1. ## GCD prof

I need a little help on this one.....Prove that GCD(a,b) = GCD(b, a+b).

2. 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

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

4. 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 $\displaystyle \begin{pmatrix} {\alpha}&{\beta}\\ {\gamma}&{\eta} \end{pmatrix} \in \mbox{SL}_2(\mathbb{Z})$.
Heu ...

5. Originally Posted by Syndicate01
Heu ...
$\displaystyle \alpha,\beta,\gamma,\eta\in\mathbb{Z},\text{ and }\alpha\eta-\beta\gamma=1$

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

7. Originally Posted by Bruno J.
$\displaystyle =\pm 1$
Special linear group - Wikipedia, the free encyclopedia

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

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