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Is there always a fraction $\frac{r}{s}$ with $\frac{p}{q}<\frac{r}{s}<\frac{p+1}{q}$ and $s<q$ for $0<p<q-1\in\mathbb{Z}$ and $r,s\in\mathbb{Z}$?

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  • $\begingroup$ It's an interesting question. If both are in lowest terms then you can always use Farey sequences to show they aren't neighbors. I can't sort out what happens when one (or both) can reduce. $\endgroup$ – lulu Jul 19 '15 at 17:45
  • $\begingroup$ Well, I tried to come up with a counter-example. Do you think it is a true statement? $\endgroup$ – user255536 Jul 19 '15 at 17:58
  • $\begingroup$ If you made me bet right now I'd say, yes it is true. Looking at the Farey sequences of modest lengths it was always easy to insert fractions the way you want. Of course that isn't a proof. Could easily be an accident of small numbers. $\endgroup$ – lulu Jul 19 '15 at 18:09
  • $\begingroup$ @user255536: You may want to add that $s\gt 0$ because you can take $r=-2q,s=-2p-1$ for the current form. $\endgroup$ – mathlove Jul 19 '15 at 18:12
  • $\begingroup$ @mathlove. Agreed. We need to require that both r and s are > 0. $\endgroup$ – lulu Jul 19 '15 at 18:13
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Since $0\lt p\lt q-1$, we have $$ \frac pq\lt\frac p{q-1}\lt\frac{p+1}q $$

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  • $\begingroup$ Good! Thanks. Aside from being overkill, my methods weren't even working! $\endgroup$ – lulu Jul 19 '15 at 18:21

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