# Thread:

1. ## Need help proving an odd number divided by 2 is never an integer

I know it's true. This is actually part of a bigger proof, but I got it down to this, but I can't just say it's true. I have to somehow show it's true by relating it to an integer.

$\frac{2k+1}{2}$

where 2k+1 is the definition of an odd integer. I can rearrange it and get k + 1/2, but I still need to prove that k + 1/2 is not an integer. I can't just say it. The only thing we have learned about that can identify integers is divisibility.
i.e. a|b means a = bk for some k that's an integer

Anyone know how I can prove this?

If it helps, the original question was:

$\forall q \in \textbf{Q}, \exists r \in\textbf{Q}$ so that q + r is not an integer

Which I figured I could do two cases where the rational number q is an integer, or when it isn't. So now that I'm at the case where it is an integer, I'm trying to prove that adding 1/2 (r value I picked) to it makes it so it's no longer an integer. Which got me to (2k+1)/2

2. If $\frac{k+1}{2}=j$ then $2k+1=2j$.
There seems to be something wrong with that. NO?

3. Okay, so what you're saying is (2k+1)/2 = some integer j
so then 2k + 1 = 2j where left is def of an odd and right is def of an even number

This proves that 2k+1 is not even, but how does it prove that it's not an integer?

4. You can prove it by contradiction. If (2k+1)/2 is an integer, then it must be either even or odd.

If it is even:
(2k+1)/2 = 2j for some integer j
2k+1 = 4j
2k+1 = 2(2j)
Since addition and multiplication are closed operations on integers, 2j is an integer, therefore 2(2j) is even (by the definition of evenness). Contradiction, since 2k+1 is odd by the definition of oddness.

If it is odd:
(2k+1)/2 = 2j + 1 for some integer j.
2k+1 = 2(2j + 1)
Again, since addition and multiplication are closed operations on integers, 2j is an integer. Therefore 2(2j + 1) is an even number by definition. 2k + 1 is an odd number by definition => contradiction.

Therefore (2k+1)/2 is not an integer.

5. Thanks! So to prove something like this, we have to break it up into two cases? And as long as you prove that its neither even nor odd, then it can't be an integer.
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