How to prove that the $\gcd(n,m)=m$? I'm struggling with how to prove that the $\gcd(n,m)=m$?

It is given that $m$ is positive and $m|n$&hellip; But isn't that the proof already? If $m$ divides $n$ and $m$ divides itself (of course) then how can I formulate this properly?

Thanks!
Let $d$ be a natural number which divides $m$. Since $m\mid n$, it follows that $d\mid n$. Therefore$$\{\text{divisors of }m\}\subset\{\text{divisors of }n\},$$and so\begin{align}\{\text{common divisors of }m\text{ and }n\}&amp;=\{\text{divisors of }m\}\cap\{\text{divisors of }n\}\\&amp;=\{\text{divisors of }m\}.\end{align}Therefore,$$\gcd(m,n)=\text{greatest divisor of }m=m.$$
This is proper way to formulate this problem. $m \mid n$, thus there is an integer $k$ such that $n=mk$, then 

$$\gcd(m, n)=\gcd(m, mk)=m \ \underbrace{\gcd(1, k)}_{=1}=m$$