Proving that a relation $(f,g)∈S$ iff $f(x)-g(x)\in\mathbb Z$ for every $x\in\mathbb R$ is an equivalence relation

Question

I have $S$ is relation on $A =\{\text{set of all functions }\mathbb R\to\mathbb R\}$ that $(f,g)∈S$ iff $f(x)-g(x)\in\mathbb Z$ for every $x\in\mathbb R$

I start to prove it this way

Reflexive: for every $x\in\mathbb R$ we have $f(x)-f(x)\in\mathbb Z$

Symmetric: for every $x\in\mathbb R$ if $f(x)-g(x)\in\mathbb Z$ then also $g(x)-f(x)\in\mathbb Z$

Transitive : for every $x\in\mathbb R$ if $f(x)-g(x)\in\mathbb Z$ and $g(x)-h(x)\in\mathbb Z$ then $f(x)-h(x)\in\mathbb Z$

Is that enough to prove equivalence relation ?

and:

How to find $|A/S|$?

Thanks

Answer 1

Your proof that $S$ is an equivalence relation is okay but completeness is disputable.

For symmetry you should say something like: $f(x)-g(x)\in\mathbb Z\implies g(x)-f(x)=-(f(x)-g(x))\in\mathbb Z$.

By transitivity something like: $\cdots\implies f(x)-h(x)=(f(x)-g(x))+(g(x)-h(x))\in\mathbb Z$.

Hint for finding cardinality of $A/S$:

$A/S$ and $[0,1)^{\mathbb R}$ - i.e. the set of functions $\mathbb R\to[0,1)$ - are isomorphic.