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

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 |