Let $R$ be a relation on $\mathbb{Z}\times\mathbb{Z}$ by $(n,m)R(n′,m′)$ iff $nm′=n′m$. Is this relation equivalence?

Question

I proved that it is symmetric and also transitive. But is it a transitive?

reflexive:

$(n,m) R (n,m) \Rightarrow nm = nm$, This is True

Symmetric:

$(n,m) R (n',m') \Rightarrow (n',m') R (n,m)$

$nm' = n'm \Rightarrow n'm = m'n $, also True

BUT

In prooving transitivity is a problem with zeros.

$(n,m) R (n',m') \wedge (n',m') R (a,b) \Rightarrow (a,b)R(n,m)$

$nm' = n'm \wedge n'b = m'a \Rightarrow nb = ma$

$n = \frac{mn'}{m'}$, $b = \frac{m'a}{n'}$

Now I'm going to substitute $n,b$, so

$\frac{mn'}{m'}\frac{m'a}{n'}=ma \Rightarrow ma = ma$, what is true.

But in the process, I can have problems with zeros in fractions. And I don't know what to do with them.