# Proof involving lcm and biconditional statement.

Suppose $a,b\in\mathbb{Z}$. Then $a = \operatorname{lcm}(a,b)$ if and only if $b\mid a$

Unsure of how to approach this problem.

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$\newcommand{\lcm}{\operatorname{lcm}}$ By definition $\lcm(a,b)=bk$ for some $k\in\Bbb Z$. So if $bk=a$ then $b\mid a$ by definition.

Conversely, $b\mid a$ implies $\lcm(a,b)\le a$ and $a\mid a$ implies $\lcm(a,b)\ge a$.

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Hint $\,\ b\mid a \iff a,b\mid a \iff {\rm lcm}(a,b)\mid a\iff {\rm lcm}(a,b) = a$

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As shown in this answer, $$\lcm(a,b)\gcd(a,b)=ab$$ Thus, we have that $$\gcd(a,b)=b$$ This means that $b$ is a divisor of $a$; i.e. $b\mid a$.

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I see that, nowadays, Bezout is deleted. This is much healthier! –  Bill Dubuque Jul 17 '14 at 23:44
@BillDubuque: ah, but I do use properties of Euclidean Division in the cited answer; very similar to your Bezout. –  robjohn Jul 17 '14 at 23:59