Bridge to Abstract Mathematics: Mathematical Proof and Structures

348 CONSTRUCTION OF NUMBER SYSTEMS Chapter 10

integers except 1 and - 1 also fail to have multiplicative inverses in Z. We

next introduce an order relation in Z.

DEFINITION 2

Let x = [(a, b)] and y = [(c, d)] be integers. We say that x < y in case

a+d<b+c.

In Exercise 4(a) you are asked to verify that the relation < is well de-

fined. Our intuitive expectation of the inequality [(a, b)] < [(c, d)] is that

a - b be less than c - d. The definition we have given amounts to the

same thing, and furthermore, involves positive integers a + d and b + c, so

that we may take advantage of the "less than" relation already defined on N.

Note that if x E Z, x = [(a, b)], then x > 0 if and only if a > b, x = 0

if and only if a = b, and x < 0 if and only if a < b. Furthermore, x > 0 if

and only if - x < 0. Denoting again by Z + the subset {[(p, q) J 1 p E N,

q E N, p > q) of Z, we can easily show that Z+ is closed under both ad-

dition and multiplication.

Once again, many properties of less than in N carry over to Z. A number

of such properties are listed in Exercise 3. We have, in addition, a number

of other properties of Z contained in Exercises 4 and 5, as well as in the

following theorem.

THEOREM 6

Let x, y, zeZ:

(a) If x < y and z> 0, then xz< yz

(b) If x < y and z < 0, then xz > yz

(c) If xz < yz and z > 0, then x < y

(d) If xz < yz and z < 0, then x > y

Proof (a) Let x = [(a, b)], y = [(c, d)], z = [(e, f)], where all variables in

parentheses represent positive integers. By hypotheses, a + d < b + c

and f < e. To show xz < yz, we note that xz = [(ae + bf, af + be)] and

yz = [(ce + df, cf + de)], so that the inequality ae + bf + cf + de <

af + be + ce + df must be derived. This inequality may be rewrit-

ten as (b + c)f + (a + d)e < (b + c)e + (a + d)f, which is equivalent to

(b + c)(f - e) + (a + d)(e - f) < 0 or [(b + c)-(a + d)](f - e) < 0.

But the latter inequality is true since f - e < 0 and (b + c) - (a + d) > 0.

The proofs of (b), (c), and (d) are similar and are left to you [Exercise

WI. 0

THE RATIONALS
In Theorem 7(d), Article 10.1, we showed that positive integers not equal to
1 do not have multiplicative inverses in N. A similar result can be estab-
lished in 2. Suppose x is an integer not equal to + 1. If x = 0, then x
has no multiplicative inverse, by Theorem 5(a). If x is positive and y is an
integer such that xy = 1, then y must be positive (since 1 > 0) and we have
b