Given the relation $(z\_1, n\_1)\sim(z\_2, n\_2) :\Leftrightarrow z\_1 \cdot n\_2 = z\_2 \cdot n\_1$ on the set $\mathbb Z \times \mathbb N$.

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> a) Prove that $\sim$ is an equivalence relation.
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Here is how I've proven it, could someone check if it's right?

* *Reflexive*: $a\sim a$: $z\_1 \cdot n\_1=z\_1 \cdot n\_1$ (is it always this trivial for equations?)
* *Symmetric*: $a\sim b \Rightarrow b\sim a$: $z\_1 \cdot n\_2=z\_2 \cdot n\_1\Rightarrow z\_2 \cdot n\_1=z\_1 \cdot n\_2$
* *Transitive*: $a\sim b \land b\sim c \Rightarrow a\sim c$.
$$z\_1 \cdot n\_2=z\_2 \cdot n\_1 \land z\_2\cdot n\_1=z\_3 \cdot n\_3 \Rightarrow z\_1 \cdot n\_2 = z\_3 \cdot n\_3 .$$ 

I think got this last one wrong because of the $z\_3 \cdot n\_3$, this is not respecting the relation that is: $(z\_1, n\_1)\sim(z\_2, n\_2) :\Leftrightarrow z\_1 \cdot n\_2 = z\_2 \cdot n\_1$. But how can I make a new pair $c$ that is not the same as $a$? Am I allowed to rewrite it to $\frac{z\_1}{n\_1}=\frac{z\_2}{n\_2}$? Then I could write $$ \frac{z\_1}{n\_1}= \frac{z\_2}{n\_2} \land \frac{z\_2}{n\_2} = \frac{z\_3}{n\_3} \Rightarrow \frac{z\_1}{n\_1}= \frac{z\_3}{n\_3} .$$

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> b) Give one bijective function of the set $M$ of equivalence classes, where $M = \{ (\widetilde{z, n}) \mid (z, n) \in \mathbb Z \times \mathbb N\}$ on the set $\mathbb Q$. A little lost here. Any directions? 
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Thanks in advance!