If $a | b$, prove that $\gcd(a,b)$=$|a|$. 
If $a | b$, prove that $\gcd(a,b)$=$|a|$.

I tried to work backwards. If $\gcd(a,b)=|a|$, then I need to find integers $x$ and $y$ such that $|a|=xa+yb$. So if I set $x=1$ and $y=0$ (if $|a|=a$) or if I set $x=-1$ and $y=0$ (if $|a|=-a$), is this good enough?

If $a|b,\ |a||b\implies |a|\le gcd(a,b)\le a\le |a|\implies |a|=gcd(a,b)$.
