Proof by contradiction problem on rational numbers

Question

Using proofs by contradiction, show that there is no smallest negative rational number and no largest positive rational number.

Assume that there is a smallest negative rational number. Therefore, the number is of the form $r = - \frac{p}{q}$, where $p$ and $q$ are positive integers. But, there is a rational number $- \frac{p}{q+1}$, which is smaller than $r$. This is a contradiction. Therefore, there is no smallest negative rational number. QED

Assume that there is a largest positive rational number. Therefore, the number is of the form $r = \frac{p}{q+1}$, where $p$ and $q$ are positive integers. But, there is a rational number $\frac{p+1}{q}$, which is larger than $r$. This is a contradiction. Therefore, there is no largest positive rational number. QED

Do you think that these are correct proofs?

Answer 1

Two less "fussy" ways:

(i).$\;x\in Q\implies x\pm 1\in Q.$

And $ \;x-1<x<x+1.$

Use $x-1<x$ when $x<0$. Use $x<x+1$ when $x>0$.

(ii). $x\in Q\implies 2x\in Q.$

And $\;x>0\implies 2x>x, $ and $\;x<0\implies 2x<x.$

Answer 2

As Henning Makholm pointed out in his comment, your proof that there is always a smaller rational number should point to the existence of $\frac{-p + 1}{q} < \frac{-p}{q}$.

Otherwise, yes: you have have supposed that your conclusion was false and shown that that supposition led to a contradiction. Therefore your conclusion must be true.