TITLE: Show that $\rm lcm(a,b)=ab \iff gcd(a,b)=1$
QUESTION [5 upvotes]: Show that 
$$\rm lcm(a,b)=ab \iff \gcd(a,b)=1.$$
My attempt: If $\gcd(a,b)=1$ then there exist two integers $r$ and $s$ such that $$ar+bs=1.$$
and then I'm stuck...
any advice?

REPLY [2 votes]: $\dfrac{a\cdot b}{gcd(a,b)}=a\cdot\left(\dfrac b{gcd(a,b)}\right)=\left(\dfrac a{gcd(a,b)}\right)\cdot b$ is a multiple of both $a$ and $b$.
As $a\cdot b$ is known to be the smallest multiple, then $gcd(a,b)=1$.