Finding how many distinct equivalence classes there are.

Question

Define a relation $R$ on the set of all integers $Z$ by $xRy$ ($x$ related to $y$) if and only if $x-y=3k$ for some integer $k$.

I have already verified that this is in fact an equivalence relation. But now I need to find how many distinct equivalence classes there are.

I am confused on how to find how many distinct equivalence classes there are.

Answer 1

All equivalence classes are congruent classes of modulo 3. $Cl(0)=\{0, 3, -3, 6, -6,\ldots\}$ $Cl(1) = \{1, -2, 4, -5, 7, \ldots\}$ $Cl(2)=\{2, -1, 5, -4,\ldots\}$

Answer 2

For any integer $x$, we can write $x$ in modulo $3$: $x = 3k + r$ where $r$ is the remainder when $x$ divided by $3$. For example: $8 = 2\cdot 3 + 2$. This means $8 - 2 = 3\cdot 2\implies 8 \sim 2$. This means also that $[8]=[2]$. In general: $x - r = 3k \implies x \sim r \implies [x] = [r]$. And we can count the distinct possibilities of $r$. With modulo $3$, there are only $3$ choices for $r$, namely: $0,1,2$. Thus there are $3$ equivalence classes, and they are: $[0] = \{3k: k \in \mathbb{Z}\}, [1] = \{3k+1: k \in \mathbb{Z}\}, [2] = \{3k+2: k \in \mathbb{Z}\}. $