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I am trying to prove that $\operatorname{lcm}(a, b) = \frac{ab}{\operatorname{gcd}(a,b)}$ for $a, b \in \mathbb{N}$.

The proof I came up with goes as follows.

  • Denote $m = \frac{ab}{\operatorname{gcd}(a,b)}$. Then $a$ and $b$ divide $m$ since $m = a \cdot \frac{b}{\operatorname{gcd}(a,b)}$ and $\frac{b}{\operatorname{gcd}(a,b)}$ is an integer by definition of $\operatorname{gcd}$. Divisibility of $m$ by $b$ obtained in the symmetrical way. Therefore, $m$ is a common multiple of $a$ and $b$.

  • Let $n < m$ be a common multiple of $a$ and $b$. Then $a = \frac{ab}{n}\cdot \frac{n}{b}$ and since $n$ is a multiple of $b$, it follows that $\frac{ab}{n}$ is a divisor of $a$. Symmetrically, we establish that $\frac{ab}{n}$ divides $b$. But since $\frac{ab}{n} > \frac{ab}{m}$, we have a common divisor of $a$ and $b$ that exceeds $\operatorname{gcd}(a,b)$, and we have arrived at a contradiction. Therefore, $m$ is the smallest multiple of both $a$ and $b$.

I feel my proof is flawed, but I am quite unable to catch any errors, so I came here for help, which I hope to receive.

Thank you.

Update: there is a requirement of not using the Fundamental theorem of arithmetic.

Update: the proof is valid after all, and is a variation of one using cofactor reflection symmetry.

Bill Dubuque
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Damian Vårfågelberg
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    Welcome. With these sorts of questions it is good to point out why you feel the proof is flawed – FShrike Jun 23 '22 at 17:03
  • I can't see anything obviously wrong with this. Its possible I am missing something though. I agree with the person above that it helps to know why you think it might be flawed as at the very least it helps you to practice knowing what you should be looking for when checking a proof. – Fishbane Jun 23 '22 at 17:05
  • Well, I suspect a flaw in the proof because another proof I found on ProofWiki makes use of the fact that $\frac{a}{\operatorname{gcd}(a,b)}$ and $\frac{b}{\operatorname{gcd}(a,b)}$ are coprime, and my proof does not. So, I believe that my logic must be broken somewhere. – Damian Vårfågelberg Jun 23 '22 at 17:06
  • Please give a link to the proof you are comparing to. – Bill Dubuque Jun 23 '22 at 17:09
  • I am comparing to this one: https://proofwiki.org/wiki/Product_of_GCD_and_LCM#Proof_2 – Damian Vårfågelberg Jun 23 '22 at 17:10
  • The proof looks good to me (assuming we are assuming only positive values... which we should). – fleablood Jun 23 '22 at 17:11
  • Yes, we are assuming $a$ and $b$ to be positive integers. – Damian Vårfågelberg Jun 23 '22 at 17:12
  • I think the cited proof is using the coprime to allow us to claim their product is the lowest common multiple (because the lowest common multiple of relative primes is the product). You avoid that by doing a proof by contradiction that if the product weren't the lowest you get a contradiction. Your proof is valid. And clever. I'd never seen it that way. – fleablood Jun 23 '22 at 17:20
  • Thank you very much for your kind answer fleablood! – Damian Vårfågelberg Jun 23 '22 at 17:21
  • Precisely how do you infer $,ab/n > \gcd(a,b)?\ \ $ – Bill Dubuque Jun 23 '22 at 17:24
  • tl;dr.... The proof you cite covers your first bullet point but doesn't touch your second bullet point at all. By using the fact the values are coprime the cited proof doesn't need your second bullet point. But your second bullet point is perfectly valid. So I could just as easy change perspective and say: Your proof covers the cited proofs construction of a common multiple but doesn't touch that its components are relatively prime at all. By constructing a contradiction your proof didn't need to. ... hmmm, I guess that wasn't so tl;dr as I thought it would be. – fleablood Jun 23 '22 at 17:26
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    Replying to the question from Bill Dubuque above: the number $m$ was defined as $m = \frac{ab}{\operatorname{gcd(a,b)}}$. Hence, $\operatorname{gcd(a,b)} = \frac{ab}{m}$. Since by assumption $n < m$ we have $ab/n > \operatorname{gcd(a,b)}$. – Damian Vårfågelberg Jun 23 '22 at 17:27
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    It's essentially the same as this proof in the linked dupe, but omits explicit mention of the innate duality (cofactor reflection). – Bill Dubuque Jun 23 '22 at 19:35
  • For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. – Bill Dubuque Oct 03 '23 at 20:10

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