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Thread: 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; Apr 20th 2010 at 07:27 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 $\displaystyle d\mid a,b$ then $\displaystyle d\mid a+b\implies d\mid a,a+b$ so $\displaystyle (b,a+b)\mid (a,b)$. Work simillarly
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    MHF Contributor Bruno J.'s Avatar
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    In general, $\displaystyle \mbox{GCD}(a,b) = \mbox{GCD}(\alpha a+\beta b,\gamma a + \eta b)$ for any matrix $\displaystyle \begin{pmatrix}
    {\alpha}&{\beta}\\
    {\gamma}&{\eta}
    \end{pmatrix} \in \mbox{SL}_2(\mathbb{Z})$.
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    Quote Originally Posted by Bruno J. View Post
    In general, $\displaystyle \mbox{GCD}(a,b) = \mbox{GCD}(\alpha a+\beta b,\gamma a + \eta b)$ for any matrix $\displaystyle \begin{pmatrix}
    {\alpha}&{\beta}\\
    {\gamma}&{\eta}
    \end{pmatrix} \in \mbox{SL}_2(\mathbb{Z})$.
    Heu ...
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Syndicate01 View Post
    Heu ...
    $\displaystyle \alpha,\beta,\gamma,\eta\in\mathbb{Z},\text{ and }\alpha\eta-\beta\gamma=1$
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    $\displaystyle \alpha,\beta,\gamma,\eta\in\mathbb{Z},\text{ and }\alpha\eta-\beta\gamma=1$
    $\displaystyle =\pm 1$
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bruno J. View Post
    $\displaystyle =\pm 1$
    Special linear group - Wikipedia, the free encyclopedia

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

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