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Math Help - Biconditional Proof

  1. #1
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    Biconditional Proof

    Hi all,

    This is my question with my thoughts about how to approach it:

    For all integers m and n, 4 divides (m2-n2) if and only if m and n are both even or m and n are both odd.

    So obviously, I have to prove it from both directions.

    --> For all integers m and n, 4 divides (m2-n2) if m and n are both even or both odd.
    <-- If m and n are both even or both odd, then 4 divides (m2-n2) for all integers m and n.

    Assuming these directions are correct, I would attempt to prove this the following way. Let m and n be even integers, say 2m and 2n. By substitution it is 4m2-4n2 and by algebra that is equal to 4(m2-n2). m2-n2 is an integer by definiton of integers while the 4 in front of the integer makes the number divisible by 4. That's for the evens. For the odds 2m+1 and 2n+1, by substitution and algebra I have 4(m2+m-n2-n). Again, that is an integer with a factor of 4 making it divisible by 4.

    In terms of the second direction...in past biconditionals I remember taking the contrapositive of the 2nd. So...

    There exists integers m and n that are both even or both odd such that 4 does not divide (m2-n2) for all integers m and n.

    At this point, I am stuck. I might have done the contrapositive wrong because I think the "or" needs to change to "and". But I could be wrong on that...
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  2. #2
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    Re: Biconditional Proof

    You're right--you need to re-formulate your statements. For the most straightforward (but not the most elegant) method, break it up into two conditions: (1) m, n both even, and (2) m, n both odd. Then prove the statement for each case. That is, first prove (both directions) that 4 divides m2-n2 if and only if m, n both even. Then show 4 divides m2-n2 if and only if m, n both odd (each direction, of course). It may involve more pencil lead, but I believe it makes the logic more transparent and less likely to be inaccurately manipulated.

    If you choose to do anything by contrapositive in these cases, the negation of "m, n both odd" becomes "either m or n is not odd," and likewise with "even".

    Also, when you want to represent m and n as generic integers, it is customary to use alternate letters. For example, let m be an even integer represented by 2r, and n an even integer represented by 2s, for all integers r, s.
    Thanks from Measure13
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