Floor Function

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November 14th 2010, 11:38 AM
dwsmith

Floor Function

Prove:
$\displaystyle\left \lfloor \frac{n}{2} \right \rfloor=\frac{n-1}{2} \ \mbox{if n is odd}$

Since n is odd, $\displaystyle n=2p+1 \ \ni \ p\in\mathbb{Z}$

$\displaystyle\left \lfloor \frac{2p+1}{2} \right \rfloor$ but I am not sure how that will help.
November 14th 2010, 11:41 AM
roninpro

As a hint, consider $\frac{2p+1}{2}=p+\frac{1}{2}$ . Then, $\lfloor p+\frac{1}{2}\rfloor=p$ .
November 14th 2010, 11:42 AM
dwsmith

I have that down as well but don't see the connection.
November 14th 2010, 11:44 AM
roninpro

Then what is $\frac{n-1}{2}$ ?

It is $\frac{n-1}{2}=\frac{(2p+1)-1}{2}=\frac{2p}{2}=p$ .

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