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 4.4 - Page 189: 17

Answer

By the result of example 4.4.5, we know that consecutive integers have opposite parity (evenness or oddness). Therefore, we divide the proof into two cases: with the first number even, and with the first number odd. Let $m$ and $m+1$ be consecutive integers such that $m$ is even. By the definition of even, $m=2k$ for some integer $k$. Therefore, $m(m+1)=2k(2k+1)=4k^{2}+2k=2(2k^{2}+1)$, which by definition is even. Now let $m$ and $m+1$ be consecutive integers such that $m$ is odd. Then by the definition of odd, $m=2k+1$ for some integer $k$. Therefore, $m(m+1)=(2k+1)[(2k+1)+1]=(2k+1)(2k+2)=4k^{2}+6k+ 2=2(2k^{2}+3k+1)$, which by definition is even. In either case, the product is even. Since these are the only two possible cases, we conclude that the product of any two consecutive integers is even.

Work Step by Step

The assertion that $2(2k^{2}+1)$ is even is justified by the fact that $2k^{2}+1$ is an integer, which is in turn justified by the fact that it is the result of adding and multiplying integers. Thus, $2(2k^{2}+1)$ is of the form $2x$ for some integer $x$, making it even by definition. These concepts are explained more fully in Section 4.2 and Appendix A.
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