on the equivalence relations of α-continued fractions

$: On the equivalence relations of α-continued fractions$
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Indagationes Mathematicae ( ) –www.elsevier.com/locate/indag

On the equivalence relations of α-continued fractions

Hitoshi Nakadaa,∗, Rie Natsuib

a Department of Mathematics, Keio University, 3-14-1 Hiyoshi, Kohoku-ku, Yokohama, 223-8522, Japanb Department of Mathematics, Japan Women’s University, 2-8-1 Mejirodai, Bunkyo-ku, Tokyo, 112-8681, Japan

Abstract

We compare the equivalence relations on real numbers arising from having eventually agreeing α-continued fraction expansion (for each 0 ≤ α ≤ 1) with those from sharing GL(2, Z)- or SL(2, Z)-orbits.We show that the α-relation and the GL(2, Z)-relation of x are identical for any α > 0 when x is ofunbounded type. On the other hand, they are not identical for x of bounded type.c⃝ 2014 Royal Dutch Mathematical Society (KWG). Published by Elsevier B.V. All rights reserved.

Keywords: Continued fractions; Diophantine approximations

1. Introduction

For a real number x , we put [x] the maximum integer which is not larger than x . Then wedefine

[x]α = [x + 1 − α]

for 0 ≤ α ≤ 1. If α = 1, then [x]1 is the same as [x] and if α = 0, then [x]0 is the minimuminteger which is not smaller than x .

For α ∈ [0, 1], we define a map Tα of [α − 1, α) onto itself by

Tα(x) =

1|x |

−

1|x |

α

if x = 0

0 if x = 0.

∗ Corresponding author.E-mail addresses: [email protected] (H. Nakada), [email protected] (R. Natsui).

http://dx.doi.org/10.1016/j.indag.2014.02.0060019-3577/ c⃝ 2014 Royal Dutch Mathematical Society (KWG). Published by Elsevier B.V. All rights reserved.

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http://dx.doi.org/10.1016/j.indag.2014.02.006

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http://dx.doi.org/10.1016/j.indag.2014.02.006

2 H. Nakada, R. Natsui / Indagationes Mathematicae ( ) –

For x ∈ [α − 1, α), we put

εn = sgn T n−1α (x), cn =

1T n−1α (x)

α

(1)

for n ≥ 1. Here we put cn = ∞ and 1cn

= 0 if T n−1α (x) = 0. Although εn and cn depend on

the choice of x and α we do not write these εn(α, x) nor cn(α, x) unless we compare these fordifferent x . Then we have the continued fraction expansion of x , which we call the α-continuedfraction expansion of x , given by

ε1

c1+

ε2

c2+ · · · +

εn

cn+ · · · .

In the case of α = 1, T1 is the Gauss map (the simple continued fraction map) which gives thesimple continued fraction expansion of x ∈ [0, 1). We simply write T in the case of α = 1.

The definition of α-continued fraction for 12 ≤ α ≤ 1 is given in [10] with its realization of the

natural extension of Tα and the explicit value of the entropy of Tα with respect to the Lebesguemeasure. Moreover, [5] showed that the α-continued fraction is given by the simple continuedfractions by the induced transformation of the natural extension of T1 when 1

2 ≤ α < 1. It wasalso shown that the notion of normal numbers is independent of each α in this area of α, see [6].Later, it turned out that this can be generalized to

√2 − 1 ≤ α < 1

2 , see [9]. The behavior of theentropy of Tα as α → 0 was discussed in [8,11,1,2,7,3]. In this paper we discuss the behavior ofthese continued fraction expansions from a different point of view.

We define three equivalence relations for an irrational x ∈ (0, 1). For fixed α ∈ [0, 1), let

z = zα =

z if z ∈ [0, α);

z − 1 otherwise.

Two real numbers x, y ∈ [0, 1) are said to be α-equivalent (denoted by x ∼α y) if there existpositive integers n and m such that T n

α (x) = T mα (y). Two real numbers x, y ∈ [0, 1) are said

to be GL(2, Z)-equivalent or SL(2, Z)-equivalent if there exists A =

a bc d

∈ GL(2, Z) or

∈ SL(2, Z) such that ax+bcx+d = y, respectively. We denote them by x ∼G y and x ∼S y, respec-

tively. We put the following

Rα(x) = {z ∈ (0, 1) : z ∼α x},

RGL(2,Z)(x) = {z ∈ (0, 1) : z ∼G x},

and

RSL(2,Z)(x) = {z ∈ (0, 1) : z ∼S x}.

In the sequel we discuss these relations as α varies. The classical theory of continued fractionssays that

R1(x) = RGL(2,Z)(x)

and

R0(x) = RSL(2,Z)(x).

H. Nakada, R. Natsui / Indagationes Mathematicae ( ) – 3

Remark. We refer to [4] for the proof of the former case. Then it is easy to prove the latter casein a similar fashion. Since there seems to be no suitable reference for this fact, we give its proofat the end of this paper.

The motivation of this paper is the following. For α ∈ (0, 1), we can consider the α-continued

fraction expansion of α since we can start with ε1 = 1, c1 =

1α

, and Tα(α) =

1α

− c1. In [10],

it was shown that Tα(α) = T 2α (α − 1) for

√5−12 < α < 1 and T 2

α (α) = T 2α (α − 1) for 1

2 < α <√

5−12 . Then in [9], it was shown that the latter also holds for

√2−1 < α < 1

2 . In general, a num-ber of values of α which have the property T n

α (α) = T mα (α −1) holds for some n ≥ 1 and m ≥ 1

were given in [11]. The set of α which have this property was characterized by [2,7] recently.In particular they showed that the set of such α is open dense in the unit interval [0, 1] and hasLebesgue measure 1, which gave an affirmative answer to a question posed by [11]. On the otherhand, [11] also had shown that {T n

α (α) : n ≥ 0}∩{T nα (α−1) : n ≥ 0} = ∅ for some α of bounded

type, here we put Tα(α) =1α

−

1α

α

. This result implicitly shows that Rα(α) = RGL(2,Z)(α).

This paper gives further description of this property from the equivalence relations point of view.The main result is the following three theorems. Key to the proof of the main result is the exis-tence of the Legendre constant of the α-continued fractions (and its proof), see [12]. Indeed, in theunbounded case, this allows us to pass sufficiently far in a regular continued fraction expansion soas to find a corresponding approximant that also realized as an α-continued fraction approximant.

To state the theorems explicitly, we need the following notions. An irrational number x ∈

(0, 1) is said to be of unbounded type if coefficients of its simple continued fraction expansionare unbounded, that is,

supn≥1

an = ∞

with

x =1

a1+

1

a2+ · · · +

1

an+ · · · .

On the other hand, x ∈ (0, 1) is said to be of bounded type if supn≥1 an < ∞.A quadratic irrational number has period k if the (n + k)-th coefficients coincide with the n-th

coefficients for sufficiently large n and k is the minimum positive integer having this property.It is well-known that any quadratic irrational number has a periodic simple continued fractionexpansion. This means that any quadratic irrational number is of bounded type.

Theorem 1. Let x be a non-quadratic irrational number of bounded type in (0, 1). Then thereexists α0 in (0, 1) such that the α-relation of x does not coincide with RGL(2,Z)(x) and does notcontain RSL(2,Z)(x), that is,

Rα(x) = RGL(2,Z)(x) and Rα(x) ⊃ RSL(2,Z)(x)

for any α < α0. On the other hand,0<α0≤1

α<α0

Rα(x) = RSL(2,Z)(x).


Theorem 2. Let x be of unbounded type in (0, 1). Then

Rα(x) = RGL(2,Z)(x)

for any α ∈ (0, 1].

Let x be a quadratic irrational number in (0, 1) whose simple continued fraction expansion hasan even period. Then the assertion of Theorem 1 also holds for x . On the other hand, the oddperiod case, the assertion of Theorem 2 holds. These are a direct consequence of the followingtheorem. A quadratic irrational number x is said to be strictly periodic if there exists a positiveinteger m such that T m(x) = x .

Theorem 3. Suppose that x is a strictly periodic quadratic irrational number. We have thefollowing:

(i) If the period of x is even, then, for any α, 0 < α ≤ min{T n(x) : n ≥ 0} we haveT (x) ∈ Rα(x).

(ii) If the period of x is odd, then, for any α, 0 ≤ α ≤ 1, we have T j (x) ∈ Rα(x) for any j ≥ 1.

In Section 2, we give some basic facts concerning the α-continued fractions and then in Section 3give proofs of the main theorems. We show some examples illustrating the main theorems inSection 4. Finally we give a proof of the fact “R0(x) = RSL(2,Z)(x)” in the Appendix.

2. Some basic facts

We fix α, 0 < α < 1. For x ∈ [α − 1, α), we have the sequenceε1c1

,

ε2c2

, . . . ,

εncn

, . . . ,

with εn, cn as in (1). The n-th iteration T nα (x) is given by a linear fractional transformation

associated to the inverse ofpα,n−1 pα,nqα,n−1 qα,n

= A1(x)A2(x) · · · An(x) (2)

with

Ak(x) =

0 εk1 ck1

for k ≥ 1. (3)

Then we write

x

=

pα,n−1T nα (x) + pα,n

qα,n−1T nα (x) + qα,n

=

pα,n−1 pα,nqα,n−1 qα,n

(T n

α (x)) (4)

and

T nα (x) =


−1

(x).

We call the matrix appeared on the right hand side the associated matrix of T n for x ∈ [α−1, α).

We should note that qα,n is always positive for n ≥ 0. The determinant of


is equal


to (−1)nε1ε2 · · · εn . Then from (4), we see

x −pα,n

qα,n=

pα,n + pα,n−1T nα (x)

qα,n + qα,n−1T nα (x)

−pα,n

qα,n=

(−1)nε1ε2 · · · εn · T nα (x)

qα,nqα,n + qα,n−1T n

α (x) . (5)

From (2), we havepα,n+1 = cn+1 pα,n + εn+1 pα,n−1qα,n+1 = cn+1qα,n + εn+1qα,n−1

(6)

for n ≥ 1. It is easy to see that

cα,n ≥ 2 whenever εn = −1. (7)

Thus we have the following.

Lemma 1. (i) {qα,n : n ≥ 1} is strictly increasing.(ii) pα,n

qα,n− x is positive if and only if (−1)n+1ε1ε2 · · · εnεn+1 = +1.

The rational number pα,nqα,n

is said to be the n-th α-convergent of x ∈ [α − 1, α). If 0 < α ≤ 1,then we see the following

Proposition 1. Suppose x ∈ (α − 1, α). For any n ≥ 1, we have

(i)

1cn+1 + α + 1

·1

q2α,n

<

x −pα,n

qα,n

, (8)

(ii)

x −pα,n

qα,n

<

1

cn+1 − 1 − (1 − α)qα,n−1

qα,n

·1

q2α,n

if cα,n+1 > 1

1

q2α,n

if cα,n+1 = 1,

(9)

(iii) in the case of the simple continued fractions (i.e. α = 1), we have

1an+1 + 2

·1

q2n

<

x −pn

qn

<1

an+1·

1

q2α,n

(10)

where we denote by an the nth partial quotient of the simple continued fraction expansionof x ∈ (0, 1) with pn = p1,n and qn = q1,n .

Proof. From (5), we havex −pα,n

qα,n

=

1

qα,n

qα,n

T nα (x)

+ qα,n−1

=

1

qα,n

εn+1(cn+1 + T n+1

α (x))qα,n + qα,n−1

=1

qα,n

(cn+1 + T n+1

α (x))qα,n + εn+1qα,n−1

. (11)


By the latter form, we get (8) and (9) since {qα,n : n ≥ 1} is an increasing sequence of positiveintegers which follows from (6) and (7). We also get (10) from this form. Here we note thatεn+1 = 1 in this case.

On the other hand, we have the following.

Theorem A ([12]). For any α, 0 < α ≤ 1, there exists a positive constant Cα such that if arational number p

q = 0, q a positive integer, satisfies the followingx −p

q

≤ Cα ·1

q2

then pq =

pα,nqα,n

for some n ≥ 1.

In the case of α = 1, 12 is the best possible value for C1, which is known as the Legendre

constant of simple continued fractions. In this sense, we call the best possible constant for Cα

the α-Legendre constant for 0 < α ≤ 1. In the case of α = 0, (5) is always negative and

p

q− x ≤

1

q2

implies pq =

p0,nq0,n

for some n ≥ 1, here x ∈ [−1, 0).

3. Proof of the main results

First of all, we note that if x, 0 < x < 1, has the following α0-continued fraction expansion

−1

c1+

−1

c2+ · · · +

−1

cn+ · · · ,

that is, εn = −1 for n ≥ 1, then the α-continued fraction expansion of x is the same for any0 ≤ α ≤ α0.

Proof of Theorem 1. We fix a non-quadratic irrational number of bounded type x, 0 < x < 1.Suppose that the simple continued fraction expansion of x is the following

1

a1+

1

a2+ · · · +

1

an+ · · · .

Since x is of bounded type, M = supn≥1 an < ∞, and infn≥0 T n(x) > 1M+1 . We choose

α < 1M+2 . Then we see x = x − 1. We claim T n

α (x) < 0 for any n ≥ 0. Indeed, if T nα (x) > 0

for some n ≥ 1, then εn+1 = 1 and by (11)(x − 1) −pα,n

qα,n

<1

(M + 2) · q2α,n

,

equivalently,x −qα,n + pα,n

qα,n

<1

(M + 2) · q2α,n

.

Since 1M+2 < 1

2 , there exists m ≥ 1 such that pα,nqα,n

=pmqm

. This contradicts (10). Hence thereexists a sequence of positive integers {cn : n ≥ 1} such that the α-continued fraction expansion


of x is

−1

c1+

−1

c2+ · · · +

−1

cn+ · · · .

By the same reasoning for

y =1

a2+

1

a3+ · · ·

we have a sequence of positive integers {c′n : n ≥ 1} such that the α-continued fraction

expansion of y is

−1

c′

1+

−1

c′

2+ · · · +

−1

c′n

+ · · · .

There are no positive integers m and n such that T mα (x) = T n

α (y) since the associated matricesof T k

α , k ≥ 1, for x and y are all of determinant 1 and y ∈ RSL(2,Z)(x). Consequently we haveRα(x) = RGL(2,Z)(x).

Next we choose a sufficiently large positive integer K so that

z =1

K+ y ∈ (α − 1, α)

i.e. z = z. Then Tα(z) = y and the α-continued fraction expansion of z is

+1

K+

−1

c′

1+

−1

c′

2+ · · · +

−1

c′n

+ · · · .

This shows z ∈ RSL(2,Z)(x) and z ∈ Rα(x). Consequently we have Rα(x) ⊃ RSL(2,Z)(x).Finally the last part of the assertion of Theorem 1 follows from the following proposition.

Proposition 2. Let x and y be non-quadratic irrational numbers of bounded type in [0, 1) suchthat x ∼G y. For

α ≤ min{ infn≥0

T n(x), infn≥0

T n(y)},

we see x ∼α y if and only if x ∼S y.

Remark. It is easy to see that if x is of bounded type, then supn≥0 T n(x) < 1. Hence for anyα, supn≥0 T n(x) < α ≤ 1, we have the α-continued fraction expansion of x is the same as itssimple continued fraction expansion. Thus under the assumption of Proposition 2, x ∼α y holdsfor any α,

max{supn≥0

T n(x), supn≥0

T n(y)} < α ≤ 1.

Proof of Proposition 2. For α ≤ infn≥0 T n(x), the α-continued fraction expansion of x is

−1

c1(x)+

−1

c2(x)+ · · · +

−1

cn(x)+ · · · ,


where the cn are independent of α. Indeed if εm > 0 (equivalently T mα (x) > 0) then T m−1

α (x) =

T n(x) for some n ≥ 1, see [12], Lemma 3(A). For y with α ≤ infn≥0 T n(y), the same holds. Thusif x ∼S y, then the classical theory of continued fractions shows that x ∼0 y, which is equivalentto x ∼α y in this case. The other direction is also easy to follow because all iterations of Tα of xand y are the linear fractional transformations defined by elements of SL(2, Z).

Proof of Theorem 2. It is obvious that y ∈ Rα(x) implies y ∈ RGL(2,Z)(x). We will show thaty ∈ Rα(x) whenever y ∈ RGL(2,Z)(x). We consider the simple continued fraction expansions ofx and y:

x =1

a1(x)+

1a2(x)

+1

a3(x)+ · · ·

y =1

a1(y)+

1a2(y)

+1

a3(y)+ · · · .

We denote their n-th convergent by pn(x)qn(x)

and pn(y)qn(y)

, respectively. We also denote the α-continuedfraction expansions of x and y as

x =ε1(x)

c1(x)+

ε2(x)

c2(x)+

ε3(x)

c3(x)+ · · ·

y =ε1(y)

c1(y)+

ε2(y)

c2(y)+

ε3(y)

c3(y)+ · · ·

and their n-th α-convergent by pα,n(x)

qα,n(x)and pα,n(y)

qα,n(y), respectively. Because y ∈ RGL(2,Z)(x), there

exist positive integers n0 and m0 such that an0+ j (x) = am0+ j (y) for any j ≥ 0, equivalently,T n0(x) = T m0(y). If there exist positive integers n1 and m1 such that εn1(x) = εm1(y) = +1,then T n1−1

α (x) and T m1−1α (y) are in (0, α). In this case, it is enough to show that T m1

α (y) ∈

Rα(T n1α (x)). So we can replace x and y by T n1

α (x) and T m1α (y), respectively. If εn(x) = −1 for

any n ≥ 1, then it turns out that

x + 1 = x =

11

+1

c1(x) − 1+

ε2(x)

c2(x)+

ε3(x)

c3(x)+ · · · if c1(x) ≥ 3

1s + 1

+1

cs+1(x) − 1+

εs+2(x)

cs+2(x)+

εs+3(x)

cs+3(x)+ · · ·

if c j (x) = 2, 1 ≤ j ≤ s and cs+1 ≥ 3

(12)

see [12, p. 497 line 7ff]. In this case we can consider x with its expansion (12) instead of theα-continued fraction expansion of x − 1. The same holds for y. Thus we can assume that

ε1(x) = ε1(y) = +1.

It is easy to see that there exists a positive integer L = L(α) such that

As+ j (w) =

0 −11 2

, 1 ≤ j ≤ L

for any w ∈ [α − 1, α) and s ≥ 0 (see (3) for the definition of As+ j ). We choose ℓ1 so that

Aℓ1(x) =

0 −11 2

. Then, by Lemma 3 and its proof of [12], there exists n ≥ 1 such that


A1(x) · · · Aℓ1(x) =

pn−1(x) pn(x)

qn−1(x) qn(x)

if εα,ℓ1+1(x) = +1

pn−1(x) pn(x) + pn−1(x)

qn−1(x) qn(x) + qn−1(x)

if εα,ℓ1+1(x) = −1.

From (4), with α = 1, we see

x =

pn−1(x)w + pn(x)

qn−1(x)x + qn(x)if εℓ1+1(x) = +1

pn−1(x)w + (pn(x) + pn−1)

qn−1(x)w + (qn(x) + qn−1)if εℓ1+1(x) = −1

with w = T ℓ1α (x). This means

w = T n(x) or T n(x) − 1. (13)

We can further restrict our choice of ℓ1 such that n = nℓ1 > n0 and an(x) is sufficiently large sothat an > L + 1 and 1

an(x)is smaller than Cα of Theorem A. Then we put m = m0 + (n − n0).

By the choice of an(x) = am(y),y −pm−1(y)

qm−1(y)

<1

am(y) · q2m−1(y)

(14)

and thus there exists ℓ2 ≥ 1 such that

pα,ℓ2−1(y)

qα,ℓ2−1(y)=

pm−1(y)

qm−1(y)

by Theorem A. Moreover if cα,ℓ2 = 2, then from (8) we havey −pα,ℓ2−1(y)

qα,ℓ2−1(y)

>14

1

q2α,ℓ2−1(y)

.

Since we have chosen an(x) sufficiently large, this contradicts (14). Thus we see Aℓ2(y) =0 −11 2

. For this ℓ2, we get

T ℓ2α (y) = T m(y) or T m(y) − 1 (15)

as we have shown in the above for x . From the fact T m0(y) = T n0(x), (13), and (15), we see

T ℓ2α (y) = T ℓ1

α (x)

which concludes the assertion of this theorem. �

Now we assume that x ∈ (0, 1) is a strictly periodic quadratic irrational number. We putx = min{T n(x) : n ≥ 0}.

Proof of Theorem 3(i). Suppose that the period of x is 2k, k ≥ 2 and let the simple continuedfraction expansion of x be

x =1

a 1

+1

b 1

+1

a 2

+1

b 2

+ · · ·1

a k

+1

b k

+1

a 1

+1

b 1

+ · · · .


If α ≤ x , then we see T jα (x) < 0 for any j ≥ 0. This follows from Lemma 3 [12] again. Indeed,

if T jα (x) > 0, then it follows that T j−1

α (x) = T ℓ(x) for some ℓ ≥ 1. From (12), we also have

x =−1

2+ · · · +

−1

2 a1−1

+−1

b1 + 2+

−1

2+ · · · +

−1

2 a2−1

+−1

b2 + 2+ · · · . (16)

By the same way we have

T (x) =−1

2+ · · · +

−1

2 b1−1

+−1

a1 + 2+

−1

2+ · · · +

−1

2 b2−1

+−1

a2 + 2+ · · · .

Here the right hand side is periodic. Hence, if T (x) ∈ Rα(x), then there exists ℓ ≥ 1 such thatb1 − 1 = aℓ − 1. In this case we have

T x =1

aℓ

+1

bℓ

+ · · · ,

that is,

(b1, a2, b2, a3, . . . , ak, bk, a1) = (aℓ, bℓ, . . . , ak, bk, a1, . . . , aℓ−1, bℓ−1).

This is impossible because of the minimality of the period 2k. This shows the assertion of (i).

Proof of Theorem 3(ii). Now we suppose that the period of x is odd, k = 2k0 − 1. First wediscuss the case 0 < α ≤ x . In this case we consider the doubled period length 2k. We write

x =1

a1+

1

b1+

1

a2+

1

b2+ · · · +

1

ak+

1

bk+

1

a1+

1

b1+ · · ·

and consider the expansion (16). This implies

bℓ =

aℓ+k0 , 1 ≤ ℓ ≤ k0 − 1aℓ−k0+1, k0 ≤ ℓ ≤ k.

For 2ℓ − 1, 1 ≤ ℓ < k, we have the α-continued fraction expansion of T 2ℓ−1(x)

T 2ℓ−1(x) =−1

2+ · · ·

−1

2 bℓ−1

+−1

aℓ+1 + 2+ · · ·

=−1

2+ · · ·

−1

2 aℓ+k0−1

+−1

bℓ+k0 + 2+ · · ·

or =

−1

2+ · · ·

−1

2 aℓ−k0+1−1

+−1

bℓ−k0+1 + 2+ · · ·

.


This shows T 2ℓ−1(x) ∈ Rα(x). On the other hand, for 2ℓ, 1 ≤ ℓ < k, we have easily that

T 2ℓ(x) =−1

2+ · · ·

−1

2 aℓ+1−1

+−1

bℓ+1 + 2+ · · · ∈ Rα(x).

Thus we get the assertion of (ii) in this case. Next we discuss the case x < α. We consider theexpansion of x .

x =1

a1+

1

b1+

1

a2+

1

b2+ · · · +

1

ak+

1

bk+

1

a1+

1

b1+ · · · .

Then we can get the α-continued fraction expansion of x by Lemma 2(i), (ii), and (iii) of [12],that is, we apply one of those when T j (x) > α, 0 < j ≤ 2k − 1. Again, by the same methodwe have T j (x) ∈ Rα(x), 1 ≤ j ≤ k − 1, which is equivalent to

T j (x) ∈ Rα(x), 1 ≤ j ≤ k − 1.

4. Some examples

Example 1 (The Case of Bounded Type). We give an example of x of bounded type, whichexplains that the α-relation does not include the SL(2, Z)-relation. We put

x =1

2+

1

1+

1

2+

1

2+

1

2+

1

1+ · · · ,

where the sequence of digits

2 2 · · · 2 2k+1

1

follows after

2 2 · · · 2 2k−1

1

for k ≥ 1. Then we have

T (x) =1

1+

1

2+

1

2+

1

2+

1

1+

1

2+

1

2+

1

2+

1

2+

1

2+ · · · ,

where the sequence of digits

1 2 2 · · · 2 2k+1

follows after

1 2 2 · · · 2 2k−1

for k ≥ 2. In this case, for α smaller than x , the α-continued fraction expansions of x − 1 andT (x) − 1 are


−1

2+

−1

3+

−1

2+

−1

4+

−1

2+

−1

3+

−1

2+

−1

4+

−1

2+

−1

4

+−1

2+

−1

3+ · · ·

and

−1

4+

−1

2+

−1

4+

−1

4+

−1

2+

−1

4+

−1

2+

−1

4+

−1

4+ · · ·

respectively. These show that 3 appears infinitely often for the expansion of x − 1 and neverappears for T (x) − 1. Thus we have y ∼S x and y �α x for

y =+1

10+

−1

4+

−1

2+

−1

4+

−1

4+

−1

2+

−1

4+

−1

2+

−1

4+

−1

4+ · · ·

and α =17 . The next example will show the above situation more clearly.

Example 2 (Quadratic Irrational Number Even Period Case). We consider x =

√15−33 . Here

we have

x =1

3+

1

2+

1

3+

1

2+ · · ·

and

x − 1 =−1

2+

−1

2+

−1

4+

−1

2+

−1

2+

−1

4+ · · · ,

on the other hand,

T (x) =1

2+

1

3+

1

2+

1

3+ · · ·

and

T (x) − 1 =−1

2+

−1

5+

−1

2+

−1

5+ · · ·

for α ≤ x(= x). This shows Rα(x) ∩ Rα(T (x)) = ∅ for 0 ≤ α ≤ x . Moreover for y =24−

√15

117 ,for example, its α-continued fraction expansion with α =

17 is

1

10+

−1

2+

−1

5+

−1

2+

−1

5+ · · · .

Then it is easy to see that y ∼S x and y �α x .

Example 3 (Quadratic Irrational Number with Odd Period). We consider x =

√37−43 . Here we

have

x =1

1+

1

2+

1

3+

1

1+

1

2+

1

3+ · · ·


and

x − 1 =−1

4+

−1

2+

−1

4+

−1

3+

−1

2+

−1

5+

−1

4+

−1

2+

−1

4+

−1

3

+−1

2+

−1

5+ · · · .

In this case, we see T (x) − 1 = T 4α (x − 1) and T 2(x) − 1 = Tα(x − 1). This shows

Rα(x) = Rα(T (x)) = Rα(T 2(x)) for 0 ≤ α ≤ x = T 2(x) =

√37−54 .

Acknowledgments

The authors would like to thank the referee for careful reading of an early version of this paperand valuable comments. This research was supported in part by the Grant-in-Aid for Scientificresearch (Nos. 24340020 and 23740088), the Japan Society for the Promotion of Science.

Appendix

Every real number x ∈ (0, 1) has the following type of continued fraction expansion.

x =1

c1−

1

c2−

1

c3− · · · , (17)

where c j is a positive integer larger than 1. If x is irrational, then this expansion is unique. It isclear that the α-continued fraction expansion of −x with α = 0 is the following.

−x = −1

c1−

1

c1−

1

c3− · · · .

We refer to [13] for the relation between SL(2, Z)-action and this type of continued fractionsconcerning to the theory of quadratic form. We put

−Pn−1 Pn−Qn−1 Qn

=

0 1

−1 c1

0 1

−1 c2

· · ·

0 1

−1 cn

.

Then we have {Qn, n ≥ 1} is strictly increasing and

−Pn

Qn=

p0,n

q0,n,

where the right hand is the n-th α-convergent of −x with α = 0. We prove the following.

Proposition 3. For any irrational numbers x and y in (0, 1), x ∼S y if and only if there existpositive integers n0 and m0 such that cn0+k(x) = cm0+k(y) for any positive integer k.

Proof. First of all, we note that this proof is the minor variant of the proof of [4, Theorem 175].Let us assume that there exist positive integers n0 and m0 such that cn0+k(x) = cm0+k(y) for anypositive integer k. We put

C =

0 1

−1 c1(x)

0 1

−1 c2(x)

· · ·

0 1

−1 cn0(x)


and

D =

0 1

−1 c1(y)

0 1

−1 c2(y)

· · ·

0 1

−1 cm0(y)

.

Then x is mapped to y by the linear fractional transformation associated to DC−1, whichis of determinant +1. Thus we have x ∼S y. Now we assume that x ∼S y, i.e. there exists

s tu v

∈ SL(2, Z) such that y =

sx+tux+v

. Since −sx−t−ux−v

=sx+tux+v

, we can assume that cx + d > 0.

We consider the expansion (17) of x . Since x is irrational and

1 =1

2−

1

2−

1

2− · · · ,

there exist infinitely many n such that cn+1 ≥ 3. We choose and fix such n. Put

Cn =

0 1

−1 c1

0 1

−1 c2

· · ·

0 1

−1 cn

and

xn =1

cn+1−

1

cn+2−

1

cn+3− · · · .

Then

y =sn xn + tnun xn + vn

with sn tnun vn

=

s tu v

Cn . (18)

It is clear that snvn − tnun = +1. We are going to estimate un and vn . We have

0 < x −Pn

Qn=

Pn − Pn−1 · xn

Pn − Pn−1 · xn−

Pn

Qn

=xn

Qn(Qn − Qn−1xn)

=1

Qn

Qnxn

− Qn−1

=

1Qn (cn+1 Qn − xn+1 Qn − Qn−1)

≤1

Q2n

whenever cn+1 ≥ 3 (note that 0 < xn+1 = cn+1 −1xn

< 1). Thus there exists δn such that0 < δn < 1 and

x −Pn

Qn=

δn

Q2n,


equivalently

Pn = Qn x −δn

Qn. (19)

If cn ≥ 3, then we also have

Pn−1 = Qn−1x −δn−1

Qn−1

for some 0 < δn−1 < 1, on the other hand, if cn = 2, then there exists 0 < δn−1 < 3 such that

Pn−1 = Qn−1x −δn−1

Qn−1. (20)

(20) follows from the fact that cn = 2 and cn+1 ≥ 3 imply xn < 23 . From (18)–(20), we have

un = −(sx + t)Qn−1 −δn−1

Qn−1

vn = (sx + t)Qn +δn

Qn.

Thus for sufficiently large n with cn+1 ≥ 3, we see

vn > 0, un < 0, and − un < vn .

We fix such n and apply a kind of Euclidean algorithm: there exists unique integer b1 ≥ 2 suchthat un < vn + b1un < 0. By this b1, we write

snb1 + tn −snunb1 + vn −un

0 1

−1 b1

=

sn tnun vn

.

Inductively, we gett∗ 1−1 bℓ

· · ·

0 1

−1 b2

0 1

−1 b1

=

sn tnun vn

.

Because all matrices appearing here are of determinant +1, t∗ has to be equal to 0. Consequentlywe have

y =

0 1

−1 bℓ

· · ·

0 1

−1 b2

0 1

−1 b1

(xn)

which is equivalent to

y =1

bℓ

−1

bℓ−1− · · · −

1

b1−

1

cn+1−

1

cn+2− · · · .

This completes the proof of this proposition.

References

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[3] C. Carminati, G. Tiozzo, Turning and plateaux for the entropy of α-continued fractions, Nonlinearity 26 (2013)1049–1070.

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Hochschultext, Springer-Verlag, Berlin, New York, 1981.











on the equivalence relations of α-continued fractions

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