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: 29

Answer

$3n + 5$ is a even number if n is odd, because n can be write as $2k + 1$, substituting in the formula, we get 2(3k + 3) that is a even integer;

Work Step by Step

If $n$ is an odd integer. $n$ can be write as: $n = 2 k + 1$ (with $k$ being one integer number) $k \in \mathbb{Z}$ Now we can develop the formula $3n + 5$: $\begin{split} 3n + 5 & = 3(2k + 1) + 5 \\ & = 6k + 1 + 5 \\ & = 6k + 6 \\ & = 2(3k + 3) \end{split}$ Since $3k + 3k$ is an integer, then $2(3k + 3)$ is an even number, by the definition of even numbers.
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