Proving Equivalence Relation for xRy: x-y is an Integer on Real Numbers

[Kamataat]

I answered this wrong on a test, but now I've come up with a different solution.
Problem: Prove that a relation xRy\Leftrightarrow x-y\in\mathbb{Z} defined on \mathbb{R} is an equivalence relation.
Solution:
1.) Reflexivity: xRx,\forall x\in\mathbb{R}
For every x we have x-x=0 which is an integer, so reflexivity holds.
2.) Symmetricity: xRy\Rightarrow yRx,\forall x,y\in\mathbb{R}
If for all x,y\in\mathbb{R} we have (x-y)\in\mathbb{Z}, then y-x=-1\cdot(x-y) (any integer multiplied by -1 is also an integer) and thus (y-x)\in\mathbb{Z} and the relation is symmetric.
3.) Transitivity: xRy\wedge yRz\Rightarrow xRz,\forall x,y,z\in\mathbb{R}
For some x,y,z\in\mathbb{R} we have (x-y)+(y-z)=x-y+y-z=x-z (the sum of two integers is also an integer) and thus (x-z)\in\mathbb{Z}. The relation is also transitive.
Is this it?
PS: sorry for the poor spelling
- Kamataat

 

[HallsofIvy]

Yes, that is precisely what you should have done!
(And don't you just hate it when that comes right after the test!)

 

[Kamataat]

Yeah, especially since it's such an elementary thing. I'd think anyone with a (future) professional interest in maths/physics should know this stuff like the back of his/her hand! Oh well, it's happened before to me, so...

/rant

Anyway, thanks!

- Kamataat