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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 
, x+y
A2. For all x, y 
, x + y = y + x.
A3. For all x, y, z 
and if x = w and y = z, then x + y = w + z. Addition
Commutative Property
, x + ( y + z) = (x + y) + z.
Associative Property
A4. There is a unique real number 0 such that x + 0 = x for all x 
.
Additive Identity
A5. For each x 
there is a unique real number  x such that x + ( x) = 0. Add. Inverse
M1. For all x, y 
, xy
M2. For all x, y 
, x  y = y  x.
M3. For all x, y, z 
and if x = w and y = z, then x  y = w  z. Multiplication
Commutative Property
, x  ( y  z) = (x  y)  z.
Associative Property
M4. There is a unique real number 1 such that 1  0 and x  1 = x for all x 
M5. For each x 
DL. For all x, y, z 
. Mult. Identity
with x  0, there is a unique real number 1/ x such that x  (1/ x) = 1.
, x  ( y + z) = (x  y) + (x  z)
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.
O1. For all x, y 
trichotomy law
, exactly one of the relations x = y, x > y, or x < y holds.
O2. For all x, y, z 
, if x < y and y < z, then x < z.
O3. For all x, y, z 
, if x < y then x + z < y + z.
O4. For all x, y, z 
, if x < y and z > 0, then x  z < y  z.
transitive property
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.
(b) x  0 = 0.
(c) – 0 = 0.
(d) (–1)  x = – x.
(e) x  y = 0 iff x = 0 or y = 0.
(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)
x + [z + (– z)] = y + [z + (– z)]
by A5 (add. inverse)
and A1 (addition)
by A3 (assoc. property)
x+0 = y+0
by A5 (add. inverse)
x = y
by A4 (add. identity)
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
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.
We have
x + (–1)  x = x + x  (–1)
by M2 (commutative)
= 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)
Thus (–1)  x = – 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 are an example of an ordered field. But there are other examples as well.
The rational numbers
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.
is the set of all quotients of polynomials. A typical element of
an x n 
bk x k 
 a1x  a0
,
 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. We can define an order on by saying that a
quotient such as above is positive iff an and bk have the same sign; that is, an  bk > 0.
For example,
3x 2  4 x  1
 0, since (3)(7) > 0.
5
7x  5
4 x5  3 x  2
 0, since (4)(–7) < 0.
But
2
3
6x  7x
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
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,
2
4
4
3
x2
3
x
21

x

x
 21
We have
or
?



 0.
7
x2
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
as we shall see later.
has a number of interesting properties,
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 +  for every  > 0. Then x  y.
Proof: We shall establish the contrapositive.
By axiom O1 (the trichotomy law), the negation of x  y is x > y. Thus we suppose
that x > y and we must show that there exists an  > 0 such that x > y + .
Question: If x > y, what positive  can we add on to y so that x > y +  ?

y
x
We could take  equal to half
the distance from x to y.
Let  = (x – y) /2. Since x > y,  > 0. Furthermore,
y   y 
x y
x y
xx


 x, as required. 
2
2
2
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
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  a  x  a. If x  0, then | x | = 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  (| x |  | y |)  x  y  | x |  | y |,
which implies that | x + y |  | x | + | y | by part (b). 
Copyright © 2013, 2005, 2001 Pearson Education, Inc.
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.
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