A Review On Recent Developments In Neutrosophic Linear Diophantine Equations

Mikail Bal, Gaziantep University, Turkey

Katy D. Ahmad, Islamic university Of Gaza, Palestine

Rozina Ali, Cairo University, Egypt

Corrsepondence: katyon765@gmail.com  

Abstract:

This paper is dedicated to present a wide review study on the recent advantages of neutrosophic linear diophantine equations and number theory.

We revise the neutrosophic linear diophantine equations, refined neutrosophic Diophantine equations, and n-refined neutrosophic linear Diophantine equations. 

Key words: Neutrosophic number theory, neutrosophic linear diophantine equation, n-refined neutrosophic integer.

1. Introduction

Neutrosophy is a new generalization of fuzzy logic presented by Smarandache [1-5]. Neutrosophic sets were very useful in the algebraic strudies such as rings [6-12], modules [13-30], matrices [31-45], and spaces [56-75]. Recently, it is used to deal with Non-Euclidean data sets using Neutrogeometry [76-90] as well as Turiyam set [91-92]. In this way the applications of Neutrosophic Number and its equations is indeed required for multi-decision process. Neutrosophic number theory is a new research field released by many authors. Where we find a study of the foundations of neutrosophic number theory [59,73], and refined number theory [80].

In the literature, the linear Diophantine equations in the neutrosophic systems was studied firstly by Sankari et. al [4], and then they were generalized to refined and n-refined neutrosophic integers [79] and its extensive properties [76-80, 90-92].

In this work, we give the interested reader a wide review on recent developments in the field of neutrosophic number theory, especially the subject of neutrosophic linear Diophantine equations.

 This work may be very useful in future, especially in defining the Turiyam number theory approaches.

Main Discussion

Definition:

Let Z(I)={a+bI; a,b∈Z} be the neutrosophic ring of integers. The neutrosophic linear Diophantine equation with two variables is defined as follows:

AX+BY=C;A,B,C∈Z(I).

Theorem :

Let Z(I)={a+bI; a,b∈Z} be the neutrosophic ring of integers. The neutrosophic linear Diophantine equation  AX+BY=C  with two variables X=x_1+x_2 I,Y=y_1+y_2 I, where

A=a_1+a_2 I,B=b_1+b_2 I is equivalent to the following two classical Diophantine equations:

(1) a_1 x_1+b_1 y_1=c_1.

(2) (a_1+a_2 )(x_1+x_2 )+(b_1+b_2 )(y_1+y_2 )=c_1+c_2.

Proof:

It is sufficient to show that AX+BY=C implies (1) and (2).

AX+BY=C is equivalent to:

(a_1+a_2 I)(x_1+x_2 I)+(b_1+b_2 I)(y_1+y_2 I)=c_1+c_2 I, by easy computing we find

[a_1 x_1+b_1 y_1 ]+[a_1 x_2+a_2 x_1+a_2 x_2+b_1 y_2+b_2 y_1+b_2 y_2 ]I=c_1+c_2 I, hence

a_1 x_1+b_1 y_1=c_1,and a_1 x_2+a_2 x_1+a_2 x_2+b_1 y_2+b_2 y_1+b_2 y_2=c_2. We can see that we get equation (1). For equation (2) we take

 a_1 x_2+a_2 x_1+a_2 x_2+b_1 y_2+b_2 y_1+b_2 y_2=c_2, by adding equation (1) to the two sides we obtain

a_1 x_1+b_1 y_1+a_1 x_2+a_2 x_1+a_2 x_2+b_1 y_2+b_2 y_1+b_2 y_2=c_1+c_2, which implies equation (2)

(a_1+a_2 )(x_1+x_2 )+(b_1+b_2 )(y_1+y_2 )=c_1+c_2.

The following theorem determines the criteria for the solvability of neutrosophic linear Diophantine equation.

Theorem:

Let Z(I)={a+bI; a,b∈Z} be the neutrosophic ring of integers. The neutrosophic linear Diophantine equation  AX+BY=C  with two variables X=x_1+x_2 I,Y=y_1+y_2 I and A=a_1+a_2 I,B=b_1+b_2 I is solvable if and only if gcd⁡(a_1,b_1 ) |c_1, gcd⁡(a_1+a_2,b_1+b_2 ) |c_1+c_2.

Example:

(a) The neutrosophic Diophantine equation (2+2I)X+(3+4I)Y=5+5I is solvable, that is because

gcd⁡(2,3) |5,and gcd⁡(4,7)|10.

(b) The neutrosophic Diophantine equation (2+3I)X+(4+5I)Y=5+I is not solvable, since

gcd⁡(2,4)=2 does not divide 5.

Now, we describe an algorithm to solve a neutrosophic linear Diophantine equation AX+BY=C.

Remark:

Let Z(I)={a+bI; a,b∈Z} be the neutrosophic ring of integers. Consider a neutrosophic linear Diophantine equation  AX+BY=C  with two variables X=x_1+x_2 I,Y=y_1+y_2 I and A=a_1+a_2 I,B=b_1+b_2 I. To solve this equation follow these steps:

(a) Check the solvability of AX+BY=C by Theorem 3.3 .

(b) Solve a_1 x_1+b_1 y_1=c_1.

(c) Solve (a_1+a_2 )(x_1+x_2 )+(b_1+b_2 )(y_1+y_2 )=c_1+c_2.

(d) Compute x_2,y_2.

Example:

The neutrosophic Diophantine equation (2+2I)X+(3+4I)Y=5+5I is solvable.

2x_1+3y_1=5 is a classical linear Diophantine equation. It has a solution x_1=4,y_1=-1.

(2+2)(x_1+x_2 )+(3+4)(y_1+y_2 )=5+5, i.e 4M+7N=10;M=x_1+x_2,N=y_1+y_2. It is a classical linear Diophantine equation with M,N as variables. It has a solution M=-1,N=2.

x_2=M-x_1=-5,y_2=N-y_1=3, thus the equation (2+2I)X+(3+4I)Y=5+5I has a solution X=4-5I,Y=-1+3I.

Definition:

Let Z(I_1,I_2)={(a,bI_1,cI_2); a,b,c∈Z} be the refined neutrosophic ring of integers. The refined neutrosophic linear Diophantine equation with two variables is defined as follows:

AX+BY=C;A,B,C∈Z(I_1,I_2).

Theorem:

Let Z(I_1,I_2)={(a,bI_1,cI_2); a,b,c∈Z} be the refined neutrosophic ring of integers,

AX+BY=C;A,B,C∈Z(I_1,I_2) be a refined neutrosophic linear Diophantine equation, where

X=(x_0,x_1 I_1,x_2 I_2 ),Y=(y_0,y_1 I_1,y_2 I_2 ),A=(a_0,a_1 I_1,a_2 I_2 ),

B=(b_0,b_1 I_1,b_2 I_2 ),C=(c_0,c_1 I_1,c_2 I_2). Then AX+BY=C is equivalent to the following three Diophantine equations:

(1) a_0 x_0+b_0 y_0=c_0.

(2) (a_0+a_2 )(x_0+x_2 )+(b_0+b_2 )(y_0+y_2 )=c_0+c_2.

(3) (a_0+a_1+a_2 )(x_0+x_1+x_2 )+(b_0+b_1+b_2 )(y_0+y_1+y_2 )=c_0+c_1+c_2.

Proof:

By replacing A,B,C,X,Y we find

AX=(a_0,a_1 I_1,a_2 I_2 )(x_0,x_1 I_1,x_2 I_2 ) =

 (a_0 x_0,[a_0 x_1+a_1 x_0+a_1 x_1+a_1 x_2+a_2 x_1 ] I_1,[a_0 x_2+a_2 x_0+a_2 x_2 ] I_2),

BY=(b_0,b_1 I_1,b_2 I_2 )(y_0,y_1 I_1,y_2 I_2 )=

(b_0 y_0,[b_0 y_1+b_1 y_0+b_1 y_1+b_1 y_2+b_2 y_1 ] I_1,[b_0 y_2+b_2 y_0+b_2 y_2 ] I_2), thus the equation

AX+BY=C implies

(*) a_0 x_0+b_0 y_0=c_0. (Equation (1)).

(**)〖 a〗_0 x_2+a_2 x_0+a_2 x_2+b_0 y_2+b_2 y_0+b_2 y_2=c_2.

(***) a_0 x_1+a_1 x_0+a_1 x_1+a_1 x_2+a_2 x_1+b_0 y_1+b_1 y_0+b_1 y_1+b_1 y_2+b_2 y_1=c_1.

By adding (*) to (**) we get (a_0+a_2 )(x_0+x_2 )+(b_0+b_2 )(y_0+y_2 )=c_0+c_2. (Equation (2)).

By adding (2) to (***) we get (a_0+a_1+a_2 )(x_0+x_1+x_2 )+(b_0+b_1+b_2 )(y_0+y_1+y_2 )=c_0+c_1+c_2 . (Equation (3)).

Theorem:

Let Z(I_1,I_2)={(a,bI_1,cI_2); a,b,c∈Z} be the refined neutrosophic ring of integers,

AX+BY=C;A,B,C∈Z(I_1,I_2) be a refined neutrosophic linear Diophantine equation, where

X=(x_0,x_1 I_1,x_2 I_2 ),Y=(y_0,y_1 I_1,y_2 I_2 ),A=(a_0,a_1 I_1,a_2 I_2 ),

B=(b_0,b_1 I_1,b_2 I_2 ),C=(c_0,c_1 I_1,c_2 I_2). Then AX+BY=C is solvable if and only if:

(a) gcd⁡(a_0,b_0 ) |c_0.

(b) gcd⁡(a_0+a_2,b_0+b_2 ) |c_0+c_2.

(c) gcd⁡(a_0+a_1+a_2,b_0+b_1+b_2 ) |c_0+c_1+c_2.

Example:

(a) Consider the refined neutrosophic linear Diophantine equation

(1,2I_1,3I_2 ).X+(3,3I_1,8I_2 )Y=(2,〖4I〗_1,I_2), we have

gcd⁡(1,3)=1|2, gcd⁡(1+3,3+8)=gcd⁡(4,11)=1|(2+1=3),

gcd⁡(1+2+3,3+3+8)=gcd⁡(6,14)=2 which does not divide 2+4+1=7, thus it is not solvable.

(b) Consider the refined neutrosophic linear Diophantine equation

(1,2I_1,3I_2 ).X+(3,3I_1,8I_2 )Y=(2,〖4I〗_1,〖2I〗_2), we have

gcd⁡(1,3)=1|2, gcd⁡(1+3,3+8)=gcd⁡(4,11)=1|(2+2=4),

gcd⁡(1+2+3,3+3+8)=gcd⁡(6,14)=2| (2+4+2=8). Thus it is solvable.

Remark:

 Let Z(I_1,I_2)={(a,bI_1,cI_2); a,b,c∈Z} be the refined neutrosophic ring of integers,

AX+BY=C;A,B,C∈Z(I_1,I_2) be a refined neutrosophic linear Diophantine equation, where

X=(x_0,x_1 I_1,x_2 I_2 ),Y=(y_0,y_1 I_1,y_2 I_2 ),A=(a_0,a_1 I_1,a_2 I_2 ),

B=(b_0,b_1 I_1,b_2 I_2 ),C=(c_0,c_1 I_1,c_2 I_2), we summarize the algorithm of solution as follows:

(a) Check the solvability condition.

(b) Solve a_0 x_0+b_0 y_0=c_0.

(c) Solve (a_0+a_2 )(x_0+x_2 )+(b_0+b_2 )(y_0+y_2 )=c_0+c_2.

(d) Compute x_2,y_2.

(e) Solve (a_0+a_1+a_2 )(x_0+x_1+x_2 )+(b_0+b_1+b_2 )(y_0+y_1+y_2 )=c_0+c_1+c_2.

(f) Compute x_1,y_1.

Example:

we found that (1,2I_1,3I_2 ).X+(3,3I_1,8I_2 )Y=(2,〖4I〗_1,〖2I〗_2) is solvable.

We consider  x_0+3y_0=2. It has a solution x_0=-1,y_0=1.

We take (1+3)(x_0+x_2 )+(3+8)(y_0+y_2 )=2+2, i.e 4M+11N=4;M=x_0+x_2,and

N=y_0+y_2, it has a solution M=1,N=0, thus x_2=M-x_0=2,y_2=N-y_0=-1.

The third equation is (1+2+3)(x_0+x_1+x_2 )+(3+3+8)(y_0+y_1+y_2 )=2+4+2, i.e

6S+14T=8;S=x_0+x_1+x_2,T=y_0+y_1+y_2. It has a solution S=-1,T=1, thus

x_1=S-x_0-x_2=-2,y_1=T-y_0-y_2=1. The solution of

(1,2I_1,3I_2 ).X+(3,3I_1,8I_2 )Y=(2,〖4I〗_1,〖2I〗_2) is X=(-1,-2I_1,2I_2), Y=(1,I_1,-I_2).

Definition:

Let Z_n (I)={t_0+t_1 I_1+⋯+t_n I_n; t_i∈Z}  be the n-refined neutrosophic ring of integers. The following equation

AX+B=C; A,B,X,C∈Z_n (I). Where A=a_0+a_1 I_1+⋯+a_n I_n , B=b_0+b_1 I_1+⋯+b_n I_n,X=x_0+x_1 I_1+⋯+x_n I_n, C=c_0+c_1 I_1+⋯+c_n I_n

is called an n-refined neutrosophic linear Diophantine equation.

Example :

Let n=3, the following equation is an 3-refined neutrosophic linear Diophantine equation

(1-I_1-I_2 )X+(2+3I_2-4I_3 )=I_2+2I_3 .

Theorem:

Let AX+B=C (*); A,B,X,C∈Z_n (I)  be an n-refined neutrosophic linear Diophantine equation. It is solvable if and only if the following system of classical linear Diophantine equations is solvable.

(1-) a_0 x_0+b_0=c_0 .

(2-) (a_0+a_n )(x_0+x_n )+(b_0+b_n )=c_0+c_n .

(3-) (a_0+a_n+a_(n-1) )(x_0+x_n+x_(n-1) )+(b_0+b_n+b_(n-1) )=c_0+c_n+c_(n-1) .

.

.

(n+1-) (a_0+a_1+⋯+a_n )(x_0+x_1+⋯+x_n )+(b_0+b_1+⋯+b_n )=c_0+c_1+⋯+c_n.

Theorem:

The sufficient and necessary condition of the solvability of Diophantine equation (*) is:

gcd⁡(a_0,b_0 ) |c_0,gcd⁡(a_0+a_n,b_0+b_n)|(c_0+c_n ),gcd⁡(a_0+a_n+a_(n-1),b_0+b_n+b_(n-1))|(c_0+c_n+c_(n-1) ),…gcd⁡(a_0+a_1+⋯+a_n,b_0+b_1+⋯+b_n ),|(c_0+c_1+⋯+c_n ).

Example:

Consider the following 3-refined neutrosophic linear Diophantine equation:

(1-I_2+I_3 )X+(I_2+2I_3 )=2-I_1+4I_3.

We have: A=1-I_2+I_3,B=I_2+2I_3,C=I_1+4I_3,i.e.a_0=1,a_1=0,a_2=-1,a_3=1.

b_0=0,b_1=0,b_2=1,b_3=2,c_0=0,c_1=1,c_2=0,c_3=4.

We have : gcd⁡(a_0,b_0 )=1|0,gc d⁡(a_0+a_3,b_0+b_3 )=gcd⁡(2,2)=2|4 ,gc d⁡(a_0+a_3+a_2,b_0+b_3+b_2 )=gcd⁡(1,3)=1|4 ,gcd⁡(a_0+a_1+a_2+a_3,b_0+b_1+b_2+b_3 )=gcd⁡(1,3)=1|5.  This implies that the previous linear Diophantine equation is solvable.

Now, we find the solution.

The equivalent system is:

a_0 x_0+b_0=c_0,thus x_0=0.(1)

(a_0+a_3 )(x_0+x_3 )+(b_0+b_3 )=c_0+c_3,thus 2(x_0+x_3 )+2=4.(2)

(a_0+a_3+a_2 )(x_0+x_3+x_2 )+(b_0+b_3+b_2 )=c_0+c_3+c_2, thus (x_0+x_3+x_2 )+3=4. (3)

 (a_0+a_3+a_1+a_2 )(x_0+x_3+x_1+x_2 )+(b_0+b_3+b_1+b_2 )=c_0+c_3+c_1+c_2, thus (x_0+x_1+x_3+x_2 )+3=5. (4)

The equation (1) has a solution x_0=0. The equation (2) has a solution x_0+x_3=1,hence x_3=1.

The equation (2), has a solution x_0+x_3+x_2=1,hence x_2=0. The equation (4) has a solution x_0+x_1+x_3+x_2=2,hence x_1=1.

The previous discussion means that the solution of the first n-refined neutrosophic linear Diophantine equation is X=I_1+I_3 .

Conclusion

In this work, we gave the interested reader a wide review on recent developments in the field of neutrosophic number theory, especially the subject of neutrosophic linear Diophantine equations.

As a future research direction, we aim to generalize the previous efforts to define and study Turiyam linear Diophantine equations.

Acknowledgements: Author thanks the editorial team for the valuable time. 

Funding :Author declares that, there is no funding for this paper. 

Conflicts of Interest: Author declares that, there is no conflict of interest.

Ethics approval: This article does not contain any studies with human or animals participants.

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[54] Abobala, M., "On Some Algebraic Properties of n-Refined Neutrosophic Elements and n-Refined Neutrosophic Linear Equations", Mathematical Problems in Engineering, Hindawi, 2021

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[58] F. Smarandache, Neutrosophic Theory and Applications, Le Quy Don Technical University, Faculty of Information technology, Hanoi, Vietnam, 17th May 2016.

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