TITLE: $R$ be a relation on $\mathbb{Z}\times\mathbb{N}$ by $(n,m)R(n',m')$ iff $nm'=n'm$
QUESTION [2 upvotes]: Let $R$ be a relation on $\mathbb{Z}\times\mathbb{N}$ by $(n,m)R(n',m')$ iff $nm'=n'm$
I've shown that this defines an equivalence relation. 
I want to describe the equivalence classes. I've noticed that the relation happens if $\frac{n}{m}=\frac{n'}{m'}$. So my guess is $[(n,m)]=\{\frac{n\cdot a}{m\cdot a}, a\in\mathbb{Z}\setminus\{0\}\}$?
Also I'm asked to give a bijection $(\mathbb{Z}\times\mathbb{N})/R\rightarrow \mathbb{Q}$. For this could we pick $[(n,m)]\mapsto \frac{n\cdot a}{m\cdot a}$
for $a\in\mathbb{Z}\setminus\{0\}$.
Also I'm a little in doubt about the notation - is $[(n,m)]$ correct to write?

REPLY [2 votes]: The notation $[(n, m)]$ is fine. 
Note that, as rational numbers, we have $\frac nm=\frac{na}{ma}$ for any nonzero $a$.
Note that here $0\notin\Bbb N$ (else everybody would be equivalent to $(0,0)$).
We have $[(n, m)]=\{(na, ma):a\in \Bbb N \}$ whenever $\gcd(n, m) =1$.
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