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I came across a Tricki article with the above title, but it was a stub that didn't say how to do that particular trick.

I've run into this issue a few times before (can't put my finger on which contexts at the moment), and I was wondering how you do it in the sense the author meant.

Alok Singh
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    There are no rationals in lowest terms with denominator smaller than 3 in the interval $(-1/3, 1/3)$, for instance. – Sort of Damocles Jan 05 '18 at 04:04
  • Please describe more fully what you are trying to do. It sort of sounds like the fact that there are no rationals in $(\frac 14,\frac 13)$ with denominator less than $7$, but that is a guess. – Ross Millikan Jan 05 '18 at 04:23
  • @Ross Millikan Yes, that's it – Alok Singh Jan 06 '18 at 04:08
  • "That's it" is not a sufficient response to my request for clarification. What is your question? You should not expect people to click through and read an article. Even if I had I would not understand what you are asking. – Ross Millikan Jan 06 '18 at 04:13
  • My question is: given an interval, how do you figure out the lowest denominator that can occur in an interval, when reduced to lowest terms? – Alok Singh Jan 06 '18 at 07:40

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Let $n$ be a fixed positive integer > 1. Then, the interval $$ I_n = \left [ \frac{n-1}n, 1 \right ) $$ contains rationals with denominator $> n$.

Proof: Suppose $\frac{a}b \in n$ and $b < n.$ Clearly $a < b$ so

$$\frac{a}{b} \le \frac{b-1}{b} = 1 - \frac{1}b \le1 - \frac{1}{n-1} = \frac{n-2}{n-1} < \frac{n-1}{n} $$ which cannot happen.

Sandeep Silwal
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