## Discrete Mathematics with Applications 4th Edition

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