Q: How do i prove that this is a equivalence relation Let
$R = \{(x, xe^n) : x \in \Bbb R,n \in \Bbb Z\}$. Prove that $R$ is an equivalence relation on the set of real
numbers.
Edit: Sorry this is my first time. I am aware that in order to prove that it is an equivalence relation, I need to prove that it is reflexive, symmetric an transitive.
For reflexive I am unsure if this is correct but I have put:
$(x,x) \in R$ so, $x = xe^n$ so if we let $n = 0$ then, $x = x$
For symmetric, since $(x,y) ∈ R \implies (y,x) \in R$
$(x,xe^n) \implies (xe^n,x)$, so
$y =xe^n$ , so $x = y/e^n$
for transitive, so far I wrote:
$(x,y) , (y,z) \implies (x,z)$
$(x,xe^n)$ and $(xe^n, e^n(xe^n)) \implies (x,e^n(xe^n))$
which can be simplified to,
$( x,xe^{2n} )$
however I do not know where to go from here.

A: To summarize the discussion in the comments:
Reflexivity:  $(x,x)\in R$ for all $x\in \mathbb R$ since $x=x\times e^0$.
Symmetry:  $(x,y)\in R\implies y=x\times e^n$ for some integer $n$. But then $x=y\times e^{-n}$ and, of course, $-n$ is still an integer.  Thus $(y,x)\in R$ as desired.
Transitivity:  $(x,y)\,\&\,(y,z)\in R$ means that there are integers $n,m$ with $y=x\times e^n$ and $z=y\times e^m$.  But in that case we have $z=x\times e^n\times e^m=x\times e^{n+m}$.  And of course $n+m$ is still an integer, so we have $(x,z)\in R$.
And we are done.
