Definition of positive in $\mathbb{Q}$ is well-defined.

Question

The definition of positivity in $\mathbb{Q}$ I'm using is:

$\frac{m}{n} \in \mathbb{Q}$ ($m,n \in \mathbb{Z}, n \neq 0$) is positive if $mn > 0$.

I'm trying to show that this definition is well-defined. Hence, if $(m,n) \sim (p,q)$ and $\frac{m}{n}$ is positive, then $\frac{p}{q}$ is positive. Here is my attempt at a proof.

Suppose $(m,n) \sim (p,q)$ and $\frac{m}{n}$ is positive. Then, $mn > 0$. It suffices to show that $pq > 0$.

Since $mn > 0$, $mn \neq 0$ by trichotomy. Hence, $m, n \neq 0$. So $m$ and $n$ must have the same sign. We have $mq = np$ by equivalence. Hence, $mq$ and $np$ must have the same sign. Since $m$ and $n$ have the same sign, it must be the case that $q$ and $p$ have the same sign. Since $q \neq 0$ by definition of $q$, $mq \neq 0$, so $np \neq 0$. Hence, $np \neq 0$. Hence, $p$ and $q$ are either both positive or both negative, so $pq > 0$.

How does this proof look? Are there any leaps of logic?

I am especially interested in feedback on this proof, but aside from that, I would love to hear if there is an easier way to prove this fact.

Answer 1

Looks good. If you are asking for easier, I have a suggestion that would cut down on the argumentation a little bit:

We have $mq = np$ by equivalence. Multiply both sides with $mq$, this gives $$(mq)^2 = (mn)(pq).$$ As $mn > 0$, both $m,n$ and thus also $p,q$ are not zero. On the left side we have a square of a non-zero number, which is positive. On the right side, $mn > 0$, hence also $pq > 0$, as they must have the same sign to give a positive number in the product.

So the main ideas are exactly the same as in your proof, only difference is that I use the "same sign" argument exactly once in the end and that in this proof, $pq$ appears in a prominent place, making it (in my eyes, this is highly subjective) a little nicer to see the important points in the proof.