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Math Help - Induction Proof - Divisibility

  1. #1
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    Induction Proof - Divisibility

    Could someone check to see if I got this right. It says 'divisible' not 'evenly divisible'. I think it proves divisibility. And is there a difference between 'divisible by ' and 'factor of'? they seem to be the same thing.

    a^{2n}-b^{2n} is divisible by (a+b) if a \ne -b . For all positive integers n.

    1. first I prove n=1 is true.

    a^{2(1)}-b^{2(1)} = a^{2}-b^{2} = (a+b)(a-b)

    2. then I assume n=k is true and solve for (k+1)th item.

    a^{2(k+1)}-b^{2(k+1)}

    a^{2k+2}-b^{2k+2}

    a^{2k+2}-a^{2k}b^{2}+a^{2k}b^{2}-b^{2k+2}

    a^{2k}(a^{2}-b^{2})+b^{2}(a^{2k}-b^{2k})
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  2. #2
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    Quote Originally Posted by skoker View Post
    Could someone check to see if I got this right. It says 'divisible' not 'evenly divisible'. I think it proves divisibility. And is there a difference between 'divisible by ' and 'factor of'? they seem to be the same thing.

    a^{2n}-b^{2n} is divisible by (a+b) if a \ne -b . For all positive integers n.

    1. first I prove n=1 is true.

    a^{2(1)}-b^{2(1)} = a^{2}-b^{2} = (a+b)(a-b)

    2. then I assume n=k is true and solve for (k+1)th item.

    a^{2(k+1)}-b^{2(k+1)}

    a^{2k+2}-b^{2k+2}

    a^{2k+2}-a^{2k}b^{2}+a^{2k}b^{2}-b^{2k+2}

    a^{2k}(a^{2}-b^{2})+b^{2}(a^{2k}-b^{2k})

    ....and?!? You already have it, can you see?

    Tonio
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  3. #3
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    I am new to doing this kind of proof and have no answer for this problem. I just wanted to make sure this is correct form someone with more skill.
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  4. #4
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    Quote Originally Posted by skoker View Post
    I am new to doing this kind of proof and have no answer for this problem. I just wanted to make sure this is correct form someone with more skill.
    You know that (a^2-b^2) is divisible by (a+b) and by the induction assumption you know that (a^{2k}-b^{2k}) is divisible by (a+b)

    CB
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  5. #5
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    ok thanks guys, I just want to make sure I have a clear understanding. and the correct statements etc.
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