How to prove $S=\{(x,y) \in \mathbb{R}\times \mathbb{R}|x - y \in \mathbb{Q} \}$ is an equivalence relation? I am really stuck with this problem, and I cannot come out with a solution.
I know that to prove a relation is an equivalence relation we have to prove that it's reflexive, symmetric and transitive, but in this case I don't know how, maybe because I don't much experience (but this exercise should be easy) 
This is the relation:
$S=\{(x,y) \in \mathbb{R}\times \mathbb{R}|x - y \in \mathbb{Q} \}$
 A: The way to prove these things, in general, is to work on each of the three parts separately: show that the relation is reflexive, symmetric, and transitive. For each of these, write what you want to show explicitly, and then follow the general method for showing that kind of statement.
For example, for symmetry in this relation, you want to prove that if $xSy$ then $ySx$. This means you want to prove that if $x - y \in \mathbb{Q}$ then $y - x \in \mathbb{Q}$. But this follows from the fact that $y - x  =  -(x-y)$ and the fact that the negative of a rational number is rational. 
The other two clauses (reflexivity and transitivity) are similar.
Also, regarding terminology: this kind of relation is an "equivalence relation". It is not correct to call it an "equivalent relation". Also, you cannot correctly say that the relation "is equivalent" or "is equivalence". You can say the relation is "an equivalence", but it is more clear to say it in full: it is "an equivalence relation".
A: This is what you need to show:


*

*reflexive: $(x,x)\in S$; so $x-x \in \mathbb{Q}$, for all $x \in \mathbb{R}$. 

*symmetric: if $(x,y)\in S$, then  $(y,x)\in S$, so if $x-y \in \mathbb{Q}$, then $y-x \in \mathbb{Q}$, for all $x, y \in \mathbb{R}$. 

*transitive if $(x,y), (y,z)\in S$, then $(x,z)\in S$;  so if $x-y, y-z \in \mathbb{Q}$, then $x - z \in \mathbb{Q}$, for all $x, y,z \in \mathbb{R}$. 
I think the actual arguments should be doable.
