LCM(a,b)=a IFF $b\mid a$ My approach:
$LCM(a,b)=a$ $\rightarrow$ $b\mid a$ 
By definition of LCM, $a=bx$ therefore $b \mid a$
$b\mid a$ $\rightarrow$ $LCM(a,b)=a$ 
Since $b\mid a$ then  $a=bk$.  Since $a$ is a multiple of $a$ and $a$ is a multiple of $b$, then $a$ is a common multiple of $(a,b)$.
But can I claim it is the lowest common multiple?
$a \times 1$ is clearly the smallest multiple of $a$ that I can generate. And since $b\mid a$, that doesn't make a smaller common multiple.  But I'm not sure if this is clearest way to prove or describe this.

Suppose $lcm(a,b)=m\neq a$. Then by the definition of $lcm(a,b)$, if $\exists d$ such that $a|d,b|d$ then $m|d$. As $a|a, b|a$ then $m|a$, but as $m=lcm(a,b)\implies a|m$. So $a|m,m|a\implies a=\pm m\implies a=lcm(a,b)$
