Math 310: Introduction to Mathematical Reasoning
Summer 2006
Homework 3 Solutions
(From An Introduction to Mathematical Reasoning, by Peter J. Eccles.)
Definition: Given two integers aand b, we say that adivides bor bis a multiple of ato mean
that there is an integer qsuch that aq =b. Written in symbols we write a|bfor adivides b.
1. Prove that for all integers a,band c,
(adivides b) and (bdivides c) =⇒(adivides c).
Proof. Suppose adivides band bdivides c. The there is a q∈Zsuch that aq =b. Likewise
there is a p∈Zsuch that bp =c. Then substituting aq for bwe have
c=bp = (aq)p=a(qp).
Since the integers are closed under multiplication, qp is an integer and hence by definition, a
divides c.
2. Prove that the square of an even integer is even. (Hint: An even integer is one that is a
multiple of 2.)
Proof. Suppose n∈Zis even. Then there is an integer kwith n= 2k. Thus
n2= (2k)2= 2(2k2).
Then by definition, n2is even.
3. Prove that 0 divides an integer aif and only if a= 0.
Proof. This is an “if and only if” statement, so we have two things to prove.
First, assume 0 divides a. Then for some integer q, 0q=a. We have already proven that for
any integer, 0q= 0. Thus
a= 0q= 0.
Now suppose a= 0. Then since 0q= 0 for every q∈Z, we see that 0 divides 0 by defintion.
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