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Thread: Contradiction 1 of 3 (I'll go through these 1 by 1 )

  1. #1
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    Contradiction 1 of 3 (I'll go through these 1 by 1 )

    Prove the following contradiction.
    If a and b are Integers then a^ 2 −4b 2
    Proposition , then a^ 2 −4b 2
    This is what I've done

    Assume

    a^ 2 −4b = 2
    a^2 = 2 + 4b
    a^2 = 2(2b+1)
    a^2 is even because if we sub anything into b it has an even answer.
    because a^2 is even it follows that a is even so a = 2k for some integer k.
    now put 2k back into original assumption
    (2k)^2-4b=2,
    so 4k^2-4b=2 now lets devide by 2
    we get 2k^2-2b=1
    ∴ 1 = 2(k^2-b)
    because c^2-b ∈ Z it follows that 1 is even ..
    so it is incorrect
    we assumed the proposition false but the proposition is true.
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  2. #2
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    Re: Contradiction 1 of 3 (I'll go through these 1 by 1 )

    A little awkwardly stated but the proof is correct.
    Thanks from bee77
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    Re: Contradiction 1 of 3 (I'll go through these 1 by 1 )

    Thanks Halls of Ivy , I think I have a better understanding of recent than the early days of the course where I was painfully frustrating to correct and didn't even know what a P and Q was ...cheers
    Last edited by bee77; Aug 23rd 2017 at 07:22 AM.
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