Relatively Prime Integers If $m$ and $n$ are relatively prime and $k\mid m$, show that $k$ and $n$ are also relatively prime.


I haven't really any idea where to start with this. I have that if k|m then m=km' but I'm not really sure where to go after that. Thanks!
There are already some good answers here. I'll try something basic.

Since $m$ and $n$ are relatively prime, $\gcd(m,n) = 1.$ Since $k \mid m$, $m = rk, \exists r \in \mathbb{Z}$. But then we can just write $\gcd(rk, n) = 1$.

This implies $\gcd(k, n) = 1.$ You can see this by contradiction.
Since the integers are a Principal Ideal Domain, we have a Bézout identity and can write $am+bn=1$ for some $a,b \in \mathbb{Z}$. Since $k\, |\, m$, we have $atk+bn=1$ for some $t \in \mathbb{Z}$. Now suppose for contradiction that $gcd(k,n) \neq 1$. Then, clearly $gcd(k,n) \, | \, atk+bn$ and moreover $gcd(k,n) \, | \, 1$. Since $gcd(k,n)$ is an integer, this implies that $gcd(k,n)=1$, a contradiction. Hence, $gcd(k,n)=1$.