# How to solve this direct proof

• Nov 8th 2012, 06:34 AM
blueaura94
How 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, 06:43 AM
ebaines
Re: 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, 06:47 AM
blueaura94
Re: 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, 07:04 AM
ebaines
Re: 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, 07:46 AM
Plato
Re: How to solve this direct proof
Quote:

Originally Posted by blueaura94
Prove that for all integers m and n, if m and m - n are odd, then n is even.

If $m=2k-1~\&~m-n=2j-1$ then $n=~?$