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 4.5 - Page 197: 28


Let $n$ be an odd integer. Then $n=2k+1$ for some integer $k$. Hence, $\lfloor\frac{n^{2}}{4}\rfloor=\lfloor\frac{(2k+1)^{2}}{4}\rfloor=\lfloor\frac{4k^{2}+4k+1}{4}\rfloor$$=\lfloor k^{2}+k+\frac{1}{4}\rfloor$. Applying Theorem 2.5.1, we get $\lfloor k^{2}+k+\frac{1}{4}\rfloor$$=k^{2}+k+\lfloor\frac{1}{4}\rfloor=k^{2}+k=k(k+1)$. Substituting $\frac{n-1}{2}$ for $k$, this gives $k(k+1)=(\frac{n-1}{2})(\frac{n-1}{2}+1)$$=(\frac{n-1}{2})(\frac{n+1}{2})$, so by the transitive property of equality, $\lfloor\frac{n^{2}}{4}\rfloor=(\frac{n-1}{2})(\frac{n+1}{2})$. Since $n$ was an arbitrary odd integer, we conclude that the result holds for all odd integers.

Work Step by Step

In applying Theorem 2.5.1, we implicitly use the associative property of addition: $\lfloor k^{2}+k+\frac{1}{4}\rfloor=\lfloor(k^{2}+k)+\frac{1}{4}\rfloor=k^{2}+k+\lfloor\frac{1}{4}\rfloor$.
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