Additive Inverse in Z mod 5

Question

I've got a small question regarding a little exercise:

$\forall z \in \mathbb{Z}_5, \exists y \in \mathbb{Z}_5$ such that z + y = 0

Of course in $\mathbb{Z}_5$ we could just go through all the possibilities but this would be impossible for $\mathbb{Z}_n$ for large enough n. So my thought was to just pick y as $y = 5-z$ so that adding leads to 5 which leads to our desired result. But am I even allowed to do this? because 5 isn't contained in our set.

Or do just have to show the uniqueness of the additive inverse?

Thanks in advance

Answer 1

Here's an answer for every $n$: if $x=0$, take $y=0$; otherwise, take $y=n-x$. Note that, although $n\notin\mathbb{Z}_n$ (I am assuming that, for you, $\mathbb{Z}_n=\{0,1,2,\ldots,n-1\}$), $n-x\in\mathbb{Z}_n$.

Answer 2

Don't forget that $\mathbf Z/5\mathbf Z$ is the set of congruence classes modulo $5$, so you may represent a congruence class by any of its representatives. Indeed $$\bigl[(5-z)+5\mathbf Z\bigr]+\bigl[z+5\mathbf Z\bigr]=\bigl[(5-z)+z+5\mathbf Z\bigr]=\bigl[5+5\mathbf Z\bigr]=\bigl[5\mathbf Z\bigr].$$ This proves the opposite of the congruence class of $z\,$ $(0\le z<5$) is the complement of $z$ to $5$.