Number Theory, Lecture 7

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Number Theory, Lecture 7 Jan Snellman

Finite continued fractions

Infinite continued fractions

Diophantine approximation Geometric interpretation Applications Periodic continued fractions

Number Theory, Lecture 7

Continued fractions

Jan Snellman¹

1Matematiska Institutionen Link¨opings Universitet

Link¨oping, spring 2019 Lecture notes availabe at course homepage http://courses.mai.liu.se/GU/TATA54/

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Summary

1 Finite continued fractions Examples, simple properties Existence and uniqueness CF as rational functions Euler’s rule

Convergents Applications

2 Infinite continued fractions Partial convergents, repetition The continued fraction process

3 Diophantine approximation 4 Geometric interpretation 5 Applications

Recognizing a rational number Calendar

Huygen’s planetarium 6 Periodic continued fractions

Golden ratio

Quadratic irrationality Lagrange’s theorem

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Summary

Golden ratio

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Summary

Golden ratio

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Summary

Golden ratio

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Summary

Golden ratio

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Summary

Golden ratio

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Examples, simple properties Existence and uniqueness

CF as rational functions Euler’s rule

Example

5 + 1

3 + ¹

11+¹₂

=5 + 1

3 +₂₃² =5 + 23 71 = 378

71

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Example

5 = 5 ≈ 5.000 5 + 1

3 = 16

3 ≈ 5.333

5 + 1

3 + 1 11

= 181

34 ≈ 5.323

5 + 1

3 + 1

11 +1 2

= 378

71 ≈ 5.324

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Example

720

164 =4 + 64

164 =4 + 1

164 64

=4 + 1 2 + ³⁶₆₄

=4 + 1 2 + 64¹

36

=4 + 1

2 + ¹

1+²⁸₃₆

=4 + 1

2 + ¹

1+₃₆¹

28

=4 + 1

2 + ¹

1+ ¹

1+ 828

=4 + 1

2 + ₁₊ ¹1 1+ 128

8

=4 + 1

2 + ₁₊ ¹1 1+ 1

3+ 48

=4 + 1

2 + ¹

1+ ¹

1+ 1 3+ 18

4

=4 + 1

2 + ¹

1+ ¹

1+ 1 3+ 12

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Example (Cont) Notation:

4 + 1

2 + ¹

1+ ¹

1+ 1 3+ 12

= [4, 2, 1, 1, 3, 2]

Convergents:

[4, ] = 4, [4, 2] = 4 +1 2 =9

2 [4, 2, 1] = 4 + 1

2 +¹₁ = 13

3, [4, 2, 1, 1] = 4 + 1 2 + ¹

1+¹₁

= 22 5 [4, 2, 1, 1, 3] = 4 + 1

2 + ¹

1+ ¹

1+ 13

= 79 18

[4, 2, 1, 1, 3, 2] = 4 + 1

2 + ¹

1+ ¹

1+ 1 3+ 12

=180

41 =180 ∗ 4 41 ∗ 4 = 720

164

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Example (Cont)

Compare with Euclides alg:

720 = 4 ∗ 164 + 64 164 = 2 ∗ 64 + 36

64 = 1 ∗ 36 + 28 36 = 1 ∗ 28 + 8 28 = 3 ∗ 8 + 4

8 = 2 ∗ 4

gcd(720, 164) = 4, ⁷²⁰₁₆₄ = ¹⁸⁰₄₁ = [4, 2, 1, 1, 3, 2].

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Example

Another example:

45 = 2 ∗ 16 + 13 16 = 1 ∗ 13 + 3 13 = 4 ∗ 3 + 1

3 = 3 ∗ 1 + 0

so 45

16= [2, 1, 4, 3] = 2 + 1 1 + ¹

4+¹₃

Geometric picture:

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Example

• ³¹₁₈ = [1, 1, 2, 1, 1, 2] = 1 + 1

1 + 1

2 + 1

1 + 1 1 +1

2

• ¹⁸₃₁ = [0; 1, 1, 2, 1, 1, 2] = 0 + 1

1 + 1

2 + 1

1 + 1 1 +1

2

• [2, 3, 4] = 2 + ₃₊¹1 4

=2 +₃₊¹1 3+ 11

= [2, 3, 3, 1]

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• If c0>0 then 1/[c0, . . . ,cn] = [0, c0, . . . ,cn]; otherwise 1/[0, c₁, . . . ,c_n] = [c₁, . . . ,c_n]

• If m > 1 then

[0, m] = 1

m = 1

m − 1 + ¹₁ = [0, m − 1, 1]

and similarly

[c₀, . . . ,c_n−1,m] = [c₀, . . . ,c_n−1,m − 1, 1]

We prefer the first form

• We also have that [c₀, . . .c_n+1] = [c₀, . . . ,c_n−1,c_n+ _c¹

n+1]

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Theorem

Let a, b be positive integers. Then there is a unique way of writing a/b as a finite continued fraction, such that

a

b = [c0,c1,c2, . . . ,cn] with c_i ∈ Z, c0 ≥ 0, c_j ≥ 1 for j ≥ 1, c_n≥ 2 if n > 0 If a < b then c0=0 and

(c1, . . . ,cn)7→ [0, c1, . . . ,cn]

is a bijection between finite sequences of positive numbers and Q ∩ [0, 1].

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Proof.

WLOG a > b.

Existence: Euclidean algorithm.

Uniqueness: if a/b = c0+ _[c ¹

1,c2,...,cn] =d0+ _[d ¹

1,d2,...,dn], then since

1

[c1,c2,...,cn] <1, it follows that c0 =ba/bc, and similarly d₀ =ba/bc. Thus c₀ =d₀.

Subtract, and consider 1

[c1,c2, . . . ,cn] = 1

[d1,d2, . . . ,dn] =⇒ [c1,c2, . . . ,cn] = [d1,d2, . . . ,dn] Then c₁ =d₁, and so on.

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Example

We tabulate the continued fraction expansion of k/13 for 1 ≤ k ≤ 12:

k CF

1 [0; 13]

2 [0; 6, 2]

3 [0; 4, 3]

4 [0; 3, 4]

5 [0; 2, 1, 1, 2]

6 [0; 2, 6]

7 [0; 1, 1, 6]

8 [0; 1, 1, 1, 1, 2]

9 [0; 1, 2, 4]

10 [0; 1, 3, 3]

11 [0; 1, 5, 2]

12 [0; 1, 12]

Note that the reverse of every CF occurs in the list (possibly written in the non-standrad form).

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We can explain this phenomena using the following result:

Lemma

Suppose that a₀>0, [a₀, . . . ,a_n] =_B^A, [a₀, . . . ,a_n−1] = ^C_D. Then

[a_n,a_n−1, . . . ,a₁,a₀] = A C

Proof.

Needs Euler’s rule.

Example

[2, 3, 7] = 2 + 1 3 +¹₇ =51

22 [2, 3] = 2 +1

3=7 3 [7, 3, 2] = 51

7

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Rational function

• [c0] =c0,

• [c₀,c₁] =c₀+ _c¹

1 = ^c⁰^c_c¹⁺¹

1

• [c0,c1,c2] = [c0,_[c¹

1,c2]] =c0+ _[c¹

1,c2] =c0+ _c ^c²

1c2+1 = ^c⁰^c¹_c^c²^+c⁰^+c²

1c2+1

• [c₀,c₁,c₂,c₃] = [c₀,_[c ¹

1,c2,c3]] =c₀+ _[c ¹

1,c2,c3] =c₀+ _c ^c²^c³⁺¹

1c2c3+c1+c3 =

c0c1c2c3+c0c1+c0c3+c2c3+1 c1c2c3+c1+c3

• In general, [c0, . . .cn+1] = [c0, . . . ,cn−1,cn+ _c¹

n+1] =c0+_[c ¹

1,...,cn+1] = [c0, [c1, . . . ,cn+1]]

• Thus [c0, . . . ,cn] = ^p_qⁿ^(c⁰^,...,cⁿ⁾

n(c0,...,cn) with pn,qn∈ Z[c0, . . . ,cn]

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Theorem (Euler’s rule)

Let Ln(c0, . . . ,cn) be the polynomial which is the sum of

• the product Tn=c0c1· · · c_n,

• all factors of T_n obtained by removing a consecutive pair c_ic_{i +1},

• all factors of T_n obtained by removing two disjoint consecutive pairs cici +1 and c_`c_`+1,

• all factors obtained by removing three disjoint consecutive pairs, and so on

Then

• [c0, . . . ,cn] = _L^Lⁿ^(c⁰^,...,cⁿ⁾

n−1(c1,...,cn)

• L_n(c₀, . . . ,c_n) =c₀L_n−1(c₁, . . . ,c_n) +L_n−2(c₂, . . . ,c_n)

• L_n(c₀, . . . ,c_n) is invariant under reversal of the order of its variables

• L_n(c₀, . . . ,c_n) =c_nL_n−1(c₀, . . . ,c_n−1) +L_n−2(c₀, . . . ,c_n−2)

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Sketch of proof.

Induction on n. By induction hypothesis,

[c₀, . . . ,c_n] = L_n(c₀, . . . ,c_n) L_n−1(c₁, . . . ,c_n) Then

[c₀, . . . ,c_n,c_n+1] =c₀+ 1 [c₁, . . . ,c_n]

=c₀+L_n−1(c₁, . . . ,c_n) L_n(c₀, . . . ,c_n)

=c₀+Ln−1(c₁, . . . ,c_n) Ln(c₀, . . . ,c_n)

= c₀L_n(c₀, . . . ,c_n) +L_n−1(c₁, . . . ,c_n) L_n(c₀, . . . ,c_n)

= L_n+1(c₀, . . . ,c_n,c_n+1) L_n(c₁, . . . ,c_n,c_n+1)

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Example

• L₀(c₀) =c₀

• L₁(c₀,c₁) =c₀c₁+1

• L₂(c0,c1,c2) =c0L₁(c1,c2) +L₀(c0) =c0(c1c2+1) + c0 = c₀c₁c₂+c₀+c₂

• [c₀,c₁,c₂] = ^c⁰^c¹_c^c²^+c⁰^+c²

1c2+1

• [3, 5, 7] = 3 + ¹

5+¹₇ = ^{3∗5∗7+3+7}_5∗7+1

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Convergents I

• [c₀, . . . ,c_n] = _L^Lⁿ^(c⁰^,...,cⁿ⁾

n−1(c1,...,cn) = ^A_Bⁿ

n

• [c₀] =c₀= ^c₁⁰ = ^A_B⁰

0

• [c0,c1] =c0+ _c¹

1 = ^c⁰^c_c¹⁺¹

1 = ^A_B¹

1

• L_n(c0, . . . ,cn) =cnL_n−1(c0, . . . ,cn−1) +L_n−2(c0, . . . ,cn−2)

• A_n =c_nA_n−1+A_n−2

• B_n =c_nB_n−1+B_n−2

• A_n Bn

=cn

A_n−1 Bn−1

+A_n−2 Bn−2

=A_n−1 A_n−2 Bn−1 Bn−2

c_n 1

• A_n An−1

B_n B_n−1

=A_n−1 An−2

B_n−1 B_n−2

c_n 1 1 0

•

cn 1

1 0

= −1

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Convergents II

•

A₀ A₁ B₀ B₁

=

c₀ c₀c₁+1

1 c₁

= −1

• AnBn−1−An−1Bn=

A_n A_n−1 Bn Bn−1

=

A₁ A₀ B1 B0

c₂ 1 1 0

· · ·

c_n 1 1 0

= (−1)ⁿ⁻¹

• Thus, if ci pos integers, gcd(Ai,Bi) =1

• Let α = α_n= [c₀, . . . ,c_n] = ^A_Bⁿ

n. Then for 1 ≤ i ≤ n, α_i − α_{i −1}= ^A_Bⁱ

i − ^A_B^{i −1}

i −1 = ⁽⁻¹⁾_B ^{i −1}

iBi −1

• Furthermore |α − α_i| < |α − α_{i −1}|

• Furthermore α0< α₂<· · · < α < · · · < α₃< α₁

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Linear Diophantine eqns

• Study ax − by = 1, gcd(a, b) = 1

• Suppose _b^a = [c₀, . . . ,c_n]

• Last convergent is ^A_Bⁿ

n = _b^a

• From AnBn−1−An−1Bn= (−1)ⁿ⁻¹ we get aBn−1−An−1b = (−1)ⁿ⁻¹

• If n odd: x = Bn−1, y = An−1

• If n even: [c0, . . . ,cn] = [c0, . . . ,cn−1, 1], do as above

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The rational number ¹³⁹³₉₇₂ ≈ 1.43312757201646 has the following convergents:

i CF conv value

0 [1] 1 1.00000000000000

1 [1,2] 3/2 1.50000000000000

2 [1,2,3] 10/7 1.42857142857143

3 [1,2,3,4] 43/30 1.43333333333333 4 [1,2,3,4,5] 225/157 1.43312101910828 5 [1,2,3,4,5,6] 1393/972 1.43312757201646

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The convergents converge to the exact value as follows:

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Example

We want to solve 474x − 151y = 1. The CF is [3, 7, 5, 4] = 474

151, and the second to last convergent is

[3, 7, 5] =113 36

Since n = 3 is odd, x = 36, y = 113 should be a solution of the linear Diophantine equation. Indeed,

474 ∗ 36 − 151 ∗ 113 = 1

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Example

We want to solve 113x − 36y = 1. The CF is [3, 7, 5] = 113

36 and the penultimate convergent is

[3, 7] = 22 7

Since n = 2 is even, x = 7, y = 22 should be a solution of the linear Diophantine equation with RHS −1. Indeed,

113 ∗ 7 − 36 ∗ 22 = −1 Writing

[3, 7, 5] = [3, 7, 4, 1], [3, 7, 4] = 91 29 gives x = 29, y = 91, which solve the original LDE.

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Partial convergents, repetition

The continued fraction process

Finite continued fractions, repetition

• The following treatment is almost verbatim from Stein, Elementary Number Theory

• Assume a₀,a₁, . . . are positive real numbers (a₀ may be zero)

• For 0 ≤ n ≤ m, the nth convergent of the continued fraction [a₀, . . . ,a_m]is c_n= [a₀, . . . ,a_n]. These convergents for n < m are also called partial convergents.

• [a0,a1, . . . ,an−1,an] = h

a0,a1, . . . ,an−2,an−1+ _a¹

n

i

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Theorem

For each n with −2 ≤ n ≤ m, define real numbers p_n and q_n as follows:

p−2 =0, p−1 =1, p0 =a0

q₋₂ =1, q₋₁ =0, q₀ =1 and for n ≥ 1,

p_n=a_np_n−1+p_n−2 q_n=a_nq_n−1+q_n−2 Then, for n ≥ 0 with n ≤ m we have

[a₀, . . . ,a_n] = p_n

q_n = a_np_n−1+p_n−2 a_nq_n−1+q_n−2 (the last equality for n ≥ 1)

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Proof.

We use induction. The assertion is obvious when n = 0, 1. Suppose the proposition is true for all continued fractions of length n − 1. Then

[a₀, . . . ,a_n] = [a₀, . . . ,a_n−2,a_n−1+ 1 a_n]

=

a_n−1+_a¹

n

p_n−2+p_n−3

a_n−1+_a¹

n

q_n−2+q_n−3

= (a_n−1a_n+1)p_n−2+a_np_n−3

(a_n−1a_n+1)q_n−2+a_nq_n−3 = a_n−1a_np_n−2p_n−2+a_np_n−3 a_n−1a_nq_n−2+q_n−2+a_nq_n−3

= a_n(a_n−1p_n−2+p_n−3) +p_n−2 a_n(a_n−1q_n−2+q_n−3) +q_n−2

= a_np_n−1+p_n−2 a_nq_n−1+q_n−2

= p_n q_n.

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Theorem

For n ≥ 0 with n ≤ m we have

p_nq_n−1−q_np_n−1 = (−1)ⁿ⁻¹ (1) and

pnqn−2−qnpn−2 = (−1)ⁿan. (2) Equivalently,

pn

q_n − pn−1

q_n−1 = (−1)ⁿ⁻¹· 1 q_nq_n−1

and pn

qn

− pn−2

qn−2

= (−1)ⁿ· an

qnqn−2

.

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Proof.

The case for n = 0 is obvious from the definitions. Now suppose n > 0 and the statement is true for n − 1. Then

p_nq_n−1−q_np_n−1= (a_np_n−1+p_n−2)q_n−1− (a_nq_n−1+q_n−2)p_n−1

=pn−2qn−1−qn−2pn−1

= −(pn−1qn−2−pn−2qn−1)

= −(−1)ⁿ⁻²= (−1)ⁿ⁻¹. This completes the proof of (1). For (2), we have

p_nq_n−2−p_n−2q_n= (a_np_n−1+p_n−2)q_n−2−p_n−2(a_nq_n−1+q_n−2)

=a_n(p_n−1q_n−2−p_n−2q_n−1)

= (−1)ⁿa_n.

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Theorem (How Convergents Converge)

The even indexed convergents c2n increase strictly with n, and the odd indexed convergents c_2n+1decrease strictly with n. Also, the odd indexed convergents c_2n+1 are greater than all of the even indexed convergents c2m.

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Proof.

The a_n are positive for n ≥ 1, so the q_n are positive. By the previous theorem, for n ≥ 2,

c_n−c_n−2= (−1)ⁿ· a_n q_nq_n−2, which proves the first claim.

Suppose for the sake of contradiction that there exist integers r and m such that c_2m+1<c_2r. The previous theorem implies that for n ≥ 1,

cn−cn−1 = (−1)ⁿ⁻¹· 1 qnqn−1

has sign (−1)ⁿ⁻¹, so for all s ≥ 0 we have c_2s+1>c_2s. Thus it is impossible that r = m. If r < m, then by what we proved in the first paragraph, c_2m+1<c_2r <c_2m, a contradiction (with s = m). If r > m, then c_{2r +1}<c_2m+1<c_2r, which is also a contradiction (with s = r ).

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The continued fraction process I

• Let x ∈ R and write

x = a0+t0

with a₀ ∈ Z and 0 ≤ t0 <1. We call the number a₀ the floor of x , and we also sometimes write a0=bxc.

• If t0 6= 0, write

1

t₀ =a₁+t₁ with a₁ ∈ Z, a1 >0, and 0 ≤ t₁ <1.

• Thus t₀ = _a ¹

1+t1 = [0, a₁+t₁], which is a continued fraction expansion of t0, which need not be simple.

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The continued fraction process II

• Continue in this manner so long as t_n6= 0 writing 1

t_n =an+1+tn+1

with an+1 ∈ Z, an+1>0, and 0 ≤ tn+1<1.

• We call this procedure, which associates to a real number x the sequence of integers a0,a1,a2, . . ., the continued fraction process.

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Example Let x = ¹⁺

√ 5 2 .Then

x = 1 +−1 +√ 5

2 ,

so a₀=1 and t₀ = ⁻¹⁺

√ 5

2 . We have 1

t0

= 2

−1 +√

5 = −2 − 2√ 5

−4 = 1 +√

5

2 ,

so a₁=1 and t₁ = ⁻¹⁺

√ 5

2 . Likewise, a_n=1 for all n. As we will see below, the following exciting equality makes sense.

1 +√ 5

2 =1 + 1

1 + ₁₊ ¹1

1+ 1

1+ 1 1+···

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Lemma

For every n such that a_n is defined, we have x = [a0,a1, . . . ,an+tn], and if tn6= 0, then x = [a0,a1, . . . ,an,_t¹

n].

Proof.

We use induction. The statements are both true when n = 0. If the second statement is true for n − 1, then

x =

a0,a1, . . . ,an−1, 1 tn−1

= [a₀,a₁, . . . ,a_n−1,a_n+t_n]

=

a0,a1, . . . ,an−1,an, 1 tn

.

Similarly, the first statement is true for n if it is true for n − 1.

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Theorem

Let a0,a1, . . .be a sequence of integers such that an>0 for all n ≥ 1, and for each n ≥ 0, set c_n= [a₀,a₁, . . .a_n]. Then lim_n_→∞c_n exists.

Proof.

For any m ≥ n, the number c_n is a partial convergent of [a₀, . . . ,a_m]. We know that the even convergents c2n form a strictly increasing sequence and that the odd convergents c_2n+1 form a strictly decreasing sequence.

Moreover, the even convergents are all ≤ c₁ and the odd convergents are all ≥ c0. Hence α0=limn→∞c2n and α1 =limn→∞c2n+1 both exist, and α₀≤ α₁. Finally,

|c2n−c_2n−1| = 1

q2n· q_2n−1 ≤ 1

2n(2n − 1) → 0, so α0= α₁.

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Theorem

Let x ∈ R be a real number. Then x is the value of the (possibly infinite) simple continued fraction [a0,a1,a2, . . .]produced by the continued fraction procedure.

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Proof

• If the sequence is finite, then some tn=0 and the result follows

• Suppose the sequence is infinite.

• Then

x = [a₀,a₁, . . . ,a_n, 1 tn

].

• By a previous result (which we apply in a case when the partial quotients of the continued fraction are not integers), we have

x =

1

tn · p_n+pn−1 1

tn · q_n+qn−1

.

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Proof (contd)

Thus, if c_n= [a₀,a₁, . . . ,a_n], then x − c_n=x −p_n

qn

=

1

tnp_nq_n+p_n−1q_n− _t¹

np_nq_n−p_nq_n−1 qn

1

tnqn+qn−1

.

= pn−1qn−pnqn−1

qn

1

tnqn+qn−1

= (−1)ⁿ

qn

1

tnqn+qn−1

.

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Proof (contd) Thus

|x − cn| = 1 qn

1

tnqn+qn−1

< 1

q_n(a_n+1q_n+q_n−1)

= 1

q_n· q_n+1 ≤ 1

n(n + 1) → 0.

In the inequality, we use that an+1 is the integer part of _t¹

n, and is hence

≤ _t¹

n.