Answer
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: $\lfloor k^{2}+k+\frac{1}{4}\rfloor=\lfloor(k^{2}+k)+\frac{1}{4}\rfloor=k^{2}+k+\lfloor\frac{1}{4}\rfloor$.