Implicit assumptions: q != 0 (p/q is a rational number).

This is a proof that mn>0 as a definition that a rational number m/n is "positive" is a well-defined with respect to equivalence classes of rational numbers. I.e. if p/q is in the same equivalence class of m/n (i.e. np=mq) and m/n is positive (mn>0) then p/q is also positive (pq>0).

First, np=mq => np-mq=0, but since mn>0 (neither m nor n is zero), and that q!=0 (by definition), then p!=0. So one of pq<0 or pq>0 is true. Assume the former towards a contradiction. First let p>0 and q<0 and let q=-q' where q' is positive. Then 0=np-mq=np+mq', but all of these are positive, so they cannot equal zero: contradiction. Same argument for p<0 and q>0, we end up with 0=-np'-mq, but the right hand side is negative: contradiction. So pq<0 cannot be true, therefore pq>0.

This is a proof that mn>0 as a definition that a rational number m/n is "positive" is a well-defined with respect to equivalence classes of rational numbers. I.e. if p/q is in the same equivalence class of m/n (i.e. np=mq) and m/n is positive (mn>0) then p/q is also positive (pq>0).

First, np=mq => np-mq=0, but since mn>0 (neither m nor n is zero), and that q!=0 (by definition), then p!=0. So one of pq<0 or pq>0 is true. Assume the former towards a contradiction. First let p>0 and q<0 and let q=-q' where q' is positive. Then 0=np-mq=np+mq', but all of these are positive, so they cannot equal zero: contradiction. Same argument for p<0 and q>0, we end up with 0=-np'-mq, but the right hand side is negative: contradiction. So pq<0 cannot be true, therefore pq>0.

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