# if n is even...

• February 24th 2010, 11:49 PM
flower3
if n is even...
for $n \in \mathbb{Z}$ prove that n is even $\iff \ n -2 [\frac{n}{2}] =0$
• February 24th 2010, 11:58 PM
Drexel28
Quote:

Originally Posted by flower3
for $n \in \mathbb{Z}$ prove that n is even $\iff \ n -2 [\frac{n}{2}] =0$

Really? $n=2z$ so that $\frac{n}{2}=z$ so that $\left\lfloor\frac{n}{2}\right\rfloor=z\implies2\le ft\lfloor\frac{n}{2}\right\rfloor=2z=n$
• February 25th 2010, 08:30 PM
chiph588@
The other way:

$n-2\lfloor \frac{n}{2} \rfloor = 0 \iff n=2\lfloor \frac{n}{2} \rfloor \implies 2\mid n \implies n$ is even.