Equivalence relation confusion Why is $\{(x,y)\mid x-y\text{ is a rational number}\}$ an equivalence relation and  and why are $\{(x,y)\mid x-y\text{ is a irrational number}\}$ and $\{(x,y)\mid x+y\text{ is an integer}\}$ not?
 A: The second one is not reflexive since $x-x$ is not irrational.  It's also not transitive since one can have $x-y$ and $y-z$ irrational while $x-z$ is rational.  That happens if $y=0$, $x=\pi$, and $z=10$.
The third one is not reflexive since $x+x$ can fail to be an integer, as, for example, when $x=1/3$.  It is also not transitive: consider what happens when $x=1/3$, $y=2/3$, and $z=1/3$.  Then $x+y$ and $y+z$ are integers and $x+z$ is not.
The first one is reflexive because $x-x$ is rational, transitive because if $x-y$ and $y - z$ are rational then so is $x-z$ (because $x-z$ is just the sum of those other two, and the sum of two rational numbers is rational).  It is symmetric since if $x-y$ is rational then so is $y-x$.
A: For the first one: 
(1)  x - x = 0 $\in \mathbb{Q} $
(2)  if $ x - y = p \in \mathbb{Q} \ then \ y - x = -p \in \mathbb{Q} $
(3)  if $ x-y = p; y-z=q; p,q \in \mathbb{Q} \ then \  x - z = x -y +y -z = p+q \in  \mathbb{Q}$
For the second one fail condition 1 and 3 (the sum of two irrationals may be rational); for the third also fail condition 1 (as TooOldForMath pointed out) and 3.
