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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?

JKnecht
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nightmarish
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2 Answers2

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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 $\;x<0\implies 2x<x.$

DanielWainfleet
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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.

Razi Shaban
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