Show given relation $R$ is equivalence relation on $S$ I will display the exact problem, then my questions.  I have searched to the extremes to figure this out and can't:
Show that the given relation $R$ is an equivalence relation on set $S$.
$S$ is the set of ordered pairs of positive integers.
Define $R$ so that $(x_1,x_2)R(y_1, y_2)$ means that $x_1 + y_2 = y_1 + x_2$.
Edit:


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*Reflexive: If $(x_1, x_2)R(y_1, y_2)$ means $x_1 + y_2 = y_1 + x_2$ then $(x_1, x_2)R(x_1, x_2)$.  Such that $x_1 + x_2 = x_1 + x_2$.  This is true $\forall$ $\in$ $S$, therefore it is reflexive.    

*Symmetric: If $(x_1, x_2)R(y_1, y_2)$ means $x_1 + y_2 = y_1 + x_2$ then $(y_1, y_2)R(x_1, x_2)$ means $y_1 +x_2 = x_1 + y_2$.  $y_1 +x_2 = x_1 + y_2$ is the equivalent as $x_1 + y_2 = y_1 + x_2$ when $\forall$ $\in$ $S$.  Therefore it is symmetric.

*Transitive: *If $(x_1, x_2)R(y_1, y_2)$ means $x_1 + y_2 = y_1 + x_2$, suppose   $α=(x_1,x_2)$ and $β=(y_1,y_2)$.  Since it is reflexive and symmetric, $α R β$ $β R α$ and $α R α$ are all true.  Since transitive is $x R y$, $y R z$, and $z R x$ we can say that it is transitive because $α R β$ $β R α$ and $α R α$ are all true.


$\infty$ number of equivalence classes of $R$
Are all my reasons here valid and correct?
Thank you!

The objects that are being related are ordered pairs. So given an object $\alpha=(x_1,x_2)$ and an object $\beta=(y_1,y_2)$, you have that $\alpha \, R \, \beta$ by the condition $x_1-x_2=y_1-y_2$ (the difference of the coordinates in the two pairs are the same. For example, $(1,3) \, R \, (2,4)$ because in both cases the difference of coordinates is $-2$. Likewise $(1,1) \, R \, (2,2)$ because the difference is $0$.
For reflexivity, you need to check if every object is related to itself, i.e. is $\alpha \, R \, \alpha$ for every $\alpha \in S$? So you are asking if
$$(x,y) \, R \,  (x,y) \, \quad \forall \, (x,y) \in S?$$
