I'm curious to know whether my proof makes sense and if there's an easier method, as I'm still trying to wrap my head around this question. My proof is as follows using the contrapositive.
Contrapositive: If n is not congruent to 1 (mod 3) AND n is not congruent to 2 (mod 3), then 3|n.
Integers of mod 3, Z₃ = {[0], [1], [2]}
(1) Assume n is not congruent to 1 (mod 3) and n is not congruent to 2 (mod 3), then we can subtract both equivalence classes from Z₃. (2) Thus |Z₃| = 1, which is the remaining equivalence class [0]. (3) Hence 3|n.
Is this right? I'm assuming I can simply subtract an equivalence class (considered a set) from a set of equivalence classes, but I'm not sure if the differing rules for addition/subtraction and multiplication of sets and equivalence classes make this not possible.
"Subtracting equivalence classes" sounds like a weird writing to me.
You can just use $n = 3k+r$, where $k \in \mathbb{Z}$ and $r \in \{0, 1, 2\}$. Now use the assumption.