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I am trying exercises from Tom M Apostol and I could not think about this problem in Chapter 5.

Problem is - Two reduced fractions $a/b$ and $c/d$ are said to be similarly ordered if $(c-a)\times(d-b)\ge0$. Prove that any two neighboring fractions $\frac{a_i}{b_i}$ and $\frac{a_{i+1}}{b_{i+1}}$ are similarly ordered.

My attempt - I tried using result - for any two consecutive Farey fractions $a/b<c/d$, $bc-ad=1$ holds and then using definition of similarly ordered fractions. But it doesn't yields result when $b\neq d$.

Can someone please help.

zipirovich
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  • Perhaps you mean $\frac{a_{i + 1}}{b_{i + 1}}$? – cip999 Dec 24 '19 at 16:54
  • @cip999 yes I edited it. Unfortunately, in my browser I can't preview my post. So, I look for any Mathjax mistake only after it's posted. –  Dec 24 '19 at 16:55

2 Answers2

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$a_{i + 1}b_i - a_ib_{i + 1} = 1$ is indeed the right identity to look at.

Note that, since $\frac{a_i}{b_i} < \frac{a_{i + 1}}{b_{i + 1}}$, the only way $(a_{i + 1} - a_i)(b_{i + 1} - b_i) \ge 0$ could possibly fail is if $a_{i + 1} \ge a_i + 1$ and $b_{i + 1} \le b_i - 1$. But then we'd have $$a_{i + 1}b_i - a_ib_{i + 1} \ge (a_i + 1)b_i - a_i(b_i - 1) \ge a_i + b_i > 1$$ which is a contradiction.

cip999
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The non-negative Farey fractions can be obtained by starting with $\frac01,\frac11,\frac21,\ldots$ (where the claim clearly holds) and then repeatedly inserting $\frac{a+c}{b+d}$ between adjacent fractions $\frac ab$ and $\frac cd$. Near an inserted fraction, we have $$ ((a+c)-a)\cdot ((b+d)-b)=cd\ge 0$$ and $$ (c-(a+c))\cdot (d-(b+d))=ab\ge0.$$

Weihnachtsmann
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