$\mathrm{lcm}(ma,mb)=m\mathrm{lcm}(a,b)$

For simplicity, let us work only with positive integers.

We want to prove that if a,b,m are integers, then

$$\mathrm{lcm}(ma,mb)=m\mathrm{lcm}(a,b).$$

First notice that any common multiple of $ma$ and $mb$ is also a multiple of $m$, so any common multiple of $ma$ and $mb$ is of the form $mk$ with some integer $k$.

Now notice that if $t=\mathrm{lcm}(a,b)$ and , it cannot happen that $a\mid u$ and $b\mid u$, since $t$ is the smallest number, So, when $a\nmid u$ then $ma\nmid mu$, and if $b\nmid u$ then $mb\nmid mu$. We conclude that $mu$ is not a common multiple of $ma$ and $mb$ when .

So far, we proved that $mt=m\mathrm{lcm}(a,b)$ is a common multiple of $ma$ and $mb$, and previous paragraph shows that there is no smaller common multiple, therefore $m\mathrm{lcm}(a,b)$ is the least common multiple of $ma$ and $mb$, in other words:

$$\mathrm{lcm}(ma,mb)=m\mathrm{lcm}(a,b).$$

Title	$\mathrm{lcm}(ma,mb)=m\mathrm{lcm}(a,b)$
Canonical name	mathrmlcmmambmmathrmlcmab
Date of creation	2013-03-22 15:03:22
Last modified on	2013-03-22 15:03:22
Owner	drini (3)
Last modified by	drini (3)
Numerical id	12
Author	drini (3)
Entry type	Theorem
Classification	msc 11-00