# GCD prof

• Apr 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).
• Apr 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 $\displaystyle d\mid a,b$ then $\displaystyle d\mid a+b\implies d\mid a,a+b$ so $\displaystyle (b,a+b)\mid (a,b)$. Work simillarly
• Apr 20th 2010, 08:10 AM
Bruno J.
In general, $\displaystyle \mbox{GCD}(a,b) = \mbox{GCD}(\alpha a+\beta b,\gamma a + \eta b)$ for any matrix $\displaystyle \begin{pmatrix} {\alpha}&{\beta}\\ {\gamma}&{\eta} \end{pmatrix} \in \mbox{SL}_2(\mathbb{Z})$. (Giggle)
• Apr 20th 2010, 08:33 AM
Syndicate01
Quote:

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

Heu ...
• Apr 20th 2010, 08:33 AM
Drexel28
Quote:

Originally Posted by Syndicate01
Heu ...

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

Originally Posted by Drexel28
$\displaystyle \alpha,\beta,\gamma,\eta\in\mathbb{Z},\text{ and }\alpha\eta-\beta\gamma=1$

$\displaystyle =\pm 1$ (Wait)
• Apr 20th 2010, 02:49 PM
Drexel28
Quote:

Originally Posted by Bruno J.
$\displaystyle =\pm 1$ (Wait)

Special linear group - Wikipedia, the free encyclopedia

$\displaystyle \pm 1\implies O_2(\mathbb{Z})$
• Apr 20th 2010, 03:09 PM
Bruno J.
Replace $\displaystyle \mbox{SL}_2$ by $\displaystyle \mbox{O}_2$ in my post! (Cool)