Discrete Mathematics with Applications 4th Edition

Published by Cengage Learning
ISBN 10: 0-49539-132-8
ISBN 13: 978-0-49539-132-6

Chapter 4 - Elementary Number Theory and Methods of Proof - Exercise Set 4.3 - Page 178: 16

Answer

Let $a$, $b$ and $c$ be integers such that $a|b$ and $a|c$. Then by the definition of divisibility, $b=ma$ and $c=na$ for some integers $m$ and $n$. But that means that $b-c=ma-na=(m-n)a$, where $m-n$ is an integer because $m$ and $n$ are integers. Therefore, by the definition of divisibility, $a|(b-c)$.

Work Step by Step

We know that $ma-na=(m-n)a$ by the distributive property, and that $m-n$ is an integer by the closure property of the integers under subtraction.
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