If then .
There seems to be something wrong with that. NO?
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.
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:
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
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.