# Discrete Mathematics Problem (Proofs)

• Oct 9th 2009, 12:29 PM
sderosa518
Discrete Mathematics Problem (Proofs)
Prove for all M and N, if M and M-N are even, then N is even.
• Oct 9th 2009, 12:38 PM
Defunkt
Quote:

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 $\displaystyle N$ is odd. Then, $\displaystyle N=2k+1$ for some $\displaystyle k \in \mathbb{N}$
We also know that $\displaystyle M = 2r$ for some $\displaystyle r \in \mathbb{N}$. Then, $\displaystyle M-N = 2r - (2k+1) = 2(r-k) + 1$ which is an odd number, thus $\displaystyle M-N$ is odd, in contradiction, and so N is even.

..

Or:

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

And the sum of two even integers is even.
• Oct 9th 2009, 02:36 PM
tonio
Quote:

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
• Oct 10th 2009, 06:52 AM
HallsofIvy
Quote:

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).