Bridge to Abstract Mathematics: Mathematical Proof and Structures

10.2 DEVELOPMENT OF THE INTEGERS AND RATIONAL NUMBERS 347

need to have a solution in N. In fact, a solution exists in N if and only
if a c b. This situation is improved upon in Z.

THEOREM 4

Let a, b E 2. Then there exists a unique integer x such that a + x = b. This x,

which is given by the formula x = b + (-a), is denoted b - a and is called b

minus a.

Proof Note first that a+x= a+(b+ (-a)) =(a+(-a)) + b=O+ b
= b. Hence x = b + (-a) is a solution. Furthermore, if b = a + x,
then (-a)+b= -a+(a+x)=(-a+a)+x=O+x=x, so that
b + (-a) is the only solution. 0

Note that the use of the symbol b - a in Theorem 4 is consistent with its
use for N in Article 10.1 (recall Definition 3). The difference, of course, is
that b - a exists in Z for any integers a and b, whereas in N, b - a exists,
for positive integers a and b, if and only if a < b. We describe this situation
by saying that Z is closed under the operation of subtraction, whereas N fails
to be closed under subtraction.
The integer 0, introduced in Theorem 3, has additional properties of in-
terest, including Theorem 5.

THEOREM 5

For any XEZ, x.0 = 0

For any x, ~EZ, if xy = 0, then either x= 0 or y= 0

For anyx,y,z~Z, if xy=xzand x#O, then y=z

(a) Let x = [(p, q)] be an arbitrary element of Z. Recalling that

• 
  
  
  
  • [(I, I)], we note that x -0 = [(p, q)]. [(I, I)] = [(p + q, q + p)] =
    [(I, 1)J = 0, as claimed.
    (b) Let x = [(p, q)] and y = [(r, s)] be integers and suppose xy = 0.
    That, is, suppose 0 = [(I, l)] = [(pr + qs, ps + qr)]. Suppose further-
    more that x # 0; that is, p # q. We claim y = 0; that is, r = s. Assume,
    with no loss of generality, that p > q. From xy = 0, we have ps + qr =
    pr + qs. Using algebraic properties of N, we arrive at the equation
    (p - 4)s = (p - q)r. Since p > q, we have p - q E N so that we may use
    multiplicative cancellation in N [recall Theorem 7(e), Article 10.11 to
    cancel p - q and arrive at the desired conclusion s = r.
    (c) If xy = xz, then xy - xz = 0 [by Exercise 4(b)(ii)]. Hence 0 =
    xy - xz = x(y - z). Since x # 0, then by (b), we must have y - z = 0,
    or y = z. O
  
  
  
  

A consequence of Theorem 5(a) is that 0 cannot possibly have a multi-

plicative inverse in Z. Because the ring Z satisfies the property in Theorem

5(b) (often referred to as "having no zero divisors"), Z is an example of

what algebraists call an integral domain. We will soon see that all other