Properties of the relation $R=\{(x,y)\in\Bbb R^2|x-y\in \Bbb Z\}$ 
$A= \Bbb R \\
R=\{(x,y)\in\Bbb R^2|x-y\in \Bbb Z\}$
Determine if the relation is (a)reflexive, (b)symmetric, (c)transitive, (d)anti-reflexive, (e)anti-symmetric, (f)asymmetric, (g)equivalence relation.

This is what I did but I'm not sure:
It is reflexive: $\forall x:(x,x)\in R:x-x=0\in\Bbb Z$
It is symmetric: $\forall x,y\in\Bbb R:xRy\in\Bbb Z\Rightarrow yRx\in\Bbb Z$
It is transitive: $\forall a,b,c\in \Bbb R:(aRb \ and \ bRc)\in \Bbb Z \Rightarrow aRc $
It isn't anti-reflexive: $(1,1)\in \Bbb R$
It is anti-symmetric: $aRa\in \Bbb Z \Rightarrow a=a$
It isn't asymmetric because it symmetric.
There is an equivalence relation.
Is it correct ? Thanks.
 A: Most of your answers are correct, but the justifications given are a little confusing. In general, you should offer a genuine proof. For example:
It is reflexive.
Proof. Let $x \in \mathbb{R}$ be fixed but arbitrary. Then $x-x=0$. Thus $x-x \in \mathbb{Z}.$ So $xRx.$
Anyway, your answers for "reflexive", "symmetric" and "transitive" are correct. 
The claim that $R$ is anti-symmetric is incorrect. Observe that $0R1$ and $1R0$, but it does not follow that $0=1$.
Also, if a relation on a non-empty domain is reflexive, then its not anti-reflexive (exercise!). So that answer is also correct. Along a similar vein, the only relation that is both symmetric and asymmetric is the always-false relation. But since $0R0$, the given relation $R$ is not always false. So it cannot be asymmetric. Therefore, that answer is also correct.
Edit. By the way, defining $R$ via set-builder notation is imo confusing. I would suggest defining $R$ as the unique subset of $\mathbb{R}^2$ such that:
$$\forall x,y \in \mathbb{R} : xRy \;\leftrightarrow\;x-y \in \mathbb{Z}.$$
From the above form, it is obvious that any time $xRy$ is written down, we may deduce $x-y \in \mathbb{Z}$, and any time $x-y \in \mathbb{Z}$ is written down, we may deduce $xRy$.
