# \lcm of more than two numbers as a formula of \gcds

It is a common elementary number theory exercise to prove that \lcm(a,b) = \frac{ab}{\gcd(a,b)}.

A student might ask what is the \lcm of three numbers. Some might think that \displaystyle \lcm(a,b,c) = \frac{abc}{\gcd(a,b,c)} It isn't.

Still, one might want a formula for the \lcm of three numbers. Of course one can say \lcm(a,\lcm(b,c)). In fact this is the common algorithm for computation. Are they ways to relate \lcm and \gcd without nesting those functions together?

Yes, but the formula is not so pretty. \displaystyle \lcm(a,b,c) = \frac{abc \gcd(a,b,c)}{\gcd(a,b)\gcd(b,c)\gcd(a,c)}

This article shows how we can prove this result, and easily infer a more general theorem. First, we see there is a group isomorphism from the naturals to it's prime factors f:\mathbb{N} \to \mathbb{N}^\infty, f(p_1^{e_1} \ldots p_n^{e^n}) = (e_1,\ldots,e_n,0,0,\ldots), where p_n is the nth prime.

It's easy to show \displaystyle \begin{aligned} \lcm(a_1,a_2,\ldots,a_n) &= f^{-1} (\max(f(a_1),\ldots,f(a_n)))\\ \gcd(a_1,a_2,\ldots,a_n) &= f^{-1} (\min(f(a_1),\ldots,f(a_n))) \end{aligned} where \max and \min are defined coordinate-wise. In fact we only need to concern with one single coordinate. So the problem become proving \displaystyle \max(a,b,c) = a+b+c+\min(a,b,c)-(\min(a,b)+\min(b,c)+\min(a,c)) , then the formula for \lcm of 3 numbers holds.

This look familiar to the inclusion-exclusion principle, and certainly we can use it to prove it and generalize! Let \mu be the Lebesgue measure, then for a finite sequence of non-negative reals \set{a_i}, \displaystyle \max(a_1,\ldots,a_n) = \mu(\bigcup_{i=1}^n [0,a_i]). It's just some standard arguments to show \max does have the inclusion-exclusion structure. It generalize to allow negative reals by simply add a large enough constant to make them positive, and subtract the constant from the result. Formulas for \min,\gcd,\lcm follows similarly.

Posted by Chao Xu on 2012-03-06.
Tags: number theory.