Why is the implication obvious?

[evinda]

Hello! (Smile)

I am looking at the proof of the follwing sentence:

Let $a,b \neq 0$.
The common multiples of $a,b$ are the same as the multiples of $[a,b]$, where $[a,b]$ is the least common multiple of $a \text{ and } b$.

• Let $a \mid m, b \mid m$
  
  $$m=q \cdot [a,b]+r , \ 0 \leq r < [a,b] (1) $$
  
  $$a \mid m, a \mid [a,b] \Rightarrow a \mid r$$
  
  $$b \mid m, b \mid [a,b] \Rightarrow b \mid r$$
  
  So, $r$ is a common multiple of $a,b$.
  
  As $[a,b]$ is the least common multiple of $a,b$, we conclude from the relation $(1)$ that $r=0$,so $m=q \cdot [a,b]$ and so we conclude that $[a,b] \mid m$
  
• $$[a,b] \mid m \Rightarrow a \mid m, b \mid m$$
  
  According to my notes the last implication is obvious. But...why is it like that? (Thinking)

 

[magneto1]

We know that $a \mid [a,b]$ and $b \mid [a,b]$ since $[a,b]$ is a common multiple of both $a$ and $b$. So if $[a,b] \mid m$, it follows immediately that $a \mid m$ and $b \mid m$. (I.e., if $a \mid b$ and $b \mid c$, then $a \mid c$.)

 

[evinda]

We know that $a \mid [a,b]$ and $b \mid [a,b]$ since $[a,b]$ is a common multiple of both $a$ and $b$. So if $[a,b] \mid m$, it follows immediately that $a \mid m$ and $b \mid m$. (I.e., if $a \mid b$ and $b \mid c$, then $a \mid c$.)

Oh yes,right! Thank you very much! (Nerd) :)