given definition of a relation $R$, prove that $R$ is an Equivalence Relation

Question

The relation is on set $\mathbb{R}^\mathbb{R}$ and the definition of the relation $R$ is:

$f \mathop{R} g \iff \exists _{y\in \mathbb{R}} \forall_{x\in \mathbb{R}}\ ((x>y)\to(f(x)-g(x)\in \mathbb{Z}))$

So I have to prove that $R$ is an equivalence relation. What I did is to prove the easier version:

$f \mathop{R} g \iff \forall_{x\in \mathbb{R}}\ (f(x)-g(x)\in \mathbb{Z})$.

As it looks like, I think both of the proofs are very similar, and this is why I proved the second version. Can I say that because the second version is an equivalence relation, the first one is also an equivalence relation?

Answer 1

If this is an exercise, I would say no, you need to give more details. It was a good idea to start with the second version, but now writing the full solution takes less space that writing your question...

The interesting part is transitivity. You have to consider three functions $f$, $g$ and $h$ such that $f \mathop{R} g$, $g \mathop{R} h$. Thus there exists $y$ and $z$ in $\mathbb{R}$ such that, for all $x > y$, $f(x) - g(x) \in \mathbb{Z}$ and for all $x > z$, $g(x) - h(x) \in \mathbb{Z}$. Can you show now that $f \mathop{R} h$?