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Thread: 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, $\displaystyle x^2 + x $ is even.

    This is what i have so far:

    Prove by contradiction,

    Suppose $\displaystyle x^2 + x $ is ODD

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

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

    $\displaystyle (2k + 1)^2 + (2k + 1) $ = ODD

    $\displaystyle 4k^2 + 4k + 1 + 2k + 1 $ = ODD

    $\displaystyle 4k^2 + 6k + 2 $ = ODD

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

    Therefore $\displaystyle 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!
    TheEmptySet's Avatar
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    Quote Originally Posted by mybrohshi5 View Post
    Prove that for every integer x, $\displaystyle x^2 + x $ is even.

    This is what i have so far:

    Prove by contradiction,

    Suppose $\displaystyle x^2 + x $ is ODD

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

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

    $\displaystyle (2k + 1)^2 + (2k + 1) $ = ODD

    $\displaystyle 4k^2 + 4k + 1 + 2k + 1 $ = ODD

    $\displaystyle 4k^2 + 6k + 2 $ = ODD

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

    Therefore $\displaystyle 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

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

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

    Why not look at it this way

    $\displaystyle x^2+x=x(x+1)$

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

    either way notice that no matter what $\displaystyle x$ is either $\displaystyle x$ or $\displaystyle x+1$ is even and so is there product.
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  3. #3
    MHF Contributor

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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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