Equivalence Class for Abstract Algebra Class Let
$$R_3= \{(a,b)\mid a,b \in \mathbb{Z}\text{ and there exists }k \in \mathbb{Z} \text{ such that }a-b=3k\}.$$
I know there is an equivalence relation but I'm not 100% on what it means to be an equivalence class for this problem. In class we got 3: $\{0,3,6,9,\ldots\}$ and $\{1,4,7,10,-2,-5,\ldots\}$ and $\{2, 5, 8, 11, -1, -4,\ldots\}$. 
I don't understand where these cells came from. Help?

This question is old, but I'd like to share my view of it. I hope this offends no one.
What you are dealing with are equivalence classes modulo an integer. That is, you are dealing with equivalence classes defined by $[a]=\{b \in \mathbb{Z}:b \cong a \pmod 3\}$ for an element $a$.
In the case of the relation of congruence modulo $n$, there are $n$ equivalence classes: $[0], [1], [2], \dots, [n-1]$. You can see that supposing $[n]$ or $[n+1]$ is silly, since $b \cong n \pmod n$ reduces to $b \cong 0\pmod 0$ and $b \cong n+1 \pmod n$ reduces to $b \cong 1 \pmod n$ (and, hence $[n]=[0]$ and $[n+1]=1$). In other words, $[0], [1], [2],\dots, [n-1]$ are all of the equivalence classes.
For congruence modulo $3$, we thusly have $[0], [1], $ and $[2]$. That is,
\begin{align}
[0]&=\{b \in \mathbb{Z}: b\cong 0 \pmod 3\}\\
&=\{3,6,9,\dots\}.\\
[1]&=\{b \in \mathbb{Z}: b\cong 1 \pmod 3\}\\
&=\{4,7,10,\dots\}.\\
[2]&=\{b \in \mathbb{Z}: b\cong 2 \pmod 3\}\\
&=\{5,8,11,\dots\}.
\end{align}
In other words, the three equivalence classes are numbers that divide $3$ with remainder $0$, all numbers that divide $3$ with remainder $1$, and all numbers that divide $3$ with remainder 2.
This is the beauty of equivalence classes: They partition the set they are defined on into disjoint sets which as a whole form the entire set. i.e. For any given equivalence relation $E$ on set $S$ and the equivalence classes $[a]=\{b \in S: aEb\}$, $$\bigcup_{a \in S}[a]=S.$$
