LCM(a,b)=a IFF $b\mid a$

Question

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.

Answer 1

${\rm lcm}(a,b) = a\iff a\Bbb Z\cap b\Bbb Z = a\Bbb Z\iff b\Bbb Z\supseteq a\Bbb Z\iff b\mid a$

Answer 2

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 $a|m,m|a\implies a=\pm m\implies a=lcm(a,b)$

Answer 3

Yes, it is the lowest common multiple, no doubt. Observe that if $b|a$, then $a=bk$ for some non-zero constant $k$ and $$\mathrm{lcm(a,b)}=\mathrm{lcm(bk,b)}=bk=a$$