If $a,b$ are integers such that $a \mid x$ and $b \mid x$ , must $\mathrm{lcm}(a,b)\mid x$?

If $a,b$ are integers such that $a \mid x$ and $b \mid x$ , must it be true that $\mathrm{lcm}(a,b)\mid x$ ? I claim that it is true but I don't know how to prove it.

Solution 1:

Suppose that it is not the case that $\mathrm{lcm}(a,b)$ divides $x$. By the division algorithm, we know that there are integers $q$ and $r$, with $0\leq r<\mathrm{lcm}(a,b)$, such that $$x=\mathrm{lcm}(a,b)q+r,$$ and because $\mathrm{lcm}(a,b)$ does not divide $x$, we have that $r\neq 0$. Rewrite this as $$r=x-\mathrm{lcm}(a,b)q.$$

Because $a$ divides $x$ and $\mathrm{lcm}(a,b)$, we see that $a$ divides $r$. Similarly, $b$ must divide $r$. Therefore $r$ is a common multiple of $a$ and $b$. But $r<\mathrm{lcm}(a,b)$, so this is a contradiction.

Solution 2:

Conceptually it is quite simple. The set M of all common multiples of $a,b$ is a nonempty set of integers closed under subtraction, therefore every element of M is a multiple of its least positive element $ = {\rm lcm}(a,b).\,$ In particular, $x$ is a multiple of $\,{\rm lcm}(a,b).$

Remark $\ $ Hence the lcm is not only least in absolute value, but also least in terms of divisibility. The latter is the property used to define lcm in more general rings, where no notion of absolute value need exist, i.e. generally $\ a,b\mid c\iff {\rm lcm}(a,b)\mid c,\ $ and, dually, $\ c\mid a,b\iff c\mid \gcd(a,b)$.

If you know ring theory, you can restate the first paragraph as: the set M of common multiples of $a,b$ forms a nonzero ideal in $\,\Bbb Z.\,$ Since $\,\Bbb Z\,$ is a PID, we infer that $\, {\rm M} = (m)\,$ is principal, generated by its least positive element $\,m = {\rm lcm}(a,b).\:$ Therefore $\, x\in {\rm M} = (m)\:\Rightarrow\: m\mid x$.