Neutrosophic Extended Triplet Group Based on Neutrosophic Quadruple Numbers

Abstract

In this paper, we explore the algebra structure based on neutrosophic quadruple numbers. Moreover, two kinds of degradation algebra systems of neutrosophic quadruple numbers are introduced. In particular, the following results are strictly proved: (1) the set of neutrosophic quadruple numbers with a multiplication operation is a neutrosophic extended triplet group; (2) the neutral element of each neutrosophic quadruple number is unique and there are only sixteen different neutral elements in all of neutrosophic quadruple numbers; (3) the set which has same neutral element is closed with respect to the multiplication operator; (4) the union of the set which has same neutral element is a partition of four-dimensional space.

1. Introduction

The notion of a neutrosophic set is proposed by F. Smarandache [1] in order to solve real-world problems and some in-depth analysis and research have been carried out [2,3,4,5]. Recently, Smarandache and Ali in [6] proposed a new algebraic system, neutrosophic triplet group (NTG), which different from classical groups. From the original definition of NTG, the neutral element is different from the classical algebraic unit element. By removing this restriction, the neutrosophic extended triplet group (NETG) is proposed in [7,8] and the classical group is regarded as a special case of NETG.

As a new algebraic structure, NTG (NETG) immediately attracted the attention of scholars and conducted in-depth research. These studies are mainly carried out by the following three aspects. Firstly, the structure properties of NTG (NETG) have been studied deeply. For examples, paper [8] has conducted an in-depth analysis of the nature of NTG, and the properties and structural features of NTG are studied by using theoretical analysis and software calculations. In paper [9], the notion of the neutrosophic triplet coset and its relation with the classical coset are proposed and the properties of the neutrosophic triplet cosets are given. The neutrosophic duplet sets, neutrosophic duplet semi-groups, and cancellable neutrosophic triplet groups are proposed and the characterizations of cancellable weak neutrosophic duplet semi-groups are established in paper [10]. In order to explore the structure of the algebraic system $(Z_n,\otimes )$, where ⊗ is the classical mod multiplication, paper [11] reveals that for each $n\in Z^{+},n\ge 2,(Z_n,\otimes )$ is a commutative NETG if and only if the factorization of n is a product of single factors. Moreover, the generalized neutrosophic extended triplet group (GNETG) is proposed in [11] and verify that for each $n\in Z^{+},n\ge 2,(Z_n,\otimes )$ is a commutative GNETG. Secondly, it is the application research on the algebraic system NET. For example, In paper [12], the distinguishing features between an NTG and other algebraic structures are investigated and the first isomorphism theorem was established for NTGs, furthermore, applications of the results on NTG to management and sports are discussed. In paper [13], NTGs and their applications to mathematical models, such as fuzzy cognitive maps model, neutrosophic cognitive maps model and fuzzy relational maps model, are discussed. Thirdly, extend the idea of NTG(NETG) to another algebraic system. For example, in paper [14,15], the extend to Abel–Grassmann groupoid (AG-groupoid) is studied. The neutrosophic triplet ring and a neutrosophic triplet field are discussed in paper [16,17]. A notion of neutrosophic triplet metric space is given and properties of neutrosophic triplet metric spaces are studied in [18]. The notion of neutrosophic triplet v-generalized metric space are introduced in [19]. Paper [20] applies the neutrosophic set theory to pseudo-BCI algebras. The idea of a neutrosophic triplet set to non-associative semihypergroups is given in paper [21]. The above results enrich the research content of the algebraic system NTG (NETG).

In neutrosophic logic, each proposition is approximated to represent respectively the truth (T), the falsehood (F), and the indeterminacy (I), where $T,I,F$ are standard or non-standard subsets of the non-standard unit interval $]0^{-},1^{+}[=0^{-}\cup [0,1]\cup 1^{+}$. The notion of neutrosophic quadruple number, which is represented by a known part and an unknown part to describe a neutrosophic logic proposition, was introduced by Florentin Smarandache in [22]. The algebra system $(NQ,\ast )$ based on neutrosophic quadruple numbers are introduced and the properties have discussed [22,23]. In this paper, we will reveal that $(NQ,\ast )$ is a NETG and some properties are discussed.

The paper is organized as follows. Section 2 gives the basic concepts. In Section 3, $(NQ,\ast )$ be a NETG is proved and some properties are discussed. In Section 4, two kinds of degradation algebra systems of $(NQ,\ast )$ are introduced and studied. Finally, the summary and future work are presented in Section 5.

2. Basic Concepts

In this section, we will provide the related basic definitions and properties of NETG and neutrosophic quadruple numbers, the details can be seen in [7,8,22,23].

Definition 1

([7,8]). Let N be a non-empty set together with a binary operation ∗. Then, N is called a neutrosophic extended triplet set if for any $a\in N$, there exists a neutral of “a” (denote by $neut\left(a\right)$), and an opposite of “a”(denote by $anti\left(a\right))$, such that $neut\left(a\right)\in N$, $anti\left(a\right)\in N$ and:

$$a\ast neut\left(a\right)=neut\left(a\right)\ast a=a,a\ast anti\left(a\right)=anti\left(a\right)\ast a=neut\left(a\right).$$

The triplet $(a,neut(a),anti(a\left)\right)$ is called a neutrosophic extended triplet.

Definition 2

([7,8]). Let $(N,\ast )$ be a neutrosophic extended triplet set. Then, N is called a neutrosophic extended triplet group (NETG), if the following conditions are satisfied:

(1) 

$(N,\ast )$is well-defined, i.e., for any$a,b\in N$, one has$a\ast b\in N$.

(2) 

$(N,\ast )$is associative, i.e.,$(a\ast b)\ast c=a\ast (b\ast c)$for all$a,b,c\in N$.

A NETG N is called a commutative NETG if for all $a,b\in N,a\ast b=b\ast a$.

Proposition 1

([8]). Let $(N,\ast )$ be a NETG. We have:

(1) 

$neut\left(a\right)$is unique for any$a\in N$.

(2) 

$neut\left(a\right)\ast neut\left(a\right)=neut\left(a\right)$for any$a\in N$.

(3) 

$neut\left(neut\right(a\left)\right)=neut\left(a\right)$for any$a\in N$.

Definition 3

([22,23]). A neutrosophic quadruple number is a number of the form $(a,bT,cI,dF)$, where $T,I,F$ have their usual neutrosophic logic meanings and $a,b,c,d\in \mathbb{R}$ or $\mathbb{C}$. The set $NQ$, defined by

$$NQ=\left\{\right(a,bT,cI,dF):a,b,c,d\in \mathbb{R}\mathrm{or}\mathbb{C}\}.$$

(1)

is called a neutrosophic set of quadruple numbers. For a neutrosophic quadruple number $(a,bT,cI,dF)$, a is called the known part and $(bT,cI,dF)$ is called the unknown part.

Definition 4

([22,23]). Let N be a set, endowed with a total order $a\prec b$, named “a prevailed by b” or “a less stronger than b” or “a less preferred than b”. We consider $a⪯b$ as “a prevailed by or equal to b” “a less stronger than or equal to b”, or “a less preferred than or equal to b”.

For any elements $a,b\in N$, with $a⪯b$, one has the absorbance law:

$$a·b=b·a=\mathrm{absorb}(a,b)=max(a,b)=b,$$

(2)

which means that the bigger element absorbs the smaller element. Clearly,

$$a·a=a^2=\mathrm{absorb}(a,a)=max(a,a)=a.$$

(3)

and

$$a_1·a_2\cdots a_n=max(a_1,a_2,\cdots ,a_n).$$

(4)

Analogously, we say that “$a\succ b$” and we read: “a prevails to b” or “a is stronger than b” or “a is preferred to b”. Also, $a⪰b$, and we read: “a prevails or is equal to b” “a is stronger than or equal to b”, or “a is preferred or equal to b”.

Definition 5

([22,23]). Consider the set $\{T,I,F\}$. Suppose in an optimistic way we consider the prevalence order $T\succ I\succ F$. Then we have: $TI=IT=max(T,I)=T,$ $TF=FT=max(T,F)=T,$ $IF=FI=max(I,F)=I,$ $TT=T^2=T,$ $II=I^2=I,$$FF=F^2=F$.

Analogously, suppose in a pessimistic way we consider the prevalence order $T\prec I\prec F$. Then we have: $TI=IT=max(T,I)=I,$ $TF=FT=max(T,F)=F,$ $IF=FI=max(I,F)=F,$ $TT=T^2=T,$ $II=I^2=I,$$FF=F^2=F$.

Definition 6

([22,23]). Let $a=(a_1,a_{2}T,a_{3}I,a_{4}F),b=(b_1,b_{2}T,b_{3}I,b_{4}F)\in NQ$, Suppose in an pessimistic way, the neutrosophic expert considers the prevalence order $T\prec I\prec F$. Then the multiplication operation is defined as following:

$$\begin{array}{cc}a\ast b=\hfill & (a_1,a_{2}T,a_{3}I,a_{4}F)\ast (b_1,b_{2}T,b_{3}I,b_{4}F)\hfill \\ & =(a_1{b}_1,(a_1{b}_2+a_2{b}_1+a_2{b}_2)T,(a_1{b}_3+a_2{b}_3+a_3{b}_1+a_3{b}_2+a_3{b}_3)I,\hfill \\ & (a_1{b}_4+a_2{b}_4+a_3{b}_4+a_4{b}_1+a_4{b}_2+a_4{b}_3+a_4{b}_4)F).\hfill \end{array}$$

(5)

Suppose in an optimistic way the neutrosophic expert considers the prevalence order $T\succ I\succ F$. Then:

$$\begin{array}{cc}a⭑b=\hfill & (a_1,a_{2}T,a_{3}I,a_{4}F)\ast (b_1,b_{2}T,b_{3}I,b_{4}F)\hfill \\ & =(a_1{b}_1,(a_1{b}_2+a_2{b}_1+a_2{b}_2+a_3{b}_2+a_4{b}_2+a_2{b}_3+a_2{b}_4)T,\hfill \\ & (a_1{b}_3+a_3{b}_1+a_3{b}_3+a_3{b}_4+a_4{b}_3)I,(a_1{b}_4+a_4{b}_1+a_4{b}_4)F).\hfill \end{array}$$

(6)

Proposition 2

([22,23]). Let $NQ=\left\{\right(a,bT,cI,dF):a,b,c,d\in \mathbb{R}\mathrm{or}\mathbb{C}\}$. We have:

(1) 

$(NQ,\ast )$is a commutative monoid.

(2) 

$(NQ,⭑)$is a commutative monoid.

3. Main Results

From Proposition 2, we can see that $(NQ,\ast )$ (or $(NQ,⭑)$) be a commutative monoid. In these section, we will show that the algebra system $(NQ,\ast )$(or $(NQ,⭑)$) is a NETG.

Theorem 1.

For the algebra system $(NQ,\ast )$, for every element $a\in NQ$, there exists the neutral element $neut\left(a\right)$ and opposite element $anti\left(a\right)$.

Proof analysis: the proof of this theorem contains two aspects. Firstly, given an element $a\in NQ,a=(a_1,a_{2}T,a_{3}I,a_{4}F)$, $a_i\in \mathbb{R},i\in \{1,2,3,4\}$. Being $a_i$ can select every element in $\mathbb{R}$, we should discuss from different cases and in each case $netu\left(a\right)$ and $anti\left(a\right)$ should given. Secondly, we should prove that all the cases discussed above include all the elements in $NQ$.

Proof. 

Let $a=(a_1,a_{2}T,a_{3}I,a_{4}F)$, we consider $a_i\in \mathbb{R},i\in \{1,2,3,4\}$ and the same results can be gotten when $a_i\in \mathbb{C}$.

Set $neut\left(a\right)=(b_1,b_{2}T,b_{3}I,b_{4}F),b_i\in \mathbb{R},i\in \{1,2,3,4\}$ and $anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F),c_i\in \mathbb{R},i\in \{1,2,3,4\}$. From Definition 1 we can get $a\ast neut\left(a\right)=a$, that is $a_1{b}_1=a_1$ should hold. So we discuss from two cases, $a_1=0$ or $a_1\ne 0$.

Case A: when $a_1=0$.

In this case, we have $a=(0,a_{2}T,a_{3}I,a_{4}F)$. From Definition 1, $a\ast anti\left(a\right)=neut\left(a\right)$, that is $0·c_1=b_1$, so we have $b_1=0$, i.e., $neut\left(a\right)=(0,b_{2}T,b_{3}I,b_{4}F)$. Moreover, from $a\ast neut\left(a\right)=a$, we have $(0,a_{2}T,a_{3}I,a_{4}F)\ast (0,b_{2}T,b_{3}I,b_{4}F)=(0,a_{2}T,a_{3}I,a_{4}F)$, so we have $a_2{b}_2=a_2$. So we discuss from $a_2=0$ or $a_2\ne 0$.

Case A1: $a_1=0,a_2=0$. That is, $a=(0,0,a_{3}I,a_{4}F),netu\left(a\right)=(0,b_{2}T,b_{3}I,b_{4}F),anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$. From $a\ast anti\left(a\right)=neut\left(a\right)$, we have $0c_1+0(c_1+c_2)=b_2$, so $b_2=0$, i.e., $netu\left(a\right)=(0,0,b_{3}I,b_{4}F)$. From $(0,0,a_{3}I,a_{4}F)\ast (0,0,b_{3}I,b_{4}F)=(0,0,a_{3}I,a_{4}F)$, we have $a_3{b}_3=a_3$. So we discuss from $a_3=0$ or $a_3\ne 0$.

Case A11: $a_1=a_2=a_3=0$, that is, $a=(0,0,0,a_{4}F),netu\left(a\right)=(0,0,b_{3}I,b_{4}F),anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$. In the same way, from $a\ast anti\left(a\right)=neut\left(a\right)$, we have $b_3=0$, i.e., $netu\left(a\right)=(0,0,0,b_{4}F)$. From $(0,0,0,a_{4}F)\ast (0,0,0,b_{4}F)=(0,0,0,a_{4}F)$, we have $a_4{b}_4=a_4$. So we discuss from $a_4=0$ or $a_4\ne 0$.

Case A111: $a_1=a_2=a_3=a_4=0$, that is, $a=(0,0,0,0)$, in this case, we can easily get $neut\left(a\right)=(0,0,0,0)$ and $anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$, $c_i$ can be chosen arbitrarily in $\mathbb{R}$.

Case A112: $a_1=a_2=a_3=0,a_4\ne 0$, being that $a_4{b}_4=a_4$ and $a_4\ne 0$, we have $b_4=1$, that is, $a=(0,0,0,a_{4}F),netu\left(a\right)=(0,0,0,F),anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$. From $(0,0,0,a_{4}F)\ast (c_1,c_{2}T,c_{3}I,c_{4}F)=(0,0,0,F)$, we have $a_4(c_1+c_2+c_3+c_4)=1$, so the opposite element of a should satisfy $c_1+c_2+c_3+c_4=\frac{1}{a_4},c_i\in \mathbb{R}$.

Case A12: $a_1=a_2=0,a_3\ne 0$. From $a_3{b}_3=a_3$ and $a_3\ne 0$, we have $b_3=1$. That is $a=(0,0,a_{3}I,a_{4}F),netu\left(a\right)=(0,0,I,b_{4}F),anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$. From $(0,0,a_{3}I,a_{4}F)\ast (0,0,I,b_{4}F)=(0,0,a_{3}I,a_{4}F)$, we have $0b_4+0b_4+a_3{b}_4+a_4(0+0+1+b_4)=a_4$, so $(a_3+a_4)b_4=0$. We discuss from $a_3+a_4=0$ or $a_3+a_4\ne 0$.

Case A121: $a_1=a_2=0$, $a_3\ne 0,a_3+a_4=0$, that is $a=(0,0,a_{3}I,-a_{3}F),neut\left(a\right)=(0,0,I,b_{4}F),anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$. From $a\ast anti\left(a\right)=neut\left(a\right)$, that is $(0,0,a_{3}I,-a_{3}F)\ast (c_1,c_{2}T,c_{3}I,c_{4}F)=(0,0,I,b_{4}F)$. So we have $a_3(c_1+c_2+c_3)=1$ and $a_3{c}_4-a_3(c_1+c_2+c_3+c_4)=b_4$ i.e., $c_1+c_2+c_3=\frac{1}{a_3}$ and $b_4=1$. Thus $neut\left(a\right)=(0,0,I,-F)$, $anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$, where $c_1+c_2+c_3=\frac{1}{a_3}$, $c_4$ can be chosen arbitrarily in $\mathbb{R}$.

Case A122: $a_1=a_2=0,a_3\ne 0,a_3+a_4\ne 0$. From $(a_3+a_4)b_4=0$, we have $b_4=0$. that is $a=(0,0,a_{3}I,a_{4}F),neut\left(a\right)=(0,0,I,0),anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$. From $a\ast anti\left(a\right)=neut\left(a\right)$, that is $(0,0,a_{3}I,a_{4}F)\ast (c_1,c_{2}T,c_{3}I,c_{4}F)=(0,0,I,0)$. So we have $a_3(c_1+c_2+c_3)=1$ and $a_3{c}_4-a_3(c_1+c_2+c_3+c_4)=0$ i.e., $c_1+c_2+c_3=\frac{1}{a_3}$ and $c_4=-\frac{a_4}{a_3(a_3+a-4)}$. Thus $neut\left(a\right)=(0,0,I,0)$, $anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$, where $c_1+c_2+c_3=\frac{1}{a_3}$, $c_4=-\frac{a_4}{a_3(a_3+a_4)}$.

Case A2: when $a_1=0,a_2\ne 0$. From $a_2{b}_2=a_2$, we have $b_2=1$, that is, $a=(0,0,a_{3}I,a_{4}F),netu\left(a\right)=(0,T,b_{3}I,b_{4}F),anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$. In the same way, from $a\ast neut\left(a\right)=a$, we have $(a_2+a_3)b_3=0$, so we discuss from $a_2+a_3=0$ or $a_2+a_3\ne 0$.

Case A21: when $a_1=0,a_2\ne 0,a_2+a_3=0$. that is, $a=(0,a_{2}T,-a_{2}I,a_{4}F),netu\left(a\right)=(0,T,b_{3}I,b_{4}F),anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$. In the same way, from $a\ast neut\left(a\right)=a$, we have $a_4+a_4(b_3+b_4)=a_4$, that is $a_4(b_3+b_4)=0$, so we discuss from $a_4=0$ or $a_4\ne 0$.

Case A211: when $a_1=0,a_2\ne 0,a_2+a_3=0,a_4=0$. that is, $a=(0,a_{2}T,-a_{2}I,0),netu\left(a\right)=(0,T,b_{3}I,b_{4}F),anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$. From $(0,a_{2}T,-a_{2}I,0)\ast (c_1,c_{2}T,c_{3}I,c_{4}F)=(0,T,b_{3}I,b_{4}F)$, so we have $a_2(c_1+c_2)=1$ and $-a_2(c_1+c_2)=b_3$, that is $b_3=-1$. In the same way, we can get $b_4=0$. Thus $neut\left(a\right)=(0,T,-I,0)$, $anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$, where $c_1+c_2=\frac{1}{a_2}$, $c_3,c_4$ can be chosen arbitrarily in $\mathbb{R}$.

Case A212: when $a_1=0,a_2\ne 0,a_2+a_3=0,a_4\ne 0$, From $a_4(b_3+b_4)=0$, we have $b_3+b_4=0$, that is, $a=(0,a_{2}T,-a_{2}I,a_{4}F)$, $netu\left(a\right)=(0,T,b_{3}I,-b_{3}F)$, $anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$. From $(0,a_{2}T,-a_{2}I,a_4)\ast (c_1,c_{2}T,c_{3}I,c_{4}F)=(0,T,b_{3}I,-b_{3}F)$, so we have $a_2(c_1+c_2)=1$ and $-a_2(c_1+c_2)=b_3$, i.e., $b_3=-1$. Thus $neut\left(a\right)=(0,T,-I,F)$, $anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$, where $c_1+c_2=\frac{1}{a_2}$, $c_3+c_4=\frac{1}{a_4}-\frac{1}{a_2}$.

Case A22: when $a_1=0,a_2\ne 0,a_2+a_3\ne 0$. From $(a_2+a_3)b_3=0$, we have $b_3=0$. that is, $a=(0,a_{2}T,-a_{2}I,a_{4}F),netu\left(a\right)=(0,T,0,b_{4}F),anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$. From $a\ast neut\left(a\right)=a$, we have $(a_2+a_3+a_4)b_4=0$, so we discuss from $a_2+a_3+a_4=0$ or $a_2+a_3+a_4\ne 0$.

Case A221: when $a_1=0,a_2\ne 0,a_2+a_3\ne 0,a_2+a_3+a_4=0$. In this case $a=(0,a_{2}T,a_{3}I,a_{4}F),netu\left(a\right)=(0,T,0,b_{4}F),anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$. From $(0,a_{2}T,a_{3}I,a_{4}F)\ast (c_1,c_{2}T,c_{3}I,c_{4}F)=(0,T,0,b_{4}F)$, so we have $a_2(c_1+c_2)=1$, $c_3=-\frac{a_3}{a_2{(}_2+a_3)}$, $(a_2+a_3+a_4)b_4+a_4(c_1+c_2+c_3)=b_4$, so we have $b_4=-1$. Thus $neut\left(a\right)=(0,T,0,-F)$, $anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$, where $c_1+c_2=\frac{1}{a_2}$, $c_3=-\frac{a_3}{a_2{(}_2+a_3)}$, $c_4$ can be chosen arbitrarily in $\mathbb{R}$.

Case A222: when $a_1=0,a_2\ne 0,a_2+a_3\ne 0,a_2+a_3+a_4\ne 0$. From $(a_2+a_3+a_4)b_4=0$, we have $b_4=0$. that is, $a=(0,a_{2}T,a_{3}I,a_{4}F),netu\left(a\right)=(0,T,0,0),anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$. From $(0,a_{2}T,a_{3}I,0)\ast (c_1,c_{2}T,c_{3}I,c_{4}F)=(0,T,0,0)$, so we have $a_2(c_1+c_2)=1$, $c_3=-\frac{a_3}{a_2{(}_2+a_3)}$, $(a_2+a_3+a_4)b_4+a_4(c_1+c_2+c_3)=0$, Thus $neut\left(a\right)=(0,T,0,0)$, $anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$, where $c_1+c_2=\frac{1}{a_2}$, $c_3=-\frac{a_3}{a_2{(}_2+a_3)}$, $c_4=-\frac{a_4}{(a_2+a_3)(a_2+a_3+a_4)}$.

Case B: when $a_1\ne 0$.

In this case, from $a_1{b}_1=a_1$ and $a_1\ne 0$, we have $b_1=1$. That is $a=(a_1,a_{2}T,a_{3}I,a_{4}F),neut\left(a\right)=(1,b_{2}T,b_{3}I,b_{4}F),anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$. From Definition 1, $a\ast neut\left(a\right)=a$, that is $a_1{b}_2+a_2+a_2{b}_2=a_2$, so $(a_1+a_2)b_2=0$. So we discuss from $a_1+a_2=0$ or $a_1+a_2\ne 0$.

Case B1: when $a_1\ne 0,a_1+a_2=0$. That is $a=(a_1,-a_{1}T,a_{3}I,a_{4}F),neut\left(a\right)=(1,b_{2}T,b_{3}I,b_{4}F),anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$. From $a\ast anti\left(a\right)=neut\left(a\right)$, we have $c_1=\frac{1}{a_1}$, $a_1{c}_2-a_1{c}_1-a_1{c}_2=b_2$, so $b_2=-1$. From $a\ast neut\left(a\right)=a$, so we have $a_3+a_3{b}_2+a_3{b}_3=a_3$, i.e., $a_3(b_2+b_3)=0$. So we discuss from $a_3=0$ or $a_3\ne 0$.

Case B11: when $a_1\ne 0,a_1+a_2=0,a_3=0$. That is $a=(a_1,-a_{1}T,0,a_{4}F),neut\left(a\right)=(1,-T,b_{3}I,b_{4}F),anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$. From $a\ast neut\left(a\right)=a$, we have $a_1{b}_4-a_1{b}_4-0b_4+a_4(1-1+b_3+b_4)=a_4$, i.e., $a_4(b_3+b_4)=a_4$. So we discuss from $a_4=0$ or $a_4\ne 0$.

Case B111: when $a_1\ne 0,a_1+a_2=0,a_3=0,a_4=0$. That is $a=(a_1,-a_{1}T,0,0),neut\left(a\right)=(1,-T,b_{3}I,b_{4}F),anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$. From $a\ast anti\left(a\right)=neut\left(a\right)$, i.e., $(a_1,-a_{1}T,0,0)\ast (c_1,c_{2}T,c_{3}I,c_{4}F)=(1,-T,b_{3}I,b_{4}F)$, we have $c_1=\frac{1}{a_1}$, $b_3=b_4=0$. Thus $neut\left(a\right)=(1,-T,0,0)$, $anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$, which satisfies $c_1=\frac{1}{a_1}$ and $c_2,c_3,c_4$ can be chosen arbitrarily in $\mathbb{R}$.

Case B112: when $a_1\ne 0,a_1+a_2=0,a_3=0,a_4\ne 0$. From $a_4(b_3+b_4)=a_4$, we have $b_3+b_4=1$. That is $a=(a_1,-a_{1}T,0,a_{4}F),neut\left(a\right)=(1,-T,b_{3}I,b_{4}F),anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$. From $a\ast anti\left(a\right)=neut\left(a\right)$, i.e., $(a_1,-a_{1}T,0,a_{4}F)\ast (c_1,c_{2}T,c_{3}I,c_{4}F)=(1,-T,b_{3}I,b_{4}F)$, we have $c_1=\frac{1}{a_1}$, $b_3=0,b_4=1$. Thus $neut\left(a\right)=(1,-T,0,F)$, $anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$, where $c_1=\frac{1}{a_1}$ and $c_2+c_3+c_4=\frac{1}{a_4}-\frac{1}{a_1}$.

Case B12: when $a_1\ne 0,a_1+a_2=0,a_3\ne 0$. From $a_3(b_2+b_3)=0$ and $a_3\ne 0$, we have $b_2+b_3=0$, i.e., $b_3=1$. That is $a=(a_1,-a_{1}T,a_{3}I,a_{4}F),neut\left(a\right)=(1,-T,I,b_{4}F),anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$. From $a\ast neut\left(a\right)=a$, we have $a_3{b}_4+a_4+a_4{b}_4=a_4$, i.e., $(a_3+a_4)b_4=0$. So we discuss from $a_3+a_4=0$ or $a_3+a_4\ne 0$.

Case B121: when $a_1\ne 0,a_1+a_2=0,a_3\ne 0,a_3+a_4=0$. That is $a=(a_1,-a_{1}T,a_{3}I,-a_{3}F)$, $neut\left(a\right)=(1,-T,I,b_{4}F)$, $anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$. From $a\ast anti\left(a\right)=neut\left(a\right)$, i.e., $(a_1,-a_{1}T,a_{3}I,-a_{3}F)\ast (c_1,c_{2}T,c_{3}I,c_{4}F)=(1,-T,I,b_{4}F)$, we have $c_1=\frac{1}{a_1}$, $c_2+c_3=\frac{1}{a_3}-\frac{1}{a_1}$, $-a_3(c_1+c_2+c_3)=b_4$, i.e., $b_4=-1$. Thus $neut\left(a\right)=(1,-T,I,-F)$, $anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$, where $c_1=\frac{1}{a_1}$, $c_2+c_3=\frac{1}{a_3}-\frac{1}{a_2}$, $c_4$ can be chosen arbitrarily in $\mathbb{R}$.

Case B122: when $a_1\ne 0,a_1+a_2=0,a_3\ne 0,a_3+a_4\ne 0$, from $(a_3+a_4)b_4=0$, we have $b_4=0$. That is $a=(a_1,-a_{1}T,a_{3}I,a_{4}F),neut\left(a\right)=(1,-T,I,0),anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$. From $a\ast anti\left(a\right)=neut\left(a\right)$, i.e., $(a_1,-a_{1}T,a_{3}I,a_{4}F)\ast (c_1,c_{2}T,c_{3}I,c_{4}F)=(1,-T,I,0)$, we have $c_1=\frac{1}{a_1}$, $c_2+c_3=\frac{1}{a_3}-\frac{1}{a_1}$, $c_4=-\frac{a_4}{a_3(a_3+a_4)}$. Thus $neut\left(a\right)=(1,-T,I,-F)$, $anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$, where $c_1=\frac{1}{a_1}$, $c_2+c_3=\frac{1}{a_3}-\frac{1}{a_1}$, $c_4=-\frac{a_4}{a_3(a_3+a_4)}$.

Case B2: when $a_1\ne 0,a_1+a_2\ne 0$, from $(a_1+a_2)b_2=0$, we have $b_2=0$. That is $a=(a_1,-a_{1}T,a_{3}I,a_{4}F),neut\left(a\right)=(1,0,b_{3}I,b_{4}F),anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$. From $a\ast neut\left(a\right)=a$, so we have $a_1{b}_3+a_2{b}_3+a_3+a_3{b}_3=a_3$, i.e., $(a_1+a_2+a_3)b_3=0$. So we discuss from $a_1+a_2+a_3=0$ or $a_1+a_2+a_3\ne 0$.

Case B21: when $a_1\ne 0,a_1+a_2\ne 0,a_1+a_2+a_3=0$. That is $a=(a_1,a_{2}T,a_{3}I,a_{4}F),neut\left(a\right)=(1,0,b_{3}I,b_{4}F),anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$. From $a\ast neut\left(a\right)=a$, so we have $(a_1+a_2+a_3)b_4+a_4+a_4{b}_3+a_4{b}_4=a_4$, i.e., $(b_3+b_4)a_4=0$. So we discuss from $a_4=0$ or $a_4\ne 0$.

Case B211: when $a_1\ne 0,a_1+a_2\ne 0,a_1+a_2+a_3=0,a_4=0$. That is $a=(a_1,a_{2}T,a_{3}I,0),neut\left(a\right)=(1,0,b_{3}I,b_{4}F),anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$. From $a\ast anti\left(a\right)=neut\left(a\right)$, i.e., $(a_1,a_{2}T,a_{3}I,0)\ast (c_1,c_{2}T,c_{3}I,c_{4}F)=(1,0,b_{3}I,b_{4}F)$, we have $c_1=\frac{1}{a_1}$, $c_2=-\frac{a_2}{a_1(a_1+a_2)}$. $a_3(c_1+c_2)=b_3$, $(a_1+a_2+a_3)c_4+0(c_1+c_2+c_3+c_4)=0$, which means $b_3=-1,b_4=0$. Thus $neut\left(a\right)=(1,0,-I,0)$, $anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$, where $c_1=\frac{1}{a_1}$, $c_2=-\frac{a_2}{a_1(a_1+a_2)}$, $c_3,c_4$ can be chosen arbitrarily in $\mathbb{R}$.

Case B212: when $a_1\ne 0,a_1+a_2\ne 0,a_1+a_2+a_3=0,a_4\ne 0$. From $(b_3+b_4)a_4=0$, we have $b_3+b_4=0$. That is $a=(a_1,a_{2}T,a_{3}I,a_{4}F),neut\left(a\right)=(1,0,b_{3}I,-b_{3}F),anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$. From $a\ast anti\left(a\right)=neut\left(a\right)$, i.e., $(a_1,a_{2}T,a_{3}I,a_{4}F)\ast (c_1,c_{2}T,c_{3}I,c_{4}F)=(1,0,b_{3}I,-b_{3}F)$, we have $c_1=\frac{1}{a_1}$, $c_2=-\frac{a_2}{a_1(a_1+a_2)}$. $a_3(c_1+c_2)=b_3$, i.e., $b_3=-1,b_4=1$. Thus $neut\left(a\right)=(1,0,-I,F)$, $anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$, where $c_1=\frac{1}{a_1}$, $c_2=-\frac{a_2}{a_1(a_1+a_2)}$, $c_3+c_4=\frac{1}{a_4}-\frac{1}{a_1+a_2}$.

Case B22: when $a_1\ne 0,a_1+a_2\ne 0,a_1+a_2+a_3\ne 0$, from $(a_1+a_2+a_3)b_3=0$, we have $b_3=0$. That is $a=(a_1,a_{2}T,a_{3}I,a_{4}F),neut\left(a\right)=(1,0,0,b_{4}F),anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$. From $a\ast neut\left(a\right)=a$, so we have $(a_1+a_2+a_3+a_4)b_4+a_4=a_4$, i.e., $(a_1+a_2+a_3+a_4)b_4=0$. So we discuss from $a_1+a_2+a_3+a_4=0$ or $a_1+a_2+a_3+a_4\ne 0$.

Case B221: when $a_1\ne 0,a_1+a_2\ne 0,a_1+a_2+a_3\ne 0,a_1+a_2+a_3+a_4=0$, That is $a=(a_1,a_{2}T,a_{3}I,a_{4}F),neut\left(a\right)=(1,0,0,b_{4}F),anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$. From $a\ast anti\left(a\right)=neut\left(a\right)$, i.e., $(a_1,a_{2}T,a_{3}I,a_{4}F)\ast (c_1,c_{2}T,c_{3}I,c_{4}F)=(1,0,0,b_{4}F)$, we have $c_1=\frac{1}{a_1}$, $c_2=-\frac{a_2}{a_1(a_1+a_2)}$. $c_3=-\frac{a_3}{(a_1+a_2)(a_1+a_2+a_3)}$, $a_4(c_1+c_2+c_3)=b_4$, so $b_4=-1$. Thus $neut\left(a\right)=(1,0,0,-F)$, $anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$, where $c_1=\frac{1}{a_1}$, $c_2=-\frac{a_2}{a_1(a_1+a_2)},c_3=-\frac{a_3}{(a_1+a_2)(a_1+a_2+a_3)}$, $c_4$ can be chosen arbitrarily in $\mathbb{R}$.

Case B222: when $a_1\ne 0,a_1+a_2\ne 0,a_1+a_2+a_3\ne 0,a_1+a_2+a_3+a_4\ne 0$. From $(a_1+a_2+a_3+a_4)b_4+a_4=a_4$, we have $b_4=0$. That is $a=(a_1,a_{2}T,a_{3}I,a_{4}F),neut\left(a\right)=(1,0,0,0),anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$. From $a\ast anti\left(a\right)=neut\left(a\right)$, i.e., $(a_1,a_{2}T,a_{3}I,a_{4}F)\ast (c_1,c_{2}T,c_{3}I,c_{4}F)=(1,0,0,0)$, we have $c_1=\frac{1}{a_1}$, $c_2=-\frac{a_2}{a_1(a_1+a_2)}$, $c_3=-\frac{a_3}{(a_1+a_2)(a_1+a_2+a_3)}$. $(a_1+a_2+a_3+a_4)c_4+a_4(c_1+c_2+c_3)=0$, i.e., $c_4=-\frac{a_4}{(a_1+a_2+a_3)(a_1+a_2+a_3+a_4)}$. Thus $neut\left(a\right)=(1,0,0,0)$, $anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$, where $c_1=\frac{1}{a_1}$, $c_2=-\frac{a_2}{a_1(a_1+a_2)},c_3=-\frac{a_3}{(a_1+a_2)(a_1+a_2+a_3)},c_4=-\frac{a_4}{(a_1+a_2+a_3)(a_1+a_2+a_3+a_4)}$.

Finally, we should show that all the above cases include each element $a\in NQ$, i.e., $a_i,i=1,2,3,4$ can take all the values on $\mathbb{R}$. It is obvious that $a_1$ can take all the values on $\mathbb{R}$ because $a_1=0$ according to case A and that $a_1\ne 0$ according to case B. Moreover, for case A, $a_2$ can take all the values on $\mathbb{R}$ because case A1 according to $a_2=0$ and case A2 according to $a_2\ne 0$. For case B, $a_2$ can take all the values on $\mathbb{R}$ because case B1 according to $a_1+a_2=0$ and case B2 according to $a_1+a_2\ne 0$. That is for each element $a=(a_1,a_2,a_3,a_4)\in NQ$, $a_1,a_2$ can select all of value in $\mathbb{R}$. We will verify that $a_3$ and $a_4$ can take all the values on $\mathbb{R}$ when case A1 or A2 or B1 or B2 respectively.

For case A1, $a_3$ can take all the value in $\mathbb{R}$ because case A11 according to $a_3=0$ and case A12 according to $a_3\ne 0$. Similarly, for case A11, $a_4$ can take all the value in $\mathbb{R}$ because case A111 according to $a_4=0$ and case A112 according to $a_4\ne 0$. For case A12, $a_4$ can take all the value in $\mathbb{R}$ because case A121 according to $a_3+a_4=0$ and case A122 according to $a_3+a_4=0$. The top left subgraph of Figure 1 shows that the four cases A111, A112, A211 and A222. The unique □ point represents the case A111, the + points represent the case A112, the ∗ points represent the case A121 and the • points represent the case A122. This explain the that for case A1, $a_3$ and $a_4$ can take all the points on the plane. For case A2, B1 or B2, we can get that $a_3$ and $a_4$ can take all the points on the plane respectively. The top right subgraph of Figure 1 represents the case A2 if we select $a_1=0,a_2=1$, the bottom left subgraph of Figure 1 represents the case B1 if we select $a_1=1,a_2=-1$ and bottom right subgraph of Figure 1 represents the case B2 if we select $a_1=1,a_2=0$. The figure intuitively illustrates that all the points $(a_1,a_2,a_3,a_4),a_i\in \mathbb{R}$ are included.

Through the above analysis, we can get that for each element $a\in NQ$, there exists the neutral element $neut\left(a\right)$ and opposite element $anti\left(a\right)$.□

For algebra system $(NQ,\ast )$,

Table 1. The corresponding neutral element and opposite elements for $(NQ,\ast )$.

The Subset of $\mathit{NQ}$	Neutral Elements	Opposite Element $({\mathit{c}}_1,{\mathit{c}}_2\mathit{T},{\mathit{c}}_3\mathit{I},{\mathit{c}}_4\mathit{F})$
$\left\{\right(0,0,0,0\left)\right\}$	$(0,0,0,0)$	$c_i\in \mathbb{R}$
$\left\{(0,0,0,a_{4}F)\right|a_4\ne 0\}$	$(0,0,0,F)$	$c_1+c_2+c_3+c_4=\frac{1}{a_4}$
$\left\{(0,0,a_{3}I,-a_{3}F)\right|a_3\ne 0\}$	$(0,0,I,-F)$	$c_1+c_2+c_3=\frac{1}{a_3},c_4\in \mathbb{R}$
$\left\{(0,0,a_{3}I,a_{4}F)\right|a_3\ne 0,a_3+a_4\ne 0\}$	$(0,0,I,0)$	$c_1+c_2+c_3=\frac{1}{a_3},c_4=-\frac{a_4}{a_3(a_3+a_4)}$
$\left\{(0,a_{2}T,-a_{2}I,0)\right|a_2\ne 0$ }	$(0,T,-I,0)\}$	$c_1+c_2=\frac{1}{a_2},c_3,c_4\in \mathbb{R}$
$\left\{(0,a_{2}T,-a_{2}I,a_{4}F)\right|a_2\ne 0,a_4\ne 0\}$	$(0,T,-I,F)$	$c_1+c_2=\frac{1}{a_2},c_3+c_4=\frac{1}{a_4}-\frac{1}{a_2}$
$\left\{(0,a_{2}T,a_{3}I,a_{4}F)\right|a_2\ne 0,a_2+a_3\ne 0,a_2+a_3+a_4=0\}$	$(0,T,0,-F)$	$c_1+c_2=\frac{1}{a_2},c_3=-\frac{a_3}{a_2(a_2+a_3)},c_4\in \mathbb{R}$
$\left\{(0,a_{2}T,a_{3}I,a_{4}F)\right|a_2\ne 0,a_2+a_3\ne 0,a_2+a_3+a_4\ne 0\}$	$(0,T,0,0)$	$c_1+c_2=\frac{1}{a_2},c_3=-\frac{a_3}{a_2(a_2+a_3)},c_4=-\frac{a_3}{(a_2+a_3)(a_2+a_3+a_4)}$
$\left\{(a_1,-a_{1}T,0,0)\right|a_1\ne 0\}$	$(1,-T,0,0)\}$	$c_1=\frac{1}{a_1},c_2,c_3,c_4\in \mathbb{R}$
$\left\{(a_1,-a_{1}T,0,a_{4}F)\right|a_1\ne 0,a_4\ne 0\}$	$(1,-T,0,F)$	$c_1=\frac{1}{a_1},c_2+c_3+c_4=\frac{1}{a_4}-\frac{1}{a_1}$
$\left\{(a_1,-a_{1}T,a_{3}I,-a_{3}F)\right|a_1\ne 0,a_3\ne 0\}$	$(1,-T,I,-F)$	$c_1=\frac{1}{a_1},c_2+c_3=\frac{1}{a_3}-\frac{1}{a_1},c_4\in \mathbb{R}$
$\left\{(a_1,-a_{1}T,a_{3}I,a_{4}F)\right|a_1\ne 0,a_3\ne 0,a_3+a_4\ne 0\}$	$(1,-T,I,0)$	$c_1=\frac{1}{a_1},c_2+c_3=\frac{1}{a_3}-\frac{1}{a_1},c_4=-\frac{a_4}{a_3(a_3+a_4)}$
$\left\{(a_1,a_{2}T,a_{3}I,0)\right|a_1\ne 0,a_1+a_2\ne 0,a_1+a_2+a_3=0\}$	$(1,0,-I,0)$	$c_1=\frac{1}{a_1},c_2=-\frac{a_2}{a_1(a_1+a_2)},c_3,c_4\in \mathbb{R}$
$\left\{(a_1,a_{2}T,a_{3}I,a_{4}F)\right|a_1\ne 0,a_1+a_2\ne 0,a_1+a_2+a_3=0,a_4\ne 0\}$	$(1,0,-I,F)$	$c_1=\frac{1}{a_1},c_2=-\frac{a_2}{a_1(a_1+a_2)},c_3+c_4=\frac{1}{a_4}-\frac{1}{a_1+a_2}$
$\left\{(a_1,a_{2}T,a_{3}I,a_{4}F)\right|a_1\ne 0,a_1+a_2\ne 0,a_1+a_2+a_3\ne 0,a_1+a_2+a_3+a_4=0\}$	$(1,0,0,-F)$	$c_1=\frac{1}{a_1},c_2=-\frac{a_2}{a_1(a_1+a_2)},c_3=-\frac{a_3}{(a_1+a_2)(a_1+a_2+a_3)},c_4\in \mathbb{R}$
$\left\{(a_1,a_{2}T,a_{3}I,a_{4}F)\right|a_1\ne 0,a_1+a_2\ne 0,a_1+a_2+a_3\ne 0,a_1+a_2+a_3+a_4\ne 0\}$	$(1,0,0,0)$	$c_1=\frac{1}{a_1},c_2=-\frac{a_2}{a_1(a_1+a_2)},c_3=-\frac{a_3}{(a_1+a_2)(a_1+a_2+a_3)},c_4=-\frac{a_4}{(a_1+a_2+a_3)(a_1+a_2+a_3+a_4)}$

gives all the subset, which has the same neutral element, and the corresponding neutral element and opposite elements.

Example 1.

For the algebra system $(NQ,\ast )$, if $a=(a_1,a_2,a_3,a_4)=(1,-T,3I,-F)$, i.e., $a_1\ne 0,a_1+a_2=0,a_3\ne 0,a_3+a_4\ne 0$, then from Table 1, we can get $neut\left(a\right)=(1,-T,I,0)$. Let $anti\left(a\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$, so $c_1=\frac{1}{a_1}=1,c_2+c_3=\frac{1}{a_3}-\frac{1}{a_1}=-\frac{2}{3},c_4=-\frac{a_4}{a_3(a_3+a_4)}=\frac{1}{6}$, so $anti\left(a\right)=(1,c_{2}T,c_{3}I,\frac{1}{6}F)$, where $c_2+c_3=-\frac{2}{3}$. Thus we can easily get the neutral element and opposite elements of each neutrosophic quadruple number. For more examples, see the following:

1. 

Let $b=(0,0,I,-F)$, then $neut\left(b\right)=(0,0,I,-F)$ and $anti\left(b\right)=(c_1,c_{2}T,c_{3}I,c_{4}F)$, where $c_1+c_2+c_3=1$, $c_4$ can be can be chosen arbitrarily in $\mathbb{R}$.

2. 

Let $c=(1,T,I,-F)$, then $neut\left(c\right)=(1,0,0,0)$ and $anti\left(c\right)=(1,-\frac{1}{2}T,-\frac{1}{6}I,\frac{1}{6}F)$.

3. 

Let $d=(0,T,I,F)$, then $neut\left(d\right)=(0,T,0,0)$ and $anti\left(d\right)=(c_1,c_{2}T,-\frac{1}{2}I,-\frac{1}{6}F)$, where $c_1+c_2=1$.

In the following, we will discuss the algebra structure properties of $(NQ,\ast )$.

Proposition 3.

For algebra system $(NQ,\ast )$, let $NS=\left\{neut\right(a\left)\right|a\in NQ\}$, we have:

(1) $NS=\left\{\right(1,0,0,0),(0,0,0,F),(0,0,I,-F),(0,0,I,0),(0,T,-I,0),(0,T,-I,F),(0,T,0,-F),(0,T,0,0)$, $(1,-T,0,0),(1,-T,0,F),(1,-T,I,-F),(1,-T,I,0),(1,0,-I,0),(1,0,-I,F),(1,0,0,-F),(1,0,0,0)\}$.

(2) $NS$ is closed with respect to operation ∗.

(3) Set $IS=\left\{a\right|a^2=a,a\in NQ\}$, which is all the set of idempotent elements of $(NQ,\ast )$, then $NS=IS$.

Proof. 

(1) Obviously.

(2) If $c,d\in NS$, that is $neut\left(a\right)=c,neut\left(b\right)=d,a,b\in NQ$. From Proposition 1, $neut\left(a\right)\ast neut\left(b\right)=neut(a\ast b)$, i.e., $c\ast d=neut(a\ast b)$, then form Theorem 1, every element in $NQ$ has neutral element, so $a\ast b$ also has neutral element, that is $neut(a\ast b)\in NS$, i.e., $c\ast d\in NS$, thus $NS$ is closed with respect to operation ∗.

(3) From Proposition 1, $neut\left(a\right)\ast neut\left(a\right)=neut\left(a\right)$, so $neut\left(a\right)$ is a idempotent element and $NS\subseteq IS$. On the other hand if a is a idempotent element, so $a\ast a=a$, that is a exists the neutral element a and the opposite element a, so a is a neutral element, that is $IS\subseteq NS$. Thus $NS=IS$.□

Proposition 4.

For algebra system $(NQ,\ast )$, let $V_c=\left\{a\right|a\in NQ\wedge neut\left(a\right)=c\}$, $V_{c\ast d}=\{a\ast b|a,b\in NQ\wedge neut\left(a\right)=c\wedge neut\left(b\right)=d\}$, we have:

(1) $V_c$ is closed with respect to operation ∗.

(2) $V_{c\ast d}$ is closed with respect to operation ∗.

Proof. 

(1) If $a,b\in V_c$, that is $neut\left(a\right)=neut\left(b\right)=c$. From Proposition 1, $neut\left(a\right)\ast neut\left(b\right)=neut(a\ast b)$, we can see that $neut(a\ast b)=neut\left(a\right)=c$, i.e., the neutral element of $a\ast b$ is the neutral element of a, so $a\ast b\in V_c$, that is $V_c$ is closed with respect to operation ∗.

(2) If $a_1\ast b_1,a_2\ast b_2\in V_{c\ast d}$, i.e., $neut\left(a_1\right)=neut\left(a_2\right)=c,neut\left(b_1\right)=neut\left(b_2\right)=d$. From Proposition 3(2), $a_1\ast a_2=a_3\in V_c$, $b_1\ast b_2=b_3\in V_d$, so $neut\left(a_3\right)=c,neut\left(b_3\right)=d$, from $(a_1\ast b_1)\ast (a_2\ast b_2)=a_3\ast b_3$, so $neut(a_1\ast a_2\ast b_1\ast b_2)=neut(a_3\ast a_4)$, that is $a_3\ast a_4\in V_{c\ast d}$, that means $a_1\ast a_2\ast b_1\ast b_2\in V_{c\ast d}$. Thus $V_{c\ast d}$ is closed with respect to operation ∗.□

Definition 7.

Assume that $(N,\ast )$ is a neutrosophic triplet group and H be a nonempty subset of N. Then H is called a neutrosophic triplet subgroup of N if;

(1) $a\ast b\in H$ for all $a,b\in H$;

(2) there exists $anti\left(a\right)\in \left\{anti\right(a\left)\right\}$ such that $anti\left(a\right)\in H$ for all $a\in H$, where $\left\{anti\right(a\left)\right\}$ is the set of opposite element of a in $(N,\ast )$.

Theorem 2.

For algebra system $(NQ,\ast )$, let $V_c=\left\{a\right|a\in NQ\wedge neut\left(a\right)=c\}$, $V_{c\ast d}=\{a\ast b|a,b\in NQ\wedge neut\left(a\right)=c,neut\left(b\right)=d\}$, we have:

(1) $V_c$ is a neutrosophic triplet subgroup of $NQ$.

(2) $V_{c\ast d}$ is a neutrosophic triplet subgroup of $NQ$.

Proof. 

(1) From Proposition 3, we can see that $V_c$ is closed with respect to operation ∗. In the following, we will prove there exists $anti\left(a\right)\in \left\{anti\right(a\left)\right\}$ such that $anti\left(a\right)\in V_c$ for all $a\in V_c$.

Proof by contradiction.

Assume that $\left\{anti\left(a\right)\right\}\cap V_c=\varnothing$. From Proposition 1 we can see that $a\ast anti\left(a\right)=c$. On the other hand, $anti\left(a\right)\in NQ$, so $anti\left(a\right)$ exists neutral element, denoted by $neut\left(anti\right(a\left)\right)$. Being $anti\left(a\right)\notin V_c$, so $neut\left(anti\right(a\left)\right)\ne c$.

From $a\ast anti\left(a\right)=c$, we have $a\ast anti\left(a\right)\ast neut\left(anti\right(a\left)\right)=c\ast neut\left(anti\right(a\left)\right)$, being $anti\left(a\right)\ast neut\left(anti\right(a\left)\right)=anti\left(a\right)$ and $a\ast anti\left(a\right)=c$, we have $c\ast neut\left(anti\right(a\left)\right)=c$, and then we have $a\ast c\ast neut\left(anti\right(a\left)\right)=a\ast c=a$, that means $a\ast neut\left(anti\right(a\left)\right)=a$, so $neut\left(anti\right(a\left)\right)$ is also a neutral element of a. This leads to the contradiction being the uniqueness of neutral element for each element. Therefore $\left\{anti\left(a\right)\right\}\cap V_c\ne \varnothing$. Thus from Definition 7, $V_c$ is a neutrosophic triplet subgroup of $NQ$.

(2) The same way we can get $V_{c\ast d}$ is a neutrosophic triplet subgroup of $NQ$.□

Theorem 3.

For algebra system $(NQ,\ast )$, let $V_c=\left\{a\right|a\in NQ\wedge neut\left(a\right)=c\}$, we have:

(1) $V_c\cap V_d=\varnothing$ if $c\ne d$.

(2) $NQ={\cup }_{c\in NS}{V}_c$. So ${\cup }_{c\in NS}{V}_c$ is a partition of $NQ$, where $NS$ is a set, which contains all the neutral elements of $(NQ,\ast )$.

Proof. 

(1) Proof by contradiction.

Assume $V_c\cap V_d\ne \varnothing$ when $c\ne d$, so exist $a\in V_c\cap V_d$, such that a has two neutral elements c and d. This leads to the contradiction being the uniqueness of neutral element. So $V_c\cap V_d=\varnothing$ if $c\ne d$.

(2) From the proof of Theorem 1, we can get $NQ={\cup }_{c\in NS}{V}_c$. So ${\cup }_{c\in NS}{V}_c$ is a partition of $NQ$.□

For the algebra system $(NQ,⭑)$, we have the similar results. We describe as following and omit the proof.

Theorem 4.

For the algebra system $(NQ,⭑)$, for every element $a\in NQ$, there exists the neutral element $neut\left(a\right)$ and opposite element $anti\left(a\right)$.

For algebra system $(NQ,⭑)$,

Table 2. The corresponding neutral element and opposite elements for $(NQ,⭑)$.

The Subset of $\mathit{NQ}$	Neutral Elements	Opposite Element $({\mathit{c}}_1,{\mathit{c}}_2\mathit{T},{\mathit{c}}_3\mathit{I},{\mathit{c}}_4\mathit{F})$
$\left\{\right(0,0,0,0\left)\right\}$	$(0,0,0,0)$	$c_i\in \mathbb{R}$
$\left\{(0,a_{2}T,0,0)\right|a_2\ne 0\}$	$(0,T,0,0)$	$c_1+c_2+c_3+c_4=\frac{1}{a_2}$
$\left\{(0,-a_{3}T,a_{3}I,0)\right|a_3\ne 0\}$	$(0,-T,I,0)$	$c_1+c_3+c_4=\frac{1}{a_3},c_2\in \mathbb{R}$
$\left\{(0,a_{2}T,a_{3}I,a_{4}F)\right|a_3\ne 0,a_2+a_3\ne 0\}$	$(0,0,I,0)$	$c_1+c_3+c_4=\frac{1}{a_3},c_2=-\frac{a_2}{a_3(a_2+a_3)}$
$\left\{(0,0,-a_{4}I,a_{4}F)\right|a_4\ne 0\}$	$(0,0,-I,F)\}$	$c_1+c_4=\frac{1}{a_4},c_2,c_3\in \mathbb{R}$
$\left\{(0,a_{2}T,-a_{4}I,a_{4}F)\right|a_2\ne 0,a_4\ne 0\}$	$(0,T,-I,F)$	$c_1+c_4=\frac{1}{a_4},c_2+c_3=\frac{1}{a_2}-\frac{1}{a_4}$
$\left\{(0,a_{2}T,a_{3}I,a_{4}F)\right|a_4\ne 0,a_3+a_4\ne 0,a_2+a_3+a_4=0\}$	$(0,-T,0,F)$	$c_1+c_4=\frac{1}{a_4},c_3=-\frac{a_3}{a_4(a_3+a_4)},c_2\in \mathbb{R}$
$\left\{(0,a_{2}T,a_{3}I,a_{4}F)\right|a_4\ne 0,a_3+a_4\ne 0,a_2+a_3+a_4\ne 0\}$	$(0,0,0,F)$	$c_1+c_4=\frac{1}{a_4},c_3=-\frac{a_3}{a_4(a_3+a_4)},c_2=-\frac{a_2}{(a_3+a_4)(a_2+a_3+a_4)}$
$\left\{(a_1,0,0,-a_{1}F)\right|a_1\ne 0\}$	$(1,0,0,-F)\}$	$c_1=-\frac{1}{a_1},c_2,c_3,c_4\in \mathbb{R}$
$\left\{(a_1,a_{2}T,0,-a_{1}F)\right|a_1\ne 0,a_2\ne 0\}$	$(1,T,0,-F)$	$c_1=\frac{1}{a_1},c_2+c_3+c_4=\frac{1}{a_2}-\frac{1}{a_1}$
$\left\{(a_1,-a_{3}T,a_{3}I,-a_{1}F)\right|a_1\ne 0,a_3\ne 0\}$	$(1,-T,I,-F)$	$c_1=\frac{1}{a_1},c_3+c_4=\frac{1}{a_3}-\frac{1}{a_1},c_4\in \mathbb{R}$
$\left\{(a_1,a_{2}T,a_{3}I,-a_{1}F)\right|a_1\ne 0,a_3\ne 0,a_2+a_3\ne 0\}$	$(1,0,I,-F)$	$c_1=\frac{1}{a_1},c_3+c_4=\frac{1}{a_3}-\frac{1}{a_1},c_2=-\frac{a_2}{a_3(a_2+a_3)}$
$\left\{(a_1,0,a_{3}I,a_{4}F)\right|a_1\ne 0,a_1+a_4\ne 0,a_1+a_3+a_4=0\}$	$(1,0,-I,0)$	$c_1=\frac{1}{a_1},c_4=-\frac{a_4}{a_1(a_1+a_4)},c_2,c_3\in \mathbb{R}$
$\left\{(a_1,a_{2}T,a_{3}I,a_{4}F)\right|a_1\ne 0,a_1+a_4\ne 0,a_1+a_3+a_4=0,a_2\ne 0\}$	$(1,T,-I,0)$	$c_1=\frac{1}{a_1},c_4=-\frac{a_4}{a_1(a_1+a_4)},c_2+c_3=\frac{1}{a_2}-\frac{1}{a_1+a_4}$
$\left\{(a_1,a_{2}T,a_{3}I,a_{4}F)\right|a_1\ne 0,a_1+a_4\ne 0,a_1+a_3+a_4\ne 0,a_1+a_2+a_3+a_4=0\}$	$(1,-T,0,0)$	$c_1=\frac{1}{a_1},c_4=-\frac{a_4}{a_1(a_1+a_4)},c_3=-\frac{a_3}{(a_1+a_4)(a_1+a_3+a_4)},c_2\in \mathbb{R}$
$\left\{(a_1,a_{2}T,a_{3}I,a_{4}F)\right|a_1\ne 0,a_1+a_4\ne 0,a_1+a_3+a_4\ne 0,a_1+a_2+a_3+a_4\ne 0\}$	$(1,0,0,0)$	$c_1=\frac{1}{a_1},c_4=-\frac{a_4}{a_1(a_1+a_4)},c_3=-\frac{a_3}{(a_1+a_4)(a_1+a_3+a_4)},c_2=-\frac{a_2}{(a_1+a_3+a_4)(a_1+a_2+a_3+a_4)}$

gives all the subset, which has the same neutral element, and the corresponding neutral element and opposite elements.

Theorem 5.

For an algebra system $(NQ,⭑)$, let $V_c=\left\{a\right|a\in NQ\wedge neut\left(a\right)=c\}$, $V_{c⭑d}=\{a⭑b|a,b\in NQ\wedge neut\left(a\right)=c,neut\left(b\right)=d\}$, we have:

(1) $V_c$ is a neutrosophic triplet subgroup of $NQ$.

(2) $V_{c⭑d}$ is a neutrosophic triplet subgroup of $NQ$.

Theorem 6.

For algebra system $(NQ,⭑)$, Let $V_c=\left\{a\right|a\in NQ\wedge neut\left(a\right)=c\}$, we have:

(1) $V_c\cap V_d=\varnothing$ if $c\ne d$.

(2) $NQ={\cup }_{c\in NS}{V}_c$. So ${\cup }_{c\in NS}{V}_c$ is a partition of $NQ$, where $NS$ is a set, which contains all the neutral elements of $(NQ,⭑)$.

4. Two Kinds of Degenerate Systems of Neutrosophic Quadruple Numbers

The neutrosophic quadruple numbers consider $(T,I,F)$ to solve real problems. In this section, we will explore two kinds of degenerate systems about neutrosophic quadruple numbers. The first system is only consider logical true, and the second system is only consider logical true and logical indeterminacy.

4.1. The Neutrosophic Binary Numbers

Definition 8.

A neutrosophic binary number is a number of the form $(a,bT)$, where T have their usual neutrosophic logic true and $a,b\in \mathbb{R}$ or $\mathbb{C}$. The set $NB$ defined by

$$NB=\left\{\right(a,bT):a,b\in \mathbb{R}\mathrm{or}\mathbb{C}\}.$$

(7)

is called a neutrosophic set of binary numbers. For a neutrosophic binary number $(a,bT)$, a is called the known part and $\left(bT\right)$ is called the unknown part.

Definition 9.

Let $a=(a_1,a_{2}T),b=(b_1,b_{2}T)\in NB$, the multiplication operation is defined as following:

$$a\ast b=(a_1,a_{2}T)\ast (b_1,b_{2}T)=(a_1{b}_1,(a_1{b}_2+a_2{b}_1+a_2{b}_2)T).$$

(8)

We have the following results similar to $(NQ,\ast )$.

Theorem 7.

For the algebra system $(NB,\ast )$, for every element $a\in NB$, there exists the neutral element $neut\left(a\right)$ and opposite element $anti\left(a\right)$.

For algebra system $(NB,\ast )$,

Table 3. The corresponding neutral element and opposite elements for $(NB,\ast )$.

The Subset	Neutral Elements	Opposite Element $({\mathit{c}}_1,{\mathit{c}}_2\mathit{T})$
$\left\{\right(0,0\left)\right\}$	$(0,0)$	$c_i\in \mathbb{R}$
$\left\{(0,a_{2}T)\right|a_2\ne 0\}$	$(0,T)$	$c_1+c_2=\frac{1}{a_2}$
$\left\{(a_1,-a_{1}T)\right|a_1\ne 0\}$	$(1,0)$	$c_1=\frac{1}{a_1},c_2\in \mathbb{R}$
$\left\{(a_1,a_{2}T)\right|a_1\ne 0$	$(1,-T)$	$c_1=\frac{1}{a_1},c_2=-\frac{a_2}{a_1(a_1+a_2)}$

gives all the subset, which has the same neutral element, and the corresponding neutral element and opposite elements.

Theorem 8.

For algebra system $(NB,\ast )$, let $V_c=\left\{a\right|a\in NB\wedge neut\left(a\right)=c\}$, $V_{c\ast d}=\{a\ast b|a,b\in NB\wedge neut\left(a\right)=c,neut\left(b\right)=d\}$, we have:

(1) $V_c$ is a neutrosophic triplet subgroup of $NB$.

(2) $V_{c\ast d}$ is a neutrosophic triplet subgroup of $NB$.

Theorem 9.

For an algebra system $(NB,\ast )$, let $V_c=\left\{a\right|a\in NB\wedge neut\left(a\right)=c\}$, we have:

(1) $V_c\cap V_d=\varnothing$ if $c\ne d$.

(2) $NB={\cup }_{c\in NS}{V}_c$. So ${\cup }_{c\in NS}{V}_c$ is a partition of $NB$, where $NS$ is a set, which contains all the neutral elements of $(NB,\ast )$.

4.2. The Neutrosophic Triple Numbers

Definition 10.

A neutrosophic triple number is a number of the form $(a,bT,cI)$, where $T,I$ have their usual neutrosophic logic meanings and $a,b,c\in \mathbb{R}$ or $\mathbb{C}$. The set $NT$ defined by

$$NT=\left\{\right(a,bT,cI):a,b,c\in \mathbb{R}\mathrm{or}\mathbb{C}\}$$

(9)

is called a neutrosophic set of triple numbers. For a neutrosophic triple number $(a,bT,cI)$, a is called the known part and $(bT,cI)$ is called the unknown part.

Definition 11.

Let $a=(a_1,a_{2}T,a_{3}I),b=(b_1,b_{2}T,b_{3}I)\in NT$, suppose in an pessimistic way, the neutrosophic expert considers the prevalence order $T\prec I$. Then the multiplication operation is defined as following:

$$\begin{array}{cc}a\ast b=\hfill & (a_1,a_{2}T,a_{3}I)\ast (b_1,b_{2}T,b_{3}I)\hfill \\ & =(a_1{b}_1,(a_1{b}_2+a_2{b}_1+a_2{b}_2)T,(a_1{b}_3+a_2{b}_3+a_3{b}_1+a_3{b}_2+a_3{b}_3)I.\hfill \end{array}$$

(10)

Suppose in an optimistic way the neutrosophic expert considers the prevalence order $T\succ I$. Then:

$$\begin{array}{cc}a⭑b=\hfill & (a_1,a_{2}T,a_{3}I)\ast (b_1,b_{2}T,b_{3}I)\hfill \\ & =(a_1{b}_1,(a_1{b}_2+a_2{b}_1+a_2{b}_2+a_2{b}_2+a_3{b}_2)T,(a_1{b}_3+a_3{b}_1+a_3{b}_3)I).\hfill \end{array}$$

(11)

Theorem 10.

For the algebra system $(NT,\ast )$, for every element $a\in NT$, there exists the neutral element $neut\left(a\right)$ and opposite element $anti\left(a\right)$.

For algebra system $(NT,\ast )$,

Table 4. The corresponding neutral element and opposite elements for $(NT,\ast )$.

The Subset	Neutral Elements	Opposite Element $({\mathit{c}}_1,{\mathit{c}}_2\mathit{T},{\mathit{c}}_3\mathit{I})$
$\left\{\right(0,0,0\left)\right\}$	$(0,0,0)$	$c_i\in \mathbb{R}$
$\left\{(0,0,a_{3}I)\right|a_3\ne 0\}$	$(0,0,I)$	$c_1+c_2+c_3=\frac{1}{a_3}$
$\left\{(0,a_{2}T,-a_{2}I)\right|a_2\ne 0,a_2+a_3=0\}$	$(0,T,-I)$	$c_1+c_2=\frac{1}{a_2},c_3\in \mathbb{R}$
$\left\{(0,a_{2}T,a_{3}I)\right|a_2\ne 0,a_2+a_3\ne 0\}$	$(0,T,0)$	$c_1+c_2=\frac{1}{a_2},c_3=-\frac{a_3}{a_2(a_2+a_3)}$
$\left\{(a_1,-a_{1}T,0)\right|a_1\ne 0\}$	$(1,-T,0)\}$	$c_1=\frac{1}{a_1},c_2,c_3\in \mathbb{R}$
$\left\{(a_1,-a_{1}T,a_{3}I)\right|a_1\ne 0,a_3\ne 0\}$	$(1,-T,I)$	$c_1=\frac{1}{a_1},c_2+c_3=\frac{1}{a_3}-\frac{1}{a_1}$
$\left\{(a_1,a_{2}T,a_{3}I)\right|a_1\ne 0,a_1+a_2\ne 0,a_1+a_2+a_3=0\}$	$(1,0,-I)$	$c_1=\frac{1}{a_1},c_2=-\frac{a_2}{a_1(a_1+a_2)},c_3\in \mathbb{R}$
$\left\{(a_1,a_{2}T,a_{3}I)\right|a_1\ne 0,a_1+a_2\ne 0,a_1+a_2+a_3\ne 0\}$	$(1,0,0)$	$c_1=\frac{1}{a_1},c_2=-\frac{a_2}{a_1(a_1+a_2)},c_3=-\frac{a_3}{(a_1+a_2)(a_1+a_2+a_3)}$

gives all the subset, which has the same neutral element, and the corresponding neutral element and opposite elements.

Theorem 11.

For an algebra system $(NT,\ast )$, let $V_c=\left\{a\right|a\in NT\wedge neut\left(a\right)=c\}$, $V_{c\ast d}=\{a\ast b|a,b\in NT\wedge neut\left(a\right)=c,neut\left(b\right)=d\}$, we have:

(1) $V_c$ is a neutrosophic triplet subgroup of $NT$.

(2) $V_{c\ast d}$ is a neutrosophic triplet subgroup of $NT$.

Theorem 12.

For an algebra system $(NT,\ast )$, let $V_c=\left\{a\right|a\in NT\wedge neut\left(a\right)=c\}$, we have:

(1) $V_c\cap V_d=\varnothing$ if $c\ne d$.

(2) $NT={\cup }_{c\in NS}{V}_c$. So ${\cup }_{c\in NS}{V}_c$ is a partition of $NT$, where $NS$ is a set, which contains all the neutral elements of $(NT,\ast )$.

Theorem 13.

For the algebra system $(NT,⭑)$, for every element $a\in NT$, there exists the neutral element $neut\left(a\right)$ and opposite element $anti\left(a\right)$.

For an algebra system $(NT,⭑)$,

Table 5. The corresponding neutral element and opposite elements for $(NT,⭑)$.

The Subset	Neutral Elements	Opposite Element $({\mathit{c}}_1,{\mathit{c}}_2\mathit{T},{\mathit{c}}_3\mathit{I})$
$\left\{\right(0,0,0\left)\right\}$	$(0,0,0)$	$c_i\in \mathbb{R}$
$\left\{(0,a_{2}T,0)\right|a_2\ne 0\}$	$(0,T,0)$	$c_1+c_2+c_3=\frac{1}{a_2}$
$\left\{(0,a_{3}T,-a_{3}I)\right|a_3\ne 0,a_2+a_3=0\}$	$(0,-T,I)$	$c_1+c_3=\frac{1}{a_3},c_2\in \mathbb{R}$
$\left\{(0,a_{2}T,a_{3}I)\right|a_3\ne 0,a_2+a_3\ne 0\}$	$(0,0,I)$	$c_1+c_3=\frac{1}{a_3},c_2=-\frac{a_2}{a_3(a_2+a_3)}$
$\left\{(a_1,0,-a_{1}I)\right|a_1\ne 0\}$	$(1,0,-I)\}$	$c_1=\frac{1}{a_1},c_2,c_3\in \mathbb{R}$
$\left\{(a_1,a_{2}T,-a_{1}I)\right|a_1\ne 0,a_2\ne 0\}$	$(1,T,-I)$	$c_1=\frac{1}{a_1},c_2+c_3=\frac{1}{a_2}-\frac{1}{a_1}$
$\left\{(a_1,a_{2}T,a_{3}I)\right|a_1\ne 0,a_1+a_3\ne 0,a_1+a_2+a_3=0\}$	$(1,-T,0)$	$c_1=\frac{1}{a_1},c_3=-\frac{a_3}{a_1(a_1+a_3)},c_2\in \mathbb{R}$
$\left\{(a_1,a_{2}T,a_{3}I)\right|a_1\ne 0,a_1+a_3\ne 0,a_1+a_2+a_3\ne 0\}$	$(1,0,0)$	$c_1=\frac{1}{a_1},c_3=-\frac{a_3}{a_1(a_1+a_3)},c_2=-\frac{a_2}{(a_1+a_3)(a_1+a_2+a_3)}$

gives all the subset, which has the same neutral element, and the corresponding neutral element and opposite elements.

Theorem 14.

For algebra system $(NT,⭑)$, let $V_c=\left\{a\right|a\in NT\wedge neut\left(a\right)=c\}$, $V_{c⭑d}=\{a⭑b|a,b\in NT\wedge neut\left(a\right)=c,neut\left(b\right)=d\}$, we have:

(1) $V_c$ is a neutrosophic triplet subgroup of $NT$.

(2) $V_{c⭑d}$ is a neutrosophic triplet subgroup of $NT$.

Theorem 15.

For an algebra system $(NT,⭑)$, let $V_c=\left\{a\right|a\in NT\wedge neut\left(a\right)=c\}$, we have:

(1) $V_c\cap V_d=\varnothing$ if $c\ne d$.

(2) $NT={\cup }_{c\in NS}{V}_c$. So ${\cup }_{c\in NS}{V}_c$ is a partition of $NT$, where $NS$ is a set, which contains all the neutral elements of $(NT,⭑)$.

5. Conclusions

In the paper, we prove that $(NQ,\ast )$(or $NQ,⭑$) is a neutrosophic extended triplet group, and provide new examples of a neutrosophic extended triplet group. We also explore the algebra structure properties of neutrosophic quadruple numbers. Moreover, we discuss two kinds of degenerate systems of neutrosophic quadruple numbers. For neutrosophic quadruple numbers, the results in the paper can be extended to general fields. In the following, we will explore the relation of neutrosophic quadruple numbers and other algebra systems [24,25,26]. Moreover, on the one hand, we will discuss the neutrosophic quadruple numbers based on some particular ring which can form a neutrosophic extended triplet group, on the other hand, we will introduce a new operation ∘ in order to guarantee $(NQ,\ast ,\circ )$ is a neutrosophic triplet ring.