Bridge to Abstract Mathematics: Mathematical Proof and Structures

308 PROPERTIES OF NUMBER SYSTEMS Chapter 9

the first two sentences of this proof. Hence 1x1 = 0 implies x = 0, so that

the conclusion "1x1 = 0 e x = 0" is justified.

(b) There are six distinct cases to check. For example, if x is pos-

itive and y is negative, then xy is negative [Exercise 4(a)(ii)J so that

lxyl = (x)(- y) = - xy. However 1x1 = x and (yl = - y so that 1x1 1 y( =

(x)(-y) = -(xy) also. Formulation and verification of the other five

cases are left as an exercise [Exercise 4(a)(iii)J.

(c) Our approach assumes the result of Exercise 4(b)(ii) and uses (b)

of this theorem. Namely, we note

Ix + yI2 = I@ + Y)~I [follows immediately from (b)]

= (x + Y)~ [from Theorem l(c) and

Definition 41

= x2 + 2xy + y2

s x2 + 2lxyl + y2 [from Exercise S(a)(iii)]

Since lx + yI2 5 (1x1 + 1~1)~. and since both lx + yl and 1x1 + lyl are

nonnegative, we conclude, from Exercise 4(b)(ii), that lx + yl 5 1x1 + lyl,

as desired.

(d) 1x1 = ly + (x - y)l 5 lyl + IX - yl, where the inequality follows

from (c). Hence 1x1 - lyl 2 IX - yl, as desired.

Note the following discussion about the proof of Theorem 5(c). In an

arbitrary field we need to be careful when using the symbol "2" to represent

1 + 1 (and 2x for x + x for any x E F, etc.), since it is possible that 1 + 1 = 0.

This is indeed the case in the field 2,. However, in an ordered field, 1 + 1

cannot equal zero, so that 2 = 1 + 1 must represent a nonzero, and, in fact,

a positive field element. In particular, 4 = 2-I has got to exist in any

ordered field (cf. Exercise 8).

The concept of absolute value in an arbitrary ordered field provides us

with a notion of distance between field elements, in a manner totally analo-

gous to the way we calculate distances between real numbers along a number

line.

DEFINITION 5

Let F be an ordered field. If x, YE F, we define the distance between x and y,

denoted d(x, y), by the rule d(x, y) = Ix - yI.

Note that d is a mapping from F x F into F.

it is left as an exercise to prove [see Exercise 7(a)],

erties listed in Theorem 6.

It is easy to show, and

that d satisfies the prop-