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How would one go about proving the following. Any ideas as to where to start?

For any integer n, the floor of n/2 equals n/2 if n is even and (n-1)/2 if n is odd.


[n/2] = n/2 if n = even

[n/2] = (n-1)/2 if n = odd

Working through it, I try to initially set n = 2n for the even case but am stuck on how to show its a floor...


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up vote 2 down vote accepted

You should set $n=2m$ for even numbers, where $m$ is an integer. Then $\frac n2=m$ and the floor of an integer is itself. The odd case is similar.

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for the odd case, if i set n = 6m+1, which is always odd and plug in for simplification...will that be sufficient for proof? I got 3m at the end – user1234440 Oct 17 '12 at 22:18
@user1234440 For generality set $n=2m+1$ or $n=2m-1$. – Américo Tavares Oct 17 '12 at 22:24
@user1234440: For the odd case, if $n=2m+1$, with $m$ integral, what is $\frac n2$ in terms of $m$ – Ross Millikan Oct 17 '12 at 22:43
for the odd case, its n-1/2 in terms of m – user1234440 Oct 17 '12 at 22:47
@user1234440: no, $\frac n2=m+\frac 12$ and the floor of that is $m$ – Ross Millikan Oct 17 '12 at 23:11

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