
Floor Function
Prove:
$\displaystyle \displaystyle\left \lfloor \frac{n}{2} \right \rfloor=\frac{n1}{2} \ \mbox{if n is odd}$
Since n is odd, $\displaystyle \displaystyle n=2p+1 \ \ni \ p\in\mathbb{Z}$
$\displaystyle \displaystyle\left \lfloor \frac{2p+1}{2} \right \rfloor$ but I am not sure how that will help.

As a hint, consider $\displaystyle \frac{2p+1}{2}=p+\frac{1}{2}$. Then, $\displaystyle \lfloor p+\frac{1}{2}\rfloor=p$.

I have that down as well but don't see the connection.

Then what is $\displaystyle \frac{n1}{2}$?
It is $\displaystyle \frac{n1}{2}=\frac{(2p+1)1}{2}=\frac{2p}{2}=p$.