Farey Fractions

U.U.D.M. Project Report 2017:24

Farey Fractions

Rickard Fernström

Examensarbete i matematik, 15 hp Handledare: Andreas Strömbergsson Examinator: Jörgen Östensson Juni 2017

Department of Mathematics Uppsala University

Farey Fractions Uppsala University

Rickard Fernstr¨om June 22, 2017

1 1 Introduction

The Farey sequence of order n is the sequence of all reduced fractions between 0 and 1 with denominator less than or equal to n, arranged in order of increasing size. The properties of this sequence have been thoroughly in- vestigated over the years, out of intrinsic interest. The Farey sequences also play an important role in various more advanced parts of number theory.

In the present treatise we give a detailed development of the theory of Farey fractions, following the presentation in Chapter 6.1-2 of the book

MNZ = I. Niven, H. S. Zuckerman, H. L. Montgomery, ”An Introduction to the Theory of Numbers”, ﬁfth edition, John Wiley & Sons, Inc., 1991, but ﬁlling in many more details of the proofs.

Note that the definition of ”Farey sequence” and ”Farey fraction” which we give below is apriori different from the one given above; however in Corollary 7 we will see that the two definitions are in fact equivalent.

2 Farey Fractions and Farey Sequences

We will assume that a fraction is the quotient of two integers, where the denominator is positive (every rational number can be written in this way). A reduced fraction is a fraction where the greatest common divisor of the 3.5 7 numerator and denominator is 1. E.g. is not a fraction, but is both 4 8 a fraction and a reduced fraction (even though we would normally say that 3.5 7 −1 0 = ). Also and are not fractions, since their denominators are 4 8 −2 −1 negative.

We will construct a table in the following way, where the fractions in each row of the table are in the speciﬁed order. The 1st row only contains the 0 1 fractions and . If the nth row has been constructed, then the (n + 1)st 1 1 row is constructed by copying the nth row and then for each pair of consec- a a0 utive fractions and that exist in the nth row that satisfy b + b0 ≤ n + 1, b b0

2 a + a0 a a0 the fraction is inserted in the (n + 1)st row between and . For ex- b + b0 b b0 ample, the 2nd row is constructed by copying the 1st row and then inserting 0 + 1 1 0 1 0 1 1 = between and , so that the 2nd row is , , . When con- 1 + 1 2 1 1 1 2 1 0 + 1 1 0 1 1 + 1 2 structing the 3rd row, = is inserted between and and = 1 + 2 3 1 2 2 + 1 3 1 1 0 1 1 2 1 is inserted between and , so that the 3rd row becomes , , , , . 2 1 1 3 2 3 1 0 + 1 1 0 1 When constructing the 4th row, = is inserted between and 1 + 3 4 1 3 2 + 1 3 2 1 1 + 1 2 and = is inserted between and , but = is not inserted 3 + 1 4 3 1 3 + 2 5 1 1 1 + 2 3 1 between and because 5 6≤ 4 and = is not inserted between 3 2 2 + 3 5 2 2 0 1 1 1 2 3 1 and because 5 6≤ 4. So the 4th row is , , , , , , . The ﬁrst six 3 1 4 3 2 3 4 1 rows in the table are: 0 1 1 1 0 1 1 1 2 1 0 1 1 2 1 1 3 2 3 1 0 1 1 1 2 3 1 . 1 4 3 2 3 4 1 0 1 1 1 2 1 3 2 3 4 1 1 5 4 3 5 2 5 3 4 5 1 0 1 1 1 1 2 1 3 2 3 4 5 1 1 6 5 4 3 5 2 5 3 4 5 6 1

Deﬁnition 1 (Farey sequence). The sequence of fractions in the nth row in the above table is called the Farey sequence of order n.

Deﬁnition 2 (Farey fraction of order n). A Farey fraction of order n is a fraction in the Farey sequence of order n.

3 Deﬁnition 3 (Farey fraction). A Farey fraction is a Farey fraction of some order, i.e. a fraction in the Farey table.

a a0 Theorem 1 (Theorem 6.1 and Corollary 6.3 in MNZ). If and are b b0 a a0 consecutive fractions in the nth row with to the left of , then a0b−ab0 = 1. b b0 The fractions in the nth row are also listed in order of their size (in strictly ascending order).

Proof. Base case (n = 1): 1 · 1 − 0 · 1 = 1 and the fractions in the ﬁrst row are clearly listed in order of their size (in strictly ascending order). a a0 Induction hypothesis: Assume for some n ∈ that if and are fractions N>0 b b0 a a0 in the nth row with to the left of , then a0b − ab0 = 1. Also assume the b b0 fractions in the nth row are listed in order of their size (in strictly ascending order). Induction step: We want to show that the fractions in the (n + 1)st row are listed in order of their size (in strictly ascending order). We know that the (n + 1)st row is constructed by copying the nth row and then for each pair of consecutive fractions in the nth row, insert 1 or 0 fractions between those fractions and we also know by the induction hypothesis that the fractions in the nth row are listed in order of their size (in strictly ascending order). So p r it’s suﬃcient to show that for each pair of consecutive fractions , in the q s p p p + r r p r nth row with to the left, < < . So let and be consecutive q q q + s s q s p fractions in the nth row with to the left. Then q

p p + r p(q + s) q(p + r) < ⇐⇒ < q q + s q(q + s) q(q + s) ⇐⇒ p(q + s) < q(p + r) ⇐⇒ rq − ps > 0.

But the last inequality holds because it follows from the induction hypoth- p p + r esis that rq − ps = 1. Hence < . Similarly it can be shown that q q + s

4 p + r r < . So the fractions in the (n + 1)st row are listed in order of their q + s s size (in strictly ascending order) and from that it follows that no fraction can appear twice in the (n + 1)st row.

a a0 a Let and be 2 consecutive fractions in the (n + 1)st row with to b b0 b a0 the left of . We want to show that a0b − ab0 = 1. If they are also con- b0 a secutive fractions in the nth row with to the left, then it follows from the b induction hypothesis that a0b − ab0 = 1. So assume they are not consecutive a fractions in the nth row with to the left. b a a0 We want to show it can’t be the case that both and exist in the nth b b0 a a0 row. Assume for contradiction that both and exist in the nth row. If b b0 they are not consecutive fractions in the nth row, then there is some frac- p tion in the nth row somewhere between them, say . But because of how q p a a0 rows are constructed, must be between and in the (n + 1)st row, q b b0 a a0 which contradicts and being consecutive fractions in the (n + 1)st row. b b0 a a0 Assume instead for contradiction that and are consecutive fractions in b b0 a0 the nth row, but with to the left. Either no fraction was added between b0 a a0 them when constructing the (n + 1)st row, in which case and will be b b0 a0 consecutive fractions in the (n + 1)st row, but with to the left. This leads b0 a a0 to a contradiction. If instead a fraction was added between and when b b0 a a0 constructing the (n + 1)st row, then and are not even consecutive frac- b b0 tions in the (n + 1)st row, which is a contradiction. So it can’t be the case a a0 that both and exist in the nth row when they are consecutive fractions b b0 a in the (n + 1)st row with to the left and they are not consecutive fractions b

5 a in the nth row with to the left. b a We want to show that it can’t be the case that neither of the fractions and b a0 exist in the nth row. Assume or contradiction that neither of the fractions b0 a a0 and exist in the nth row. Since they exist in the (n + 1)st row, it must b b0 be the case that they were both inserted between fractions in the nth row. But because at most 1 fraction is inserted between each pair of consecutive fractions, they must have been inserted between distinct pairs of fractions in a the nth row. But then it’s clear that some fraction must exist between and b a0 in the (n + 1)st row, which contradicts them being consecutive fractions b0 a in the (n + 1)st row. So it’s impossible that neither of the fractions and b a0 exist in the nth row when they are consecutive fractions in the (n + 1)st b0 a row with to the left. b The remaining case is when one of the fractions exist in the nth row but a the other doesn’t. Assume it’s the left fraction that exists in the nth row b (if instead it’s the right fraction that exists in the nth row, then the proof a0 can be done analogously). The right fraction must have been inserted in b0 a the (n + 1)st row between and the fraction in the nth row directly to the b a p right of , say . Then the nth and (n + 1)st row will look something like b q a p Row n: ...... b q this: a a + p a0 p . Row n + 1: ... = ... b b + q b0 q Then a0b − ab0 = (a + p)b − a(b + q) = pb − aq = 1. {by induction hypothesis}

6 a Corollary 2 (Corollary 6.2 in MNZ). Every fraction in the table is in b reduced form, i.e. gcd(a, b) = 1. a Proof. Let n ∈ . We want to show every fraction in the nth row has N>0 b gcd(a, b) = 1. a Let be a fraction in the nth row. Since there is more than 1 fraction in b p a the nth row, there is a fraction, say next to in the nth row. Assume q b p a a is to the right of (if it is to the left of , then the proof can be done q b b analogously). By Theorem 1, pb−aq = 1. Consider the diophantine equation bx + ay = 1. It has a solution iﬀ gcd(a, b) = 1. But x = p, y = −q is a solution, so gcd(a, b) = 1.

a a0 Theorem 3. ∀n ∈ : if and are consecutive fractions in the nth N>0 b b0 row, then b + b0 ≥ n + 1.

Proof. For n = 1, 1 + 1 = 2 ≥ 1 + 1, so it’s true for n = 1. Induction hypothesis: Assume it’s true for some n ∈ N>0. Induction step: We want to show it’s true for n + 1. a a0 a Let and be 2 consecutive fractions in the (n+1)st row with to the left. b b0 b If they are also consecutive fractions in the nth row then by the induction hypothesis, b + b0 ≥ n + 1. It can’t be the case that b + b0 = n + 1, because if that were the case then b + b0 ≤ n + 1, so in the (n + 1)st row a fraction a a0 a a0 should have been inserted between and , but then and can’t be b b0 b b0 consecutive fractions in the (n + 1)st row, a contradiction. So b + b0 > n + 1, i.e. b + b0 ≥ n + 2. Otherwise if they are not consecutive fractions in the nth row, exactly 1 of the a fractions appear in the nth row like in Theorem 1. Say appears in the nth b p a a0 row and is the fraction in the nth row directly to the right of . Then is q b b0 a p the fraction ”between” and , so b0 = b + q. By the induction hypothesis, b q b+q ≥ n+1. And hence b+b0 = b+b+q ≥ 1+b+q ≥ 1+(n+1) = n+2.

7 p Theorem 4. Let n ∈ and let be a fraction in the nth row. Then q ≤ n N>0 q p with equality iff does not exist in any previous row. q p Proof. We will begin by showing the first part, i.e. ∀n ∈ : if is a N>0 q fraction in the nth row, then q ≤ n. 0 1 For n = 1 the only fractions are and , and 1 ≤ 1. 1 1 p Induction hypothesis: Assume for some n ∈ that for every fraction in N>0 q the nth row, q ≤ n. p Induction step: Let be a fraction in the (n + 1)st row. We want to show q that q ≤ n + 1. p Case 1 : exists in the nth row as well. Then by induction hypothesis, q q ≤ n < n + 1. p p Case 2 : does not exist in the nth row. Then must lie between 2 fractions q q a a0 that exist in the nth row, say and . Then q = b + b0 and because of the b b0 rules of how rows are constructed, b + b0 ≤ n + 1, so q ≤ n + 1. p Now we want to prove the second part, i.e. ∀n ∈ : if is a fraction N>0 q p in the nth row, then q = n iff does not exist in any previous row. q p For n = 1, we see that q = 1 and does not exist in any previous row, so q the theorem holds for n = 1. p p Say n > 1 and let be a fraction in the nth row. If exists in a previous q q p row, then since rows are constructed by copying the previous row, must q exist in the (n − 1)st row. This means we are in case 1 of the proof of the previous part of this theorem, so q < n, i.e. q 6= n. p If instead doesn’t exist in a previous row, then we are in case 2 of the proof q p a of the previous part of this theorem, i.e. lies between the fractions and q b a0 that exist in the (n − 1)st row. Then q = b + b0 ≤ n. But because of b0 8 Theorem 3, b + b0 ≥ n. Since q ≤ n and q ≥ n we must have that q = n.

Lemma 5. Let n ∈ and let x ∈ be such that 0 ≤ x ≤ n + 1 and N>a0 N gcd(x, n + 1) = 1. Let be the greatest fraction in the nth row that is less b x a0 than and let be the smallest fraction in the nth row that is greater n + 1 b0 x than . Then a + a0 = x and b + b0 = n + 1. n + 1 x Proof. It can’t be the case that is fraction in the nth row, since that n + 1 a a0 would contradict Theorem 4 (n + 1 6≤ n). So then and are consecutive b b0 a fractions in the nth row with to the left and b a x a0 < < . b n + 1 b0 a x We see that < ⇐⇒ bx − a(n + 1) > 0 ⇐⇒ bx − a(n + 1) ≥ 1. b n + 1 Consider the diophantine equation

bu − av = k, where k := bx − a(n + 1) and u and v are the unknowns. It has a particular solution hu, vi = hka0, kb0i since

bka0 − akb0 = k(a0b − ab0) = k. by Theorem 1

We have that gcd(a, b) = 1 because of Corollary 2, so it has the general 0 0 solution hu, vi = hka + ae, kb + bei, e ∈ Z. But we also know it has a 0 particular solution hu, vi = hx, n + 1i, so ∃m ∈ Z : x = ka + am ∧ n + 1 = 0 kb + bm. Let m ∈ Z be such that kb0 + bm = n + 1 (1) and ka0 + am = x. (2)

9 We want to show that k = m = 1. Since 0 ≤ x ≤ n + 1 and gcd(x, n + 1) = 1, it must be the case that x > 0. We have that x a0 n + 1 b0 < ⇐⇒ > n + 1 b0 x a0 ⇐⇒ (n + 1)a0 − b0x > 0 ⇐⇒ (n + 1)a0 − b0x ≥ 1 ⇐⇒ (kb0 + bm)a0 − b0(ka0 + am) ≥ 1 {by (1) and (2)} ⇐⇒ (a0b − ab0)m ≥ 1 ⇐⇒ m ≥ 1. by Theorem 1

This shows that m ≥ 1. We also know that k ≥ 1. If we assume for contradiction that k > 1 or m > 1, then

b + b0 < bm + kb0 = n + 1, {by (1)} but that’s impossible, since by Theorem 3, b + b0 ≥ n + 1. This shows that k ≤ 1 and m ≤ 1. Since also m ≥ 1 and k ≥ 1, it follows that k = m = 1. Equation (1) now becomes b + b0 = n + 1 and equation (2) becomes a + a0 = x.

Theorem 6 (Theorem 6.5 in MNZ). If 0 ≤ x ≤ y, gcd(x, y) = 1, then the x fraction appears in the yth and all later rows. y x Proof. It is clear that if appears in the yth row then it also appears in all y x later rows, so it is enough to show that appears in the yth row. y 0 1 If y = 1 then either x = 0 or x = 1. Both and appear in the 1st row, so 1 1 the theorem holds for y = 1. We want to show the theorem holds when y ≥ 2, i.e. when y = n + 1 for some n ∈ N>0. Let n ∈ N>0, 0 ≤ x ≤ n + 1 and gcd(x, n + 1) = 1. If x = 0 or x = n + 1 then gcd(x, n + 1) = n + 1 > 1, so it must be the case that 0 < x < n + 1.

10 x a We want to show that appears in the (n + 1)st row. Let be the n + 1 b x a0 greatest fraction in the nth row that is less than and let be the n + 1 b0 x smallest fraction in the nth row that is greater than . By Lemma 5 we n + 1 know that a + a0 = x and b + b0 = n + 1. From the ﬁrst few lines of the proof x a of Lemma 5 we know that does not exist in the nth row, so and n + 1 b a0 a + a0 x must be consecutive fractions in the nth row. Because = b0 b + b0 n + 1 x and how rows are constructed, we know that must be inserted in the n + 1 a a0 (n + 1)st row between and . b b0

Corollary 7 (Corollary 6.6 in MNZ). The nth row consists exactly of all a a reduced fractions such that 0 ≤ ≤ 1 and 0 < b ≤ n. The fractions are b b listed in order of their size. a a Proof. Let n ∈ and let be a reduced fraction such that 0 ≤ ≤ 1 N>0 b b a and 0 < b ≤ n. We want to show that appears in the nth row. Clearly b a 0 ≤ a ≤ b ≤ n. By Theorem 6 we have that appears in the bth and all b a a later rows, so appears in the nth row. So every reduced fraction such b b a that 0 ≤ ≤ 1 and 0 < b ≤ n exist in the nth row, and they are listed b in order of their size by Theorem 1. We want to show that the nth row p doesn’t contain any other fractions than these. Let be a fraction in the q p nth row. The fraction must be reduced by Corollary 2. By Theorem 4 q we have that q ≤ n and q > 0 by the deﬁnition of fractions. So 0 < q ≤ n. Because rows are constructed by copying the previous row and then inserting fractions between the fractions and the fact that rows are ordered by size, we 1 can never get a row with a fraction that is strictly greater than or strictly 1 0 p less than . So 0 ≤ ≤ 1. 1 q

11 a a0 Theorem 8. If and are Farey fractions with a0b − ab0 = 1 (and thus b b0 a a0 a a0 < ), then and are consecutive Farey fractions of some order. b b0 b b0 a a0 Proof. Let and be Farey fractions with b b0 a0b − ab0 = 1 (3) a a0 c a c a0 (and thus < ). Let be a Farey fraction such that < < . We b b0 d b d b0 a c c a0 want to show that d > max(b, b0). From < and < it follows that b d d b0 ( cb − da > 0 da0 − cb0 > 0.

Let k := cb − da and l := da0 − cb0. Then k and l are positive integers and ( cb − da = k (4) da0 − cb0 = l.

Consider the diophantine equation bx − ay = k. (5) We know that b(ka0) − a(kb0) = k(a0b − ab0) = k. {by (3)} Hence hx, yi = hka0, kb0i is a particular solution to (5). Because gcd(a, b) = 1 (this follows from Corollary 2), the general solution to (5) is hx, yi = 0 0 0 0 hka + am, kb + bmi = kha , b i + mha, bi, m ∈ Z. Consider the diophantine equation −b0x + a0y = l. (6) Since −b0(al) + a0(bl) = l(a0b − ab0) = l {by (3)}

12 it follows that hx, yi = hal, bli is a particular solution to (6). We have that gcd(a0, b0) = 1 because of Corollary 2, so the general solution to (6) is 0 0 0 0 hx, yi = hal + a n, bl + b ni = lha, bi + nha , b i, n ∈ Z. We know that a a0 , is a basis for the vector space 2, since b b0 R a a0 det = ab0 − a0b b b0 = −(a0b − ab0) = −1 {by (3)} 6= 0. u u But then every vector ∈ 2 can be written in a unique way as = v R v a a0 c +c , where c , c ∈ . From (4) it follows that hx, yi = hc, di is 1 b 2 b0 1 2 R a particular solution to both (5) and (6). By looking at the general solution c a0 a of (5) we see that = k + m for some m ∈ . By looking d b0 b Z c a a0 at the general solution of (6) we see that = l + n for some d b b0 c n ∈ . By comparing these 2 ways of writing and using the fact that Z d a a0 , is a basis for 2 we conclude that l = m and k = n. Thus b b0 R c a0 a = k + l . But then d b0 b d = kb0 + lb ≥ b0 + b > max(b, b0) =: m. In the above equations we used the fact that k and l are positive integers. c By Corollary 7 it follows that does not exist in the mth row, since d > m. d c a a0 But was an arbitrary Farey fraction lying strictly between and . So d b b0 a a0 no Farey fraction lying strictly between and exist in the mth row, but b b0 a a0 a a0 both and exist in the mth row. So and must be consecutive Farey b b0 b b0 fractions in the mth row.

13 a a0 a a0 Corollary 9. Let and be Farey fractions such that < . Then b b0 b b0 a a0 a0b − ab0 = 1 iﬀ and are consecutive Farey fractions of some order. b b0 Proof. ” =⇒ ”: Assume a0b − ab0 = 1. Then it follows from Theorem 8 that a a0 and are consecutive Farey fractions of some order. b b0 a a0 ” ⇐= ”: Assume and are consecutive Farey fractions of some order. b b0 Then it follows from Theorem 1 that a0b − ab0 = 1.

a a0 Theorem 10 (Theorem 6.4 in MNZ). If and are consecutive fractions b b0 a p in the nth row (with to the left), then among all fractions such that b q a p a0 p a + a0 < < , = is the unique fraction with smallest denominator. b q b0 q b + b0 a a0 Proof. Let n ∈ and let and be consecutive fractions in the nth row, N>0 b b0 a with to the left. There are 2 possibilities. b Case 1 : b + b0 ≤ n + 1. Then b + b0 ≥ n + 1 by Theorem 3, so b + b0 = n + 1. a + a0 a a0 We have that will be inserted in the (n + 1)st row between and . b + b0 b b0 Case 2 : b + b0 > n + 1. Then for all integers k such that n ≤ k < b + b0: a a0 a + a0 and will be consecutive fractions in the kth row and will be b b0 b + b0 a a0 inserted between and in the (b + b0)th row. This is because the kth row b b0 is constructed by copying the (k − 1)st row and inserting fractions between a a0 the fractions. If and are consecutive fractions in the (k − 1)st row and b b0 a + a0 was not inserted between them in the kth row (i.e. b + b0 6≤ k), then b + b0 a a0 a a0 and will be consecutive fractions in the kth row as well. If and b b0 b b0 are consecutive fractions in the (k − 1)st row then a fraction is only inserted between them in row k if b + b0 ≤ k.

14 a a + a0 a0 In both cases we get that , and are 3 consecutive fractions in the b b + b0 b0 a a0 (b + b0)th row and that and are consecutive fractions in the (b + b0 − 1)st b b0 row. x a x a0 a a0 Let be a reduced fraction such that < < . Since and are y b y b0 b b0 a fractions in the nth row, it follows from Corollary 7 that 0 ≤ ≤ 1 and b a0 a x a0 x 0 ≤ ≤ 1. Since < < , we have that 0 ≤ ≤ 1. If we assume for b0 b y b0 y x contradiction that y < b+b0, then by Theorem 6, appears in the (b+b0−1)st y row. But that’s impossible since we know from Theorem 1 that fractions are a a0 listed in order of their size and there is no fraction strictly between and b b0 in the (b+b0 −1)st row. So y ≥ b+b0, i.e. a reduced fraction that lies strictly a a0 x between and can’t have a denominator smaller than b + b0. Suppose b b0 y x has minimal denominator, i.e. y = b + b0. By Theorem 6, exists in the y a a + a0 a0 (b + b0)th row. But since , and are 3 consecutive fractions in the b b + b0 b0 (b+b0)th row and the (b+b0)th row is ordered by size by Theorem 1, the only a a0 a + a0 fraction in the (b + b0)th row that lies strictly between and is . So b b0 b + b0 x a + a0 must be the fraction . y b + b0 p a a0 If you have a fraction that lies strictly between and that is not re- q b b0 duced, then it can’t be the case that q ≤ b + b0, since then you could simplify p into a reduced fraction where the denominator is strictly less than b + b0, q but that’s impossible. r a a0 Say you have a fraction that lies strictly between and . Then s ≥ b+b0, s b b0 r r since if is not reduced, then s > b + b0 and if is reduced, then s ≥ b + b0. s s a + a0 Also since exists in the (b + b0)th row it must be a reduced fraction b + b0 r because of Corollary 2. So the smallest denominator can have is b + b0, s

15 r and in that case must be a reduced fraction. So if you have a fraction that s a a0 lies strictly between and with smallest denominator, then it must be a b b0 reduced fraction with denominator b+b0 (since b+b0 is the smallest denominator). But the only reduced fraction with denominator b + b0 that lies strictly a a0 a + a0 between and is . So among all fractions that lie strictly between b b0 b + b0 a a0 a + a0 and , is the unique fraction with smallest denominator. b b0 b + b0

Proposition 11 (Problem 6.1.1 in MNZ). Let n be a positive integer such a a0 that n > 1 and let and be the Farey fractions immediately to the left b b0 1 and the right of respectively in the Farey sequence of order n. Then b = 2 n − 1 b0 = 1 + 2 · , i.e. b is the greatest odd integer ≤ n. It is also true 2 that a + a0 = b. 1 Proof. By Corollary 7 we know that exists in the Farey sequence of order 2 n since n ≥ 2. We will prove the Proposition by induction. a Base case (n = 2): Then we see from the Farey table on page 3 that is b 0 a0 1 and is . We have that 1 = 1, 1 is the greatest odd integer ≤ 2 and 1 b0 1 0 + 1 = 1. a a0 Induction hypothesis: Assume for some n ≥ 2 that if and are the Farey b b0 1 fractions immediately to the left and the right of the fraction respectively 2 in the Farey sequence of order n, then b = b0 and b is the greatest odd integer ≤ n and a + a0 = b. a a0 Induction step: Let and be the Farey fractions immediately to the left b b0 1 and the right of the fraction respectively in the Farey sequence of order n. 2 c c0 Let and be the Farey fractions immediately to the left and the right of d d0 1 the fraction respectively in the Farey sequence of order n + 1. We want to 2

16 show that d = d0 and that d is the greatest odd integer ≤ n+1 and c+c0 = d. Case 1: b + 2 = b0 + 2 = n + 1 (b = b0 by induction hypothesis). Then when a + 1 a constructing the (n + 1)st row the fraction is inserted between and b + 2 b 1 a0 + 1 1 a0 and also the fraction is inserted between and . Then it must 2 b0 + 2 2 b0 c a + 1 c0 a0 + 1 be the case that is the fraction and is the fraction , so then d b + 2 d0 b0 + 2 d = b + 2 and d0 = b0 + 2. Also from the induction hypothesis we see that b and b0 are odd, and since b + 2 = n + 1 we have that n = b + 1, i.e. n is even. From the induction hypothesis we know that b is the greatest odd integer ≤ n. Since d = b + 2 we need to show that b + 2 is the greatest odd integer ≤ n + 1, but this is obvious, since we know that n is even. Since d = b + 2 and d0 = b0 + 2 and b = b0 by induction hypothesis, it follows that d = d0, so d = d0 and d is the greatest odd integer ≤ n + 1. We also see that

c + c0 = a + 1 + a0 + 1 {c = a + 1 and c0 = a0 + 1} = b + 2 {by induction hypothesis} = d.

Case 2: b + 2 = b0 + 2 > n + 1 (b = b0 by induction hypothesis). Then a 1 1 a0 no fraction will be inserted between and or between and when b 2 2 b0 c a c0 constructing the (n + 1)st row. So is the fraction and is the fraction d b d0 a0 . We know that b0 n + 1 < b + 2 ≤ n + 2, {b ≤ n by Theorem 4} so it follows that b + 2 = n + 2, i.e. b = n. But b is odd by the induction hypothesis, so n must be odd. We know from the induction hypothesis that b is the greatest odd integer ≤ n and d = b. We need to show that b is the greatest odd integer ≤ n + 1 But that is obvious.

17 Since b = b0 by induction hypothesis, b = d and b0 = d0 it follows that d = d0. So d = d0 and d is the greatest odd integer ≤ n + 1. We also see that

c + c0 = a + a0 {a = c and a0 = c0} = b {by induction hypothesis} = d.

There is no case where b + 2 = b0 + 2 < n + 1 because of Theorem 3.

a a Theorem 12 (Problem 6.1.9 in MNZ). For each Farey fraction let C b b a denote the circle in the plane of radius (2b2)−1 and center , (2b2)−1 . b These circles are called the Ford circles. The interior of a Ford circle contains a a0 no point of any other Ford circle and two Ford circles C and C b b0 a a0 are tangent if and only if and are consecutive Farey fractions of some b b0 order. a a0 Proof. First assume that and are distinct Farey fractions such that b b0 a a0 a a0 < . We want to show that neither of the Ford circles C and C b b0 b b0 a contain a point that is an interior point of the other Ford circle. C has b radius 1 r := 2b2 and center a 1 a , = , r . b 2b2 b

18 a0 C has radius b0 1 s := 2(b0)2 and center a0 1 a0 , = , s . b0 2(b0)2 b0

a a0 Let d be the distance between the center of C and the center of C . b b0 Then a0 a2 (a0b − ab0)2 d2 = − + (s − r)2 = + s2 + r2 − 2rs. (7) b0 b b2(b0)2

a a0 In order to show that the Ford circles C and C don’t contain any b b0 interior points of the other Ford circle, it is suﬃcient to show that d ≥ r + s, i.e. d2 ≥ r2 + s2 + 2rs. Using (7) we see that we must show that

(a0b − ab0)2 + s2 + r2 − 2rs ≥ r2 + s2 + 2rs b2(b0)2 (a0b − ab0)2 1 ⇐⇒ ≥ 4rs = b2(b0)2 b2(b0)2 ⇐⇒ (a0b − ab0)2 ≥ 1. (8)

It can’t be the case that a0b − ab0 = 0, since then

a0b − ab0 = 0 ⇐⇒ a0b = ab0 a0 a ⇐⇒ = , {divide both sides by bb0} b0 b which contradicts the Farey fractions being distinct. But then a0b − ab0 is a non-zero integer, hence its square must be greater than or equal to 1, i.e. (8) a a0 is true. So the Ford circles C and C don’t contain any interior b b0

19 points of the other Ford circle and d ≥ r + s.

a a0 a a0 We want to show that C and C are tangent iﬀ and are b b0 b b0 a a0 consecutive Farey fractions of some order. If and are the same Farey b b0 a a0 fraction, then the Ford circles C and C are the same and are b b0 a a0 therefore not tangent and and are not consecutive Farey fractions of b b0 a a0 some order, so the Theorem holds in that case. So assume and are dis- b b0 a a0 tinct. Without loss of generality assume that < like before. We know b b0 a a0 that C and C are tangent if and only if d = r + s or d = |r − s|. b b0 Since r and s are positve, it follows that |r − s| ≤ max(r, s) < r + s. But before we showed that d ≥ r + s, so it can’t be the case that d = |r − s|. a a0 Therefore C and C are tangent if and only if d = r + s. But b b0 d = r + s iﬀ d2 = r2 + s2 + 2rs. Using (7) we see that

d2 = r2 + s2 + 2rs (a0b − ab0)2 ⇐⇒ + s2 + r2 − 2rs = r2 + s2 + 2rs b2(b0)2 (a0b − ab0)2 1 ⇐⇒ = 4rs = b2(b0)2 b2(b0)2 ⇐⇒ (a0b − ab0)2 = 1 ⇐⇒ a0b − ab0 = ±1.

a a0 But it can easily be shown that < iﬀ a0b−ab0 > 0, and since we assumed b b0 a a0 that < , it must be the case that a0b − ab0 > 0. Therefore b b0 a0b − ab0 = ±1 ⇐⇒ a0b − ab0 = 1. a a0 From Corollary 9 it follows that a0b − ab0 = 1 ⇐⇒ and are consecutive b b0

20 a a0 Farey fractions of some order. So C and C are tangent if and only b b0 a a0 if and are consecutive Farey fractions of some order. b b0

a a0 Remark 1. We will now use an alternative method to show that if and b b0 a a0 are consecutive Farey fractions of some order, then C and C are b b0 tangent. When using this method we will ﬁnd explicitly the coordinates for the intersection of the Ford circles. a a0 a a0 Proof. Let and be consecutive Farey fractions of order n, with < . b b0 b b0 a a0 We want to show that C and C are tangent. The equation for b b0 a C is b

a2 1 2 1 x − + y − = (9) b 2b2 4b4

a0 and the equation for C is b0

a0 2 1 2 1 x − + y − = . (10) b0 2(b0)2 4(b0)4 a Then the x-coordinate of center of C is less than (i.e. to the left of) the b a0 x-coordinate of the center of C . Let L be the line that goes through b0

21 a a0 the center of C and the center of C . It’s slope is b b0

1 1 a0 a − / − 2(b0)2 2b2 b0 b b2 − (b0)2 a0b − ab0 = / 2b2(b0)2 bb0 b2 − (b0)2 1 = / {by Theorem 1} 2b2(b0)2 bb0 b2 − (b0)2 = . 2bb0 1 b2 − (b0)2 a0 The equation for L is y− = x − . We want to see where 2(b0)2 2bb0 b0 a0 L intersects C , so we plug the equation for L into (10) and get b0

a0 2 b2 − (b0)2 2 a0 2 1 x − + x − = b0 2bb0 b0 4(b0)4 ! b2 − (b0)2 2 a0 2 1 ⇐⇒ 1 + x − = 2bb0 b0 4(b0)4 4b2(b0)2 (b2)2 − 2b2(b0)2 + ((b0)2)2 a0 2 1 ⇐⇒ + x − = 4b2(b0)2 4b2(b0)2 b0 4(b0)4 b2 + (b0)2 2 a0 2 1 ⇐⇒ x − = 2bb0 b0 4(b0)4 a0 2 b2 ⇐⇒ x − = b0 (b0)2 (b2 + (b0)2)2 s a0 b2 ⇐⇒ x = ± b0 (b0)2(b2 + (b0)2)2 a0 b ⇐⇒ x = ± . b0 b0(b2 + (b0)2)

a0 b We only care about the left intersection, i.e. where x = − . b0 b0(b2 + (b0)2)

22 If we plug this x-value into the equation for L and solve for y we get

1 b2 − (b0)2 −b y = + · 2(b0)2 2bb0 b0(b2 + (b0)2) 1 b2 − (b0)2 = − 2(b0)2 2(b0)2(b2 + (b0)2) b2 + (b0)2 b2 − (b0)2 = − 2(b0)2(b2 + (b0)2) 2(b0)2(b2 + (b0)2) 1 = . b2 + (b0)2

a0 b 1 So x = − and y = is the left intersection of L and b0 b0(b2 + (b0)2) b2 + (b0)2 a0 a C . We want to show that this point also belongs to C . Plugging b0 b

23 the x-value and y-value into the left hand side of (9) gives a0 b a2 1 1 2 − − + − b0 b0(b2 + (b0)2) b b2 + (b0)2 2b2 a0b − ab0 b 2 = − bb0 b0(b2 + (b0)2) 2b2 b2 + (b0)2 2 + − 2b2(b2 + (b0)2) 2b2(b2 + (b0)2) 1 b 2 b2 − (b0)2 2 = − + {by Theorem 1} bb0 b0(b2 + (b0)2) 2b2(b2 + (b0)2) b2 + (b0)2 b2 2 = − bb0(b2 + (b0)2) bb0(b2 + (b0)2) b2 − (b0)2 2 + 2b2(b2 + (b0)2) 2b(b0)3 2 b2(b0)2 − (b0)4 2 = + 2b2(b0)2(b2 + (b0)2) 2b2(b0)2(b2 + (b0)2) 4b2(b0)6 + b4(b0)4 − 2b2(b0)6 + (b0)8 = 4b4(b0)4(b2 + (b0)2)2 (b0)4(b2 + (b0)2)2 = 4b4(b0)4(b2 + (b0)2)2 1 = , 4b4 a0 b 1 a a0 which means that − , belongs to C , C b0 b0(b2 + (b0)2) b2 + (b0)2 b b0 a a0 and L, so it must be the case that C and C are tangent. b b0

a Theorem 13 (Problem 6.1.2 in MNZ). The number of Farey fractions of b n a X order n satisfying the inequalities 0 ≤ ≤ 1 is 1 + φ(j) and their sum is b j=1 exactly half this value. Proof. We want to prove the first part of the theorem. Let n be a posi- a tive integer. Every Farey fraction of order n must satisfy the inequalities b 24 a a 0 ≤ ≤ 1 by Corollary 7, so the Farey fractions of order n satisfying the b b a inequalities 0 ≤ ≤ 1 are precisely the Farey fractions of order n. b By Corollary 7 the Farey fractions of order n are precisely the reduced frac- a a tions such that 0 ≤ ≤ 1 and 0 < b ≤ n. For integers j such that b b 1 ≤ j ≤ n, define a a A(j) := the number of reduced fractions such that 0 ≤ ≤ 1. j j Also let F (n) := the number of Farey fractions of order n. n X Then clearly F (n) = A(j). It is clear that A(j) is the number of Farey j=1 fractions of order n with denominator j. We have that A(1) = 2, because 0 1 and are the only reduced fractions with denominator 1 that lie between 1 1 0 and 1. The definition of A(j) can be rewritten by multiplying the last inequalities by j and using the definition of reduced fraction to obtain a A(j) = the number of fractions such that gcd(a, j) = 1 and 0 ≤ a ≤ j. j a Different values for a give different fractions , so we can rewrite A(j) as j A(j) = the number of integers a such that gcd(a, j) = 1 and 0 ≤ a ≤ j. For j ≥ 2 we have that gcd(0, j) = j 6= 1, so for j ≥ 2 we can rewrite A(j) as A(j) = the number of integers a such that gcd(a, j) = 1 and 1 ≤ a ≤ j = φ(j). So n n n X X X F (n) = A(j) = A(1) + A(j) = 2 + φ(j) j=1 j=2 j=2 n X = 1 + φ(1) + φ(j) j=2 n X = 1 + φ(j). j=1

25 We want to prove the second part of the theorem. Let S(n) be the sum of the n ! 1 X Farey fractions of order n. We want to show that S(n) = 1 + φ(j) . 2 j=1 X a S(n) = . b a a b | b is a Farey fraction of order n a a b − a If is a Farey fraction of order n, then 1− = is also a Farey fraction b b b a of order n. This is because if is Farey fraction of order n then by Corollary b 7 we have that 0 ≤ a ≤ b, so 0 ≤ b − a ≤ b and if gcd(a, b) = 1, then b − a gcd(b − a, b) = 1 and hence by Corollary 7, is also a Farey fraction b a b − a a of order n. If b ≥ 3, then it can’t be the case that = if is a b b b Farey fraction of order n, since that would imply 2a = b, i.e. a|b, but also b a = > 1, so that a is a common divisor of a and b that is strictly greater 2 than 1, which means that gcd(a, b) 6= 1. So X a X a S(n) = + . b b a a a a b | b is a Farey fraction of order n and b≤2 b | b is a Farey fraction of order n and b>2 a b − a The terms in the right sum can be paired up as , . The terms in b b each pair add up to 1. If we assume n ≥ 2 then the only Farey fractions of n 0 1 1 X order n with denominator ≤ 2 are , and , so there are 1 + φ(j) − 3 1 1 2 j=1 n ! 1 X terms being added in the right sum, so there are φ(j) − 2 pairs being 2 j=1 n ! 1 X summed in the second sum. Thus the right sum equals φ(j) − 2 . 2 j=1 3 The left sum equals , so 2 n ! n ! 3 1 X 1 X S(n) = + φ(j) − 2 = 1 + φ(j) . 2 2 2 j=1 j=1

26 1 ! 0 1 1 X If n = 1, then S(n) = + = 1 = 1 + φ(j) , so the theorem holds 1 1 2 j=1 for all n.

3 Rational Approximations a c Theorem 14 (Theorem 6.7 in MNZ). Let and be Farey fractions of b d order n such that no other Farey fraction of order n lies between them. Then

a a + c 1 1 − = ≤ b b + d b(b + d) b(n + 1) and c a + c 1 1 − = ≤ . d b + d d(b + d) d(n + 1) Proof. For the ﬁrst formula we have

a a + c a(b + d) − b(a + c) − = b b + d b(b + d) |ad − bc| = b(b + d) 1 = by Theorem 1 b(b + d) 1 ≤ . {since b + d ≥ n + 1 by Theorem 3} b(n + 1)

The second formula is obtained in a similar way.

Theorem 15 (Theorem 6.8 in MNZ). Let n ∈ N>0 and x ∈ R. Then there a is a rational number such that 0 < b ≤ n and b a 1 x − ≤ . b b(n + 1)

27 Proof. Let n ∈ N>0 and x ∈ R. Let k be the unique integer such that 0 ≤ x + k < 1. There are 2 cases. p Case 1 : x+k has the same value as some Farey fraction of order n. Let be q the Farey fraction of order n that x+k simpliﬁes to. Let the rational number a p − k · q p be . Since is a Farey fraction of order n, q ≤ n by Theorem 4. b q q a So is a rational number such that 0 < b ≤ n and b

a p − k · q p − k · q k · q p p x − = x − = x + k − − = − b q q q q q = 0 1 ≤ . b(n + 1)

Case 2 : x + k does not have the same value as some Farey fraction of order 0 n. It can’t be the case that x + k = 0, since is a Farey fraction of order 1 p n. So 0 < x + k < 1. Let be the greatest Farey fraction of order n that is q r smaller than x + k and let be the smallest Farey fraction of order n that is s p greater than x + k. There is no Farey fraction of order n that lies between q r p r and . Since and are Farey fractions of order n, it follows from Theorem s q s 4 that q ≤ n and s ≤ n. There are 2 cases now. p + r a r − k · s Case 2.1 : ≤ x + k. Let the rational number be . We have q + s b s

28 a that is a rational number such that 0 < b ≤ n and b

a r − k · s r x − = x − = x + k − b s s r = − (x + k) s r p + r p + r r ≤ − since ≤ x + k < s q + s q + s s 1 ≤ by Theorem 14 s(n + 1) 1 = . b(n + 1)

p + r a p − k · q Case 2.2 : > x + k. Let the rational number be . We have q + s b q a that is a rational number such that 0 < b ≤ n and b

a p − k · q p x − = x − = x + k − b q q p + r p p p + r ≤ − since < x + k < q + s q q q + s

p p + r = − q q + s 1 ≤ by Theorem 14 q(n + 1) 1 = . b(n + 1)

Theorem 16 (Theorem 6.9 in MNZ). If ξ ∈ R \ Q, then there are inﬁnitely a many distinct rational numbers such that b a 1 ξ − < . b b2

29 Proof. Let ξ ∈ R \ Q. For each n ∈ N>0, let a a 1 A := ∈ 0 < b ≤ n ∧ ξ − ≤ . n b Q b b(n + 1)

Let X := {An| n ∈ N>0} By Theorem 15, each An is non-empty, so that X is a set of non-empty sets. By the axiom of choice, there exists a choice function f : X → ∪X such that ∀A ∈ X : f(A) ∈ A. So let f : X → ∪X be an a function such that ∀A ∈ X : f(A) ∈ A. For each n ∈ N>0, let be the bn rational number f(An). Then

an 1 ξ − ≤ bn bn(n + 1) 1 < 2 {since n + 1 > bn} bn for all n ∈ . N>0 a We want to show there are infinitely many distinct n . Assume for contra- b a n diction there are only finitely many distinct n . Then there are only finitely b n an an many distinct values for ξ − . Let d := min ξ − . Clearly d ≥ 0 bn n∈N>0 bn ak and d 6= 0, since d = 0 would imply that ξ = for some k ∈ N>0, but that bk contradicts ξ being irrational. So d > 0. Then for all n ∈ N>0 we have that

an 1 d ≤ ξ − ≤ bn bn(n + 1) 1 ≤ . {since b ≥ 1} n + 1 n 1 1 1 But if we let n be suﬃciently large (n ≥ will work), then < ≤ d. d n + 1 n But then d < d, a contradiction.

Lemma 17 (Lemma 6.10 in MNZ). If x and y are positive integers then not 1 1 1 1 1 1 1 1 both of the inequalities ≥ √ + and ≥ √ + xy 5 x2 y2 x(x + y) 5 x2 (x + y)2 can hold.

30 Proof. We will rewrite the inequalities. We see that

1 1 1 1 ≥ √ + xy 5 x2 y2 1 1 y2 + x2 ⇐⇒ ≥ √ · xy 5 (xy)2 √ n √ o ⇐⇒ 5xy ≥ y2 + x2 multiply both sides by 5(xy)2 and 1 1 1 1 ≥ √ + x(x + y) 5 x2 (x + y)2 1 1 (x + y)2 + x2 ⇐⇒ ≥ √ · x(x + y) 5 (x(x + y))2 √ n √ o ⇐⇒ 5x(x + y) ≥ (x + y)2 + x2. multiply both sides by 5(x(x + y))2

Assume for contradiction that both √ 5xy ≥ y2 + x2 (11) and √ 5x(x + y) ≥ (x + y)2 + x2 (12) are true. By adding the inequalities (11) and (12) we get √ 5(x2 + 2xy) ≥ 3x2 + 2xy + 2y2 √ ⇐⇒ 3x2 + 2xy + 2y2 ≤ 5(x2 + 2xy) √ √ ⇐⇒ (3 − 5)x2 − 2( 5 − 1)xy + 2y2 ≤ 0 √ √ ⇐⇒ (5 − 2 5 + 1)x2 − 4( 5 − 1)xy + 4y2 ≤ 0 {multiply by 2} √ ⇐⇒ (2y − ( 5 − 1)x)2 ≤ 0 √ ⇐⇒ 2y − ( 5 − 1)x = 0 {since squares are non-negative} √ 2y + x n √ o ⇐⇒ 5 = , solve for 5 x √ but that contradicts 5 being irrational.

31 a c a Lemma 18. If and are consecutive Farey fractions of order n with b d b c a 1 to the left, then − ≤ . d b n Proof. The fractions 0 1 2 n − 1 n , , , ··· , , (13) n n n n n are Farey fractions of order n if they are simplified by Corollary 7 (the simplified fractions lie between 0 and 1 and the denominator of each simplified a j c j fraction lies between 1 and n). It can’t be the case that < and > for b n d n j j some in the sequence (13), since then would be a Farey fraction of order n n a c n (if simplified) that lies strictly between and , which would contradict b d a c a c and being consecutive Farey fractions of order n. So and must both b d b d a c 1 lie between 2 consecutive elements in (13), which means − ≤ , since b d n 1 the distance between consecutive elements of (13) is always . n

Theorem 19 (Theorem 6.11 in MNZ). Given any ξ ∈ R \ Q, there exist h inﬁnitely many diﬀerent rational numbers such that k

h 1 ξ − < √ . k 5k2 Proof. Let µ be the unique integer such that 0 ≤ ξ + µ < 1. It can’t be the case that ξ + µ = 0, since that would imply that ξ is rational, which is not the case. So 0 < ξ + µ < 1. Let λ := ξ + µ. Then 0 < λ < 1. Let n be a a positive integer. Let n be the greatest Farey fraction of order n that is less b c n than λ and let n be the smallest Farey fraction of order n that is greater d a n c than λ. Then n and n are consecutive fractions in the nth row. We want bn dn

32 a c a + c h to show that at least one of the numbers n , n and n n will work as bn dn bn + dn k in the theorem (when ξ = λ). Assume for contradiction this is not the case. Then a 1 λ − n ≥ √ , (14) 2 bn 5bn c 1 n − λ ≥ √ (15) 2 dn 5dn and a + c 1 λ − n n ≥ √ . (16) 2 bn + dn 5(bn + dn) a + c Case 1 : n n < λ. Then (16) becomes bn + dn a + c 1 λ − n n ≥ √ . 2 bn + dn 5(bn + dn) By adding the inequalities (14) and (15) we get

c a 1 1 1 n n √ − ≥ 2 + 2 dn bn 5 bn dn b c − a d 1 1 1 n n n n √ ⇐⇒ ≥ 2 + 2 bndn 5 bn dn 1 1 1 1 √ ⇐⇒ ≥ 2 + 2 by Theorem 1 (17) bndn 5 bn dn and by adding the inequalities (15) and (16) we get c a + c 1 1 1 n n n √ − ≥ 2 + 2 dn bn + dn 5 dn (bn + dn) c (b + d ) − d (a + c ) 1 1 1 n n n n n n √ ⇐⇒ ) ≥ 2 + 2 dn(bn + dn) 5 dn (bn + dn) b c − a d 1 1 1 n n n n √ ⇐⇒ ≥ 2 + 2 dn(bn + dn) 5 dn (bn + dn) 1 1 1 1 √ ⇐⇒ ≥ 2 + 2 . by Theorem 1 dn(bn + dn) 5 dn (bn + dn) (18)

33 But not both the inequalities (17) and (18) can be true by Lemma 17 (if we let x = dn and y = bn in the lemma), so we get a contradiction. a + c Case 2: n n > λ. Then (16) becomes bn + dn a + c 1 n n − λ ≥ √ . 2 bn + dn 5(bn + dn) Just like in case 1 we can add the inequalities (14) and (15) to get (17). By adding the inequalities (14) and (16) we get a + c a 1 1 1 n n n √ − ≥ 2 + 2 bn + dn bn 5 bn (bn + dn) b (a + c ) − a (b + d ) 1 1 1 n n n n n n √ ⇐⇒ ≥ 2 + 2 bn(bn + dn) 5 bn (bn + dn) b c − a d 1 1 1 n n n n √ ⇐⇒ ≥ 2 + 2 bn(bn + dn) 5 bn (bn + dn) 1 1 1 1 √ ⇐⇒ ≥ 2 + 2 . by Theorem 1 bn(bn + dn) 5 bn (bn + dn) (19)

But not both inequalities (17) and (19) can be true by Lemma 17 (if we let x = bn and y = dn in the lemma), so we get a contradiction. a c a + c h Thus one of the numbers n , n and n n will work as in the the- bn dn bn + dn k hn an orem (when ξ = λ). So for each n ∈ N>0, let be one of the fractions , kn bn c a + c n and n n such that dn bn + dn h 1 λ − n < √ , 2 kn 5kn a c where n is the greatest Farey fraction of order n that is less than λ and n bn dn is the smallest Farey fraction of order n that is greater than λ. h We want to show that for every > 0 there exists a Farey fraction n (as kn

hn deﬁned above) such that λ − < . Let > 0 and let n be a positive in- kn 1 1 a c teger such that < (i.e. n > ). We know that n and n are consecutive n bn dn c a 1 fractions of order n, so we know by Lemma 18 that n − n ≤ < . Clearly d b n n n an cn an an + cn cn λ − < and λ − < . Also we know that < < , so bn dn bn bn + dn dn

hn hn an cn an + cn λ − < , since is one of the fractions , and . kn kn bn dn bn + dn h We want to show that there are infinitely many n with distinct values. kn h Assume for contradiction that there are only finitely many n . Then let k n hn hn A := n ∈ N>0 and := min λ − . But as we’ve shown there is a k hn k n k ∈A n n hn hn Farey fraction such that λ − < , which contradicts being minimal. kn kn h So there are infinitely many n with distinct values. kn

For every n ∈ N>0: h 1 λ − n < √ 2 kn 5kn h − µ · k 1 ⇐⇒ λ − µ − n n < √ 2 kn 5kn h − µ · k 1 ⇐⇒ ξ − n n < √ 2 kn 5kn h − µ · k and there are inﬁnitely many n n with distinct values since there are kn h h − µ · k h inﬁnitely many n with distinct values. This is because n n = n −µ kn kn kn and the function g : R → R, x 7→ x − µ is bijective.

35 √ Theorem 20 (Theorem 6.12 in MNZ). The constant√ 5 in Theorem 19 is the best possible, i.e. Theorem 19 does not hold if 5 is replaced by any larger value. √ Proof. It’s enough to ﬁnd one ξ such that√ 5 cannot be replaced by any 1 + 5 larger value in Theorem 19. Let ξ := . We see that 2 √ ! √ ! √ ! 1 − 5 1 + 5 1 − 5 (x − ξ) x − = x − x − = x2 − x − 1. 2 2 2 So if we let h and k be integers with k > 0, then

h h √ − ξ − ξ + 5 (20) k k √ ! h h 1 − 5

= − ξ − k k 2 2 h h = − − 1 k2 k

1 2 2 = h − hk − k . (21) k2 h h √ We know that (20) can’t be zero since that would imply ξ = or = ξ− 5, √ k k 2 2 but neither ξ nor ξ − 5 are rational. We know that h − hk − k is a non- 2 2 negative integer. It can’t be the case that h − hk − k = 0, since then 2 2 equation (21) is zero and hence equation (20) is zero. Since h − hk − k ≥ 1 2 2 1 1 it must be the case that h − hk − k ≥ and because equation (21) k2 k2 is equal to equation (20) it follows that

h h √ 1 − ξ − ξ + 5 ≥ . (22) k k k2 Let m be a positive real number and assume we have an inﬁnite sequence of hj rational numbers , kj > 0 such that kj

hj 1 − ξ < 2 . (23) kj mkj

36 If we multiply by kj on both sides of (23) we get 1 |hj − kjξ| < . mkj From this we can see that 1 1 kjξ − < hj < kjξ + , mkj mkj h so that for each rational number j in the sequence there can only be finitely kj many other fractions in the sequence with the same denominator, but different numerator. It follows that lim kj = ∞. j→∞ We see that 1 2 kj √ hj hj ≤ − ξ − ξ + 5 {by (22)} kj kj √ 1 hj < 2 − ξ + 5 {by (23)} mkj kj √ 1 hj ≤ 2 − ξ + 5 {by ∆-inequality} mkj kj 1 1 √ < 2 2 + 5 . {by (23)} mkj mkj 2 Multiplying by mkj we get 1 √ m < 2 + 5. mkj Therefore 1 √ √ m ≤ lim + 5 = 5. lim kj = ∞ j→∞ 2 j→∞ mkj √ h So if m > 5 then there is no infinite sequence of rational numbers j , kj kj > 0 such that

hj 1 − ξ < 2 kj mkj

37 h for every j in the sequence. kj