Farey Series

Let m1/n1 and m2/n2 be two successive terms of the Farey series FN. Then

(1)

m2n1 - m1n2 = 1

For three successive terms m1/n1, m2/n2, and m3/n3, the middle term is the mediant of the other two

(2)

m2/n2 = (m1 + m3)/(n1 + n3)

To prove that (1) implies (2), assume we have two identities

(3)

m2n1 - m1n2 = 1
m3n2 - m2n3 = 1

Subtract one from the other and recombine the terms:

(4)

(m3 + m1)n2 = m2(n3 + n1)

which leads to (2). Manipulating linear equalities should be working both ways. I'll try to reverse the flow of reasoning from (2) to (1). To prove that (2) implies (1), I'll use mathematical induction.

Assume we are given a series of fractions mi/ni, i = 1,2,..., with the condition that any three consecutive terms satisfy

(2')

(m3 + m1)n2 = m2(n3 + n1)

I want to show that for such a series (1) also holds. But (1) contains only two fractions. Therefore, let's assume that for i = 2 (1) indeed holds. This is clearly true for FN which starts with 0/1, 1/N, ... Assume that for some i > 2

(1')

mini- 1 - mi- 1ni = 1

Rewrite (2') as

(4')

(mi+1 + mi- 1)ni = mi(ni+1 + ni- 1)

Subtracting (4') from (1') easily gives

(1'')

mi+1ni - mini+1 = 1

which would appear to complete our inductive proof. Does it mean that the Farey series is infinite? After all, this is what mathematical induction is good for: establishing properties of infinite sequences. Let's try to continue the Farey series beyond 1/1.

Consider, for example, F5 whose last two terms are 4/5 and 1/1. We are looking for a fraction m/n such that (4 + m)/(5 + n) = 1/1. The solution is obviously 5/4. The next fraction should solve the equation (1 + m)/(1 + n) = 5/4. The solution is 4/3. A pattern seems to emerge. The new numbers are the very terms of the Farey series F5 written upside down and in the reverse order.

0/1, 1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5, 1/1, 5/4, 4/3, 3/2, 5/3, 2/1, 5/2, 3/1, 4/1, 5/1, 1/0

The fact becomes obvious when we notice that equations (3)-(4) are symmetric with respect to the indices 1 and 3. They may be solved for m1/n1 as well as for m3/n3. Equations (3)-(4) also do not change if we turn the fractions involved upside down. Once the first step beyond 1/1 is made, all others are forced to follow symmetrically their left-side-of-1/1 counterparts.

1/0 is a forbidden fraction. However, since it follows ... (N-1)/1, N/1 at the end of the Farey series FN, there is a good reason to agree that no harm will come from calling it infinity: 1/0 = . This is a big number, indeed. Can we carry our inductive process beyond infinity? No, the process breaks down. Trying to solve (N + m)/(1 + n) = 1/0 leads to n = -1 and an arbitrary m. Is anything wrong with our induction? How come we did not get an infinite sequence of terms?

To answer these questions we have to analyze the proof step by step.

Copyright © 1996-2017 Alexander Bogomolny

(4') implies (2') only when ni≠0. Division by 0 is employed in other arithmetic "paradoxes".

So the induction did not go through. Does this mean that our proof was wrong? No, it does not. We did proof that, for the Farey series (a series that starts with 0/1 and 1/N), condition (2) implies (1). However, our inductive reasoning (carrying from step "i" to the next step "i+1") broke down just after a finite number of steps. The upshot is that it applies to every Farey series FN. An additional insight comes from looking down the Stern-Brocot tree.

Copyright © 1996-2017 Alexander Bogomolny

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