The gcd divides the lcm

Statement

Let $a, b$ be positive integers. Prove that $\gcd(a,b) \mid \operatorname{lcm}(a,b)$.

Solution

Set $d = \gcd(a,b)$ and $\ell = \operatorname{lcm}(a,b)$.

By the definition of greatest common divisor, $d \mid a$. By the definition of least common multiple, $a \mid \ell$ (the lcm is a common multiple, in particular a multiple of $a$).

By transitivity of divisibility, $d \mid a$ and $a \mid \ell$ give $d \mid \ell$, i.e. $\gcd(a,b) \mid \operatorname{lcm}(a,b)$. $\blacksquare$