Let $n \in \mathbb{N}$. Let $R$ be the equivalence relation $\equiv \pmod{n}$. Suppose $[a], [b] \in \mathbb{Z}_n$ and $[a]\cdot[b] = [0]$. Is it necessarily true that either $[a] = [0]$ or $[b] = [0]$?
Proof:
Premises:
$ 1.\, [a]\cdot[b] = [0]\\ 2. \forall x\forall y\forall k(x \equiv y \pmod{n} \to x \equiv ky \pmod{n}) $
Assume $[a] \neq [0]$:
$ \begin{align*} &\langle\text{Assumption} \rangle\\ &c \in [b]\\ \iff&\langle\text{Def. of $[b]$} \rangle\\ & cRb\\ \iff&\langle\text{Def. of $R$} \rangle\\ & c \equiv b \pmod{n}\\ \iff&\langle\text{By Premise 2} \rangle\\ & c \equiv ab \pmod{n}\\ \iff&\langle\text{Def. of $R$} \rangle\\ & cRab\\ \iff&\langle\text{Def. of $[ab]$} \rangle\\ & c \in [ab]\\ \iff&\langle\text{$[a][b]=[0]$} \rangle\\ &c \in [0]\\ \end{align*} $
$[b] \subseteq [0]$
- Is my proof heading in the right direction ?
- Is the addition of Premise 2 correct? How can I prove it to use it in my proof ?
As pointed out in the comments, the statement is false as well as the "Premise 2".
But you can try to show this: if $p$ is prime, then $[a]_p[b]_p=[0]_p$ implies $[a]_p=[0]_p$ or $[b]_p=[0]_p$. There is more or less nothing to prove apart from correctly using the definition of congruence.
In particular in $\mathbb{Z}/p\mathbb{Z}$ there are no zero-divisors. On the contrary, you can also show that if $n$ is composite then there exist zero-divisors in $\mathbb{Z}/n\mathbb{Z}$.