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Math Help - Prove x^2 + x is even?

  1. #1
    Member mybrohshi5's Avatar
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    Prove x^2 + x is even?

    Prove that for every integer x,  x^2 + x is even.

    This is what i have so far:

    Prove by contradiction,

    Suppose  x^2 + x is ODD

     x = 2k + 1 (where k is any integer)

    now plug in (2k + 1) for  x^2 + x is ODD

     (2k + 1)^2 + (2k + 1) = ODD

     4k^2 + 4k + 1 + 2k + 1 = ODD

     4k^2 + 6k + 2 = ODD

    This is False!, ex. when k = 1,  4k^2 + 6k + 2 is EVEN

    Therefore  x^2 + x is EVEN

    Does this all seem correct? i am very new to proofs. any criticism is appreciated

    Thank you
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by mybrohshi5 View Post
    Prove that for every integer x,  x^2 + x is even.

    This is what i have so far:

    Prove by contradiction,

    Suppose  x^2 + x is ODD

     x = 2k + 1 (where k is any integer)

    now plug in (2k + 1) for  x^2 + x is ODD

     (2k + 1)^2 + (2k + 1) = ODD

     4k^2 + 4k + 1 + 2k + 1 = ODD

     4k^2 + 6k + 2 = ODD

    This is False!, ex. when k = 1,  4k^2 + 6k + 2 is EVEN

    Therefore  x^2 + x is EVEN

    Does this all seem correct? i am very new to proofs. any criticism is appreciated

    Thank you
    I don't think what you have follows. If I read it right you assume by way of contradiction that

    x^2+x is odd this implies that
    x^2+x=2k+1 for some k \in \mathbb{Z}

    This does not imply that x=2k+1.

    Why not look at it this way

    x^2+x=x(x+1)

    Now just look at two cases
    what x is even x=2k
    and what if x is odd x=2k+1

    either way notice that no matter what x is either x or x+1 is even and so is there product.
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  3. #3
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    Here is another way.
    For any two consecutive integers, one is even the other is odd.
    Even times an odd is even.
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