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How would I prove there is no smallest positive rational number? what is the best method to prove this statement?

tonix
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    Hint: suppose by contradiction $a$ is the smallest positive rational number, and consider $\frac{1}{2}a$. – Crostul Mar 02 '15 at 10:01

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Assume that $q$ is the smallest positive rational number. But then clearly $\frac{q}{2}$ is also rational and $$0 < \frac{q}{2} < q.$$

iiivooo
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Let $\epsilon > 0$ be given. There exists $n \in \mathbb{N}$ such that $n > \frac{1}{\epsilon}$. Then $0 < \frac{1}{n} < \epsilon$. Hence there can exist no positive lower bound on positive rational numbers.

Tom
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  • So in plain English, whatever you come up with the smallest rational number, I can show another smaller number > 0. I wonder where this theorem useful at all other than being an example for proof by contradiction. – nehem Oct 23 '20 at 16:34
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Let $x$ be the smallest positive rational number. Then $y = \frac{x}{2} \in \mathbb{Q}$ but $0 < y < x$. Contradiction.

Empiricist
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A proof by contradiction is rather simple:

Assume that the smallest rational number exists and is of the form: $a/b$

Then note that we can define $a/(b+1)$, which is rational as it is the quotient of 2 integers, and is strictly smaller than $a/b$ as its divisor is greater. Hence we contradict our initial statements that $a/b$ is the smallest possible rational number, so by this contradiction, we know that there is no smallest rational number.

So since we can always define a smaller rational number than the one we have, there can be no smallest rational number.

Just_a_fool
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Since You're proving by contradiction, the statement "There is no smallest rational number." becomes "There is a smallest rational number."

We can prove this by saying:

Let r be any rational number. Since r is a rational number we know that r/2 is also rational. Because r/2 is rational, we can assume r/2 < r, therefore there is no smallest rational number.

Jason
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Use proof by contradiction.

Let $q$ be the smallest positive rational number. Then $\frac q2$ is rational, as rationals are closed under division. As $2>1$, dividing a positive real (let alone rational) number strictly yields a lower positive number. So $\frac q2$ is smaller than $q$, contradicting the original premise.

A similar argument can be used for positive irrational numbers, as well as as positive real algebraics and positive real transcendentals.

Michael Ejercito
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