Neutrosophic Duplets of {Z_pn,×} and {Z_pq,×} and Their Properties

Abstract

The notions of neutrosophy, neutrosophic algebraic structures, neutrosophic duplet and neutrosophic triplet were introduced by Florentin Smarandache. In this paper, the neutrosophic duplets of $Z_{p^n}$, $Z_{pq}$ and $Z_{p_1{p}_2\dots p_n}$ are studied. In the case of $Z_{p^n}$ and $Z_{pq}$, the complete characterization of neutrosophic duplets are given. In the case of $Z_{p_1\dots p_n}$, only the neutrosophic duplets associated with $p_i$s are provided; $i=1,2,\dots ,n$. Some open problems related to neutrosophic duplets are proposed.

1. Introduction

Real world data, which are predominately uncertain, indeterminate and inconsistent, were represented as neutrosophic set by Smarandache [1]. Neutrosophy deals with the existing neutralities and indeterminacies of the problems. Neutralities in neutrosophic algebraic structures have been studied by several researchers [1,2,3,4,5,6,7,8]. Wang et al. [9] proposed Single-Valued Neutrosophic Set (SVNS) to overcome the difficulty faced in relating neutrosophy to engineering discipline and real world problems. Neutrosophic sets have evolved further as Double Valued Neutrosophic Set (DVNS) [10] and Triple Refined Indeterminate Neutrosophic Set (TRINS) [11]. Neutrosophic sets are useful in dealing with real-world indeterminate data, which Intuitionistic Fuzzy Set (IFS) [12] and Fuzzy sets [13] are incapable of handling accurately [1].

The current trends in neutrosophy and related theories of neutrosophic triplet, related triplet group, neutrosophic duplet, and duplet set was presented by Smarandache [14]. Neutrosophic duplets and neutrosophic triplets have been of interest and many have studied them [15,16,17,18,19,20,21,22,23,24]. Neutrosophic duplet semigroup were studied in [19] and the neutrosophic triplet group was introduced in [8]. Neutrosophic duplets and neutrosophic duplet algebraic structures were introduced by Smarandache.

In the case of neutrosophic duplets, we see $ax=a$ and $x=neut\left(a\right)$, where, as in L-fuzzy sets [25] as per definition is a mapping from $A:X\to L$, L may be semigroup or a poset or a lattice or a Boolean $\sigma$-ring; however, neutrosophic duplets are not mapping, more so in our paper algebraic properties of them are studied for $Z_n$ for specific values of n. However, in the case of all structures, the semigroup or lattice or Boolean $\sigma$-ring or a poset, there are elements which are neutrosophic duplets. Here, we mainly analyze neutrosophic duplets in the case of $Z_n$ only number theoretically.

In this paper, we investigate the neutrosophic duplets of $\{Z_{p^n},\times \}$, where p is a prime (odd or even) and $n\ge$ 2. Similarly, neutrosophic duplets in the case of $Z_{pq}$ and $Z_{p_1{p}_2\dots p_n}$ are studied. It is noted that the major difference between the neutrals of neutrosophic triplets and that of neutrosophic duplets is that in the former case they are idempotents and in the latter case they are units. Idempotents in the neutrosophic duplets are called trivial neutrosophic duplets.

This paper is organized as five sections, Section 1 is introductory in nature and Section 2 provides the important results of this paper. Neutrosophic duplets in the case of $Z_{p^n}$; p an odd prime are studied in Section 3. In Section 4, neutrosophic duplets of $Z_{pq}$ and $Z_{p_1{p}_2\dots p_n}$, and their properties are analyzed. Section 5 discusses the conclusions, probable applications and proposes some open problems.

2. Results

The basic definition of neutrosophic duplet is recalled from [8].

Consider U to be the universe of discourse, and D a set in U, which has a well-defined law #.

Definition 1.

Consider $⟨a,neut\left(a\right)⟩$, where a, and $neut\left(a\right)$ belong to D. It is said to be a neutrosophic duplet if it satisfies the following conditions:

1. 

$neut\left(a\right)$ is not the same as the unitary element of D in relation with the law # (if any);

2. 

$a\#$$neut\left(a\right)$ = neut(a) # a = a; and

3. 

$anti\left(a\right)\notin$ D for which a # anti(a) = anti(a) # a = neut(a).

Here, the neutrosophic duplets of $\{Z_{p^n},\times \}$, p is a prime (odd or even) and $n\ge$ 2 are analyzed number theoretically. Similarly, neutrosophic duplets in the case of $Z_{pq}$ and $Z_{p_1{p}_2\dots p_n}$ are studied in this paper.

The results proved by this study are:

• The neutrals of all nontrivial neutrosophic duplets are units of $\{Z_{p^n},\times \}$, $\{Z_{pq},\times \}$ and $\{Z_{p_1{p}_2\dots p_n},\times \}$.
• If p is a prime in anyone of the semigroups ($\{Z_{p^n},\times \}$ or $\{Z_{pq},\times \}$ or $\{Z_{p_1{p}_2\dots p_n},\times \}$) as mentioned in 1, then $mp$ has only p number of neutrals, for the appropriate m.
• The neutrals of any $m{p}^t$ for a prime p; $(m,p)=1$ are obtained and they form a special collection.

3. Neutrosophic Duplets of $\{Z_{p^n},\times \}$ and its Properties

Neutrosophic duplets and neutrosophic duplet algebraic structures were introduced by Florentin Smarandache in 2016. Here, we investigate neutrosophic duplets of $\{Z_{p^n},\times \}$, where p is a prime (odd or even) and $n\ge$ 2. First, neutrosophic duplets in the case of $Z_{2^4}$ and $Z_{3^3}$ and their associated number theoretic properties are explored to provide a better understanding of the theorems proved. Then, several number theoretical properties are derived.

Example 1.

Let $S=\{Z_{16},\times \}$ be the semigroup under × modulo 16. $Z_{16}$ has no idempotents. The units of $Z_{16}$ are $\{1,3,5,7,9,11,13,15\}$. The elements which contribute to the neutrosophic duplets are $\{2,4,6,8,10,12,14\}$. The neutrosophic duplet sets under usual product modulo 16 are:

$$\left\{\right\{2,1\},\{2,9\left\}\right\},\left\{\right\{4,1\},\{4,5\},\{4,9\},\{4,13\left\}\right\},$$

$$\left\{\right\{6,1\},\{6,9\left\}\right\},\left\{\right\{8,1\},\{8,3\},\{8,5\},\{8,7\},\{8,9\},\{8,11\},\{8,13\},\{8,15\left\}\right\},$$

$$\left\{\right\{10,1\},\{10,9\left\}\right\},\left\{\right\{12,1\},\{12,5\},\{12,9\},\{12,13\left\}\right\},\left\{\right\{14,1\},\{14,9\left\}\right\}$$

The observations made from this example are:

1. 

Every non-unit of $Z_{16}$ is a neutrosophic duplet.

2. 

Every non-unit divisible by 2, viz. $\{2,6,10,14\}$, has only $\{1,9\}$ as their neutrals.

3. 

Every non-unit divisible by 4 are 4 and 12, which has $\{1,5,9,13\}$ as neutrals.

The biggest number which divides 16 is 8 and all units act as neutrals in forming neutrosophic duplets. Thus, $A=\{1,3,5,$$7,9,11,13,15\}$, which forms a group of order 8, yields the 8 neutrosophic duplets; $8\times i=8$ for all $i\in A$ and A forms a group under multiplication modulo 16; and $\{1,9\}$ and $\{1,5,9,13\}$ are subgroups of A.

In view of this, we have the following theorem.

Theorem 1.

Let $S=\{Z_{2^n},\times \}$, be the semigroup under product modulo $2^n$, $n\ge 2$.

(i) 

The set of units of S are $A=\{1,3,5,\dots ,2^n-1\}$, forms a group under × and $\left|A\right|=2^{n-1}$.

(ii) 

The set of all neutrosophic duplets with $2^{n-1}$ is A; neutrals of $2^{n-1}$ are A.

(iii) 

All elements of the form $2m\in Z_{2^n}$ (m an odd number) has only the elements $\{1,2^{n-1}+1\}$ to contribute to neutrosophic duplets (neutrals are $1,2^{n-1}+1$).

(iv) 

All elements of the form $m{2}^t\in Z_{2^n};1<t<n-1$; m odd has its neutrals from $B=\{1,2^{n-t}+1$, $2^{n-t+1}+1$, $2^{n-t+2}+1$, $\dots ,2^{n-1}+1,2^{n-t}+2^{n-t+1}+1,\dots ,2^{n-t}+2^{n-1}+1,\dots ,$$1+2^{n-t}+2^{n-t+1}+\dots +2^{n-1}\}$.

Proof. 

(i)

Given $S=\{Z_{2^n},\times \}$ where $n\ge 2$ and S is a semigroup under product modulo $2^n$. $A=\{1,3,5,7,\dots ,2^n-1\}$ is a group under product as every element is a unit in S and closure axiom is true by property of modulo integers and $\left|A\right|=2^{n-1}$. Hence, Claim (i) is true.

(ii)

Now, consider the element $2^{n-1}$; the set of duplets for $2^{n-1}$ is A for $2^{n-1}\times 1=2^{n-1}$; $2^{n-1}\times 3=2^{n-1}[2+1]$ = $2^n+2^{n-1}=2^{n-1},\dots ,2^{n-1}\left(m\right);$ (m is odd) will give only $m{2}^{n-1}$. Hence, this proves Claim (ii).

(iii)

Consider $2m\in Z_{2^n}$; we see $2m\times 1=2m$ and $2m(2^{n-1}+1)=m{2}^n+2m=2m$. $(2m,2^{n-1}+1$) is a neutrosophic duplet pair; hence, the claim.

(iv)

Let $m{2}^t\in Z_{2^n}$; clearly, $m{2}^t\times x=m{2}^t$ for all $x\in B$.

☐

Next, we proceed onto describe the duplet pairs in $S=\{Z_{3^3},\times \}$.

Example 2.

Let $S=\{Z_{3^3},\times \}$ be a semigroup under product modulo $3^3$. The units of S are $A=\{1,2,4,5,7,8,$$10,11,13,14,16,17,19,20,22,23,25,26\}$. Clearly, A forms a group under a product. The non-units of S are $\{3,6,9,12,15,18,21,24\}$. Zero can be included for $0\times x=0$ for all $x\in S$, in particular for $x\in A$. The duplet pairs related to 3 are $B_1=\{\{3,1\},\{3,10\},\{3,19\}\}$. The duplet pairs related to 6 are $B_2=\{\{6,1\},\{6,10\},\{6,19\}\}$. The duplet pairs related to 9 are

$$B_3=\{\{9,1\},\{9,4\},\{9,7\},\{9,13\},\{9,10\},\{9,16\},\{9,19\},\{9,22\},\{9,25\}\}.$$

The neutrosophic duplets of 12 are $B_4=\{\{12,1\},\{12,10\},\{12,19\}\}.$ The neutrosophic duplets of 15 are $B_5=\{\{15,1\},\{15,10\},\{15,19\}\}.$ Finally, the neutrosophic duplets of 18 are

$$B_6=\{\{18,1\},\{18,4\},\{18,7\},\{18,13\},\{18,10\},\{18,16\},\{18,19\},\{18,22\},\{18,25\}\}.$$

The neutrosophic duplets associated with 21 are $B_7=\{\{21,1\},\{21,10\},\{21,19\}\}$ and 24 are $B_8=\{\{24,1\},\{24,10\},$$\{24,19\}\}.$ Now, the trivial duplet of 0, which we take is

$$B_0=\{\{0,1\},\{0,4\},\{0,7\},\{0,13\},\{0,10\},\{0,16\},\{0,19\},\{0,22\},\{0,25\}\}.$$

We see $L=\{B_0\cup B_1\cup B_2\cup \dots \cup B_8\}$ forms a semigroup under product modulo 27 and $o\left(L\right)=45$.

We have the following result.

Theorem 2.

Let S = $\{Z_{p^n},\times \}$, where p is an odd prime, $n\ge 2$ is a semigroup under ×, and product modulo is $p^n$. The units of S are denoted by A and non-units of S are denoted by B. The neutrosophic duplets of S associated with B are groups under product and are subgroups of A. The neutrals of $t{p}^s=b\in B$ are of the form $D=\{1,1+p^{n-s},1+p^{n-s+1},1+p^{n-s+2},\dots ,1+p^{n-1},1+p^{n-s}+p^{n-s+1}$, $1+p^{n-s}+p^{n-s+2},\dots ,1+p^{n-1}+p^{n-s},\dots 1+p^{n-s}+\dots +p^{n-1}\}$; $1\le t<m,p/m;1<s<n$.

Proof. 

Let $t{p}^s\in Z_{p^n}$ all elements which act as neutrosophic duplets for $t{p}^s$ are from the set D. For any $x\in D$ and $t{p}^s\in Z_{p^s}$, we see $xt{p}^s=t{p}^s$; hence, the claim.  □

It is important to note that $S=\{Z_{p^n},\times \}$ has no non-trivial neutrosophic triplets as $Z_{p^n}$ has no non-trivial idempotents.

Next, we proceed to finding the neutrosophic duplets of $Z_{pq}$; p and q are distinct primes.

4. Neutrosophic Duplets of $Z_{pq}$ and $Z_{p_1{p}_2\dots p_n}$

In this section, we study the neutrosophic duplets of $Z_{pq}$ where p and q are primes. Further, we see $Z_{pq}$ also has neutrosophic triplets. The neutrosophic triplets in the case of $Z_{pq}$ have already been characterized in [23]. We find the neutrosophic duplets of $Z_{2p}$, p a prime. We find the neutrosophic duplets and neutrosophic triplets groups of $Z_{26}$ in the following.

Example 3.

Let $S=\{Z_{26},\times \}$ be the semigroup under product modulo 26. The idempotents of S are 13 and 14. We see 13 is just a trivial neutrosophic triplet, however only 14 contributes to non-trivial neutrosophic triplets. We now find the neutrosophic duplets of $Z_{26}$. The units of $Z_{26}$ are $A=\{1,3,5,7,9,11,15,17,19,21,23,25\}$ and they act as neutrals of the duplets. The non-units which contribute for neutrosophic duplets are $B=\{2,4,6,8,10,12,13,14,16,18,20,22,24\}$. 0 is the trivial duplet as $0\times x=0$ for all $x\in A$. Consider $2\in B$ the pairs of duplets are $\{2,1\},2\times 14=2$ but 14 cannot be taken as $anti\left(2\right)=20$ and $anti\left(2\right)$ exists so 2 is not a neutrosophic duplet for $(2,14,20)$ is a neutrosophic triplet group.

Consider $4\in B;\{4,1\}$ is a trivial neutrosophic duplet. Then, $4\times 14=4$ and $(4,14,16)$ are again a neutrosophic triplet as $anti\left(4\right)=16$ so 4 is not a neutrosophic duplet. Thus, 16 and 20 are also not neutrosophic duplets. Consider $6\in B$; we see $\{6,1\}$ is a non-trivial neutrosophic duplet. In addition, $(6,14,10)$ are neutrosophic triplet groups so 6 and 10 are not non-trivial neutrosophic duplets. Consider $8\in B,(8,14,18)$ is a neutrosophic triplet group. hence 8 and 18 are not neutrosophic duplets. Then, $(12,14,12)$ is also a neutrosophic triplet group. Thus, 12 is not a neutrosophic duplet. Let $22\in B$ be such that $(22,14,24)$ is a neutrosophic triplet group, hence 22 and 24 are not neutrosophic duplets.

Consider $13\in B$; we see the neutrals are $\{1,3,5,7,9,11,15,17,19,21,23,25\}$. We see the collection of neutrosophic duplets associated with $13\in Z_{26}$ happens to yield a semigroup under product if 13 is taken as the trivial neutrosophic duplets, as it is an idempotent in $Z_{26}$, and, in all pairs, it is treated as semigroup of order 13, where $(13,1)$ and $(13,13)$ are trivial neutrosophic duplets.

In view of this, we have the following theorem.

Theorem 3.

Let $S=\{Z_{2p},\times \}$ be a semigroup under product modulo $2p$; p an odd prime. This S has only p and $p+1$ to be the idempotents and only p contributes for a neutrosophic duplet collection with all units of $Z_{2p}$ and the collection $B=\left\{(p,x)\right|x\in Z_{2p}\}$, x is a unit in $Z_{2p}$ forms a commutative semigroup of order p which includes 1 and p which result in the trivial duplets pair $(p,1)$ and $(p,p).$

Proof. 

Given $S=\{Z_{2p},\times \}$ is a semigroup under × and p is an odd prime. We see from [23] p and $p+1$ are idempotents of $Z_{2p}$. It is proven in [23] that $p+1$ acts for the neutrosophic triplet group of $Z_{2p}$ (formed by elements $2,4,6,\dots ,2p-2$) as the only neutral. $(p,p,p)$ is a trivial neutrosophic triplet. However, $Z_{2p}$ has no neutrosophic duplet other than those related with p alone and $p\times x=p$ for all x belonging to the collection of all units of $Z_{2p}$ including 1. If x is a unit in $Z_{2p}$, two things are essential: x is odd and $x\ne p$. Since x is odd, we see $x=2y+1$ and $p\left(x\right)=p(2y+1)=2yp+p=p$, hence $(p,x)$ is a neutrosophic duplet. The units of $Z_{2p}$ are $(p-1)$ in number. Further, $(p,p)$ and $(p,1)$ form trivial neutrosophic duplets. Thus, the collection of all neutrosophic duplets $B=\left\{\right(p,x\left)\right\}$, x is a unit and $x=p$ is also taken to form the semigroup of order p and is commutative as the collection of all odd numbers forms a semigroup under product modulo $2p$; hence, the claim.  □

It is important and interesting to note that, unlike $Z_{p^n}$, p is a prime and $n\ge 2$. We see $Z_{2p}$ has both non-trivial neutrosophic triplet groups which forms a classical group [23] as well as has a neutrosophic duplet which forms a semigroup of order p.

Next, we study the case when $Z_{pq}$ is taken where both p and q are odd primes first by an example.

Example 4.

Let $S=\{Z_{15},\times \}$ be a semigroup under product. The idempotents of $Z_{15}$ are 10 and 6. However, 10 does not contribute to non-trivial neutrosophic triplet groups other than $\{5,10,5\}$, $\{10,10,10\}$. The neutrosophic triplet groups associated with 6 are $(3,6,12),(12,6,3),$$(9,6,9)$ and $(6,6,6)$. The neutrosophic duplets of $Z_{15}$ are contributed by $\left\{5\right\},\left\{10\right\}$ and $\{3,12,6,9\}$ in a unique way.

$$D_1=\{\{5,1\},\{5,4\},\{5,7\},\{5,13\},\{5,10\}\},$$

$$D_2=\{\{10,13\},\{10,7\},\{10,1\},\{10,4\},\{10,10\}\},$$

$$D_3=\{\{3,11\},\{3,1\},\{3,6\},\{12,11\},\{12,1\},\{12,6\},\{6,11\},\{6,1\},\{6,6\},\{9,11\},\{9,1\},\{9,6\}\}$$

All three collections of duplets put together is not closed under ×; however, $D_2$ and $D_3$ form a semigroup under product modulo 15. If we want to make $D_1$ a semigroup, we should adjoin the trivial duplets $\{0,4\},$$\{0,7\},\{0,13\},\{0,1\},\{0,6\},\{0,10\}$ as well as $D_2$. Further, we see $D_1\cup D_2\cup D_3$ is not closed under product.

Thus, the study of $Z_{pq}$ where p and q are odd primes happens to be a challenging problem. We give the following examples in the case when $p=5$ and $q=7$.

Example 5.

Let $S=\{Z_{35},\times \}$ be a semigroup of order 35. The idempotents of $Z_{35}$ are 15 and 21. The neutrosophic triplets associated with 15 are $\left\{\right(15,15,15),(5,15,10),(25,15,30)$, $(20,15,20),(30,15,25),$$(10,15,5)\}$, a cyclic group of order six. The cyclic group contributed by the neutrosophic triplet groups associated with 21 is as follows: $\left\{\right(21,21,21),(7,21,28),(28,21,7),(14,21,14\left)\right\}$, which is of order four. The neutrosophic duplets are tabulated in Table 1. Similarly, the neutrosophic duplets associated with $S=\{Z_{105},\times \}$ are tabulated in Table 2.

Table 1. Neutrosophic Duplets of $\{Z_{35},\times \}$.

Neutrals for duplets	Neutrals for duplets
5, 10, 15, 20, 25, 30	7, 14, 21, 28
1, 8, 15, 22, 24	1, 6, 11, 16, 21, 26, 31

Table 2. Neutrosophic Duplets of $\{Z_{105},\times \}$.

Neutrals for duplets	Neutrals for duplets
3, 6, 9, 12, 18, 21, 24, 27,	5, 10, 20, 25, 40, 50,
33, 36, 39, 48, 51, 54, 57, 66,	55, 65, 80, 85, 95, 100
69, 78, 81, 87, 93, 96, 99, 102
1, 36, 71	1, 22, 43, 64, 84
Neutrals for duplets	Neutrals for duplets
7, 14, 28, 49, 56, 77, 91, 98	15, 30, 45, 60, 75, 90
1, 16, 31, 46, 61, 76, 91	1, 8, 15, 22, 29, 36, 43, 50,
	57, 64, 71, 78, 85, 92, 99
Neutrals for duplets	Neutrals for duplets
21, 42, 63, 84	35, 70
1, 6, 11 16, 21, 26, 31, 36,	1, 4, 7, 10, 13, 16, 19, 22, 25, 28,
41, 46, 51, 56, 61, 66, 71,	31, 34, 37, 40, 43, 46, 49, 52, 55,
76, 81, 86, 91, 96, 101	58, 61, 64, 67, 70, 73, 76, 79,
	82, 85, 88, 91, 94, 97, 100, 103

Theorem 4.

Let $\{Z_n,\times \}$ be a semigroup under product modulo n; $x\in Z_n\setminus \left\{0\right\}$ has a neutral $y\in Z_n\setminus \left\{0\right\}$ or is a non-trivial neutrosophic duplet if and only if x is not unit in $Z_n$.

Proof. 

$x\in Z_n\setminus \left\{0\right\}$ is a neutrosophic duplet if $x\times y=x\left(modn\right)$ and y is called the neutral of x. If $x^2=x$, then we call the pair $(x,x)$ as trivial neutrosophic duplet pair. We see $x\times y=x$, if x is a unit in $Z_n$, then there exists a $z\in Z_n$ such that $z\times (x\times y)=z\times x,$ so that $y=1$ as $z\times x=1\left(modn\right)$; so $y=1$ gives trivial neutrosophic duplets. Thus, x is not a unit if it has to form a non-trivial neutrosophic duplet pair; $x\times y=x$ and $y\ne 1$ then if x is a unit we arrive at contradiction; hence, the theorem.  □

Theorem 5.

Let $S=\{Z_{pq},\times \}$ be a semigroup under product modulo $pq$, p and q distinct odd primes. There is p number of neutrosophic duplets for every $p,2p,3p,\dots ,(q-1)p$. Similarly, there is q number of neutrosophic duplets associated with every $q,2q,\dots (p-1)q.$ The neutrals of $sq$ and $tp$ is given by $1+nq$ for $1\le t\le q-1,0\le n\le p-1$ and that of sq is given by $1+mp$; $1\le s\le p-1,0\le m\le q-1$.

Proof. 

Given $\{Z_{pq},\times \}$ is a semigroup under product modulo $pq$ (p and q two distinct odd primes). The neutrals associated with any $tp;1\le t\le q-1$ is given by the sequence $\{1+q,2q+1,3q+1,\dots ,(p-1)q+1\}$ for every $tp\in \{p,2p,\dots ,(q-1\left)p\right\}.$ We see, if $tp\in Z_{pq}$,

$$tp\times (1+nq)=tp+tpnq$$

$$=tp+tnpq=tp\left(modpq\right).$$

A similar argument for $sq$ completes the proof; hence, the claim.  □

Theorem 6.

Let $S=\{Z_{p_1{p}_2\dots p_n},\times \}$ be the semigroup under product modulo $p_1{p}_2\dots p_n$, where $p_1,p_2,\dots ,p_n$ are n distinct primes. The duplets are contributed by the non-units of S. The neutrosophic duplets associated with $A_i=\{p_i,2p_i,\dots ,(p_1{p}_2\dots p_{i-1}{p}_{i+1}\dots p_n-1)p_i\}$ are $\{1+\left(p_1{p}_2\dots p_{i-1}{p}_{i+1}\dots p_n\right)t\}$ where $t=1,2,\dots ,p_i-1$; and $i=1,2,\dots ,n$. Thus, every element $x_i$ of $A_i$ has only $p_i-1$ number of elements which neutralizes $x_i$; thus, using each $x_i$, we have $p_i-1$ neutrosophic duplets.

Proof. 

Given $S=\{Z_{p_1{p}_2\dots p_n},\times \}$ is a semigroup under product modulo $p_1\dots p_n$, where $p_i$s are distinct primes, $i=1,2,\dots ,n.$ Considering $A_i=\{p_i,2p_i,\dots ,(p_1{p}_2\dots p_{i-1}{p}_{i+1}\dots p_n-1)p_i\}$, we have to prove that, for any $s{p}_i$, $s{p}_i\times [1+\left(p_1{p}_2\dots p_{i-1}{p}_{i+1}\dots p_n\right)t]=s{p}_i$; $1\le t\le p_{i-1}.$

Clearly,

$$s{p}_i\times [1+\left(p_1{p}_2\dots p_{i-1}{p}_{i+1}\dots p_n\right)t]=s{p}_i+s{p}_i\left[\left(p_1{p}_2\dots p_{i-1}{p}_{i+1}\dots p_n\right)t\right]$$

$$=s{p}_i+st\left[\left(p_1{p}_2\dots p_{i-1}{p}_i{p}_{i+1}\dots p_n\right)\right]=s{p}_i$$

as $p_1{p}_2\dots p_n=0\left(mod\left(p_1{p}_2\dots p_n\right)\right).$ Hence, the claim.  □

Thus, for varying t and varying s given in the theorem, we see

$$\{s{p}_i,(1+\left(p_1{p}_2\dots p_{i-1}{p}_{i+1}\dots p_n\right)t)\}$$

is a neutrosophic duplet pair $1\le t\le p_i-1$; $1\le s\le p_1{p}_2\dots p_{i-1}{p}_{i+1}\dots p_n$ and $i=1,2,\dots ,n$.

5. Discussions and Conclusions

This paper studies the neutrosophic duplets in the case $Z_{p^n}$, $Z_{pq}$ and $Z_{p_1{p}_2\dots p_n}$. In the case of $Z_{p^n}$ and $Z_{pq}$, a complete characterization of them is given; however, in the case $Z_{p_1\dots p_n}$, only the neutrosophic duplets associated with $p_i$s are provided; $i=1,2,\dots n$. Further, the following problems are left open:

• For $Z_{pq},$p and q odd primes, how many neutrosophic duplet pairs are there?
• For $Z_{p_1\dots p_n}$, what are the neutrals of $p_i{p}_j,p_i{p}_j{p}_k,\dots ,p_1{p}_2\dots p_{i-1}{p}_{i+1}\dots p_n$?
• The study of neutrosophic duplets of $Z_{p_1^{t_1}{p}_2^{t_2}\dots p_n^{t_n}}$; $p_1,\dots ,p_n$ are distinct primes and $t_i\ge 1;1\le i\le n$ is left open.

For future research, one can apply the proposed neutrosophic duplets to SVNS, DVNS or TRINS. These neutrosophic duplets can be applied in problems where neutral elements for a given a in $Z_{p^n}$ or $Z_{pq}$ happens to be many. However, the concept of $anti\left(a\right)$ does not exist in the case of neutrosophic duplets. Finally, these neutrosophic duplet collections form a semigroup only when all the trivial neutrosophic duplet pairs $(0,a)$ for all appropriate a are taken. These neutrosophic duplets from $Z_{p^n}$ and $Z_{pq}$ can be used to model suitable problems where the $anti\left(a\right)$ under study does not exist and many neutrals are needed. This study can be taken up for further development.