Bridge to Abstract Mathematics: Mathematical Proof and Structures

9.2 ORDERED FIELDS 307

THEOREM 4

Let a, b, x, and y be elements of an ordered field F:

(a) If x < y, then a + x < a + y

(b) If a < b and x < y, then a + x < b + y

(c) If x < y and a > 0, then ax < ay

(d) If x < y and a < 0, then ax > ay

Partial proof We leave (a) and (c) as exercises [see Exercise 3(b)]. For (b),

assume a < b and x < y. To prove a + x < b + y, we must show that

(b + y) - (a + x) E 9. Now, by assumption, b - a E 9 and y - x E 9.

Hence (b - a) + (y - x) E 9 by (a) of Definition 1. But we have the

equation (b - a) + (y - x) = (b + y) - (a + x) (Verify!) so that the latter

element of F is in 9, as desired.

To prove (d), suppose x < y and a < 0. To prove ax > ay, we must

show that ax - ay E 9. By assumption, y - x E 9 and -a E 9. Hence

( - a)(y - x) E 9 by (a) of Definition 1. But ( - a)(y - x) = ax - ay

(Verify!) so that the latter is contained in 9, as we wished to show.

We next consider extensions of the notion of absolute value of a real num-

ber and distance between two real numbers to the context of an arbitrary

ordered field.

DEFINITION 4

Let F be an ordered field. If x E F, we define the absolute value of x, denoted

1x1, by the rule

We issue the usual caveat that in a general ordered field, as in the familiar
ordered field R, we must be careful to avoid false conclusions such as " - x
is negative for any x E F' and ''1 -x( = x for any x E F." A number of
properties of absolute value are contained in the exercises (see Exercise 6);
several that are especially familiar are listed in the following theorem.

THEOREM 5

Let F be an ordered field with x, y E F. Then:

Proof (a) If x > 0, then 1x1 = x > 0. If x < 0, then 1x1 = -x z 0. If x = 0,

then 1x1 = x = 0. If x # 0, then either x > 0 or x < 0 so that 1x1 # 0, from