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This is from I. M. Gelfand's Algebra book.

Fractions $\displaystyle\frac{a}{b}$ and $\displaystyle\frac{c}{d}$ are called neighbor fractions if their difference $\displaystyle\frac{ad - bc}{bd}$ has numerator of $\pm 1$, that is, $ad - bc = \pm 1$. Prove that

(a) in this case neither fraction can be simplified (that is, neither has any common factors in numerator and denominator);

(b) if $\displaystyle\frac{a}{b}$ and $\displaystyle\frac{c}{d}$ are neighbor fractions, then $\displaystyle\frac{a+c}{b+d}$ is between them and is a neighbor fraction for both $\displaystyle\frac{a}{b}$ and $\displaystyle\frac{c}{d}$; moreover,

(c) no fraction $\displaystyle\frac{e}{f}$ with positive integer $e$ and $f$ such that $f < b + d$ is between $\displaystyle\frac{a}{b}$ and $\displaystyle\frac{c}{d}$.

Parts (a) and (b) weren't too difficult, but I'm stuck on part (c). I've included (a) and (b) in case they're related to the solution to (c).

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    Okay, I'll do that. Thanks!2011-06-05

1 Answers 1

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Assume $\frac{e}{f}$ is (strictly) between $\frac{a}{b}$ and $\frac{c}{d}$. Then $\left|\frac{a}{b}-\frac{e}{f}\right| + \left|\frac{e}{f}-\frac{c}{d}\right| = \left|\frac{a}{b}-\frac{c}{d}\right| = \frac{1}{bd}$

But $\left|\frac{a}{b}-\frac{e}{f}\right| \geq \frac{1}{bf}$ and $\left|\frac{e}{f}-\frac{c}{d}\right|\geq \frac{1}{df}$. So $\frac{1}{bf} + \frac{1}{df} \leq \frac{1}{bd}$. Multiply both sides by $bdf$ and we get that $b+d\leq f$.

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    In this case, I wrote "strictly' precisely because I wanted to be clear that I meant \frac{a}{b}<\frac{e}{f}<\frac{c}{d} or \frac{c}{d}<\frac{e}{f}<\frac{a}{b}. @ArshJhaj The theorem is not true if you allow $e=c,f=d$. The first part of my line is true for $\frac{e}{f}=\frac{a}{b}$, but the next step: $\left|\frac{a}{b}-\frac{e}{f}\right|\geq \frac{1}{bf}$ is obviously not2015-07-08