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

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 $u, 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 $u.

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 $\mathrm{lcm}(ma,mb)=m\mathrm{lcm}(a,b)$ mathrmlcmmambmmathrmlcmab 2013-03-22 15:03:22 2013-03-22 15:03:22 drini (3) drini (3) 12 drini (3) Theorem msc 11-00