# Thread:

1. ## Discrete Mathematics Problem (Proofs)

Prove for all M and N, if M and M-N are even, then N is even.

2. Originally Posted by sderosa518
Prove for all M and N, if M and M-N are even, then N is even.
Assume, by contradiction, that $N$ is odd. Then, $N=2k+1$ for some $k \in \mathbb{N}$
We also know that $M = 2r$ for some $r \in \mathbb{N}$. Then, $M-N = 2r - (2k+1) = 2(r-k) + 1$ which is an odd number, thus $M-N$ is odd, in contradiction, and so N is even.

..

Or:

$N = M + (-)(M - N)$

And the sum of two even integers is even.

3. Originally Posted by sderosa518
Prove for all M and N, if M and M-N are even, then N is even.

Do you know that sum or substraction of even numbers is even? If you can't use (or don't know) this then prove it: it's trivial when we characterize an even number as 2k , where k is an integer.

Well, since N = M - (M-N) we're done.

Tonio

4. Originally Posted by sderosa518
Prove for all M and N, if M and M-N are even, then N is even.
Or simply, if M is even then M= 2k for some integer k. If M-N is even, the M-N= 2j for some integer j. N= M-(M-N)= 2k- 2j= 2(j-k).
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