I was given this question, suppose that $x\equiv y(\mod mn)$, show that $x\equiv y(\mod m)$ and $x\equiv y(\mod n)$, is this proof correct, do we need to have $\gcd(m,n)=1$?
We have $x\equiv y \mod m\times n\Rightarrow mn|(x-y)\Rightarrow \exists k\in \mathbb{Z}$ such that $x-y=mnk\Rightarrow x-y=m(nk)\Rightarrow m|(x-y)$ and $x-y=n(mk)\Rightarrow n|(x-y)$, as such $x\equiv y \mod m$ and $x\equiv y \mod n$.
I know that the reverse of this implication is only true given that $m$ and $n$ are coprime.
