Relation $ (x,y) \in \rho \Leftrightarrow (\exists k \in \mathbb{Z})\mid x- y=3k$ I know that there is a similar question here, but it's about classes of equivalence of this relation. I would like to know how to prove that this is an equivalence relation. It seems simple, but the first part of the definition of the relation confuses me. It says:
The ordered pair $(x, y)$, being in $\rho$ is the same as the difference between the two elements of the ordered pair can be written as $3k$, with  $k$ is a whole number.
Now, considering the definition of binary relation, $\rho$ is a subset of the Cartesian product $X\times Y$ where both $X$ and $Y$ are subsets of the set of whole numbers. We can say that an ordered pair is in this set only if the difference between them is divisible by three. I would like to know if I understood this correctly.  
Now, considering all the  above, I will try to prove that this is an equivalence relation:
Reflexivity:
$(\forall x \in \mathbb{Z}) x\rho x \Leftrightarrow x-x=3k$
This is correct since when I subtract the number by itself I will get $0$, which can be written as $3k$ where $k=0$.
Symmetry:
$(\forall x, y \in \mathbb{Z}) x\rho y \Leftrightarrow x-y=3k \Leftrightarrow y -x=-3k \Leftrightarrow y-x = 3k_1, k_1=-k \in \mathbb{Z}$
Transitivity:
$(\forall x, y, z \in \mathbb{Z}) x\rho y \wedge  y\rho z \Leftrightarrow x-y=3k \wedge y-z=3k \Rightarrow x-z=6k $
Now, I am not quite sure how this is transitive. I would appreciate explanation here, and explanation on the classes of equivalence that I mentioned before, since I don't understand that.
 A: Your proofs for reflexivity and stymmetry are somewhat correct. I say "somewhat" because you mean to say the right thing, but I'm inclined to say you natation is incorrect. More on the notation at the end of this answer.
To prove transitivity need to do something like this:
$$(x,y)\in \rho\wedge(y,z)\in\rho\implies x-y=3k_1\wedge y-z=3k_2\implies x-z=3(k_1+k_2),$$ where $k_1,k_2\in \mathbb Z$. 
This proves transitivity.
The the use of two (possibly different) integers $k_1$ and $k_2$ is vital here.
The classes of equivalence now consist of numbers with the same remainder when divided by $3$. Or in symbols: $x\sim y\iff x\equiv y\bmod 3$.
Now for the notation. As it is, $\rho$ is a subset of $\mathbb Z\times \mathbb Z$ presumably. Though it could also be defined as a subset of $\mathbb R\times \mathbb R$. It gets awkward when you write $x\rho y$. What would this mean? Instead you best stick to writing $(x,y)\in \rho$. Once you've proved that it indeed is an equivalence relation, it is conventional to write $x\sim y$ for equivalent $x$ and $y$, as I did above. 
