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Math Help - chain of equivalants

  1. #1
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    chain of equivalants

    Im on my last problem and i i know the first one implies the 2nd, the 2nd implies the 3rd and the 3rd implies the first but i keep screwing up somewhere.

    i : x2(squared) is odd
    ii: 1-x is even
    iii: x2 +1 is even
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  2. #2
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    Quote Originally Posted by sprint10s View Post
    Im on my last problem and i i know the first one implies the 2nd, the 2nd implies the 3rd and the 3rd implies the first but i keep screwing up somewhere.

    i : x2(squared) is odd
    ii: 1-x is even
    iii: x2 +1 is even

    Well, if you already "know" that i ==> ii --> iii ==> i then you're done since this is what you've got to do!
    What's it that you think you've screwed?

    Tonio
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  3. #3
    Member oldguynewstudent's Avatar
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    defintion of odd even

    if x^2 is odd then x is odd; let x = 2k +1

    then x^2 = (2k+1)(2k+1) = 4k^2 + 4k + 1

    1 - (4k^2 + 4k + 1) = -4k^2 - 4k = 2(-2k^2 - 2k) which is of the form 2k which proves it's even

    (4k^2 + 4k + 1) + 1 = 4k^2 + 4k + 2 = 2(2k^2 + 2k + 1) which again is of the form 2k which proves it's even.
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