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Math Help - GCD prof

  1. #1
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    GCD prof

    I need a little help on this one.....Prove that GCD(a,b) = GCD(b, a+b).
    Last edited by Syndicate01; April 20th 2010 at 08:27 AM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Syndicate01 View Post
    I need a little help on this one.....Prove that GCP(a,b) = GCD(b, a+b).
    Suppose that d\mid a,b then d\mid a+b\implies d\mid a,a+b so (b,a+b)\mid (a,b). Work simillarly
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    In general, \mbox{GCD}(a,b) = \mbox{GCD}(\alpha a+\beta b,\gamma a + \eta b) for any matrix \begin{pmatrix}<br />
{\alpha}&{\beta}\\ <br />
{\gamma}&{\eta}<br />
\end{pmatrix} \in \mbox{SL}_2(\mathbb{Z}).
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  4. #4
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    Quote Originally Posted by Bruno J. View Post
    In general, \mbox{GCD}(a,b) = \mbox{GCD}(\alpha a+\beta b,\gamma a + \eta b) for any matrix \begin{pmatrix}<br />
{\alpha}&{\beta}\\ <br />
{\gamma}&{\eta}<br />
\end{pmatrix} \in \mbox{SL}_2(\mathbb{Z}).
    Heu ...
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Syndicate01 View Post
    Heu ...
    \alpha,\beta,\gamma,\eta\in\mathbb{Z},\text{ and }\alpha\eta-\beta\gamma=1
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    \alpha,\beta,\gamma,\eta\in\mathbb{Z},\text{ and }\alpha\eta-\beta\gamma=1
    =\pm 1
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bruno J. View Post
    =\pm 1
    Special linear group - Wikipedia, the free encyclopedia

    \pm 1\implies O_2(\mathbb{Z})
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  8. #8
    MHF Contributor Bruno J.'s Avatar
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    You're right! My bad.

    Replace \mbox{SL}_2 by \mbox{O}_2 in my post!
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