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Prove $\frac{ab}{m} = \gcd(a,b)$ when $m= \operatorname{lcm}(a,b)$ for all natural numbers $a$ and $b$. I should be able to prove this using only basic rules of $\gcd$ and lcm.

I instead let $m$ be a common divisor of $a$ and $b$ (not necessarily the $\gcd(a,b)$). We know $ab$ is a common multiple of $a$ and $b$, but we also know that $\frac{ab}{m}$ is a common multiple of $a$ and $b$ as well since $m|a$ and $m|b$, $\Rightarrow m|ab$.

Since $m\in \mathbb{N},\;\frac{ab}{m} < ab$, and the larger $m$ gets, the smaller $\frac{ab}{m}$ becomes. Thus when $m$ is $\gcd(a,b)$, $\frac{ab}{m} = \operatorname{lcm}(a,b)$.

But how can one prove $\frac{ab}{m} = \gcd(a,b)$when $m$ is known to be the $\operatorname{lcm}(a,b)$ from the start?

amWhy
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Reis DeSantis
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    (https://math.stackexchange.com/questions/349858/easiest-and-most-complex-proof-of-gcd-a-b-times-operatornamelcm-a-b-a/350465#350465) – Yadati Kiran Nov 14 '18 at 10:49
  • $\gcd (a,b) \times \operatorname{lcm} (a,b) =ab.$ You can see https://math.stackexchange.com/questions/349858/easiest-and-most-complex-proof-of-gcd-a-b-times-operatornamelcm-a-b-a – 1ENİGMA1 Nov 14 '18 at 10:50
  • It'd be nice to see it proved only using definitions lcm, gcd, and basic divisibility. No Lemmas, no factorisation, no group theory.... just elementary number theory 101. My proof I wrote above was apparently not precise enough – Reis DeSantis Nov 14 '18 at 11:19
  • See here for a proof highlighting the innate duality – Bill Dubuque Nov 14 '18 at 15:58

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