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I am wondering if there is a proof validating the definition of the LCM. I know that the least common multiple of two integers, say $a, b$ is just the smaller number $n$, such that $a|n$ and $b|n$, but is there a proof that goes along with this?

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    $\begingroup$ If you want to check that $n$ is indeed $LCM(a,b)$, compute $GCD(n/a,n/b)$. It must be $1$. $\endgroup$
    – user65203
    Oct 21 '15 at 21:15
  • $\begingroup$ This should be an answer. $\endgroup$
    – Niklas
    Oct 21 '15 at 21:17
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No, there is not a proof of this. This is because LCM is just defined that way, just like how we define operations like multiplication, addition, subtraction, etc.

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I think what you are looking for is the following statement:

Let $a,b\in\mathbb N$. Let $c = \frac{ab}{(a,b)} =: \operatorname{lcm}(a,b).$ Then:

  • $c$ is a natural number
  • $a|c$ and $b|c$
  • For all $n,$ if $a|n$ and $b|n$ then $n\ge c.$
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  • $\begingroup$ Is there a way to prove the last bullet point is true? $\endgroup$
    – Ben Knoll
    Oct 23 '15 at 4:45
  • $\begingroup$ Yes. You can prove all 3 using well known theorems about gcd. $\endgroup$ Oct 23 '15 at 8:42
  • $\begingroup$ In fact, you can also replace $n\ge c$ with $c|n$ and the result will still hold $\endgroup$ Oct 23 '15 at 8:43

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