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Math Help - proof by contradiction

  1. #1
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    proof by contradiction

    how would you prove by contradiction that if n is even then n^2 is even??
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  2. #2
    Super Member craig's Avatar
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    To prove by contradiction you assume the opposite of what is stated, ie that n^2 is odd.

    As you are starting with an even number, then instead of n, lets use 2n.

    (2n)^2 = 4N, therefore if n \in Z(intergers), n^2 is always even.

    This is how I would do it anyway, I am sure there are different methods out there.

    Craig

    ps does anyone know how to do the 'member of' sign in your maths formatting thing?

    Edit: solved
    Last edited by mr fantastic; January 5th 2009 at 04:09 AM. Reason: Merge
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  3. #3
    Member Last_Singularity's Avatar
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    Quote Originally Posted by craig View Post
    To prove by contradiction you assume the opposite of what is stated, ie that n^2 is odd.

    As you are starting with an even number, then instead of n, lets use 2n.

    (2n)^2 = 4N, therefore if n \in Z(intergers), n^2 is always even.

    This is how I would do it anyway, I am sure there are different methods out there.

    Craig

    ps does anyone know how to do the 'member of' sign in your maths formatting thing?

    Edit: solved
    And of course, once you assume that n^2 is odd, the only way that is true is if n itself is also odd. But that contradicts the original assumption that you made about it being even. So that completes the proof.
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  4. #4
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    Quote Originally Posted by hmmmm View Post
    how would you prove by contradiction that if n is even then n^2 is even??
    This statement is equivalent to:
    If n^2 is odd, then n is odd

    By law of excluded middle: n is odd or n is even. Assume n is even, then n = 2m where m \in \mathbb{Z}. But n^2 = 4m^2 = 2(2m^2), which is a contradiction.
    Last edited by Chop Suey; January 5th 2009 at 06:02 AM.
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