1. Chapter 3
The Real Numbers

2. Section 3.2
Ordered Fields

3. 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  and if x = w and y = z, then x + y = w + z.
Addition
A2. For all x, y  , x + y = y + x.
Commutative Property
A3. For all x, y, z  , 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  and if x = w and y = z, then x  y = w  z.
Multiplication
M2. For all x, y  , x  y = y  x.
Commutative Property
M3. For all x, y, z  , 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  .
Mult. Identity
M5. For each x  there is a unique real number 1/ x such that x  (1/ x) = 1.
Mult. Inverse
DL. For all x, y, z  , x  ( y + z) = (x  y) + (x  z)
Distributive Law
These 11 axioms are called the field axioms.

4. 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  , 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.
transitive property
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.
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).

5. 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)]
by A3 (assoc. property)
x + 0 = y + 0
by A5 (add. inverse)
x = y
by A4 (add. identity)

6. 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. 

7. 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 be the set of all rational functions.
That is, is the set of all quotients of polynomials.
A typical element of 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, since (3)(7) > 0.
But since (4)(–7) < 0.

8. If p /q and f /g are rational functions, then we say that
That is,
Practice 3.2.7*
Which is larger,
We have
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.

9. 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 +  ? 
We could take  equal to half
the distance from x to y. y x
Since x > y,  > 0.
Let  = (x – y) /2.
Furthermore,
as required. 

10. Recall the definition of absolute value from Section 1.4.
If x  , then the absolute value of x, denoted by | x |, is defined by
The basic properties of absolute value are summarized in the following theorem.
Theorem
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).

11. Theorem 3.2.10 (b) Theorem 3.2.10 (d) | x |  a iff  a  x  a.
Let x, y  and let a > 0. Then
Proof:
Since x = | x | or x =  | x |, it follows that  | x |  x  | x |.
If | x |  a, then we have
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 (d)
| x + y |  | x | + | y |.
Let x, y  and let a > 0. Then
Proof:
As in part (b), we have
Adding the inequalities together, we obtain
which implies that | x + y |  | x | + | y | by part (b). 

12. 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.