# 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 $d\mid a,b$ then $d\mid a+b\implies d\mid a,a+b$ so $(b,a+b)\mid (a,b)$. Work simillarly
• Apr 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)
• Apr 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 ...
• Apr 20th 2010, 08:33 AM
Drexel28
Quote:

Originally Posted by Syndicate01
Heu ...

$\alpha,\beta,\gamma,\eta\in\mathbb{Z},\text{ and }\alpha\eta-\beta\gamma=1$
• Apr 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)
• Apr 20th 2010, 02:49 PM
Drexel28
Quote:

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

Special linear group - Wikipedia, the free encyclopedia

$\pm 1\implies O_2(\mathbb{Z})$
• Apr 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)