Equivalence classes of relation $\rho: (x,y)\in \rho \Leftrightarrow (\exists k \in \mathbb{Z})x-y=3k$

Question

I don't understand how equivalence classes are $$C(1)=\{3k+1:k\in \mathbb{Z}\}$$ $$C(2)=\{3k+2:k\in \mathbb{Z}\}$$ $$C(3)=\{3k:k \in \mathbb{Z}\}$$

Could someone explain?

Answer 1

The idea here is that when you divide an integer by another, via the euclidean algorithm, there's a unique remainder. So, dividing by 3 gives us a unique remainder that is 0, 1, or 2. The equivalence relationship here is that two numbers have the same remainder. Here's the reason:

$x=3q_1+r_1$, $y=3q_2+r_2$, where $q_1,q_2$ are the quotients and $r_1, r_2$ are the remainder. Well, what does it mean to have the same remainder? Then $r_1=r_2$, so $x-y=3q_1+r_1-3q_2-r_2=3q_1-3q_2=3(q_1-q_2)$. Call $q_1-q_2=k$, and thus we have the two numbers have the same remainder if there exists a k such that $x-y=3k$.

Now, the equivalence classes are just saying numbers that have 0, 1, 2 as the remainder and any $k$ as the quotient. So $C(3)$ are the multiples of 3, $C(1)$ are the ones that leave a remainder of 1, and $C(2)$ are the ones that leave a remainder of 2.

Make sense?

Answer 2

$$\begin{array}{ccc} 3k&3k+1&3k+2\\ 0&1&2\\ 3&4&5\\ 6&7&8\\ 9&10&11\\ 12&13&14\\ 15&16&17 \end{array}$$ Got the idea? The columns are the equivalence classes.