Patrick from Woodbridge School proved the first part of this question:
If 0<bd<ac<1 then bc<ad . Adding bd to both sides of the inequality gives b(c+d)<d(a+b), while adding ac to both sides gives c(a+b)<a(c+d). So it follows that
bd<a+bc+d<ac.
We know F1=01,11 and F2=01,12. The mediant of 01 and 12 is 13 and the mediant of 12 and 11 is 23. So
F3=01,13,12, 23,11.
The mediant of 01 and 13 is 14. The mediant of 13 and 12 is 25 and so is not included in F4. The mediant of 12 and 23 is 35 and so is not included in F4. The mediant of 23 and 11 is 34. Thus
F4=01,14,13,12,23,34,11.
The mediants of F4 are : 15, 27 (not included in F5), 25, 35 , 57 (not included in F5) and 45. Therefore
F5=01,15,14,13,25,12,35,23,34,45,11.
Edd from Sidcot school did the first part in a similar way and used mathematical induction to prove that, in every Farey sequence, if bd and ac are consecutive terms (said to be Farey neighbours) then |ad−bc|=1.
First F1=01,11 and
,so the result is true for n=1.
Assuming the result for Fn we next prove that the result follows for Fn+1.
We have Fn=…bd,ac… and |ad−bc|=1.
If d+c≤n+1 then Fn+1=…bd,(b+a)(d+c)ac… and |d(b+a)−b(d+c)|=|ad−bc|=1. The only other possibility is that d+c>n+1 in which case Fn+1=…bd,ac… with |ad−bc|=1, so the result is true for n+1.
Hence, by the axiom of induction the result is true for all positive integer values of n
.
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