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  "body": "\n## Proof\n\nWe show that $ab / \\gcd(a, b)$ is the least positive common multiple, using Bézout's identity.\n\n1. Let $a$ and $b$ be natural numbers.\n2. If $a = 0$ or $b = 0$, then $\\operatorname{lcm}(a, b) = 0$ by the convention for the [[least common multiple|09S5SN35DJ]]. So both sides of the equation are $0$.\n3. Now let $a \u003e 0$ and $b \u003e 0$. Put $d = \\gcd(a, b)$, the [[greatest common divisor|02WF878BH1]]. Then $d \\ge 1$, because $1$ is a common divisor.\n4. Since $d \\mid a$ and $d \\mid b$, write $a = d a'$ and $b = d b'$ with positive integers $a'$ and $b'$.\n5. Put $m = a' b$. Then $m$ is a positive integer, and $d m = d a' b = a b$.\n6. Also $m = a' d b' = a b'$. So $a \\mid m$ and $b \\mid m$, and $m$ is a positive common multiple of $a$ and $b$.\n7. By [[Bézout's identity|0DQVHMYAGA]], there are integers $x$ and $y$ with $a x + b y = d$.\n8. Let $c$ be a positive common multiple of $a$ and $b$. Write $c = a u$ and $c = b v$ with integers $u$ and $v$.\n9. Then $c d = c a x + c b y = (b v) a x + (a u) b y = a b (v x + u y)$.\n10. By step 5, $a b = d m$, so $c d = d m (v x + u y)$. Since $d \\neq 0$, we get $c = m (v x + u y)$.\n11. Here $c \u003e 0$ and $m \u003e 0$, so the integer $v x + u y$ is at least $1$. Hence $c \\ge m$.\n12. By steps 6 and 11, $m$ is the smallest positive common multiple, so $\\operatorname{lcm}(a, b) = m$.\n13. Therefore $\\gcd(a, b) \\cdot \\operatorname{lcm}(a, b) = d m = a b$.\n\n## Idea\n\nDividing $ab$ by the gcd removes the common part of $a$ and $b$ once, and the result is still a multiple of both. Bézout's identity writes the gcd as $ax + by$, and this shows that every common multiple is a multiple of $ab/\\gcd(a,b)$.\n",
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