Show that a divisor multiplied by an absolute value is still a divisor I need to show that: 
If $s \neq 0$ then $(sa,sb)=|s|·(a,b)$
I did the following but the instructor said it's not the best path:
Let $d = (a, b)$. Then, $dn = a$ and $dm = b$. It follows that $sdn = sa$ and $sdm = sb$. But I don't know how to take my proof to $|s|·(a,b)$. 
I hope someone can help with this, thanks. 
 A: Let us put $$ d \colon= (a, b). $$
Then
(1) $d$ is a positive integer,
(2) $d$ divides both $a$ and $b$, and 
(3) if $c$ is any integer that divides both $a$ and $b$, then $c$ divides $d$ as well.
Moreover, we can find integers $x$ and $y$ such that
$$ d = ax + by. \tag{1} $$
Therefore, upon multiplying both sides by $s$, we get
$$ sd = (sa)x + (sb)y. \tag{2} $$
Note that as $d$ is a positive integer and as $s$ is a non-zero integer, so $|s|d$ is also a positive integer.
Now let us put
$$ d^\prime \colon= (sa, sb). $$
Then $d^\prime$ is a positive integer such that $d^\prime$ divides both $sa$ and $sb$, and so from (2) it follows that $d^\prime$ divides $sd$ and hence also divides $|s|d$.
Conversely, as $d$ divides both $a$ and $b$ and as $s$ is a non-zero integer, so $|s|d$ divides both $sa$ and $sb$, which implies that $|s|d$ divides $d^\prime$.
From the preceding two paragraphs, we obtain
$$ d^\prime = \pm |s|d. $$
But as $d^\prime > 0$ and as $|s|d > 0$, so we must have
$$ d^\prime = |s|d, $$
which is the same as
$$ (sa, sb) = |s|(a, b). $$
Hope this helps.
