I am trying to prove the fact that the $0$ divisors of $(Z\_n,+,\times)$ are the non-zero elements $m\not=0$ that are NOT coprime with $n$. I see this in the following way: 

-If $m$ is a $0$ divisor, then $m$ not coprime with $n$ 

or equivalently: 

-If $m$ coprime with $n$, then $m$ is not a $0$ divisor 

I chose the contrapositive. 

By definition, $m,n$ coprime if $gcd(m,n)=1$. 

By definition $m$ is a $0$ divisor if for $m\not=0$ and $m$ in $(Z\_n,+,\times)$ there exists $q\not=0$ and $q$ in $(Z\_n,+,\times)$ such that $mq=0$. 

Suppose $m,n$ coprime ($gcd(m,n)=1$), that means that $|m|=n$ (the order of $m$ is $n$), which we know from Group Theory (As $(Z\_n,+)$ is cyclic, and using Lagrange). So only $m\*(k\*n)=0$ for $k$ in $(Z\_n,+,\times)$ in $(Z\_n,+,\times)$, but $k\*n=0$ by definition in $(Z\_n,+,\times)$, so $m$ is not a divisor of $0$ as there doesn't exist a $q$ satisfying the conditions above as $q=k\*n=0$. Fraleigh gives a different proof, but I like to practice proving stuff by myself, and I have serious doubts about the way I proceeded with this. Any help is immensely appreciated!