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I got asked to prove this which makes no sense to me

“Proof if a and b are two arbitrary numbers then (a,b) are divisors of lcm(a;b)”

Isn’t this a part of the definition of lcm? What is there to prove?

  • Does $(a,b)$ mean greatest common divisor? – Laars Helenius Oct 15 '20 at 18:44
  • If you are defining the lcm as the least common multiple of $a$ and $b$ then, yes...it certianly has to be a multiple of both $a$ and $b$. – lulu Oct 15 '20 at 18:45
  • It just says (a,b) nothing more but @lulu in that case there is nothing to prove, though...:/ –  Oct 15 '20 at 18:47
  • Even if $(a,b)$ means $\gcd(a,b)$ the claim follows at once from the definition. – lulu Oct 15 '20 at 18:48
  • Maybe they just want a trivial proof. gcd(a,b) | a. a | lcm(a,b). So by transitivity of divisibility gcd(a,b)| lcm(a,b). – Ameet Sharma Oct 15 '20 at 18:50
  • Immediate consequence of LCM & GCD universal properties $,\gcd(a,b)\mid a,b\mid {\rm lcm}(a,b),$ by transitivity of divisibility, see the linked dupes. In general domains LCM & GCD are defined by this property, but that's not always the case in $\Bbb Z$, so one has to prove them, e.g. as in the links. – Bill Dubuque Oct 15 '20 at 22:49

2 Answers2

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Given the title of your question, I suspect that you misunderstood the notation $(a,b)$, which actually means $\gcd(a,b)$. Thus the question is to prove that the greatest common divisor of $a$ and $b$ divides the least common multiple of $a$ and $b$. Not really difficult to prove, but still needs a short argument.

J.-E. Pin
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    Here is the short argument: $\gcd(a,b)$ divides $a$ and $a$ divides $\text{lcm}(a,b)$. By transitivity of the division, $\gcd(a,b)$ divides $\text{lcm}(a,b)$. – J.-E. Pin Oct 15 '20 at 19:07
  • even for arbitrary numbers? so potentially on R? –  Oct 15 '20 at 19:45
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If $p_1,...,p_n$ are the prime numbers that appear in the fatorization of $a$ or $b$ then

  1. $a=p_1^{\alpha_1}\times...\times p_n^{\alpha_n}$ and $b=p_1^{\beta_1}\times...\times p_n^{\beta_n}$ with $\alpha_i,\beta_i\in\mathbb{N}\cup\{0\}$ for $i=1,...,n$

  2. by definition $gcd(a,b)$ we have that $gcd(a,b)=p_1^{c_1}\times...\times p_n^{c_n}$ with $c_i=min\{\alpha_i,\beta_i\}$ for $i=1,...,n$

  3. by definition $lcm(a,b)$ we have that $lcm(a,b)=p_1^{d_1}\times...\times p_n^{d_n}$ with $d_i=max\{\alpha_i,\beta_i\}$ for $i=1,...,n$

As $c_i\leq d_i$ then $p_i^{c_i}|p_i^{d_i}$ and finally $gcd(a,b)|lcm(a,b)$.

  • Overly complicated, I am afraid... – J.-E. Pin Oct 15 '20 at 19:08
  • How would you do? The definition of $gcd$ and $lcm$ are simple enough to get the result, however I think this is a detailed way of proving it. –  Oct 15 '20 at 19:13
  • Look at the comment to my answer. – J.-E. Pin Oct 15 '20 at 19:41
  • It is indeed simpler! Nice :) –  Oct 15 '20 at 19:43
  • does this work for arbitrary numbers to just argue with transitivity? –  Oct 15 '20 at 20:01
  • @bymathformath the notion of "highest common factor" only makes sense if we are talking about integers. A factor pair of $x\in\Bbb Z$ is a pair $x_1,x_2\in\Bbb Z : x_1x_2=x$. If we allowed $x_1,x_2\in \Bbb R$, I could claim that $9$ is a factor of $10$ and so the $hcf(9,10)=9$, which is absurd. – Rhys Hughes Oct 15 '20 at 20:10