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Math Help - How to solve this direct proof

  1. #1
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    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.
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  2. #2
    MHF Contributor ebaines's Avatar
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    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?
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    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?
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  4. #4
    MHF Contributor ebaines's Avatar
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    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.
    Last edited by ebaines; November 8th 2012 at 06:07 AM.
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    Re: How to solve this direct proof

    Quote Originally Posted by blueaura94 View Post
    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=~?
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