GCD prof

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April 20th 2010, 07:04 AM
Syndicate01

GCD prof

I need a little help on this one.....Prove that GCD(a,b) = GCD(b, a+b).
April 20th 2010, 07:25 AM
Drexel28

Quote:

Originally Posted by Syndicate01

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
April 20th 2010, 08:10 AM
Bruno J.

In general, $\mbox{GCD}(a,b) = \mbox{GCD}(\alpha a+\beta b,\gamma a + \eta b)$ for any matrix $\begin{pmatrix} {\alpha}&{\beta}\\ {\gamma}&{\eta} \end{pmatrix} \in \mbox{SL}_2(\mathbb{Z})$ . (Giggle)
April 20th 2010, 08:33 AM
Syndicate01

Quote:

Originally Posted by Bruno J.

In general, $\mbox{GCD}(a,b) = \mbox{GCD}(\alpha a+\beta b,\gamma a + \eta b)$ for any matrix $\begin{pmatrix} {\alpha}&{\beta}\\ {\gamma}&{\eta} \end{pmatrix} \in \mbox{SL}_2(\mathbb{Z})$ . (Giggle)

Heu ...
April 20th 2010, 08:33 AM
Drexel28

Quote:

Originally Posted by Syndicate01

Heu ...

$\alpha,\beta,\gamma,\eta\in\mathbb{Z},\text{ and }\alpha\eta-\beta\gamma=1$
April 20th 2010, 01:50 PM
Bruno J.

Quote:

Originally Posted by Drexel28

$\alpha,\beta,\gamma,\eta\in\mathbb{Z},\text{ and }\alpha\eta-\beta\gamma=1$

$=\pm 1$ (Wait)
April 20th 2010, 02:49 PM
Drexel28

Quote:

Originally Posted by Bruno J.

$=\pm 1$ (Wait)

Special linear group - Wikipedia, the free encyclopedia

$\pm 1\implies O_2(\mathbb{Z})$
April 20th 2010, 03:09 PM
Bruno J.

You're right! My bad. (Thinking)

Replace $\mbox{SL}_2$ by $\mbox{O}_2$ in my post! (Cool)

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