Analysis with an introduction to proof 5th by steven lay ch02d

Chapter 3
The Real Numbers

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 3.2, Slide 1


Section 3.2
Ordered Fields

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 3.2, Slide 2


The real numbers

are an example of an ordered field.

We have two operations, + and ⋅ , called addition and multiplication, such that
the following properties apply:
A1. For all x, y ∈
A2. For all x, y ∈

, x+y∈

and if x = w and y = z, then x + y = w + z.

, x + y = y + x.

A3. For all x, y, z ∈

Commutative Property

, x + ( y + z)

Associative Property

= (x + y) + z.

A4. There is a unique real number 0 such that x + 0 = x for all x ∈
A5. For each x ∈
M1. For all x, y ∈
M2. For all x, y ∈

, x⋅y∈

, x ⋅ ( y ⋅ z)

Additive Identity

.

there is a unique real number − x such that

Multiplication

Commutative Property
Associative Property

= (x ⋅ y) ⋅ z.

M4. There is a unique real number 1 such that 1 ≠ 0 and x ⋅ 1 = x for all x ∈
M5. For each
DL. For all x, y, z ∈

Add. Inverse

x + (− x) = 0.

and if x = w and y = z, then x ⋅ y = w ⋅ z.

, x ⋅ y = y ⋅ x.

M3. For all x, y, z ∈

Addition

, x ⋅ ( y + z)

x∈

with x ≠ 0, there is a unique real number 1/ x

= (x ⋅ y) + (x ⋅ z)

.

Mult. Identity

such that x ⋅ (1/ x) = 1.

Distributive Law

Mult. Inverse

These 11 axioms are called the field axioms.

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 3.2, Slide 3


In addition to the field axioms, the real numbers also satisfy four order axioms.
These axioms indentify the properties of the relation “<”.
Note: a > b means b < a.

trichotomy law

O1. For all x, y ∈
O2. For all x, y, z ∈
O3. For all x, y, z ∈

, exactly one of the relations x = y, x > y, or x < y holds.
, if x <

, if x <

O4. For all x, y, z ∈

transitive property

y and y < z, then x < z.

y then x + z < y + z.

, if x <

y and z > 0, then x ⋅ z < y ⋅ z.

Our first theorem shows how the axioms may be used to derive some familiar algebraic
properties.

Theorem 3.2.1

Let

x, y, and z be real numbers.

(a) If x + z = y + z, then x = y.
(c) – 0 = 0.
(e) x ⋅ y = 0 iff x = 0 or y = 0.

(b) x ⋅ 0 = 0.
(d) (–1) ⋅ x = – x.
(f) x < y iff – y

<

– x.

(g) If x < y and z < 0, then x ⋅ z > y ⋅ z.
We illustrate the proofs by doing parts (a) and (d).
Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 3.2, Slide 4


Theorem 3.2.1 Let x, y and z be real numbers.
(a) If x + z = y + z, then x = y.

Proof:
If x + z = y + z,
then (x + z) + (– z) = ( y + z) + (– z)

by A5 (add. inverse)
and A1 (addition)

x + [z + (– z)] = y + [z + (– z)]

x+0 = y+0

x = y

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

by A3 (assoc. property)

by A5 (add. inverse)

by A4 (add. identity)

Section 3.2, Slide 5


Theorem 3.2.1 Let x, y and z be real numbers.
(d) For any real x, (–1) ⋅ x = – x.
Question: What is –1?

Answer: –1 is the number which when added to 1 gives 0.

Question: What is – x?

Answer: – x is the number which when added to x gives 0.

Question: So how do we show that (–1) ⋅ x =

– x?

Answer: We show that (–1) ⋅ x satisfies definition of – x.
Namely, when (–1) ⋅ x is added to x, the result is 0.

Proof:
We must show that x + (–1) ⋅ x

= 0.

x + (–1) ⋅ x

We have

Thus (–1) ⋅ x

by M2 (commutative)

= x + x ⋅ (–1)

= x ⋅ (1) + x ⋅ (–1)

by M4 (mult. identity)

= x ⋅ [1 + (–1)]

by DL (distributive law)

= x⋅0

by A5 (add. inverse)

= 0

by part (b)

= – x by the uniqueness of – x in A5. ♦

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 3.2, Slide 6

Any mathematical system that satisfies these fifteen axioms is called an ordered field.
The real numbers
The rational numbers

But there are other examples as well.

are an example of an ordered field.
are another example.

Example 3.2.6
For a more unusual example of an ordered field, let
That is,

be the set of all rational functions.
A typical element of

is the set of all quotients of polynomials.

an x n + L + a1x + a0
,
k
bk x + L + b1x + b0

looks like

where the coefficients are real numbers and bk ≠ 0.

Using the usual rules for adding, subtracting, multiplying, and dividing polynomials,
it is not difficult to verify that

is a field.

quotient such as above is positive iff

We can define an order on

by saying that a

an and bk have the same sign; that is, an ⋅ bk > 0.

3x 2 + 4 x − 1
> 0,
For example,
since (3)(7) > 0.
5
7x + 5
4 x5 + 3x − 2
< 0,
2But
3
6x − 7x

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

since (4)(–7) < 0.

Section 3.2, Slide 7


If

p /q and f /g are rational functions, then we say that

That is,

p f
>
q g

iff

p f
>
q g

iff

p f
− > 0.
q g

pg − fq
> 0.
qg

Practice 3.2.7*

Which is larger,

3

x2
or
?
7
x2

We have

3 x2
21 − x 4
− x 4 + 21
−
=
=
< 0.
x2 7
7 x2
7 x2

3
x2
So, 2 <
.
7
x
The verification that “> ” satisfies the order axioms is Exercise 11.

It turns out that the ordered field

has a number of interesting properties,

as we shall see later.

There is one more algebraic property of the real numbers to which we give special
attention because of its frequent use in proofs in analysis, and because it may not familiar.

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 3.2, Slide 8


Theorem 3.2.8

Let x, y ∈

such that

x ≤ y + ε for every ε > 0. Then x ≤ y.

Proof: We shall establish the contrapositive.
By axiom O1 (the trichotomy law), the negation of

that x > y and we must show that there exists an ε
Question: If

Thus we suppose

x ≤ y is x > y.

> 0 such that x > y + ε.

x > y, what positive ε can we add on to y so that x > y + ε ?

ε

We could take ε
y

Let ε =

(x – y) /2.

x

Since x > y, ε > 0.

y +ε = y +

the distance from x to y.

Furthermore,

x− y
x+ y
x+x
=
<
= x,
2
2

2

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

equal to half

as required. ♦

Section 3.2, Slide 9


Recall the definition of absolute value from Section 1.4.

Definition 3.2.9
If x ∈

, then the absolute value of x, denoted by | x |, is defined by

 x, if x ≥ 0,
| x| =
− x, if x < 0.
The basic properties of absolute value are summarized in the following theorem.

Theorem 3.2.10
Let

x, y ∈

and let a > 0. Then

(a) | x | ≥ 0,

(b) | x | ≤ a iff − a ≤ x ≤ a,

(c) | x y | = | x | ⋅ | y |,

(d) | x + y | ≤ | x | + | y |.

We will prove parts (b) and (d).

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 3.2, Slide 10


Theorem 3.2.10 (b)

Let

x, y ∈

and let a > 0. Then

| x | ≤ a

iff − a ≤ x ≤ a.

Proof:
Since x = | x | or x = − | x |, it follows that − | x | ≤ x ≤ | x |.

If | x | ≤ a, then we have

− a ≤ − | x | ≤ x ≤ | x | ≤ a.
Conversely, suppose that

−

If x ≥ 0, then | x | = x ≤ a.

a ≤ x ≤ a.

And if x < 0, then | x | = − x ≤ a.

In both cases, | x | ≤ a.

Theorem 3.2.10 (d)

Let

x, y ∈

and let a > 0. Then

| x + y | ≤ | x | + | y |.

Proof:
As in part (b), we have

− | x | ≤ x ≤ | x | and − | y | ≤ y ≤ | y | .

Adding the inequalities together, we obtain
which implies that | x + y |

≤|

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

− (| x | + | y |) ≤ x + y ≤ | x | + | y |,

 x | + | y | by part (b). ♦

Section 3.2, Slide 11


Part (d) of Theorem 3.2.10 is referred to as the triangle inequality:

| x + y | ≤ | x | + | y |.

Its name comes from its being used with vectors in the plane, where

| x | represents

the length of vector x.

x+

y
y

x

It says that the length of one side of a triangle is less than or equal to the sum of the
lengths of the other two sides.

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 3.2, Slide 12