0

I am trying to prove $(ma, mb) =m(a,b)$ where $m$ is a positive integer. Should I start like this :

$$(ma, mb) =d\\ d \mid ma, \quad d \mid mb\\ d \mid max + may\\ d \mid m(ax+by)$$

Then what should I do?

Théophile
  • 24,627
Pragna
  • 77
  • As general advice, your reasoning will be more clear (not only for the reader but for yourself also) if you use words to explain your steps and introduce the variables you're using. For example: "Let $d = (ma,mb)$. Then $d \mid ma$ and $d \mid mb$" etc. – Théophile Jul 01 '17 at 17:19

1 Answers1

2

Let $d$ be the gcd of $a,b$. By Bézout's identiy there exist integers $x,y$ such that $$ ax+by=d. $$ Then $$ (ma)x+(mb)y=md. $$ The above equation shows that if $e$ is a common divisor of $ma$ and $mb$, then $e\mid md$. Show that $md$ divides both $ma$ and $mb$ and you are done.

Théophile
  • 24,627
Sri-Amirthan Theivendran
  • 31,116