Find equivalence classes (Solution with questions) I have to find the relation properties and the equivalence classes.

$$X = \mathbb{R}^{2}$$ $$(x,y) \sim  (u,v) \Leftrightarrow x - y = u -
> v$$

Showing the relation properties of the equivalence relation was easy I think:
Reflexive:
$$(x,y) \sim  (x,y) \Leftrightarrow x - y = x-y$$
Symmetric: 
$(x, y) \sim (u, v)$. By definition, $x - y = u - v$. By symmetry of equality: $u - v = x - y$. By definition of $\sim$, we get that
$(u, v) \sim (x, y)$.
Transitive:
$$(x,y) \sim  (u,v) \wedge (u,v) \sim  (a,b)\Rightarrow (x,y) \sim  (a,b)$$
$$x-y=u-v \wedge u-v=a-b \Rightarrow x-y = a-b$$
(by transitivity of equality)
But I'm not sure how I can show the equivalence classes.
First I tried to make a set which has all the relation properties, so I can easily read the equivalence classes.
$$ R= \left \{ [(x,y),(u,v)],[(u,v),(x,y)],[(x,y),(x,y)],[(u,v),(u,v)] \right \}$$
This should work.
But now I'm not sure whether I should use the tuples like this because of R² or look at the variables separately. Also I completely ignored the right part of the equivalence (x - y = u - v).
Also I don't know whether I should even use the variables, or work with real numbers.
The equivalence classes read from above would be:
$$[(x,y)] = \left \{ (x,y),(u,v) \right \}$$
$$[(u,v)] = \left \{ (x,y),(u,v) \right \}$$
They would also apply to this sentence:
$$a \sim b \Rightarrow [a] = [b]$$
Is this right or wrong?
EDIT:
Another equivalence classes approach:  
$(x,y)$ equivalent to $(0,0)$
Pairs of $(x,y)$ where $x-y = 0-0$, i.e. $x=y$
$(x,y)$ equivalent to $(0,1)$
Pairs of $(x,y)$ where $x-y = 0-1$, i.e. $x=y-1$
$(x,y)$ equivalent to $(1,0)$
Pairs of $(x,y)$ where $x-y = 1-0$, i.e. $x=y+1$
$(x,y)$ equivalent to $(1,1)$
Pairs of $(x,y)$ where $x-y = 1-1$, i.e. $x=y$
But what does that tell me and how do I write it down formally?
 A: You were sort of OK up until you looked for equivalence classes. 
Let me fixc up your "SYMMETRY" proof. You need to show that IF $(x, y) \sim (u, v)$, THEN
$(u, v) \sim (x, y)$. Your "proof" says something about what happens if both of these are true. Here's a repaired version: 
Suppose that $(x, y) \sim (u, v)$. Then by definition, $x - y = u - v$. By symmetry of equality, we have $u - v = x - y$. And then by the definition of $\sim$, we get that
$(u, v) \sim (x, y)$. 
Do you see the difference? Can you fix up the transitivity "proof" as well? Hint: start by assuming that $(x, y) \sim (u, v)$ and $(u, v) \sim (a, b)$, and find a sequence of statements that leads to the conclusion that $(x, y) \sim (a, b)$. The best route? Write out what each of these means without the "$\sim$" and see whether you can fill in the (few) gaps. 
On to finding the classes:
Let's look for things equivalent to $(0,0)$. Those are ordered pairs $(x, y)$ where $x - y = 0 - 0$, i.e., where $x = y$. So this equivalence class is a straight line of slope 1 through the origin. 
Can you do the same thing for the point $(0, 1)$? Once you've done that, you'll see a pattern, I expect. 
As a general hint, when you're facing things with notions you don't quite understand, making them as specific as possible (e.g., trying to work out the $(0,0)$ example rather than some generic $(x, y)$ example) can often help a lot. 
