Answer
True.
Work Step by Step
True.
Since $ab|c$, let $c=kab$ (where $k$ is an integer), we have $c=(kb)a$ and $c=(ka)b$. We can see that $kb$ and $ka$ are also integers, thus $a|c$ and $b|c$.
Discrete Mathematics with Applications 4th Edition
by Epp, Susanna S.
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: 26
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