Why Read This Book? - Index of

2.4 The Division Algorithm 57
mn = m(2k1) = 2(mk1), so that we may let k2 = mk1 (which is an integer) to have
mn = 2k2. Thus mn is even.
Proof 2. Pick integers m and n, and suppose (without loss of generality) that n
is even. Then there exists integer k1 such that n = 2k1. Let k2 = mk1, which by
closure is also an integer. Then
Thus mn is even.
mn = m(2k1) = 2(mk1) = 2k2
Corollary 2.4.3 If n is an even integer, then n 2 is even.
Proof. Let m = n in Theorem 2.4.2.
EXERCISE 2.4.4 If m and n are odd integers, then mn is odd.
Corollary 2.4.5 If n is an odd integer, then n 2 is odd.
EXERCISE 2.4.6 Prove the following.
(a) The sum of two even integers is even.
(b) The sum of two odd integers is even.
(c) The sum of an even integer and an odd integer is odd.
(2.12)
At this point, you’re likely thinking that Definition 2.4.1 addresses the only two
possible situations that can happen in the integers. After all, isn’t every integer
either even or odd, and not both? Well, maybe, but how do you know that? How
do you know that every integer can be written either in the form 2k or 2k + 1, but
not both? This, and a whole lot more, is addressed by the division algorithm.
Ever since elementary school, you have been familiar with the idea of dividing
an integer b by another integer a>0 to produce a quotient q and remainder r. One
way you could express the results of your division calculation in an equation is to
write
b = aq + r (2.13)
where 0 ≤ r