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My question from an easy problem.

$p,q$ are positive integers such that $$ \frac{5}{9}<\frac{p}{q}<\frac{4}{7} $$ find $p,q$ such that $q$ is the smallest number that satisfies this inequality.

Draw the line of $ y<\frac{9}{5}x$ and $y>\frac{7}{4}x$ , we can "observe" that $\frac{9}{16}$ is such number.

However, if the question becomes

$a,b,c,d$ are positive integers such that $$\frac{a}{c}<\frac{b}{d} $$ find $p$,$q$ such that $q$ is the smallest number that satisfies the inequality

$$\frac{a}{c}<\frac{p}{q}<\frac{b}{d}$$

No idea about this.

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  • $\begingroup$ now try denominators from 2 to 15. $\endgroup$ – Will Jagy Oct 20 at 13:48
  • $\begingroup$ $\frac59\lt\frac pq\lt\frac47$ is not an equation. An equation has an $=$ sign in it. $\endgroup$ – bof Oct 20 at 14:00
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    $\begingroup$ Those who are familiar with Farey Sequences will solve this immediately. If $a,b,c,d\in \Bbb Z^+$ with $a/b<c/d$ and $ ***\; (bc-ad)=1 \; ***\; $ then every $x\in \Bbb Q\cap (a/b,c/d)$ is expressed in lowest terms as $x=(ma+nc)/(mb+nd)$ for unique $m,n \in \Bbb Z^+.$ So $mb+nd\ge b+d,$ with equality iff $m=n=1.$ So in the Q, $p/q=(5+4)/(9+7)=9/16.$ $\endgroup$ – DanielWainfleet Oct 20 at 14:23
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    $\begingroup$ @DanielWainfleet great! Farey sequence can perfectly be applied to this question. Did not spot that before lol. $\endgroup$ – camhunter Oct 20 at 14:29
  • $\begingroup$ @camhunter I think you'll find it's the same solution. $\endgroup$ – Matthew Daly Oct 20 at 15:05
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What do we get with continued fractions?

$\dfrac{5}{9}=\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{4}}}$

$\dfrac{4}{7}=\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{3}}}$

The fractions are identical until we get to the last layer where one has a $4$ and the other has a $3$. Were there an integer between $3$ and $4$ we could replace the last layer with the smallest such integer, following the accepted answer here.

We don't have such an integer between $3$ and $4$ so this does not appear to work. But we can force the issue by rendering

$4=3+\dfrac{1}{1}$

$3=3+\dfrac{1}{M}$

where $M$ is taken as approaching infinity. Then we have

$\dfrac{5}{9}=\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{3+\dfrac{1}{1}}}}$

$\dfrac{4}{7}=\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{3+\dfrac{1}{M}}}}$

Now we put $2$ as the smallest integer between $1$ and $M$ to get an intervening fraction with a minimal denominator. Thus

$\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{3+\dfrac{1}{2}}}}=\dfrac{9}{16}$

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    $\begingroup$ Thanks! This is a brilliant method! $\endgroup$ – camhunter Oct 20 at 14:27
  • $\begingroup$ Apparently not brilliant enough for some people. $\endgroup$ – Oscar Lanzi Nov 7 at 15:01

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