Thread: Discrete Mathematics Problem (Proofs)

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

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 is odd. Then, for some
We also know that for some . Then, which is an odd number, thus is odd, in contradiction, and so N is even.

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Or:



And the sum of two even integers is even.

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

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).