Showing that $\frac{m}{n}$ is positive if $mn > 0$

Question

We are given that $\frac{m}{n} = \frac{p}{q} \iff qm = pn$

We define $\frac{m}{n}$ as positive if $mn > 0$

Show that the definition of $\frac{m}{n}$ is well defined.

Attempt:

What I need to show is that for any fraction equivalent to $\frac{m}{n}$ that the definieiton is also satisfied. So let's take $\frac{p}{q}$ from the same equivalence class as $\frac{m}{n}$. Since they are from the same equivalence class this means: $$qm = pn$$

I'm not sure what kind of algebraic manipulation I should try. I know that I would like to arrive at $pq > 0$ and I feel that it just involves getting $qm = pn \implies mn = pq$.

But how to get this?

Answer 1

Suppose $mn>0$ and $m/n=p/q$. Then $mq=np$ and so $m^2q^2=mnpq$. If $pq<0$, then $mnpq<0$: a contradiction with $m^2q^2>0$.

To add some details, note that $m\ne0$, otherwise $mn=0$, which is impossible. Since $q\ne0$, we have $mq\ne0$ and therefore $(mq)^2>0$.

Answer 2

Hint:

$$\frac{p}{q}=\frac{pq}{q^2}$$

Answer 3

$pqn^2 = q^2 mn\ge 0$.

If $pqn^2 = 0$ $\Leftrightarrow$ $p=0$ or $q=0$ $\Leftrightarrow$ $pq = 0$