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 4.1 - Page 162: 46

Answer

$x = 2k$ $\begin{split} x.y & = (2k)y \\ & = 2(k.y) \\ \end{split}$ $x.y$ is even.

Work Step by Step

Let $x$ be a even integer, and $y$ just a integer, $x$ can be rewriten: $x = 2k$ (with $k$ being a integer) $k \in \mathbb{Z}$ Now we can multiply $x.y$: $\begin{split} x.y & = (2k)y \\ & = 2(k.y) \\ \end{split}$ Since $k.y$ is integer, then x.y is even, by the definition of even numbers.
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