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This question is from Apostol modular functions and Dirichlet series in number theory.

It is related to this problem - When are two neighbouring fractions in Farey sequence are similarly ordered

Apostol in 2 nd part of this exercise asks to prove that any two 2nd neighborhoods $\frac {a_i }{ b_i} $ and $\frac { a_{i+2 }} {b_{i +2}} $ are similarly ordered.

My attempt - There exists 2 cases if the fractions are not similarly ordered.

Case1 -$a_{i+2}$ < $a_i$ and $b_{i+2}$ > $b_i$ . It is easy to obtain contradiction in this case and I obtained it.

But in Case2 - $a_{i+2} $> $a_i$ and $b_{i+2} $< $b_i$ I cannot obtain any contradiction.

Can somebody please help.

1 Answers1

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If $\frac{a}{b} < \frac{c}{d}$ ($a,b,c,d$ positive integers [$a = 0$ is also admissible]) and $d \geqslant b$, then it immediately follows that $$c > d\cdot \frac{a}{b} = \frac{d}{b}\cdot a \geqslant a\,,$$ so the fractions are similarly ordered. This holds generally, no Farey sequence properties required. (You already know this part, but it's here for completeness.)

So let's look at the case $b_{i+2} < b_i$ for the second neighbours in a Farey sequence. We want to deduce $a_{i+2} \leqslant a_i$. Suppose it weren't so. Then $$\frac{a_{i+2}}{b_{i+2}} > \frac{a_{i+2} - 1}{b_{i+2}} \geqslant \frac{a_i}{b_{i+2}} \geqslant \frac{a_i}{b_i-1} > \frac{a_i}{b_i}$$ and since the two fractions are second neighbours the three fractions in the middle must all be equal, that is, we must have $a_{i+2} = a_i + 1$ and $b_{i+2} = b_i - 1$.

But we also know that the fraction between the two is their mediant, $$\frac{a_{i+1}}{b_{i+1}} = \frac{a_{i} + a_{i+2}}{b_i + b_{i+2}}\,,$$ and that means we must have $$\frac{a_{i}}{b_{i}-1} = \frac{2a_{i} + 1}{2b_{i} - 1} \iff \frac{2b_{i}-1}{b_{i}-1} = \frac{2a_{i}+1}{a_{i}} \iff \frac{1}{b_i-1} = \frac{1}{a_i} \iff a_i = b_i - 1\,.$$ This however means $$\frac{a_{i+2}}{b_{i+2}} = \frac{a_i+1}{b_i-1} = \frac{b_i}{b_i-1} > 1\,,$$ which contradicts the assumption that $0 < \frac{a_{i+2}}{b_{i+2}} \leqslant 1$.

Therefore $\frac{a_i}{b_i}$ and $\frac{a_{i+2}}{b_{i+2}}$ must be similarly ordered also when $b_{i+2} < b_i$.

Daniel Fischer
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  • Thank you very much for answering. But I am really busy and could not read the answer. So, I am awarding bounty as bounty period is about to end. If I have any questions, I will ask in comments. –  Jan 30 '20 at 13:18
  • @Daniel Fischer I am really sorry to ask you in this way but can you please take a look at this question asked by me if you have some spare time. I am badly struck on it. I am asking you as you are one of the few consistent person in analytic number theory tag who helps beginners in their questions even if some questions are trivial to expert. Can you please help if you got some spare time. https://math.stackexchange.com/questions/3530801/a-question-in-paper-one-of-odd-zeta-values-from-zeta5-to-zeta25-is –  Feb 02 '20 at 21:00
  • @Daniel Fischer can you please tell how did you cancelled $ 2b_i -1 $ and $ 2 a_i +1 $ in the fractions in 9 th line from below!! I am not getting it. –  Apr 12 '20 at 17:21
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    I guess you meant the "$\frac{2b_i-1}{b_i-1} = \frac{2a_i+1}{a_i} \iff \frac{1}{b_i-1} = \frac{1}{a_i}$". I'm just subtracting $2$ from each fraction, $\frac{2b_i-1}{b_i-1} = \frac{2(b_i-1)+1}{b_i-1} = 2 + \frac{1}{b_i-1}$, and $\frac{2a_i+1}{a_i} = 2 + \frac{1}{a_i}$. If you meant something else, please tell what you meant. – Daniel Fischer Apr 24 '20 at 21:00