Prove that for all integers m and n, if m and m - n are odd, then n is even.

I'm not really sure where to start.

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- Nov 8th 2012, 05:34 AMblueaura94How to solve this direct proof
Prove that for all integers m and n, if m and m - n are odd, then n is even.

I'm not really sure where to start. - Nov 8th 2012, 05:43 AMebainesRe: How to solve this direct proof
I would start by considering that an odd number is always equal to an even number plus 1. So adding two odds is the same as adding two evens plus (1+1). We're almost done - all you need do now is show that adding two evens plus 2 always yields an even - can you take it from here?

- Nov 8th 2012, 05:47 AMblueaura94Re: How to solve this direct proof
From what im getting at is that m would be defined as 2k1+1 and n would be defined as 2k1.

So I am right now at (2k1+1)-(2k1) am I doing this correct? - Nov 8th 2012, 06:04 AMebainesRe: How to solve this direct proof
There is no requirement that n = m-1, so if m = 2k1+1, you don't know whether n = 2k1 or not.

I'm afraid I misread the original question - I thought it was asking to prove that the sum of two odds is even. But we can still use some of what I wrote in my first post. You know that m-n = odd, so that means n = m - odd, and you are told that m is odd. Now prove that subtracting two odds yields an even, in a similar manner as I suggested previously. - Nov 8th 2012, 06:46 AMPlatoRe: How to solve this direct proof