Download Irrational Numbers with Generalized Bounded Partial Quotients

Thai Journal of Science and Technology  ปี ที่ 2  ฉบับที่ 3  กันยายน - ธันวาคม 2556
Irrational Numbers with
Generalized Bounded Partial Quotients
Ratchanai Kaikeaw* and Charinthip Hengkrawit
Department of Mathematics and Statistics, Faculty of Science and Technology,
Thammasat University, Rangsit Centre, Khlong Nueng, Khlong Luang, Pathum Thani 12120
Abstract
The definition of irrational numbers with generalized bounded partial quotients is defined. We
will investigate some relation between irrational numbers with bounded partial quotients in simple
continued fraction and in generalized continued fraction. Furthermore, using three distance theorem
of Slater, we also find the result of three distance theorem for generalized continued fraction in some
kind of sequence.
Keywords: continued fraction; bounded partial quotients; three distance theorem.
1. Introduction
A continued fraction is an expression of
1
the form  
, where  n and
0
1 
2
2 
continued fractions expansion of any real
number is finite if and only if that number is
rational.
1
For examples, 19  2 
,
8
2
 n may be any kind of numbers, variables, or
functions. The number of term can be either
finite or infinite. If 0  ,  n  and n  1
for all n  1 , then the expression is called a
simple continued fraction; otherwise, it is called
a generalized continued fraction. For simplicity,

1
we denote the term
by   n .
1 
2
2 
n1  n
It is also known that all irrational
numbers can be uniquely expressed as a
simple continued fraction. Moreover, simple
*Corresponding author: [email protected]
13
1

,
95 7  1
1
3
4
and   3 
7
2  1
1
15 
1
1
1
1
1
1
2
2
1
1
2
,
1
2
.
Thus, we can determine that a fixed
irrational number  has infinite simple
continued fraction expansion:
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Thai Journal of Science and Technology
1
1
,
 a0 
1
n1 an
a1 
a2 
and an  for all n  1 . We start


only if the sequence  d ( N )  is bounded
 d ' ( N )  N 1
  a0  
by using a theorem proved by Slater (1967) as
follows:
Theorem 1.4 (three distance theorem)
(I) For all N  , N can be uniquely
represented as N  rqk  qk 1  s , for
uniquely integer k  0 , 1  r  ak 1 ,
and 0  s  qk .
(II) For the partition P ( N ) , represent N
as N  rqk  qk 1  s in part (I), then
there are (r 1)qk  qk 1  s 1 subintervals
of length k , s  1 subintervals of
length k 1  rk , and qk  ( s  1)
subintervals of length k 1  (r 1)k . 
Because  can be expressed as an infinite
generalized continued fraction:
where a0 
with introducing some definition about simple
continued fraction.
Definition 1.1 We say that  has
bounded partial quotients if and only if

sup an   .
n1
Definition 1.2 Define integers pn and
q n by: p1  1 , p0  a0 , pn  an pn1  pn2 for
n  1 , q1  0 , q0  1 , qn  anqn1  qn2 for n  1
and define n  qn  pn for all n  1 .

Remark: For all n  1 , pn is called nth
qn
n
convergent to  and satisfy pn  a0   1 ;
i 1 ai
qn
p
in fact, lim n   .
n q n
n
1
0 

n1  n
1  2
2 

 0  
Definition 1.3 For any real number x , let
 x  denote the largest integer less than or
equal to x and let { x} denote the fractional part
of x , i.e., { x}  x   x  .
For N  , consider the sequence of
distinct points {1 },{2 },...,{ N } in [0,1] , arranged
in increasing order: 0 {k1 } {k2 }  {k N }  1,
where k j {1,2,..., N } for j  1,2,..., N . Then, interval
[0,1] is partitioned into N  1 subintervals. A set
of these subintervals are a partition of [0,1] . Let
P ( N ) denote an aforementioned partition.
The maximum and minimum lengths of the
subintervals of P ( N ) are denoted, respectively,
by d ( N ) and d ' ( N ) .

In 1998, Mukherjee and Karner proved
that  has bounded partial quotients if and
,
the definition of irrational numbers with
bounded partial quotients may be established.
In this paper, we attend in case that  0 is an
integer,  n and  n are positive integer for all
n  1 . However, for a given sequence  n n1
of positive integer,  may have more than one
generalized continued fraction expansion. For
example, let   2 , n  3 for all n  1 , we
get
2 1
232
3
6
2
3
, 2 1
3
3
6
2
3
7
12 
3
8
,
3
3
12 
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2 1
3
5
1
3
8
T hai Journal of Science and Technology
, etc.

 
 n  n1  Pn1   n Pn2

n1 
 n1 
P
0   j  
 n1 .
j 1  j
Qn1

 n1 
n 
 Qn1   nQn2
 n1 

We conclude that Pn   0  n  j and
j 1  j
Qn
3
3
3
Hence, we must obviate this ambiguousness
for aforementioned purpose.
Qn Pn1  Qn1 Pn  ( 1)n 12 n for all n  1 . 
2. Numbers with generalized bounded
partial quotients
Lemma 2.2 For a given sequence
 Bn n1 of positive integer,  can be uniquely
 B
B1
written as   A  
n
A 
,
In this section, we provide the definition
of numbers with bounded partial quotients in
generalized continued fraction form and
investigate some basic property related with
numbers of simple continued fraction form.
First of all, we start with proving two lemmas.
Lemma 2.1 For any sequences  n n0 ,
 n n1 of complex number, where n , n  0
for all n  1 . If we define complex numbers Pn
and Qn by: P1  1 , P0  0 , Pn n Pn1  n Pn2
for n  1 , Q1  0 , Q0  1 , Qn nQn1  nQn2
0
where A0  , An 
 ( 1)n 12
Proof.

 n n 0
  A0 

n n1

j n
Aj
0    ;
A0    ,
Let
B1
A1 
B2
A2 
,
for all n 1 .
Bn
An  n
We can define integers Pn and Qn as in
Lemma 2.1, then we have
n B
P A P B P
A0   j  n  n n1 n n2 . By replacing
of complex number. We see that P1   0  1
1
and Q1 P0  Q0 P1  1 . By mathematical induction,
AnQn1  BnQn2
An by An n , we obtain   ( An n ) Pn1  Bn Pn2
( An n )Qn1  BnQn2
P  P
 n n n1 , for all n  1 . By Lemma 2.1, we
Qn  nQn1
j 1 A j
assume that Pn   0  n  j and Qn Pn1  Qn1Pn 
Qn
and 0   B j  1 for all
induction, we have
Proof. For any sequences   ,  
Q1
B2
A2 
 B 
 B 
An   n  ,  n   n  for all n  1 . We see
 n1 
 n1 
that A0 is an integer, An are positive integer,
0 n1  1 and Bn  An for all n 1 . By mathematical
j 1  j
n for all n  1 .
A1 
n 1 .
for n  1 , then Pn   0  n  j and Qn Pn1  Qn1Pn
Qn
0
n1 An
j 1  j
( 1)n 12 n , we obtain that Qn1 Pn  Qn Pn1 
n1 (Qn Pn1  Qn1 Pn )  ( 1)n1 12 n1
Since  n Pn1  n Pn2  Pn   0  n  j ,
j 1  j
 nQn1   nQn2 Qn
replace the term  n by  n   n1 we obtain
 n1
Qn
get   Pn  Pn n Pn1  Pn  n  Qn Pn1  Qn1 Pn 
Qn
233
Qn nQn1 Qn
Qn (Qn nQn1 )
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
B1 B2 Bn n . By mathematical
Qn (Qn  nQn1 )

a form in the Lemma. So we have   a0   1
Bn , where a  , A  , a  ,
0
0
n
n1 An
 B
An  and 0   j  1 for all n  1 . Then
j n A j
induction,
we have 0  B1 B2 Bnn  1 for all n 1 . Since
 A0  
Qn
1
lim
 0 and by the squeeze theorem,
n Qn  nQn1
there is the definition of nth convergent to  for
generalized continued fraction which as in
Lemma 2.1.
Definition 2.5 Define integers Pn and
Qn by: P1  1 , P0  A0 , Pn  An Pn1  Bn Pn2 for
n  1 , Q1  0 , Q0  1 , Qn  AnQn1  BnQn2 for
n  1 and define n  Qn  Pn for all n  1 . 
Now we can define numbers with
bounded partial quotients in generalized
continued fraction form and get some
preliminary fact as follows:
Definition 2.3 We say that  has
bounded  Bn n1 ’s partial quotients if and only
if sup An   .

we have lim   Pn  0 . Thus,   lim Pn 
n Qn
Qn
n
n B

lim  A0   j
n
j 1 A j


B1
 ,i.e.,   A0 
B2

A1 
A2 
.
By mathematical induction, we obtain
 B
0   n1   j  1 for all n  1 . Hence,  can
j n A j
be written as   A0   Bn , where 0   B j  1


n1 An
j n A j
n  1 . For uniqueness, let
Bn , where C  , C  and
  C0  
n
0
n1 C n
 B
0   j  1 for all n  1 . Since C 0 is an integer
j n C j

and 0   Bn  1 , we have C0     A0 and
n1 C n
 B
 n  0 . By mathematical induction, assume
n1 C n


that Cn  An and  B j n . Then Cn1   B j  Bn1 .
j n1 C j
j n 2 C j  n

Since C n1 is an integer and 0   B j  1 ,
j n 2 C j
for
all

n1
Theorem 2.4 For all n  1 , Bn  An and if
the sequence  Bn n1 is not bounded, then 
has not bounded  Bn n1 ’s partial quotients.
Proof. Let n  1 . Since  B j  1 and

j n A j
 B
Bj ,
 1 Bn  An   j  An  1 . Since An
j n1 A j
j n1 A j


and Bn are integer, Bn  An . Hence, if the
sequence  Bn n1 is not bounded, then  An n1
is not bounded i.e.  has not bounded  Bn n1
’s partial quotients.

If  has bounded partial quotients. We
want to find necessary conditions and sufficient
conditions for sequence  Bn n1 of positive
we obtain C n1   Bn1   An1 and  B j 

 n 
n1 an

j n 2 C j
 Bn1 

  0 . Hence, Cn  An for all n  0 . 

 n 
By Lemma 2.2, let  Bn n1 be a fixed
sequence of positive integer, we can write  as
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integer such that  also has bounded  Bn n1
’s partial quotients. Theorem 2.4 give we a
necessary condition. In this paper, we provide
a basic example of sufficient condition for
 Bn n1 such that  has bounded  Bn n1 ’s
partial quotients.
From now on we attend in case that the
sequence  Bn n1 satisfying B2 n1  B2 n  a2 n for
all n  1 . Before reach to the next lemma, for
Theorem 2.7.  has bounded partial
quotients if and only if  has bounded  Bn n1
’s partial quotients.
Proof. First, we will show that if 
has bounded partial quotients, then  has
bounded  Bn n1 ’s partial quotients. Assume
that  has bounded partial quotients. Then
there exists a such that an  a for all n  1 .
By Lemma 2.6, we get An  Bnan  a 2 for all
n  1 , i.e.,  has bounded  Bn n1 ’s partial
quotients. Conversely, assume that  has
bounded  Bn n1 ’s partial quotients. Then there
exists A such that An  A , for all n  1 . By
Lemma 2.6, an  An  A for all n  1 i.e.  has
bounded partial quotients.

Lemma 2.8 For all n  1 ,
n
0 ; n is even
simplicity, we define f ( n)  1  ( 1)  
2
1 ; n is odd
for all integer n .
Lemma 2.6 For all n  0 ,
 an ; n is even
An  
  f ( n)( Bn  1)  1  an .
B
a
;
n
is
odd
 nn
Proof. For n  0 , we have a0     A0
.
Furthermore,
we
see
that
 B0 B2
Qn  
 B0 B2
 B0 B2
Pn  
 B0 B2
 B0 B2
n  
 B0 B2
1 ; n is even
for all n 1 .
f ( n)( Bn 1)  1  
B
;
n
is
odd
n

 1
Consider   a0   , for all n  1 , we multiply
n1 an
 1
the term 
by B2 n1  1 . Then we
i 2 n1 a i
B2 n
Bnqn ; n is even
Bn1qn ; n is odd
Bn pn ; n is even
 B0 B2
Bn  f ( n ) q n ,
 B0 B2 Bn f ( n ) pn ,
Bn1 pn ; n is odd
Bnn ; n is even
 B0 B2 Bn f ( n )n ,
Bn1n ; n is odd
where B0  1 .
 1

Proof. First, we have Q1  B0 q1 and
Bn
  a0    A0  
.
n1 an
n1  f ( n )( Bn  1)  1  an
Q0  B0 q0 . By mathematical induction, assume
B B B q ; n is even
Since B2 n1  B2 n  a2 n for all n  1 ,
that Qn   0 2 n n
 B0 B2 Bn f ( n ) qn

Bj
B0 B2 Bn1qn ; n is odd

for
all
.
But
n

1
0 
1
j n  f ( j )( B j  1)  1  a j
B B B q ; n-1 is even
and Qn1   0 2 n1 n1
 B0 B2 Bn1 f (n1)qn1 .
 B
B
B
B
q
;
n-1
is
odd
0
2
n
n

1
 is uniquely written as   A0   n , where

n1 An
 If n is even, then Qn1  An1Qn  Bn1Qn1 . By
 B
Lemma 2.6, we have Qn1  Bn1an1Qn  Bn1Qn1
0   j  1 for all n  1 . Hence, An 
obtain
j n
Aj
 f (n)( Bn 1)  1 an for all
n0.
 Bn1an1 ( B0 B2

 B0 B2
Bnqn )  Bn1 ( B0 B2
Bnqn1 )
Bn2 ( an1qn  qn1 ) ; since n 1 is
odd, Bn1  Bn2 ,  B0 B2 Bn2 qn1
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 If n is odd, then Qn1  An1Qn  Bn1Qn1 . By
(II) For the partition P ( N ) , represent N as
Qk
Qk 1
N r

 s in
Lemma 2.6, we have Qn1  an1Qn  Bn1Qn1
B0 B2
Bk 1 f ( k 1)
Qk
part (I), then there are ( r  1)
B0 B2 Bk  f ( k )
Qk 1
subintervals

 s 1
B0 B2 Bk 1 f ( k 1)
 an1 ( B0 B2 Bn1qn )  Bn1 ( B0 B2 Bn1qn1 )
 B0 B2 Bn1 ( an1qn  qn1 )  B0 B2 Bn1qn1
B B B q ; n+1 is even
Hence, Qn1   0 2 n1 n1
 B0 B2 Bn2 qn1 ; n+1 is odd
 B0 B2 B4 Bn1 f ( n1) qn1 . We conclude
of
that Qn  B0 B2 B4 Bn f ( n)qn for all n  1 .
Similarly, we obtain
and n 
Pn  B0 B2 B4 Bn f ( n ) pn

B0 B2 B4 Bn f ( n )n for all n  1 .
Finally, we get the result of three
distance theorem for generalized continued
fraction, which is our main result, for the
sequence  Bn n1 which satisfied B2 n1 
B2 n  a2 n for all n  1 .
Theorem 2.9 (three distance theorem for
generalized continued fraction).
(I) For all N  , N can be uniquely
Qk
represented as N  r

B0 B2
Bk  f ( k ) B0 B2
k
length
B0 B2
subintervals of length
r
k
B0 B2
Bk  f ( k )
, and
subintervals of length
( r  1)
k
B0 B2
Bk  f ( k )
,
s 1
k 1

Bk  f ( k )
B0 B2
Bk 1 f ( k 1)
Qk
 ( s  1)
B0 B2 Bk  f ( k )
k 1
B0 B2
Bk 1 f ( k 1)

.
Proof. The proof of this theorem follows
straight from Theorem 1.4 and Lemma 2.8. 
3. References
Mukherjee, M. and Karner, G., 1998, Irrational
numbers of constant type – A new
characterization, J. Math. 4: 31-34.
Slater, N.B., 1967, Gaps and steps for the
sequence nθ mod 1, Proc. Camb. Phil.
Soc., 63: 1115-1123.
Bk  f ( k )
Qk 1
 s , for uniquely integer
B0 B2 Bk 1 f ( k 1)
Ak 1
, and
k 0 , 1 r 
f ( k  1)( Bk 1  1)  1
Qk
.
0s
B0 B2 Bk  f ( k )
236