The Multiplication and Division of Simple Continued Fractions

International Journal of Innovation in Science and Mathematics Volume 8, Issue 3, ISSN (Online): 2347–9051

Souad I. Mugassabi 1* and Somia M. Amsheri 2 1 University of Benghazi, Benghazi, Libya. 2 University of Elmergib, Alkhums, Libya.

Date of publication (dd/mm/yyyy): 22/06/2020 Abstract – The finite and infinite simple continued fractions for (rational and irrational) are considered. The multiplication of two simple continued fractions are discovered. Also, the multiplicative inverse and the division of the simple continued fractions are showed. The most important that we will do in this paper, we exploring how the simple continued fractions can be used to calculate the numbers operations on roots. Many definitions and examples that we used of these low are presented.

Keywords – Simple Continued Fractions, Multiplicative, Multiplicative Inverse, Division.

I. INTRODUCTION

There are many applications of continued fractions: combine continued fractions with the concepts of golden ratio and Fibonacci numbers, Pell equations and calculation of fundamental units in quadratic fields, reduction of quadratic forms and calculation of class numbers of imaginary quadratic field [7]. 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].

The simple continued fractions have been studied in mathematical (Diophantine Equation, congruence 푎푥 ≡ 푏 (푚표푑 푚) and Pell's equations) and physical (gear ratio) [4]. The analytic of continued fractions for the real and complex values have been studied in [3, 9]. However, [2, 4, 6] studied the continued fractions for the integer values. There are many applications of simple continued fractions (Gosper's batting average problem [8], Cryptography …). In [5, 6] we show that, any number, rational or irrational, can be expression as a finite or infinite continued fraction. Also, we can solve any Diophantine Equation or congruence 푎푥 ≡ 푏 (푚표푑 푚).

The most important, that we did in [5, 6], we define the addition and subtraction of the simple continued fractions. Also, we showed that, how can we know which simple continued fraction is greater than of the other. In this paper, we will define the multiplication and division of the simple continued fractions and we will be exploring how continued fractions can be used to multiply the numbers 푎 ∙ 푏 . We start with some definitions and theorems that we used to defined the multiplication of two simple continued fractions.

II. PRELIMINARIES

2.1. Definition

푏0 An expression of the form 푎0 + 푏 is called a continued fraction 푎0, 푎1, 푎2, … and 푏0, 푏1,푏2,… can be 푎 + 1 1 푏2 푎2+ 푎3+⋯ real or complex, and their numbers can be either finite or infinite.

2.2. Definition

A continued fraction (The above fraction) is called a finite simple continued fraction if 푏푛 = 1 for all n, that is 1 푎0 + 1 where an is positive integer for all 푛  1, 푎0 can be any integer number. This fraction is 푎1+ 1 푎2+ 1 ⋱ + 푎푛 sometime represented by [푎0; 푎1, 푎2, … , 푎푛 ] for finite simple continued fraction and [푎0; 푎1, 푎2, … ] for infinite simple continued fraction. In this paper we will use the symbol (S.C.F.) for the simple continued fraction.

2.1. Theorem

A number is rational if and only if it can be expressed as a finite S.C.F. [4].

2.1. Example

20 2 1 1 = 6 + = 6 + 3 = 6 + 1 = [6; 1,2]. 3 3 1+ 2 2

2.1. Remark

푎 푏 3 If 푎 > 푏 > 0 and = [푎 ; 푎 , 푎 , … , 푎 ] then = [0; 푎 , 푎 , 푎 , … , 푎 ]. For example = [0; 2, 3]. 푏 0 1 2 푛 푎 0 1 2 푛 7

2.3. Definition

퐾푛 −1 푎2 퐾푛 +1 푎0 The S.C.F. [푎0; 푎1, … , 푎푛 ] can be defined by 푎0 ;푎1 , … , 푎푛 = 푎0 + or 푎0 ;푎1 , … , 푎푛 = , 퐾푛 푎1 퐾푛 푎1 where

퐾0 푎표 = 1 퐾0 푎1 = 1

퐾1 푎0 = 푎0, 퐾1 푎1 = 푎1

퐾2 푎0 = 푎0푎1 + 1 퐾2 푎1 = 푎1푎2 + 1 ⋮ ⋮

퐾푖 푎0 = 푎푖−1퐾푖−1 푎0 + 퐾푖−2 푎0 퐾푖 푎1 = 푎푖 퐾푖−1 푎1 + 퐾푖−2 푎1

In general, 퐾푖 푎푗 = 푎푖+ 푗 −1 퐾푖−1 푎푗 + 퐾푖−2 푎푗 , 푖 = 1,2, … , 푛, 푗 = 0 , 1 , … , 푛 and 퐾−푖 푎푗 = 0, 퐾0 푎푗 = 1.

2.2. Example

Evaluate the S.C.F. [2; 2, 2].

Solution:

퐾1 푎2 푎2 Let [2; 2,2] = 푎0 ;푎1 , 푎2 , then n = 2. From definition 2.3 we get: 푎0 ;푎1 , 푎2 = 푎0 + = 푎0 + , 퐾2 푎1 푎1푎2+1 2 12 therefore 2; 2,2 = 2 + = . (2∙2+1) 5

2.1. Lemma a) [푎0; 푎1, … , 푎푛 ] = [푎0; 푎1, … , 푎푛 − 1,1]. b) [푐0; 푐1, … , 푐푗 −1, 0, 푐푗 +1, … , 푐푛 ] = [푐0; 푐1, … , 푐푗 −1 + 푐푗 +1, … , 푐푛 ]. c) [푐0; 푐1, … , 푐푗 −1, 0,0, 푐푗 +2, … , 푐푛 ] = [푐0; 푐1, … , 푐푗 −1, 푐푗 +2, … , 푐푛 ].

III. THE MULTIPLICATION OF SIMPLE CONTINUED FRACTIONS

3.1. Definition

Let [푎0; 푎1, … , 푎푛 ] and [푏0; 푏1, … , 푏푛 ] be two S.C.F., 푎0, 푏0 are non-negative, we define their multiplication by: (1) If m  n then

[푎0; 푎1, … , 푎푛 ] × [푏0; 푏1,… , 푏푛 ] = [푑0; 푑1, … , 푑푛 ] (3.1a)

Where

푑0 = 푎0푏0

푎1푏1 푑1 = 푎0푎1+푏0푏1+1

푎0푏2 푎1푎2+1 +푎2푏0 푏1푏2+1 +푎2푏2 푑2 = 푎1푎2+1 푏1푏2+1 −푑1 푎표 푏2 푎1푎2+1 +푎2푏0 푏1푏2+1 +푎2푏2

⋮ ⋮ ⋮

퐾푖 푎1 퐾푖 푏1 퐾푖−3 푑2 − 푎0퐾푖 푎1 퐾푖−1 푏2 +푏0퐾푖−1 푎2 퐾푖 푏1 +퐾푖−1 푎2 퐾푖−1 푏2 퐾푖−2 푑1 푑푖 = , if 푖 odd, and 푎0퐾푖 푎1 퐾푖−1 푏2 +푏0퐾푖−1 푎2 퐾푖 푏1 +퐾푖−1 푎2 퐾푖−1 푏2 퐾푖−1 푑1 −퐾푖 푎1 퐾푖 푏1 퐾푖−2 푑2

푎0퐾푖−1 푏2 퐾푖 푎1 +푏0퐾푖−1 푎2 퐾푖 푏1 +퐾푖−1 푎2 퐾푖−1 푏2 퐾푖−2 푑1 −퐾푖 푎1 퐾푖 푏1 퐾푖−3 푑2 푑푖 = , if i even. 퐾푖 푎1 퐾푖 푏1 퐾푖−2 푑2 − 푎0퐾푖−1 푏2 퐾푖 푎1 +푏0퐾푖−1 푎2 퐾푖 푏1 +퐾푖−1 푎2 퐾푖−1 푏2 퐾푖−1 푑1 for i = 2,3,…,n. The last term 푑푛 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 [ ┤] means the greatest integer number.

(2) If m  n (suppose that m < n) then

[푎0; 푎1, … , 푎푚 ] × [푏0; 푏1, … , 푏푚 , 푏푚+1, … , 푏푛 ] = [푑0; 푑1, … , 푑푛 ] (3.1b)

Where 푑′푗 = 푑푗 for j = 1, 2,…,m, and 푑푗 as we did for case m = n while 푑′푗 = 푑푗 ,푗 −푚

퐾푚 푎1 퐾푗 푏1 퐾푗−3 푑2 − 푎0퐾푚 푎1 퐾푗 −1 푏2 +푏0퐾푚 −1 푎2 퐾푗 푏1 +퐾푚 −1 푎2 퐾푗−1 푏2 퐾푗−2 푑1 푑푗 ,푗 −푚 = , if j is odd, and 푎0퐾푚 푎1 퐾푗 −1 푏2 +푏0퐾푚 −1 푎2 퐾푗 푏1 +퐾푚 −1 푎2 퐾푗−1 푏2 퐾푗−1 푑1 −퐾푚 푎1 퐾푗 푏1 퐾푗−2 푑2

푎0퐾푗 −1 푏2 퐾푚 푎1 +푏0퐾푚 −1 푎2 퐾푗 푏1 +퐾푚 −1 푎2 퐾푗−1 푏2 퐾푗−2 푑1 −퐾푚 푎1 퐾푗 푏1 퐾푗−3 푑2 푑푗 ,푗 −푚 = , for j = m+1, m+2, 퐾푚 푎1 퐾푗 푏1 퐾푗−2 푑2 − 푎0퐾푗 −1 푏2 퐾푚 푎1 +푏0퐾푚 −1 푎2 퐾푗 푏1 +퐾푚 −1 푎2 퐾푗−1 푏2 퐾푗−1 푑1 …, n. Also, the last term is to be treated as a simple continued fraction.

3.1. Example

Find [1; 2]  [1; 3].

Solution:

Let 1; 2 = 푎0 ;푎1 and 1; 3 = 푏0 ;푏1 , we get m  n 1. From Equation (3.1a) we have 1; 2 × 1; 3 =

푎0 ;푎1 × 푏0 ;푏1 = 푑0 ;푑1 ,

푎1푏1 2∙3 6 Where 푑1 is the last term and 푑0 = 푎0푏0 = 1 ∙ 1 = 1, 푑1 = = = = 1. 푎0푎1+푏0푏1+1 1∙2+1∙3+1 6

3 4 Therefore [1; 2] [1; 3] = [1; 1] = 2 (from lemma 2.1a). To check, we have 1; 2 × 1; 3 = × = 2. 2 3

3.2. Example

Find [2; 1, 4][0; 2, 3].

Solution:

Let 2; 1,4 = 푎0 ;푎1, 푎2 and 0; 2,3 = 푏0; 푏1,푏2 , m  n  2. From Equation (3.1a), we have 2; 1, 4 ×

0; 2, 3 = 푎0; 푎1, 푎2 × 푏0; 푏1, 푏2 = 푑0; 푑1, 푑2 , where 푑2 is the last term and 푑0 = 푎0푏0 = 2 ∙ 0 = 0, 푑1

푎1푏1 1∙2 2 푎0푏2 푎1푎2+1 +푎2푏0 푏1푏2+1 +푎2푏2 6 = = = = 0, 푑2 = = = 1; 5 , 푎0푎1+푏0푏1+1 2∙1+0∙2+1 3 푎1푎2+1 푏1푏2+1 −푑1 푎표 푏2 푎1푎2+1 +푎2푏0 푏1푏2+1 +푎2푏2 5 therefore [2; 1, 4] [0; 2, 3] = [0; 0, 1, 5] = [1; 5] (from lemma 2.1c).

14 3 6 1 5+1 6 To check, we have [2;1,4] [0;2,3]= × = and [1; 5]= 1 + = = . 5 7 5 5 5 5

3.3. Example

Find [2; 1, 4] × [1; 2].

Solution:

Let [1; 2] = 푎0 ; 푎1 and [2 ; 1, 4] = 푏0 ; 푏1,푏2 , m = 1, n = 2, m  n, from Equation (3.1b), we get

[1; 2] [2; 1,4] = 푎0 ; 푎1 × 푏0 ; 푏1, 푏2 = 푑0 ; 푑1, 푑′2 , where 푑′2 is the last term and 푑0 = 푑′0 = 푎0푏0 = 2 ∙ 1 = 2.

푎1푏1 2∙1 2 푑1 = 푑′1 = = = = 0. 푎0푎1+푏0푏1+1 1∙2+2∙1+1 5

푎0푏2푎1+푏0 푏1푏2+1 +푏2 11 푑′2 = = = 2 ; 5 , therefore 1 ; 2 2 ; 1,4 = [2 ; 0, 2, 5] = [4 ; 5] 푎1 푏1푏2+1 −푑′1 푎표 푏2푎1+푏0 푏1푏2 +1 +푏2 5 (from lemma 2.1b).

3 14 21 1 21 To check, we have 1; 2 2; 1,4 = × = and 4; 5 = 4 + = . 2 5 5 5 5

3.4. Example

Find 1 ; 4 2 ; 3, 1, 2, 4 .

Solution:

Let [1; 4] = 푎0 ; 푎1 and 2 ; 3, 1, 2, 4 = 푏0 ; 푏1,푏2, 푏3, 푏4 , we get m = 1, n = 4, m  n, from Equation ′ ′ (3.1b) we have [1; 4] × 2; 3,1,2,4 = 푎0 ;푎1 × 푏0 ;푏1, 푏2,푏3, 푏4 = 푑0 ;푑1, 푑 2, 푑 3, 푑′4 , where 푑′4 is the last term and 푑0 = 푑′0 = 푎0푏0 = 1 ∙ 2 = 2

푎1푏1 4∙3 푑1 = 푑′1 = = = 1 푎0푎1+푏0푏1+1 1∙4+2∙3+1

푎0푏2푎1+푏0 푏1푏2+1 +푏2 4+8+1 13 푑′2 = = = = 4 푎1 푏1푏2+1 −푑′1 푎표 푏2푎1+푏0 푏1푏2+1 +푏2 16−13 14

푎1퐾3 푏1 −[푎표 푎1퐾2 푏2 +푏0퐾3 푏1 +퐾2 푏2 ]푑′1 44−37 73 푑′3 = = = = 0 [푎표 푎1퐾2 푏2 +푏0퐾3 푏1 +퐾2 푏2 ]퐾2 푑′1 −푎1퐾3 푏1 푑′2 185−176 9

[푎표 푎1퐾3 푏2 +푏0퐾4 푏1 +퐾3 푏2 ]퐾2 푑′1 −푎1퐾4 푏1 푑′2 37 푑′4 = = = [1; 5, 6], therefore [1 ; 4] × 2 ; 3, 1, 2, 4 = [2 ; 1, 푎1퐾4 푏1 퐾2 푑′1 −[푎표 푎1퐾3 푏2 +푏0퐾4 푏1 +퐾3 푏2 ]퐾3 푑′1 21 4, 0, 1, 5, 6] = [2 ; 1, 5, 5, 6] (from lemma 3.1b)

5 109 545 1 1 To check, we have [1 ; 4] × 2 ; 3, 1, 2, 4 = × = and 2 ; 1, 5, 5, 6 = 2 + 1 = 2 + 1 = 4 48 192 1+ 1 1+ 6 5+ 5+ 1 31 5+ 6 1 545 2 + 31 = . 1+ 192 161

IV. THE DIVISION OF SIMPLE CONTINUED FRACTIONS

4.1. Definition

1 Let [푎0; 푎1, … , 푎푛 ] be a S.C.F., then we define the multiplicative inverse of [푎0; 푎1, … , 푎푛 ] as = [푎0;푎1,…,푎푛 ]

[0; 푎0, 푎1, … , 푎푛 ].

4.1. Example

Find the multiplicative inverse of [0; 1,1,4].

Solution:

1 From definition 4.1 we have: = 0; 0, 1, 1, 4 = [1; 1, 4] (from lemma 2.1c). [0; 1,1,4]

1 1 9 1 4 9 To check, we have = 5 = and 1; 1,4 = 1 + 1 = 1 + = . [0;1,1,4] 5 1+ 5 5 9 4

4.2. Definition

Let [푎0 ; 푎1, … , 푎푚 ] and [푏0 ; 푏1,… , 푏푛 ] be two S.C.F. we define the Division by [푎0 ; 푎1, … , 푎푚 ] ÷ 1 1 [푏0 ; 푏1, … , 푏푛 ] = [푎0 ; 푎1, … , 푎푚 ] × where = [0; 푏0, 푏1,… , 푏푛 ]. [푏0;푏1,…,푏푛 ] [푏0 ;푏1,…,푏푛 ]

4.2. Example

Find 35; 1, 2, 2 ÷ [3; 1, 1, 3].

Solution:

1 Let 35 ; 1, 2, 2 = 푎 ; 푎 , 푎 , 푎 and = 0 ; 3, 1, 1, 3 = 푏 ; 푏 , 푏 , 푏 ,푏 , we get m = 3, n = 4, m  0 1 2 3 [3;1,1,3] 0 1 2 3 4 n, from Equation (3.1b), we have

35 ; 1, 2, 2 × 0 ; 3, 1, 1, 3 = 푎0 ; 푎1, 푎2, 푎3 × 푏0 ; 푏1,푏2,푏3, 푏4 = 푑0 ; 푑1, 푑2, 푑3, 푑′4 , where 푑′4 is the last term and 푑0 = 푑′0 = 푎0푏0 = 35 ∙ 0 = 0,

푎1푏1 1∙3 3 푑1 = 푑′1 = = = = 0, 푎0푎1+푏0푏1+1 35∙1+0∙3+1 36

푎0푏2(푎1푎2+1)+푎2푏0 푏1푏2+1 +푎2푏2 35∙3+2 107 푑2 = 푑′2 = = = = 8, (푎1푎2+1) 푏1푏2+1 −푑′1 푎표 푏2 (푎1푎2+1)+푎2푏0 푏1푏2+1 +푎2푏2 3∙4 12

퐾3 푎1 퐾3 푏1 퐾0 푑′2 −[푎0퐾3 푎1 퐾2 푏2 +푏0퐾3 푏1 퐾2 푎2 +퐾2 푎2 퐾2 푏2 ]퐾1(푑′1) 7.7 49 푑′3 = = = = 0, [푎0퐾3 푎1 퐾2 푏2 +푏0퐾3 푏1 퐾2 푎2 +퐾2 푎2 퐾2 푏2 ]퐾2 푑′1 −퐾3 푎1 퐾3 푏1 퐾1(푑′2) 35∙7∙2+5∙2 −7∙7∙8 108

[푎0퐾3 푎1 퐾3 푏2 +퐾2 푎2 퐾3 푏2 ]−퐾3 푎1 퐾4 푏1 퐾1(푑′2) 350 푑′4 = = =2, therefore 퐾3 푎1 퐾4 푏1 퐾2(푑′2) 175

35 ; 1, 2, 2 ÷ 3 ; 1, 1, 3 = 35 ; 1, 2, 2 × 0 ; 3, 1, 1, 3 = 0 ; 0, 8, 0, 2 = 8, 0, 2 (from lemma 2.1c) = 10

(from lemma 2.1b).

250 25 To check, we have 35 ; 1, 2, 2 ÷ 3 ; 1, 1, 3 = ÷ = 10. 7 7

4.3. Example

Find 1; 2,2 ÷ 1; 3 .

Solution:

1 Let [1; 2, 2]= 푎 ; 푎 ,푎 = and = 0; 1,3 = 푏 ;푏 ,푏 , we get m = n = 2, and from Equation (3.1a), 0 1 2 [1;3] 0 1 2 we have 1; 2, 2 × 0; 1, 3 = 푎0; 푎1, 푎2 × 푏0; 푏1,푏2 = 푑0; 푑1, 푑2 , where 푑2 is the last term and 푑0 =

푎0푏0 = 1 ∙ 0 = 0

푎1푏1 2∙1 2 푑1 = = = = 0 푎0푎1+푏0푏1+1 1∙2+0∙1+1 3

푎0푏2 푎1푎2+1 +푎2푏0 푏1푏2+1 +푎2푏2 21 푑2 = = = 1; 20 , therefore 1; 2,2 ÷ 1; 3 = 1; 2,2 × 푎1푎2+1 푏1푏2+1 −푑1 푎표 푏2 푎1푎2 +1 +푎2푏0 푏1푏2+1 +푎2푏2 20 0; 1,3 = 0; 0,1,20 = 1; 20 (from lemma 2.1c).

7 4 21 1 21 To check, we have 1; 2,2 ÷ 1; 3 = ÷ = and 1; 20 = 1 + = . 5 3 20 20 20

4.1. Theorem

Let 푎 = 훼0 be an irrational number and define the sequence 푎0, 푎1, 푎2, … recursively by 푎푘 = [훼푘 ], 훼푘+1 = 1 for 푘 = 0,1,2, …. Then  is the value of infinite S.C.F. [푎0; 푎1, 푎2, … ]. 훼푘 −푎푘

For example 3 = [푎0; 푎1, 푎2, 푎3, 푎4, 푎5, … ] = 1; 1, 2, 1, 2, … = [1; 1 , 2 ]. We can use the same operations of finite S.C.F. for infinite S.C.F.

4.4. Example

Find [1; 1 , 2 ] × [2; 4 ].

Solution:

Let 1 ; 1 , 2 = 1 ; 1,2,1,2, … = [푎0; 푎1, 푎2, 푎3, 푎4, … ] and 2; 4 = 2; 4, 4, 4, 4, … = [푏0; 푏1,푏2,푏3, 푏4, … ] from (3.1a) we have [푎0; 푎1, 푎2, 푎3, 푎4, … ] × [푏0 ; 푏1, 푏2,푏3,푏4, … ] = [푑0 ; 푑1, 푑2, 푑3, 푑4, … ], Where

푑0 = 푎0푏0 = 1 ∙ 2 = 2,

푎1푏1 1∙4 4 푑1 = = = = 0, 푎0푎1+푏0푏1+1 1∙1+2∙4+1 10

푎0푏2 푎1푎2+1 +푎2푏0 푏1푏2+1 +푎2푏2 88 푑2 = = = 1, 푎1푎2+1 푏1푏2+1 −푑1 푎표 푏2 푎1 푎2+1 +푎2푏0 푏1푏2+1 +푎2푏2 51

퐾3 푎1 퐾3 푏1 288 푑3 = = = 1, 푎표 퐾3 푎1 퐾2 푏2 +푏0퐾2 푎2 퐾3 푏1 +퐾2 푎2 퐾2 푏2 −퐾3 푎1 퐾3 푏1 푑2 263

푎표 퐾4 푎1 퐾3 푏2 +푏0퐾3 푎2 퐾4 푏1 +퐾3 푎2 퐾3 푏2 −퐾4 푎1 퐾4 푏1 2893 푑4 = = = 6, 퐾4 푎1 퐾4 푏1 퐾2 푑2 − 푎표 퐾4 푎1 퐾3 푏2 +푏0퐾3 푎2 퐾4 푏1 +퐾3 푎2 퐾3 푏2 462



1; 1 , 2 × 2; 4 = [2; 0,1,1,6, … ] = [3; 1,6, … ] (from lemma 2.1b), To check, we have: 1; 1 , 2 × 2; 4 = 3 × 5 = 15, and 1; 1 , 2 × 2; 4 = 3; 1,6, … = 15.

V. CONCLUSION

This paper is the Second part for the operations of the simple continued fractions. In the first part [6] we discovered the definitions of addition and subtractions of simple continued fractions.

In this part we defined the multiplication, multiplicative inverse and the division of the simple continued fractions.

REFERENCES

[1] Barbeau E. “Pell's Equation”. Springer, 2003. [2] Euler L. and John D.B. “Introduction to Analysis of the Infinite”. New York U. a.: Springer, 1988. [3] Kline M. “Mathematical thought from Ancient to Modern times”. New York: Oxford UP. 1972. [4] Moore C.G. “An Introduction to Continued Fractions". The National Council of Teachers of Mathematics: Washington, D.C. 1964. [5] Mugassabi S. and Mistiri F. “Simple Continued Fractions”. (Unpublished work style), 2014. [6] Mugassabi S. and Mistiri F. “The Elementary Arithmetic Operators of Continued Fractions”. Am-Euras. J. Sci. Res., 2015, 10 (5): 251 - 263. [7] Rosen K.H. “Elementary number theory and its applications”. Addison-Wesley Pub. Co. 2005. [8] Viader C., Pelegri, Jaume B. and Lluis B. “On the Concept of Optimality Interval”. UPF Economics & Business Working. 2000, Paper 466. [9] Wall H. S. "Analytic theory of continued fractions". New York: D. Van Nostrand Co. 1948. AUTHOR’S PROFILE

First Author Dr. Souad I. Mugassabi., MSc of Mathematics from department of Mathematics, University of Benghazi, Benghazi, Libya. phD of Quantum information from University of Bradford, UK.

Second Author Dr. Somia M. Amsheri, University of Elmergib, Alkhums, Libya.