Chapter 2
The Structure of R
The purpose of this chapter is to explain to the reader why the set of real
numbers is so special. By the end of this chapter, the reader should understand
the di¤erence between R and Q. We will see that R is complete while Q is
not. We will also see that Q is countable while R is not. We will explain the
consequences these di¤erences have.
2.1
The set of Real Numbers
Though it is possible to construct the set of real numbers from scratch (as did
Cantor and Dedekind) and derive its properties from the fundamental axioms of
set theory, this process is too big a task and beyond the scope of an introductory
course in real analysis. Instead, we will make certain assumptions and derive all
the important properties of R from these assumptions. Most of the properties
we will derive are already known to you. The goal here is to see how they can
be proven using only our assumptions and what has already been proven.
More precisely, we assume we are given the following:
1. The set of real numbers R.
2. Two binary operations (addition, denoted + and multiplication, denoted
:)
3. An order relation < : See section 2.2
4. We will also assume a set of properties (see section 2.1.2). From these assumptions, we will prove most of the important results about real numbers,
needed for this class. Finally, we will assume the axiom of completeness
(see section 2.4). When we want to talk about the real numbers together
with the two binary operations, we will use the notation (R; +; :). Before
we do this though, let us give an intuitive presentation of the real numbers.
31
32
2.1.1
CHAPTER 2. THE STRUCTURE OF R
Intuitive De…nition of the Real Numbers
The concept of number is fundamental to our way of life. The concept of natural
numbers, or the numbers we use to count is grasped as an early age. This set,
is denoted N. Thus,
N = f1; 2; 3; :::g
Very soon one realizes that the natural numbers are not enough. If we think
of numbers as used in counting to represent things one owns, then how does
one represent what one owes? The answer is by putting a negative sign. If we
combine the natural numbers, their negative and zero, we get the set of integers.
This set is denoted by Z. Thus
Z = f0; 1; 2; 3; :::g
As one learns about multiplication and division, one realizes that Z is not
enough. For example, if we try to divide 2 by 3, the answer is not an integer.
Thus, we are quickly forced to introduce new numbers, the rational numbers.
The set of rational numbers is denoted by Q. Its de…nition is
o
nm
: m 2 Z, n 2 Z and n 6= 0
Q=
n
Then, we ask the question: do we have all the numbers we’ll ever need with the
rational numbers? The answer is no, and it has been known for a long time.
Around 500 BC, Pythagora knew that not every quantity could be expressed
as a rational number. Consider for example a right triangle in which the length
of the sides around the right angle is 1. Then, the length x of the hypotenuse
satis…es 12 + 12 = x2 . In other words x2 = 2. What is the number x such that
x2 = 2? We know thispnumber exists because we can
p construct such a triangle.
We call this number 2: Pythagora proved that 2 is not a rational number.
It is not the only one.
is not rational, neither is e. If p is a prime number,
p
one can show that p is not rational. When a number is not rational, we say
that it is irrational. There are more irrational numbers than rational ones. If
we represent the rational numbers as dots on a line, there will be more empty
spaces than dots. There are several categories of numbers which are irrational.
We will not discuss those here.
p
Before we look at the properties of the real numbers, we prove that 2 is
irrational using Pythagora’ proof. We begin with a lemma which proof is left
as an exercise.
Lemma 79 Let n be an integer. If n2 is even then n must also be even.
Proof. See exercises.
We are now ready for Pythagora’ proof. We state the result as a theorem
and give its proof.
Theorem 80 There does not exist a rational number r such that r2 = 2.
2.1. THE SET OF REAL NUMBERS
33
Proof. We do a proof by contradiction. Suppose there exists integers a and
a
a
b 6= 0 such that r = , r2 = 2 and assume
is already in lowest terms that is
b
b
a2
a and b do not have any common factors. Then, 2 = 2 or a2 = 2b2 . Thus a2
b
is even. It follows that a is even from lemma 79. Thus there exists a number k
such that a = 2k. It follows that 2b2 = 4k 2 or b2 = 2k 2 . Hence, b2 is even, thus
a
b is even. Which contradicts the fact that was in lowest terms.
b
Let us now look at the essential properties which de…ne the set of real numbers.
2.1.2
Algebraic Properties of R
(R; +; :) is a …eld that is the following properties are satis…ed:
Closure R is closed under both operations, that is if a 2 R and b 2 R then a+b 2 R
and ab 2 R.
(A1) 8a; b 2 R, a + b = b + a. (+ is commutative)
(A2) 8a; b; c 2 R, (a + b) + c = a + (b + c). (+ is associative)
(A3) There exists an element denoted 0 in R with the property: 8a 2 R, a+0 =
0 + a = a. 0 is called the additive identity element.
(A4) For each element a in R, there exists an element denoted a, called the
additive inverse of a with the property: a + ( a) = ( a) + a = 0.
(M1) 8a; b 2 R, a:b = b:a. (: is commutative)
(M2) 8a; b; c 2 R, (a:b) :c = a: (b:c). (: is associative)
(M3) There exists an element denoted 1 6= 0 in R with the property: 8a 6= 0 2 R,
a:1 = 1:a = a. 1 is called the multiplicative identity element.
(M4) For each element a 6= 0 in R, there exists an element denoted 1=a, called
the multiplicative inverse of a with the property: a: (1=a) = (1=a) :a = 1.
(D) 8a; b; c 2 R, a: (b + c) = (a:b) + (a:c) and (b + c) :a = (b:a) + (c:a). (multiplication is distributive over addition)
Remark 81
1. Properties (A2)
(A4) mean that (R; +) is a group.
2. Adding property (A1) means that (R; +) is a commutative group also
called an abelian group.
34
2.1.3
CHAPTER 2. THE STRUCTURE OF R
Important Basic Properties of Real Numbers
The properties which follow do not depend on the fact that we are dealing with
real numbers. They depend on the fact that R is a …eld. Thus, they do not really
belong to a real analysis class. Students should have studied (or will study) in
a modern algebra class these properties and how to prove them. They are given
here because they are essential to the everyday manipulations we perform when
we work with real numbers. However, we will not spend time on them. We take
many of these properties for granted. Yet, without them, many of the things we
do with real numbers would not be possible. First, we establish uniqueness of
the identity element as well as uniqueness of the inverse under both operations.
This fact is a direct consequence of the properties of each operation; it has
nothing to do with the fact that we are working with real numbers.
Theorem 82 The additive and multiplicative identity are unique.
Proof. We only prove uniqueness of the additive identity. For this, we will
prove that if x is an element of R such that x + a = a for every a 2 R, then
x = 0.
x=x+0
= x + (a + ( a))
= (x + a) + ( a)
= a + ( a)
=0
Theorem 83 The additive and multiplicative inverses are unique.
Proof. We only prove uniqueness of the additive inverse. For this, we will
prove that if a and b are elements of R such that a + b = 0 then b = a.
b=0+b
= (( a) + a) + b
= ( a) + (a + b)
= ( a) + 0
=
a
Remark 84 Properties (A4) and (M4) guarantee the possibility of solving the
equations a + x = 0 and a:x = 1. Theorem 83 guarantees the uniqueness of the
solution. We now generalize this result.
Many problems in mathematics involve solving one or more equations. The
next two theorems play an important role in …nding the solutions of equations.
2.1. THE SET OF REAL NUMBERS
35
Theorem 85
1. Let a and b be arbitrary elements of R. then the equation
a + x = b has the unique solution x = ( a) + b
2. Let a 6= 0 and b be arbitrary elements of R. then the equation a:x = b has
the unique solution x = b: (1=a)
Proof. In both cases, we have to show that there is a solution. This can be
done easily by verifying that the given solution is indeed a solution. Then, we
have to show that it is unique. The details of the proof are left as an exercise.
Theorem 86 (Cancellation Laws) Suppose that a, b and x are real numbers.
Then the following is true:
1. a + x = b + x =) a = b
2. If in addition x 6= 0 then a:x = b:x =) a = b
Proof. We prove each part separately.
(a) Choose y such that x + y = 0. Then,
a+x
=
b + x =) (a + x) + y = (b + x) + y
=) a + (x + y) = b + (x + y)
=) a + 0 = b + 0
=) a = b
(b) Since x 6= 0, choose y such that xy = 1. Then
a:x
=
b:x =) (a:x) :y = (b:x) :y
=) a: (x:y) = b: (x:y)
=) a:1 = b:1
=) a = b
We now look at various miscellaneous properties of the set of real numbers.
The order in which we list them is important in the sense that the proof of some
of these properties depend on other properties.
Theorem 87 If a and b are any elements of R then
1. a:0 = 0
2.
a = ( 1) :a
3.
(a + b) = ( a) + ( b)
4.
( a) = a
36
CHAPTER 2. THE STRUCTURE OF R
5. ( 1) : ( 1) = 1
Proof.
1. Since 0+0 = 0, we have (0 + 0) :a = 0:a. Using the distributive law we can
write it as 0:a+0:a = 0:a. This in turn can be written as 0:a+0:a = 0:a+0.
Finally, using the cancellation law for addition, we obtain 0:a = 0.
2. Theorem 83 establishes the fact that every non-zero real number a has a
unique additive inverse a. If we can show that ( 1) :a has the properties
of the additive inverse of a, the result will follow.
a + ( 1) :a = 1:a + ( 1) :a
= (1 + ( 1)) :a
= 0:a
=0
3. Left as an exercise
4. a + ( a) = 0 implies that a is the additive inverse of a. But the additive
inverse of a is, by (A4) ( a). Since the additive inverse of a element
is unique, it follows that ( a) = a.
5. The details are left as an exercise. Combine parts 2 and 4 .
Theorem 88
1. If a 2 R and a 6= 0, then 1=a 6= 0 and 1= (1=a) = a
2. Suppose that a 2 R, b 2 R . Then, a:b = 0 () a = 0 or b = 0.
3. Suppose that a 2 R, b 2 R . Then ( a) : ( b) = a:b
4. If a 2 R and a 6= 0, then 1= ( a) =
Proof.
(1=a)
1. Since a 6= 0, 1=a exists by (M4). If we had 1=a = 0, we would have
1 = a: (1=a)
= a:0
=0
which is a contradiction. Thus, 1=a 6= 0.
Since a: (1=a) = 1, a is the multiplicative inverse of (1=a). But, by (M4),
the multiplicative inverse of 1=a is 1= (1=a). Since the multiplicative inverse of an element is unique, 1= (1=a) = a.
2. Left as an exercise.
2.1. THE SET OF REAL NUMBERS
37
3. Left as an exercise.
4. Since a 6= 0,
a 6= 0, so 1= ( a) exists . Furthermore,
1 = a: (1=a)
= ( a) ( (1=a)) by part 3 of this theorem
Thus, -(1=a) is the multiplicative inverse of a. But by (M4), the multiplicative inverse of a should be 1= ( a). By uniqueness of the multiplicative inverse, it follows that 1= ( a) = (1=a).
From this point on, we will drop the use of the dot to denote multiplication.
We will write:
ab instead of a:b
a2 for aa, a3 for a2 a = aaa, ... an+1 for (an ) a for n 2 N.
b
a
a for b + ( a) or ( a) + b
1
for 1=a and a
n
for 1= (an )
a
or a=b for a (1=b)
b
2.1.4
Exercises
As you do these problems, keep in mind that you already know all these results.
The goal is to prove them using all the assumptions we have made as well as
what you have already proven. When I ask you to prove a result stated in the
notes, you can only use all the assumptions we have made as well as the results
proven up to the result I am asking you to prove. If the question is a stand
alone question, then you can use all the results stated in the notes as well as all
the problems up to the question you are working on.
1. Finish proving theorem 82.
2. Finish proving theorem 83.
3. Finish proving theorem 85.
4. Finish proving theorem 87.
5. Finish proving theorem 88.
6. Prove that a + a = 0 =) a = 0 for any arbitrary real number a.
7. Prove that if n is an integer and n is even, then n2 is also even.
8. Prove lemma 79.
38
CHAPTER 2. THE STRUCTURE OF R
9. Prove that if n2 is an odd integer, then n must also be odd.
p
10. Prove that 6 is not a rational number.
11. Prove that if x is irrational and r is rational, then x + r is irrational. Also,
show that if r is a non-zero rational then xr is irrational.
12. Show by example that if x and y are irrational, then x + y and xy may be
rational.
2.2. ORDER AXIOMS AND ORDER PROPERTIES OF R
2.2
39
Order Axioms and Order Properties of R
In this section, we introduce the order properties of R. These properties are very
important and will be used heavily throughout the remainder of this course.
Like in the previous section, we do not develop these properties from scratch.
We assume a few of them, then we derive most of the properties from our
assumptions. The properties we assume will be called axioms.
2.2.1
Order Axioms for R and De…nitions
On R there is a relation, denoted <, which satis…es the following axioms:
Axiom 89 (O1) For all a and b in R, exactly one of the following holds:
1. a < b
2. a = b
3. b < a
Axiom 90 (O2) For all a, b and c in R, if a < b and b < c then a < c.
Axiom 91 (O3) For all a, b and c in R, if a < b then a + c < b + c.
Axiom 92 (O4) For all a, b and c in R, if a < b and 0 < c then ac < bc.
Remark 93 The …rst axiom, axiom 89, is known as the law of trichotomy.
Remark 94 The second axiom, axiom 90, is simply the statement of the property called transitivity.
Remark 95 If we represent the set of real numbers as a line stretching to in…nity in both directions (the real line), then by convention if a < b, then a is
positioned on the line to the left of b.
De…nition 96 We can now de…ne some well known terms. Let a and b be in
R.
1. If a < b, we say that a is less than b.
2. If a < b then we also write b > a and say b greater than a .
3. If a < b or a = b then we write a
b and say a less than or equal to b
4. Similarly, we can de…ne greater than or equal to.
5. If a > 0, we say that a is positive. If a < 0, we say that a is negative.
6. If a
0, we say that a is non-negative. If a
non-positive.
0, we say that a is
40
CHAPTER 2. THE STRUCTURE OF R
Remark 97 If we have a < b and b < c, it is convenient to combine the two
relations and write a < b < c. The same applies for the other inequalities.
Remark 98 A …eld together with an order that is a relation satisfying axioms
89 - 92 is called an ordered …eld. So, (R; +; :; <) is an ordered …eld.
Remark 99 The opposite of the statement a < b is a
opposite of the statement a > b is a b.
2.2.2
b. Similarly, the
Properties of the Order
We now establish the properties the relations we just de…ned satisfy. Most of
these properties are already known to the reader. We will prove a few of them.
The remaining proofs will be assigned as exercises.
Theorem 100 For every xin R, the following is true:
1. If x < 0 then
x > 0.
2. If x > 0 then
x < 0.
Proof. We only prove the …rst part. The second part is left as an exercise. If
x < 0 then by O3 x + ( x) < 0 + ( x). It follows that 0 < x which is the
same as x > 0.
Theorem 101 Let a 6= 0 2 R.
1. a2 > 0
2. 1 > 0
Proof. We prove each part separately.
1. Since a 6= 0, either a > 0 or a < 0. If a > 0, then a2 = aa > 0:a = 0,
hence a2 > 0. If a < 0, then a > 0. Hence, ( a) ( a) > 0: ( a) = 0.
But ( a) ( a) = aa = a2 . So, a2 > 0.
2. Follows from part 1 and the fact that 1 = 12
Theorem 102 Let a, b, c be elements of R.
1. a < b () a
b<0
2. a > b and c < 0 ) ac < bc
3. a > 0 ) 1=a > 0
4. a < 0 ) 1=a < 0
2.2. ORDER AXIOMS AND ORDER PROPERTIES OF R
41
Proof. left as an exercise.
1
Theorem 103 If a and b are any real numbers, then a > b =) a > (a + b) >
2
b
Proof. left as an exercise.
Remark 104 Theorem 103 implies (by setting b = 0) that given any strictly
1
positive number a, there exists another smaller strictly positive number ( a).
2
Thus there is no smallest strictly positive number.
Theorem 105 Let a and b be arbitrary real numbers. If a > 0 and b > 0 then
ab > 0
Proof. If a > 0 and b > 0 then by O4 ab > 0:b = 0. Therefore, ab > 0.
Theorem 106 If ab > 0 then we either have a > 0 and b > 0 or we have a < 0
and b < 0.
Proof. If ab > 0, then ab 6= 0, hence neither a = 0 nor b = 0. By the law
1
1
of trichotomy, either a > 0 or a < 0. If a > 0, then b = b a
= (ba) =
a
a
1
(ab) > 0 (why?) Hence, b > 0. The proof is similar if we assume that a < 0.
a
Corollary 107 If ab < 0 then we either have a > 0 and b < 0 or we have a < 0
and b > 0.
Proof. Left as an exercise.
2.2.3
Exercises
1. Finish proving theorem 100.
2. Finish proving theorem 102.
3. Finish proving theorem 103.
4. Finish proving theorem 106.
5. Finish proving corollary 107.
6. Prove that If a
numbers.
b and b
a then a = b where a and b are arbitrary real
7. Prove the following where a; b; c and d denote arbitrary elements of R.
(a) a < 0 and b > 0 =) ab < 0
(b) a < 0 and b < 0 =) ab > 0
(c) 0 < a < 1 =) a2 < a
(d) 1 < a =) a2 > a
42
CHAPTER 2. THE STRUCTURE OF R
(e) a < a + 1
(f) 0 < a < b =) a2 < b2
(g) a
b and c
d =) a + c
b+d
8. If a; b 2 R and a2 + b2 = 0, show that a = b = 0.
2.3. METRIC PROPERTIES OF R - ABSOLUTE VALUE
43
Metric Properties of R - Absolute Value
2.3
2.3.1
De…nition and Properties
The trichotomy property assures that if a 6= 0 then either a > 0 or a < 0. Thus,
we de…ne:
De…nition 108 (Absolute value) The absolute value of a real number a, denoted jaj is de…ned by:
a if a 0
jaj =
a if a < 0
The domain of this function is R, its range is the set of non-negative real
numbers, and it maps the elements a and a into the same element.
Geometrically, the absolute value of a number x can be interpreted as how
far from 0 x is. Similarly, jx aj can be interpreted as the distance between x
and a and therefore, one can think of the absolute value as a tool to measure
distances between numbers. In particular, for a …xed number a, the set de…ned
by fx 2 R such that jx aj < g denotes the set of numbers within a distance
of a. The inequality jx aj < is equivalent to
< x a < or a
<x<
a + that is x 2 (a
; a + ). This will be used a lot throughout this class.
Remark 109 With a = 0 above, we see that
jxj < . The same is true if we replace < by .
< x <
is equivalent to
Students should review how to solve equations and inequalities involving
absolute values as they too will be used a lot in this class.
Theorem 110 The following properties hold:
1. jaj = 0 , a = 0
2. j aj = jaj for all a in R
3. jabj = jaj jbj for all a; b in R
4.
1
1
=
for all a 6= 0 in R
a
jaj
5.
a
jaj
=
for all a; b 6= 0 in R
b
jbj
6. If c
7.
8.
p
jaj
0, then jaj
a
jaj
a2 = jaj
Proof. See problems
c,
c
a
c
44
CHAPTER 2. THE STRUCTURE OF R
The next result is often used in Mathematics.
Theorem 111 (Triangle inequality) If a and b are any real numbers, then
jjaj
jbjj
ja + bj
jaj + jbj
Proof. First, we prove the right side. Since jaj a jaj and jbj b jbj
it follows that (jaj + jbj) a + b jaj + jbj. Thus, ja + bj jaj + jbj by remark
109 or part 6 of theorem 110. Next, we prove the left side.
jaj = ja + b
bj
ja + bj + j bj
= ja + bj + jbj
Thus,
jaj
jbj
ja + bj
Similarly,
jbj =
=
ja + b
aj
ja + bj + j aj
ja + bj + jaj
Thus
jbj
jaj
jaj
jbj
ja + bj
ja + bj
Combining the two inequalities, we get
ja + bj
jaj
jbj
It follows by part 6 of theorem 110 that jjaj
ja + bj
jbjj
ja + bj
An application of the above theorem, often used when dealing with absolute
values and inequalities is given by the corollary below.
Corollary 112 If a; b and c are any given real numbers then ja cj ja bj +
jb cj
Proof. If we let a c = (a b) + (b c) and use the triangle inequality, we
have
ja
cj = j(a
ja
b) + (b
bj + jb
c)j
cj
2.3. METRIC PROPERTIES OF R - ABSOLUTE VALUE
45
Corollary 113 If a1 ; a2 ; :::an are any n real numbers, then
ja1 + a2 + ::: + an j
ja1 j + ja2 j + ::: + jan j
Proof. See exercises.
We …nish with a theorem which appears simple in statement but whose proof
can sometimes give a hard time to beginners. The theorem simply states that
the only non-negative real number less that every positive number is 0.
Theorem 114 Let x be a real number. If jxj <
Proof. See exercises.
for each
> 0 then x = 0.
Remark 115 This result is often used to show two numbers are equal. If a and
b are two real numbers, to show a = b, it is enough to show that ja bj < for
each > 0. By the theorem, it will follow that a b = 0 or a = b. Geometrically,
we are simply saying that if the distance between a and b is less than any positive
real number then a = b.
2.3.2
Equations and Inequalities Involving Absolute Values
The purpose of this class is not to learn how to solve equations and inequalities
involving absolute values. However, they will be used a lot throughout this
class. Students should make sure they review them. We remind students of the
basic principles used when solving equations and inequalities involving absolute
values. For what follows, let a and b denote any expression.
The equation jaj = b is equivalent to the two equations a = b and a =
However, solutions must be checked, some may not be valid.
b.
The equation jaj = jbj is equivalent to the two equations a = b and a =
However, solutions must be checked, some may not be valid.
b.
The inequality jaj < b is equivalent to
b<a<b
The inequality jaj
b
b is equivalent to
a
b
The inequality jaj > b is equivalent to a > b or a <
b
The inequality jaj
b
2.3.3
b is equivalent to a
b or a
Notion of a Metric Space
As we noted earlier, the absolute value can be used to measure distances between
numbers. jx yj denotes the distance between x and y on the real line. For the
rest of these notes, we will use the absolute value every time we need to measure
distances between two real numbers. However, if the set in which our elements
come from were not R, the absolute value may not be appropriate to measure
46
CHAPTER 2. THE STRUCTURE OF R
distances. For example, you already know that in R2 ,q
the distance between two
2
2
points of coordinates (x1 ; y1 ) and (x2 ; y2 ) is given by (x2 x1 ) + (y2 y1 ) .
In more general settings, we replace the absolute value by a function we call a
distance function or a metric. We give a formal de…nition for a metric.
De…nition 116 (metric) Let X be a set. A metric or a distance on X is a
function d : X X ! R which satis…es the properties below. In the properties
below, we let x; y and z be elements of X.
1. d (x; y)
0 (non-negativity)
2. d (x; y) = 0 () x = y
3. d (x; y) = d (y; x) (symmetry)
4. d (x; z)
d (x; y) + d (y; z) (triangle inequality)
De…nition 117 (metric space) A space X together with a metric d is called
a metric space. We say that (X; d) is a metric space.
Example 118 It is easy to see that the absolute value function is a metric for
R, hence we say that (R; j:j) is a metric space.
2.3.4
Exercises
1. Prove theorem 110.
2. Prove the second corollary of theorem 111.
3. Show that ja + bj = jaj + jbj , ab
0.
4. Given real numbers a and b, prove that jaj
jbj
5. Given real numbers a and b, prove that jjaj
times called the reverse triangle inequality.
jbjj
6. Prove theorem 114.
7. Solve j2x
3j = x.
8. Solve j2x
3j = 3x.
9. Solve j2x + 5j > 11.
10. Solve j2x + 5j < 11.
ja
ja
bj.
bj. This is some-