# A proof

1. Nov 3, 2005

### 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

2. Nov 3, 2005

### HallsofIvy

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

3. Nov 3, 2005

### 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