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I found this proof:

If $a<b$, there is a rational $\frac{p}{q}$ with q fixed, and p any integer.

Consider the set of numbers of the form $\frac{p}{q}$ with $q$ fixed, and $p$ any integer. Assume that there are no such numbers between $a$ and $b$. Let $\frac{p}{q}$ be the number immediately before $a$. Then $\frac{p+1}{q}$ is the number immediately after $b$. We necessarily have

$$\frac{p+1}{q} - \frac{p}{q} \geq b - a \Leftarrow\Rightarrow \frac{1}{q} \geq b-a$$

If we choose q sufficiently large, then the above inequality is wrong. Then there is at least one rational number between $a$ and $b$.

For example:

I have $a=1$ and $b=3$, $\frac{p}{q}= \frac{1}{1}$ and $\frac{p+1}{q}=\frac{1+1}{q}= \frac{2}{1} $ and $\frac{2}{2} \ngeq 3$.

How is it possible that this works for $a=1$ and $b=3$?

skinnyBug
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2 Answers2

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The spirit of the proof is this: if you dilate the interval $[a,b]$ by a factor $q$ sufficiently large that $qb-qa>1$, then the interval will contain at least one integer $p$, and the fraction $\dfrac pq$ is a rational in $[a,b]$.

With $a=1,b=3,q=1$, the condition is fulfilled (we have the three rationals $\dfrac11,\dfrac21,\dfrac31$).

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You have changed the value of $q$ in your example. It should be $\frac{p+1}{q} = \frac{1+1}{1} = 2$ instead of 1. Apart from that mistake, the proof says "If we choose $q$ sufficiently large". Obviously $q=1$ is not large enough.

Jose Pérez Cano
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  • But $q=1$ is large enough in this case, $2/1$ is between $1$ and $3$. – Conifold Nov 10 '21 at 19:39
  • But $\frac{p+1}{q}$ should be greater than 3, that's why isn't large enough. – Jose Pérez Cano Nov 11 '21 at 16:48
  • It shouldn't, just the opposite. "If we choose $q$ sufficiently large, then the above inequality is wrong. Then there is at least one rational number between a and b". When $q$ is large enough we do not blow past $b$ when adding $1$ to $p$, and get an intermediate rational. – Conifold Nov 11 '21 at 19:09