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Activity 1.9
Employ a truth table to determine whether the following argument is valid or invalid.
Indicate which rows are critical rows.
3.2 Rules of Inference
A rule of inference is a valid argument consisting of a sequence of premises followed by
a single conclusion. Rules of inference can be considered as general building blocks of
more complex arguments, where there can be intermediate conclusions deduced logically
from preceding statements, and where the intermediate conclusions themselves become
premises of further arguments and lead to other conclusions.
We now examine some important and commonly used rules of inference.
a.
Modus Ponens
If
then .
.
.
We have already seen Modus Ponens in Section 3.1. The term modus ponens in
Latin means "method of affirming". In this case, the conclusion of the argument
is the required affirmation. As a rule of inference, Modus Ponens is based on the
fact if is
true, and is true, then must also be true. These observations are
then assembled to form the rule of inference Modus Ponens.
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b.
Modus Tollens
If
then .
.
.
The term modus tollens means "method of denying". In this case, we wish to
deny or refute statement , that is, we wish to be able to conclude that
As a rule of inference, Modus Tollens is based on the fact that
equivalent to
. If
is true, and
is true, then
is false.
is logically
must also be
true. These observations are then assembled to form the rule of inference Modus
Tollens.
Modus Ponens and Modus Tollens are also known as syllogisms because they are rules
of inference consisting of exactly two premises. The first premise is known as the major
premise and the second is known as the minor premise.
c.
Generalisation (also known as Addition)
Generalisation takes two forms:
i.
.
.
ii.
.
.
Of course, it can easily be seen that these two forms are really the same rule of
inference, because
is logically equivalent to
by the Commutative rule.
An example of generalisation is as follows: Suppose we know that the statement
"John is married" is true. Then the statement "Either John is married or John has
bought a new house" is also true. Although the more general statement seems
to provide less information to the reader in that it opens up two possibilities,
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leaving the reader uncertain as to which statement (or both) is true, it is
nonetheless a logical consequence of the first statement.
d.
Specialisation (also known as Simplification)
Specialisation takes two forms:
i.
.
.
ii.
.
.
Specialisation can be thought of as the opposite of generalisation. As an example,
if we know that the statement "John is married and John has bought a new house"
is true, then the statement "John has bought a new house" must be true. The latter
statement provides less information because it only tells the reader one piece of
information, but it is nonetheless a logical consequence of the first statement.
e.
Elimination (or Disjunctive Syllogism)
Elimination takes two forms:
i.
.
.
.
ii.
.
.
.
Elimination works on the principle that if you are presented with two choices,
or , and one of the choices can be eliminated, then you will be forced to conclude
that the other choice must be correct.
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f.
Transitivity (or Hypothetical Syllogism)
.
.
.
Many arguments in mathematics take the form of sequences or chains of
conditional statements. From the fact the first statement implies the second, and
the second statement implies the third, we can deduce by transitivity that the first
statement must imply the third. For example, let be a specific integer. Given the
premises
"If is odd, then
"If
is odd, then
is odd."
is divisible by 8."
we can draw the conclusion
"If is odd, then
g.
is divisible by 8."
Proof by division into cases
.
.
.
.
If we know that either statement
know that
or statement (possibly both) is true, and we
implies , and implies , then we conclude is true.
For example, suppose we know that
"Today is either Saturday or Sunday."
"Saturday is a public holiday."
"Sunday is a public holiday."
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then we can conclude that
"Today is a public holiday."
h.
Contradiction rule
The contradiction rule takes two forms:
i.
.
.
ii.
.
.
The contradiction rule is the logical basis of proof by contradiction, a technique of proving
mathematical statements in which we prove that
is true by assuming that
is false and
then showing how a contradiction follows from that assumption.
For example, consider the theorem "There is no smallest rational number that is strictly
greater than 0". We can prove this theorem by using a proof by contradiction.
Suppose that there is a smallest rational number that is strictly greater than 0. Let this
rational number be . Then
is also a rational number that is greater than 0, and
.
This contradicts the hypothesis that is the smallest rational number that is strictly greater
than 0. We therefore conclude that the theorem must be true.
A summary of the rules of inference is laid out in the following table:
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Table 1.22 A summary of the rules of inference
We can use the rules of inference in Table 1.22 to build up mathematical arguments.
For example, suppose we start off with the collection of premises
.
.
.
.
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We claim that from these premises, we can draw the conclusion . In other words, if we
assume that the above premises are all true, then we can use the rules of inference to
deduce that the statement is also true. An argument for the validity of can be constructed
as follows:
Step
Reason
1.
.
2.
.
Specialisation using step (1).
3.
.
4.
Modus Tollens using (2) and (3).
.
Premise.
.
7.
8.
Premise.
.
5.
6.
Premise.
Modus Ponens using (4) and (5).
.
Premise.
.
Modus Ponens using (6) and (7).
When we construct an argument using the rules of inference, we should write each step
clearly, and state explicitly the reason for each step.
As another example, we construct an argument with premises
and the conclusion
.
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Step
Reason
1.
.
Premise.
2.
.
3.
.
4.
5.
Contrapositive of (1).
Premise.
.
Transitivity using (2) and (3).
.
6.
Premise.
.
Transitivity using (4) and (5).
In our argument, we are allowed to deduce a statement that is logically equivalent to any
of premises or previously derived statements. For example, in step (2) above, we deduced
the contrapositive of the premise stated in step (1).
Activity 1.10
Using the rules of inference, construct an argument which assumes the following
premises
and which leads to the conclusion
.
Do not use truth tables.
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Activity 1.11
Using the rules of inference, construct an argument which assumes the following
premises
and which leads to the conclusion
.
Do not use truth tables.
3.3 Logical Fallacies
A logical fallacy, also known as a formal fallacy, is an error in reasoning due to employing
incorrect logic. There are many different types of logical fallacies. For example, appeal
to probability refers to taking a statement to be true simply because it is believed to be
probably true or likely true. Affirming a disjunct refers to using the validity of
and
to incorrectly conclude that is false, that is, employing the following (invalid) argument:
.
.
.
An instance of employing this incorrect line of reasoning is as follows:
"John will either pass his job interview today, or he will go home to weep. It was observed
that John went home to weep. Therefore, John must have failed his job interview today."
This line of reasoning neglects to consider that John could have passed his job interview
today, but went home to weep for a completely different reason (he was forced to agree
to a low pay, for example).
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The above logical fallacy can be avoided by remembering that in mathematics, the phrase
“ or ” means that we accept either
or
or both. This is distinct from the concept
of exclusive-or (which is used widely in computer science). Exclusive-or means that we
accept either , or , but not both. When the compound statement “ or ” is used, we
must thus remember that it is possible for both
and to be true.
A comprehensive list of logical fallacies can be found at the following Wikipedia page:
List of fallacies. URL: https://en.wikipedia.org/wiki/List_of_fallacies
(Accessed 13 July 2019.)
Activity 1.12
Employ a truth table to demonstrate that the following argument, known as affirming
a disjunct, is invalid. Indicate which rows are critical rows.
.
.
.
In mathematics, we are particularly careful not to commit two kinds of logical fallacies:
the converse error, and the inverse error.
The converse error goes as follows:
.
.
.
An instance of committing the converse error is as follows:
"If John has a valid British passport, then John must be a British national. John is a British
national. Therefore, John must have a valid British passport."
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The converse error is committed when we assert that the hypothesis of a conditional
statement is true based on the truth of the conclusion. This is also known as the "fallacy
of affirming the consequent".
The inverse error goes as follows:
.
.
.
An instance of committing the inverse error is as follows:
"If interest rates are going up, then stock prices will go down. Interest rates are not going
up. Therefore, stock prices will not go down."
The inverse error is committed when we assert that the conclusion of a conditional
statement is false based on the falsity of the hypothesis. This is also known as the "fallacy
of denying the antecedent".
Activity 1.13
Employ truth tables to illustrate the converse error and the inverse error.
Do not confuse the notions of "truth" and "validity". An argument may be valid in the
logical sense, but the conclusion need not be true because one of the premises is false.
For example, the argument below is valid by Modus Ponens.
"If 5 is a prime number, then 5 < 0."
"5 is a prime number."
"Hence, 5 < 0."
In this case, the conclusion is false because the major premise is false.
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Definition
An argument is said to be sound if and only if it is a valid argument, and all of its premises
are true. If an argument is invalid, or if at least one of its premises is false, then we say
that the argument is unsound.
The argument we have just described is an example of a valid, but unsound argument. In
mathematics, we are only interested in constructing sound arguments.
Activity 1.14
Examine the following argument and classify it into valid/invalid, as well as sound/
unsound.
"If 6 is a prime number, then 6 > 0."
"6 is a prime number."
"Hence, 6 > 0."
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Summary
In this Study Unit, we introduced the notion of logical connectives and compound
statements, and the use of truth tables.
We used both truth tables as well as sequences of logical equivalences to determine
whether two compound statements are logically equivalent.
We examined what constitutes valid and invalid mathematical arguments, the rules of
inference used in mathematics, as well as some common logical fallacies to be avoided.
We also used truth tables to determine whether a given argument is valid. Finally, we
introduced the concept of sound and unsound arguments.
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Formative Assessment
1.
What is the name given to the following logical fallacy? Also classify the argument
into valid/invalid, as well as sound/unsound.
"If 1/2 is an integer, then 1/2 is a rational number."
"1/2 is not an integer."
"Hence, 1/2 is not a rational number."
2.
What is the name given to the following logical fallacy? Also classify the argument
into valid/invalid, as well as sound/unsound.
"If 3 is a prime number, then 3 is a positive integer.
"3 is a positive integer."
"Therefore, 3 is a prime number."
3.
Examine the following argument and classify it into valid/invalid, as well as sound/
unsound. Here, the letter
denotes a specific integer (could be positive or negative
or 0).
"If is a prime number, then is a positive integer."
"If is positive integer, then
is a negative integer."
"Hence, if is a prime number, then
4.
Let ,
and
is a negative integer."
be statements. Construct a truth table for the compound statement
.
5.
Let ,
and
be statements. Construct a truth table for the compound statement
.
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6.
Let
Propositional Logic
and be statements. Employ a truth table to determine whether the compound
statements
7.
and
Let , and be statements. Employ a truth table to determine whether the compound
statements
8.
are logically equivalent.
Let
and
and
are logically equivalent.
be statements. Use a chain of logical equivalences to show that
is a tautology. You are allowed to use any of the logical
equivalences listed in Table 1.11 without further justification.
9.
Let
and be statements. Use a chain of logical equivalences to show that
is logically equivalent to
You are allowed to use any of the logical equivalences listed in Table 1.11 without
further justification.
10. Employ a truth table to determine if the following argument is valid or invalid.
Indicate which rows are the critical rows.
11. Using the rules of inference, construct an argument which assumes the following
premises
and which leads to the conclusion
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.
Do not use truth tables. (There may be more than one possible solution.)
12. Construct a chain of logical equivalences to demonstrate that
is logically equivalent to
Do not use truth tables in this question.
13. Use the rules of inference to prove the validity of the following argument form. Do
not use truth tables in this question.
(Premise)
(Premise)
(Premise)
(Premise)
(Premise)
(conclusion)
14. Use the rules of inference to prove the validity of the following argument form. Do
not use truth tables in this question.
(Premise)
(Premise)
(Premise)
(conclusion)
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Propositional Logic
15. Employ a truth table to determine whether or not the following argument form is
valid.
(Premise)
(Premise)
(conclusion)
16. Use the rules of inference to prove the validity of the following argument form. Do
not use truth tables in this question.
.
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Propositional Logic
Solutions or Suggested Answers
Activity 1.1
a.
“
”
True statement
b.
“
”
False statement
c.
“Mathematics is a difficult subject”
Not a statement. This is called an opinion.
d.
“This statement is false”
Not a statement. Notice that it cannot have a well-defined truth value.
Activity 1.2
a.
T
T
T
F
T
T
T
T
F
T
T
T
T
F
T
F
F
F
T
F
F
T
T
T
F
T
T
F
T
F
F
T
F
T
T
F
F
F
T
F
F
F
F
F
F
T
T
F
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Propositional Logic
b.
T
T
T
F
T
T
F
T
T
F
F
T
T
F
T
F
T
T
T
T
F
T
F
F
T
T
T
F
F
T
T
F
F
T
F
F
T
F
F
F
F
T
F
F
T
T
T
T
F
F
F
F
T
T
T
F
Activity 1.3
a.
T
T
T
F
F
F
F
T
F
F
T
F
T
T
F
T
F
T
T
F
T
F
F
F
T
T
T
T
T
T
T
F
F
F
F
T
F
T
F
F
T
F
F
T
T
F
T
F
F
b.
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Propositional Logic
F
F
F
T
T
T
T
Activity 1.4
a.
Tautology. Notice that the statement is of the form
.
b.
Contradiction. Notice that the statement is of the form
c.
“
.
”
Neither a tautology nor a contradiction. If we are working in the decimal system
and + represents the usual addition of numbers, then this is a true statement. But
true statements are different from tautologies, as explained in Section 1.3.
d.
“
”
Neither a tautology nor a contradiction. See the explanation given in part (c).
e.
“
or
”
Neither a tautology nor a contradiction.
f.
“
and
”
Neither a tautology nor a contradiction.
g.
“
”
Tautology. Notice that the statement is of the form
h.
“
.
”
Contradiction. Notice that the statement is of the form
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.
MTH105
Propositional Logic
Activity 1.5
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Propositional Logic
Activity 1.6
Activity 1.7
Truth Table:
T
T
F
F
T
T
T
F
F
T
T
T
F
T
T
F
F
F
F
F
T
T
T
T
Chain of logical equivalences:
by Double Negation Law
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Propositional Logic
Activity 1.8
T
T
F
F
T
T
T
F
F
T
F
F
F
T
T
F
F
F
F
F
T
T
T
T
Activity 1.9
The above truth table shows that the conclusion
critical rows. Hence, the argument is invalid.
Activity 1.10
(There may be more than one possible solution.)
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attains the value of F in one of the
MTH105
Propositional Logic
Step
Reason
1.
.
2.
Premise.
.
3.
Premise.
.
Specialization from (2).
4.
.
Elimination from (1) and (3).
5.
.
Specialization from (2).
6.
.
Conjunction from (4) and (5).
Activity 1.11
(There may be more than one possible solution.)
Step
Reason
1.
2.
3.
Logically equivalent to (1).
.
.
4.
Premise.
.
Logically equivalent to (3).
5.
Transitivity from (2) and (4).
.
6.
Logically equivalent to (5).
.
7.
8.
Premise.
.
Logically equivalent to (6).
.
.
Specialisation from (7).
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Propositional Logic
An alternative approach is as follows:
Step
Reason
1.
Premise.
.
2.
Logically equivalent to (1).
.
3.
.
Specialisation from (2).
4.
.
Premise.
5.
.
Logically equivalent to (4).
6.
.
Logically equivalent to (3).
7.
.
Transitivity from (5) and (6).
8.
.
Logically equivalent to (7).
Activity 1.12
The above truth table shows that the conclusion
attains the value of F in one of the
critical rows. Hence, the argument (affirming a disjunct) is invalid.
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Propositional Logic
Activity 1.13
Illustration of converse error:
The above truth table shows that the conclusion attains the value of F in one of the critical
rows. Hence, the argument is invalid. This illustrates the converse error.
Illustration of inverse error:
The above truth table shows that the conclusion
attains the value of F in one of the
critical rows. Hence, the argument is invalid. This illustrates the inverse error.
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Activity 1.14
The argument is valid by Modus Ponens. It is unsound because its minor premise is false:
the number 6 is not a prime number. We reject the argument as unsound even though its
conclusion is a true statement (6 is indeed greater than 0, but not for the reason given in
the argument). Note that the statement "If 6 is a prime number, then 6 > 0" is in fact a true
statement as well (it is vacuously true).
Formative Assessment
1.
What is the name given to the following logical fallacy? Also classify the argument
into valid/invalid, as well as sound/unsound.
"If 1/2 is an integer, then 1/2 is a rational number."
"1/2 is not an integer."
"Hence, 1/2 is not a rational number."
Answer:
Inverse error, or fallacy of denying the antecedent. (Both answers acceptable.) Refer
to Section 3.3.
The argument is invalid and unsound.
2.
What is the name given to the following logical fallacy? Also classify the argument
into valid/invalid, as well as sound/unsound.
"If 3 is a prime number, then 3 is a positive integer.
"3 is a positive integer."
"Therefore, 3 is a prime number."
Answer:
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Propositional Logic
Converse error, or fallacy of affirming the consequent. (Both answers acceptable.)
Refer to Section 3.3.
The argument is invalid and unsound. This is despite the fact the all the premises
and the conclusion itself are true statements (when we work within the usual
number system).
3.
Examine the following argument and classify it into valid/invalid, as well as sound/
unsound. Here, the letter
denotes a specific integer (could be positive or negative
or 0).
"If is a prime number, then is a positive integer."
"If is positive integer, then
is a negative integer."
"Hence, if is a prime number, then
is a negative integer."
Answer:
This argument is valid by Transitivity. The argument is also sound because in
addition to being valid, all its premises are true statements.
4.
Let ,
and
be statements. Construct a truth table for the compound statement
.
Answer:
T
T
T
T
F
F
T
T
F
T
T
T
T
F
T
F
F
T
T
F
F
F
T
T
F
T
T
T
F
F
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5.
Propositional Logic
F
T
F
T
T
T
F
F
T
T
F
F
F
F
F
T
T
T
Let ,
and
be statements. Construct a truth table for the compound statement
.
Answer:
6.
Let
T
T
T
F
F
F
T
T
F
F
F
F
T
F
T
F
T
T
T
F
F
F
T
F
F
T
T
T
T
T
F
T
F
T
T
F
F
F
T
T
T
T
F
F
F
T
T
F
and be statements. Employ a truth table to determine whether the compound
statements
and
are logically equivalent.
Answer:
T
T
F
T
F
T
F
T
F
T
T
F
T
F
F
T
F
F
T
T
F
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Propositional Logic
F
F
T
T
T
F
T
The above truth table shows that the compound statements
and
are not logically equivalent.
7.
Let , and be statements. Employ a truth table to determine whether the compound
statements
and
are logically equivalent.
Answer:
T
T
T
T
T
F
F
T
T
T
F
T
F
F
F
F
T
F
T
F
T
F
T
T
T
F
F
F
T
F
T
T
F
T
T
F
T
T
F
T
F
T
F
F
T
T
F
T
F
F
T
F
T
T
T
T
F
F
F
F
T
T
T
T
The above truth table shows that the compound statements
and
are logically equivalent.
8.
Let
and
be statements. Use a chain of logical equivalences to show that
is a tautology. You are allowed to use any of the logical
equivalences listed in Table 1.11 without further justification.
Answer:
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9.
Let
Propositional Logic
and be statements. Use a chain of logical equivalences to show that
is logically equivalent to
You are allowed to use any of the logical equivalences listed in Table 1.11 without
further justification.
Answer:
10. Employ a truth table to determine if the following argument is valid or invalid.
Indicate which rows are the critical rows.
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Propositional Logic
Answer:
Premise
Premise
Conclusion
Critical
row
T
T
T
F
T
F
F
T
F
T
T
T
T
✓
F
F
T
T
T
✓
The above truth table shows that the argument is valid.
11. Using the rules of inference, construct an argument which assumes the following
premises
and which leads to the conclusion
.
Do not use truth tables. (There may be more than one possible solution.)
Answer:
Step
1.
Reason
.
2.
3.
Premise.
.
Premise.
.
Modus Ponens from (1) and (2).
4.
5.
.
Premise.
.
Logically equivalent to (4).
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Propositional Logic
6.
.
Modus Ponens from (3) and (5).
An alternative answer is as follows:
Step
Reason
1.
.
Premise.
2.
.
Logically equivalent to (1).
3.
.
Premise.
4.
.
Transitivity from (2) and (3).
5.
.
Premise.
6.
.
Modus Ponens from (4) and (5).
12. Construct a chain of logical equivalences to demonstrate that
is logically equivalent to
Do not use truth tables in this question.
Answer:
13. Use the rules of inference to prove the validity of the following argument form. Do
not use truth tables in this question.
(Premise)
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Propositional Logic
(Premise)
(Premise)
(Premise)
(Premise)
(conclusion)
Answer:
Step
Reason
1.
.
2.
.
3.
Modus Tollens from (1) and (2).
.
Elimination from (3) and (4).
.
Premise.
.
8.
Modus Ponens from (3) and (6).
.
9.
10.
Premise.
.
6.
7.
Premise.
.
4.
5.
Premise.
Conjunction from (5) and (7).
Premise.
.
.
Modus Ponens from (8) and (9).
14. Use the rules of inference to prove the validity of the following argument form. Do
not use truth tables in this question.
(Premise)
(Premise)
(Premise)
(conclusion)
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Propositional Logic
Answer:
Step
1.
Reason
.
Premise.
2.
.
Logically equivalent to (1).
3.
.
4.
Generalization from (2).
.
Tautology.
5.
.
Logically equivalent to (4).
6.
Conjunction from (3) and (5).
.
7.
Distributive law from (6).
.
8.
De Morgan’s law from (7).
.
9.
Logically equivalent to (8).
.
10.
.
Premise.
11.
.
Transitivity from (9) and (10).
12.
.
Premise.
13.
.
Transitivity from (11) and
(12).
15. Employ a truth table to determine whether or not the following argument form is
valid.
(Premise)
(Premise)
(conclusion)
Answer:
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Propositional Logic
The conclusion is F in at least one critical row, so the argument form is invalid.
16. Use the rules of inference to prove the validity of the following argument form. Do
not use truth tables in this question.
.
Answer:
Step
Reason
1.
.
Premise.
2.
.
Logically equivalent to (1).
3.
.
Premise.
4.
.
Logically equivalent to (3).
5.
.
Transitivity from (2) and (4).
6.
.
Logically equivalent to (5).
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Propositional Logic
References
Epp, S. S. (2019). Discrete mathematics with applications (5th ed.). Boston, MA: Brooks/
Cole Cengage Learning.
SU1-68
Study
Unit
Predicate Logic and Quantified
Statements
2
MTH105
Predicate Logic and Quantified Statements
Learning Outcomes
By the end of this unit, you should be able to:
1.
Determine the truth set of a predicate given the domain of the predicate variable.
2.
Determine whether a quantified statement of the form
or
is
true or false. If a universal statement is false, provide a counter-example. If an
existential statement is true, construct the truth set of the predicate.
3.
Determine whether a universal conditional statement is true, vacuously true, or
false. Note that vacuously true statements are true statements.
4.
Give a proof in words demonstrating that two quantified statements are logically
equivalent.
5.
Determine the truth or falsity of multiply quantified statements.
6.
Construct the negation of quantified statements, including multiply quantified
statements.
7.
Use the rules of inference to construct an argument involving quantified
statements.
8.
Identify and name common logical fallacies involving quantified statements.
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Predicate Logic and Quantified Statements
Overview
In Study Unit 1, we discussed propositional logic, which deals with the truth values of
compound statements such as
,
, and so on.
In this Study Unit, we will consider predicates, which are sentences containing variables
and so on. These variables can be substituted with mathematical objects such as
numbers, matrices, functions, etc. For this Study Unit, we will focus on numbers. For
instance,
is a predicate depending on the variables
and , and the predicate
becomes a true statement if we substitute numbers such as
. Of course, if we
try to substitute nonsensical values of and such as
, then the equation
becomes a false statement. When we substitute specific values of and into the
equation, we either turn it into a true statement, or into a false statement.
We then introduce the concept of quantified statements, such as "for all
or "there exists
for every possible value of
". The first statement (the "for all" statement) asserts that
and , the equation
holds. Clearly this statement
is false, because there are many (in fact, infinitely many) possible values of
which the equation
",
and
for
fails to hold (that is, the equation fails to be true). On the
other hand, the second statement (the "there exists" statement) asserts that there are some
and (possibly more than one pair) such that the equation
true. The equation is valid for certain values of and (for instance,
holds. This much is
), but
the statement does not assert that the equation must be valid for every possible and .
This Study Unit will examine quantified statements and lay the foundation for subsequent
theory concerning mathematical proofs.
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Predicate Logic and Quantified Statements
Chapter 1: Predicates and Quantifiers
1.1 Predicates
Definition
A predicate is a sentence containing a finite number of variables which are usually
denoted by the symbols , , , , , and so on. These variables are known as predicate
variables.
Each predicate variable can be substituted with specific values. The domain of a predicate
variable is the set of all possible values that can be substituted for that variable.
Let us examine some examples.
The sentence "
If we substitute
" is a predicate. The symbol
in the sentence is a predicate variable.
with the number 3, then the sentence becomes "
statement. But if we substitute
", which is a true
with the number , then the sentence becomes "
",
which is a false statement.
The sentence "If
is a prime number, then
predicate variables. If we substitute
is odd" is a predicate. Here,
with the number 3 and
and
are
with the number 5, then
the sentence becomes "If 3 is a prime number, then 5 is odd", which is a true statement. If
we substitute with the number 3 and with the number 6, then the sentence becomes "If
3 is a prime number, then 6 is odd", which is a false statement. If we substitute with the
number 10 and with the number 6, then the sentence becomes "If 10 is a prime number,
then 6 is odd", which is a true statement. In this case the statement is said to be vacuously
true because it is true only by virtue of the fact that the hypothesis "10 is a prime number"
is false.
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Predicate Logic and Quantified Statements
Important Note
As can be seen from the above examples, when specific values of the predicate variables
are substituted in, we sometimes get a true statement and sometimes a false statement.
The predicate itself with the variable left open (that is, not substituted with any specific
value) cannot by itself have a well-defined status of being either true or false. Hence,
predicates are not considered statements or propositions. Predicates can only be called
sentences or formulae. The reader will note that in Study Unit 1, whenever we considered
statements that had symbols like n or m in them, we were careful to state beforehand that
those symbols denoted specific numbers or specific integers.
Consider the predicate "
". If the domain of the predicate variable is the positive
real numbers, then the equation
be true no matter what value of
has no solution, and so the predicate can never
is substituted in. But if the domain of
numbers, then the equation
has two solutions: 0 and
is all the real
. The predicate becomes
a true statement if we substitute in either the value of 0 or the value of
for the variable
. This example tells us that we need to know the domain of the predicate variable in order
to determine whether or not the predicate can be turned into a true statement.
Definition
Let
be a nonempty set. If
denotes a predicate, and
of
is the set of all elements of
of
is the set
that make
In the above notation, the expression
of
true. In other words, the truth set
means " is a member of
", and it can be simply read as " in
" or " is an element
". The vertical stroke | can be read as "such
that". Thus, the expression "
that
has domain , the truth set
" can be read as " is an element of
such
is true". The curly braces { } is the notation for a set. It is used when we are
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Predicate Logic and Quantified Statements
defining a set by specifying what types of elements are found in the set, or what property
an element has to satisfy in order to be a member of the set.
For example,
is the predicate "
then the truth set of
The number
", and the domain of is the rational numbers,
is the set of all rational numbers that are at least as large as
is not itself a member of the truth set because
However, if we change the domain of
.
is not a rational number.
to be all the real numbers, then the truth set of
is now the set of all real numbers that are at least as large as
. The number
is
now a member of the truth set.
Thus, it can be seen that the truth set of a predicate
depends on the domain of the
predicate variable .
Certain sets of numbers are used so frequently that they are assigned or denoted by special
symbols. These are summarised in the table below:
Table 2.1 Commonly used sets and the symbols denoting them
Symbol
Set
The set of real numbers
The set of positive real numbers
The set of rational numbers
The set of positive rational numbers
The set of integers
The set of positive integers, also known
as natural numbers
Note: In some texts,
denotes the set of non-negative integers (which includes the
number 0). This is especially true in texts on logic or computer science.
SU2-6
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Predicate Logic and Quantified Statements
Activity 2.1
Determine the truth set of each of the following predicates. The domain of the
predicate variable is indicated along with each predicate.
a.
"
". Domain of is
b.
" has a factor of 12". Domain of is
c.
" is a factor of 12". Domain of is
1.2 Quantifiers:
.
.
.
and
Definition
Let
be the domain of . A universal statement is a statement
be a predicate and let
of the form
The symbol
is called the universal quantifier and is read as "for all". The above
statement reads "for all ,
is true". It is implicitly assumed that
is only allowed to
take values within the domain . To be precise, we could also read the statement as "for
all
,
is true", and we could equivalently write the statement as
The universal statement
in
is true if
for the predicate variable .
SU2-7
is true when we substitute any element
MTH105
Predicate Logic and Quantified Statements
The statement
into
is false if there exists some element in
makes
into a false statement. A value of
statement is known as a counter-example for
For example, let
if we substitute
is false. A counter-example is
The statement
that makes
into a false
.
. The statement
be checked that
that when substituted
is a true statement. It can
, 2 or 3. However, the statement
, because
is not strictly greater than .
is true. The square of any real number is greater than or
equal to 0. The statement
is also true. Even though we know that the
square of any real number can never be negative, it is nonetheless still true that
for all real numbers . The statement may be imprecise, but that does not change the fact
that it is a true statement.
Logical connectives such as
can also be used within universal
statements, or more generally, within quantified statements (we will later examine another
quantifier, the existential quantifier
For example,
integers,
).
is the statement which reads "for all
implies
For instance,
. It is a false statement. A counter-example would be
is not equal to 1, and yet
example would be
in the set of
.
is strictly less than 4. Another counter-
. In order to explain why a universal statement is false, it is
sufficient to give just one counter-example.
The statement
is true. If the square of an integer is larger than 1,
the square of the same integer must be at least 4.
In the universal statement
it is understood that the domain of the predicate variable
is . We also allow for more
general statements containing more than one variable. For example,
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MTH105
Predicate Logic and Quantified Statements
is a universal statement where
is a predicate containing the variables
and . In
this case, both variables and have the same domain .
The statement
reads "for all , in the set of natural
numbers, if
, then
and
". The reader can easily verify that this is a true
statement.
Activity 2.2
Determine whether the following universal statements are true or false. If they are
false, give a counter-example to demonstrate that they are false.
a.
.
b.
.
c.
.
Definition
Let
be a predicate and let
be the domain of . An existential statement is a
statement of the form
The symbol
is called the existential quantifier and is read as "there exists". The above
statement reads "there exists ,
is true". It is implicitly assumed that is only allowed
to take values within the domain
. To be precise, we could also read the statement as
"there exists
,
is true", and we could equivalently write the statement as
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Predicate Logic and Quantified Statements
The existential statement
substituted into
The statement ∃
makes
is true if there is some specific value of
that when
makes it true.
is false if there is no element in
that when substituted into
into a true statement.
For example,
make
is a true statement. If we substitute the value
true. In other words, the predicate
By contrast, the statement
, we
is satisfied by the value
is false. The equation
.
is not satisfied
for every real number . It is only satisfied for certain real numbers , in this case, only by
. Thus, the universal version is false, and the existential version is true.
As a related example, consider the statement
is false because there is no natural number
. This existential statement
satisfying the equation
. Remember
that the set of natural numbers is the set of positive integers. An existential statement is
false if no element in the domain of the predicate variable satisfies the predicate or turns
the predicate into a true statement.
This is an example of an existential statement utilising logical connectives:
. This existential statement is true, and the predicate
is satisfied by the integers
compactly as
.
If we change the domain of
now
predicate
. It is customary to write the predicate more
to the set of rational numbers, that is, the statement is
, then the existential statement is still a true statement, and the
is now satisfied by all the rational numbers strictly greater than
and strictly less than 3. The truth set of the predicate is now an infinite set.
But if we change the domain of
to the set
existential statement is recast as
, which means that the
, then it is now a false statement,
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Predicate Logic and Quantified Statements
because there is no element in the set
makes the predicate
that satisfies the predicate
or which
into a true statement.
Similar to universal statements, we also allow for existential statements in which there is
more than one variable. For example,
is an existential statement where
is a predicate containing the variables and . In
this case, both variables and have the same domain .
The statement ∃
reads "there exists , in the set of natural numbers,
". The reader can easily verify that this is a true statement. The equation
is satisfied by
and also by
.
Activity 2.3
Determine whether the following existential statements are true or false. If they are
true, give the truth set of the predicate.
a.
.
b.
.
c.
.
1.3 Universal Conditional Statements
Special attention is drawn to universal statements in which the logical connective
or
is used. A statement of the form
or of the form
is called a universal conditional statement.
The statement
; in other words,
is true if for every in the domain ,
is true whenever
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is true.
implies
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The statement
is true if for every in the domain ,
and is implied by
; in other words,
is true if and only if
implies
is true.
Variants of Universal Conditional Statements
Consider a statement of the form
.
• Its contrapositive is the statement
.
• Its converse is the statement
.
• Its inverse is the statement
Recall that if
.
is a conditional statement, it is logically equivalent to its contrapositive
. The same is therefore true when the universal quantifier is present. A
statement of the form
is logically equivalent to its contrapositive
.
The converse
and the inverse
are
logically equivalent to each other, but neither is logically equivalent to the original
statement
Let
.
be the set { 2, 3, 4, 5, 6 } and consider the statement
. The
universal conditional statement reads as "for all in the domain , if is negative, then is
even. But there is no number in the set
which is negative. The statement
is said to be vacuously true for every element in .
Definition
A universal conditional statement of the form
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is said to be vacuously true if
is false for every in its domain .
Activity 2.4
Classify the following universal conditional statements into true, vacuously true,
and false. Note that vacuously true statements are also true statements. For false
statements, give a counter-example.
a.
.
b.
.
c.
.
d.
.
Definition
The following are all equivalent ways of expressing the statement
a.
For all ,
is a sufficient condition for
is sufficient for
b.
For all ,
.
is a necessary condition for
is necessary for
c.
For all ,
d.
For all , if
, or more simply phrased,
, or more simply phrased,
.
only if
.
then
.
Usually to emphasise that we are only interested in the situation when
is
true (or when we wish to ignore the vacuously true situation), we would say for
all , if
is true, then
is true.
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e.
For all ,
implies
f.
For all ,
if
.
.
Definition
The following are all equivalent ways of expressing the statement
a.
For all ,
is a necessary and sufficient condition for
phrased,
is necessary and sufficient for
b.
For all ,
if and only if
c.
For all ,
implies and is implied by
, or more simply
.
.
.
1.4 Formal versus Informal Language
In the process of conveying mathematical ideas to other people, it is important
to understand the differences between formal mathematical language and informal
language. To avoid confusion or misunderstanding, the language we use must be precise.
We must also understand the various nuances that can arise when translating quantified
statements (statements involving
For example,
or ) into English and vice-versa.
can be translated in informal English in these ways:
a.
For all real numbers , the square of is at least 0.
b.
The square of is at least 0 for all real numbers .
c.
The square of is at least 0 for any real number .
d.
The square of any real number is nonnegative.
(Here, there is no mention of any predicate variable .)
e.
If is a real number, then the square of is nonnegative.
f.
Any real number has the property that its square is nonnegative.
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As another example, the statement
can be translated into informal English
in these ways:
a.
There exists some real number such that
.
b.
The square of is equal to for some real number .
c.
There is at least one real number for which
d.
Some real number has the property that it is equal to its square.
.
(Here, there is no mention of any predicate variable .)
e.
Some real number satisfies the condition that it is equal to its square.
f.
We can find a real number satisfying
.
1.5 Some Equivalent Forms of Quantified Statements
The universal statement
is equivalent to
In particular, if
is a subset of
statement
, then the statement
is equivalent to the
.
An example of this would be when
of squares, and
denotes the set of rectangles,
denotes the set
is the predicate " has four sides of equal length". The statement
reads "for all squares , has four sides of equal length", and is equivalent
to the statement
which reads "for all rectangles , if is a square,
then has four sides of equal length".
The existential statement
is equivalent to
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In particular, if
is a subset of
statement
, then the statement ∃
.
An example of this would be when
set of prime numbers, and
∃
is equivalent to the
denotes the set of natural numbers,
denotes the
is the predicate " is an even integer". The statement
reads "there exists a prime number such that is even", and is equivalent
to the statement
which reads "there exists a natural number such
that is in the set of prime numbers and is even".
1.6 Implicit Quantification
Consider the sentence "if
is an integer, then
is a rational number". It is customary to
regard this as a statement by adding a universal quantifier at the beginning of the sentence
so that it becomes "
".
This brings us back to the situation discussed in Section 1.4, in which we stated that
the statement of the form
statement
would be considered equivalent to the
.
The original sentence "if is an integer, then is a rational number" is considered to be a
universal statement even though when phrased using informal language, no quantifier is
mentioned. This is an example of implicit quantification.
Our example above involves the universal quantifier. Existential quantification can also
be implicit. For instance, the sentence "21 can be expressed as the product of two prime
numbers" is equivalent to
, where
. Here,
is the set of prime numbers, or
is short for
In mathematics, it is customary to use double arrows
.
and
to symbolically indicate
the presence of implicit quantification. For instance,
means the universal statement
. In this case, the domain of the predicate
variable must be stated separately.
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Definition
Let
and
be a predicates, where the variable in both predicates have the same
domain .
The notation
means the statement
.
The notation
means the statement
.
Activity 2.5
Determine whether the following statements are true or false. If they are false, give a
counter-example to demonstrate that they are false. For all three parts, the domain of
is the set of real numbers
a.
.
.
b.
.
c.
d.
.
.
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Chapter 2: Properties of Quantified Statements
2.1 Logical Equivalence of Quantified Statements
We say that two quantified statements are logically equivalent if they have the same truth
value no matter what predicates are used and regardless of the domains of the predicate
variables.
For example, the statement
, where
is logically equivalent to the statement
is the negation of
. Here, it is assumed that the variable
in both quantified statements share the same domain. We can justify this logical
equivalence as follows:
Suppose that the statement
is true. Then the statement
The statements reads "for all ,
there exists some such that
is false.
is true". If this statement is false, it must mean that
is false. Therefore, the statement
true. We have thus shown that if the statement
must be
is true, then the statement
is also true.
Conversely, suppose that the statement
such that
statement
is true. This means that there is some
is true, that is, there is some
such that
is false. Therefore, the
must be false, and so the statement
thus shown that if the statement
is true. We have
is true, then the statement
is also true.
is true if and only if the statement
Our conclusion is that the statement
is true. This completes our proof that the statement
logically equivalent to the statement
is
.
We will revisit the above ideas in the next Section, Section 2.2, on negations of quantified
statements.
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As another example, let
be any set. Then the statements
and
are logically equivalent. This justification is as
follows:
Suppose that the statement
domain of . Then
is true. Let
be an element in the
is true, so by propositional logic, both
and
are
true. (In particular, we are using Specialisation (see Table 1.22 of Study Unit 1).) Since
was arbitrarily chosen, so both the statements
Therefore,
and
are true.
must be true by propositional logic
(in particular, using Conjunction).
Conversely, suppose that the statement
Then both the statements
in the domain of . Then
is true.
and
and
are true. Let be any element
are both true, so
element was arbitrarily chosen, so the statement
is true. Since the
is true.
Activity 2.6
Show
that
the
statements
are logically equivalent.
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and
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2.2 Negation of Quantified Statements
Negation of Universal Statements
Let
be a predicate. The negation of the statement
is logically equivalent to the statement
where
is the negation of
. Here, it is assumed that the variable
in both
quantified statements share the same domain.
For example, the negation of the statement
. The first statement asserts that
is the statement
is greater than 0 for every real
number . It is clearly a false statement. The negation asserts that
to 0 for some real number . That is true. For example, take
The statement
asserts that
is less than or equal
.
is a true statement for every value of
within
the domain of . This universal statement is false if and only if there exists some value
of
within the domain of
such that
is false. Such a value of
and thereby demonstrating the falsity of the universal statement
a counter-example to the universal statement.
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making
false,
, is known as
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Negation of Existential Statements
Let
be a predicate. The negation of the statement
is logically equivalent to the statement
where
is the negation of
. Here, it is assumed that the variable
in both
quantified statements share the same domain.
For example, the negation of the statement
is the statement
. The first statement asserts that
for some real number . It is
clearly a false statement. The negation asserts that
for every real number . That
is clearly true: the square of any real number is never negative.
Negation of Universal Conditional Statements
Let
and
be predicates in which the variable
has the same domain for both
predicates. The negation of the statement
is logically equivalent to the statement
The above rule is, of course, a special case of the more general principle that the negation
of
is logically equivalent to
.
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As an example, consider the statement "people who have blonde hair also have blue eyes".
This is an implicitly quantified statement that can be more formally expressed as
or we can use double arrow notation:
If we wish to demonstrate that the universal conditional statement is true, we have to
show that all people with blonde hair have blue eyes. But if we wish to demonstrate that
the statement is false, we just have to produce a counter-example, namely, some person
having blonde hair but no blue eyes. The negation of the statement is:
Activity 2.7
Write down the negation of each of the following statements:
a.
.
b.
.
c.
.
d.
.
2.3 Statements with Multiple Quantifiers
The statement
is the statement that for every , and for every ,
is
satisfied.
If
and
share the same domain , then we can abbreviate the statement using a single
quantifier:
.
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However, if
and
have different domains, we have to separate the quantifiers:
.
It does not matter whether we write
or
. The order of the
quantifiers can be interchanged because they are the same quantifier. The statement
is logically equivalent to
.
As an example, consider the statement
. The statement reads
"for every natural number , for every positive real number ,
Here,
refers to
is greater than 0".
raised to the power of . The statement is logically equivalent to
.
The
discussion
above
also
holds
for
existential
quantifiers.
is the statement that there exists some
, such that
The
statement
, there exists some
is satisfied. The order of the quantifiers can be interchanged. We
now consider the situation when the quantifiers are different.
Consider the statement
. This statement reads "there exists some
such that for all
that there is some
,
such that
is true.” In other words, the statement is asserting
is true for all values of
. To interpret the
statement correctly, we can place parentheses as follows:
The quantified statement is read "from outside in", that is, we read the quantifier
first, followed by whatever appears within the bracket.
As an example, consider
. To be clear, we can also place a set of
brackets:
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and remind ourselves to read the statement "from outside in". The statement asserts that
there exists some integer such that for every real number , we have
statement is true. The integer
, we have
. Clearly this
satisfies the requirements, in that for every real number
. The following argument demonstrates the validity of universal modus
tollens:
Consider the statement
exists some
,
,
. This statement reads "for all
is true. In other words, the statement is asserting that for every
is true for some value of
.
As an example, consider
. The statement asserts that for every
natural number , there exists some rational number
statement is true. Given
, let
such that
. Clearly, the
. Then is a rational number and
In the statement
, different values of
given value of
, if we are
, then we have to choose
presented with
, then we must choose
is unique. For
.
may have to be used for
different values of . For instance, in the statement
presented with
, there
in order to have
. If we are
. In this example, the choice of for each
, no rational number
except for
will make
true. The uniqueness of the value of for each given value of is a consequence of
the fact that we can "solve" the equation
to obtain
(by dividing the equation
throughout by ).
When
the
quantifiers
are
different,
their
order
cannot
is not logically equivalent to
For instance, while the statement
be
interchanged.
.
is true, the statement
is false. The second statement asserts that there is some rational
number such that for all natural numbers , we have
. To show that this statement is
false, we can argue as follows: Suppose that is a rational number satisfying the required
property. We can write
we let
or
or
where
is a positive integer, and
, etc., will be a natural number, but
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is an integer. Then if
cannot possibly be equal to
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1 for all these possible values of
, etc. This shows that
cannot hold for all
natural numbers . In fact, there can only be at most one (possibly no such)
satisfying
. This is an example of a proof by contradiction – assume the statement you want to
show is true, and derive a contradiction from there.
We can also have statements involving more than two quantifiers. As a simple example,
consider
. The statement reads "for all positive real
numbers , there exists an integer such that for every real number , we have
show that this statement is true, suppose that a positive real number
. To
is given. It then
suffices to come up with a specific integer that satisfies the requirements. It is not difficult
to check that
does the job.
Activity 2.8
Determine with justification which of the following statements are true or false:
a.
.
b.
.
c.
.
d.
.
e.
.
f.
.
The following table summarises the approach to multiply quantified statements:
Table 2.2 The approach to multiply quantified statements
Statement:
.
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Typical form of an argument demonstrating the statement is true: Let
and let
be given. Show that
work no matter which
is true. Our proof that
or
be given
is true must
was given.
Statement:
.
Typical form of an argument demonstrating the statement is true: Choose some
and choose some
at our own discretion. Show that
needs to work for our choice of
and
is true. Our proof only
. It does not have to work for every
possible and .
Statement:
.
Typical form of an argument demonstrating the statement is true: Choose some
and show that
is true no matter what
is subsequently picked. Our proof
. It does not have to work for every possible
only needs to work for our choice of
. However, the proof must work for any
once we have made our choice of
.
Statement:
.
Typical form of an argument demonstrating the statement is true: Let any
given. We need to choose some
the case that the choice of
such that
be
is true. It is often (but not always)
depends on the value of .
2.4 Negations of Multiply Quantified Statements
In order to show that the statement
every and for every ,
is true, we have to demonstrate that for
is true. To show that the statement is false, we only need
to devise a counter-example, that is, exhibit specific values of
and
that would make
false.
For example, to demonstrate that the statement
need to provide a specific counter-example, such as
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is false, we only
.
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In order to show that the statement
is true, we have to demonstrate that
there is some such that for every ,
is true. If we want to show that the statement
is false, we can do so by showing that for all , we can find some value of making
false. In other words, we prove the statement
.
For example, we can show that the statement
that the statement
given. Then pick
is false by proving
is true. First suppose that a real number is
. This makes
no matter what value of was chosen in the first
place.
Negation of Multiply Quantified Statements
Let
be a predicate involving the variables and .
a.
is equivalent to
.
b.
is equivalent to
.
c.
is equivalent to
.
d.
is equivalent to ∃
.
Similar rules apply for statements involving more than two quantifiers.
When negating a quantified statement, the universal quantifier
existential quantifier , and similarly, the existential quantifier
changes to the
changes to the universal
quantifier
. In addition to the above rules, the predicate is also negated.
Consider
again
the
example
where
we
showed
that
the
statement
is false using the method of proof by contradiction, that is, by
assuming there is some rational number
satisfying the condition, and then deriving a
contradiction. We can also show the statement is false by directly proving the negation:
. Suppose that a rational number
up with some natural number such that
is a positive integer, and
is an integer. Let
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is given. We have to come
. Again as before, we write
. Then
, where
. Regardless
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of the value of , the number
cannot be equal to 1 (because
is an integer). Thus we
have demonstrated that the statement
is true. It follows that
the original statement
must be false.
Activity 2.9
Let
and let
. Determine which statement is
true or false, and write negations for each statement.
a.
.
b.
.
c.
.
Table 2.3 The negation of multiply quantified statements
Statement:
.
To prove it is true: We need to show that for all
and for all
To prove it is false: We need to show that there exists some
that
,
and some
is true.
such
is false.
Statement:
.
To prove it is true: We need to show that there is some
and some
such that
is true.
To prove it is false: We need to show that for all
Statement:
and for all
is false.
.
To prove it is true: We need to show that there is some
for all
,
.
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such that
is true
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To prove it is false: We need to show that for all
, there exists some
such that
, there exists some
such that
is false.
Statement:
.
To prove it is true: We need to show that for all
is true.
To prove it is false: We need to show that there exists some
false for all
.
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is
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Chapter 3: Arguments with Quantified Statements
3.1 Rules of Inference for Quantified Statements
We now describe some important rules of inference for quantified statements. These
rules are used throughout mathematics but often without being explicitly mentioned.
Even though they are used instinctively or even intuitively, it is nonetheless crucial that
we define and specify them precisely so that all the subsequent work on mathematical
reasoning and mathematical proof is built on a rigorous foundation.
The following table lists down some essential rules of inference for quantified statements.
In the table, it is understood that is an element of the domain of .
Table 2.4 Rules of inference for quantified statements
Rule of Inference
Name
Universal instantiation (UI)
.
.
Universal generalisation (UG)
is true for any in the domain of .
Existential instantiation (EI)
.
is true for some in the domain of .
is true for some in the domain of .
Existential generalisation (EG)
.
Universal instantiation is the rule of inference that says that if the statement
true, and is an element of the domain of , then the statement
is
must be true. As an
example of this, consider the statement "all tropical trees carry out photosynthesis", which
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we are told is true. With this knowledge, if we now know that is a tropical tree, then we
can conclude that carries out photosynthesis.
Universal generalisation is the rule of inference that states that if we know that
true for every
in the domain of , then the statement "
is
" is true. This can be
thought of as the opposite process of universal instantiation.
Suppose that we are asked to prove that the statement
is true. Using the concept
of universal generalisation, we construct an argument that begins with "suppose
is true for any element
of the given domain", and then we proceed to show that
is true. Our proof must work regardless of the value of
initially chosen, as long as
is
within the domain of . Upon successful completion of our proof of the validity of
we then conclude that
is true for all , that is, we make the conclusion "
Existential instantiation is the rule of inference that allows us to conclude that
for some in the domain of
,
if we know that the statement
conclude that any arbitrary in the domain of
".
is true
is true. We cannot
is true, but only some . It is sometimes
the case that we do not know what the specific value is of
that makes
true, only
that such a value of exists.
An example of existential instantiation occurs when we start off by asserting that "there
exists a real number whose decimal expansion differs from the decimal expansion of
every
digit", and then we proceed to let
at
denote one such number. Of course, we
cannot possibly write down what we think should be. We only know that such a must
exist. The idea is that beginning with the existential statement
further to a particular for which
, we specialise
happens to be true.
Existential generalisation is the rule of inference that allows us to conclude that the
statement
is true if we know that
is true for some in the domain of . It
can be thought of as the opposite process of existential instantiation.
Universal instantiation and modus ponens are often used together in a combination which
is called universal modus ponens. The argument form is as follows:
.
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.
.
Here, it is understood that is an element of the domain of .
As an illustration of universal modus ponens, suppose we know for a fact that for all
natural numbers , if
is greater than 4, then
. In formal notation, this means
. The domain of in this case is the set of natural numbers. Now, 10
is a natural number which satisfies
that
. Hence, by universal modus ponens, we deduce
.
To demonstrate the validity of universal modus ponens, that is, to show that universal
modus ponens is a valid argument form, we can construct the following argument which
utilises the rules of inference (both for quantified statements as well as for unquantified
statements):
Step
Reason
1.
.
2.
Premise.
Universal instantiation from step (1).
3.
.
Premise.
4.
.
Modus pollens using (2) and (3).
Here, we use the rules of inference laid out in Table 1.22 of Study Unit 1, as well as Table
2.3.
Universal instantiation and modus tollens are also often used together in a combination
which is called universal modus tollens. The argument form is as follows:
.
.
.
Again, it is understood that is an element of the domain of .
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As an illustration of universal modus tollens, consider again the following statement:
. Since
is a natural number such that
so by universal modus tollens, we conclude that
example, we already know trivially that
, that is,
, that is,
. Of course, in this
without having to use any argument to
justify it.
The following argument demonstrates the validity of universal modus tollens:
Step
Reason
1.
Premise.
.
2.
Universal instantiation from step (1).
.
3.
.
Premise.
4.
.
Modus tollens using (2) and (3).
Activity 2.10
Use the rules of inference to construct an argument demonstrating the validity of
universal transitivity, which has the following argument form:
.
.
.
The variable is assumed to have the same domain for all three statements.
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,
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Activity 2.11
Use the rules of inference to construct an argument demonstrating the validity of the
following argument form:
.
.
.
The variable is assumed to have the same domain for all three statements.
3.2 Logical Fallacies and other Flawed Arguments Involving
Quantified Statements
Common logical fallacies involving quantified statements include the usual converse error
and inverse error.
For quantified statements, the converse error is made when an argument of the following
form is attempted:
.
.
.
Here, is an arbitrary element of the domain of .
An example of the converse error is:
"If a woman wears a black skirt, she will also wear a blue dress."
"Jane (a specific woman) is seen wearing a blue dress."
"Therefore, Jane must have also worn a black skirt."
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Implicit quantification was used in the major premise "If a woman wears a black skirt, she
will also wear a blue dress." Although the English sentence is phrased to avoid using the
universal quantifier, the statement is in fact a universal statement: "For all women , if
wears a black skirt, then will also wear a blue dress."
(It is not necessary to know whether or not in real life, the major premise is in fact true.
The line of reasoning is incorrect regardless.)
The inverse error is made when an argument of the following form is attempted:
.
.
.
Here, is an arbitrary element of the domain of .
An example of the inverse error is:
"Everyone who goes to the party will eat chicken wings tonight."
"John (a specific person) did not go to the party."
"Therefore, John will not eat chicken wings tonight."
Other flawed arguments involving quantified statements can arise due to mislabelling of
elements. For instance, the following is an attempt (a flawed one) to justify the argument
form:
.
.
Step
Reason
1.
. Premise.
2.
.
3.
for some
4.
Specialisation from (1).
Existential instantiation from (2).
Specialisation from (1).
.
SU2-35
MTH105
Predicate Logic and Quantified Statements
5.
Existential instantiation from (4).
for some
6.
Conjunction from (3) and (5).
.
7.
Existential generalisation from (6).
.
What went wrong in the above argument is that in steps (3) and (5), when we used
existential instantiation to deduce
letter for both statements.
for some
and
, we should not have used the same
is true for some in the domain of , and
in the domain of , but
and
is true
need not be the same element. The argument
should have been written as follows:
Step
Reason
1.
. Premise.
2.
.
3.
for some
4.
5.
If
for some
.
is true and
Existential instantiation from (2).
Specialisation from (1).
.
6.
Specialisation from (1).
Existential instantiation from (4).
Conjunction from (3) and (5).
is true, where we recognise the possibility that
be different, then we can no longer use conjunction to assert
unable to deduce
, then we cannot deduce
We can only use Conjunction to deduce the statement
and
could
. And if we are
.
, where and
could
possibly be different elements in the domain of . From here, we cannot proceed any
further and our argument comes to a dead end. The statement
only be deduced from
can
, where the same is used for both predicates and no
other symbol appears.
SU2-36
MTH105
Predicate Logic and Quantified Statements
Summary
In this Study Unit, we introduced predicates and quantified statements using the
quantifiers
and ∃. An existential statement is a statement of the form
universal statement is a statement of the form
while a
.
We learnt how to interpret quantified statements and how to negate quantified statements.
Multiply quantified statements were also discussed.
Finally, we learnt rules of inferences that involve quantified statements, and looked at
some common logical fallacies and flawed arguments that must be avoided when dealing
with quantified statements.
SU2-37
MTH105
Predicate Logic and Quantified Statements
Formative Assessment
1.
Determine the truth set of each of the following predicates. The domain of the
predicate variable is indicated along with each predicate.
2.
a.
. The domain of is
.
b.
. The domain of is
.
c.
is a multiple of both 3 and 5. The domain of is
d.
is a factor of 24. The domain of is
e.
. The domain of
is
.
f.
. The domain of
is
.
g.
. The domain of
is
.
.
.
Determine the truth set of each of the following predicates. The domain of the
predicate variable is indicated along with each predicate.
a.
. The domain of is
b.
is a multiple of 7)
c.
is a factor of 18)
d.
3.
(
(
.
). The domain of is
). The domain of is
. The domain of is
.
.
.
Determine whether the following statements are true or false. For universal
statements that are false, provide a counter-example.
a.
.
b.
.
c.
.
d.
.
e.
if is a prime number, then is an odd integer.
f.
is a prime number and is not an odd integer.
g.
Let
.
.
SU2-38
MTH105
4.
Predicate Logic and Quantified Statements
h.
Let
.
i.
Let
.
j.
Let
.
k.
Let
.
.
.
.
.
Classify the following universal conditional statements into true, vacuously true, or
false. Note that vacuously true statements are true statements. For false statements,
give a counter-example.
a.
5.
6.
.
b.
.
c.
.
d.
.
Determine whether the following statements are true or false:
a.
.
b.
.
c.
.
d.
.
e.
.
f.
.
Let
. Determine whether the following statements are true or
false.
a.
.
b.
.
c.
.
d.
.
e.
.
SU2-39
MTH105
7.
Predicate Logic and Quantified Statements
Use the rules of inference to construct an argument demonstrating the validity the
following argument form:
.
.
.
The variable is assumed to have the same domain for all three statements. Provide
the reason for each step in the argument.
8.
Let
and let
. Determine which
statement is true or false, and write negations for each statement.
a.
.
b.
.
c.
.
d.
.
e.
f.
9.
is even.
is even.
Identify the logical fallacies in the following arguments:
a.
"Everyone who has over 80 marks will obtain an A grade."
"John obtained an A grade."
"Therefore, John must have gotten over 80 marks."
b.
"There is some woman out there who likes cheese cakes."
"There is some woman out there who likes chocolate fudge."
"Jane (a specific woman) does not like cheese cakes."
"Hence, Jane must like chocolate fudge."
c.
"All lights will be automatically shut off at 9am."
"It is not yet 9am."
"The light at the conference room must still be switched on."
d.
"Jane missed her bus yesterday."
SU2-40
MTH105
Predicate Logic and Quantified Statements
"Jenny was late for her lecture yesterday."
"Hence, there was at least one person who missed their bus and was late for
their lecture yesterday."
10.
a.
Let
and suppose that the predicate
variable has domain . Determine the truth set of the predicate
b.
Let
and suppose that the predicate variable
has domain . Determine the truth set of the predicate
11.
a.
b.
Let
denote the following quantified statement:
i.
Provide the negation of the statement
ii.
Determine whether the statement
Let
.
is true or false.
denote the following quantified statement:
i.
Provide the negation of the statement
ii.
Give a counterexample to show that statement
SU2-41
.
is false.
MTH105
Predicate Logic and Quantified Statements
Solutions or Suggested Answers
Activity 2.1
a.
b.
.
Set of all integers that are multiples of 12, that is,
or more simply expressed as:
The latter notation means the set of all numbers
in the set
c.
, as varies over all elements
.
{ 1, 2, 3, 4, 6, 12 }.
Activity 2.2
a.
True.
b.
False. A counter-example would be a negative number, for instance,
Then
, which is not greater than 0. There is another thing wrong with
the statement. The reciprocal of 0 is not defined. If
so we cannot say whether
c.
.
,
is not even a number,
is greater than 0 or not.
False. A counter-example would be
.
Activity 2.3
a.
True. The truth set of is {2}.
b.
True. The truth set of is {0,2}. Compare with part (a). The domain of the variable
affects the truth set of .
SU2-42
MTH105
Predicate Logic and Quantified Statements
c.
False.
Activity 2.4
a.
True.
b.
Vacuously true (and thus, also true). Remember that
denotes the set of positive
integers. The product of three positive integers will never be negative.
c.
False. A counter-example is
d.
Vacuously true.
,
.
is never equal to 2 if
is a rational number. The square-root
of 2 is an irrational number.
Activity 2.5
a.
True.
b.
True.
c.
True.
d.
False. If we let
, we still get
. It is thus possible for the hypothesis
to be true but the conclusion to be false.
Activity 2.6
Suppose that the statement
such that
is true. Then there is some element
is true. We have to consider two cases: Either
is true, or
is true.
Suppose first that
is true. Then the statement
is true. Therefore by
propositional logic (in particular, using Generalisation), the statement
SU2-43
MTH105
Predicate Logic and Quantified Statements
is true.
Similarly, if we suppose that
is true, then the statement
is true, and
just like in the previous case, the statement
is true by Generalisation.
Conversely, suppose that the statement
is true.
We again have to split our argument into two cases: Either
is true, or
is true.
Suppose that
is true. Then there is some
such that
proposition logic (in particular, using Generalisation),
statement
is true. By
is true, and so the
is true.
Similarly, if we suppose that
is true, then there is some
such that
is true. By proposition logic (in particular, using Generalisation),
so the statement
is true, and
is true.
Activity 2.7
a.
.
Note that the negation of this universal statement is false (because the universal
statement itself is true). Every rational number
satisfy
b.
satisfying
must also
.
.
The negation of the existential statement is true (because the existential statement
itself is false). Remember that
c.
denotes the set of positive integers.
.
SU2-44
MTH105
Predicate Logic and Quantified Statements
The negation of the universal statement is false (because the universal statement
itself is true).
d.
.
The negation of the existential statement is false (because the existential
statement itself is true).
Activity 2.8
a.
True. Given any natural number
, let
. Then
. Note that since
the domain of the variable is the set of integers, is allowed to be 0 or negative.
b.
False. Let
. Then there does not exist any natural number such that
c.
False. Suppose that a real number
. Then is a real number, but
for all real numbers , we have
d.
False. Consider
. Certainly,
real number such that
.
exists which satisfies the condition. Let
. This shows that it is not true that
. Thus we get a contradiction.
is a real number. But there does not exist any
. This shows that the statement is false – there is a
real number that does not satisfy the condition.
e.
True. Let
f.
True. Let
. Then for all natural numbers , we have
and
.
be given. Then choose
. Note that we are
able to divide by because is non-zero. For this choice of , is a real number
and
.
Activity 2.9
a.
True.
The negation of the statement is
b.
True. Choose
.
.
The negation of the statement is
SU2-45
.
MTH105
c.
Predicate Logic and Quantified Statements
False. If
, then there does not exist any
for which
The negation of the statement is
.
.
Activity 2.10
First, let be any element of the domain of . This will be needed in the argument.
Step
Reason
1.
.
2.
Premise.
Universal instantiation from step (1).
.
3.
Premise.
4.
.
Universal instantiation from step (3).
5.
.
Transitivity from (2) and (4).
6.
.
Universal generalisation from (5), because the element
can be any element in the domain of .
Activity 2.11
First, let be any element of the domain of . This will be needed in the argument.
Step
Reason
1.
.
2.
.
3.
Universal instantiation from step (1).
Premise.
4.
5.
Premise.
.
.
Universal instantiation from step (3).
Specialisation from (4).
SU2-46
MTH105
Predicate Logic and Quantified Statements
6.
.
Modus Ponens from (2) and (5).
7.
.
Specialisation from (4).
8.
Conjunction from (6) and (7).
.
9.
Universal generalisation from (8), because can
.
be any element of the domain of .
Formative Assessment
1.
Determine the truth set of each of the following predicates. The domain of the
predicate variable is indicated along with each predicate.
a.
. The domain of is
.
b.
. The domain of is
.
c.
is a multiple of both 3 and 5. The domain of is
d.
is a factor of 24. The domain of is
e.
. The domain of
is
.
f.
. The domain of
is
.
g.
. The domain of
is
.
.
.
Answer:
a.
From
, we get
, which means
Therefore, the truth set of
is
. Note that
or
.
is rejected because it
is not an integer. We must be careful only to accept values of within the
stated domain.
b.
The truth set of
answer
c.
is
. In this question, we also accept the
because the domain of is now the set of rational numbers.
The truth set of is the set of all positive multiples of 15, that is,
SU2-47
MTH105
Predicate Logic and Quantified Statements
d.
The truth set of is
. Note that we leave
out negative factors of 24 because the domain of
is the set of natural
numbers. If we change the domain to the set of all integers, then we would
also allow answers like
e.
The truth set of
.
is
. Note that we exclude 2 and
9 because of the strict inequality.
2.
f.
The truth set of
is
g.
The truth set of
is
.
.
Determine the truth set of each of the following predicates. The domain of the
predicate variable is indicated along with each predicate.
a.
. The domain of is
b.
is a multiple of 7)
c.
is a factor of 18)
d.
(
(
.
). The domain of is
). The domain of is
. The domain of is
.
.
.
Answer:
a.
b.
.
The truth set of is
. has to be a multiple of 7 and also has
to be strictly less than 20.
c.
The truth set of is
. A number is included in the
truth set of if it is either a positive factor of 18, or less than or equal to 7.
d.
The truth set of
is the set of rational numbers
rational numbers satisfy the condition
3.
. This is because all
.
Determine whether the following statements are true or false. For universal
statements that are false, provide a counter-example.
a.
.
b.
.
c.
.
SU2-48
MTH105
Predicate Logic and Quantified Statements
d.
.
e.
if is a prime number, then is an odd integer.
f.
is a prime number and is not an odd integer.
g.
Let
.
.
h.
Let
.
.
i.
Let
.
j.
Let
.
k.
Let
.
.
.
.
Answer:
a.
True.
b.
False. Consider the case where
. Therefore
. Certainly,
, but we also have
is a false statement, and so the universal
statement
is a false statement. Here, we use
as a counter-example.
c.
False. The only real numbers satisfying the equation
and
are
. Both are not natural numbers. So there are no natural numbers
satisfying the equation
.
d.
True. Both 0 and
are rational numbers satisfying the equation.
e.
False. 2 is a prime number, but 2 is even. (2 is the only even prime number)
f.
True. 2 is a prime number that is not an odd integer. Notice that statement
(f) is the negation of statement (e).
g.
False.
h.
True.
i.
False.
j.
True.
k.
True.
.
is even, but 6 is not less than 5.
SU2-49
MTH105
4.
Predicate Logic and Quantified Statements
Classify the following universal conditional statements into true, vacuously true, or
false. Note that vacuously true statements are true statements. For false statements,
give a counter-example.
a.
.
b.
.
c.
.
d.
.
Answer:
a.
True.
b.
Vacuously true. There is no natural number
such that
. The
hypothesis of the conditional statement cannot be satisfied.
5.
c.
False.
d.
False.
, but
is not equal to 30.
, but certainly
.
Determine whether the following statements are true or false:
a.
.
b.
.
c.
.
d.
.
e.
.
f.
.
Answer:
a.
True. Given any natural number , it is always possible to choose a big
enough natural number so that exceeds
b.
.
False. Suppose that such an exists, that is, suppose that there exists some
natural number
. Substituting
such that
for all natural numbers . Consider
into the inequality, we get
SU2-50
, which implies
MTH105
Predicate Logic and Quantified Statements
that
, and so
. This is a contradiction because
is supposed to
be a natural number.
c.
True. Let
. Then
, so certainly,
for every natural
number .
d.
True.
e.
False. Consider
such that
f.
6.
. Then
, so there is no natural number
. Here, we are using
as a counter-example.
True.
Let
. Determine whether the following statements are true or
false.
a.
.
b.
.
c.
.
d.
.
e.
.
Answer:
a.
True. Suppose
is given to us. If is even, we choose to be any odd
number in . If is odd, we choose to be any even number in . By doing
so, we can guarantee that
is odd. Note that the set
even and at least one odd number (in fact,
numbers). Hence, no matter what
in
b.
such that
number in
has three even and four odd
is given to us, we can always find a
is odd.
False. Suppose that such an
some
has at least one
such that for all
, that is, letting
exists, that is, suppose that there exists
,
is odd. But letting
to be any even
to be either 2,4 or 6, we have
is even,
because the product of an even integer with any integer is always even.
So it is not true that
is odd for any
in
. Since we have obtained a
contradiction, we conclude that the statement is false.
SU2-51
MTH105
Predicate Logic and Quantified Statements
False. Consider, for instance,
c.
. Then
words, there does not exist any
is even for any
making
. In other
odd. Here, we are using
as a counter-example.
d.
True. Let
e.
True. Given any
that
. Then certainly for any
, simply choose
is a multiple of
if
, we have
.
. Then is a multiple of . (Recall
for some integer . For instance, 15 is a
multiple of 3 and 5. As another example, 24 is a multiple of 1,
, 2,
.
Of course, there are other integers of which 24 is a multiple – can you list
them?)
7.
Use the rules of inference to construct an argument demonstrating the validity the
following argument form:
.
.
.
The variable is assumed to have the same domain for all three statements. Provide
the reason for each step in the argument.
Answer:
Step
Reason
1.
Premise.
.
2.
for some element
(1).
in the domain of .
3.
Premise.
.
4.
Existential instantiation from step
, where this is the
Universal instantiation from step
(3).
same element as in step (2).
5.
.
Specialisation from (2).
6.
.
Modus Ponens from (4) and (5).
SU2-52
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Predicate Logic and Quantified Statements
7.
Specialisation from (2).
.
8.
9.
8.
Conjunction from (6) and (7).
.
Existential generalisation from (8).
.
Let
and let
. Determine which
statement is true or false, and write negations for each statement.
a.
.
b.
.
c.
.
d.
.
e.
is even.
f.
is even.
Answer:
a.
True.
Negation is:
b.
True. Choose
.
.
Negation is:
c.
False. If we let
.
, then there does not exist any
Negation is:
d.
True. Choose
.
.
Negation is:
e.
f.
.
is odd.
True. For instance, we can choose
Negation is:
9.
.
False. As a counter-example, choose
Negation is:
such that
.
is odd.
Identify the logical fallacies in the following arguments:
a.
"Everyone who has over 80 marks will obtain an A grade."
SU2-53
.
MTH105
Predicate Logic and Quantified Statements
"John obtained an A grade."
"Therefore, John must have gotten over 80 marks."
b.
"There is some woman out there who likes cheese cakes."
"There is some woman out there who likes chocolate fudge."
"Jane (a specific woman) does not like cheese cakes."
"Hence, Jane must like chocolate fudge."
c.
"All lights will be automatically shut off at 9am."
"It is not yet 9am."
"The light at the conference room must still be switched on."
d.
"Jane missed her bus yesterday."
"Jenny was late for her lecture yesterday."
"Hence, there was at least one person who missed their bus and was late for
their lecture yesterday."
Answer:
a.
Converse error.
The original premise is: Above 80 marks ⇒ Obtain A grade.
b.
Affirming a disjunct.
c.
Inverse error.
The original premise is: 9am
Lights automatically shut off
(The conference room light could have been manually switched off.)
10.
d.
Mislabelling of elements.
a.
Let
and suppose that the predicate
variable has domain . Determine the truth set of the predicate
b.
Let
and suppose that the predicate variable
has domain . Determine the truth set of the predicate
SU2-54
MTH105
Predicate Logic and Quantified Statements
Answer:
a.
b.
11.
a.
b.
Let
denote the following quantified statement:
i.
Provide the negation of the statement
ii.
Determine whether the statement
Let
.
is true or false.
denote the following quantified statement:
i.
Provide the negation of the statement
ii.
Give a counterexample to show that statement
.
is false.
Answer:
a.
(i) The negation of statement
(ii) The statement
Reason: Let any
Then
b.
is:
.
is true.
be given. Choose any
.
. And thus
is vacuously true.
(i) The negation of statement (Q) is :
.
(ii) The statement
is false.
A counterexample to statement
SU2-55
is:
,
.
MTH105
Predicate Logic and Quantified Statements
References
Epp, S. S. (2019). Discrete mathematics with applications (5th ed.). Boston, MA: Brooks/
Cole Cengage Learning.
SU2-56
Study
Unit
Methods of Proof
3
MTH105
Methods of Proof
Learning Outcomes
By the end of this unit, you should be able to:
1.
State definition and theorems correctly using precise mathematical language.
2.
Show certain mathematical statements by rigorous mathematical arguments.
3.
Give counter-examples to disprove certain mathematical statements.
4.
Prove mathematical statements using indirect proofs such as proof by
contradiction or proof by contraposition.
5.
Define prime numbers, rational and irrational numbers.
6.
Define and use the concept of divisibility.
7.
Define and use the concept of the absolute value of a real number.
SU3-2
MTH105
Methods of Proof
Overview
Thus far, we have learnt about propositional logic as well as predicate logic which involves
the use of quantified statements.
In this Study Unit, we examine how to state definitions and theorems using precise
mathematical language, in particular, using the concepts from propositional logic and
predicate logic that we have learnt so far.
We will then examine various methods of proving statements in mathematics. These
include direct proofs, proofs by contraposition, proofs by contradiction, and proofs
by division into cases. We will also examine how to disprove existential or universal
statements that are false.
Some fundamental mathematical concepts that will also be covered include even and odd
integers, prime numbers, rational and irrational numbers, the concept of divisibility, and
the absolute value of a real number.
SU3-3
MTH105
Methods of Proof
Chapter 1: Introduction to Some Terminology and
Concepts
1.1 Some Terminology
Formally, a theorem is a statement that can be shown (or proven) to be true starting
from simpler assumptions or facts. In mathematical writing, the term "theorem" is usually
reserved for statements that can be considered important or crucial in the development of
the topic under discussion. Less important theorems are often called propositions.
A theorem may be expressed in the form
, or if implicit quantification
is employed, a theorem may be simply stated in the form
term
the hypothesis and
. As usual, we
the conclusion. Of course, there are theorems
that are expressed using the existential quantifier, or even using multiple quantifiers.
More generally, a theorem is presented as a series of premises from which we derive a
conclusion using the rules of inference:
Premise 1
Premise 2
.........
Premise
Conclusion
The premises used in a theorem can include axioms or postulates. These are statements
that we assume to be true often without giving an explicit proof anywhere in our writing.
For example, properties of real numbers such as
or
are usually
regarded as axioms. Mathematicians have devised a system in which all these axioms can
be logically justified, but for everyday use, it is not necessary to justify them as that would
be too repetitious.
SU3-4
MTH105
Methods of Proof
Besides axioms or postulates, the premises used in a theorem can also include results that
have been previously derived or proven. We build up subject by proving a sequence of
results or theorems, each leading to the next. This process will be made clear when we
embark on the study of Elementary Number Theory and Functions.
Very often, a result used primarily to aid in the proof of a more significant theorem is called
a lemma. It is a good practice to break down complicated proofs by first proving a series
of lemmas rather than squeezing everything together into a single argument or a single
train of thought. The lemmas proved can then be utilised in proving other theorems.
A corollary is a result that can be derived as a direct consequence of a theorem just proved.
Very often, a corollary is a special case of a theorem, singled out for special mention
because it is of mathematical interest in itself.
1.2 Understanding Definitions
In order to evaluate the truth or falsity of a statement, we must know the precise meaning
of each term used in the statement. Likewise, in order to correctly state and prove a
theorem, we must phrase the theorem using the correct terminology. In this section, we
will look at some essential definitions that apply to Elementary Number Theory.
Definition
Let be an integer. We say that
odd if
is even if
for some integer , and we say that
is
for some integer .
Definitions in mathematics are usually presented in a fashion similar to the above. In the
above definition, we are specifying exactly what we mean for an integer to be called even
or odd. Using this definition, we can see that the integers
and
. The
and 10 are even, because
symbol here is the usual multiplication of numbers.
On the other hand, 5 is odd because
.
SU3-5
, and
is also odd because
MTH105
Methods of Proof
Implicit in the above definition is that the notion of "even" or "odd" only applies to integers.
These terms do not apply to proper or improper fractions like
numbers like
or
, or to irrational
or . When we used the sentence "Let be an integer", it indicates we are
restricting the domain of to be the set of integers.
The above definition can also be recast in the form of a universal statement:
or
Here, the definition is expressed in the form of a universal conditional statement. Of
course, when defining what it means for an integer to be even, we really mean to express
the biconditional statement
In definitions such as the one above, despite using a predicate of the form
the reader should understand that the author always means
,
. The
biconditional phrase "if and only if" is abbreviated to the conditional phrase "if... then..."
only to make the language more concise and sound less pedantic. It could become slightly
tedious to keep using the phrase "if and only if" or "implies and is implied by" every time
we make a new definition.
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Activity 3.1
Let
be integers. Determine whether the following integers are even or odd or
impossible to determine without further knowledge of and .
a.
b.
c.
Definition
An integer is prime if
and cannot be written as a product of two positive integers
each strictly less than . In other words,
and
and
cannot be written as
, where
.
An integer is composite if
and is not prime.
Again, in definitions such as the one stated above, any conditional statements used to
define the meaning of the terms are understood to be, in fact, biconditional in nature. Thus,
an integer
and
the form
is called prime if and only if cannot be expressed in the form
is called composite if and only if can be expressed in
. An integer
, where
, where
and
.
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Activity 3.2
Classify the following numbers into prime, composite or neither:
1.3 Stating Theorems
In Section 1.1, we described the general form of a theorem as comprising of a set of
premises followed by a conclusion that can be logically deduced from those premises
using the rules of inference:
Premise 1
Premise 2
.........
Premise
Conclusion
For a start, we will not be stating theorems that involve a complex web of postulates and
premises. For now, we focus on theorems that directly assert that a certain property holds
for all elements in a given domain, or that there exists some element in a given domain
that satisfies a special property. The first kind of theorem involves a universal statement,
while the second kind of theorem involves an existential statement.
In general, we will be stating theorems using quantified or multiply quantified statements
like those discussed from Study Unit 2 all the way up to this point.
Example 1
For instance, a theorem can be stated as follows:
"For all positive real numbers and , if
, then
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."
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The above theorem is stated using the universal quantifier
. We can, of course, state the
above theorem similarly as:
"If we are given positive real numbers and satisfying
OR: "Suppose
. Then
, then
."
."
The above phrasing clearly reveals the Premises-Conclusion format. The premises of the
theorem are that
theorem is that
and
are positive real numbers, and that
. The conclusion of the
.
All of the above methods are equivalent ways of stating the same theorem, which in this
case involves a universal conditional statement.
Example 2
This is an example of a theorem involving an existential statement: "There exists an integer
such that can be expressed as a product of three distinct prime numbers."
An equivalent way of expressing the above theorem is: "We can find an integer that is a
product of three different primes."
For theorems that take the form of existential statements, it is more difficult to interpret
them using the Premises-Conclusion format. In the above example, we have in fact
expressed the theorem without any premises, while the existential statement itself is the
conclusion.
How is it possible to derive a conclusion if we have not stated premises then? This
is possible because we have assumed certain basic facts and properties about the set
of integers, such as the existence of addition and multiplication, as well as rules of
manipulating integers such as
or
. These basic facts and properties
can be logically developed from a set of fundamental mathematical axioms, and they
do not have to be stated every time we formulate a new theorem because it is regarded
as "assumed knowledge". In other words, think of them as "hidden premises" which
accompany each and every theorem that we state. It would be impossibly tedious to restate all the axioms and basic facts every time we formulate a new theorem.
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Further Examples
These are further examples of theorems that will be explored in this course:
"There are infinitely many prime numbers."
"
"If
is an irrational number."
is a prime number, then
is an irrational number."
"For any real number , there exists an integer
"If
is a one-to-one mapping from a set
satisfying
to a set , then
."
has at least as many elements
as ."
"If the integers
and
have no common positive divisors apart from 1, then the lowest
common multiple of and
is equal to the product
."
For examples such as "there is infinitely many prime numbers", it is more difficult to
understand where universal or existential quantification comes in, or how the theorem
fits into the Premises-Conclusion format. However, as we introduce more definitions and
concepts in mathematics (particularly, in number theory), and develop new techniques of
proof, the reader will be able to comprehend such theorems.
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Chapter 2: Direct Proofs and Usage of Counter-Examples
2.1 Proving Existential Statements
In order to prove that the existential statement
is true, we have to demonstrate that there exists some in the domain of
statement
such that the
is true. In other words, we can use the following direct approach: Find
some such that when is substituted for the variable , we obtain a statement
that
is true.
Example 1
Prove the following: There exists a positive integer such that can be written as a sum
of two squares (that is, there exists
Answer: Let
. Then
such that
. Here, we set
).
.
Of course, the reader will be able to come up with many more examples.
Example 2
Prove the following: There exists an integer such that
Answer: We first make
for some integer .
the subject of the above equation. This means rearranging the
terms of the equation so that
alone appears on the left and everything else is shifted to
the right hand side. The reader can easily work out that
Now we can substitute in
and obtain
. This demonstrates the existence of an
integer satisfying the above equation. In this example, different even integers
will yield
different values of , but each of them will be a correct answer to the question. Note also
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that if we try to substitute in odd integers , it will not give us an integer , because
would then be odd, and so
will not be an integer.
In both the above examples, the two statements that we are asked to prove are in fact
multiply quantified statements.
In the first example, the statement can be written as
In the second example, the statement can be expressed as
We consider both of these to be existential-type of statements because the "outer-most"
quantifier is the existential quantifier .
Example 3
Prove the following: There exists a real number such that
for any natural number
.
Answer: Let
for any
. It can be verified that
. In fact any real number
for any natural number , because
will satisfy the given condition.
In the above example, we have the multiply quantified statement
To demonstrate the validity of the above statement, it suffices to provide a specific real
number that satisfies the condition:
.
In the above examples, we are using the rule of inference known as Existential
Generalisation (EG). This rule, which was mentioned in Study Unit 2, states that if we
can show that
is true for some element
in the domain of , then the quantified
statement
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must be true. Of course, this easily generalises to multiply quantified existential
statements like
. The reader might want to revise Section 3.1 of Study Unit
2.
Activity 3.3
Prove the following existential statements using a direct approach:
a.
There exist integers
b.
There are distinct integers
and
and
such that
is an integer.
(distinct means
) such that
is
an integer.
c.
There are nonnegative real numbers
d.
There is a nonnegative integer such that
such that
.
is a prime number.
Fermat Primes
Fermat primes are an interesting case study in which mathematicians explore the existence
of certain objects (in this case, prime numbers taking on a particular form), and in which
there has been a great deal of research interest.
A Fermat prime is a prime number of the form
integer. The numbers
, where
is a non-negative
are known as Fermat numbers. It is easy to verify that
is prime for =0,1,2,3,4.
A remarkable fact is that as of the year 2018, the only Fermat numbers known to be prime
are
and
. It is not known whether
is prime for any other positive integer
. To date, no one has been able to prove or disprove this existential statement:
Fermat numbers and their primality have attracted the attention of mathematicians
throughout the centuries. Leonhard Euler factorized
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=641*6700417 in the year 1732. As
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of 2018,
is the largest Fermat number that has been completely factorized (that is, all
their prime factors are known). The largest prime factor of
is a number with 564 digits!
That is why it can take so much computational power to fully factorize a composite Fermat
number.
Source: https://en.wikipedia.org/wiki/Fermat_number (Accessed 14 July 2019)
2.2 Disproving Universal Statements using a Counter-Example
In order to prove that a universal statement
is false, that is, in order to disprove the above universal statement, it suffices to provide
a counter-example, that is, it suffices to come up with an element
in the domain of
is false. Here, we are using the fact that the negation of the statement
such that
is the statement
. Thus, to show that statement
direct approach would be to show that
that
is false, a
is true by providing an element such
is true.
Example 1
Disprove the following: For all real numbers
Answer:
, but
, if
is not greater than
, then
.
.
In the above example, we are asked to disprove a universal statement of the form
In order to do so, it suffices to provide a counter-example
in the domain of
above example, the domain is the set of real numbers) such that
equivalently,
is true and
is false, or
is false at the same time. The above example uses two
predicate variables and rather than just one, but the principle is the same.
Example 2
Disprove the following: If
(in the
is an integer, then is an integer.
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Answer:
is not an integer, yet
is an integer.
In the above example, we have used implicit quantification. In our answer, we provided
the counter-example
, which belongs to the set of real numbers. Thus we have,
without being told to do so, taken the initiative to assume that should belong to a set that
comprises real numbers such as surds (square roots of integers, or more generally, square
roots of rational numbers).
Activity 3.4
Disprove the following universal statements by providing a counter-example:
a.
For all integers , if is odd, then
b.
For all integers , if is even, then
c.
For all integers
d.
For all prime numbers ,
,
is also odd.
is also even.
is odd.
is also prime.
2.3 Proving Universal Statements
Many theorems in mathematics are stated in the form of universal statements. In fact, it
is very frequently the case that theorems are stated in the form of a universal conditional
statement, namely,
For the most general situation where we are asked to prove a universal statement
we may use a direct approach as follows: Pick any element in the domain of the variable
. The important point to note is that
has to be truly arbitrary: there must not be any
element in the domain of consideration that can potentially be excluded. We then show
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that
is true. If we can do this, then we would have proven the universal statement
. The rule of inference that we are using here is Universal Generalisation (UG).
Once again, the reader might want to revise Section 3.1 of Study Unit 2.
If we asked to prove a universal statement
we may use the following direct approach: Pick any element in the domain of such that
is true. Then demonstrate that
is true. Once we have shown that
is true for any element, then we can conclude using Universal Generalisation that
must be true.
Example 1
Prove that the sum of any two even integers is even.
Answer: Let
express
be any two even integers. By the definition of even integers, we can
, and
also an integer,
for some integers
. Then
. Since
is
is an even integer. This completes the proof.
In the above proof, we also used the rule of inference Existential Instantiation (EI) when
we wrote
if (and only if)
,
. Recall the definition of an even integer: An integer is called even
for some integer . In other words,
If we start off with the assumption that
is even, then the definition asserts that
. We can then specifically write
, where
takes the place of . Our
ability to discard the existential quantifier is due to the usage of Existential Instantiation.
Our proof could not proceed if we did not discard the existential quantifier.
Here, we in fact start off with two even integers,
Instantiation on both of them to get
the mistake of writing
,
,
. We can use Existential
. We must be very careful not to make
, where the symbol
and . It is wrong to use the same for both
and
is (wrongfully) used for both
because we have not assumed that
Our proof must allow for the possibility that
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.
are different integers. Such a fallacious
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argument (using the same for both
) is called mislabelling of elements, as discussed
in Section 3.2 of Study Unit 2.
Definition
Let be an integer. We say that
is a perfect square if
for some integer . Similarly,
we say that is a perfect cube, or a perfect 4th power, or in general, a perfect
, or
, or
power, if
respectively for some integer .
Example 2
Prove that if and
are perfect squares, then the product
Answer: Suppose that and
integers
is also a perfect square.
are perfect squares. We can write
. It follows that
. Since
and
for some
is also an integer, we conclude that
is a perfect square. The proof is complete.
Activity 3.5
Give a direct proof of the following statements:
a.
If is an odd integer, then
is also odd.
b.
If is an odd integer, then is the difference of two perfect squares.
c.
The product of any four consecutive integers is one less than a perfect square.
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2.4 Disproving an Existential Statement
Suppose that we wish to prove that an existential statement
is false, that is, to disprove the above existential statement. A direct approach is to prove
the universal statement
statement
is the statement
. This approach works because the negation of the
.
Example 1
Disprove the following: There exists an integer such that
Answer: Let
or
be any integer. Then
is a prime number.
. Exactly one of the factors
is an even number. If either factor
or
is equal to 2, then
certainly the other factor cannot be equal to 1. Similarly, if either factor
is equal to
, then certainly the other factor cannot be equal to
. Hence,
or
is
an even integer (because it can be factorised into a product of integers, one of which is
even) that is not equal to 2. We therefore deduce that
cannot be a prime number.
Activity 3.6
Disprove the following existential statements using a direct approach:
a.
There exists an integer
such that
b.
There exists an integer
such that
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is prime.
is prime.
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2.5 Common Mistakes in Mathematical Proofs
Mistake 1: Arguing from examples.
Although it is very often helpful to look at examples to assess the validity of a theorem or to
assist in the formulation of a new theorem, it is a mistake to think that a general universal
statement can be proven by demonstrating it for one or a few particular examples.
Here is an example of this kind of mistake:
Result: The sum of two even integers is even.
Erroneous argument:
is even.
is also even. Hence, the sum of two even
integers is even.
In this example, we are asked to prove that for all even integers and ,
The domain of
is also even.
and is the set of all integers. Hence, our proof "needs to work" for any
in the set of integers, not just a few specific values.
In some mathematical results where the domain of the variables is restricted to a small
, it might be possible to construct a proof by exhaustion by
finite set, say,
simply considering each element one by one until we have covered all elements within
the domain of consideration. We will examine this more closely later in this Study Unit.
Mistake 2: Using the same symbol to mean two different things, or to refer to two
different objects.
We have already covered this under the fallacy "mislabelling of elements". For example,
if
are two odd integers which may not be the same, it is a mistake to write
and
using the same symbol . It is also wrong to write, for instance,
and
, because even though we are not equating
dependency
of on each other that may not be valid.
Mistake 3: Jumping to a conclusion.
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to
, we are introducing a
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To jump to a conclusion means to make a particular assertion or to conclude that a result
has been established without adequate justification.
For example: Suppose we are asked to prove the theorem that there are infinite number
of prime numbers that can be expressed in the form
for some integer .
Erroneous argument: Since is allowed to be any integer, there are an infinite number of
integers of the form
the form
. Hence, there must be an infinite number of prime numbers of
.
While the above mathematical result is in fact true, there is no way that the above proof
can be treated as valid. It is impossible to justify leaping from the recognition that "there
are an infinite number of integers of the form
" to the conclusion that "there must be
an infinite number of prime numbers of the form
". Far too many details and steps
have been skipped.
Mistake 4: Circular reasoning.
A prime example of circular reasoning is to assume what is to be proved even before it
has been successfully demonstrated.
For instance: Show that the product of two odd integers is odd.
Erroneous proof: Let
Also, since
and
be odd integers. Since
is odd,
are odd, we can write
and
Thus
for some integer .
for some integers
.
.
Note how our alleged proof is actually very nearly correct, and in fact can be easily
"repaired" by removing the offending assumption right at the beginning (that the
conclusion already holds true) and perhaps tidying up the last equation just a little bit.
Another example: Show that the square of any real number can never be negative.
Erroneous proof: Suppose
is a real number. Then
. Since we have observed that
, we therefore conclude that cannot be negative.
In the above example, we did not even attempt to make any mathematical argument. We
simply leapt a particular conclusion right away and then paraphrased it to make it read
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exactly like the original statement to be proven. This example can also be said to be an
example of jumping to a conclusion.
Mistake 5: Misuse of the word "any" when the word "some" should be used.
There are a few situations in which the words "any" and "some" can be used
interchangeably. For instance, in starting a proof that the square of any odd integer is odd,
we could correctly write either "Suppose
is any odd integer" or "Suppose
is some odd
integer." But this kind of interchangeability is not valid all the time.
For example, it is wrong to write the following: "Suppose
integer. Then
is an arbitrarily chosen odd
for any integer ."
In the second sentence it is incorrect to say that "
for any integer " because the
integer , once chosen at the beginning, is treated as fixed. In fact, solving
shows that
, so
specific integer the moment
To claim that "
is uniquely determined by . Therefore
for
has to be some
is chosen, and cannot be "any" integer.
for any integer " would also quite ridiculously imply that
can
take on an infinite number of different values (one of each value of ).
Mistake 6: Misuse of the word "if" when "because" is actually meant.
Consider the following proof fragment: "Suppose
is a prime number. If is prime, then
cannot be written as a product of two smaller positive integers each of which is greater
than 1."
The use of the word "if" in the second sentence is inappropriate (though it is a highly
common mistake). It suggests that the status of
as a prime number is in doubt. But
is
known to be prime by the first sentence. Its status as a prime number cannot be in doubt
because the first sentence has put forth that assumption about .
Here is a correct version of the proof fragment: "Suppose
is prime,
is a prime number. Because
cannot be written as a product of two smaller positive integers each of which
is greater than 1."
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Chapter 3: More Proof Techniques, Indirect Proofs
3.1 Rational Numbers
We have used the terms "rational number" and "irrational number" without qualification
as we assumed the reader understands what the terms mean. Here, we give a proper
definition.
Definition
Let be a real number. We call
a rational number if there exist integers
such that
. Note that necessarily
. If there does not exist any integers
such that
, then we call an irrational number.
Thus for instance,
are all rational numbers. The set
numbers is a subset of the set
of
of real numbers. Likewise, the set
of rational
of integers is a subset
. Every integer is a rational number, because every integer can easily be expressed as
. But of course, not every rational number is an integer, as can be seen in the examples
we have provided.
Numbers like
, are irrational numbers. It can be shown that they cannot
be expressed in the form
approximate
as 3.14 or as
, where
are integers. It is common, for example, to
. The reader should understand that these are merely
rational approximations to an irrational number, and these approximations are not exactly
equal to
itself.
Theorem: The sum of two rational numbers is rational.
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Proof: Suppose that
and
are rational numbers, where
are necessarily non-zero. Then the sum is
an integer and
are integers. Note that
is rational because
is
is a non-zero integer.
In the above proof, we were able to assert that
is an integer by appealing to the
fact that the sum or the product of integers must again be an integer. This particular
fact, among other basic properties of integers such as
, can be rigorously
deduced from a fundamental set of axioms that apply to all of mathematics. We shall not
prove such results in this course.
The following activity should be attempted now.
Activity 3.7
Prove the following statements about rational numbers:
a.
The product of two rational numbers is rational.
b.
If is a non-zero rational number, then
is also a non-zero rational number.
3.2 Divisibility, Divisors, and Multiples
Definition
Let
be integers with
, if
. We say that divides , or that
is a multiple of , written
for some integer .
We can also express the relation
by saying that is a divisor of , or is a factor of , or
is divisible by . Sometimes when there is a need for emphasis, we may even say that b
exactly divides , or
is exactly divisible by . All these various expressions are widely
used, so the reader should be familiar with them.
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For example, 15 is divisible by 5, because
. Similarly, 15 is also divisible by 3.
Negative numbers should not be neglected as well. 15 is likewise divisible by
as by
. We write
,
,
, and
as well
.
As another example, 10 is a divisor of 40, or 10 divides 40. Similarly,
are also all divisors of 40. We can equivalently say that 40
is a multiple of 10, or that 10 is a factor of 40.
Every integer is divisible by
only two divisors: and
. In particular, the integers 1 and
each have
. The number 0 is of special consideration. 0 is never a divisor
of any integer by the definition above. However, any integer
. As strange as it may appear, we are allowed to write
is a divisor of 0, because
.
Let us revisit even, odd integers and prime numbers. An integer is even if and only if
is a multiple of 2, or equivalently, 2 divides . Using the same kind of terminology, we can
say that an integer is prime if and only if
itself. An integer
and has only two positive divisors, 1 and
is composite if and only if
and
has more than two different
positive divisors.
A word of caution must be made.
is used to denote the sentence " divides ", whereas
is not a sentence at all but instead represents the number obtained when is divided by
. In the latter case,
need not even be integers, even though for the purpose of division,
we have to insist that
integers with
. In the sentence
or " divides ", we insist that both
.
An important fact to take note of is: is divisible by if and only if
is an integer.
Activity 3.8
a.
Is
a multiple of ? Is
a multiple of
b.
Write down all the divisors (both positive and negative) of
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? Is
a multiple of ?
.
are
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Theorem: Transitivity of divisibility. If
are non-zero integers such that divides ,
and divides , then divides .
Proof: Suppose that
are non-zero integers such that divides , and divides . Then
by the definition of divisibility, there exist integers
follows that
. Since
such that
, then we have
Proof: Suppose that
are positive integers such that
.
are positive integers such that
of divisibility, there exist integers
and thus,
. It then
is an integer, we conclude that divides .
Theorem: Anti-commutativity of divisibility. If
and
and
such that
and
and
. Then by the definition
. It follows that
. From this equation, we deduce that either
hypothesis is that is a positive integer. Hence
possibilities available for
are that
assumed to be positive, and
and
, and so
. The
. Therefore, the only
are each either equal to 1 or
, it must be that
or
,
. But since
. Thus,
are
.
More results concerning divisibility of the integers can be found in the Formative
Exercises.
We will now return to our discussion on proof techniques in mathematics.
3.3 Proof by Contraposition
A proof by contraposition is based on the logical equivalence of a conditional statement
and its contrapositive
. In order to demonstrate that
is true, we can choose to show that
is true. It is sometimes
easier or more convenient to prove the contrapositive form of a conditional statement
rather than giving a direct proof of the original conditional statement. By the logical
equivalence of
and its contrapositive
if and only if the second version is true.
Example 1
Prove that for all integers , if
is odd, then is odd.
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, one version is true
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Answer: We prove the contrapositive of the statement: If
Suppose that
is an even integer. Then
. Since
is even, then
for some integer
is an integer, we can conclude that
is even.
. It follows that
is even. Hence, by proving
the contrapositive form, we have successfully demonstrated that if
is odd, then is odd.
Example 2
Prove that if
are positive real numbers such that
, then either
Answer: We again use the method of contraposition. Suppose that
numbers such that
to conclude that
and
. Then
or
.
are positive real
, so in particular, we are forced
.
Example 2 above effectively tells us that if we want to test whether a given integer
is prime by checking if
divisible by integers
such that
is divisible by 2,3,4,5,etc., we only need to check whether
satisfying
. If we can find some integer
, then we can conclude that
integer with
such that
is
with
is not prime. Similarly, if we cannot find any
, then we can conclude that has to be prime. Once
we have explored all integers satisfying
, there is no need to further test integers
.
This is due to the result just established, that should we be able to write
integers
, then one of or must be at most
. (Our result was established for positive
real numbers, so it applies equally to the case where
are restricted to the domain of
positive integers.) As a consequence, we can observe that if
then necessarily
of ,
has some divisor
for positive
satisfying
is also a divisor of , and either one of or
Activity 3.9
Prove the following statements by contraposition:
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has a divisor
satisfying,
. To be precise, for any divisor
must be
.
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a.
Let be a real number. If
b.
Let
c.
For all integers
d.
For all integers , if
is irrational, then is irrational.
be non-zero integers. If does not divide
, if
is even, then
, then does not divide .
are either both even or both odd.
is even, then is even.
A proof by contraposition can be said to be an "indirect proof" because we are not
directly working on the original statement to be demonstrated, but instead proving the
contrapositive of the conditional statement, and then appealing to the logical equivalence
of a conditional statement and its contrapositive.
3.4 Proof by Contradiction
Proof by contradiction is based on the rule of inference known as the Contradiction Rule:
Reminder: The letter c above denotes a contradiction in the sense of Propositional Logic
(see Study Unit 1).
In order to show that a statement is true, we begin by assuming that it is false, and
subsequently derive a contradiction. Having obtained a contradiction, we then conclude
that the original statement must be true. A proof by contradiction can also be said to be
an "indirect proof".
Example 1
Prove that there is no greatest integer.
Answer: We argue by contradiction. Suppose instead that there is a greatest integer. Let
this integer be
. But then
produced an integer larger than
would also be an integer, and
. Thus we have
. This contradicts our initial assumption that
greatest integer. Hence, we conclude that there is no greatest integer.
SU3-27
is the
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Methods of Proof
Example 2
Prove that an integer cannot be simultaneously odd and even.
Answer: Suppose
write
is an integer that is both odd as well as even. Since
for some integer . Since is odd, we can also write
This gives us
, or
is an integer and yet
is even, we can
for some integer .
. This is a contradiction because
, but
is not an integer. Thus, no integer can be simultaneously even as
well as odd.
Example 3
Prove that if is a rational number and is an irrational number, then
is irrational.
Answer: Suppose that is a rational number and is an irrational number, and yet
rational. Since and
are both rational,
is
is also rational. Hence we have
obtained a contradiction because we assumed is irrational.
Activity 3.10
Prove the following statements by contradiction:
a.
There is no such thing as a smallest positive rational number.
b.
If
is a non-zero rational number and
is an irrational number, then
is
an irrational number.
c.
For any integer ,
is not divisible by 4.
3.5 Proof by Division into Cases, the Absolute Value of a Real
Number
Sometimes, it is cumbersome or not feasible to produce a single argument that will "work"
for all cases under consideration. Proof by division into cases is a proof strategy that allows
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Methods of Proof
us to work on each case separately, and then make the desired conclusion once all cases
have been covered.
In a proof by division into cases, also called a proof by exhaustion, we appeal to the
following rule of inference which in Study Unit 1 we also called "proof by division into
cases":
Suppose we know that at least one of the statements
that all the conditional statements
,
,...,
is true. If we can show
are true, then we can conclude
that has to be true.
Another way of understanding proof by exhaustion is to recognise that the statement
is logically equivalent to
Hence, to show that is true whenever at least one of the statements is true, it is necessary
and sufficient to show that
is true.
Example 1
Prove that if is an integer, then
.
Answer: We prove this by dividing into cases. First, note that
that
, or that
.
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is equivalent to stating
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If
Methods of Proof
, then certainly the inequality is true, because
Similarly, if
Suppose that
.
, then the inequality is again true, because
is an integer other than
terms and
are negative.
Likewise, if
, then
. If
.
, then
because both the
because both the terms and
are positive.
In all of the cases examined above, we reach the conclusion
. Therefore, we have
successfully employed a proof by exhaustion to demonstrate
that for all integers .
Remark: The factorisation
cases
and
gave us the hint that we should consider the
separately.
Definition
Let be a real number. The modulus or absolute value of is defined to be the positive part
of , and is denoted by
. The absolute value of is obtained simply by discarding any
negative sign that exists. For instance,
We can formally define
to be equal to
,
whenever
,
,
, and equal to
.
whenever
.
If is negative, then
require that
If
would be positive, and so to discard the negative sign, we would
for negative.
is nonnegative, then there is no negative sign to be discarded, so
nonnegative.
We can express the formal definition of
as:
Some important basic properties of the absolute value symbol include:
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for
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Methods of Proof
for all real numbers , equality holding if and only if
. (Positive definite
property)
for all real numbers .
for all real numbers
. (Property of homogeneity)
for any real numbers
. This is known as the triangle inequality,
which we will demonstrate below.
Example 2: The triangle inequality
Prove that
for any real numbers
.
Answer: Again, we prove this statement by division into cases.
Suppose that
. Then by definition,
have
. Since
and
, we then
.
Suppose that
. Then by definition,
we then have
. Since
and
,
.
Remark: The positive definite property, the property of homogeneity, and the triangle
inequality of the absolute value symbol collectively imply that the absolute value operator
is what mathematicians term a norm on the set of real numbers
.
Other important properties of the absolute value symbol include:
If
, where is a positive real number, then
If
, where is a positive real number, then either
For any real numbers
,
if and only if
.
or
.
.
Example 3
Show that there are no integers and satisfying the equation
Answer: If
, then
, and so
equation can never be satisfied if
.
for any value of . Hence, the given
.
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Methods of Proof
We therefore restrict our attention to
be satisfied for some integers
Consider the case
. Suppose the equation
.
Then
and hence
. This is a contradiction because
would be an improper fraction when in fact
Consider the case
because
Then
has to be an integer.
, and so
. This is again a contradiction
would be an improper fraction when in fact
Similarly, consider the case
because
can
. Then
has to be an integer.
and hence
. This is a contradiction
would be an improper fraction when in fact
has to be an integer.
Thus, in all the cases considered, we obtain a contradiction. This shows that it is impossible
for the equation
to be satisfied by integers
Notation: For real numbers
and let
, let
.
denote the larger of the two values,
denote the smaller of the two values
simply denote the common value of
or . If
, then
or ,
and
.
Example 4
For all real numbers
Answer:
If
,
, prove that
.
then
.
Certainly
in
this
case,
.
Suppose without loss of generality that
In this case,
. Then
and
.
as well.
Having considered all cases, our proof is complete.
Without Loss Of Generality (WLOG)
In the proof in Example 4, we proved the required result for the case
, and then
announced the completion of the proof, because by symmetry, the proof for this case
would also effectively cover the case
. If
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, then
and
.
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Methods of Proof
The roles of and are simply reversed, and it is common to say that we adapt our proof
for the case
mutatis mutandi to the case
[ Mutatis mutandis is a Medieval Latin phrase meaning "the necessary changes having
been made" or "once the necessary changes have been made". ]
In general, when the phrase “without loss of generality” is used in a proof (often
abbreviated as WLOG), we assert that by proving one case of a theorem, no additional
argument is required to prove other specified cases.
Activity 3.11
Prove the following by division into cases:
Let
be real numbers. Then
.
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Summary
In Study Unit 1, we learnt about propositional logic, and in Study Unit 2, we learnt
about predicate logic and quantified statements. In this Study Unit, we put these concepts
together to learn how to formulate definitions and theorems precisely.
We also looked at various commonly used techniques to prove statements or to prove
theorems in mathematics.
We considered the direct approach of proving an existential statement, proving a universal
statement, disproving a false existential statement, and using counter-example to disprove
a false universal statement. Another method of proof examined was the proof by
exhaustion or by division into cases. Common mistakes in proof were also discussed.
Indirect methods of proof that we examined include the method of proving by
contraposition, and proving by contradiction.
The "without loss of generality" concept was also discussed. Basic topics in mathematics
such as the absolute value of a real number, and the concept of divisibility, were also
explored.
In the next Study Unit, we develop these techniques further when we explore other
fundamental topics in mathematics such as sequences, induction, and the Euclidean
Algorithm.
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Methods of Proof
Formative Assessment
1.
Prove the following existential statements:
a.
There is a real number such that
b.
There is a positive integer
and
such that
.
can be written as the sum of two
perfect cubes in two different ways.
2.
Disprove the following (untrue) statements by giving a counter-example:
a.
The average of any two odd integers is odd.
b.
For all integers ,
c.
For all positive integers ,
d.
Every positive integer can be expressed as a sum of three or fewer perfect
is not prime.
is a prime number.
squares (which are not necessarily distinct).
(Incidentally, it is actually true that every positive integer can be expressed
as a sum of four or fewer perfect squares.)
3.
Prove the following statements concerning divisibility:
a.
For all non-zero integers
, if
and
, then
b.
For all non-zero integers
, if
and
, then
integers
4.
.
for any
.
Prove that for all real numbers
we have
. You may divide into cases
and use the "without loss of generality" concept in your argument.
5.
Prove the following using the method of contradiction. That is, assume the statement
is false and derive a contradiction, thereby allowing one to conclude that the
statement must in fact be true.
a.
The square root of an irrational number is also an irrational number.
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Methods of Proof
b.
If and are rational numbers with
, and is an irrational number, then
is irrational.
c.
If
are integers with
, then at least one of the integers
must
be even.
6.
Prove the following using the method of contraposition. That is, prove the
contrapositive of the conditional statement instead of directly proving the original
statement. Since the contrapositive of a conditional statement is logically equivalent
to the original statement, such a proof would suffice to demonstrate that the original
statement is in fact true.
a.
Let be an integer. If
b.
Let
is even, then has to be even.
be positive integers. If
is a prime number, then either or
must
be equal to .
c.
Let
be real numbers. If
is less than
, then either
or
must be
less than .
d.
Let
be integers. If
and
does not divide , then
does not divide
.
7.
8.
Disprove the following untrue existential statements.
a.
There exists an integer
such that
is a prime number.
b.
There exists an integer
such that
is a prime number.
There are 100 rooms labelled 1 to 100 and arranged in a row in ascending order:
1,2,3,...,100. One day, a man and a woman are each randomly placed into different
rooms. Each day henceforth, the woman has to choose either to walk into the room
on her left, or into the room on her right. The man can choose either to stay put in the
room he is in, or walk into the room on his left, or into the room on his right. Each
person must make exactly one move on each day. If a person reaches the end of the
row (that is, room 1 or room 100), that he or she can only choose to walk to room 2
or to room 99 respectively (the man has the option to stay put if he desires, but the
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Methods of Proof
woman must walk). Prove that it is possible for the man to walk in such a way that
he will eventually meet the woman in a year.
9.
Let
be a real number. Show using a rigorous mathematical argument that if
irrational for some positive integer , then is irrational.
10. Show that if is any positive integer, then
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is not a prime number.
is
MTH105
Methods of Proof
Solutions or Suggested Answers
Activity 3.1
a.
is always even because the expression is of the form
integer. Note that if
b.
are integers, then the sum
c.
are integers, then the product
, then the expression is odd. On the other hand, if
the expression is
whether
about
where
is an
must also be an integer.
may be even or odd depending on the values of
if
is an
must also be an integer.
is always odd because the expression is of the form
integer. Note that if
where
. For example,
, then
, which is even. Hence, we are unable to determine
is even or odd if we are not provided with further information
.
Activity 3.2
7 and 13 are prime.
, so 9 is composite.
, so 45 is composite.
, so 15 is composite.
are all integers that are less than or equal to 1, so by definition, they are
neither prime nor composite.
2 is prime.
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Methods of Proof
Activity 3.3
a.
Let
b.
Let
c.
Let
d.
Let
.
.
. OR we can let
. OR we can let
.
.
Remark: It is an interesting problem in number theory to classify all the integers
such that
is a prime number.
Activity 3.4
a.
1 is odd, but
b.
2 is even, but
c.
Let
d.
Let
is even.
is odd.
. Then
is even.
. Then even though
is prime,
is not prime, because
.
Activity 3.5
a.
Suppose that
is an odd integer. Write
Then
conclude that
b.
for some integer
. Since
is an integer, we
is odd.
Suppose that is an odd integer. Write
for some integer . Then we have
, and so is the difference of two perfect squares.
c.
.
Let the four consecutive integers be denoted by
Then their product is equal to
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.
MTH105
Methods of Proof
The above expression is indeed one less than a perfect square.
Activity 3.6
a.
Suppose
is an integer such that
the factors
and
.
. Since
are each at least 2. Hence,
, both
cannot be a prime
number.
b.
Suppose is an integer such that
Since
.
, both the factors
and
are each at least 2. Hence,
cannot be a prime number.
Activity 3.7
a.
Suppose that
and
are rational numbers, where
are necessarily non-zero. Then the product
is a non-zero integer, and
b.
are integers. Note that
is also rational because
is an integer.
Suppose that
is a non-zero rational number, where
integers. Then
, and so
are non-zero
is also a non-zero rational number.
Activity 3.8
a.
43 is not a multiple of 8 because
because
such that
is not an integer. 43 is a multiple of
. 43 is not a multiple of 0 because there is no integer
. By definition, we are never allowed to claim that any integer
is a multiple of 0.
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Methods of Proof
b.
Activity 3.9
a.
Suppose that
is rational. Then
for some integers
is also rational. Thus, rational implies
b.
Suppose that
. Then
. Hence
c.
Suppose that
rational.
for some integer
. We conclude that
. It follows that
. It follows that
implies
.
are integers with
of opposite parity, that is, one of them is
even and one of them is odd. Then
must be odd since the sum of an even
integer and an odd integer is odd.
d.
Suppose that is an odd integer. Then
for some integer . It follows that
, and so
is an integer. Hence, odd implies
is also odd because
odd.
Activity 3.10
a.
Suppose that is the smallest positive rational number. But since is rational,
is rational. Furthermore,
Hence,
is a positive number because is a positive number.
is a positive rational number that is less than . We have obtained a
contradiction.
b.
Suppose that is a non-zero rational number, is an irrational number, and
rational number. Since
are rational with
,
is a
is a rational number.
This contradictions our assumption that is irrational.
c.
Suppose is an integer such that
integer . This gives
therefore write
is divisible by 4. Then
. Therefore,
is even, and so
for some integer . It follows that
SU3-41
for some
is even. We can
.
MTH105
Methods of Proof
This leads to
, or
is an integer and yet
. This is a contradiction because
is not an integer.
Activity 3.11
Let
be real numbers. We proceed by division into cases.
Suppose
that
.
Then
and
and
Suppose
that
.
that
.
Thus,
.
Thus,
.
Thus,
as well.
Then
and
and
Suppose
.
as well.
Then
and
and
as well.
We leave it to the reader to check the remaining cases:
,
, and
.
Formative Assessment
1.
Prove the following existential statements:
a.
There is a real number such that
b.
There is a positive integer
such that
and
.
can be written as the sum of two
perfect cubes in two different ways.
Answer:
2.
a.
Let
b.
Let
.
will also work (answer is by no means unique).
. Observe that
.
Disprove the following (untrue) statements by giving a counter-example:
a.
The average of any two odd integers is odd.
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Methods of Proof
b.
For all integers ,
c.
For all positive integers ,
d.
Every positive integer can be expressed as a sum of three or fewer perfect
is not prime.
is a prime number.
squares (which are not necessarily distinct).
(Incidentally, it is actually true that every positive integer can be expressed
as a sum of four or fewer perfect squares.)
Answer:
a.
The average of 3 and 9 is 6, but 6 is not odd.
b.
A counter-example is
c.
A counter-example is
d.
7 cannot be expressed as three or fewer perfect squares. (As a side note, 7
.
is prime.
.
is not prime.
can be expressed as four perfect squares:
3.
.
Prove the following statements concerning divisibility:
a.
For all non-zero integers
, if
and
, then
b.
For all non-zero integers
, if
and
, then
integers
.
for any
.
Answer:
a.
Let
be non-zero integers with
such that
b.
Let
integers
,
be non-zero integers such that
such that
4.
. Then there are integers
. It follows that
,
have
integers
and
,
and
and so
.
. Then there are
. It follows that for any integers
. Hence
, we
for any
.
Prove that for all real numbers
we have
. You may divide into cases
and use the "without loss of generality" concept in your argument.
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Methods of Proof
Answer:
Let
be real numbers. We divide into cases.
Suppose
and
have
. Thus,
Suppose
and
have
. Then
and
. Since
.
. Then
and
. Since
. Thus,
Suppose
and
have
. Thus,
is nonnegative, we also
is nonpositive, we also
.
. Then
and
. Since
is positive, we also
.
Since we have checked the case where
and
, without loss of generality, we
also conclude that the statement holds true for the case
and
.
Thus, all possible cases have been covered, and we conclude that the statement is
true.
5.
Prove the following using the method of contradiction. That is, assume the statement
is false and derive a contradiction, thereby allowing one to conclude that the
statement must in fact be true.
a.
The square root of an irrational number is also an irrational number.
b.
If and are rational numbers with
, and is an irrational number, then
is irrational.
c.
If
are integers with
, then at least one of the integers
must
be even.
Answer:
a.
Suppose that
express
is an irrational number but
, where
are integers with
is rational. Then we can
. We then have
.
But this would imply that is rational. We therefore obtain a contradiction
to our assumption that is irrational.
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Methods of Proof
b.
Suppose that and are rational numbers with
number such that
is rational. We write
for suitable integers
Since
, and is an irrational
. Then
is an integer and
,
,
, which implies that
is necessarily a non-zero integer, this
means is rational. Hence we obtain a contradiction.
(For the purposes of assessment and examination, you are allowed to
use the fact that the sum, difference, and product of rational numbers is
rational without proof. In the above working, we proved the statement
from scratch.)
c.
Suppose that
Since
are integers with
are odd,
is even. But then
, and that
and
are all odd.
is even. Thus we
have derived a contradiction.
6.
Prove the following using the method of contraposition. That is, prove the
contrapositive of the conditional statement instead of directly proving the original
statement. Since the contrapositive of a conditional statement is logically equivalent
to the original statement, such a proof would suffice to demonstrate that the original
statement is in fact true.
a.
Let be an integer. If
b.
Let
is even, then has to be even.
be positive integers. If
is a prime number, then either or
must
be equal to .
c.
Let
be real numbers. If
is less than
, then either
or
must be
less than .
d.
Let
be integers. If
and
.
Answer:
SU3-45
does not divide , then
does not divide
MTH105
Methods of Proof
a.
Suppose that is an odd integer. Then
Hence, odd implies
b.
Suppose
Let
is not prime.
are real numbers such that
, so
d.
and
. Suppose that
. Then
. Hence, we have demonstrated that if
too. We have thus successfully proven that if
statements "
. Then
.
be integers such that
divide
are greater than .
is a product of two integers each greater than . By the
definition of prime numbers,
Suppose that
is odd.
odd.
are positive integers such that both
Then the integer
c.
is also odd, and so
must
, then
, then the
" and " does not divide " cannot simultaneously be true
as well.
7.
Disprove the following untrue existential statements.
a.
There exists an integer
such that
is a prime number.
b.
There exists an integer
such that
is a prime number.
Answer:
a.
Suppose that
is an integer. We have the following factorisation:
. Since
are each at least . Hence,
b.
Suppose that
, both the factors
and
cannot be a prime number.
is an integer. Observe that
.
Since
, both the factors
and
.
Hence
cannot be a prime number.
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are each at least
MTH105
8.
Methods of Proof
There are 100 rooms labelled 1 to 100 and arranged in a row in ascending order:
1,2,3,...,100. One day, a man and a woman are each randomly placed into different
rooms. Each day henceforth, the woman has to choose either to walk into the room
on her left, or into the room on her right. The man can choose either to stay put in the
room he is in, or walk into the room on his left, or into the room on his right. Each
person must make exactly one move on each day. If a person reaches the end of the
row (that is, room 1 or room 100), that he or she can only choose to walk to room 2
or to room 99 respectively (the man has the option to stay put if he desires, but the
woman must walk). Prove that it is possible for the man to walk in such a way that
he will eventually meet the woman in a year.
Answer:
Label the man's room as
and the woman's room as . Let the man walk to the
leftmost end (that is, room 1), and then walk to his right on each day. If
is even, then it will always remain even on his journey to the right, and he will
eventually meet the woman. Otherwise,
is odd, and will always remain odd
on his journey to the right, and the man will reach room 100 without meeting the
woman. Let him stay put in room 100 for exactly one day, and then resume his
walk, walking one room to the left on each day. Because the man stayed put exactly
one day,
is now changed to even. On his journey to the left, the man will
eventually meet the woman. The total steps taken is at most 300. That is definitely
within a year.
9.
Let
be a real number. Show using a rigorous mathematical argument that if
irrational for some positive integer , then is irrational.
Answer:
We show the contrapositive.
SU3-47
is
MTH105
Suppose that
Methods of Proof
is a rational number, where
are integers with non-zero.
Then for any positive integer , we have
is rational as well.
Hence we prove the statement by demonstrating the contrapositive.
10. Show that if is any positive integer, then
is not a prime number.
Answer:
Since is a positive integer, both the terms
Hence,
is composite.
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and
are greater than 1.
MTH105
Methods of Proof
References
Epp, S. S. (2019). Discrete mathematics with applications (5th ed.). Boston, MA: Brooks/
Cole Cengage Learning.
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SU3-50
Study
Unit
Sequences, Induction, the
Euclidean Algorithm
4
MTH105
Sequences, Induction, the Euclidean Algorithm
Learning Outcomes
By the end of this unit, you should be able to:
1.
Understand the definition of a sequence in terms of either a closed-form formula,
or via a recurrence relation.
2.
Understand and use sigma and product notation.
3.
State and explain the property of an arithmetic and a geometric sequence.
4.
Find the nth term and the sum of the first n terms of an arithmetic sequence and
a geometric sequence.
5.
State the Well-Ordering Property.
6.
Demonstrate proofs to prove certain mathematical statements by mathematical
induction or the well ordering principle.
7.
State and prove the Division Property of integers.
8.
Use the Division Property to prove results concerning divisibility of integers and
other useful properties.
9.
Define the greatest common divisor of two integers.
10.
Prove properties concerning the greatest common divisor and find the gcd of
two integers.
11.
Implement the Euclidean Algorithm to express the gcd of two integers as an
integral linear combination of the two integers.
12.
State and explain the concept that every integer n≥2 has a unique factorisation
into a product of primes.
13.
Prove statements involving prime numbers and divisibility using the proof
techniques and theorems learnt.
SU4-2
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Sequences, Induction, the Euclidean Algorithm
Overview
In this Study Unit, we explore another important technique of proof known as induction,
or mathematical induction. Induction is used to prove that a certain mathematical
statement holds for all natural numbers, or holds for all integers at least as great as a given
certain integer.
To provide many examples of the use of induction, we first discuss sequences, as well as
the usage of sigma and product notation. We then discuss the Well-Ordering Property and
derive the principle of mathematical induction from the Well-Ordering Property. We then
discuss many examples of induction, particularly in proving statements about sequences,
and statements concerning divisibility of integers.
In this Study Unit, we will also learn about the Division Property of the integers, the
notions of greatest common divisor and lowest common multiple, and learn how to
implement the Euclidean Algorithm.
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Sequences, Induction, the Euclidean Algorithm
Chapter 1: Sequences
1.1 Introduction to Sequences
A sequence of real numbers is an ordered collection of real numbers
In the above, we have denoted the
term of the sequence,
member of the sequence by
is the second term,
. Thus,
is the first
is the third term, and so on.
A sequence of numbers can either be a finite or infinite sequence. A finite sequence
terminates after a finite number of terms, whereas an infinite sequence keeps on going
forever. The sequence
has precisely
sequence. We denote this sequence as
number of terms and
terms in total, and is a finite
or more precisely as
is a dummy variable that runs from
to
, where
is the
inclusive. An infinite
sequence
has an infinite number of terms and it is impossible to list down every member of the
sequence. In this case, we usually list down only the first several terms. We can denote
this sequence as
or as
. The symbol
is the infinity symbol.
A general formula (or a closed-form formula) of a sequence
such that for every , the
term of the sequence,
, is given by
applies to both finite and infinite sequences. Here are some examples:
a.
b.
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is a function
. This definition
MTH105
Sequences, Induction, the Euclidean Algorithm
c.
(note that starts from and
)
d.
(note that starts from and
Here, the function
)
is used to produce terms that alternate in sign.
e.
We can define a sequence in two ways. The first way is to provide a closed-form formula
for the
term of the sequence. It is understood that where sequences are concerned, is
only allowed to be a whole number. For example:
a.
Let
for all
.
Then the terms of the sequence
b.
Let
for all
.
Then the terms of the sequence
c.
For
, let
are:
are:
be defined in a piece-wise manner as:
Then the terms of the sequence
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are
.
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Sequences, Induction, the Euclidean Algorithm
Activity 4.1
Write down the first five terms of each of the sequences below which are defined via
a closed-form formula:
a.
for all
b.
for all
c.
for all
d.
e.
.
.
.
for all
.
For
The second way to define a sequence is via a recurrence relation. A recurrence relation
is an equation that expresses
as a function as some of the preceding terms like
Let us examine a famous example known as the Fibonacci sequence. This sequence is
defined by the recurrence relation
The first two terms of the sequence are explicitly given. All the subsequent terms, that
is all the terms
for
, are generated by the recurrence relation
. Thus,
,
If the recurrence relation relates
to its preceding term
, and so on.
, then in order to fully define
the entire sequence, we need to provide an explicit value of the initial term
recurrence relation relates
to its immediate two preceding terms
need to provide explicit values of the first two terms
and
. If the
, then we
. This of course generalises to
more complicated recurrence relations involving more terms. The explicit values of the
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Sequences, Induction, the Euclidean Algorithm
initial term or initial terms are needed to "start" the sequence, after which the recurrence
relation or the recurrence formula is responsible for generating the rest of the terms.
Activity 4.2
Write down the first five terms of each of the sequences below which are defined via
a recurrence relation:
a.
for
,
b.
c.
for
for
d.
,
,
,
for
,
,
1.2 Sigma and Product Notation
The purpose of this section is to introduce the Sigma or the Summation notation
as well as the Product notation
,
, two very commonly used mathematical notations
to denote a sum and to denote a product respectively.
Let be a positive integer and let
In other words,
be a sequence of real numbers. Then
denotes the sum of the terms
to
inclusive. The symbol
is
a letter of the Greek alphabet called sigma (here, sigma is used in the capitalised form).
We call the above notation of expressing a sum the Sigma or Summation notation. It is
possible to choose a starting point other than 1. For instance,
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The number at the bottom of the sigma symbol is the starting value of the dummy variable
(the running variable) and is called the lower limit of the summation. The number at the
top of the sigma symbol is called the upper limit of the summation. The following notation
expresses summing the terms of the sequence all the way to infinity:
For example,
, while
. An example of
an infinite summation is
For any given sequence
of real numbers, let
the sequence, that is, let
sequence of partial sums of
. The
denote the sum of the first terms of
also form a sequence which we call the
.
It can be observed that for all integers
, we have
Now, let us introduce the product notation. Let
sequence of real numbers. Then
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, in other words,
be a positive integer and let
be a
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Sequences, Induction, the Euclidean Algorithm
In other words,
denotes the product of the terms
to
inclusive. Let
the product of the first terms of the sequence, that is, let
. The
form a sequence which we call the sequence of partial products of
It can be observed that for all integers
, we have
denote
also
.
, in other words,
Activity 4.3
Compute the following sums or products:
a.
b.
c.
d.
e.
1.3 Arithmetic and Geometric Sequences
In this section, we highlight two special kinds of sequences known as arithmetic and
geometric sequences.
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An arithmetic sequence is a sequence in which the difference between successive terms
is constant. To be precise, we call
an arithmetic sequence (also commonly known as
an arithmetic progression, or AP) if there is some real number
for all integers
. We call
such that
the common difference of the arithmetic sequence. For
example,
is an arithmetic sequence with common difference 2. The sequence
is an arithmetic sequence with common difference -4. If the common difference is positive,
the terms keep increasing. If the common difference is negative, the terms keep decreasing
as we go further along the sequence.
It is customary to denote the first term of an arithmetic sequence by the letter . The
term of an arithmetic sequence with first term and common difference
Thus, for instance, in the sequence,
,..., the
is given by
term is given by
.
The sum of the first terms, or the
partial sum
, is given by the formula
Activity 4.4
Find the sum of the first 30 terms of the arithmetic sequence
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An geometric sequence is a sequence in which the ratio between successive terms is
constant. To be precise, we call
a geometric sequence (also commonly known as an
geometric progression, or GP) if there is some real number such that
for all integers
. We call the common ratio of the geometric sequence. For example,
is a geometric sequence with common ratio . The sequence
is a geometric sequence with common ratio
.
It is also customary to denote the first term of a geometric sequence by the letter . The
term of a geometric sequence with first term and common ratio
Thus, for instance, in the sequence
, the
is given by
term is given by
.
The sum of the first terms, or the
for
partial sum
.
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is given by the formula
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Sequences, Induction, the Euclidean Algorithm
Reflect 4.1
Why do we insist that
in the above formula for the sum of the first
geometric sequence?
Activity 4.5
Find the sum of the first 8 terms of the geometric sequence
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terms of a
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Sequences, Induction, the Euclidean Algorithm
Chapter 2: The Well-Ordering Property and Induction
2.1 The Well-Ordering Property of the Natural Numbers
The set of natural numbers is the set of positive integers and is denoted by
by
or equivalent
. In this section, we introduce the Well-Ordering Property of the natural numbers.
Well-Ordering Property of the natural numbers
be any nonempty subset of the set of natural numbers
Let
. In other words, let
be a
nonempty collection of natural numbers. The set may be a finite set or an infinite set.
The Well-Ordering Property of the natural numbers asserts that
has a least element,
or a minimal element. That is, there is some unique number
such that
for all
.
There is a very useful generalisation of the above principle. First, we define what we mean
for a non-empty set of real numbers to be bounded above or bounded below.
Definition
Let
be a non-empty set of real numbers. We say that
some real number
such that
a member of . We call
for all
is bounded above if there is
. Note that the number
itself need not be
an upper bound of . Upper bounds of sets of real numbers, if
they exist, are never unique.
Similarly, we say that
for all
. We call
is bounded below if there is some real number such that
a lower bound of . Lower bounds of sets of real numbers, if they
exist, are also never unique.
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Theorem: Every non-empty set
of integers that is bounded below contains a minimal
element or least element, that is, there is some element
Similarly, every non-empty set
such that
.
of integers that is bounded above contains a maximal
element or greatest element, that is, there is some element
Proof: Let
for all
such that
be a lower bound of . Then by definition,
for all
.
for all . Without loss of
generality, we can assume that is an integer, else we just replace with any integer less
than . Define the set
That is, the set
contains all those numbers of the form
vary over all possible elements of . Here,
, where is allowed to
denotes the absolute value of the number
.
Then is a set of positive integers, because the condition
that
for all
Let
Let
, and
for all
for all
. This implies that
.
. Then is the least or minimal element of
that
implies
. By the Well-Ordering Principle of the natural numbers,
contains a least element, say . That is,
for all
for all elements
implies that
for all
because
and the result
.
be a set of integers that is bounded above. Then
is a set of integers that is bounded below, and which therefore contains a minimal element,
say . Let
. Then is the maximal or greatest element of .
For the sake of generality, we will also call the above theorem the well-ordering property.
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Well-Ordering Property
Every non-empty set of integers that is bounded below contains a least element, and every
non-empty set of integers that is bounded above contains a greatest element.
2.2 Induction
Mathematical induction, or more simply, induction, is a vital technique of proof in
mathematics. The idea is that we are asked to prove that a certain mathematical statement
is true for all natural numbers . If we can show that the statement is true for
that is, the statement
is true, and we can also show that
natural numbers , then we would have shown that
. To be precise,
implies
,
for all
is true for all natural numbers
is actually a predicate (refer to Study Unit 2) if we do not substitute
in any specific value of . For the sake of brevity, we shall assume that some value of
has been substituted in, so we can always call
a statement. Let us make the notion
of mathematical induction clear:
Principle of Induction
Let be a set of natural numbers (positive integers), that is, let be a subset of
that
FI1:
has the following properties:
.
FI2: For every
Then
. Suppose
,
is also a member of . That is,
must be the entire set of natural numbers, that is,
.
We can demonstrate that the Principle of Induction is valid by using the Well-Ordering
Property.
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Our proof uses the method of contradiction discussed in Study Unit 3. Suppose that is a
collection of natural numbers endowed with the properties FI1 and FI2 listed above, but
is not equal to the set of natural numbers. This means that there is at least one natural
number, say, , such that
Let
(this notation means is not a member of the set ).
be the set of natural numbers that are not members of , in other words,
By our assumption,
is non-empty. Thus, by the Well-Ordering Principle,
element, say . Since is assumed to be in
. Then
FI2,
by hypothesis FI1,
is a positive integer and
has a least
, which implies that
is an element of . However, by hypothesis
must also be a member of , which contradicts the fact that is a member
of . This completes our proof that
.
Implementing the Principle of Induction
The Principle of Induction is also known as the Principle of Mathematical Induction. If
is a mathematical statement depending on the positive integer, then to prove that
is true for all
by induction, it is necessary and sufficient to prove both conditions
FI1 and FI2, where is defined to be the set of all positive integers for which the statement
is true. The proof of FI1 is known as the basis of the induction, while the proof of
FI2 is known as the inductive step. The assumption that
is true for some is known
as the inductive hypothesis. To summarise:
Basis Step: Prove that
is true, that is, prove that
Inductive Hypothesis: Assume that
is true for some
is true for
.
.
Inductive Step: Use the inductive hypothesis and other mathematical facts to prove that
is also true.
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Conclusion: Once the above steps have been accomplished, conclude that
is true for
all positive integers .
Example 1
A sequence of real numbers is defined by the recurrence relation
Prove using induction that
for all positive integers .
Answer:
Let
be the statement that
.
For the basis step, we have to prove that
. Thus,
is true. Since
by definition, clearly
is true.
For the inductive hypothesis, we assume that
For this value of , we have
is true for some value of
.
. Observe that
The above result can be obtained, by say, long division. Another way is to do some
algebraic manipulation:
Since
,
, and so
because we know that
is a positive number. It follows that
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and so
which leads us to deduce that
as well, and hence
completed the inductive step, which is to show that
we conclude that
is also true. Thus we have
implies
. By induction,
is true for all positive integers .
Example 2
This is a typical example involving summation. Prove using induction that
Answer:
Let
be the statement that
. Here, the integer appears as the upper
limit of the summation. We are therefore said to be performing induction on the upper
limit of the summation. In the summation expression
, the integer
is also the
number of terms that appear in the expansion. Hence, we can also be said to be performing
induction on the number of terms in the summation expansion.
For the basis step, we have to prove that
The left hand side (LHS) of the equation is
(RHS) of the equation is
that
is true. The statement
reads:
, which is equal to 1. The right hand side
, which is also equal to 1. Since LHS=RHS, we deduce
is true.
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Sequences, Induction, the Euclidean Algorithm
For the inductive hypothesis, we assume that
using the letter
again because
is true for some integer
. (Avoid
has already been used as a dummy variable in the
summation notation.) This means we are assuming that
for some specific but unknown positive integer . For the inductive step, we are required
to use the above assumption to demonstrate the validity of the statement
Observe that
where the terms LHS and RHS again refer to the left hand side and right hand
side respectively of the equation to be proven. In an induction question involving
summation, the correct proof method to work from LHS step by step to RHS and
thereby show that they are equal. Now, we have proven that
by induction, we conclude that
is true. Therefore,
is true for all positive integers .
Caution!
A very common mistake made is when performing the inductive step, that is, proving
, one writes down
as if it is something that is already
known to be true rather than something that is to be proven to be true using the inductive
assumption made earlier on.
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Sequences, Induction, the Euclidean Algorithm
This fallacious argument is called begging the question.
To avoid falling into the trap of unwittingly begging the question, be very strict about
starting from LHS and working to RHS without assuming anything other than the
truth
.
Caution!
Another common mistake is to write
essentially equating the statement
wrong notation because
to the numerical quantity
. This is the
is a statement, not a numerical quantity. The colon :
punctuation symbol should be used instead:
Factorial notation
Let be a positive integer. We define
Thus, for instance,
define
,
,
,
to be the product
,
, and so on. For convenience, we also
to be equal to 1.
Here are some useful observations concerning the factorial notation:
a.
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b.
Example 3
Prove by induction that
Answer:
Let
be the statement
.
For the basis step, we have to prove that
is true. The statement
reads:
The LHS of the equation is
. The RHS of the equation is equal
to
is true.
. Thus, LHS=RHS and so we deduce that
For the inductive hypothesis, we assume that
is true for some integer
.
Thus we assume that the following is true:
To perform the inductive step, we have to use the above assumption to prove the statement
We can observe that
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Sequences, Induction, the Euclidean Algorithm
Now, we have proven that
is true. Therefore, by induction, we conclude that
is true for all positive integers .
Activity 4.6
Prove using induction that:
for all positive integers .
Example 4 (Geometric Series)
Let be a real number such that
. Prove by induction that
for all positive integers .
Answer:
The statement to be proven can be expressed using sigma notation as follows:
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Note that the lower limit of the summation is
the term
, in order to allow the sum to start from
. For the basis step, we have to prove that
is true. The statement
reads:
The LHS of the equation is
, which is equal to
. The RHS of the equation
is equal to
Hence, LHS=RHS, and we deduce that
is true.
For the inductive hypothesis, we assume that
is true for some integer
. Thus
we assume that the following is true:
for some integer
.
To perform the inductive step, we have to use the above assumption to prove the statement
Observe that
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Sequences, Induction, the Euclidean Algorithm
Now, we have proven that
is true. Therefore, by induction, we conclude that
is true for all positive integers .
2.3 Generalisations of Mathematical Induction
There is a generalisation of mathematical induction known as the principle of strong
induction.
Principle of Strong Induction
Let be a set of natural numbers (positive integers), that is, let be a subset of
that
SI1:
has the following properties:
.
SI2: For any integer
Then
. Suppose
, if
are all elements of , then
must be the entire set of natural numbers, that is,
.
.
The essential difference between this version of induction known as strong mathematical
induction, and the version discussed earlier in Section 2.2, is that for the inductive
hypothesis, we are assuming that all the integers
merely assuming that the integer is a member of .
Implementing the Principle of Strong Induction
SU4-24
are members of , instead of
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The Principle of Strong Induction is also known as the Principle of Strong Mathematical
Induction. Let
be a mathematical statement depending on the positive integer . We
wish to prove that
is true for all positive integers .
Basis Step:Prove that
is true, that is, prove that
is true for
Inductive Hypothesis: Assume that there is some positive integer
for all integers
.
such that
is true
.
Inductive Step: Use the inductive hypothesis and other mathematical facts to prove that
is also true.
Conclusion: Once the above steps have been accomplished, conclude that
is true for
all positive integers .
Example 1
Define a sequence of positive integers
Prove that
via the recurrence relation
for all positive integers .
Answer:
This is a special case of a recurrence relation where, even though the term
as a function of the terms
term
is expressed
, we only need to provide the value of the initial
in order to successfully define the entire sequence. For instance, by the recurrence
relation given, we have
,
,
,
,
and so on.
Let
be the statement that
true. The statement
. For the basis step, we have to prove that
reads:
SU4-25
is
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Sequences, Induction, the Euclidean Algorithm
The LHS of the equation is
is
, which is defined to be equal to 2. The RHS of the equation
, which is also equal to . Thus, LHS=RHS. We deduce that
is true.
For the inductive hypothesis, we assume that there is some positive integer
are all true. Thus, we are assuming that
satisfying
such that
for all integers
.
To perform the inductive step, we have to use the above assumption to prove the statement
is also true. The LHS of the equation is
equal to
, which by the recurrence relation given, is
. Using the inductive hypothesis, we then observe that
Now, we have proven that
is true. Therefore, by induction, we conclude that
is true for all positive integers .
The starting point of the induction does not always have to be
are asked to prove that a certain statement
is true for all
integer. Then for the basis step, we prove that
we assume that
is true for some integer
. Suppose that we
, where is a given
is true. For the inductive hypothesis,
, and use this assumption to prove that
is also true. The conclusion then is that
is true for all integers
. This
generalises the Principle of Induction stated in Section 2.2.
Implementing the Generalised Principle of Induction
The Generalised Principle of Induction is also known as the Generalised Principle of
Mathematical Induction. Let
be an integer. Let
depending on the integer . We wish to prove that
SU4-26
be a mathematical statement
is true for all
.
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Basis Step: Prove that
is true, that is, prove that
Inductive Hypothesis: Assume that
is true for
is true for some
.
.
Inductive Step: Use the inductive hypothesis and other mathematical facts to prove that
is also true.
Conclusion: Once the above steps have been accomplished, conclude that
all
is true for
.
For the inductive hypothesis, we can also use the strong version, which would involve
assuming that there is some integer
such that
. In the inductive step, we then prove that
is true for all integers
satisfying
is true.
Theorem:
Every integer
can be expressed as a product of prime numbers.
Proof:
This is a classic application of the principle of strong mathematical induction.
Let
be the statement that can be expressed a a product of prime numbers.
For the basis step, we have to prove that
is true.
is the statement that 2 can be expressed as a product of prime numbers. Since
by definition a prime number,
is trivially true.
For the inductive hypothesis, we assume that there is some integer
is true for all satisfying
is
such that
. This means that every integer satisfying
can
be expressed as a product of prime numbers (of course, each number would be expressed
as a product of primes in different ways).
To perform the inductive step, we have to prove that
prove that
If
can also be expressed as a product of primes.
is itself a prime number, that
is true.
SU4-27
is true, that is, we have to
MTH105
If
Sequences, Induction, the Euclidean Algorithm
is not a prime number, then
. Thus, we can express
where
and
is divisible by some positive integer satisfying
as
. By inductive hypothesis, both the integers
expressed as a product of prime numbers. Since
, it follows that
expressed as a product of prime numbers. Thus,
Therefore, by induction, we conclude that
and
can be
can also be
is true.
is true for all positive integers . The proof
is thus complete.
Consider the sequence
defined using the following recurrence relation:
Suppose that we are asked to prove that
example, because the term
terms, namely
and
for all positive integers . In this
in the recurrence relation is dependent on the previous two
, so to prove the statement using induction, the basis step must
involve proving that the statement is true for both
as well as
. It is not enough
simply to demonstrate the statement for
in the basis step because the recurrence
formula can only be invoked for the terms
, and beyond. Thus, in this situation, our
induction technique will be as follows:
Basis Step: Prove that
and
and
are true, that is, prove that
is true for
.
Inductive Hypothesis: Assume that
and
are true for some
.
Inductive Step: Use the inductive hypothesis and other mathematical facts to prove that
is also true.
Conclusion: Once the above steps have been accomplished, conclude that
all positive integers .
We will leave this example to be attempted by the reader as an Activity.
SU4-28
is true for
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Activity 4.7
A sequence
is defined by the recurrence relation
Prove using induction that
for all positive integers .
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Chapter 3: Greatest Common Divisor and the Euclidean
Algorithm
3.1 The Division Property of the Integers
Theorem:
Let
. There exist unique integers
be integers with
where
. Here,
the remainder of
of integers
such that
the quotient and
denotes the absolute value of . We call
divided by . Uniqueness means that we can only find one such pair
that satisfy the above property.
Proof:
Our proof uses the Well-Ordering Property introduced in Section 2.1 of this Study Unit.
Let
be a set defined as follows:
Thus, we define the set
to be the set of all integers of the form
such that
. This definition tells us that only nonnegative integers are allowed to be
members of . Observe also that the set
is nonempty, because we can always assign
extremely negative values to so as to make
nonnegative. Thus, is a nonempty
set of integers that is bounded below (a lower bound can be given by 0, but of course, 0
itself may not be a member of ). The Well-Ordering Property allows us to deduce that
must contain a minimal element, say,
Choose an integer such that
is defined. Then
. In particular,
for all
.
. This choice is possible because of the way the set
. If
, then
, and we simply have
, which
is the equation that we wish to obtain. Otherwise, suppose that
. Then
we have
with , we obtain the
. Again, by simply replacing
required equation.
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, and
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We observe further that
This can be proven by contradiction. Suppose we have
the situation where
. Then
But this contradicts the definition of the set , which only contains integers of the form
satisfying
. Similarly, suppose that
Hence we have found an integer
Since
. We can write
such that
and
is nonnegative, it is also an element of the set . This contradicts the
minimality of
.
We are now left to demonstrate the uniqueness of and
Suppose that in addition to the
values of and found earlier, there are also other integers
and
such that
Then we have
which implies that
Hence, we obtain the equation
by applying the absolute value operator to both sides. Since
we have
. This implies that
which in turn implies that
, and so
, and so
. We then further deduce that
. This completes the proof for uniqueness of the integers
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.
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Reflect 4.2
In which places in the above proof did we use the assumption that
?
Let us illustrate the division property with several examples:
divided by
Division Property yields
Quotient
Remainder
divided by
divided by
divided by
divided by
divided by
divided by
divided by
divided by
divided
by
Let us revisit the notion of divisors and multiples that we discussed in the previous Study
Unit.
An integer is said to be divisible or exactly divisible by if
for some integer . With
the concept of quotient and remainder, we can paraphrase this as: An integer
is said to
be divisible or exactly divisible by if leaves a remainder of 0 when is divided by .
An integer that leaves a remainder of 0 upon division by 2 is even, while an integer that
leaves a remainder of 1 upon division by 2 is odd. Thus, an even integer can always be
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written in the form
the form
for some integer , while an odd integer can always be written in
for some integer .
Example 1
Show that the square of any integer can be written in the form
or
for some
integer .
Answer:
If is even, then
for some integer , and we obtain
If is odd, then
.
for some integer , and we obtain
Example 2
Using Example 1, show that the square of any even number can be written in the form
or
for some integer .
Answer:
Let
be an even integer. Then
The integer
itself can be either even or odd. If
we obtain
If
for some integer
is odd, then
is even, then
.
for some , and so
Example 3
SU4-33
, and we obtain
.
for some , and
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Prove that the square of any odd integer can be written in the form
for some
integer .
Answer:
Let be an odd integer. Then
If
is even , then
If
is odd, then
for some integer , so
.
for some , and so
for some , and so
The Division Property can be used to produce a plethora of results along a similar vein.
Example 4
The square of any integer is either of the form
or
.
Answer:
By the Division Property, any integer is of the form
case,
,
, or
. In the first
. In the second case,
In the third case,
The key point is that given a positive integer , any other integer can always be expressed
in one of the forms:
. Furthermore, the expression
is unique.
Thus, every integer can be uniquely expressed in one of the forms
Every integer can also be uniquely expressed in one of the forms
.
SU4-34
,
, or
,
,
.
, or
MTH105
Sequences, Induction, the Euclidean Algorithm
Example 5
The square of any integer is of the form
,
, or
.
Answer:
By the Division Property, any integer is of the form
,
,
,
, or
.
We reach the conclusion by examining each of the cases
Activity 4.8
Prove that for any integer ,
is also an integer.
Interesting Divisibility Results
Here are some interesting tests for divisibility:
a.
An integer is divisible by 2, that is, an integer is even, if and only if its decimal
representation terminates with an even digit. For example, 406, 318, 1110 are
divisible by 2, but 77, 83, 1001 are not.
b.
An integer is divisible by 3 if and only if the sum of its digits is divisible by 3.
Thus, 363, 5256 are divisible by 3, but 10021 is not.
c.
An integer is divisible by 4 if and only if its last 2 digits form a number that is
divisible by 4. Thus, 364, 708, 1124 are divisible by 4, but 98, 142, 670 are not.
d.
An integer is divisible by 5 if and only if its final digit is 0 or 5.
SU4-35
MTH105
e.
Sequences, Induction, the Euclidean Algorithm
An integer is divisible by 8 if and only if its last 3 digits form a number that is
divisible by 8.
f.
An integer is divisible by 9 if and only if the sum of its digits is divisible by 9.
g.
An integer is divisible by 10 if and only if its final digit is 0.
h.
An integer is divisible by
if and only if it terminates in a sequence of zeros
or possibly more zeros (but not less).
We will not prove these tests of divisibility.
3.2 Solving Divisibility Problems Using Induction
In this section, we explore problems in the divisibility of integers that can be solved using
the principle of mathematical induction.
Example 1
Prove that
is divisible by
for all positive integers .
Answer:
Let
be the statement
The statement
is the statement that
Since
.
is true.
is true for some integer
is divisible by
.
is divisible by
,
Assume that the statement
that
that is divisible by
. Thus we have assumed
.
For the inductive step, we have to prove the validity of the statement
Observe that
SU4-36
MTH105
Sequences, Induction, the Euclidean Algorithm
Since
is divisible by
multiple of
, and the term
, we conclude that
have proven that
By induction,
is also clearly a
is also divisible by
. Hence we
is true.
is true for all .
Example 2
Prove that
is divisible by
for all positive integers .
Answer:
Let
be the statement that
The statement
by
is divisible by
is the statement that
.
is divisible
.
Since
,
Assume that the statement
is true.
is true for some integer
is divisible by
. Thus we have assumed
.
For the inductive step, we have to prove the validity of the statement
We
work
out
that
and
We take the second expression minus the first expression to obtain
SU4-37
MTH105
Sequences, Induction, the Euclidean Algorithm
We now divide into cases.
Suppose that
is even. Then each of the terms
is even. Then
Suppose that
is divisible by
is odd. Then
follows that
is even, and so the sum
.
are both odd terms, and therefore
is even, and so
is even. It
is likewise divisible by
in the case of odd.
Thus we have shown that for any integer , the difference between the expression
and
is
always
divisible
by
.
that
Since
is
the
inductive
divisible
is also divisible by
by
hypothesis
24,
it
assumes
follows
. Hence we have proven that
is true.
By induction,
is true for all .
Activity 4.9
Use induction to prove that
is divisible by for every positive integer .
Activity 4.10
Use induction to prove that
is divisible by for every positive integer .
SU4-38
that
MTH105
Sequences, Induction, the Euclidean Algorithm
3.3 Greatest Common Divisor and Lowest Common Multiple
Definition
Let
be integers, where at least one of
Suppose that there is a positive integer
GCD1.
is non-zero (that is,
are not both zero).
satisfying the following:
and
GCD2. Whenever is any other integer such that
We call such an integer
and
, then
.
the greatest common divisor of and . It is the largest positive
integer that divides both and and we abbreviate it as
divisor of any two given integers
there is one and only one
. The greatest common
(not both zero) is always unique. This means that
for any pair of integers
with
not both zero.
Here are some examples:
a.
b.
c.
d.
e.
f.
g.
Some essential properties of gcd that one should bear in mind:
a.
The gcd notation is commutative, because to say that
integer dividing both
swap
is the largest positive
is a symmetrical statement; one could just as easily
without altering the meaning of the statement.
SU4-39
MTH105
Sequences, Induction, the Euclidean Algorithm
b.
for any integer .
c.
for any non-zero integer .
d.
for any non-zero integers
The purpose of putting the integer within the absolute value symbol is due to
the fact that we define the gcd to be a positive integer.
e.
If
are positive integers and
, then
More generally, one could say that if
.
are non-zero integers and
, then
.
Definition
be non-zero integers. Suppose that there is a positive integer
Let
satisfying the
following:
LCM1.
and
LCM2. Whenever is any other integer such that
We call such an integer
and
, then
.
the lowest common multiple of and . It is the smallest positive
integer that is a multiple of both
and
and we abbreviate it as
common multiple of any two given non-zero integers
that there is one and only one
is always unique. This means
for any pair of non-zero integers
Here are some examples:
a.
b.
c.
d.
e.
SU4-40
. The lowest
.
MTH105
Sequences, Induction, the Euclidean Algorithm
f.
Some essential properties of lcm that one should bear in mind:
a.
b.
for any integer non-zero .
c.
for any non-zero integer .
d.
for any non-zero integers
e.
If
are positive integers and
, then
More generally, one could say that if
.
.
are non-zero integers and
, then
.
Theorem
Let
be non-zero integers. Then there exist integers
Furthermore, whenever
is an integer such that
is an integral linear combination of
such that
for some integers
, then
, that is,
divides .
Proof:
We again have the chance to use the Well-Ordering Property.
Let
, that is,
, where
if we let
is the set of all integers of the form
are integers and
. The set is non-empty, because for instance,
, then
, the latter inequality holding because
and
are both assumed to be non-zero. This means that
is an element of . Thus,
non-empty set of positive integers. By the Well-Ordering Property, there exist integers
such that
Let
is the minimal element of .
be the gcd of
(because
and . Then certainly,
divides both
divides
and ), and thus in particular,
for any pair of integers
for the values of
found by the Well-Ordering Property in the previous paragraph. Therefore,
SU4-41
.
is a
MTH105
Sequences, Induction, the Euclidean Algorithm
We now apply the Division Property. There exist unique integers
where
. Suppose
. Then is a positive integer, and since
is also an integral linear combination of
also have
and
such that
and . This implies that
, and this contradicts the minimality of
. However, we
. Thus we conclude that
. Similarly, we can perform an argument like the one above to show that
for some integer . Thus, we have that
and
. Since is a common divisor of both
and , divides as well. Then we now know that
integers, we conclude that
and
. Since
are both positive
. We have therefore proven that there exist integers
such that
For the second part of the theorem, suppose that
for some integers
, that is,
common divisor
is an integer such that
is an integral linear combination of
divides both
, we have
. Since the greatest
. The proof is complete.
Integral linear combinations of and
The above theorem tells us that whenever
then
is a multiple of
. Conversely, since
and , it also follows that whenever
, and thus
is an integral linear combination of
for some integers
is a multiple of
is an integral linear combination of
combination of and if and only if is a multiple of
SU4-42
and ,
,
. Hence,
.
is a multiple of
is an integral linear
MTH105
Sequences, Induction, the Euclidean Algorithm
Definition
Let
are relatively prime, or coprime, if
be non-zero integers. We say that
.
Theorem
Let
if and only if there exist integers
be non-zero integers. Then
that
such
.
The above result does not apply for the case where
1, then
if and only if
. If the gcd is larger than
is a multiple of the gcd. There is no guarantee that
has
to be equal to the gcd.
Activity 4.11
Let
be non-zero integers. Suppose that
. Prove that
divides
.
Theorem
Let
be integers with
particular, if
. Then for any integer ,
, then
. In
.
Proof:
Let
and
be an integer. Let
, so
divides
divides
and let
. Thus,
divides both
.
SU4-43
. Since
and . Consequently,
MTH105
Sequences, Induction, the Euclidean Algorithm
Since
and
deduce that
, so
divides
divides
. Since
. Again, we can
and
are positive integers that divide
each other, they are equal to each other. Hence,
. The
proof is complete.
The above theorem tells us, in particular, that when we use the Division Property to write
for
, the gcd of
and
is equal to the gcd of
and the remainder
. We can also express the result as:
latter equation holds not just for the unique quotient
any integer ; in other words,
of
;
. The
divided by , but in fact for
does not have to be the remainder of divided by .
Example 1
Let be any integer. Then
Since
is odd,
does not share any common divisors with apart from the usual
divisor 1. Hence,
for all integers .
Example 2
Let
be any integer. Prove that
if
even.
Answer:
Let be any integer. Then
SU4-44
is odd, and if
is
MTH105
Sequences, Induction, the Euclidean Algorithm
If is odd, then
If is even, then
is divisible by but not by 4. In this case,
is divisible by 4. In this case,
.
.
Activity 4.12
.
3.4 The Euclidean Algorithm
The Euclidean Algorithm is a systematic method of finding the greatest common divisor
of two integers
, not both 0, and then expressing
in the form
are integers to be determined. Since
that
, where
, we can assume
are non-negative.
The Euclidean Algorithm involves repeated application of the Division Property. First,
apply the Division Property to the division of by to get
If
Suppose
If
, then
, and
.
. Then apply the Division Property to the division of by
, we stop; else we proceed as before to obtain
SU4-45
to get
MTH105
Sequences, Induction, the Euclidean Algorithm
The division process continues in this fashion, where at stage , we take the remainder
obtained two steps ago,
. The
, and divide it by the remainder obtained in the previous step,
step is thus
The algorithm continues until a remainder of 0 appears, which necessarily happens
in a finite number of steps because the sequence of remainders
is strictly
decreasing. The final three steps would be of the form:
where
. We now have
, the final non-
zero remainder obtained, because
We now use the results obtained above to express
suitable integers
. Starting from the second last equation, we write
linear combination of
Then make
and substitute
and
in the form
and
, for
as an integral
:
the subject of the formula in the preceding equation:
into first equation to obtain
as an integral linear combination of
:
Continuing with this backward substitution, we eventually express
combination of and .
Example 1
SU4-46
as an integral linear
MTH105
Sequences, Induction, the Euclidean Algorithm
Implement the Euclidean Algorithm to find
and express
as an integral
linear combination of 50 and 14.
Answer:
Apply the Division Property repeatedly until a remainder of 0 is obtained:
Hence, the
.
Now use backward substitution:
Therefore,
Example 2
Implement the Euclidean Algorithm to find
and express
linear combination of 84 and 15.
Answer:
Apply the Division Property repeatedly until a remainder of 0 is obtained:
Hence, the
.
Now use backward substitution:
SU4-47
as an integral
MTH105
Sequences, Induction, the Euclidean Algorithm
Activity 4.13
Implement the Euclidean Algorithm to find
and express
as an
integral linear combination of 102 and 32.
3.5 Prime Numbers
We devote this section to a brief introduction to some basic facts and properties of prime
numbers. Recall that a positive integer is said to be prime if
, and has no positive
divisors apart from 1 and itself.
Theorem:
If is a prime number and
are non-zero integers such that
, then either
or
.
Proof:
If
then the result is true.
Suppose that
does not divide . We show that
must divide . Since the only positive
divisors of are 1 and itself, and does not have as one of its divisors, we observe that
. Hence, there are integers and such that
Multiply the above equation throughout by
to obtain
we can also find some integer such that
Then we have
SU4-48
. Since
divides
,
MTH105
Sequences, Induction, the Euclidean Algorithm
Since
is an integer and
, we conclude that
In conclusion, we must have that either
or
.
.
Reflect 4.3
What rule of inference did we utilise in the proof of the above theorem? Refer to Study
Units 2 and 3.
The above theorem can be generalised to any products of integers. If
non-zero integers such that a prime number divides the product
one of the integers
(
). It could be the case that
are all
, then divides
divides more than one of the
integers listed, but at the very least, has to divide at least one of them.
The fundamental theorem of arithmetic states that every integer
can be expressed
as a product of primes, and this expression is unique up to rearrangement of terms.
Uniqueness means that disregarding trivial swapping of terms like
every integer
is either itself a prime, or can be expressed as a product of primes in
one and only one way. A prime factorisation of an integer
, where
is written in the form
are distinct primes, and
are positive
integers. The ordering of the prime powers does not matter.
If
and
are prime factorisations of the integers
, where we allow for some of the powers
and
where
to be equal to zero,
form a common set of primes in the above expressions, then
for each ,
, and
SU4-49
MTH105
Sequences, Induction, the Euclidean Algorithm
where
for each ,
.
We will not prove the above results here (even though what has been covered in this
Study Unit is sufficient for such a proof), but will instead focus our attention on two other
interesting properties of prime numbers.
Theorem:
Let
be a prime number. Then
is irrational.
Proof:
Our proof is by the method of contradiction. Suppose that
is rational. Then
where without loss of generality, we can assume that
with
. We are
able to employ the "without loss of generality" concept because if
, we
can express the fraction
,
as
thereby reducing the fraction to its lowest form (where the numerator and denominator
have no positive common divisors other than 1). Then
divides
. By the previous theorem in this Section,
integer . It follows that
divides
, and so
, and so
and so
divides . Therefore,
. Then
for some
. This equation implies that
. But we now have a contradiction, because we have shown
common divisor of both and , yet we previously assumed that
is a
. The proof
is complete.
Theorem:
There are infinitely many prime numbers.
Proof:
Again, this theorem is proved by contradiction. Suppose instead that there are only finitely
many prime numbers. We can list down the complete collection of primes:
SU4-50
.
MTH105
Sequences, Induction, the Euclidean Algorithm
Let
. Since we have assumed there are no other primes in the list
, and
Then
is strictly larger than all of them, it must be that
has some prime factor, say,
, so
. Since
divides
divides
and
is composite.
also divides the product
. Of course, saying that the prime
divides 1
is a contradiction. The proof is complete.
Activity 4.14
Let be a positive integer such that
. Prove that if is of the form
for some
integer , then is either itself a prime number, or is divisible by a prime number of
the form
.
SU4-51
MTH105
Sequences, Induction, the Euclidean Algorithm
Summary
In this Study Unit, we introduced new method of proof called mathematical induction.
Induction is applied when we wish to prove that a certain mathematical statement is true
for all positive integers, or for all integers above a certain value. Induction consists of a
basis step, the inductive hypothesis, the inductive step, and finally the conclusion.
We applied induction to prove that a certain property holds for a sequence, or that a certain
expression involving an integer is always divisible by a given integer.
The principle of mathematical induction was derived from the Well-Ordering Property,
which applies to sets of integers that are bounded below.
In the third chapter, we extended our knowledge of elementary number theory by
introducing the Division Property of the integers, the notion of greatest common divisor
and lowest common multiple, and the Euclidean Algorithm.
SU4-52
MTH105
Sequences, Induction, the Euclidean Algorithm
Formative Assessment
1.
Prove by induction that for all positive integers, we have
2.
A sequence of positive real numbers
with
is defined by the recurrence relation
. Prove using the method of induction that for all positive integers , we
have
3.
a.
Prove by induction that
b.
Then use the above result to implement a proof by induction that for all
is divisible by
for all integers
.
positive integers ,
is divisible by
4.
.
In Section 3.1, Example 1, we showed that the square of any integer can be written in
the form
for some integer . Use this result to prove that no integer in the
sequence
can possibly be a perfect square.
5.
For any integer , prove that one of the integers ,
.
SU4-53
, or
is divisible by
MTH105
6.
Sequences, Induction, the Euclidean Algorithm
In Section 3.1, Example 3, we showed that that the square of any odd integer can be
written in the form
for some integer . Use this result to prove that if is an odd
integer, then
is divisible by
7.
Prove that
8.
Implement the Euclidean Algorithm to find
for all integers .
integral linear combination of
9.
.
and
Prove that any prime of the form
and express
as an
.
for some integer
must also be of the form
for some integer .
10. Suppose that
is a sequence of real number in which the
partial sum is given
by the formula
Prove that
is an arithmetic sequence and find its common difference.
11. An arithmetic sequence has 10 terms. The first term is 2 and the last term is 29. Find
the common difference and the sum of the terms of the sequence.
12. A sequence of real numbers
, the sum of the first
Prove that
13. Suppose that
satisfies the property that for all positive integers
terms,
, is given by
is a geometric sequence and state its common ratio.
are positive integers, and that
, then does not divide
.
14. Demonstrate using mathematical induction that
SU4-54
. Show that if does not divide
MTH105
Sequences, Induction, the Euclidean Algorithm
is divisible by 9 for any positive integer .
15. Demonstrate using mathematical induction that for all positive integers ,
16. Implement the Euclidean Algorithm to find
and express
as an
integral linear combination of 780 and 182.
17. Implement the Euclidean Algorithm to determine
an integral linear combination of 240 and 135
SU4-55
and express as
MTH105
Sequences, Induction, the Euclidean Algorithm
Solutions or Suggested Answers
Activity 4.1
a.
The terms of the sequence
are
b.
The terms of the sequence
are
c.
The terms of the sequence
are
d.
The terms of the sequence
are
e.
The terms of the sequence
are
Activity 4.2
a.
The terms of the sequence
are
b.
The terms of the sequence
are
c.
The terms of the sequence
are
d.
The terms of the sequence
are
Activity 4.3
a.
b.
c.
SU4-56
MTH105
Sequences, Induction, the Euclidean Algorithm
d.
e.
Activity 4.4
For the arithmetic sequence
the first term is
and the common difference is
. Hence, the sum of the first 30
terms is given by
Activity 4.5
For the geometric sequence
the first term is
and the common ratio is
is given by
SU4-57
. Hence, the sum of the first 8 terms
MTH105
Sequences, Induction, the Euclidean Algorithm
Activity 4.6
Let
be the statement that
For the basis step, we have to prove
.
that is true. The statement
The left hand side (LHS) of the equation is
, which is equal to
side (RHS) of the equation is
Since LHS=RHS, we deduce that
reads:
. The right hand
, which is also equal to 1.
is true.
For the inductive hypothesis, we assume that
is true for some integer
. This
means we are assuming that
for some specific but unknown positive integer . For the inductive step, we are required
to use the above assumption to demonstrate the validity of the statement
Observe that
SU4-58
MTH105
Sequences, Induction, the Euclidean Algorithm
Now, we have proven that
is true. Therefore, by induction, we conclude that
is true for all positive integers .
Activity 4.7
Let
be the statement that
.
For the basis step, we have to prove that
The statement
and
states that
are true.
. The LHS of the equation is
definition is equal to 3. The RHS of the equation is
, which by
, which is of course also equal
to 3.
The statement
states that
. The LHS of the equation is
definition is equal to 5. The RHS of the equation is
, which by
, which is of course also equal
to 5.
For the inductive hypothesis, we assume that
and
are true for some integer
.
This means we are assuming that
and
for some integer
.
For the inductive step, we are required to use the above assumption to demonstrate the
validity of the statement
By the recurrence relation given, the LHS is equal to
hypothesis, we have
SU4-59
. Thus, by the inductive
MTH105
Sequences, Induction, the Euclidean Algorithm
Now, we have proven that
is true. Therefore, by induction, we conclude that
is true for all positive integers .
Activity 4.8
By the Division Property, any integer
is of the form
,
, or
. In the first
case,
which is clearly also an integer. In the second case,
and in the third case,
Activity 4.9
Let
be the statement that
is divisible by .
For the basis step, we have to prove that
divisible by . Since
is true.
, the statement
For the inductive hypothesis, we assume that
This means we are assuming that
is the statement that
is true.
is true for some integer
is divisible by .
For the inductive step, we are required to prove the validity of the statement:
SU4-60
is
.
MTH105
Sequences, Induction, the Euclidean Algorithm
is divisible by .
Observe that
Since
is divisible by ,
is also divisible by , and this implies that
is divisible by . Now, we have proven that
induction, we conclude that
is true. Therefore, by
is true for all positive integers .
Activity 4.10
Let
be the statement that
is divisible by .
For the basis step, we have to prove that
is divisible by . Since
is true.
, the statement
For the inductive hypothesis, we assume that
This means we are assuming that
is the statement that
is true.
is true for some integer
.
is divisible by .
For the inductive step, we are required to prove the validity of the statement:
is divisible by .
First,
,
while
. Taking the first expression minus the second expression gives us
Here, we divide into cases.
Suppose
so
is even. Then each of the terms
is divisible by .
SU4-61
are even. Hence,
is even and
MTH105
Suppose
Sequences, Induction, the Euclidean Algorithm
is odd. Then both the terms
are odd. It follows that
is again even. Likewise,
is even, and so
is divisible by when
is even.
Since the first expression minus the second expression is always divisible by
integer
, and the inductive hypothesis assumes that
is divisible by , so
is also divisible by . Now, we have proven that
Therefore, by induction, we conclude that
for any
is true.
is true for all positive integers .
Activity 4.11
Let
We conclude that
and suppose
. Since
divides
,
too. Hence,
.
Activity 4.12
Let be any integer.
Activity 4.13
Apply the Division Property until a remainder of 0 is obtained:
SU4-62
divides both
and .
MTH105
Sequences, Induction, the Euclidean Algorithm
Use backward substitution:
Activity 4.14
Suppose that
is a positive integer such that
and
is of the form
for some
integer .
We prove by contradiction.
Suppose that is not a prime number and that every prime divisor of is not of the form
.
Now, by the Division Property, every integer can be uniquely written in the form
or
,
,
.
All the prime divisors of are greater than 3 (or else would be divisible by 3 and it would
then be impossible to have of the form
).
Hence in this situation, every prime divisor of must be of the form
Therefore, there exist integers
.
such that
When we algebraically expand out the above expression on the RHS, we obtain an
expression of the form
for some integer . This contradicts our assumption that
an integer of the form
for some integer .
Therefore, we conclude that
number of the form
is either itself a prime number, or
.
SU4-63
is
is divisible by a prime
MTH105
Sequences, Induction, the Euclidean Algorithm
Formative Assessment
1.
Prove by induction that for all positive integers, we have
Answer:
Let
be the statement
Basis Step: Prove the validity of
LHS is
RHS is
Hence, LHS=RHS and so
is true.
Inductive Hypothesis:
Assume that
is true for some integer
Inductive Step:
We are required to prove the validity of the statement
Execution:
SU4-64
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Sequences, Induction, the Euclidean Algorithm
Hence
2.
is also true. By induction,
A sequence of positive real numbers
with
is true for all positive integers .
is defined by the recurrence relation
. Prove using the method of induction that for all positive integers , we
have
Answer:
Let
be the statement
.
Basis Step: Prove the validity of
LHS is
, which is by definition equal to
RHS is
Hence, LHS=RHS, and thus
is true.
Inductive Hypothesis:
Assume that
is true for some integer
SU4-65
.
MTH105
Sequences, Induction, the Euclidean Algorithm
Inductive Step:
We are required to prove the validity of the statement
Execution:
Using the recurrence relation provided,
. Then from the
inductive hypothesis, we have:
Hence
3.
is also true. By induction,
is true for all positive integers .
a.
Prove by induction that
b.
Then use the above result to implement a proof by induction that for all
is divisible by
for all integers
.
positive integers ,
is divisible by
.
Answer:
a.
Let
be the statement that
is divisible by
.
Basis Step:
We prove the validity of the statement
Since
:
, and is divisible by
is divisible by
,
Inductive Hypothesis:
Assume that the statement
is divisible by
is true, that is, assume that
for some integer
SU4-66
.
is true.
.
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Sequences, Induction, the Euclidean Algorithm
Inductive Step:
We are required to prove that
is also true, that is, to prove
is also divisible by
.
Execution:
Since
is divisible by
divisible by
, so
Hence
by inductive hypothesis, and
is divisible by
is also true. By induction,
is also
.
is true for all positive
integers .
b.
Let
be the statement that
is divisible by
.
Basis Step:
We prove the validity of the statement
divisible by
is
.
Since
,
is true.
Inductive Hypothesis:
Assume that the statement
is divisible by
is true, that is, assume that
for some integer
.
Inductive Step:
We are required to prove that
is also divisible by
Execution:
SU4-67
is also true, that is, to prove that
.
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Sequences, Induction, the Euclidean Algorithm
Since
is divisible by
is also divisible by
is also divisible by
by inductive hypothesis and
by the previous result,
. Hence
is also true. By induction,
is
true for all positive integers .
4.
In Section 3.1, Example 1, we showed that the square of any integer can be written in
the form
for some integer . Use this result to prove that no integer in the
sequence
can possibly be a perfect square.
Answer:
Every number in the sequence is of the form
. For example,
and so on. The integer
exactly divisible by
is
(refer to tests for divisibility in Section 3.1). Hence, every
number in the sequence
is of the form
for some integer
. Thus, by Example 1 of Section 3.1, none of them are perfect squares.
5.
For any integer , prove that one of the integers ,
, or
is divisible by
.
Answer:
By
the
integer
Division
. If
Property,
, then
,
is divisible by
SU4-68
or
. If
for
some
, then
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Sequences, Induction, the Euclidean Algorithm
is
divisible
by
.
If
,
then
is divisible by .
6.
In Section 3.1, Example 3, we showed that that the square of any odd integer can be
written in the form
for some integer . Use this result to prove that if is an odd
integer, then
is divisible by
.
Answer:
We can write
is divisible by
7.
for some integer . Thus,
.
Prove that
for all integers .
Answer:
Let be an integer.
8.
Implement the Euclidean Algorithm to find
integral linear combination of
and
and express
.
Answer:
Apply the Division Property until a remainder of is obtained.
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as an
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Sequences, Induction, the Euclidean Algorithm
Use backward substitution:
Hence
9.
.
Prove that any prime of the form
for some integer
must also be of the form
for some integer .
Answer:
It is easy to directly verify the result for the prime number
that
is a prime number such that
Suppose that
and , so we assume
.
for some integer . By the Division Property, is of one of the
forms
. Since is a prime, the possibilities
are all ruled out, because in each of those possibilities,
would have a positive divisor of
or . (That was why we disposed of those two
trivial cases first.) Hence, is of the form
, else
form
or
. But cannot be of the form
would then be of the form
. Thus,
10. Suppose that
, and not of the
for some integer .
is a sequence of real number in which the
by the formula
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partial sum is given
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Sequences, Induction, the Euclidean Algorithm
Prove that
is an arithmetic sequence and find its common difference.
Answer:
For
, the
term of the sequence is given by
For
, the 1st term of the sequence is given by
Hence, for every integer
checked that for every
Hence,
, the
term of the sequence
. It can then be
, we have
is an arithmetic sequence with common difference equal to .
11. An arithmetic sequence has 10 terms. The first term is 2 and the last term is 29. Find
the common difference and the sum of the terms of the sequence.
Answer:
Using
so
, we have
.
The sum of terms is
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Sequences, Induction, the Euclidean Algorithm
12. A sequence of real numbers
, the sum of the first
Prove that
satisfies the property that for all positive integers
terms,
, is given by
is a geometric sequence and state its common ratio.
Answer:
Hence for all
, we have
Clearly
as well. Hence, the sequence is a geometric sequence with common ratio equal to
3.
13. Suppose that
are positive integers, and that
, then does not divide
. Show that if does not divide
.
Answer:
Suppose that divides .
Suppose that divides
Then divides
too.
.
By a contrapositive argument, we infer that if
divide
.
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does not divide , then
does not
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Sequences, Induction, the Euclidean Algorithm
14. Demonstrate using mathematical induction that
is divisible by 9 for any positive integer .
Answer:
Let
be the statement that
is divisible by 9.
Base case:
is divisible by 9
and 9 is divisible by 9
So
is true.
Inductive hypothesis:
Assume
true for some
is divisible by 9
Inductive step:
To show
Since
so
This proves
true.
is divisible by 9 by the inductive hypothesis,
. is divisible by 9 as well.
.
And completes the proof by induction.
15. Demonstrate using mathematical induction that for all positive integers ,
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Sequences, Induction, the Euclidean Algorithm
Answer:
Let
be the statement that
Base case:
LHS =
RHS =
LHS = RHS, so
is true.
Inductive hypothesis:
Assume
true for some
Inductive step:
To show
true.
by the inductive hypothesis
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Sequences, Induction, the Euclidean Algorithm
Hence,
is also true.
This completes the proof by induction.
16. Implement the Euclidean Algorithm to find
and express
as an
integral linear combination of 780 and 182.
Answer:
Hence
We have
17. Implement the Euclidean Algorithm to determine
an integral linear combination of 240 and 135
Answer:
SU4-75
and express as
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Sequences, Induction, the Euclidean Algorithm
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Sequences, Induction, the Euclidean Algorithm
References
Epp, S. S. (2019). Discrete mathematics with applications (5th ed.). Boston, MA: Brooks/
Cole Cengage Learning.
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Sequences, Induction, the Euclidean Algorithm
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Study
Unit
Set Theory
5
MTH105
Set Theory
Learning Outcomes
By the end of this unit, you should be able to:
1.
Define a set using curly bracket
2.
Define and explain what it means for a set
3.
Determine when an element is contained in a set.
4.
Determine whether or not a given set
5.
Define and use various set constructions such as unions, intersections, and
notation.
to be a subset of a set .
is a subset of a given set .
complements.
6.
Use a Venn diagram to illustrate various set constructions.
7.
Prove set identities and set relations using both the element method and the chain
of set identities method.
8.
Disprove an alleged set property by giving an appropriate counter-example.
9.
Use
the
Addition
Principle,
the
Multiplication
Principle,
and
the
Complementation Principle in counting problems.
10.
Differentiate between permutations and combinations, and use the quantities
and
11.
whenever appropriate.
Use the classical definition of probability and basic results concerning probability
in problem solving.
12.
Define mutually exclusive events as well as independent events, and solve
problems in probability involving these types of events.
13.
Determine the winner of an election using the Approval Method, the Plurality
Method, the two-round system, the Plurality with Elimination Method, the
method of pairwise comparisons, and the Borda Count Method.
14.
Define the majority criterion, the Condorcet criterion, the Monotonicity criterion,
and the Independence of Irrelevant Alternatives criterion as measures of fairness
of a given method of determining an election result.
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Set Theory
15.
Provide examples to show when a given method of determining the winner of
an election violates a certain criterion.
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Set Theory
Overview
In this Study Unit, we formally introduce the notion of a set and establish the basic
properties of sets.
The notion of sets is a fundamental concept in mathematics. In every branch of
mathematics, definitions and theorems are stated using the language of set theory.
In the previous Study Units, we have already used the notions of sets and subsets. Here,
we will make these ideas rigorous and demonstrate various set properties by using proof
techniques such as the element method or by establishing a chain of set identities.
In this Study Unit, we also introduce two interesting applications of the mathematical
knowledge we have acquired thus far. The first application is Counting and Probability.
We will learn how to use counting principles like the Addition Principle, the
Multiplication Principle, the Complementation Principle, as well as the concept of
permutations and combinations, to tackle various counting problems. We will also
investigate the basic concepts of probability and use the counting methods learnt to solve
questions in probability.
The second application we will introduce is the Mathematics of Voting. We will study
various ways of determining the outcome of an election. We will also examine four
different criteria to measure the fairness of each method of determining election outcomes.
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Set Theory
Chapter 1: Introduction to the Concept of a Set
1.1 Definition of a Set
The reader has already been informally introduced to the notion of a set. A set is simply
defined to be a collection of objects. In mathematics, a set can contain numbers, variables,
functions, or even other sets.
Let be a set. For an element
we write
to mean that is a member of, or an element
of, the set . We can also phrase this as saying that belongs to , or that is contained in
. We write
to mean that is not a member of the set .
The numbers 2,4,6,8,0 all belong to the set of even numbers. But none of them belong to
the set of odd numbers. The number
belongs to the set of rational numbers, but not to
the set of integers. Similarly, the number
belongs to the set of real numbers, but not to
the set of rational numbers. As the reader can recall, we call
A set is defined using curly braces
an irrational number.
. We can explicitly give each member of the set,
for example,
. Or, we can also define a set by specifying
a property or a group of properties that determine whether or not an element belongs to
the set. Let
be a predicate that involves the variable . The set
is the set that comprises all those elements
elements that make
Then
that satisfy
; in other words, those
true. For instance, define
contains precisely those elements
an element
that are divisible by 6. For the set
is a member of
biconditional requirement should be carefully noted.
SU5-5
if and only if
is true. The
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Set Theory
At this junction, we must caution the reader that to properly define a set in this manner
using predicates, we really should be writing
Here,
denotes some set that contains all elements under considerations. The above
notation means that
is precisely the set of elements
of
for which
satisfies
.
Hence, we are drawing the elements only from the set . For instance, let
Then the set
comprises all those natural numbers
such that
two different prime numbers. We only allow the elements
is divisible by at least
to be drawn from the set of
natural numbers. Thus, for instance, we have excluded all the negative integers.
From now on, when we define a set using the above notation, that is, making use of some
predicate as opposed to explicitly listing down every single member of the set, we have to
invoke some universal set
from which we are drawing all the required elements. Thus,
is permissible, but
is not permissible because we have not specified from which set we are drawing those
elements . It is acceptable to define a set as follows:
because we have moved the specification
to the right hand side instead of placing
it on the left. The bottom line is that there must be a clear indication somewhere in the
definition that some universal set
is involved from which we draw all the required
elements .
We can also utilise set definitions of the following form:
SU5-6
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Set Theory
For example, let
of the form
. The set
, where
is defined to be the set of all elements
is allowed to vary over the set of natural numbers
are elements of , but
. Thus,
are not elements of .
1.2 Subsets
Let
be sets. We say that
is a subset of
We can also phrase it as follows:
if every element of
is a subset of
is contained within
if and only if
Here, we have used implicit quantification. (See Study Unit 2.)
If
is a subset of , we write
means that
. Sometimes, we also write
is a superset of , meaning that
this is equivalent to saying that
contains every element of . Of course,
is a subset of .
If
is not a subset of , we write
of
if and only if there exists some element
is called a proper subset of
. The latter notation
, or sometimes we also write
if
such that
is a subset of , and
does not belong to . We also say in this case that
.
is not a subset
; in other words,
contains some element
is a proper superset of . In other
words,
We can diagrammatically depict
by drawing two ovals, one within another:
SU5-7
that
MTH105
Set Theory
Here, the bigger (outer) oval denotes the set , while the inner circle denotes the set .
If we are asked to prove that
, we must demonstrate that
we pick an arbitrary element of the set
and explain why it has to be in the set . Note
that we do not do the reverse process, that is, picking something in
is in . We only need to go from
. To prove this,
and then proving it
into .
Example
Define sets
as follows:
We show that
is a subset of
Let
by employing the following argument:
be given. Then
for some integer . We can write
is an integer, we conclude that
therefore shown that whenever
for some integer , and so
, we also have
. Since
. We have
. This completes the proof that
is a subset of .
Note that is an element of a set
if and only if
is a subset of .
1.3 Equality of Sets
Two sets are said to be equal if they contain exactly the same elements. This means that
if and only if every element in
can be found in , and also every element of
be found in . Observe that this means that
if and only if we have
and
can
.
In other words,
How then do we go about proving that two given sets,
arbitrary element
and prove that
and prove that
and , are equal? We choose any
. Similarly, we also choose an arbitrary element
.
SU5-8
MTH105
Set Theory
Example
Define sets
as follows:
We prove that
Let
using the following argument:
be given. Then
integer, is of the form
Conversely, let
write
for some integer . Now,
for some integer . Thus,
be given. Then
, and so
. Since
is an
.
for some integer . This means that we can
.
We have thus shown that
. Our conclusion is that
.
There is an important point to take note of regarding the way we list down the elements
within a set. The sets {1, 3, 5} and {3, 5, 1} are regarded as equal, because they have exactly
the same elements. The manner in which we listed the elements of each set did not matter.
In other words, the arrangement of the numbers did not matter. It also does not matter if
an element of a set is listed more than once. Thus, for instance, {1, 3, 3, 3, 5, 5, 5, 5} is the
same as the set {1, 3, 5} because they have the same elements.
1.4 The Empty Set
There is a special set called the empty set. The empty set is denoted by the symbol
and
it is defined as the set that contains no elements. Formally, we can say that
The above statement may seem strange, but it makes perfect sense. If is an actual object,
of course
. Since an element is in
if and only if
,
cannot contain anything
at all because no object can ever satisfy the stated condition.
The empty set is a subset of every set. Thus, if
is any given set, we always have
We can understand this assertion in the following manner: To say that
that
. However, since no element satisfies
SU5-9
.
is to say
, the conditional statement
MTH105
Set Theory
is therefore vacuously true. It is therefore vacuously true that
set . In particular,
.
The reader is cautioned on the following: To say that
saying that
for any
is entirely different from
. The latter statement means something entirely different; it means that
is an actual element of the set . For example, let
The set
comprises the elements
member of . Of course,
. The element
will still be true as
both
as well as
then
. However, we still have that
is itself a
is a set. Hence in this case we have
. On the other hand, if we define
.
1.5 Containment of Elements
Careful distinction must be made between the notation that is an element of , and
is
a subset of . For instance, consider the set
The numbers 1,2,3,8,9 are all elements of . But they themselves are not subsets of . We
can write
, but of course, the sentence
is patently false.
On the other hand, 6,7 are not elements of . Instead, it is only true that the set
is an element of . This is an example of a set being an element of another set. It is also
wrong to say that
is itself not a subset of , because the
. The set
numbers 6,7 are not members of . It is only true that
We can say that
is a subset of
because every element of
. The set
.
, expressed
. This is
is also an element of . But we cannot claim that
is itself not a member of .
SU5-10
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Set Theory
1.6 Venn Diagrams
Sets can be represented graphically using Venn diagrams. In Venn diagrams, we usually
specify a universal set
which contains all the objects under consideration. The universal
set is represented by a rectangle, and all other sets are drawn within the rectangle.
For instance,
the above Venn diagram depicts a universal set
. The sets
are subsets of
, within which is contained three sets
.
1.7 The Size of a Set
If a set contains n elements, we say that the size of the set is n. Thus, the size of the set
{ 1,2,9,10 } is 4, while the size of the set { 1,2,3,{6,7},8,9 } is 6. The element { 6,7 } is considered
as a single element. Thus in this example, { 6,7 } is a single object which is itself a set that
happens to contain the numbers 6 and 7.
If there are exactly n distinct elements in a set S, where n is some nonnegative integer (the
empty set has exactly 0 elements), then we say that S is a finite set. If not, then we say that
S is an infinite set. For an infinite set, there is no way we can list down all the elements
of the set. For instance, the set of integers, the set of rational numbers, and the set of real
numbers, are all infinite sets.
If S is a finite set, then its size is uniquely determined by the number n of elements that
it has. We write |S|=n if a set has exactly n elements. The notation |S| denotes the size
SU5-11
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Set Theory
of the set S. A finite set can have only one size. It is not possible, for instance, for a set to
contain 6 elements, and at the same time, 10 elements.
Activity 5.1
List the members of these sets.
a.
b.
c.
Activity 5.2
Suppose
that
,
,
and
. Determine which of these sets are subsets of which other of these
sets.
Activity 5.3
For each of the following sets, determine whether 2 and whether
of that set. Determine also whether
is a subset of that set.
SU5-12
is an element
MTH105
Set Theory
a.
b.
c.
d.
a prime number
e.
f.
g.
h.
Activity 5.4
Write down the sizes of the following sets:
a.
b.
c.
d.
e.
f.
g.
h.
Activity 5.5
Let
SU5-13
MTH105
Set Theory
Determine which of these sets are subsets of which other of these sets.
Activity 5.6
Draw a Venn diagram to illustrate that
.
SU5-14
MTH105
Set Theory
Chapter 2: Set Theoretic Operations
2.1 Interval Notation
Let
be real numbers with
. Define:
The round bracket appearing as either ( or ) denotes that the set does not contain the
end point or the end point respectively. Hence, a round bracket means excluding that
particular end point.
The square bracket appearing as either [ or ] denotes that the set contains the end point
or the end point respectively. Hence a square bracket means inclusive of that particular
end point.
We call
the closed interval from to . We call
the open interval from to .
The other two intervals with one end closed and the other end open are called half-open
intervals. This nomenclature does not really specify which is the open half. The symbols
and
are used to indicate intervals that are unbounded either on the right or on the
left:
Examples
SU5-15
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Set Theory
We have
. We also have
and
. However,
element of
is not a subset of
but 1.01 is an element of
. For example, 1.01 is not an
.
2.2 Unions, Intersections, Complements
Definition
Let
be sets, and let
The union of
denote some universal set.
and , denoted by
(read this as " union
") is the set containing all
the elements in either , or , or both. Note that in mathematics, when we use the phrase
"
or ", we mean " or or both".
The intersection of
and , denoted by
(read this as " intersect
") is the set
containing all the elements that can be found in both and .
is some universal set containing all the sets under consideration, the complement of
If
in
set
denoted either by
) is the set of elements found in
The difference of
Note that
Also,
(if we wish to suppress mention of the universal
but outside , in other words, not in .
minus , or the relative complement of
is simply the relative complement of
in
in , is the set
, where
is some universal set.
.
The symmetric difference of
or
or by
, denoted by
but not both.
SU5-16
, is the set of elements found in either
MTH105
Set Theory
If we use the phrase "
or ", we means " or or both". Thus, to specify "
but
, we use the term "symmetric difference".
Figure 5.1 Depiction of
,
,
, and
using Venn diagrams
Activity 5.7
Let the universal set be the set
and let
and
find
. Find
,
,
, and
. Also
.
Definition
Sets
Let
are said to be disjoint if
, that is, if
be sets. We say that the collection of sets
or pairwise disjoint if
SU5-17
is the empty set.
is mutually disjoint
MTH105
Set Theory
for any pair of subscripts (commonly called indices)
distinct pair of
with
. In other words, every
are disjoint.
Examples
The sets
are pairwise disjoint.
However, the sets
are not pairwise disjoint because the
first and the second sets have an element (the number 1) in common.
2.3 Arbitrarily Indexed Collections of Sets
We now consider unions and intersections not just of two sets, but of arbitrarily indexed
collections of sets.
Let
be some non-empty set which we shall call an indexing set. Suppose that for all
, there is a uniquely defined set
. For each pair of sets
,
, they may be disjoint,
or identical, or they may intersect partially. There does not have to be any restriction on
the relationship between each pair of sets, unless specified by some mathematical context
under consideration (which will vary depending on what we are discussing).
Let
be some universal set. We define:
A sequence of sets is an ordered collection of sets
of the collection of sets
In this case, we can think
as being indexed by the indexing set
natural numbers. For any positive integer , we have:
SU5-18
, the set of
MTH105
Set Theory
Activity 5.8
Let
be an infinite sequence of sets (indexed by the natural numbers).
Determine the following sets:
a.
b.
c.
d.
e.
f.
2.4 Partitions of Sets
Definition
be a collection of non-empty sets indexed by a set . Let
Let
such that
for all
and the sets
with
forms a partition of
. We say that
are pairwise disjoint, that is,
.
SU5-19
be a set
if:
for any pair of indices
MTH105
Set Theory
Example
The sets
partition of
, or
,
,
form a
. By the Division Property, every integer can be expressed in the form
, or
, and furthermore, the representation is unique because the remainder of
any integer upon division by is unique. Thus, {
forms a partition of
.
2.5 Power Sets
Definition
Let
be a set. The power set of , denoted by
, is defined to be the set of all subsets
of . In other words,
The power set definition
is one of the rare instances in
mathematics where we have special permission to define a set in the form
without specifying a universal set
. This is because the existence of the power set of any
given set is one of the axioms in Axiomatic Set Theory, a branch of mathematics that lays
the set-theoretic foundations of mathematics (upon which all mathematics is built).
For instance,
The power set of the empty set is
itself. The power set of any set
. This is because the empty set
always contains the empty set
as can be seen in the above examples. This is because
SU5-20
is a subset of
as one of its elements,
for any set . Furthermore,
MTH105
Set Theory
for any set , we have
, that is, every set is a subset of itself. Hence
for any
set .
Activity 5.9
Let
be a set of size . Prove that the size of
is
.
2.6 Cartesian Products of Sets
An -tuple is defined to be an ordered array of elements written in the form
where we assume that the elements
universal set
if and only if
indicated above all belong to some
. We say that
,
,...,
. In other words, two -tuples are equal if and only if
their corresponding component entries are equal. Thus, for instance,
if and only if
Let
.
be sets. The Cartesian product
, where
and
Note that in general,
Note that if one of the sets
is defined to be the set of all 2-tuples
.
. Similarly, for sets
,
is the empty set ∅, then
the empty set.
SU5-21
is also
MTH105
Set Theory
Example
Let
The
,
, and
-fold Cartesian product of a set
Cartesian product
. Then
with itself, denoted by
, is defined to be the
, where there are terms appearing in the product. Thus,
SU5-22
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Set Theory
Chapter 3: Properties of Sets
3.1 Set Identities and Relations
The following table consists of set identities, which are equations that are true for all sets
in some universal set. In this table, let
denote a universal set to which all sets belong.
Table 5.1 An important collection of set identities
Commutative
Laws
Associative
Laws
Distributive
Laws
Identity Laws
Complement
Laws
Double
Complement
Law
Idempotent
Laws
Universal
Bound Laws
De Morgan’s
Laws
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Set Theory
Absorption
Laws
Complements
of
and
Difference
Law or
Relative
Complement
Law
Symmetric
Difference
Law
Compare the above table of set identities to the following table of logical equivalence (in
the context of Proposition Logic) that was introduced in Study Unit 1:
Table 5.2 An important collection of logical equivalences
Commutative
Laws
Associative
Laws
Distributive
Laws
Identity Laws
Negation
Laws
SU5-24
MTH105
Set Theory
Double
Negation Law
Idempotent
Laws
Universal
Bound Laws
De Morgan’s
Laws
Absorption
Laws
Negations of t
and c
There is a clear parallel between the two tables. The intersection
to the conjunction
disjunction
of two statements. The union
of two sets corresponds
of two sets corresponds to the
of two statements. The complement of a set corresponds to the negation of
a statement. The universal set
corresponds to a tautology. The empty set corresponds
to a contradiction.
We also have the following properties of sets involving the subset relation:
Inclusion of Intersection:
For all sets
and
Inclusion in Union:
For all sets
and
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Set Theory
Reflexive Property of Subsets:
For all sets , we have
Anti-Symmetric Property of Subsets:
For all sets
if
and
, then
Transitive Property of Subsets:
For all sets
if
and
, then
Equality of Union and Intersection:
Let
be sets. Then the following statements are equivalent:
i.
ii.
iii.
The Associative Laws and De Morgan's Laws for sets that are listed in Table 5.1 can also
be generalised to arbitrary unions and intersections:
3.2 Proving Set Identities and Relations
We will introduce two ways of proving set identities and relations. The first is by the
element method, and the second is using a chain of set identities listed in Table 5.1.
Recall that two sets
are equal if and only if
and
. Hence, to show that
,
we will pick an arbitrary element in and show that it is in , and we will pick an arbitrary
element in
and show that it is in . This is the element method.
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Example 1
Prove using the element method that
.
Answer:
We first prove that
.
Suppose that
If
. Then
, then we have
or
and
On the other hand, if
. Thus,
, then
Hence,
.
and
.
. Thus,
and
.
We have therefore proven that
.
We now prove that
.
Suppose that
and also, (
If
. Then
or
and
. Hence, (
or
),
).
, then
. If is not in , then must be an element of
element of . In this case,
. Thus,
.
We have therefore proven that
.
Since
and
that
and also an
we conclude
. The proof is complete.
Example 2
Prove using the element method
.
Answer:
We first prove that
.
Suppose that
member of , or
. Then is not a member of
is not a member of . This is equivalent to saying that
Therefore,
. We have thus shown that
We now prove that
Suppose that
. It follows that either is not a
.
. Then
or
.
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.
or
.
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If
Set Theory
, then is not a member of . If
that either
is not a member of
, and so
or
, then is not a member of . Thus, we have
is not a member of . Thus,
. We have thus proven that
Since
and
is not a member of
.
, we can therefore conclude that
.
Remark
Notice how in Example 2 above, the proof of the reverse inclusion
was essentially the reverse of the proof of the inclusion
, repeated almost
word for word.
We can also prove that two sets are equal by employing a chain of set identities. Unless
otherwise stated, you are allowed to assume that all sets belong to some universal set
Example 3
Show by employing a chain of set identities that
Answer:
by De Morgan's Laws
again, by De Morgan's Laws
Example 4
Show that
.
Answer:
by Distributive Laws
by Complement Laws
by Identity Laws
Example 5
.
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.
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Answer:
by Difference Laws
by Distributive Laws
by Difference Laws
Certain subset relations can also be demonstrated using the element method.
Example 6
Prove
that
for
all
sets
Answer:
Suppose that
Let
.
. Then
we have
and
. Therefore,
previously derived that
. Since
, we also have
is not a member of
. And since
,
. This is a contradiction, because we
. Hence, we conclude that
Example 7
Let
be non-empty sets. Prove that if
Answer:
Suppose that
Let
Since
.
. We can express as
, we have
. Since
for some elements
, we have
conclude therefore that
.
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. Hence,
and
.
. We
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Activity 5.10
Prove the converse of Example 7 by showing that if
that
, then
are non-empty sets such
.
Activity 5.11
Prove that
for any sets
using (i) the element method,
and (ii) using a chain of set identities.
3.3 Disproving an Alleged Set Property
Suppose we wish to prove that an alleged set property is false. Set properties are stated in
the form of a universal statement:
or
Thus, to prove that an alleged set property is false, we only need to come up with a
counter-example.
Example
Disprove the alleged set property
.
Answer:
Let
,
,
. Then
,
, it is therefore untrue that
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, and
. Since
.
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Activity 5.12
Disprove the following alleged set property:
For all sets , , and , if
is not a subset of , and
a subset of .
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is not a subset of , then
is not
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Chapter 4: Counting and Probability
This chapter is based on Living with Mathematics by Leong (2011).
4.1 Two Basic Counting Principles
In our everyday lives, we are often faced with the need to count, or enumerate objects, and
also determine how many ways they can be arranged. Examples of counting problems
that we might face include:
• Determining how many ways there are to arrange 5 boys and 7 girls in a row, so
that no two girls are adjacent;
• Determining the number of ways of dividing a group of 9 people into three groups
comprising of 4,3, and 2 people respectively, if there is no need to arrange the people
within each group.
In this section, we state two fundamental principles of counting.
The Addition Principle (AP)
Let be a positive integer. Assume that there are:
to occur are pairwise disjoint, then the
If the ways for the different events
number of ways for at least one of the events
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to occur is
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Example:
Suppose that we can travel from city A to city B by sea, air and road. There are 4 ways to
travel by sea, 5 ways to travel by air, and 8 ways to travel by road, and these methods of
travel are mutually exclusive. Then there would be a total of
ways to travel from city A to city B.
In set-theoretic terms, the Addition Principle can be stated as follows:
The Addition Principle (AP) – Set Theoretic Version
Let
be a positive integer. Assume that
that is,
for
are sets which are pairwise disjoint,
Then
The second basic counting principle is the Multiplication Principle.
The Multiplication Principle (MP)
Let
be a positive integer. Suppose that an event
events
, and that there are:
Then the total number of ways for event
to occur is
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can be decomposed into
ordered
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Example:
Suppose that there are 8 ways to travel from city A to city B, 4 ways to travel from city B
to city C, and 9 ways to travel from city C to city D. Then the number of journeys that take
you from A to B to C to D in this order
A–B–C–D
is equal to
8 * 4 * 9 = 288
In set-theoretic terms, the Multiplication Principle can be stated as follows:
The Multiplication Principle (MP) – Set Theoretic Version
Let be a positive integer. Assume that
are sets. Then
Example:
Suppose that
is a positive integer, and that
is the prime factorization of . Each prime power
namely,
has exactly
positive divisors,
. Therefore, the total number of positive divisors of is
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Example:
is called a k-ary sequence if each number
An (ordered) sequence of numbers
is some non-negative integer from 0 to
inclusive (repetition of numbers in the
sequence is allowed). The total number of k-ary sequences of length is equal to
Very often, solving a more complicated counting problem involves the use of both the
Addition Principle and the Multiplication Principle together.
Example:
Let
Our task is to find
and let
.
The problem may be divided into disjoint cases by considering the possible values of
, that is,
. For each possible value of
possible values of , and also
possible values of . Therefore, for each value of
, there are, by the Multiplication Principle,
satisfy the required criteria
in the stated range, there are
possible ordered pairs
that
. Then by the Addition Principle,
4.2 Permutations
Let
be a set of distinct objects. An r-permutation of is an ordered
sequence of r elements of :
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where
Set Theory
are distinct integers chosen from the index set
.
To paraphrase the above concept, an -permutation of is a collection of distinct elements
of
arranged in a particular order. The word permutation indicates that the ordering of
the elements is important. We can think of a permutation of objects as an arrangement
of the objects in a row. Note that this is to be distinguished from the idea of a circular
permutation, which is the arrangement of objects in a circle. In this syllabus, we will not
be discussing circular permutations. Hence, the reader should always assume that when
we talk about permutations, it always refers to arrangements in a row.
Let
denote the number of -permutations of a set
We can determine the value of
We have to choose
of distinct elements.
in the following manner:
distinct objects from the set
and then arrange them in a row. The
arrangement can be achieved as follows: Place boxes in a row as shown below, and then
put one object from
into each box, going from left to right.
In the first box starting from the left, we can pick any of the
objects from . Thus there
are choices of objects that can be placed in the first box.
After the first box is filled, we go on to the second box. Since one object from the set
has been used, there are
objects remaining to choose from, and thus, there are
choices of objects that can be put into the second box.
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After the first and second boxes are filled, we go on to the third box. Since two objects
from the set
have been used, there are
there are
choices of objects that can be put into the third box.
Similarly, there are
There are
objects remaining to choose from, and thus,
choices of objects that can be put into the fourth box.
choices of objects that can be put into the
And finally, there are
th box.
choices of objects that can be put into the th box, which is
the last box on the right.
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By the Multiplication Principle, the number of -permutations of the set
objects, that is, the value of
of
distinct
, is given by
The reader is reminded that for any positive integer ,
(pronounced factorial) is defined
to be
and 0! is defined to be 1.
Special Case:
If
is a set of distinct objects, then the number of -permutations of
is given by
Therefore, n! is the number of ways of arranging n distinct objects in a row. Given a nonempty set , an -permutation of will be simply called a permutation of . The set of all
permutations of
is simply the set of all possible arrangements of the elements of
in a
row.
Example:
Find the number of 5-letter codewords that can be formed from the English alphabet (all
letters are assumed to be lowercase) such that the first and last letters are distinct vowels,
and the middle 3 letters are distinct consonants.
Answer:
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There are 5 vowels and 21 consonants to choose from. We can construct a 5-letter codeword
of the required properties by first choosing 2 distinct vowels and placing them at the first
and last positions, and then choosing 3 distinct consonants and then arranging them in
the middle three positions. In other words, the construction is equivalent to choosing a
2-permutation of the 5 vowels, and then choosing a 3-permutation of the 21 consonants.
Hence, the number of valid codewords is
Note that in the construction of a codeword, the arrangement of the letters is important.
Example:
There are 7 boys and 3 girls in a gathering. How many ways can they be arranged so that:
i.
the 3 girls form a single block (that is, there is no boy between any two of the
girls)?
ii.
no two girls are adjacent to each other
Answer:
i.
Since the 3 girls are together, we can treat them as a single entity. Inclusive
of the boys, there are therefore 8 entities which are to be arranged in a row.
The number of ways of doing so is . The 3 girls can also be arranged amongst
themselves. Hence, the total number of permutations with the desired properties
is
ii.
We first arrange the boys in a row. The number of ways of doing so is 7!. Now,
fixing the arrangement of the boys, we place the girls into some of the empty gaps
between each boy, inclusive of the possibility of placing them at the start and at
the end of the sequence. There are 8 empty slots that each girl can be placed into,
with at most one girl per slot. Since there are 3 girls present, the number of ways
of placing the girls is equal to
(we are choosing 3 of the slots to be occupied,
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and then arranging one girl into each of the 3 slots with order taken into account).
Hence, the total number of arrangements with the desired properties is
Example:
How many even integers are there between 2000 and 7000 such that no digit is repeated?
Answer:
We exclude the case 7000 because no digit is allowed to be repeated. Let abcde denote an
integer with the required properties. The first digit "a" can be 2,3,4,5 or 6. The last digit "e"
can be 0,2,4,6 or 8. This is illustrated in the following diagram:
Here, it is important to note that there is some overlap in the choices for the first digit "a"
and the choices for the last digit "e". The digits in common are:
We have to treat the cases 2,4,6 separately from the cases 3,5 because it is required that in
our number, no digit can be repeated.
Case (A): Assume that the first digit "a" is either 2,4, or 6. Then due to the rule that no digit
is allowed to be repeated, the last digit "e" must be one of the digits 0,2,4,6,8 that have not
been used by "a". In this case, there are 3 choices for "a" and 4 choices for "e". The digits
"b", "c", and "d" can be other digits apart from the digits that have already been used for
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"a" and "e". Hence by the Multiplication Principle, the number of permutations is given
by
Case (B): Assume that the first digit "a" is either 3 or 5. Then the last digit "e" can be
0,2,4,6 or 8. Again, the digits "b", "c", and "d" can be other digits apart from the digits that
have already been used for "a" and "e". By the Multiplication Principle, the number of
permutations is given by
Since case (A) and case (B) are mutually exclusive, so by the Addition Principle, there are
4032 + 3360 = 7392 possibilities.
4.3 Combinations
Let
be a set of
distinct objects. An r-combination of
is an
unordered subcollection of elements of :
where
are distinct integers chosen from the index set
. In a
combination, also known as a selection, the ordering of the elements is not taken into
consideration.
We can derive the formula for the number of -combinations of elements of as follows:
Assuming that
permutations of
is a set of
distinct objects, the number of
-
is given by
We can construct an -permutation in the following way:
a.
Choose
elements of without taking order into account. In other words, first
select an -combination of , say,
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.
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b.
Set Theory
After part (a) has been done, arrange the elements
If we denote the number of -combination of by
in a row.
, then by the Multiplication Principle,
we have
Hence,
An alternative way of writing
for any
. The term
is
. We have
is also known as a binomial coefficient, because these terms
appear in the binomial expansion of
, namely,
Example:
How many ways can a committee of 5 be formed from a group of 11 people comprising
of 4 teachers and 7 students if
i.
There is no restriction in the selection?
ii.
The committe must include exactly 2 teachers?
iii.
The committe includes at most 2 students?
iv.
A particular student and a particular teacher cannot both be in the committee?
Answer:
i.
There are
ways of choosing 5 people out of 11 without any restriction.
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ii.
If there are exactly 2 teachers chosen, then there are exactly 3 students chosen.
There are
ways of choosing 2 teachers, and
ways of choosing 3 students.
Hence by the Multiplication Principle, there are
combinations in which there are exactly 2 teachers chosen.
iii.
To include at most two students means that 0, 1, or 2 students may be chosen.
Now, there cannot be 0 students chosen because there would not be enough
teachers to make up a committee of 5 people. If there is 1 student chosen, then it
means all 4 teachers are also chosen. The number of ways of achieving this is
If there are 2 students chosen, then 3 teachers are also chosen. The number of
ways of doing so is
By the Addition Principle, the number of ways is 7 + 84 = 91.
In order to answer part (iv) of this question, we introduce another basic principle of
counting known as the Complementation Principle.
The Complementation Principle (CP)
Suppose that
where
is a subset of , and
is the complement of
is a finite set. Then
in , that is, the set of elements in
that are not in .
The combinatorial interpretation of CP is as follows: Suppose that there are ways for an
event
to occur, and
ways for an event
to occur, and suppose that whenever occurs,
also occurs. Then the number of ways for
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to occur without
occurring is
.
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iv.
Let the particular student be denoted by and the particular teacher be denoted
by . We first count the number of ways in which both and are chosen for the
committee. Since and
have to be in the committee, we only need to choose 3
people out of the remaining 9 people. There are
ways of doing so. The number
of ways of constructing the committee without any restriction is
, as stated in
part (i). Hence by the Complementation Principle, the number of ways in which
,
are not both inside the committee together is
In this question, we could have avoided using CP by counting the number of
ways of excluding either and from being in the committee, as explained below:
Let
and
be sets. Then
In combinatorial terms, this means that if we want to determine the number of ways for
either event
or event
to occur, we calculate:
(number of ways for event
of ways for event
and
to occur) + (number of ways for event
to occur) – (number
to occur together)
Let us apply the above concept to answer part (iv). We observe that there are
ways for student to be excluded from the committee,
be excluded, and
ways for teacher to
ways for both of them to be excluded. Hence, the number
of ways for , to not be in the committee together, which is equal to the number
of ways for either or
to be excluded from the committee, is given by
Example:
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i.
Determine the number of ways of dividing a group of 9 people into three groups
comprising of 4,3, and 2 people respectively, if there is no need to arrange the
people within each group.
ii.
Determine the number of ways of dividing a group of 10 people into three groups
comprising of 4,3, and 3 people respectively, if there is no need to arrange the
people within each group.
iii.
Determine the number of ways of dividing a group of 9 people into three groups
each comprising of 3 people, if there is no need to arrange the people within each
group.
In this question, assume that the groups are unlabelled.
Answer:
i.
We first choose 4 of the people to be placed in one group, and then choose 3 of
the remaining 5 people to be placed in another group. After that, the remaining 2
people will be automatically placed in a third group, and there is obviously only
one way that can happen. Hence, the number of possible groupings is
ii.
Since two of the groups have the same number of people, it is helpful to first
suppose that the groups are labelled A,B,C, where group A is intended to contain
4 people, and groups B,C are intended have 3 people each. We choose 4 people
out of the 10 people to be placed in group A, and then choose 3 people out of
the remaining 6 people to be placed in group B. The remaining 3 people are then
automatically assigned to group C. The number of ways this can be done is
In the question, the groups are unlabelled. When we remove the labels A,B,C,
we observe that we end up with the same groupings even if those people
earlier assigned to C had instead been placed in B, and those assigned to B
had instead been placed in C. In other words, the labels B,C could have been
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Set Theory
interchanged without affecting the final (unlabelled) groupings. The number of
ways of forming unlabelled groups comprising 4,3,3 people is therefore
In this instance, dividing by 2 ! removes the overcounting that would have
occurred if we had simply removed the labels without accounting for the fact
that two of the groups each have the same number of people. We are dividing by
the number of permutations of the labels B,C, which is 2 !.
iii.
Again, we suppose that the groups are labelled A,B,C. The number of ways of
assigning 3 people to each of the labelled groups is
Since all three groups have the same number of people, the number of ways of
performing the same grouping with unlabelled groups is
Here, we are dividing by the number of permutations of the labels A,B,C, which
is 3! .
Reflect 5.1
In part (i) of the above example, why was there no need to divide the answer 1260 by
the number of arrangements of the group labels, as we did in parts (ii) and (iii)?
4.4 Probability
If we toss a coin and take note of what appears on the uppermost face (the side of the
coin visible to us), the result or the outcome is either a head (H) or a tail (T). If we toss
an ordinary 6-sided die and observe the number that shows up on top, then the possible
outcomes are 1,2,3,4,5 or 6.
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In both of these examples, we are performing what are termed random experiments.
Usually in such experiments, the outcomes cannot be predicted with certainty. We
therefore need the notion of probability to as a way of measuring the likelihood or chance
of each event happening.
In a random experiment, the set of all possible outcomes is called the sample space. In
this section, we will denote the sample space of a random experiment by .
The concept of sample space is similar to the concept of the universal set we had earlier
when discussing set theory. Any subset of the sample space is called an event. The sample
space
itself is an event. The empty set
is also an event, because
is a subset of any
set .
These are examples of random experiments and their sample spaces.
Random
Sample Space
An example of an event
Experiment
Toss an unbiased
At least one
coin twice
head is obtained:
Observe a
The player we observe
particular chess
has drawn the game:
player's result at
the end of a game
Throw an ordinary
The sum of the scores
6-sided die twice
observed exceeds 10:
Tell a group of
3 students
is the set of all permutations of
the letters
, in other words,
to stand in a row
and observe the
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Student a stands in front:
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Set Theory
Random
Sample Space
An example of an event
Experiment
order in which
they stand
There are 5 green
Balls of the same colour are
balls (G) and 6
drawn:
yellow balls (Y)
in a bag. Balls of
the same colour
are regarded as
indistinguishable.
Draw two balls at
random without
replacement.
The Classical Definition of Probability
Suppose that
is a non-empty, finite sample space. Let
a subset of . If each outcome in
happening, denoted by
be an event, that is, let
be
is equally likely, then the probability of the event
is defined to be
For example, if we throw an unbiased 6-sided die, then the outcomes 1,2,3,4,5,6 are equally
likely. The sample size is 6. The probability of obtaining a number that is at least 5 is
The probability of obtaining an even number is
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.
.
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Suppose that two unbiased 6-sided dies are thrown. The sample size comprises of all
ordered pairs
, where each of
are positive integers between 1 and 6 inclusive. Since
the dies are unbiased, each of the 36 outcomes are equally likely. Suppose that
is the
event that a total score of at least 10 is obtained. Then
The classical definition of probability tells us that if is a non-empty, finite sample space,
then
If
always, and
always, where
are two subsets of the sample space , and
however, is not necessarily true.
denotes the empty set.
, then
. The converse,
does not necessarily imply that
. Can
you come up with a simple example of this phenomenon?
If
is a sample space, and
are two events, then:
•
is the event that both
occur.
•
is the event that either
•
is the event that
or
(or both) occur.
does not occur.
Example:
A deck has ten cards numbers 11 to 20 inclusive. A card is picked at random. Define events
as follows:
: the number is divisible by 5.
: the number is divisible by 3.
: the number is prime.
Then
because none of the cards are prime and also divisible by 5.
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Some Properties and Results concerning Probability
Let
be a non-empty, finite sample space.
• For any event ,
.
• The total probability of all the outcomes is 1, that is,
• The probability of the empty set is 0, that is,
• For any event ,
• For any events
.
.
.
,
.
Mutually Exclusive Events:
We term events
mutually exclusive if
. More generally, we term events
pairwise mutually exclusive if
for
.
In our syllabus, we are restricting ourselves to finite sample spaces. In finite sample spaces,
two events are mutually exclusive if and only if their intersection is empty, that is,
mutually exclusive if and only if
If
are
.
are mutually exclusive, then
. More generally, if
are pairwise mutually exclusive, then
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Independent Events:
We term
If
independent events if
.
are independent events, then
.
Example:
Suppose that two fair die are thrown and the number that appears on top of each die is
recorded. Assume that the scores on each die are independent of each other.
Let
be the event that the total score of the two die is at most 5.
Let
be the event that both die are odd.
Let
be the event that one die is even and one die is odd.
Then
Since the scores on each die are independent of each other,
On the other hand,
It is important to note that in the calculation of
, we have to add up two mutually
exclusive cases: (i) first die even and second die odd, (ii) first die odd and second die
even. In this calculation, we are assuming that the two die are distinct. In other words, we
assume that the die are labelled, say, as X,Y. If we do not assume that the die are distinct,
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then the sample space will be more difficult to determine, and probabilities harder to
calculate.
For the calculation of
, there was only one case to consider due to the symmetry of
the situation: To say that die X is even and die Y is even is exactly the same as saying that
die Y is even and die X is even.
We have
.
Since
, the events
Also, the events
are not independent.
are mutually exclusive: It is impossible for both die to be odd, and at
the same time, for one to be even and the other odd.
If two events are mutually exclusive and their respective probabilities are non-zero, then
the two events cannot be independent.
Example:
A bag contains 6 white balls, 7 black balls, and 7 green balls. Balls of the same colour are
considered indistinguishable.
i.
If 3 balls are drawn without replacement, find the probability that all are the
same colour.
ii.
If 5 balls are drawn without replacement, find the probability that 2 are white
and 3 are black.
iii.
If 3 balls are drawn without replacement, find the probability that they are of
different colours.
iv.
If 3 balls are drawn with replacement, find the probability that they are of
different colours.
Answer:
i.
The number of ways of selecting 3 white balls out of the 6 available white balls
is
. The number of ways of selecting 3 black balls out of the 7 available black
balls is
. The number of ways of selecting 3 green balls out of the 7 available
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green balls is
. Hence by the Addition Principle, the total number of ways of
selecting 3 balls of different colours out of the 20 available balls is
Since the total number of ways of selecting 3 balls out of 20 without restriction
is given by
, by the classical definition of probability, the probability
that 3 balls drawn without replacement are of the same colour is
ii.
The number of ways of selecting 2 white balls out of the 6 available white balls
is
. The number of ways of selecting 3 black balls out of the 7 available black
balls is
. Hence, the probability of selecting 2 white balls and 3 black balls is
equal to
iii.
To get three balls of different colours, we must have one white, one black, and
one green. The probability of obtaining this is equal to
iv.
If balls are being drawn with replacement, the probability of obtaining a white
ball on each draw is constant at 6/20, the probability of obtaining a black ball on
each draw is constant at 7/20, and the probability of obtaining a green ball on
each draw is constant at 7/20.
Since the probabilities are constant, we prefer to use a multiplicative approach
rather than a combinatorial approach. Instead of counting the number of ways
of drawing 3 balls of different colours with replacement, and then determining
the proability by dividing this answer by the total number of ways of drawing
3 balls with replacement and without any restriction, we prefer to the following
more direct approach:
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The probability of obtaining 3 balls of different colours when 3 balls are drawn
is equal to
The reason for multiplying by the term 3! in the above working is because there
are ways in which the colours could have been drawn – there are 3! permutations
of the three colours: white, black, green. The probability of obtaining each of
the colour sequences is
.Hence, the total probability of obtaining three
different colours in any order is
Let
and
be two events that may occur simultaneously. Suppose that we know
happens, or that we can assume
is
.
will happen. The set of outcomes in which
, because the sample space has now been restricted to the set
also occurs
– in other words,
the set of all possible outcomes is now the set .
Hence, if we know that
occurs, the probability that
also occurs is equal to
In the above formula, we divide the cardinality of the set
set
because the sample space has been restricted to the set .
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by the cardinality of the
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The conditional probability that
occurs given that
occurs is denoted by
, and
is defined as
From the above formula, we can write
This result can be understood as follows: The probability that both
denoted by
that
Recall that we say that
has already occurred. Since
are independent events if
, then
only if
occur, which is
, is equal to the probability that occurs multiplied by the probability
occurs given we know
If
and
.
are independent if and only if
. In other words, two events
, we also have
, if and
with non-zero probability are
independent if and only if the chance of one event happening is not affected by whether
or not the other event happens.
Example:
A fair die and a fair disc are tossed. The fair disc has 2 dots on one face and 3 dots on the
other face. The score is defined to be the product of the number of dots on the die and
the disc. You can assume that the die and the disc are independent of each other. Find the
probability that
i.
the score is odd.
ii.
the score is at most 6.
iii.
the score is at most 6 given that it is odd.
iv.
the score is even given that it is at most 6.
Determine also whether the events "the score is odd" and "the score is at most 6" are
independent.
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Answer:
i.
The score is odd if and only if both the die and the disc display an odd number
of dots. We have P (score odd) = P (both die and disc are odd)
Since the die and the disc are independent,
P (both die and disc are odd) = P (die is odd) P (disc is odd)
Therefore, P (score odd) = P (both die and disc are odd)
= P (die is odd)P(disc is odd)
.
ii.
P (score is at most 6)
= P (die=1) + P (die=2) + P (die=3 and disc=2)
Since the die and the disc are independent,
P (die=3 and disc=2) = P (die=3) P (disc=2)
Hence,
P (score is at most 6)
= P (die=1) + P (die=2) + P (die=3)P (disc=2)
.
iii.
We use the formula for conditional probability.
P (score is at most 6 given score is odd)
= P (score is at most 6 and is odd) / P (score is odd)
The score is at most 6 as well as odd only when the die is 1 and the disc is 3.
Hence,
P (score is at most 6 and is odd)
= P (die=1 and disc=3)
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= P (die=1) P (disc=3) =
.
It follows that
P (score is at most 6 given score is odd)
.
iv.
P (score is even given it is at most 6)
= P (score is at even and at most 6) / P (score is at most 6)
The score is at most 6 as well as even only when the die is 1 and the disc is 2, or
the die is 2, or when the die is 3 and the disc is 2. Hence,
P (score is even and at most 6)
= P (die=1 and disc=2) + P (die=2) + P (die=3 and disc=2)
= P (die=1) P (disc=2) + P (die=2) + P (die=3) P (disc=2)
=
.
It follows that
P (score is even given it is at most 6)
.
The score is odd and is at most 6 only when the die is 1 and the disc is 3. Hence,
P (score is odd and is most 6) = P (die=1 and disc=3) = 1/12.
On the other hand,
P (score is odd) P (score is at most 6) =
.
Since P (score is odd and is at most 6)
is not equal to
P (score is odd) P (score is at most 6),
we conclude that the events "the score is odd" and "the score is at most 6" not independent.
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Example:
A bag contains 6 white balls, 7 black balls, and 7 green balls. Balls of the same colour are
considered indistinguishable.
i.
If 3 balls are drawn without replacement, find the probability that all green given
that all are the same colour.
ii.
If 3 balls are drawn with replacement, find the probability that all white given
that all are the same colour.
Answer:
i.
P (all balls are green and all are same colour) = P (all balls are green), because if
the three balls are all green, then they are definitely the same colour – the event
"all three balls are green" is a subset of the event "all three balls are the same
colour".
For drawing without replacement, P (all balls are green) =
.
P(all balls are same colour) =
Hence, P (all balls green given all balls same colour)
= P (all balls green and all same colour) / P (all balls same colour)
= P (all balls green) / P (all balls same colour)
= (7/228) / (3/38) = 7/18.
ii.
For drawing with replacement, P(all balls are white) =
P(all balls are same colour)
= P(all white) + P(all black) + P(all green)
=
= 451/4000.
Hence, for drawing with replacement,
P (all balls white given all balls same colour)
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= P (all balls white and all same colour) / P (all balls same colour)
= P (all balls white) / P (all balls same colour)
= (27/1000) / (451/4000) = 108/451.
Reflect 5.2
In part (ii) of the above example, when calculating the probability of drawing all white
balls with replacement, we did not have to multiply
by
. However, if we
calculate the probability of drawing 3 balls of different colours with replacement, the
answer is
.
Why did we not have to multiply by
when all three balls are white?
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Chapter 5: The Mathematics of Voting
This chapter is based on The Mathematics of Politics by Robinson and Ullman (2017).
5.1 Why study the mathematics of voting?
Very often, a group of people has to make a decision as to which leader to appoint, or
which course of action to take. However, being made up of fallible or contentious human
beings, it is possible that the group cannot be expected to unanimously decide on a single
outcome.
In such a situation, we would like to find a way to give everyone a fair voice in the final
decision. The most obvious solution is via polling. We ask everyone on their preferred
choice, and then make the decision based on which outcome is favoured by the majority.
In real life however, things are not that simple. There may be several candidates to
choose from, with no single candidate obtaining over 50 percent of the votes. Additionally,
different people may rank candidates differently.
The mathematics of voting is concerned with studying various methods of deciding the
final outcome based on a survey of the opinions or the preferences of the decision makers,
and devising criteria to determine how fair is a given method of determining election
outcomes.
5.2 Non-Preferential Voting
Non-preferential voting means that the voters do not rank the candidates in order of
preference. Voters only indicate which candidate or candidates they want to elect.
The approval voting method is the main example of non-preferential voting. In approval
voting, each voter specifies his desire, or his approval, for either a single candidate, or for
a few candidates. Each voter does not rank his choices in order of preference -- he is only
allowed to state which his choices are. There is no restriction as to how many candidates
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a voter is allowed to give his approval to. The winner of the election is the candidate with
the highest approval count, that is, the candidate which has received the most number of
approvals amongst all candidates.
Example:
In Table 5.3 below, the approvals received by each of four candidates A,B,C,D, are
indicated using ticks (✓).
Table 5.3 A schedule showing the approvals each candidate has received
Candidate
3 voters
A
1 voter
2 voters
✓
D
2 voters
✓
B
C
1 voter
✓
✓
✓
✓
✓
✓
3 voters have each given approval to candidate C, and to no other candidate.
1 voter has given approval to candidates A and D.
2 voters have each given approval to candidates B and C.
1 voter has given approval to candidate D, and to no other candidate.
2 voters have each given approval to candidates A and B.
There are a total of 9 voters. No other combination of approvals have been recorded.
In this example:
Candidate A has received 1+2=3 votes.
Candidate B has received 2+2=4 votes.
Candidate C has received 3+2=5 votes.
Candidate D has received 1+1=2 votes.
The winner is therefore candidate C.
If there is a tie between two or more candidates, some other method must be used to
decide the outcome.
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5.3 Preferential Voting
In preferential voting, each voter has to rank all the candidates in order of preference.
Each voters casts what is known as a preference ballot. If there are a total of candidates,
each voter has to rank the candidates from 1 to inclusive, with 1 being the most preferred
and
being the least preferred. After voting has been carried out, we then construct a
preference schedule, which is a table listing down each of the candidates and how they
have been ranked by the voters.
Example:
Table 5.4 A preference schedule showing the rankings each candidate has received
Candidate
3 voters
2 voters
4 voters
5 voters
2 voters
A
2
1
3
2
4
B
4
2
2
4
1
C
3
4
4
1
3
D
1
3
1
3
2
In the above example:
3 voters have ranked the candidates A,B,C,D as 2-4-3-1 respectively.
2 voters have ranked the candidates A,B,C,D as 1-2-4-3 respectively.
4 voters have ranked the candidates A,B,C,D as 3-2-4-1 respectively.
1 voter has ranked the candidates A,B,C,D as 2-4-1-3 respectively.
2 voters have ranked the candidates A,B,C,D as 4-1-3-2 respectively.
5.3.1 The Plurality Method
In the Plurality Method, the candidate with the most first-place votes wins.
Thus, in the above example, D is the winner with a total of 7 first-place votes.
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The preference schedule illustrated in Table 5.4 can also be presented as follows:
Table 5.5 A preference schedule showing the various ranking combinations
Rank
3 voters
2 voters
4 voters
5 voters
2 voters
First
D
A
D
C
B
A
B
B
A
D
C
D
A
D
C
B
C
C
B
A
Choice
Second
Choice
Third
Choice
Fourth
Choice
If there is a tie between two or more candidates with the most first-place votes, then some
other method must be used to determine the winner.
The Plurality Method is the simplest possible method of deciding the winner under
preferential voting. Under the Plurality Method, the winner may not have 50 percent or
more of the first-place votes, as the above example illustrates.
5.3.2 Two-Round System
From now on, we make the following assumptions concering preference ballots:
Preference Assumptions
i.
If a voter ranks candidate X higher than candidate Y, and the voter subsequently
has to vote again with some other candidates excluding X or Y removed from
the list of choices, the voter would continue to rank X higher than Y.
ii.
The order of preference is not changed if one or more of the candidates is
eliminated. Note that this is simply a restatement of point (i) above.
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Defintion
A candidate is said to have a majority of the votes if the candidate has obtained more than
50 percent of the votes.
Similarly, a candidate is said to have been ranked first-place by a majority of the voters if
over 50 percent of the voters had ranked the candidate as their first choice. The definition
can, of course, be easily extended to the second-place ranking, the third-place ranking,
and so on.
We now describe the two-round system. Preference ballots are used by voters to indicate
their ranking of all the candidates. If one candidate has received a majority of first-place
votes, that candidate is declared the winner.
Now suppose that no candidate has received a majority of first-place votes. In this case, we
eliminate all the candidates except the top two candidates with the largest number of firstplace votes. All ranking information pertaining to the eliminated candidates is removed,
and the ordering of the top two candidates by each voter is preserved in accordance with
the preference assumptions listed above.
Now, recount the first-place votes. The winner is determined using the pluarity method
of Section 5.3.1, provided there is no tie. If there is a tie, then another method has to be
used.
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Example:
Suppose that the following preference schedule is obtained after voting is completed.
Table 5.6 A preference schedule showing various ranking combinations
Rank
1 voter
2 voters
2 voters
4 voters
2 voters
First
D
A
D
C
B
A
B
B
A
D
C
D
A
D
C
B
C
C
B
A
Choice
Second
Choice
Third
Choice
Fourth
Choice
In the above preference schedule, no candidate has obtained a majority of the firstplace votes. Since the top two candidates with the largest number of first-place votes
are candidates C and D (with four top-place and three top-place votes respectively), we
eliminate all the other candidates (in this case, A and B) to obtain the following revised
preference schedule:
Table 5.7 Revised preference schedule
Rank
7 voters
4 voters
First Choice
D
C
Second Choice
C
D
The winner of the election by the two-round system is thus candidate D.
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In the above example, had we used the Plurality Method, candidate C would have been
the winner instead, because C was the candidate was the most number of first-place votes
in the original preference schedule.
5.3.3 Plurality with Elimination
The Plurality with Elimination method, or Pluraility with last-player Elimination method,
is executed as follows:
i.
If there is a candidate with a majority of first-place votes, that candidate wins.
ii.
If there is no candidate with a majority of first-place votes, then the candidate
or group of candidates with the least number of first-place votes is eliminated,
and the rankings of the remaining candidates redistributed according to the
preference assumptions listed above in Section 5.3.2.
iii.
Now, recount the first-place votes. If there is a candidate with a majority of firstplace votes, then he or she is the winner, else repeat step (ii).
iv.
If all the remaining candidates have the same number of first-place votes, then
some other method has to be used.
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Example:
Suppose that the following preference schedule is obtained after voting is completed:
Table 5.8 A preference schedule showing various ranking combinations
Rank
2 voters
2 voters
4 voters
8 voters
7 voters
First
D
A
D
C
B
A
B
B
A
D
C
D
A
D
C
B
C
C
B
A
Choice
Second
Choice
Third
Choice
Fourth
Choice
In this example, no candidate has a majority of the first-place votes. Hence, we eliminate
the candidate with the least number of first-place votes, namely, candidate A, to obtain
the following revised schedule:
Table 5.9 A revised preference schedule
Rank
2 voters
9 voters
4 voters
8 voters
First Choice
D
B
D
C
Second Choice
C
D
B
D
Third Choice
B
C
C
B
Again, there is no candidate with a majority of first-place votes. We eliminate candidate
D – the candidate with the fewst first-place votes in the revised schedule. The following
schedule is thus obtained:
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Table 5.10 A revised preference schedule
Rank
10 voters
13 voters
First Choice
C
B
Second Choice
B
C
The winner of the election is candidate B.
In the Plurality Method we only need to make one comparison. In the above example, C
would have been the winner.
In the two-round system, we only need to make at most two comparisons. Here, B would
be the winner, which is the same outcome as Plurality with Elimination. However, in the
Plurality with Elimination method, we might need to make three or more comparisons
depending on the preference schedules obtained.
5.3.4 Pairwise Comparison (or the Copeland Method)
In the method of Pairwise Comparison (or the Copeland Method), every candidate is
matched head-to-head against every other candidate. Each of these head-to-head matches
is called a pairwise comparison. If there are
candidates, then there are a total of
pairwise comparisons.
For each pairwise comparison made, the winner gets 1 point while the loser gets 0 points;
in the case of a tie, each candidate gets 1/2 point.
The winner of the entire election is the candidate with the most points after taking into
account all pairwise comparisons. If there is a tie in terms of the number of points obtained,
then some other method has to be used.
Example:
Suppose that the following preference schedule is obtained after voting is completed:
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Table 5.11 A preference schedule showing various ranking combinations
Rank
14 voters
10 voters
8 voters
4 voters
1 voter
First
A
C
D
B
C
B
B
C
D
D
C
D
B
C
B
D
A
A
A
A
Choice
Second
Choice
Third
Choice
Fourth
Choice
There are
pairwise comparisons that have to be made. There are 37 voters in total.
Table 5.12 Result of Pairwise Comparison
Comparison
Winner of Comparison
Number of votes in
favour of winner
A versus B
B
23
A versus C
C
23
A versus D
D
23
B versus C
C
19
B versus D
B
28
C versus D
C
25
C has won 3 comparisons, and thus gets 3 points. C is the winner of this election.
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Defintion
A Condorcet candidate is a candidate who wins every pairwise comparison with the other
candidates.
In Table 5.12, we can see that for this example, candidate C is a Condorcet candidate,
because C has won every pairwise comparison with the other candidates A,B,D.
Example:
Suppose that the following schedule is obtained:
Table 5.13 A preference schedule showing various ranking combinations
Rank
10 voters
13 voters
15 voters
First Choice
A
B
C
Second Choice
B
C
A
Third Choice
C
A
B
This is the corresponding table of pairwise comparisons:
Table 5.14 Result of Pairwise Comparison
Comparison
Winner of Comparison
Number of votes in
favour of winner
A versus B
A
25
A versus C
C
28
B versus C
B
23
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In this case, there is no Condorcet candidate. There is also no Pairwise Comparison winner,
because no candidate has won more pairwise comparisons compared to other candidates.
In this case, some other method has to be used to determine the outcome.
In the above example (Table 5.13), the winner using the Plurality Method is candidate
C, and the winner using either the Plurality Method with Elimination or the two-round
system is candidate B.
5.3.5 The Borda Count Method
In the Borda Count Method, each candidate gets 1 point for each first-place vote received,
2 points for each second-place vote, etc., all the way up to points for each last-place vote,
where is the total number of candidates. The candidate with the smallest total wins the
election.
Example:
Table 5.15 A preference schedule showing various ranking combinations
Rank
10 voters
15 voters
25 voters
10 voters
First Choice
A
B
C
A
Second Choice
B
C
A
C
Third Choice
C
A
B
B
In this example:
Candidate A has a Borda Count of 10 + 15(3) + 25(2) + 10 = 115.
Candidate B has a Borda Count of 10(2) + 15 + 25(3) + 10(3) = 140.
Candidate C has a Borda Count of 10(3) + 15(2) + 25(1) + 10(2) = 105.
Hence, candidate C is the Borda Count winner.
In the above example, the winner using the Plurality Method as well as the Plurality
Method with Elimination is also candidate C.
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5.4 Fairness criteria for various methods of determining election
outcomes under preferential voting
There are four criteria with which we can measure the fairness of a method for determining
election outcomes.
Defintion
i.
A method of determining an election outcome is said to satisfy the majority
criterion if whenever a candidate has more than 50 percent of the first-place
votes, then the method would always name that candidate as the winner of the
election.
ii.
A method of determining an election outcome is said to satisfy the Condorcet
criterion if whenever a candidate is a Condorcet candidate, the method would
always name that candidate as the winner of the election.
iii.
A method of determining an election outcome is said to satisfy the Monotonicity
criterion if the following condition is satisfied: Suppose an election is held and
candidate X is the winner under the given method. Suppose that now, the voters
are allowed to cast their votes again, and they do so in a way such that each voter
gives X the same or higher rank compared to the previous preference ballot. Then
X will remain the winner in the revote.
iv.
A method of determining an election outcome is said to satisfy the Independence
of Irrelevant Alternatives criterion if the following condition is satisfied:
Suppose that X is the winner of the election under the given method. Suppose
that now, one or more of the losing candidates is eliminated, and the rankings
preserved in accordance with the preference assumptions stated in Section 5.3.2.
Then X will remain the winner in the recount.
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Table 5.16 A tabulation of the various fairness criteria satisfied by each method of determining an election
result
Method
Majority
Condorcet
Monotonicity
Independence
Criterion
Criterion
Criterion
of Irrelevant
Alternatives
Criterion
Plurality
Satisfies
Violates
Satisfies
Violates
Two-Round
Satisfies
Violates
Violates
Violates
Satisfies
Violates
Violates
Violates
Satisfies
Satisfies
Satisfies
Violates
Violates
Violates
Satisfies
Violates
system
Plurality with
elimination
Pairwise
comparison
Borda Count
Example:
We give an example to show that the Borda Count Method violates the majority criterion
as well as the Condorcet criterion. Consider the preference schedule below:
Table 5.17 A preference schedule showing various ranking combinations
Rank
10 voters
12 voters
20 voters
18 voters
First Choice
A
C
C
A
Second Choice
C
B
A
B
Third Choice
B
A
B
C
In this example:
Candidate A has a Borda Count of 10 + 12(3) + 20(2) + 18 = 104.
Candidate B has a Borda Count of 10(3) + 12(2) + 20(3) + 18(2) = 150.
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Candidate C has a Borda Count of 10(2) + 12 + 20 + 18(3) = 106.
Candidate A is the Borda Count winner, but clearly, C has obtained a majority of the
first-place votes (32 out of 60 first-place votes). Thus, it can be seen that the Borda Count
Method violates the majority criterion in this particular example.
This is the corresponding table of pairwise comparisons for the above example.
Table 5.18 Result of Pairwise Comparison
Comparison
Winner of Comparison
Number of votes in
favour of winner
A versus B
A
48
A versus C
C
32
B versus C
C
42
In this case, C is the Condorcet candidate, but C does not win based on the Borda Count
Method. Thus, it can be seen that the Borda Count Method violates the Condorcet criterion
in this specific example.
Note that in order to show that a particular voting method violates a given criterion, we
only have to come up with a counter-example, that is, we only have to produce a specific
example where using the given voting method does not produce the winner required by
the given criterion.
Example:
We give examples where the Plurality with Elimination Method violates the Condorcet
criterion as well as the Monotonicity criterion. Suppose that the following preference
schedule has been obtained:
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Table 5.19 A preference schedule showing various ranking combinations
Rank
10 voters
7 voters
5 voters
5 voters
4 voters
First
A
D
B
C
B
C
B
C
D
C
B
A
A
A
D
D
C
D
B
A
Choice
Second
Choice
Third
Choice
Fourth
Choice
This is the corresponding table of pairwise comparisons for the above example.
Table 5.20 Result of Pairwise Comparison
Comparison
Winner of Comparison
Number of votes in
favour of winner
A versus B
B
16
A versus C
A
17
A versus D
D
16
B versus C
B
16
B versus D
B
19
C versus D
C
20
In this example, B is the Condorcet candidate. However, the Plurality with Elimination
Method would produce D as the winner. Hence, in this example, the Plurality with
Elimination Method violates the Condorcet criterion.
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Suppose now that a re-vote is taken, and voters cast their preference ballots according to
the following schedule:
Table 5.21 A preference schedule showing various ranking combinations
Rank
15 voters
7 voters
5 voters
4 voters
First Choice
A
D
C
A
Second Choice
C
B
D
C
Third Choice
B
A
A
D
Fourth Choice
D
C
B
B
The change is that each voter in the 3rd and 5th column of Table 5.19 interchange their
preferences for A and B. Each voter in the election has not changed their ranking for
candidate D. Yet, candidate D is no longer the winner under Plurality by Elimination –
the new winner is candidate A (who is also the winner by the Plurality Method). Thus,
Plurality with Elimination Method violates the Monotonicity criterion in this example.
The same example in Table 5.19 shows also that the Plurality Method violates the
Condorcet criterion – the Plurality Method would have produced A as the winner in Table
5.19, not the Condorcet candidate which is candidate B.
However, unlike Plurality with Elimination, the (simple) Plurality Method does not
violate the Monotonicity criterion.
Indeed in any re-election in which each voter alters his rankings in a manner that does not
disfavour the previous winner, there can only be equal or more first-place votes for the
candidate that already had most of the first-place votes in the original election. Therefore,
the Plurality Method satisfies the Monotonicity criterion.
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Reflect 5.3
The Independence of Irrelevant Alternatives criterion has been heavily criticized by
mathematicians for being too strict. Why is this so?
Can you devise a less strict replacement to the Independence of Irrelevant Alternatives
criterion so that at least one of the methods we have learnt in this chapter would satisfy
it?
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Set Theory
Summary
In this Study Unit, we formally introduced the concept of a set and defined various set
constructions such as subsets, power sets, the union, intersection and complement of sets,
and Cartesian products of sets.
We also used the element method and the method of establishing a chain of set identities
to prove various properties concerning sets.
We also used appropriate counter-examples to disprove certain alleged set properties.
In this Study Unit, we introduced two applications of the mathematical knowledge we
have acquired. The first application is Counting and Probability, in which we learnt how
to use counting principles like the Addition Principle, the Multiplication Principle, the
Complementation Principle, as well as the concept of permutations and combinations, to
tackle various counting problems. We also investigated the basic concepts of probability
and used the counting methods learnt to solve various questions in probability.
The second application is the Mathematics of Voting. We studied various ways of
determining the outcome of an election amd examined four different criteria to measure
the fairness of each method.
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Formative Assessment
1.
Let
,
, and
.
Prove or disprove each of the following statements:
2.
a.
.
b.
.
c.
.
Let
,
, and
.
Find each of the following:
a.
b.
c.
d.
e.
f.
g.
h.
3.
Let the universal set be the set R of all real numbers and let
,
, and
.
Write down each of the following sets using interval notation, and as simply as
possible:
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Set Theory
a.
b.
c.
d.
e.
f.
4.
Suppose that
and
. Find the following sets:
a.
b.
c.
5.
Prove that
by using (i) the element method, and
(ii) establishing a chain of set identities.
6.
Use
the
element
method
to
show
that
.
7.
Let
8.
In how many ways can a committee of 4 women and 5 men be formed from a group
be sets. Prove that
if and only if
.
of 7 women and 10 men?
The youngest of the women is Alice and the youngest of the men is James. If it is
decided that the committee must include at most one of Alice or James, in how many
ways can the committee be formed?
9.
Find the number of ways in which 4 girls and 5 boys can stand in a line under each
of the following conditions separately:
a.
No 2 girls may stand side by side.
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Set Theory
b.
All 4 girls must stand next to each other.
c.
The first and last positions are occupied by boys.
10. I want to use some of all the notes in my wallet to tip a waitress. How many possible
amounts can I tip her if I have five $2 notes, one $5 note, and three $10 notes?
11. How many non-empty subsets are there of a set of elements?
12. A man can construct his restraunt desert by choosing some or all of 4 different
flavours of chocolate fudge, some or all of 5 different types of buttered cream, and
some or all of 6 different varieties of nuts. How many different deserts are there?
13. Eight students are seated on a bench in a row. What is the probability that two of
them, Denise and Ellen, are seated side by side?
14. Two boxes are labelled A and B. Box A contains 3 blue discs and 2 white discs. Box B
contains 2 blue discs and 3 white discs. A random sample of 2 discs is drawn without
replacement from each box. What is the probability that all discs are the same colour?
15. Two boxes are labelled A and B. Box A contains 3 blue discs and 4 white discs. Box
B contains 2 blue discs and 5 white discs. First, a box is selected. We select Box A
with probability 0.3, and Box B with probability 0.7. Next, we pick two discs from our
chosen box without replacement. What is the probability that all discs are the same
colour?
16. The following preference schedule was obtained after an election.
Rank
11 voters
7 voters
7 voters
3 voters
9 voters
5 voters
First
A
D
C
C
B
D
B
B
A
A
C
C
Choice
Second
Choice
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Set Theory
Rank
11 voters
7 voters
7 voters
3 voters
9 voters
5 voters
Third
C
A
B
D
A
A
D
C
D
B
D
B
Choice
Fourth
Choice
Determine the winner under the Plurality Method, the two-round system, and the
Plurality Method with Elimination.
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Solutions or Suggested Answers
Activity 5.1
a.
Elements are
b.
Elements are
c.
Elements are
Activity 5.2
,
We have
,
,
, and
.
. These are the ONLY subset relations that exist.
Activity 5.3
a.
2 is an element of the set. Hence
is a subset of the set.
is not an element
of the set.
b.
2 is not an element of the set. Hence
is not a subset of the set.
is not an
is not a subset of the set.
is not an
element of the set.
c.
2 is not an element of the set. Hence
element of the set.
d.
is a prime number
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Set Theory
2 is an element of the set. Hence
is a subset of the set. {2} is not an element
of the set.
e.
2 is an element of the set. Hence {2} is a subset of the set. {2} is also an element of
the set. Hence, { {2} } is a subset of the set.
f.
is not an element of the set. Hence
element of the set. Hence,
is not a subset of the set.
is an
is a subset of the set.
g.
is not an element of the set. Hence
element of the set. Hence,
h.
is not a subset of the set.
is an
is a subset of the set.
{{{2}}}
is not an element of the set. Hence
element of the set.
Activity 5.4
a.
Size: 1
b.
Size: 1
c.
Size: 2
d.
Size: 3
e.
Size: 0
f.
Size: 1
SU5-84
is not a subset of the set.
is not an
MTH105
Set Theory
g.
Size: 2
h.
Size: 3
Activity 5.5
We have
,
,
. These are the ONLY subset relations that exist.
Activity 5.6
Here, U simply denotes some universal set containing
.
Activity 5.7
Universal set
;
SU5-85
;
.
MTH105
Set Theory
Activity 5.8
a.
b.
c.
(the set containing only the number 1)
d.
(the set containing only the number 1)
e.
f.
Activity 5.9
Suppose that a set
in
has elements. Consider a subset
or outside . We can assign an element
element
the number 0 if
of . Every member in
the number 1 if
. For a given subset
is either
, and assign the
of , the binary assignment of
1s and 0s to each
is uniquely determined by . Conversely, given any assignment of
1s and 0s to each
, we can construct a subset
of
that is completely and uniquely
determined by the given binary assignment. Hence, the total number of subsets
of
is equal to the total number of different possible binary assignments of 1s and 0s to the
members of . We can view each element
both). Thus, the number of subsets of
is
as having 2 choices (either 0 or 1, but not
.
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Activity 5.10
To prove: If
are non-empty sets such that
, then
.
Suppose that
Let
.
be given, and let
also be given. Then
, so
. This implies that
therefore that
. Since
and
. We conclude
.
Activity 5.11
To prove:
.
Element method:
Let
If
If
Then either
, then
or
.
.
, then is in
but not in . Since
, so
.
Hence we have shown that
.
Conversely, suppose that
If
. Then either
, then
Suppose that
or
.
.
. Then we must have
but not in . Hence,
because either
. We deduce that
We have shown that
or
is in
.
Thus,
Chain of set identities method:
by Difference Laws
by Distributive Laws
by Complement Laws
by Identity Laws
SU5-87
. Thus,
.
MTH105
Set Theory
Activity 5.12
Let
,
,
. Then
is not a subset of
and
is not a subset of . But
.
Formative Assessment
1.
Let
,
, and
.
Prove or disprove each of the following statements:
a.
.
b.
.
c.
.
Answer:
is not a subset of .
a.
For example,
is an element of
because
. However, 12 is not
an element of .
b.
Yes,
is a subset of .
If
, then
for some integer . Then,
, and so
for some integer . Then
, and so
.
c.
Yes,
.
If
, then
.
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Set Theory
Conversely, if
, then
, and so
2.
Let
for some integer
.
,
, and
Find each of the following:
a.
b.
c.
d.
e.
f.
g.
h.
Answer:
a.
b.
c.
d.
e.
f.
g.
h.
3.
. Then
Let the universal set be the set R of all real numbers and let
,
, and
.
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.
MTH105
Set Theory
Write down each of the following sets using interval notation, and as simply as
possible:
a.
b.
c.
d.
e.
f.
Answer:
a.
b.
c.
d.
e.
f.
4.
Suppose that
and
. Find the following sets:
a.
b.
c.
Answer:
a.
b.
c.
SU5-90
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5.
Set Theory
Prove that
by using (i) the element method, and
(ii) establishing a chain of set identities.
Answer:
Element method:
First we show that
.
Let
. Then either
Suppose
. Then is in
Thus,
or
.
but not in . It follows that
and
.
but not in . It follows that
and
.
.
Suppose
. Then is in
Thus,
.
We have therefore shown that
.
Conversely, we show that
Let
.
be given. Then
), or (
and
but
. It follows that either (
). Hence,
and
.
We have therefore shown that
.
We can therefore conclude that
.
Chain of set identities method:
by Difference Laws
by Distributive Laws
6.
Use
the
element
method
to
show
.
Answer:
We first show that
.
SU5-91
that
MTH105
Set Theory
Suppose that
. Then
for some elements
and
.
If
, then
.
If
, then
.
Hence,
.
Therefore, we have shown that
.
Conversely, we show that
.
Let
. Then either
If
, then
for some
. Thus,
If
or
and
. Since
, we have
and
. Since
, we have
.
, then
for some
. Thus,
.
We have therefore shown that
.
Hence,
7.
Let
.
be sets. Prove that
if and only if
.
Answer:
Suppose that
. Let
we can deduce that
be given. Then
. Hence,
We have shown that
Conversely, suppose that
Let
be given. Then
subset of . Hence,
It follows that
Hence, we have proven that
8.
,
.
.
, the set containing just the single element , is a
. Since
, and so
. Since we also have
, we deduce that
. We have thus shown that
if and only if
.
.
.
In how many ways can a committee of 4 women and 5 men be formed from a group
of 7 women and 10 men?
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Set Theory
The youngest of the women is Alice and the youngest of the men is James. If it is
decided that the committee must include at most one of Alice or James, in how many
ways can the committee be formed?
Answer:
If a committee of 4 women and 5 men be formed from a group of 7 women and 10
men without any further restriction, the number of ways of doing so is
.
Suppose that we impose the restriction that tthe committee must include at
most one of Alice or James. We can count the number of ways using the
Complementation Principle. The number of ways of including both Alice and
James is given by
because after including both of them in the committee, we only need to choose
another 3 women out of the remaining 6 women, and another 4 men out of the
remaining 9 men. Therefore, by the Complementation Principle, the number of
ways of including at most one of Alice or James is given by 8820 – 2520 = 6300.
9.
Find the number of ways in which 4 girls and 5 boys can stand in a line under each
of the following conditions separately:
a.
No 2 girls may stand side by side.
b.
All 4 girls must stand next to each other.
c.
The first and last positions are occupied by boys.
Answer:
a.
We first arrange the 5 boys in a row. The number of ways is 5!=120.
Now, imagine there are 5+1=6 empty spaces in between the boys, inclusive
of the empty spaces at the sides of the row of boys (to the left of the first
boy, and to the right of the last boy). We can insert one girl into each
SU5-93
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Set Theory
empty space. By doing so, we ensure that the girls are separated, that is,
no two girls are side-by-side. The number of ways of doing so is given by
. (Note that permutation is used here because ordering of the 4
girls is taken into account).
By Multiplication Principle, the number of ways in which no 2 girls are
side-by-side is given by 120*360=43200.
b.
We group the 4 girls as a single unit since they all have to stand together.
We treat each of the 5 boys as a single unit as well. Hence, there are a total
of 5+1=6 units. First, we arrange these 6 units in a row. The number of
ways is 6!=720.
We must also arrange the girls amongst themselves. Since there are 4 girls
to be arranged, the number of ways is 4!=24.
By Multiplication Principle, the number of ways in which all 4 girls stand
next to each other is given by 720*24=17280.
c.
Imagine placing one boy at each of the first and last positions. The boys
placed there can be any 2 out of the 5 boys available. We must take into
account which of the 5 boys are placed there, as well as the arrangement of
these 2 boys (the 2 boys can swap places with each other – in other words,
they can permute with each other). The number of ways of selecting 2 out
of 5 boys to be placed in those positions and then arranging them is given
by
.
In the middle, there are 4 girls and 3 boys to be arranged. The number
of arrangements of these 7 people without any restriction is given by 7!
=5040.
By Multiplication Principle, the total number of ways is 20*5040=100800.
10. I want to use some of all the notes in my wallet to tip a waitress. How many possible
amounts can I tip her if I have five $2 notes, one $5 note, and three $10 notes?
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Set Theory
Answer:
We must divide into cases because it is possible to use all the $2 notes to make up
ten dollars, or to use a single $10 note to make up ten tollars.
Case (i): Some or all of the $2 notes are used.
We can choose all, some, or none of the remaining notes. Regardless of how many
$2 notes was used, each of our choices of the remaining notes will give us different
dollar amounts. We can choose the $5 note or not -- this gives 2 choices. We can also
choose 0,1,2 or 3 $10 notes -- this gives 4 choices. Hence, the number of choices of
the remaining notes (other than the $2 notes) is 2*4=8.
Now, we have also to decide how many $2 notes to use. Here, we have five
options, because we must use at least one of the $2 notes. Hence, by Multiplication
Principle, the number of different tip amounts in which some or all of the $2 notes
are used is given by 5*8=40.
Case (ii): None of the $2 notes are used and none of the $10 notes are used.
We have only one choice here – to use the $5 note.
Case (iii): None of the $2 notes are used and at least one the $10 notes is used.
In this case, regardless of how many $5 and $10 notes are chosen, it will overlap
with one of the dollar amounts considered in Case (i), because one of the $10 notes
chosen can be simply replaced with five $2 notes. To avoid double-counting, we
must therefore not take into account any combinations of notes under this case,
even though it is certainly valid to give the waitress a combination of $5 and $10
notes without including any $2 notes.
The above three cases are exhaustive.
By the Addition Principle, the total number of dollar amounts that the waitress can
receive is given by 40+1=41. Note that this excludes the case of zero dollars.
11. How many non-empty subsets are there of a set of elements?
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Answer:
Let
be a set of distinct elements. We can count the total number of subsets of
inclusive of the empty set as follows: Each element has two options – to be included
or excluded in the subset. Therefore by the Multiplication Principle, there are are
total of
subsets of , inclusive of the empty set. (This method of counting was
mentioned in the solution to Question 10 as well).
If we impose the restriction that the empty set cannot be chosen, then by the
Complementation Principle, there are
possible non-empty subsets of .
12. A man can construct his restraunt desert by choosing some or all of 4 different
flavours of chocolate fudge, some or all of 5 different types of buttered cream, and
some or all of 6 different varieties of nuts. How many different deserts are there?
Answer:
The number of choices of flavours of chocolate fudge is
without any
restriction, inclusive of the cases where no flavour of chocolate fudge is chosen.
Since he has to select at least one flavour, the number of choices is
Similarly, the number of choices of buttered cream is
choices of varieties of nuts is
.
, and the number of
.
By the Multiplication Principle, the number of deserts is 15*31*63=29295.
13. Eight students are seated on a bench in a row. What is the probability that two of
them, Denise and Ellen, are seated side by side?
Answer:
The total number of ways of arranging 8 students in a row without restriction is
given by
.
The number of ways of arranging them in which Denise and Ellen sit side by
side is computed as follows: Place Denise and Ellen together as a single unit, and
SU5-96
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Set Theory
treat each of the remaining 6 people as single units each. There are 7 units in
total. To arrange the 7 units in a row, there are
permutations. To arrange
Denise and Ellen between themselves, there are
permutations. Hence by the
Multiplication Principle, there are 5040*2=10080 permutations in which Denise
and Ellen sit side by side.
The probability of finding Denise and Ellen side by side is
14. Two boxes are labelled A and B. Box A contains 3 blue discs and 2 white discs. Box B
contains 2 blue discs and 3 white discs. A random sample of 2 discs is drawn without
replacement from each box. What is the probability that all discs are the same colour?
Answer:
Treat all the 10 discs as distinct (we can imagine that we label them with the letters
A to J).
There are a total of
ways of drawing two discs of each of
the boxes A and B without replacement, and without any further restriction.
To obtain all discs of the same colour, we can either have all blue or all white. The
number of ways of drawing all blue is
ways of drawing all white is
ways. The number of
ways. Hence there are 6 ways of
obtaining all the same colour.
The probability that all discs are the same colour is
.
15. Two boxes are labelled A and B. Box A contains 3 blue discs and 4 white discs. Box
B contains 2 blue discs and 5 white discs. First, a box is selected. We select Box A
with probability 0.3, and Box B with probability 0.7. Next, we pick two discs from our
chosen box without replacement. What is the probability that all discs are the same
colour?
SU5-97
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Set Theory
Answer:
P (choose A and choose all same colour)
= P (choose A) P (choose all same colour given A was chosen)
=
P (choose B and choose all same colour)
= P (choose B) P (choose all same colour given B was chosen)
=
The probability of having all same colour is
.
16. The following preference schedule was obtained after an election.
Rank
11 voters
7 voters
7 voters
3 voters
9 voters
5 voters
First
A
D
C
C
B
D
B
B
A
A
C
C
C
A
B
D
A
A
D
C
D
B
D
B
Choice
Second
Choice
Third
Choice
Fourth
Choice
Determine the winner under the Plurality Method, the two-round system, and the
Plurality Method with Elimination.
Answer:
The winner under the Plurality Method is Candidate D, with 12 first-place votes.
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Set Theory
In the two-round system, we eliminate all candidates except the top two candidates,
A and D. The following revised preference schedule is obtained:
Rank
30 voters
12 voters
First Choice
A
D
Second Choice
D
A
The winner under the two-round system is candidate A.
In the Plurality Method with Elimination, we first eliminate the candidate with
the least number of first-place votes. Here, we get rid of candidate B who obtained
only 9 first-place votes.
The following revised preference schedule is obtained:
Rank
11 voters
7 voters
19 voters
5 voters
First Choice
A
D
C
D
Second
C
A
A
C
D
C
D
A
Choice
Third Choice
We now eliminate candidate A, who has the least number of first-place votes under
the above revised preference schedule.
Rank
30 voters
12 voters
First Choice
C
D
Second Choice
D
C
Hence, under Plurality with Elimination, the winner is candidate C.
SU5-99
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Set Theory
References
Epp, S. S. (2019). Discrete mathematics with applications (5th ed.). Boston, MA: Brooks/
Cole Cengage Learning.
Leong, Y. K. (2011). Living with mathematics (3rd ed.). Singapore: McGraw-Hill
Education.
Robinson E.A., J., & Ullman, D. H. (2017). The mathematics of politics (2nd ed.). CRC Press.
SU5-100
Study
Unit
Relations and Functions
6
MTH105
Relations and Functions
Learning Outcomes
By the end of this unit, you should be able to:
1.
Define a relation and give its domain and range.
2.
Prove that a given relation is an equivalence relation.
3.
Describe the equivalence classes of a given equivalence relation.
4.
Apply the concept that congruence classes are equivalence classes, and prove
results in divisibility using congruency modulo .
5.
Determine whether a given function is injective or surjective, or both.
6.
Calculate the domain and range of a real-valued function.
7.
Construct the inverse of a one-to-one onto function.
8.
Verify whether a given composite function can be defined.
9.
Compute the secret key corresponding to a public key pair
of the RSA
Algorithm, given the prime factorization of .
10.
Encrypt and decrypt plaintext and ciphertext messages respectively using the
RSA Algorithm.
SU6-2
MTH105
Relations and Functions
Overview
In this Study Unit, we study two more important mathematical concepts – the concept of
a relation, and the concept of a function. Although functions are a special kind of relation,
they are so fundamental to mathematics that most people who encounter mathematics
usually encounter the concept of a function before they encounter the concept of a relation.
We will examine the concept of a relation and pay particular attention to an important
class of relations known as equivalence relations.
Congruence classes modulo an integer are a key example of equivalence classes. We will
discuss the concept of congruency modulo
and use the notion of congruency to prove
interesting results in divisibility.
Having discussed relations in details, we discuss functions, which are a special kind of
relation. We will learn about the important properties of functions such as injectivity and
surjectivity, and study how to find the domain and range of functions.
In this Study Unit, we also introduce an application of congruence arithmetic – the RSA
Algorithm, which is used for encrypting data so that it does not fall into the wrong hands.
The RSA Algorithm relies heavily on modular exponentiation which will be discussed in
this Study Unit, as well as the concepts of divisibility and the Euclidean Algorithm that
we discussed in Study Unit 4.
SU6-3
MTH105
Relations and Functions
Chapter 1: Relations and Congruency
1.1 Definition of a Relation
be non-empty sets. A relation from
Let
. Recall that
. If
to
is a subset
is the set of all ordered pairs, or
of the Cartesian product
-tuples
, where
and
, we say that is related to via , or simply that is related to if it is
understood what the relation
writing
is. Whenever
, we often abbreviate this fact by
.
A relation can be defined by simply listing down all its elements, which in this case are
ordered pairs
.
Example 1
Suppose that
and
. An example of a relation
from
to
is
In this example, is related to , so we can write
In an ordered pair
, We also have
,
, the ordering of the letters matter, so
,
,
,
.
is not the same as
. Thus, it may be the case that under a certain relation , we have
, but not
.
We can also define a relation by stating a necessary and sufficient condition for any
ordered pair
to be a member of the relation.
Example 2
Let
be the set of rational numbers, and let
relation
from
to
be the set of integers. We can define a
by declaring that for any rational number
and integer ,
if
and only if has a denominator of when it is written as a fraction expressed in its lowest
form.
SU6-4
MTH105
Relations and Functions
To write a fraction in its lowest form, or to reduce a fraction to its lowest form, is to write
it as
, where
. Then in this case, the integers
are uniquely determined
up to sign (that is, uniquely determined up to positive or negative).
For instance,
,
reduce the fraction
, but
. This is because when we
to its lowest form, it becomes
either on the numerator or denominator), so the denominator of
form is or
(the positive sign can go
written in its lowest
, and not .
Activity 6.1
In Example 2 above, prove the following:
a.
For any rational number ,
b.
Let be a rational number. Then is an integer if and only if either
or
.
.
Definition
Let
be a set. If
is a relation from
to , in other words, if
, we say that
is a
relation on .
The definition of a relation provided earlier can be generalised. Given a positive integer ,
and a collection of sets
, an -ary relation is a subset of the Cartesian product
We will not be concerned with general
-ary relations in this course. We will only be
concerned about relations involving the Cartesian product of two sets, that is,
SU6-5
MTH105
Relations and Functions
to a set . We call such relations binary relations.
in other words, relations from a set
They really are just
-ary relations, going by the definition in the previous paragraph.
Thus, binary relations are subsets of a certain Cartesian product
under what circumstances an element
that declare precisely
is to be considered related to an element
.
Example 3
Let
be a positive integer, and let
For any integers
be the relation from
to
defined as follows:
,
Then under this relation,
if and only if both and leave the same remainder when
divided by . For instance,
,
, and
.
Definition
Let
be a relation from a set
defined to be the set of all possible
The range of , denoted
for some
We have that
to a set . The domain of , denoted
such that
for some
, is
.
, is defined to be the set of all possible
such that
.
, and
.
Example 4
Suppose that
and
. Define a relation
SU6-6
from
to
by
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Relations and Functions
Then
and
.
Activity 6.2
Suppose that
and
. Define a relation
by declaring that for all
and
Determine the elements of
and determine the domain and range of .
from
to
,
We can also define relations by giving an algebraic equation that specifies the relationship
between two variables
and , where
belongs to a set
and belongs to a set
are allowed to be the same set).
For instance, we can define a relation
from
Figure 6.1 Graph of
SU6-7
to
as follows:
( and
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In this example, the points
,
,
are elements of . We can visualise
this relation by sketching a graph on the - plane in which a point
if and only if
. Thus, a point
is on the graph
is plotted on the graph if and only if
.
Activity 6.3
Define a relation
from
to
as follows:
Determine the domain and range of
.
1.2 Depicting Relations using Ovals and Dots
Relations can be depicted using ovals and dots. Let
. Assume that
and
be a relation from a set
to a set
are finite sets (even though relations can very well be between
infinite sets). We can draw an oval to represent , another oval to represent , and place
dots in both ovals to represent the individual elements of
line from a dot in
and
to a dot in
for which
and . We also draw straight
if and only if the dots respectively represent elements
.
For instance, let us again consider the relation
from the set
given by
to the set
as
provided in Example 1 of Section 1.1. We can depict the above relation
dots as follows:
SU6-8
using ovals and
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Figure 6.2 Depicting relation
1.3 Inverse Relations
Definition
Let
be a relation from a set
to a set . The inverse relation
from
to
is
defined as:
Example 1
Again, consider the relation
from the set
by
to the set
. The inverse relation
the relation
In this example,
can be illustrated using the following dot-and-oval diagram:
SU6-9
given
is
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Figure 6.3 Depicting relation
Example 2
Consider the relation
The inverse relation
equation. Thus,
from
to
defined as follows:
can be obtained by interchanging the roles of and in the above
is a relation from
to
defined by the rule
Theorem:
Let
be a relation from a set
to a set
. Then
and
. In other words, the domain and range of a relation
and its inverse
have their roles interchanged.
Proof:
The relationship between
and its inverse
. Hence, for any element
follows that
is that
,
if and only if
. Similarly, for any element
. It follows that
if and only if
.
SU6-10
,
. It
if and only if
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Activity 6.4
Define a relation
from
to
as follows:
Determine the domain and range of
as well as
.
1.4 Equality of Relations
Suppose that
are subsets of
and are both relations from a set
. The relation
to a set . In other words, both
is equal to, or identical to the relation if and only if
as sets. In other words, an ordered pair
is a member of
is a member of . This condition is equivalent to writing
If two relations
and
are equal, then
if and only if
if and only if
and
.
.
1.5 Equivalence Relations - Definition
An equivalence relation is a special kind of relation, but one that has widespread
applications. An equivalence relation has three properties: reflexive, symmetric, and
transitive.
Let
be a relation on a set . In other words,
.
a.
is said to be reflexive if for any element
b.
is said to be symmetric if for any elements
In other words, if
c.
, then
is said to be transitive if for any elements
SU6-11
, we have
,
.
,
.
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In other words, if
A relation
on a set
and
, then
.
that is reflexive, symmetric, and transitive is termed an
equivalence relation.
Example 1
Consider the relations on a set
The relation
.
is reflexive and transitive. However,
is not symmetric.
but
.
The relation
is transitive. But
must have
for every element
is not reflexive and not symmetric. To be reflexive, we
. However, it can be observed that
is not an element of . is also not symmetric because, for instance,
The relation
is symmetric, but not reflexive nor transitive.
for example,
and
, but
.
.
is reflexive, symmetric, and transitive. Hence,
The relation
is reflexive and symmetric, but not transitive. For example,
The relation
but
is not transitive because,
The relation
, but
, for instance,
is an equivalence relation.
and
.
is reflexive, symmetric, and transitive. Hence,
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is an equivalence relation.
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Relations and Functions
The relation
is reflexive, symmetric, and transitive. Hence,
is an equivalence relation.
Activity 6.5
Let
and define relations , , and
on
as follows:
Determine, for each relation above, which (if any) of the properties: reflexive,
symmetric, transitive, are true for the relation.
Example 2
Define a relation
on
defined as follows:
is reflexive because for any real number , we have
is not symmetric because in general,
.
does not imply that
. For example,
but certainly is not less than or equal to .
is transitive because for any real numbers
Hence,
, if
and
, then
.
is reflexive and transitive, but not symmetric.
Activity 6.6
Give an example of a relation
that is symmetric and transitive but not reflexive.
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1.6 Congruency Modulo
We now provide an important example of an equivalence relation on the set of integers
Let
be a positive integer. For integers
, we say that is congruent to modulo if
and leave the same remainder when divided by .
Thus, for instance,
is congruent to modulo because
and both leave a remainder
of upon division by .
is congruent to 9 modulo 6 because both
and
leave a remainder of
when
divided by .
If
and
leave the same remainder when divided by , we say that
modulo , and write
is congruent to
. By the symmetry of this definition, we can also write
. Indeed, to say that and leave the same remainder when divided by
is equivalent to saying that and leave the same remainder when divided by .
Thus, for instance,
and
.
Theorem:
Let
be a positive integer, and let
if and only if divides
Note that divides
be integers. Then
.
if and only if divides
therefore symmetrical in the variables
.
SU6-14
. The statement to be proved is
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Relations and Functions
Proof:
Suppose that
. Then by the definition of congruency modulo ,
and
leave the same remainder when divided by . It follows that we can find integers
such that
and
where is the common remainder when either or is divided by . The integer satisfies
the inequality
Therefore,
by the Division Property. It follows that
.
Conversely, suppose that
. Then there exists an integer such that
Again, we use the Division Property to write
and
where
are the unique remainders when and are respectively divided by .
We have inequalities
and
. Then
As a result,
SU6-15
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Since
, we have
But since
is an integer, we conclude that
, which implies
.
Hence, and leave the same remainder when divided by .
Theorem:
Let
be a positive integer. Then congruency modulo is an equivalence relation on the
set of integers
.
This means that if we define a relation
on
by declaring that
if and only if
then the relation
is an equivalence relation on the set of integers.
Proof:
To show that the relation
show that
is an equivalence relation on the set of integers, we need to
is reflexive, symmetric, and transitive. This means that we have to prove the
following:
a.
For all integers , we have
b.
For all integers and , if
c.
For
all
integers
.
, then
,
if
.
and
,
then
.
Statement (a) is true because for any integer , it is trivially true that and leave the same
remainder when divided by . Or, we can observe that
, and hence
.
SU6-16
is clearly divisible by
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Statement (b) is also trivially true by our earlier remarks, in which we stated that to say
that and leave the same remainder when divided by is equivalent to saying that and
leave the same remainder when divided by . Or, we can observe, again quite trivially,
that divides
if and only if divides
.
For statement (c), we suppose that
divides
divides
and
and
divides
. Hence,
divides
. It follows therefore that
The equivalence relation
on
. This implies that
, and so
.
defined by declaring that
if and only if
will henceforth be called the relation of congruency modulo .
The relation of congruency modulo
obeys some (but not all) the properties of
manipulating equations involving integers. These are some key properties of congruency
modulo :
a.
If
and
, then
b.
If
and
, then
c.
If
and is any non-negative integer, then
.
.
The above is by no means an exhaustive list of all the properties satisfied by congruency
modulo . Properties (a) and (b) have appropriate generalisations:
If
and
are integers such that
then
SU6-17
for all ,
,
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Relations and Functions
Example 1
We have
and
. Hence,
and
Also,
Clearly,
if and only if
is a multiple of . We can exploit this property in
various ways, such as in the following example:
Example 2
Show that
divides
.
Answer:
We consider congruency modulo
. The reason for choosing the number
are asked to verify divisibility by
.
is because we
Our strategy is to start with the number , and repeatedly take the square modulo
.
Thus, we compute
We do not take any further squares because the power to which is raised would exceed
. From the above computations, we can now deduce that
SU6-18
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Hence, we reach the conclusion that
or equivalently,
divides
.
Example 3
Find the remainder when
is divided by .
Answer:
We consider congruency modulo . Our strategy here is to start with the number
repeatedly take the square modulo . Firstly however,
with the number .
From the above, we now deduce that
Hence,
leaves a remainder of when divided by .
SU6-19
, and
. So we can start
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Relations and Functions
Activity 6.7
Find the remainder when
is divided by .
Activity 6.8
Prove that
.
Then use this to find the remainder when the sum
is divided by
Recall that
.
(read factorial) is defined to be the product
Theorem:
This theorem is commonly called Euclid's Lemma. It is often used as a critical component
in proofs of other number-theoretic results.
If
are non-zero integers such that divides
and
, then divides .
be non-zero integers such that divides
and
. Since
Proof:
Let
there are integers
. Since
and
such that
. Multiply by
, we can also find an integer such that
SU6-20
,
throughout to obtain
. Then we have
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Relations and Functions
Since
is an integer, we can conclude that divides .
Lemma:
Suppose that
are non-zero integers such that
that
are non-zero integers because
and
. Then
. Note
is a common divisor of and .
Proof:
Since
, there exist integers
It follows that
such that
. Then
.
Theorem:
Suppose that
are non-zero integers, and
where
. Note that
. Suppose that
is a positive integer because
. Then
is a divisor of .
Proof:
Since
Since
, there exists some integer such that
, there also exist integers
and so dividing by
yields
SU6-21
such that
and
. Hence,
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Thus,
Relations and Functions
divides
Euclid's Lemma,
. Since
divides
by the above lemma, so by
. Hence
is congruent to
modulo
; in other
words,
Corollary:
Suppose that
are non-zero integers, and
, then
. Suppose that
and
.
We can restate the theorem just proved as follows:
Corollary:
Given
are non-zero integers with
are both divisible by , and that
positive such that
. Suppose that
. Then we have
Note the difference in the dominators of each fraction!
Example 4
Since
are both divisible by , and we also have
, it follows that
Example 5
Since
are both divisible by
, and we also have
SU6-22
, it follows that
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Relations and Functions
Activity 6.9
Give an example to show that
need not necessarily imply that
.
1.7 Equivalence Classes
Recall that if is a set, and
is a collection of subsets of indexed by an indexing
forms a partition of
set , then we say that
a.
is the union of the sets
,
if
, that is,
and
b.
The collection
for any
Let
is a pairwise disjoint collection of sets, that is,
with
.
be an equivalence relation on a set . Let an element
be given. The equivalence
class containing , denoted by
or denoted simply by
if it is clearly understood what equivalence relation
we are
considering, is defined to be
In other words, the equivalence class containing
related to via . We call the sets
class
is given by
is the set of all elements of
that are
as the equivalence classes under . If an equivalence
, then we call a class representative of .
SU6-23
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Example 1
Consider the relation of congruency modulo , where
integer , the equivalence class containing
is a positive integer. Given any
is the set of integers that are congruent to
modulo , that is,
Thus, for instance, when
,
Example 2
Consider the equivalence relation
Then the equivalence classes under
Let
given by
are
be an equivalence relation on a set . Since the relation
, we have
class
on the set
. Hence, for any
,
, that is, is a member of the equivalence
. Or, to put it in another way, every element of
class for which it is a representative – if
From the above observation, it follows that
the equivalence relation
is reflexive, for any element
, then
is a member of the equivalence
.
is the union of its equivalence classes under
:
SU6-24
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In fact, more is true. The set of equivalence classes under
. Not only is
forms a partition of the set
the union of the equivalence classes under
equivalence classes
and
, either
for
, but also, for any two
.
Theorem:
Let
be an equivalence relation on a set . Then for any two equivalence classes
, either
or
and
.
Proof:
Suppose that
. Then there is an element
. Pick any element
have
. We also have
we have
element
. By the definition of the equivalence class
because
, and therefore,
element
and
conclude that
, since
because
. This proves
, we can show that
of
. We show that
,
. Thus,
, we
. Thus by transitivity,
. Similarly, if we pick any
by a similar (symmetrical) argument: For any
, and
, we have
by transitivity. We likewise
.
The above theorem tells us that when we consider the relation of congruency modulo ,
the collection of equivalence class
is a pairwise disjoint collection that forms a partition of ℤ. Every integer
belongs to
one and only one of the above equivalence classes. From now on, we will refer to these
equivalence classes as congruence classes modulo .
SU6-25
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For every integer
, there are exactly
represented as
congruence classes modulo , which can be
. An integer
is in the congruence class
if and
only if
For a fixed integer , the fact that every integer
classes modulo
belongs to exactly one of the congruence
is simply a restatement of the known result that every integer
unique remainder when divided by , where
belongs can be represented by
has a
. The equivalence class to which
, where is the aforementioned remainder.
Activity 6.10
Find the equivalence classes of the set
under the equivalence relation
1.8 Application of Congruence Arithmetic: The RSA Algorithm
This section is based on Living with Mathematics by Leong (2011).
Cryptography refers to the science of modifying the data being stored in a retrieval system
or transmitted over a communications network in a way so as to prevent unauthorized
people from understanding what is being stored or transmitted.
Preventing information from being seen, interpreted, and utilized by the wrong hands
is a pivotal component of computer and network security, military defence, and the
maintainence of the integrity of the financial markets and banking system.
In this section, we examine a very popular cryptographic method called the RSA
Algorithm. The acryonym RSA stands for Rivest, Shamir, Adleman, the names of the
algorithm's three designers.
SU6-26
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All modern computers communicate with each other in binary, that is, the messages
comprise of the symbols "0" and "1". All information exchange between sender and
recipient must use binary format. When people communicate with each other, they pass on
information such as English words, diagrams, pictures, etc. All these information must be
converted to binary format before transmission via a computer system. When the recipient
receives the message from the computer system, he or she must then convert the binary
message back into human-readable information.
In cryptography, the original message being transmitted from the sender to the recipient
is called the plaintext. The process of modifying the plaintext so that unauthorized
people cannot undertand it is called encryption, and the resultant message is called the
ciphertext. When the recipient receives the ciphertext, he or she must then convert it back
to plaintext so as to understand what has been sent. This process is called decryption.
When a computer system is used for data transmission, both the plaintext and ciphertext
will be sent and received in binary form.
In practical terms, when a very long binary string is transmitted by the sender, it is first
broken into blocks of identical size. The enryption algorithm is then applied to each
plaintext block before transmission. When the recipient receives the encrypted blocks, he
or she then applies the decryption algorithm to convert each ciphertext block back into a
plaintext block.
In this section, we will use the technique of congruence arithmetic to describe and apply
the RSA algorithm. Since we are used to working with decimal numbers (that is, numbers
in base 10), we will assume that each binary string has been converted to a decimal
number, and we will perform all our calculations in decimal format.
1.8.1 The RSA Algorithm
Let
be distinct odd prime numbers. In order to make the RSA Algorithm secure,
should each be large. Both the prime numbers
SU6-27
are kept secret.
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Let
. In the discussion that follows, we assume that all messages being transmitted
are integers between 2 and
inclusive, and we will use modulo arithmetic to encrypt
and decrypt plaintext and ciphertext messages respectively.
Let
denote the number of positive integers between 1 and
inclusive that are
coprime to . This is called the Euler totient function of .
In the RSA algorithm, we have
, where
are distinct primes. For this case, we have
.
Choose an integer
satisfying
such that
is coprime to
, that is
, The integer as well as the integer are made known to the public. We
the public key.
call the ordered pair
Let
be the unique positive integer between 1 and
The integer
such that
is known as the multiplicative inverse of modulo
. It can be found
using the Euclidean Algorithm that we studied in SU4. Using the Euclidean Algorithm,
we find integers
such that
This is possible because
so
Else, we let
. If
is an integer between 1 and
be the unique remainder when
residue of modulo
The integer
Then we can observe that
, and
, then we simply let
is divided by
. We call
.
the unique
.
is called the secret key, and as the name suggests,
public. It is known only to the recipient. The value of
is kept secret from the
is also kept secret from the
public.
Encryption:
For a given plaintext , where
, the ciphertext is computed as
SU6-28
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Decryption:
For a given plaintext , the plaintext
can be recovered by computing
1.8.2 Why the RSA Algorithm works
Let us now give a detailed explanation of why the RSA Algorithm outlined above works.
First, we state and prove a theorem by Euler.
Theorem (Euler):
Let
be positive integers such that
. Then
Proof (Outline):
Let
be the set of positive integers strictly less than
coprime to . We multiply the integer to each of the numbers
that are
to obtain the
set
Since
, the residues of each of the integers
arrangement of the integers
in some order. Hence,
and so
SU6-29
modulo
is an
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(End of Proof)
In the RSA Algorithm,
, and so
a ciphertext message be given, where
Suppose first that
for some integer . Let
.
Then we have
by Euler's Theorem. It
follows that
Hence,
successfully recovers the plaintext .
Now suppose that
be that
. Since
is equal to one of the primes
Since
and
are distinct primes,
, and
, it must
. Suppose without loss of generality that
.
, by Euler's Theorem, we have
Therefore,
and so
Also, since
Therefore,
, we have
is divisible by both
and
, and thus
It follows that
and so
successfully recovers the plaintext .
SU6-30
is divisible by
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1.8.3 A Toy Example
We give a toy example of the RSA Algorithm. In real life, the prime numbers used cannot
be so small. When expressed in binary form, they have to be anywhere from 1024 to 4096
digits long.
Let
and
Then
.
, and
.
We choose a number that is strictly between 1 and
The public key is the ordered pair
. Suppose that we choose
.
.
We need to determine the private key. To do so, we execute the Euclidean Algorithm to
determine integers
such that
.
264 = 8(31) + 16
31 = 1(16) + 15
16 = 1(15) + 1
We have
1 = 16-15 = 16-(31-16)
= (-1)(31)+(2)(16)
= (-1)(31)+(2)(264-8(31))
= (-17)(31)+(2)(264)
Since
, the integer
is the private key. This private key
is kept secret. It is known only to the recipient. The value of
is also kept secret.
Suppose that a member of the public wishes to transmit the plaintext message
the recipient. The member first encrypts the plaintext
The recipient upon receiving the ciphertext
then decrypts it as follows:
The plaintext is successfully recovered by the recipient.
SU6-31
as follows:
to
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1.8.4 The Security of the RSA Algorithm
The security of the RSA Algorithm relies on the difficulty of factoring large composite
numbers of the form
, where
If a hacker is able to factorize
are distinct large primes.
, then
can be easily computed using the result
, and from there it would also be straightforward to compute the
secret key , which is the multiplicative inverse of modulo
. Thus, the hacker would
be able to decrypt any message that he or she is able to intercept.
As the processing power of computers continue to grow, we must create public-private
key pairs based on larger and larger prime numbers
, so that
remains difficult
to factorize. Based on current computing technology, a key length of 2048 bits is the
minimum, and an ideal amount of security is provided by a key length of 4096 bits.
SU6-32
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Chapter 2: Definition and Basic Properties of Functions
2.1 Definition and Examples of Functions
Recall that a relation
from a set
to a set
is a subset
of the Cartesian product
function is a special kind of relation. For a relation
.A
to be considered a function,
we impose two constraints:
a.
For every
, there exists some
(not necessarily unique) such that
Thus, the domain of the relation,
b.
, is the whole of the set .
It is forbidden to have two different
such that
In other words, it is forbidden for one element
elements in
and
.
to be related to two different
via . Another equivalent way of phrasing this restriction is that
and
implies that
.
Example 1
Figure 6.4 A non-example of a function
The above example depicts a relation from
not every element in
to . It is a non-example of a function because
is related to an element in . In particular,
any element of .
SU6-33
is not related to
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Example 2
Figure 6.5 A non-example of a function
The above example depicts a relation from
because there is some element of
particular,
to . It is again a non-example of a function
that is related to more than one element of
. In
is related to both and in .
Example 3
Figure 6.6 A non-example of a function
The above example depicts a relation
from
to . Every element of
from
to . This is now an example of a function
is related to some element of , and no element in
is related to more than one element of . Note that it does not matter that
preimage, that is, there is no element
such that
SU6-34
.
has no
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Example 4
Define a relation
Then
on the interval
is not a function on
is an example of a relation
element from
by the following:
because, for instance,
from the set
and
to a set
(the number ) is related to two different elements in
. This
in which one
(the numbers
and ) via .
If we plot the graph of the relation
on the
plane, we get a circle of radius
centred at the origin:
Figure 6.7 Graph of
The above graph does not depict a function because it fails the vertical line test: A relation
fails to be function if we can draw a vertical line on the graph such that the line intersects
the graph in two distinct points. Note that the horizontal axis must be the -axis and the
vertical axis must be the -axis.
SU6-35
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Figure 6.8 Failing the vertical line test
If
is a function from
to , we write
, and if
via the function , then we write
If
, we say that
, that is,
is related to
.
is a mapping that sends
to , or that maps to .
We say that is the image of under .
We also say that is a preimage of under .
Using the above terminology, we say that a function
For all
one
, there exists
such that
must satisfy the following:
, and furthermore, there can only be
such that
.
We say that every element in
has a unique image. However, note that not every element
in needs to have a preimage. It is acceptable for there to be some
for all
. It is also acceptable for some
such that
to have more than one preimage under .
Take for example the following function
SU6-36
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Figure 6.9 An example of a function
We have
,
The element
The element
,
,
, and
.
has no preimage under .
has more than one preimage under
, because we have
,
.
If
is a function, we call
codomain, denoted
the domain, denoted by
, and we call
the
. Note that the codomain can be different from the range,
. The range of is the set of all elements
for which
for some
. If there is some
such that
, that is, the set of all images of elements of
for all
under
, then the element is a member of
the codomain but not a member of the range.
In the example just described, the domain of
the set
, and the range is
is the set
, the codomain is
.
SU6-37
MTH105
Relations and Functions
Activity 6.11
The above dot-and-oval diagram depicts a function .
a.
What is the domain and range?
b.
What are the images of each of the elements
c.
What are the preimages of each of the elements
?
?
2.2 Defining Functions using Algebraic Formulae
A very important way of defining functions is by the use of algebraic formulae. The reader
has already seen examples of this during this course.
Let
denote an algebraic expression involving the variable . We can regard
a function that maps the value of to the value of
usually be specified alongside the algebraic function.
For example, define a function
as follows:
SU6-38
as
. The domain of the function must
MTH105
Relations and Functions
In the above specification, the domain of the function
numbers
such that
. In interval notation,
function is the algebraic formula
The function
. The rule of the
. In this example, the rule of the function is
is therefore a function that maps to
and so on. The variable
is taken to be the set of real
. Thus, for instance, we have
is called the independent variable because we can allow
take on any real number as long as it is at least
value within the specified domain
, that is,
to
is allowed to take on any
. We can write
The variable is then called the dependent variable, because is computed based on the
value of .
In the above terminology, the set of values that can be assumed by the independent
variable is the domain of , and the set of values that can be assumed by the dependent
variable
is the range of . The codomain can be arbitrarily chosen. In this case, if we
simply set the codomain of the function to be
, then
It is straightforward to work out that the range is
.
As another example, define
In this example, the domain of the function
open interval
is all the positive real numbers, that is, the
. We can also provide a sketch of the function on the
SU6-39
plane. The
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Relations and Functions
dependent variable is always placed on the vertical axis, and the independent variable
is placed on the horizontal axis.
Figure 6.10 Graph of
2.3 Equality of Functions
Suppose that
equal, that is,
are functions with the same domain and codomain. The functions are
, if and only if
for all in their common domain.
Example
Suppose that the domain and codomain of
and
are set to be equal to
define
and
where in this case, we take the unique remainder modulo , so that the functions
and
always attain values within the set
SU6-40
.
. We
MTH105
Relations and Functions
In this example, it turns out that
For
for all
.
, we have
and
. To demonstrate this, we have to verify that
, so taking the unique
remainder modulo yields
For
.
, we have
modulo yields
For
, so taking the unique remainder
. Similarly,
, so
, we have
, so
, so
for all
.
. We also have
. We have therefore shown that
,and so we conclude that
.
2.4 The Identity Function
Let
be any non-empty set. Define a function
We call
by
the identity function on , and usually denote it by
the function that maps each element
. The identity function is
to itself.
2.5 Relationship between Functions and Sequences
Suppose that we have a sequence
We can define a function
given by an ordered collection of real numbers
by setting
Every sequence of real numbers can therefore be uniquely defined by a function from the
natural numbers to the real numbers.
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Relations and Functions
2.6 The Image and Preimage of a Function acting on a Specific
Subset
Suppose that
of the set
is a function from
under the function
to . Suppose that
under
is the set
is the set of images of elements of the subset
is the set of preimages of elements of the subset
If
is the empty set, then so is
. The image
is the set
The preimage, also known as the inverse image, of the set
In other words,
and
. Likewise, if
of , while
of .
is the empty set, then so is
.
Example
Consider the function
depicted by the oval-and-dot diagram as follows:
Figure 6.11 Depiction of a function
In this example, if we let
, then
.
SU6-42
. If we let
, then
MTH105
Relations and Functions
If we let
, then
.
Activity 6.12
Let
be a function defined by:
a.
What is the domain and codomain of the function?
b.
What is the range of the function?
c.
What is the image of the set
d.
What is the preimage of the set
?
?
Theorem:
Suppose that we have a function
, and that
are subsets of . Then
Proof:
Suppose that
. Then we can express
for some element
. We have that
Since
, we have both
and
.
as well as
.
Therefore, we are able to conclude that
.
We have thus proven that
.
SU6-43
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Relations and Functions
Activity 6.13
Give an example of a function
and subsets
of
such that
Theorem:
Suppose that we have a function
, and that
are subsets of . Then
Proof:
First, we prove that
.
Suppose that
. Then we can express
for some element
. We have that
or
.
Suppose that
. Then since
, we have
.
Suppose that
. Then since
, we have
.
Therefore, either
or
. It follows that
Hence, we have shown that
.
We now prove that
.
Suppose that
. Then either
Suppose that
Since
Suppose that
.
. Then
, so
or
for some
.
for some
.
.
. Then
SU6-44
.
MTH105
Since
Relations and Functions
, so
.
We therefore conclude that in all cases, we have
We have thus proven that
.
.
Theorem:
Suppose that we have a function
, and that
are subsets of . Then
Proof:
Activity 6.14
Suppose that we have a function
, and that
SU6-45
are subsets of . Then
MTH105
Relations and Functions
Chapter 3: Injective, Surjective, Bijective Functions, and
Composition of Functions
3.1 Injective, Surjective, Bijective Functions, and Inverse Functions
Let
be a function. We say that
elements
of
is injective, or one-one, if no two different
map to the same element
, we have
. In other words, for any
with
.
The contrapositive of the above condition, of course, is:
If a function
is specified using an algebraic formula as detailed in Section 2.2,
then such a function is injective if and only if it passes the horizontal line test: When the
graph of
is plotted on the
plane (as usual, we always place the dependent
variable on the vertical axis and the independent variable on the horizontal axis), then
any horizontal line intersects the graph on at most one point.
Thus, for example,
,
, fails the horizontal line test and is therefore not
injective, as can be seen from the graph below:
SU6-46
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Relations and Functions
Figure 6.12 Graph of
,
The term "one-to-one" is often used as well to denote injectivity.
Example 1
Let
and
Define
as follows:
Then
is injective while
. Define
is not injective. For instance,
as follows:
, so
sends two
different elements in the domain to the same element in the range.
Example 2
Define a function
The domain of the function is all real numbers except for the point
algebraically that
is an injective function:
SU6-47
. We can verify
MTH105
Relations and Functions
Suppose that
, where
are elements of the domain. Then we have
Cross-multiplying the two fractions yields
which implies
This simplifies to
, which leads to
condition for injectivity, that
. We have therefore proven the required
implies
Take note that the condition for a relation
that
and
implies that
from
.
to
to be a legitimate function is
, or, if we insist on using the notation
to denote an
element in the range that is related to , we would have to write
or equivalently
This is different from the condition for a given function to be injective.
Let
be a function. We say that
codomain , there exists some element
is surjective, or onto, if for every element in the
such that
Hence, a function is surjective if and only if the range of
.
is equal to the codomain.
Example 3
Let
and
. Define
follows:
Define
as follows:
SU6-48
as
MTH105
Then
Relations and Functions
is not surjective while is surjective.
fails to be surjective when there is some element
A function
for every
such that
.
Remark:
If a function is defined using an algebraic formula but without explicitly stating the
codomain, for instance in the following matter:
(here, only the domain and formula of the function have been explicitly stated), it is
customary to regard the range of the function as being identical to the codomain, so
that surjectivity is assumed, or rather, imposed, by default. This is often done when the
codomain is not important within the mathematical context under discussion.
, with the domain
But if a function is defined as
and codomain
explicitly
stated, then we do not assume that the range is equal to the codomain. In this case, it is
possible for a given function
to be not surjective.
Example 4
Define a function
by the rule
Then is not surjective. For instance, we can prove that there is no element in the domain
such that
Suppose that
.
for some
. Then
, which implies that
. But
this would contradict the fact that is an integer.
In Example 4, the function is a mapping from the set of all integers onto the set of odd
integers. Therefore, no even number is within the range of the function, and consequently,
the function is not surjective.
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MTH105
Relations and Functions
A function that is both injective and surjective is termed bijective. A bijective function is
often called a one-to-one onto function, or a one-to-one correspondence.
The identity function on any non-empty set is the simplest example of a bijective
function.
Figure 6.13 An example of a function that is injective but not
surjective
Figure 6.14 An example of a function that is surjective but not
injective
SU6-50
MTH105
Relations and Functions
Figure 6.15 An example of a function that is neither injective nor
surjective
Figure 6.16 An example of a function that is both injective and
surjective
Example 5
Let
Then
be the function from
to
with
is a bijective function.
A bijective function from a set
to itself is also called a permutation of .
SU6-51
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Relations and Functions
For any bijective function
, we can define a function from
inverse function of , denoted by
back to
called the
. As we will see, the inverse function
uniquely determined by the function , and
is
will also be a bijective function.
Proposition:
For a given bijective function
is not just a relation from
, the definition
to , but in fact a legitimate function from
to . The inverse
function is also bijective.
Proof:
We first prove that the definition
produces a legitimate
function, not just a relation. Firstly, let
be given. Then since is surjective, there exists
some
such that
. This implies that
for all
, there exists
such that
Now suppose that
and
we have
have
. Hence, we have shown that
are elements of
; in other words,
. Hence,
. Then by the definition of
. Since
is a legitimate function from
,
is injective, we
to . Very often, such a proof that a
given relation is indeed a function is termed proving that the function is well-defined,
or proving the well-definedness of a function.
We now prove injectivity. Suppose that
. Then
legitimate function,
Hence,
, so
. Since
. Write
is a
.
We now prove surjectivity. Let
such that
for elements
. Thus,
be given. Then since
, and so
is surjective. We have therefore proven that
SU6-52
is a function, there exists some
We deduce that
is bijective.
.
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Relations and Functions
The above can be summarised as follows: Let
be a given bijective function. Let
be given. Since is surjective, there exists some
is injective, this element
. We then define
,
such that
. And because
is unique: there will not be any other
. This produces a function
is equal to the unique element
defining characteristic for the inverse function
For a bijective function
such that
, where for any
such that
. We have the following
:
, we have
and
Example 6
Let
Show that
is injective.
Assume that the codomain of
setting the codomain of
the inverse function
has been defined so as to make
surjective as well (by
to be the set of real numbers excluding the point
.
Answer:
Suppose that
are elements in the domain such that
. Then
Cross multiplying yields
which leads to
Hence we have
, so
. We have shown that
SU6-53
is injective.
. Derive
MTH105
Relations and Functions
Assume now that
is bijective. This can be done by setting the codomain of
set of real numbers excluding the point
to be the
. To derive the inverse function, let
We rearrange the equation to make the subject of the formula. Again, we cross-multiply
to obtain
which implies that
Hence,
The inverse function is then given by
In the above example, the range of was
. Hence, the domain of the function
is the same.
A quadratic function is a function of the form
where
If
are real numbers called coefficients with
, then the graph
.
of the quadratic function has a shape which is termed
concave upwards, and looks like this:
SU6-54
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Relations and Functions
Figure 6.17 Graph of
If
, then the graph
with
of the quadratic function has a shape which is termed
concave downwards, and looks like this:
Figure 6.18 Graph of
with
For a general quadratic function
if
, then the graph
the graph
of the function has a minimum point, and if
, then
of the function has a maximum point. The minimum point or the
maximum point always occurs at
SU6-55
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Relations and Functions
As can be seen from the sketch of the quadratic graphs, a general quadratic function is
never injective if the domain is the set of real numbers
. However, if we change the
domain (or restrict the domain) to either the interval
or the interval
then the function will be injective.
Example 7
Let
Compute the maximum value of
so that the function will be injective for a domain of
the form
For the domain
, derive the inverse function assuming the function
is bijective,
stating both its domain and its range.
Answer:
For the quadratic function
, the leading coefficient is negative, so the
graph has a maximum point. The maximum point will occur at
Hence, the maximum value of
domain of the form
is
in order for the function to be injective within a
. To derive the inverse function, let
SU6-56
MTH105
Relations and Functions
and make the subject of the formula.
We do so by the technique of "completing the square":
The range of
is thus observed to be
This above working leads to
Since the domain of the function is now given by
at the point
,
itself, so we should take the negative square root:
Hence,
with the range of
given by
.
SU6-57
will be negative except
MTH105
Relations and Functions
Activity 6.15
Determine if the following functions
are injective or surjective or both:
a.
b.
c.
Activity 6.16
Let
Compute the minimum value of
so that the function will be injective for a domain
of the form
For the domain
, derive the inverse function assuming the function
stating both its domain and its range.
SU6-58
is bijective,
MTH105
Relations and Functions
Activity 6.17
Determine whether
is onto if
a.
.
b.
.
c.
.
d.
.
e.
.
Activity 6.18
Let
Compute the minimum value of
so that the function will be injective for a domain
of the form
For the domain
, derive the inverse function assuming the function
is bijective,
stating both its domain and its range.
3.2 Composite Functions
Let
and
be functions. The composite function
can be defined when the range of
is a subset of the domain of , in other words, when
SU6-59
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Relations and Functions
The composite function
is the function that maps a real number
We shall follow the convention that the composite
other words, the composite function
to
.
is performed "right to left". In
performs or applies the function
first, and then
followed by the function .
Example 1
Let
be functions given by
and
Then the composite function
The composite function
is given by
can also be determined.
is the function given by
As can be seen from the above example, the order in which the composition is done affects
the end result. In general, the composite function
function
is different from the composite
.
Example 2
The functions
and are defined as follows:
Which of the composite functions
or
exists? Give the formulae of the composite
functions that exist.
Answer:
SU6-60
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Relations and Functions
We have:
Since
, so the composite function
exists and
is given by the
formula
Note
for all real numbers .
The composite function
does not exist because
is not a subset of
.
Theorem:
If
and
are injective functions, then the composite
is also an
injective function.
Proof:
Suppose that
are given, and that
Since is injective, so
And because
. Then
.
is injective, so
. Hence, we have shown that
is also an
injective function.
Theorem:
If
and
are surjective functions, then the composite
surjective function.
Proof:
SU6-61
is also a
MTH105
Let
Relations and Functions
be given. Since
since
is surjective, there exists some
is surjective, there exists some
Hence, for any
shown that
such that
such that
, we can always find some
. It follows that
such that
. Hence, we have
is also a surjective function.
Recall that the identity map on a non-empty set
is the map that sends every
itself.
If
denotes the identity map
for any
Similarly, if
for any
. Hence, the functions
, and
and are equal. We can write
denotes the identity map
. Hence, the functions
is any function, then
, and
is any function, then
and are equal. We can thus write
Theorem:
Let
. And
be a bijective function. Then
and
Proof:
SU6-62
to
MTH105
Relations and Functions
We use the relationship
Let
be given and let
Therefore,
Now, let
. Then
for every
be given and let
Therefore,
for every
, so
. It follows that
. Then
. It follows that
, so
.
Activity 6.19
If
and
are functions and
is onto, must
be onto?
Prove or give a counter-example.
Activity 6.20
The functions
and are defined as follows:
Which of the composite functions
or
exists? Give the formulae of the composite
functions that exist.
SU6-63
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Relations and Functions
Summary
We defined relations, equivalence relations, congruence classes modulo , and used
congruence arithmetic to prove interesting results in divisibility.
For functions, we learnt about domain and range, injective, surjective and bijective
functions, and how to construct the inverse of a bijective function. Lastly, we also studied
composite functions.
In this Study Unit, we also introduces an application of congruence arithmetic – the RSA
Algorithm, which is used for encrypting data so that it does not fall into the wrong hands.
SU6-64
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Relations and Functions
Formative Assessment
1.
Refer to Example 2 of Section 1.1. Let
be the set of rational numbers, and let
the set of integers. We can define a relation
rational number
and integer ,
from
if and only if
to
be
by declaring that for any
has a denominator of
when it
is written as a fraction expressed in its lowest form.
Let be a rational number and let be an integer. Prove that
, and whenever
2.
Define a relation
Prove that
3.
on
on the set of integers
is an equivalence relation on
form the elements of the equivalence class
4.
, then divides .
as follows:
is an equivalence relation on
Define a relation
Prove that
is any integer such that
if and only if
.
as follows:
and determine in the simplest possible
.
Let
Compute the minimum value of so that the function will be injective for a domain of
the form
For the domain
, derive the inverse function assuming the function
is bijective, stating both its domain and its range.
5.
Determine if the function
is injective, surjective, or neither.
SU6-65
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6.
Relations and Functions
If
and
are functions and
is one-to-one, must be one-
to-one? Prove or give a counter-example.
7.
If
and
are functions and
is onto, must
Prove or give a counter-example.
8.
Let
and
. Let
be the public key
i.
Compute the private key .
ii.
Encrypt the plaintext message
iii.
Decrypt the ciphertext message
SU6-66
.
.
be onto?
MTH105
Relations and Functions
Solutions or Suggested Answers
Activity 6.1
a.
For any rational number ,
.
Proof: When we express a rational number as a fraction
denominator
Let
.
be a rational number. Then
or
, the
must be non-zero. It is forbidden to divide by zero. Hence, for
any rational number ,
b.
for integers
is an integer if and only if either
.
Proof: If is an integer, then
and
are fractions in their lowest form, because
. Hence, both
suppose that
as
and
. Conversely,
. Then by definition of the relation ,
for some integer . It follows that
argument can be applied to show that if
can be expressed
and thus is an integer. A similar
,
must likewise be an
integer.
Activity 6.2
,
,
,
,
,
,
,
. Hence in this example,
, and
. Only the element is not inside the range of .
Activity 6.3
From the graph of
and
we can tell that
is the set of all nonnegative real numbers,
is the set of all real numbers. Indeed, we can argue algebraically as follows:
if and only if
, in other words,
if and only if
or
. The
variable is only allowed to take on nonnegative real values because we have to take the
SU6-67
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Relations and Functions
square root of . Another way to argue this point is to observe that
is the square of some real number, and so
if and only if
can only be allowed to be nonnegative. On
the other hand, there is no restriction on the values of , as can be seen from the fact that
or
. Hence, the range of
must be all the real numbers.
Activity 6.4
The relation
Since
is defined by:
for all
achieved by setting
Hence,
, we have
for all
. Equality can be
, that is, we have
is the set of all real numbers
. We can write
using the interval notation defined in Study Unit 5. The variable
can take on any real
number – we can substitute any value of into the equation
to obtain a certain value for . Since there is no restriction on the values that
we have
can take,
.
As for the relation
, the roles of the domain and range are interchanged. Hence we have
, and
.
Activity 6.5
is reflexive and symmetric, but not transitive. For example,
.
is transitive, but not reflexive and not symmetric.
SU6-68
and
, but
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Relations and Functions
is transitive, but not reflexive and not symmetric. In this example, it is vacuously
true that if
and
, then
. The reason why we say that the
conditional statement is vacuously true is because the hypothesis
and
is in fact never satisfied. Therefore the conditional statement is true by default.
Activity 6.6
One possible answer:
.
Activity 6.7
Hence,
Hence,
leaves a remainder of upon division by .
Activity 6.8
which is divisible by
For any integer
Hence,
,
, so
.
is divisible by
because
for any integer
. It follows that
SU6-69
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Relations and Functions
Hence,
leaves a remainder of when divided by
Activity 6.9
but is not congruent to
modulo
Activity 6.10
The equivalence classes are
Activity 6.11
,
,
Preimage of is ; preimage of is ; preimages of are
Activity 6.12
a.
b.
c.
d.
Activity 6.13
Let
be a function defined by
SU6-70
.
.
MTH105
Relations and Functions
,
Let
and let
Then
.
while
.
Activity 6.14
Activity 6.15
a.
Both injective and surjective.
b.
Neither injective nor surjective.
c.
Injective but not surjective.
Activity 6.16
For the quadratic function
, the leading coefficient is positive, so the
graph has a minimum point. The minimum point will occur at
Hence the minimum value of
domain of the form
is
in order for the function to be injective within a
. To derive the inverse function, let
and make the subject of the formula. By completing the square, we have
This gives us
SU6-71
MTH105
Since
Relations and Functions
,
, so we take the positive square root.
Activity 6.17
a.
Onto
b.
Not onto
c.
Onto
d.
Onto
e.
Not onto
Activity 6.18
For the quadratic function
, the leading coefficient is positive, so the
graph has a minimum point. The minimum point will occur at
Hence, the minimum value of
domain of the form
is
in order for the function to be injective within a
. To derive the inverse function, let
and make the subject of the formula. We have
SU6-72
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Relations and Functions
Since
,
, so we take the positive square root.
Activity 6.19
Let
,
Define
Then
,
and
by
is onto (in fact, bijective), but is not onto.
Activity 6.20
Since
, so
exists.
We have
,
SU6-73
.
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Relations and Functions
is not a subset of the domain of , so
does not exist.
Formative Assessment
1.
Refer to Example 2 of Section 1.1. Let
be the set of rational numbers, and let
the set of integers. We can define a relation
rational number
and integer ,
from
if and only if
to
be
by declaring that for any
has a denominator of
when it
is written as a fraction expressed in its lowest form.
Let be a rational number and let be an integer. Prove that
, and whenever
is any integer such that
if and only if
, then divides .
Answer:
Suppose that
. Then by definition of the relation , there exists an integer
such that
and
is a fraction in its lowest form. Thus,
Furthermore, suppose that
is an integer such that
. Let
. Then by cross-multiplying the fractions, we obtain
that divides
. Since
(because
.
. Then
, so we deduce
is a fraction in its lowest form), it
follows from Euclid's Lemma (see Question 1) that divides . We have therefore
proven that whenever
is any integer such that
Conversely, suppose that
then divides . Let
, and whenever
. Then
in its lowest form. Let
Let
. Then
By hypothesis, we have
2.
. Since
is a divisor of ,
is an integer because
. Hence,
Define a relation
as follows:
is an integer.
is also a divisor of .
is a fraction in its lowest form.
is an equivalence relation on
SU6-74
.
,
is a fraction
is an integer, and so 1/
, and so
Prove that
is any integer such that
. We have to prove that
therefore conclude that
on
, then divides .
. We
MTH105
Relations and Functions
Answer:
Clearly, for all natural numbers
the relation
, we have
because
. Hence
is reflexive.
Suppose
are natural numbers, and suppose that
Of course this trivially implies that
. Then
. Hence
.
. So the relation
is symmetric.
Suppose that
Since
and
. Then
is positive, we can divide by
. Hence, the relation
equivalence relation on
3.
Define a relation
Prove that
to obtain
and
. Then
. This implies that
is transitive. We conclude that
is an
.
on the set of integers
is an equivalence relation on
form the elements of the equivalence class
as follows:
and determine in the simplest possible
.
Answer:
Suppose that is an integer. Then
. So
is reflexive.
Suppose that
Hence
Suppose
. Hence,
. So
that
. Then
. Trivially, we also have
is symmetric.
and
.
. Since
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Then
and
MTH105
Relations and Functions
we have
It follows that
. So
is transitive. We conclude that
equivalence relation. Suppose
. Then
is an
so we have
. Hence,
. Since
, we
have
As a result,
is
. Under congruency modulo , the only possibility for
, that is, can be any odd number. So
consists of all the odd
integers.
4.
Let
Compute the minimum value of so that the function will be injective for a domain of
the form
For the domain
, derive the inverse function assuming the function
is bijective, stating both its domain and its range.
Answer:
For the quadratic function
, the leading coefficient is negative,
so the graph has a maximum point. The maximum point will occur at
Hence the minimum value of
is
in order for the function to be injective
within a domain of the form
. To derive the inverse function, let
and make the subject of the formula. We have
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MTH105
Relations and Functions
This gives us
Since
5.
, we take the positive square-root
Determine if the function
is injective, surjective, or neither.
Answer:
Put
For all
,
. Then
as well as
. So
, we have
and also for all such that
So this proves
is not injective.
, we have
for all
particular there does not exist such that
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, so
cannot be surjective because in
, for instance.
MTH105
6.
Relations and Functions
If
and
are functions and
is one-to-one, must be one-
to-one? Prove or give a counter-example.
Answer:
Let
,
,
Define
Then
7.
and
by
is one-to-one (in fact, bijective), but is not one-to-one.
If
and
are functions and
is onto, must
be onto?
Prove or give a counter-example.
Answer:
Yes, is onto. For any
Then
8.
Let
, there exists
. This proves that for any
and
. Let
such that
, there exists
. Let
such that
.
.
be the public key
i.
Compute the private key .
ii.
Encrypt the plaintext message
iii.
Decrypt the ciphertext message
.
.
Answer:
Part (i):
such that
. Execute the Euclidean Algorithm determine integers
.
180 = 3(49) + 33
49 = 1(33) + 16
33 = 2(16) + 1
We have
1 = 33-2(16) = 33-2(49-33)
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MTH105
Relations and Functions
= (-2)(49)+(3)(33)
= (-2)(49)+(3)(180-3(49))
= (-11)(49)+(3)(180)
Since
, the integer
is the private key.
Part (ii):
The plaintext
The
following
is encrypted to the ciphertext
is
the
working
for
.
reducing
modulo
Hence,
Part (iii):
The ciphertext
is deciphered to the plaintext
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.
209:
MTH105
Relations and Functions
References
Epp, S. S. (2019). Discrete mathematics with applications (5th ed.). Boston, MA: Brooks/
Cole Cengage Learning.
Leong, Y. K. (2011). Living with mathematics (3rd ed.). Singapore: McGraw-Hill
Education.
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