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I'm going through the Harvey Mudd Real Analysis lectures. In the second lecture the professor gives a little problem to check knowledge:

if $m/n$ is a positive quotient, then $mn > 0$

How would you show that this is well defined?

I understand that if $m/n = p/q$ and $mn > 0$ therefore $pq > 0$ but am unsure where to proceed from there.

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    Can you prove that $\;\frac mn>0\iff (m,n>0\;\;\text{or}\;\;m,n<0)\;$ ? – 2017-02-28
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    Take cases, based on the signs of $m,n$. If $m = 0$, then what? If $n=0$, then what? If $m$ is positive, $n$ is negative, then what? If $m$ is negative, $n$ is positive, then what? – 2017-02-28
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    If you already proved that all nonzero squares are positive, you can simply argue that $$mn = \frac{m}{n} \cdot n^2$$ is a product of positive numbers. – 2017-02-28
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    @DonAntonio is showing that each case $ m/n = 0 \Rightarrow m=0; m/n > 0 \Rightarrow m,n > 0 \lor (-1)m,(-1)n . 0$ sufficient? – 2017-02-28
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    @supremus_01 I can't really say as that depends on what axioms and assumptions you did/are allowed to assume. – 2017-02-28
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    @DonAntonio up to the second lectures, no axioms for negative number have been given, only basic arithmetic for quotients(proof that quotients are equivalent using definition $m/n = p/q \Rightarrow pn = qm$, and addition, multiplication of quotients). integer $m0$ have been given as well – 2017-02-28
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    @Crostul i can't use this as no axiom for squares has been given yet – 2017-02-28
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    @supremus_01 Fine. Then how did you people define *positive number*? I mean, what does $\;r>0\;$ mean for you? – 2017-02-28
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    @DonAntonio yeah sorry should've included that. $(m < n) \Rightarrow (n-m > 0)$ – 2017-02-28
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    @supremus_01 That doesn't answer my question... – 2017-02-28
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    @DonAntonio other than stating that integers are an ordered set satisfying trichotomy and transitivity, and the definition above, no other rigorous definition has been given for positive integers. – 2017-02-28
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    @supremus_01 I see (I think...), but then it seems to be you must work with the usual, "intuitive" definition of positive numbers, and thus the exercise is completely trivial as the rules of signs for fractions are the same as for multiplication... – 2017-02-28
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    @DonAntonio i just checked through the lecture. subtraction is defined as the addition of a negative value, multiplication of negative integers is assumed – 2017-02-28

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This is used in construction of $\mathbb Q$ from $\mathbb Z$ as definition of positive.

This condition tells us when two fractions represent the same rational number (belong to the same equivalent class): $$\frac mn = \frac pq \qquad\Leftrightarrow\qquad mq=pn.$$

So under the assumptions that $mn=pq$, $n\ne0$, $q\ne0$ and $m,n,p,q\in\mathbb Z$ we want to show that $mn>0$ implies $pq>0$. We are allowed to use known properties of $\mathbb Z$.

Let us have a look at $mn\ge0$ instead. If we multiply both sides of the inequality by $(mq)^2$, the inequality does not change, since $(mq)^2\ge0$. And we will then use that $(mq)^2=mqpn$, which follows from $mq=pn$. \begin{align*} mn&\ge0\\ mn\cdot(mq)^2&\ge0\\ mn\cdot mqpn &\ge 0\\ m^2n^2pq&\ge0\\ pq&\ge 0 \end{align*}

This solves similar problem with strict inequality. I.e., in this way we can show $mn\ge0 \Leftrightarrow pq\ge0$.

So it remains to check when a fraction is equal to zero and what happens in that case.

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    and thanks for getting back to me in the chat as well. I'll make sure to note the necessary information next time i ask. – 2017-03-05