American-Eurasian Journal of Scientific Research 10 (5): 251-263, 2015
ISSN 1818-6785
© IDOSI Publications, 2015
DOI: 10.5829/idosi.aejsr.2015.10.5.22697
The Elementary Arithmetic Operators of Continued Fraction
S. Mugassabi and F. Mistiri
Department of Mathematics, University of Benghazi, Benghazi, Libya
Abstract: The finite and infinite simple continued fractions are considered. The addition, subtraction and
equality of two simple continued fractions are presented. The applications of continued fractions are also
studied.
Key words: Simple continued fraction
Negative rational number
INTRODUCTION
The continued fractions have been studied in
mathematical (Diophantine and Pell's equations) and
physical (gear ratio) [1-4]. Our simple continued fractions
that we study are reminiscent of various other continued
fractions problems. However they are very different from
ours. One is analytic of continued fractions [3-8], which
has been discussed for the real and complex values.
Another study is the continued fractions for the integer
values [2, 4, 6].
Currently, continued fractions have many practical
uses in mathematics. Forinstance, we can express any
number, rational or irrational, as a finite or
infinitecontinued fraction expression. We can also solve
any Diophantine congruence thatis any equivalence of
the form ax b (mod m). In other words, in most realworld applications of mathematics, continued fractions
are rarely themost practical way to solve a given set of
problems as decimal approximations are much more
useful (and computers can work with decimals at a much
faster rate). However, some interesting observations can
still be made using continued fractions. Namely, in this
paper, we will be exploring how continued fractions can
be used to add and subtract the numbers a  b and
a
 c . We start with some definitions and theorems that
b
we
used to defined the addition and the subtraction of
two simple continued fractions.
Definition 1: An expression of the form.
b0
a0 +
b1
a1 +
b2
a2 +
a3 + 
Multiplicative inverse
is called a continued fraction. The values a0, a1, a2 and
b0,b1b2,... can be either real or complex and their numbers
can be either finite or infinite.
Definition 2: A finite simple continued fraction is a
continued fraction (definition 1) in which bn= 1 for all n,
that is:
a0 +
1
a1 +
1
a2 + 
+
1
an
wherean is positive integer for all n 1, a0 can be any
integer number. The above fraction is sometime
represented by [a1;a1,a2,...,an] for finite simple continued
fraction and [a00;a1,a2] for infinite simple continued
fraction. In this paper we will use the symbol (S.C.F.) for
the simple continued fraction.
1+
is a finite complex valued continued
1
2
2+
3i
fraction.
3+
4+
1
6+
7+
1
2+
3+
1
is finite S.C.F..
[3,6,7,9,17]
=
1
9+
1
1
17
is an infinite S.C.F..
[4;2,3,3,]
=
1
3 +
Corresponding Author: S. Mugassabi, Department of Mathematics, University of Benghazi, Benghazi, Libya.
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Am-Euras. J. Sci. Res., 10 (5): 251-263, 2015
Theorem 1: A number is rational if and only if it can be expressed as a finite S.C.F. [4].
For example
62
16
1
1
1
1
1
=2 +
=2 +
=2 +
=2 +
=2 +
=2 +
=[2;1,2,3,2].
23
7
1
1
1
23
23
1+
1+
1+
1+
16
2
1
16
16
2+
2+
1
7
7
3+
2
Remark 1:
To expand a rational number b , (a > b > 0) into S.C.F. we write
a
b
1
= 0+ ,
a
a
b
and then we use the same techniques as in theorem 1 for a .
b
b
To expand a negative rational number − (a, b, > 0) into S.C.F. we take the greatest integer number
a
first term of S.C.F.
 b
 −  for the
 a
 
that is,
 b
 −  = − a′
0
 a
 
where
− a0′ ≤ −
b
< − a0′ + 1.
a
We write
−
b
1
=
−a0′ + ,
b′
a
a′
and then we use the same techniques as in theorem 1 to get the remaining terms for b′ . That is, if b′ = [a′ ; a′ ,..., a′ ] then
1 2
n
b
a′
a′
− =−
[ a0′ ; a1′ ,..., an′ ] .
a
Example 1:
3
1
(1) 7= 0 +
1 = [0, 2, 3]
2+
3
(2)
−
27
1
=−6 +
1
5
1+
1
1+
2
= [-6, 1, 1, 2].
Lemma 1: Any finite S.C.F. [a0; a1,...,an] can be also represented by [a0;a1,...,an – 1,1] [5].
Definition 3:
The S.C.F. [a0; a1,...,an] can be defined by
[a0 ; a1,..., an=
] a0 +
K n −1 ( a2 ) or
K (a )
,
[a0 ; a1,..., an ] = n +1 0 .
K n (a1)
K n (a1)
where,
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Am-Euras. J. Sci. Res., 10 (5): 251-263, 2015
K 0 ( a0 ) 1=
K 0 ( a1 ) 1
K1 (a0 )
a=
K1 (a1 ) a1
0
K 2 ( a0 ) =
a0 a1 + 1
K 3 (=
a0 ) a0 a1a2 + a0 + a2
K 2 (a1 ) =
a1 a2 + 1
K3 (=
a1 ) a1 a2a3 + a1 + a3


K i (a0 ) =
ai −1 K i −1 ( a0 ) + K i − 2 ( a0 )
Ki ( a1 ) =
ai Ki −1(a1 ) + Ki − 2 (a1)
In general
K i ( a j ) =ai + ( j −1) K i −1 ( a j ) + K i − 2 ( a j ),
K −i ( a j )
i =1, 2,..., n,
j =0,1,..., n
0=
K 0 (a j ) 1
Lemma 2: If Cj = 0 in, [c0; c1,...,cj–1, cj, cj+1,...cn] for some j with 0 < j < n, then we can replace cj– 1by cj–1 = cj–1 + cj+1 and
delete cj, cj+1 from the simple continued fraction expansion, without changing the value of S.C.F.
Lemma 3: If cj = cj+1 0 in [ c0 ; c1,..., c j −1, c j , c j +1, c j + 2 ,..., cn ], for some j with 0 < j < n then we can delete cj, cj+1 from the
simple continued fraction expansion, without changing the value of S.C.F..
Definition 4:
Let [a0; a1,..., an ], [b0; b1,..., bn ] be two simple continued fractions, then,
[a0; a1,..., an ] = [b0 ; b1,...,
=
bn ] if ai b=
for i 0,1, 2,..., n.
i
[a=
0 ; a1,..., an −1, an ] [a0 ; a1,..., an −1, an − 1,1].
[a0 ;..., a j − 2 , a j −1,0, a=
j +1, a j + 2 ,..., an ] [a0 ;..., a j − 2 , a j −1 + a j +1 , a j + 2 ,..., an ].
Theorem 2: Let [a0 ; a1,..., am ] and [b0; b1,..., bn ] be two S.C.F. with ai
and [a0 ; a1,..., am ] ≠ [b0 ; b1,..., bn ] then
If a0
b0, let a0 > b0, then
[a0 ; a1,..., am ] > [b0 ; b1,..., bn ].
If ai = bi for all i=0,1,2,...k, k< min (m, n)
let ak+1 > bk+1, with k +1
min (m, n)
then [a0 ; a1,..., am ] > [b0 ; b1,..., bn ] if k is odd
and [a0 ; a1,..., am ] < [b0; b1,..., bn ] if k is even.
If ai = bi for all i = 0,1,2,...k. k = min (m, n)
letk = m, then [a0 ; a1,..., am ] > [b0 ; b1,..., bn ] if k is odd
and [a0 ; a1,..., am ] < [b0; b1,..., bn ] if k is even.
253
0 for all i=1,2,...m and, bj
0 for all j = 1,2,...n
Am-Euras. J. Sci. Res., 10 (5): 251-263, 2015
For examples
[3;2, 2] > [2;1, 4], since a0 > b0.
[3;1, 1, 3]> [3;7], since a0 = b0 and b1 > ai.
[1;2, 3] < [1;2, 3, 4], since ai = bi for all i = 0, 1, 2 and k = m = 2 (even).
[1;2, 3, 4] > [1;2, 3, 4, 5], since ai = bi for all i= 0, 1, 2, 3 and k = m = 3 (odd).
Definition 5: Let [a0;a1,...,am] and [b0;b1,...,bm] be two S.C.F., we define addition by
(1) If m=n then
[a0; a1,..., an ] + [b0 ; b1,..., bn ] =
[c0 ; c1,..., cn ]
(5a)
where,
c0 = a0 + b0 = K1 (a0 ) + K1(b0 )
 a b   K (a ) K (b ) 
11  
1 1 1 1 
c1 =
=
 

 a1 + b1   K1 ( a1) + K1 (b1 ) 


a2 (b1b2 + 1) + b2 ( a1a2 + 1)

c2 = 

 (a1a2 + 1)(b1b2 + 1) − c1[a2 (b1b2 + 1) + b2 ( a1a2 + 1)] 


K1 (a2 ) K 2 (b1 ) + K1 (b2 ) K2 ( a1)

= 

 K 2 ( a1) K 2 (b1) − c1[ K1 (a2 ) K 2 (b1 ) + K1 (b2 ) K2 ( a1)] 
 K (a ) K (b ) K (c ) − [ K (a ) K (b ) + K (b ) K (a )]K (c ) 
2 2
3 1
2 2
3 1
1 1 
c3 =  3 1 3 1 0 2

 [ K 2 (a2 ) K3 (b1 ) + K 2 (b2 ) K3 ( a1 )]K 2 (c1 ) − K 3 ( a1 ) K3 (b1 ) K1 (c2 ) 



  Ki (a1)K i (b1 ) K i −3 (c2 ) − [ Ki −1 (a2 ) Ki (b1 ) + Ki −1 (b2 )K i ( a1 )]K i − 2 (c1 ) 

 , i odd
  [ Ki −1 ( a2 )Ki (b1 ) + Ki −1 (b2 )K i (a1 )]K i −1(c1) − Ki (a1 ) K i (b1 ) Ki − 2 (c2 ) 

ci = 


 [ Ki −1 ( a2 ) Ki (b1 ) + K i −1 (b2 )K i (a1 )]K i − 2 (c1) − Ki (a1) K i (b1 ) Ki −3 (c2 )  , i even
 Ki (a1)K i (b1 ) K i − 2 (c2 ) − [ Ki −1 (a2 ) Ki (b1 ) + Ki −1 (b2 )K i (a1 )]K i −1(c1) 

for i = 2, …, n.
The last term cn of the resulting S.C.F. is to be expanded again as a S.C.F. if necessary and not to be treated as the
greatest integer number as the preceding terms have been treated.
(2) If m n then
Suppose that m<n then
′ , cm
′ +1,..., cn′ ]
[a0; a1,..., am ] + [b0 ; b1,..., bm , bm+1,..., bn ] =
[c0′ ; c1′ ,..., cm
where
c’j = cj for j = 0, 1, 2, …, m ,
and cj as we did for case m=n, while c′j = c j , j − m and
254
(5b)
Am-Euras. J. Sci. Res., 10 (5): 251-263, 2015
c j, j −m
  K m ( a1 ) K j (b1 ) K j −3 (c2′ ) − [ K m −1 ( a2 ) K j (b1 ) + K j −1(b2 ) K m (a1)]K j − 2 (c1′ ) 

 , j odd
  [ K m −1 (a2 ) K j (b1) + K j −1 (b2 ) K m (a1)]K j −1 (c1′ ) − K m ( a1 ) K j (b1 ) K j − 2 (c′2 ) 

=


 [ K m −1 (a2 ) K j (b1) + K j −1(b2 ) K m (a1)]K j − 2 (c1′ ) − K m ( a1 ) K j (b1 ) K j −3 (c2′ )  , j even
 K ( a ) K (b ) K (c′ ) − [ K (a ) K (b ) + K (b ) K (a )]K (c′ ) 
m −1 2
j 1
j −1 2
m 1
j −1 1 

 m 1 j 1 j − 2 2
forj = m+1, m +2, …, n.
Also, the last term of the resulting S.C.F. is to be treated as mentioned earlier.
Remark 2:
If the number of the S.C.F. term’s for [a0;a1,...,a n] and [b 0;b 1,...,b n] are n, then it is not necessary for the number of
terms of the resulting S.C.F. [a0;a1,...,an] + [b0;b1,...,bn] to be n.
If [a0] and [b0;b1,...,bn] are two S.C.F. then [a0] + [b0;b1,...,bn] = [c0b0;b1,...,bn] where c0 is given in equation (5a).
Example 2:
Find [1;2,5] + [4;3,2].
Solution:
Let [1;2, 5] = [a0;a1,a2] and [4;3, 2] = [b0b1,b2], we get m=n= 2.
From equation (5a) we have
[1;2,5] + [4;3,2] = [a0;a1,a2] + [b0;b1,b2 = [c0;c1,c2] ,
where c2 is the last term and
5
c=
0 a 0 + b0 = 1 + 4 =
 a b   2 ⋅3  6
  =
 =
 1
=
c1  1 1=

 a1 + b1   2 + 3   5 
a2 (b1b2 + 1) + b2 (a1a2 + 1)
c2 =
(a1a2 + 1)(b1b2 + 1) − c1[a2 (b1b2 + 1) + b2 (a1a2 + 1)]
5(3 ⋅ 2 + 1) + 2(5 ⋅ 2 + 1)
(3 ⋅ 2 + 1)(5 ⋅ 2 + 1) − 1[5(3 ⋅ 2 + 1) + 2(5 ⋅ 2 + 1)]
57
57
5(7) + 2(11)
=
=
= = [2;1,5,1, 2],
(7)(11) − [35 + 22] 77 − 57 20
=
then [1;2,5] + [4;3,2] = [5;1,2,1,5,1,2]
Example 3:
Find [17;7] + [18;1,1,3].
Solution:
Let [17;7] = [a0; a1] and [18;1, 1, 3] = [b0, b1,b2,b3], we get m= 1, n= 3 and m <n.
From equation (5b) we have,
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Am-Euras. J. Sci. Res., 10 (5): 251-263, 2015
[17;7] + [18;1,1,3] = [a0;a1] + [b0;b1,b2,b3] = [c0; c1, c′2, c′3]
where c′3 is the last term and
c0 = c0′ = a0 + b0 = 17 + 18 = 35
 a b   7 ⋅1   7 
 11 =
 
  =
 0
=
c=

1 c=
1′ 
 a1 + b1   7 + 1  8 
 [ K (a ) K (b ) + K (b )K (a )]K (c′ ) − K (a ) K (b ) K (c′ ) 
 0 2 2 1
1 2 1 1
0 1
1 1 2 1 −1 2 
=
c′2 c=

2,1 
 K1 (a1) K 2 (b1 ) K 0 (c2′ ) − [ K 0 (a2 ) K 2 (b1 ) + K1 (b2 )K1 ( a1 )]K1 (c1′ ) 
 [1 ⋅ (b b + 1) + b a ] ⋅ 1 − a ⋅ (b b + 1) ⋅ 0 

1 2
2 1
1
1 2
= 

 a1 (b1b2 + 1) − [(b1b2 + 1) + b2a1 ] ⋅ c1′ 
 (1 ⋅ 1 + 1) + 1 ⋅ 7   9 
 =

= 
=
  14  0
7⋅2

  
K1 (a1 ) K3 (b1 ) K 0 (c′2 ) − [ K 0 ( a2 ) K3 (b1 ) + K 2 (b2 ) K1 ( a1 )]K1 (c1′ )
=
c3′ c=
3,2
[ K 0 (a2 ) K3 (b1 ) + K 2 (b2 ) K1 (a1 )]K 2 (c1′ ) − K1 (a1) K3 (b1 ) K1 ( c2′ )
a1 (b1b2b3 + b1 + b3 ) − [(b1b2b3 + b1 + b3 ) + a1 (b2b3 + 1)]c1′
=
[(b1b2b3 + b1 + b3 ) + a1 (b2b3 + 1)](c1′c′2 + 1) − a1 (b1b2b3 + b1 + b3 ) ⋅ c2′
7 ⋅ (1 ⋅ 1⋅ 3 + 1 + 3) − [(1 ⋅ 1 ⋅ 3 + 1 + 3) + 7 ⋅ (1 ⋅ 3 + 1)] ⋅ 0
[(1 ⋅ 1⋅ 3 + 1 + 3) + 7 ⋅ (1 ⋅ 3 + 1)] − 7 ⋅ (1 ⋅ 1⋅ 3 + 1 + 3) ⋅ 0
7 ⋅ (1 ⋅ 1⋅ 3 + 1 + 3)
=
[(1 ⋅ 1⋅ 3 + 1 + 3) + 7 ⋅ (1 ⋅ 3 + 1)]
7⋅7
7
7
=
= = [1;2,2, ]
=
7 + 7 ⋅ 4 1+ 4 5
=
Therefore [17;7] + [18;1,1,3] = [35;0, 0, 1, 2, 2]= [35;1, 2, 2].
Example 4:
Find [4] + [1;1, 2].
Solution:
Let [4] = [a0] and [1;1, 2] = [b0;b1,b2] we get m= 0, n= 2, m<n.
From remark 2(ii) we have,
[4] + [1;1, 2] = [a0] + [b0,b1,b2] = [c0b1,b2] where
c0 = a0 + b0 = 4 + 1 = 5.
Therefore [4] + [1, 1, 2] = [5, 1, 2].
Corollary 1: If c is a non-zero integer then c[a , a , a , a ,...] = [ca , a1 , ca , a3 ,...] which is not necessary a S.C.F..
0 1 2 3
0
2
c
c
Example 5:
Find –2[1;2, 2].
Solution:
From corollary 1 we have
2
−2[1; 2, 2] = [ −2(1); , −2(2)] = [ −2; − 1, − 4].
−2
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Am-Euras. J. Sci. Res., 10 (5): 251-263, 2015
Remark 3: Note that the multiplication of S.C.F. by a non-zero integer
seen in example 5.
1 does not necessarily lead to S.C.F. as we have
Definition 6: Let [a0;a1,...,an] be a S.C.F., then we can define additive inverses by −[a0; a1,..., an ] =
[b0′ ; bn′ ]
where,
b0′ =−1 − a0 =d 0
bn′ =
K n (a1 )
K n (a1 ) − K n −1 (a2 )
(to be treated as S.C.F. )
= [d1; d 2 ,..., dl ]
therefore,
−[a0; a1,...,
=
an ] [b=
0′ ; bn′ ] [d0 ; d1, d 2 ,..., dl ].
Definition 7: If [a0 ; a1,..., am ] and [b0; b1,..., bn ] are two S.C.F. then we define Subtraction [a0 ; a1,..., am ] - [b0; b1,..., bn ] by
the addition, [a0 ; a1,..., am ] + [d0 ; d1,..., dl ] where [d0; d1,..., dl ] given by definition 6.
That is;
If m=l then
(7a)
[a0; a1,..., am ] + [d 0 ; d1,..., dm ] =
[c0 ; c1,..., cm ]
where c0,c1,...,cm as given in equation5a.
If m l, l<m then
′ ]
[a0; a1,..., al , al +1,..., am ] + [d0; d1,..., dl ] =
[c0′ ; c1′ ,..., cl′ , cl′+1,..., cm
′ as given in equation 5b.
where c0′ , c1′ ,..., cm
The motivations of our definitions and the analytic prove are published in [5].
Example 6:
Find [1; 2, 2,3,5,3,1, 2, 4,9] + [1;1,8, 4, 2,1,3,5,3, 2, 2] and −[1;1,8, 4, 2,1,3,5,3, 2, 2] .
Solution:
To find [1; 2, 2,3,5,3,1, 2, 4,9] + [1;1,8, 4, 2,1,3,5,3, 2, 2] .
Let [1;2, 2,3,5,3,1, 2,4,9] = [a0 ; a1, a 2, a 3, a 4, a 5, a6, a7 , a8 , a9 ], m= 9
and [1;1,8, 4, 2,1,3,5,3,2,2] = [b0 ; b1, b2, b3, b4, b5, b6 , b7 , b8 , b9 , b10 ], n= 10, n m.
From equation (5b) we have
[ a0 ; a1, a 2, a 3, a4, a 5 , a6 , a7 , a8 , a9 ] + [b0 ; b1, b2, b3, b4 , b5 , b6 , b7 , b8 , b9 , b10 ]
′ ]
= [c0 ; c1, c 2, c 3, c4 , c5 , c6 , c7 , c8 , c9 , c10
257
(7b)
Am-Euras. J. Sci. Res., 10 (5): 251-263, 2015
where c’10 is the last term and
c0 = c0′ = a0 + b0 = 1 + 1 = 2
 a b   2
  =
 0
′  1 1 =
c1= c1=

 a1 + b1   3 


K1(a2 )K 2 (b1) + K1(b2 ) K2 ( a1)


c=

2 c=
2′ 
−
+
K
a
K
b
c
K
a
K
b
K
b
K
a
(
)
(
)
[
(
)
(
)
(
)
(
)]
 2 1 2 1
1 1 2 2 1
1 2 2 1 
 2(9) + 8(5)   58 
 =
= 
=
   1
 (9)(5)   45 
 K (a ) K (b ) K (c ) − [ K (a ) K (b ) + K (b ) K (a )]K (c ) 
 3 1 3 1 0 2
2 2 3 1
2 2 3 1 1 1 
c=

3 c=
3′ 
 [ K 2 (a2 ) K3 (b1) + K 2 (b2 ) K3 (a1)]K 2 (c1) − K3 (a1) K3 (b1)K1(c2 ) 


K3 (a1) K3 ( b1)

= 

 K 2 (a2 ) K3 (b1) + K 2 (b2 ) K3 ( a1) − K3 (a1) K3 (b1) K1(c2 ) 


17(37)

= 
=

 (33)(17) + (7)(37) − (17)(37) 
 629 
=

 191  3


 [ K ( a ) K (b ) + K (b ) K ( a )]K (c ) − K (a ) K (b ) K (c ) 
 3 2 4 1
3 2 4 1 2 1
4 1 4 1 1 2 
c=

4 c=
4′ 
 K 4 (a1) K 4 (b1 ) K 2 (c2 ) − [ K3 ( a2 ) K 4 (b1) + K3 (b2 ) K 4 ( a1 )]K3 (c1 ) 
 [(37)(83) + (47)(90)] − (83)(90)(1)   2261 
 =
 3
= 
=
 

  687 
(83)(90)(4) − [9731](3)
 K (a ) K (b ) K (c ) − [ K (a )K (b ) + K (b ) K (a )]K (c ) 
 5 1 5 1 2 2
4 2 5 1
4 2 5 1 3 1 
c=

5 c=
5′ 
 [ K 4 (a2 )K 5 (b1) + K 4 (b2 ) K5 (a1)]K 4 (c1) − K5 ( a1) K5 (b1) K3 ( c2 ) 
 (287)(120)(4) − [(118)(120) + (107)(287)](3)   3153 
 =
 3
= 
=
 

(44869)(100) − [34440](13)

  970 
 [ K ( a )K (b ) + K (b ) K (a )]K (c ) − K (a ) K (b ) K (c ) 
5 2 6 1 4 1
6 1 6 1 3 2 
 5 2 6 1
c=

6 c=
6′ 
 K6 (a1) K6 (b1) K 4 (c2 ) − [ K5 (a2 ) K 6 (b1 ) + K5 (b2 ) K6 (a1)]K5 (c1) 
 [(377)(395) + (443)(155)](10) − (377)(443)(13) 

= 

(167011)(43) − [217580](33)


 4657 
 3
= =

 1333 
 K ( a ) K (b ) K (c ) − [ K ( a ) K (b ) + K (b ) K ( a )]K (c ) 
 7 1 7 1 4 2
6 2 7 1
6 2 7 1 5 1 
c=

7 c=
7′ 
 [ K6 (a2 )K 7 (b1) + K6 (b2 ) K7 (a1)]K 6 (c1) − K7 (a1) K7 (b1 ) K5 (c2 ) 
 (1041)(2335)(43) − [(1041)(2082) + (2335)(428)](33) 

= 

(3166742)(109) − [2430735](142)


 19119 
 1
= =

 10508 
 [ K (a )K (b ) + K (b ) K (a )]K (c ) − K (a ) K (b ) K (c ) 
 7 2 8 1
7 2 8 1 6 1
8 1 8 1 5 2 
c=

8 c=
8′ 
 K8 (a1) K8 (b1) K6 (c2 ) − [ K7 (a2 ) K7 (b1 ) + K7 (b2 ) K8 (a1)]K 7 (c1) 
 [(4541)(6641) + (7448)(1857)](109) − (4541)(7448)(142) 

= 

(33821368)(6106) − [66331717](142)


 4227522897 
 0
= =

 197094169194 
 K ( a ) K (b ) K (c ) − [ K ( a )K (b ) + K (b ) K (a )]K (c ) 
 9 1 9 1 6 2
8 2 9 1
8 2 9 1 7 1 
c=

9 c=
9′ 
 [ K7 (a2 )K 9 (b1) + K8 (b2 ) K9 (a1)]K8 (c1) − K9 ( a1 ) K9 (b1) K7 (c2 ) 
 (41910)(17231)(185) − [(41410)(15364) + (17231)(17231)](142) 

= 

(940812601)(109) − [722151210](142)


 2584508 
 0
= =

 3101709 
[ K ( a ) K (b ) + K 9 (b2 ) K9 ( a1)]K8 (c1′ ) − K9 (a1) K10 (b1 ) K 7 (c2′ )
′ = 8 2 10 1
c10
K9 (a1) K10 (b1) K8 (c2′ ) − [ K8 ( a2 ) K10 (b1) + K9 (b2 ) K9 (a1)]K 9 (c1′ )
[(41910)(37369) + (41910)(17231)](109) − (41910)2 (142)
(1756448100)(185) − [2288286000](142)
7543800
=
6286500
6
= = [1;5]
5
=
258
Am-Euras. J. Sci. Res., 10 (5): 251-263, 2015
therefore,
[1, 2, 2, 3, 5, 3, 1, 2, 4, 9] + [1;18, 4, 2, 1, 3, 5, 3, 2, 2] = [2;0, 1, 3, 3, 3, 3, 1, 0, 0, 1, 5]
= [3;3, 3, 3, 3, 1, 0, 0, 1, 5]= [3;3, 3, 3, 3, 1, 1, 5]
To check, we have
59141 79279
+
41910 41910
138420 4614
=
41910
1397
[1;2,2,3,5,3,1,2,4,9] + [1;18,4,2,1,3,5,3,2,2]=
=
and
[3;3, 3, 3, 3, 1, 1, 5]= 3 +
which is true.
K 6 ( a2 )
K7 (a1 )
423 4614
3+
=
=
1397 1397
To find –[1;1,8,4,2,1,3,5,3,2,2].
Let [1;18, 4, 2, 1, 3, 5, 3, 2, 2] = [a0 ; a1, a 2, a 3, a4, a5 , a6 , a7 , a8 , a9 , a10 ] , we get: n= 10,
From definition 6 we have
′ ],
−[1;1,8,4,2,1,3,5,3,2,2] =
[b0′ ; b10
where,
b0′ = −1 − a0 = −1 − 1 = −2
′ =
b10
K10 (a1 )
K10 (a1 ) − K9 (a2 )
K (a )
79279
Since [1;1,8, 4, 2,1,3,5,3, 2,2]
=
= a0 + 9 2 , (by definition 3)
41910
K10 ( a1 )
and K10(a1) = 41910,
K 9=
( a2 ) 79279 − a0 K10 ( a1 )
= 79279 − 1(41910) = 79279 − 41910 = 37369
Therefore,
′
=
b10
=
K10 ( a1 )
41910
=
K10 (a1 ) − K 9 ( a2 ) 41910 − 37369
41910
4541
= [9;4, 2, 1, 3, 5, 3, 2, 2].
Therefore –[1;1,8,4,2,1,3,5,3,2,2] = [-2;9, 4, 2, 1, 3, 5, 3, 2, 2].
Theorem 3: Let
ak
=
=
k , k +1
k
= 0 be an irrational number and define the sequence a0,a1,a2,... recursively by
1
for k =0, 1, 2, … .Then is the value of infinite S.C.F. [a0;a1,a2].
− ak
259
Am-Euras. J. Sci. Res., 10 (5): 251-263, 2015
For example
=
6 [ a0 ; a1, a 2, =
a 3, a4 , a5 ,...] [2;
=
2, 4, 2, 4...] [2; 2,4] .
We can use the same operations of finite S.C.F. for infinite S.C.F..
Example 7:
Find [ 1;2] + [1;1, 2 ]
Solution:
Let [1;2] [1;
=
=
2, 2, 2, 2,...] [a0 ; a1, a 2 , a 3 , a 4 ,...]
and [1;1,2] [1;1,
=
=
2,1, 2,...] [b0 ; b1, b2, b3 , b4 ,...].
[c0 ; c1, c 2, c3, c4 ,...]
From equation (5a) we have [a0; a1, a 2, a3 , a4 ,...] + [b0 ; b1, b2 , b3 , b4 ,...] =
where,
c0 = a 0 + b0 = 1 + 1 = 2
 a b   2 ⋅1 
1 1  =
 0
=
c1  =
 

 a1 + b1   2 + 1


a2 (b1b2 + 1) + b2 (a1a2 + 1)

c2 = 

 (a1a2 + 1)(b1b2 + 1) − c1[a2 (b1b2 + 1) + b2 (a1a2 + 1)] 
 2(3) + 2(5) 

= 

 (3)(5) 
16 
 1
= =

 15 


K3 ( a1) K3 ( b1 )

c3 = 

 K 2 (a2 ) K 3 (b1 ) + K 2 (b2 ) K3 ( a1 ) − K3 ( a1 ) K 3 (b1 ) K1 (c2 ) 


12(4)

= 

 5(4) + 3(12) − 12(4) 
 48 
 6
= =

8
 [ K ( a ) K (b ) + K (b ) K ( a )]K (c ) − K (a ) K (b ) K (c ) 
3 2
4 1
2 1
4 1 4 1 1 2 
c4 =  3 2 4 1

 K 4 ( a1) K 4 (b1) K 2 (c2 ) − [ K 3 ( a2 ) K 4 (b1) + K3 (b2 ) K 4 (a1)]K3 (c1 ) 
 [(12)(11) + (8)(29)] − (29)(11)(1) 

= 



(29)(11)(7) − [364](6)
 45 
 0
= =

 49 
 K (a ) K (b ) K (c ) − [ K (a ) K (b ) + K (b ) K (a )]K (c ) 
4 2
5 1
4 2
5 1
3 1 
c5 =  5 1 5 1 2 2

 [ K 4 (a2 ) K5 (b1 ) + K 4 (b2 ) K5 (a1)]K 4 (c1 ) − K5 (a1 ) K5 (b1 ) K3 (c2 ) 
 (70)(15)(7) − [(29)(15) + (11)(70)](6) 

= 



(1205)(1) − [1050](1)
120 
 0
= =

 155 
260
Am-Euras. J. Sci. Res., 10 (5): 251-263, 2015
 [ K ( a ) K (b ) + K (b ) K ( a )]K (c ) − K (a ) K (b ) K (c ) 
5 2
6 1
4 1
6 1 6 1 3 2 
c6 =  5 2 6 1

 K 6 ( a1 ) K6 (b1) K4 (c2 ) − [ K5 (a2 ) K 6 (b1) + K5 (b2 ) K 6 (a1 )]K5 (c1 ) 
 [(70)(41) + (30)(169)](1) − (169)(41)(1) 

= 



(6929)(7) − [7940](6)
1011

= 
=1
 863 
 K (a ) K (b ) K (c ) − [ K (a ) K (b ) + K (b ) K ( a )]K (c ) 
6 2
7 1
6 2
7 1
5 1 
c7 =  7 1 7 1 4 2

 [ K 6 (a2 ) K 7 (b1 ) + K 6 (b2 ) K 7 ( a1 )]K 6 (c1) − K 7 (a1 ) K7 (b1 ) K5 ( c2 ) 
 (408)(56)(7) − [(169)(56) + (41)(408)](6) 

= 

(26192)(7) − [22848](8)


 2784 
=4
= 

 560 
 [ K (a ) K (b ) + K (b ) K ( a )]K (c ) − K (a ) K (b ) K ( c ) 
7 2 8 1
6 1
8 1 8 1 5 2 
c8 =  7 2 8 1

 K8 ( a1 ) K8 (b1 ) K 6 (c2 ) − [ K 7 (a2 ) K 7 (b1 ) + K 7 (b2 ) K8 ( a1 )]K 7 (c1 ) 
 [(408)(153) + (112)(985)](7) − (985)(153)(8) 

= 

(150705)(39) − [172744](34)


 3568 
 0
= =

 4199 

Therefore [1;2] + [1;1, 2] = [2;0, 1, 6, 0, 0, 1, 4, 0, …] = [3;6, 0, 0, 1, 4, 0, …] = [3;6, 1, 4, 0, …]
Example 8:
Find [1;2] + [2;1,1,1, 4]
Solution:
Let [1;2] = [a0 ; a1] and [ 2;1,1,1,4] = [2;1,1,1, 4,1,1,1, 4,... ] = [b0; b1, b2 , b3 , b4 ,...]
[c0 ; c1, c′2, c′3, c′4 ,...] where
From (5b) we have [a0 ; a1] + [b0 ; b1, b2, b3 , b4 ,...] =
′
=
+
=
1
+
2
=
3
c=
c
a
b
0
0
0
0
 a b   2 ⋅1 
 11 =
 
 0
=
c=
 
1 c=
1′ 

 a1 + b1   2 + 1
 K (b ) + K ( b ) a   2 + 2 
1 =
2 1  =
 1
=
c′2  2 1
 

  4 
a1K 2 (b1)


a1K 3 (b1 )c2′

c3′ = 

 [ K3 (b1 ) + K 2 (b2 ) a1 ] − a1K 3 (b1 )c2′ 

 6
2(3)
  =6
= =
  
 3 + 2(2) − 2(3)   1 


[ K 4 (b1 ) + K 3 (b2 ) a1 ] − a1K 4(b1)c′2

c4′ = 

 a1K 4 (b1 ) K 2 (c2′ ) − [ K 4 (b1 ) + K3 (b2 )a1 ]K 3 (c1′ ) 
 [(14) + (9)(2)] − (2)(14) 

= 

 (2)(14)(7) − [32](6) 
 4
 1
= =

 4
261
Am-Euras. J. Sci. Res., 10 (5): 251-263, 2015
 a K (b ) K (c′ ) − [ K (b ) + K (b )a ]K (c′ ) 
5 1
4 2 1 3 1 
c5′ =  1 5 1 2 2

 [ K 5 (b1 ) + K 4 (b2 )a1 ]K 4 (c1′ ) − a1K 5 (b1 ) K3 (c2′ ) 
 (2)(17)(7) − [(17) + (11)(2)](6) 

= 



(39)(7) − (34)(8)
4
 4
= =

1
 [ K (b ) + K (b ) a ]K (c′ ) − a K (b ) K (c′ ) 
5 2 1 4 1
1 6 1 3 2 
c6′ =  6 1

 a1K 6 (b1 ) K 4 (c2′ ) − [ K 6 (b1 ) + K5 (b2 )a1 ]K5 (c1′ ) 
 [(31) + (20)(2)](7) − (2)(31)(8) 

= 

(62)(39) − (71)(34)


1
 0
= =

 4
 a K (b ) K (c′ ) − [ K (b ) + K (b ) a ]K (c′ ) 
7 1
6 2 1 5 1 
c7′ =  1 7 1 4 2

 [ K 7 (b1) + K 6 (b2 )a1 ]K 6 (c1′ ) − a1K 7 (b1 ) K5 (c2′ ) 
 (2)(48)(39) − [(48) + (31)(2)](34) 

= 

(110)(7) − (96)(8)


 4
 2
= =

 2

Therefore [1;2] + [2;1,1,1,4] = [3;0, 1, 6, 1, 4, 0, 2, …] = [4;6, 1, 6, …].
There are many applications of continued fractions: combine continued fractions with the concepts of golden ratio
and Fibonacci numbers, Pell equation and calculation of fundamental units in quadratic fields, reduction of quadratic
forms and calculation of class numbers of imaginary quadratic field.There is a pleasant connection between Chebyshev
polynomials, the Pell equation and continued fractions, the latter two being understood to take place in real quadratic
function fields rather than the classical case of real quadratic number fields [1, 5].
The very nice elementary application of simple continued fractions is Gosper's batting average problem which is,
if a baseball player's (3-digit rounded) batting average is 0.334, what's the smallest number of at-bats that player could
have? (Batting average is computed as number of hits . The solution proceeds by noting that a rounded average of
(
at − bats
)
0.334 corresponds to an actual number in the range (3335,3345), finding the continued fractions for these values yields
and
. This implies that the 'simplest' number within the range
667
669
0.3335
=
is
= [0; 2,1,666]
2000
[0, 2,1,95]
=
0.3345
=
= .[0; 2,1,94,1,1,3]
2000
[7].
69
≈ 0.334495
287
This paper is the first part for the operations of the simple continued fractions. The second part will define the
multiplication, multiplicative inverse and the powers of the simple continued fractions.
CONCLUSION
are much more useful (and computers can work with
decimals at a much faster rate). However, some interesting
observations can still be made using continued
fractions.Namely, in this paper, we will be exploring how
continued fractions can be used to add and subtract the
numbers a  b and a  c .
We can also solve any Diophantine congruence that
is any equivalence of the form ax b(mod m). In other
words, in most real-world applications of mathematics,
continued fractions are rarely the most practical way to
solve a given set of problems as decimal approximations
b
262
Am-Euras. J. Sci. Res., 10 (5): 251-263, 2015
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1.
2.
3.
4.
5.
Barbeau, E., 2003. Pell's Equation, Springer.
Euler, L. and D.B. John, 1988. Introduction to
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Kline, M., 1972. Mathematical Thought from Ancient
to Modern times, New York: Oxford UP.
Moore, C.G., 1964. An Introduction to Continued
Fractions, The National Council of Teachers of
Mathematics: Washington, D.C.
6.
7.
8.
263
Mugassabi, S. and F. Mistiri, 2014. Simple Continued
Fractions, In press.
Rosen, K.H., 2005. Elementary number theory and its
applications, Addison-Wesley Pub. Co.
Viader, C., Jaume B. Pelegrí and B. Lluís, 2000. On the
Concept of Optimality Interval, UPF Economics &
Business Working Paper, pp: 466.
Wall, H.S., 1948. Analytic theory of continued
fractions, New York: D. Van Nostrand Co.