Let n be a positive integer and consider $[a]$ in $Zn$. Prove that $[a]$ is a zero divisor in $Zn$ if and only if it does not have an inverse in $Zn$.
An element $[a]$ of $Zn$ has a multiplicative inverse in $Zn$ if and only if $(a,n)=1$ in $Zn$ where n is prime this will always be true and the theorem holds.
i start with for all $[a] $s.t $(a,n)$$> 1$ and $[a]$ is not $$ i state that there exists a $[b]$ s.t $[a][b] = $ and $[b] \neq $
from this i attempt to draw the conclusion that for all $[a]$ that satisfied $(a,n)>1$ are not invert-able as $$ is not invert-able. then using the theorem when $(a,n)=1$ then for all of these class's $[a]$ they are invert-able
i feel like i am missing half of my proof because i haven't proved if its not invert-able then it is a $$ divisor. i also haven't proved that $$ is not invert-able.
Perhaps i can say [x]= for all [x] thus $($($^-1$)$) \neq $ therefor $$ is not invert-able?
Please help :)