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If $n$ is odd, prove that $n/2 + 1/2$ is even.

Context: I'm a Statistician and the term $n/2 + 1/2$ showed up in the index of a summation when deriving the pdf of some Order Statistic:

$$ \sum_{j = (n+1)/2}^{n}... $$

I realized that $n/2 + 1/2$ is always even if $n$ is odd, but I couldn't prove the result to myself (well, I have no training in Number Theory).

What I've tried:

Suppose $n$ is odd. Then $n + 1$ is even (by the successor function?). Then $n + 1 = 2k$ for $k \in \mathbb{N} \Rightarrow (n+1)/2 = k $. But this doesn't show that $k$ is even.

Bill Dubuque
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Sigma
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    Statistically, your claim holds $50%$ of the time, or so. –  Sep 29 '20 at 20:06
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    It doesn't hold when $n=1, 5, 9, ...$ – J. W. Tanner Sep 29 '20 at 20:11
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    "I realized that n/2+1/2 is always even if n is odd" why would you think that? $\frac n2 + \frac 12 = \frac {n+1}2$ and if $n$ is odd then $n+1$ is even so $\frac {n+1}2$ is an integer. That's all you can can conclude. "(n+1)/2=k. But this doesn't show that k is even" Of course it doesn't. And you should have testes $\frac {n+1}2 = 3\implies n+1 =6\implies n = 5$ is a counter example. – fleablood Sep 29 '20 at 20:21
  • @fleablood "you should have testes"? – 2'5 9'2 Sep 29 '20 at 20:22
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    I truly appreciate the snarky answers and comments, everyone. Thank you. – Sigma Sep 29 '20 at 20:23

4 Answers4

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No.

Consider $n=1$.

I add this sentence because my answer is too short.

TheSilverDoe
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If $n$ is odd, then $n=2k+1$ for some integer $k$. So $$\frac{n+1}{2} = \frac{2k+1+1}{2} = k+1.$$So your statement is false. E.g., for $n=1$ we have $(n+1)/2=1$ odd and for $n=3$ we have $(n+1)/2=2$ even.

Ivo Terek
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You are right indeed by $n=2k+1$ we obtain

$$\frac n2 + \frac12= \frac{2k+1}2+\frac12=k+1$$

which is even only if $k$ is odd that is only if $n=2(2h+1)+1=4h+3$.

user
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If $n$ is odd then $n$ is of the form $2k - 1$.

The number $\frac n2 +\frac 12$ is simply $\frac {n+1}2 = \frac {2k-1 +1}2 = \frac {2k}2 = k$. The expression $\frac n2 + \frac 12$ is simply the the index of which odd number $n$ is.

If $n = 1,3,5,7,9,......, 2k -1 ,.....$ then $\frac n2 + \frac 12 = 1,2,3,4,5,.....,k,.....$ and $k$ is even only if $n$ is the in an even position on the list and odd if $n$ is in an odd position on the list.

....

There are $2$ types of odd numbers. Those where $n = 2k-1$ and $k$ is even, and those where $n=2k-1$ and $k$ is odd.

If $k = 2m$ is even then $n= 2(2m)-1 = 4m -1$ and $\frac n2 +\frac 1n = \frac {2k-1}2 + \frac 12 = k = 2m$ is even.

If $k = 2m -1$ then $n = 2(2m-1) -1 = 4m -3$ and $\frac n2 + \frac 1n = \frac {2k-1}2 +\frac 12= k = 2m-1$ is odd.

So $k$ will be even if $n$ is of the form $4m -1$ or $4m+3$ so for example if $n = 3,7,11,15,....$ then $\frac n2 + \frac 12$ will be even.

But $k$ will be odd if $n$ is of the form $4m -3$ or $4m + 1$ so for example if $n=1,5,9,13,17,..$ then $\frac n2 + \frac 12$ will be even.

fleablood
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