Let $S = \mathbb{Z}$ and $a ∼ b ⇐⇒ 3|(a + 2b)$.
- Show that $∼$ is an equivalence relation.
- Determine all equivalence classes for $∼$.
Part $1$ was pretty simple to prove, however I can't figure out how to proceed with part $2$. I'm thinking of either doing classes by the coefficient $m$ such that $3m = a+2b$ (ex: $cl(1) = (1,1) , (-1,2), (-3,3)... $ for $(a,b)$ ), by the sum of $a+2b$ (ex: $cl(0) = (-2,1)$ because $-2+2(1) = 0$ ), or just a single class where this relation exists (ex: all the values of $(a,b)$ where $3|(a + 2b)$. The last one wouldn't partition the set $S$ obviously, because it would only take the values divisible by $3$.
As a follow up thought, all $3$ of my methods seem incorrect to me, as the set $S$ is only $\mathbb{Z}$, which implies that we are only looking for single integer values in each class, not the $(a,b)$ coordinates that I put in my classes. Any help is appreciated. Thank you!
Hint:
$$a\sim b\iff a\equiv b\pmod 3$$
In other words $3|a+2b$ iff $3|a-b$