StudyontheAlgebraicStructureofRefined NeutrosophicNumbers
QiaoyanLi 1,YingcangMa 1,*,XiaohongZhang 2 andJuanjuanZhang 1
1 SchoolofScience,Xi’anPolytechnicUniversity,Xi’an710048,China
2 SchoolofArtsandSciences,ShaanxiUniversityofScience&Technology,Xi’an710021,China
* Correspondence:mayingcang@126.com
Received:28May2019;Accepted:22July2019;Published:27July2019
Abstract: Thispaperaimstoexplorethealgebrastructureofrefinedneutrosophicnumbers.Firstly, thealgebrastructureofneutrosophicquadruplenumbersonageneralfieldisstudied.Secondly, Theadditionoperator ⊕ andmultiplicationoperator ⊗ onrefinedneutrosophicnumbersareproposed andthealgebrastructureisdiscussed.Werevealthatthesetofneutrosophicrefinednumberswith anadditiveoperationisanabeliangroupandthesetofneutrosophicrefinednumberswitha multiplicationoperationisaneutrosophicextendedtripletgroup.Moreover,algorithmsforsolving theneutralelementandoppositeelementsofeachrefinedneutrosophicnumberaregiven.
Keywords: neutrosophicextendedtripletgroup;neutrosophicquadruplenumbers;refined neutrosophicnumbers;refinedneutrosophicquadruplenumbers;neutrosophicset
1.Introduction
ThenotionofneutrosophicsetwasproposedbyF.Smarandache[1],whichisanextensionoffuzzy setandinordertosolvereal-worldproblems.Aneutrosophicsethasthreemembershipfunctions, andeachmembershipdegreeisarealstandardornon-standardsubsetofthenonstandardunitinterval ]0 ,1+ [= 0 ∪ [0,1] ∪ 1+ .
Inrecentyears,theideaofneutrosophicsethasbeenapplicableinrelatedalgebraicstructures. Amongthesealgebraicstructures,SmarandacheandAli[2]proposedthealgebraicsystemneutrosophic tripletgroup(NTG),whichisanextensionoftheclassicalgroupbuttheneutralelementisdifferent fromtheclassicalalgebraicunitelement.Toregardtheunitelementasaspecialneutralelement, theneutrosophicextendedtripletgroup(NETG)hasbeenproposed[3,4]andtheclassicalgroupis regardedasaspecialcaseofaNETG.Moreover,someresearchpapershavecarriedoutin-depth researchbasedonNTG(NETG).Forexample,theinclusionrelationsofneutrosophicsets[5], neutrosophictripletcoset[6],neutrosophicdupletsemi-groups[7],generalizedneutrosophicextended tripletgroup[8],AG-neutrosophicextendedtripletloops[9,10],theneutrosophicsettheoryto pseudo-BCIalgebras[11],neutrosophictripletringandaneutrosophictripletfield[12,13],neutrosophic tripletnormedspace[14],neutrosophicsoftsets[15],neutrosophicvectorspaces[16]andsoonhave beenstudied.
AsanexampleofNETG,Ma[8]revealedthatforeach n ∈ Z+ , n ≥ 2, (Zn, ⊗) isacommutative NETGifandonlyifthefactorizationof n isaproductofsinglefactors.Asanotherexample,Ma[17] showedthatthesetofneutrosophicquadruplenumberswithamultiplicationoperationisaNETG. Theconceptofneutrosophicnumbersoftheform a + bI,where I istheindeterminacywith In = I,and, a andarerealorcomplexnumbers.If I intomanytypesofindeterminacies I1, I2, , Iq, in[18],Smarandacheextendedtheneutrosophicnumbers a + bI intorefinedneutrosophicnumbers oftheform a + b1 I1 + b2 I2 + + bn In,where a, b1, b2, , bn arerealorcomplexnumbersand consideredtherefinedneutrosophicsetbasedontheserefinedneutrosophicnumbers.Thenotionof Symmetry 2019, 11,954;doi:10.3390/sym11080954 www.mdpi.com/journal/symmetry
neutrosophicquadruplenumber,whichhasform: NQ = a + bT + cI + dF where a, b, c, d arereal(or complex)numbers;and T isthetruth/membership/probability; I istheindeterminacy;and F isthe false/membership/improbabilityarecalledNeutrosophicQuadruple(Real,respectively,Complex) Numbers.“a”iscalledtheknownpartof NQ,while bT + cI + dF iscalledtheunknownpartof NQ. Similartorefinedneutrosophicnumbers,if T canbesplitintomanytypesoftruths, T1, T2, , Tp, I intomanytypesofindeterminacies, I1, I2, , Ir,and F intomanytypesoffalsities, F1, F2, , Fr,we cangettherefinedneutrosophicquadruplenumbers.Weknowthatthesetofneutrosophicquadruple numberswithamultiplicationoperationisaNETG.Inthispaper,weexplorethealgebrastructure ofrefinedneutrosophicnumbers(refinedneutrosophicquadruplenumbers)andgivenewexamples ofNETG.Infact,thesolvingmethodoftheneutralelementandoppositeelementsforeachrefined neutrosophicnumberisdifferentfromthesolvingmethodforeachneutrosophicquadruplenumber. Thepaperisorganizedasfollows.Section 2 givesthebasicconcepts.InSection 3,weshowthat thesetofneutrosophicquadruplenumbersonthegeneralfieldwithamultiplicationoperationalso consistsofaNETG.InSection 4,thealgebrastructureofrefinedneutrosophicnumbersandrefined neutrosophicquadruplenumbersarestudied.Finally,thesummaryandfutureworkispresentedin Section 5
2.BasicConcepts
Inthissection,weprovidetherelatedbasicdefinitionsandpropertiesofNETG,neutrosophic quadruplenumbers,andrefinedneutrosophicnumbers(fordetails,see[3,4,18–20]).
Definition1 ([3,4]). Let N beanon-emptysettogetherwithabinaryoperation ∗.Then, N iscalleda neutrosophicextendedtripletsetif,forany a ∈ N,thereexistsaneutralof“a”(denoteby neut(a)),andan oppositeof“a”(denotebyanti(a)),suchthatneut(a) ∈ N,anti(a) ∈ Nand: a ∗ neut(a)= neut(a) ∗ a = a, a ∗ anti(a)= anti(a) ∗ a = neut(a) Thetriplet (a, neut(a), anti(a)) iscalledaneutrosophicextendedtriplet.
Definition2 ([3,4]). Let (N, ∗) beaneutrosophicextendedtripletset.Then, N iscalledaneutrosophic extendedtripletgroup(NETG),ifthefollowingconditionsaresatisfied:
(1) (N, ∗) iswell-defined,i.e.,foranya, b ∈ N,onehasa ∗ b ∈ N. (2) (N, ∗) isassociative,i.e., (a ∗ b) ∗ c = a ∗ (b ∗ c) foralla, b, c ∈ N.
ANETGNiscalledacommutativeNETGifforalla, b ∈ N, a ∗ b = b ∗ a.
Proposition1 ([4]). (N, ∗) beaNETG.Wehave:
(1) neut(a) isuniqueforanya ∈ N.
(2) neut(a) ∗ neut(a)= neut(a) foranya ∈ N. (3) neut(neut(a))= neut(a) foranya ∈ N.
Definition3 ([18,19]). Aneutrosophicnumberisanumberoftheform (a, bI),where I istheindeterminacy with I2 = I,and a and b arerealorcomplexnumbers.Arefinedneutrosophicnumberisanumberoftheform (a0, a1 I1, a2 I2, ··· , an In ),where I1, I2, ··· , In aredifferenttypesofindeterminacies,and a0, a1, a2, ··· , an are realorcomplexnumbers.ThesetNNdefinedby
NN = {(a, bI)|a, b ∈ R or C}.(1) iscalledaneutrosophicsetofneutrosophicnumbers.ThesetRNNdefinedby
RNN = {(a0, aI1, a2 I2, , an In )|a0, a1, a2, , an ∈ R or C}.(2)
iscalledaneutrosophicsetofrefinedneutrosophicnumbers.
Definition4 ([18,20]). Aneutrosophicquadruplenumberisanumberoftheform (a, bT, cI, dF),where T, I and F havetheirusualneutrosophiclogicmeanings,i.e.,truth,indeterminacyandfalse,respectively,and a, b, c, d ∈ R or C.ThesetNQdefinedby
NQ = {(a, bT, cI, dF)|a, b, c, d ∈ R or C}.(3) iscalledaneutrosophicsetofquadruplenumbers.Foraneutrosophicquadruplenumber (a, bT, cI, dF), a is calledtheknownpartand (bT, cI, dF) iscalledtheunknownpart.ThesetRNQdefinedby
RNQ = {(a, b1T1, b2T2, , bp Tp, c1 I1, c2 I2, , cq Iq, d1 F1, d2 F2, , dr Fr )| a, b1, b2, , bp, c1, c2, , cq, d1, d2, , dr ∈ R or C}. (4) iscalledaneutrosophicsetofrefinedneutrosophicquadruplenumbers.
Definition5 ([18,20]). Let N beaset,endowedwithatotalorder a ≺ b,named“a prevailedby b”, “aless strongerthan b”or“a lesspreferredthan b”.Weconsider a b as“a prevailedbyorequalto b”,“a less strongerthanorequaltob”,or“alesspreferredthanorequaltob”.
Foranyelementsa, b ∈ N,witha b,onehastheabsorbancelaw:
a · b = b · a = absorb(a, b)= max(a, b)= b.(5) whichmeansthatthebiggerelementabsorbsthesmallerelement.Clearly,
a · a = a2 = absorb(a, a)= max(a, a)= a.(6) and a1 a2 an = max(a1, a2, , an ).(7)
Analogously,wesaythat“a b”andweread:“aprevailstob”,“aisstrongerthanb”or“aispreferred tob”.Inaddition, a b,andweread:“aprevailsorisequaltob”,“aisstrongerthanorequaltob”,or“ais preferredorequaltob”.
Definition6 ([18,20]). Considertheset {T, I, F}.Supposeinanoptimisticwayweconsidertheprevalence order T I F.Then,wehave: TI = IT = max(T, I)= T, TF = FT = max(T, F)= T, IF = FI = max(I, F)= I, TT = T2 = T, II = I2 = I, FF = F2 = F.
Analogously,supposeinapessimisticwayweconsidertheprevalenceorder T ≺ I ≺ F.Then,wehave: TI = IT = max(T, I)= I, TF = FT = max(T, F)= F, IF = FI = max(I, F)= F, TT = T2 = T, II = I2 = I, FF = F2 = F.
Definition7 ([18,20]) Let a =(a1, a2T, a3 I, a4 F), b =(b1, b2T, b3 I, b4 F) ∈ NQ.Suppose,inanpessimistic way,theneutrosophicexpertconsiderstheprevalenceorder T ≺ I ≺ F.Then,themultiplicationoperationis definedasfollows:
a ∗ b =(a1, a2T, a3 I, a4 F) ∗ (b1, b2T, b3 I, b4 F) =(a1b1, (a1b2 + a2b1 + a2b2)T, (a1b3 + a2b3 + a3b1 + a3b2 + a3b3)I, (a1b4 + a2b4 + a3b4 + a4b1 + a4b2 + a4b3 + a4b4)F).
(8)
SupposeinanoptimisticwaytheneutrosophicexpertconsiderstheprevalenceorderT I F.Then, a b =(a1, a2T, a3 I, a4 F) (b1, b2T, b3 I, b4 F) =(a1b1, (a1b2 + a2b1 + a2b2 + a3b2 + a4b2 + a2b3 + a2b4)T, (a1b3 + a3b1 + a3b3 + a3b4 + a4b3)I, (a1b4 + a4b1 + a4b4)F).
(9)
Proposition2 ([18,20]). LetNQ = {(a, bT, cI, dF) : a, b, c, d ∈ R or C}.Wehave:
(1) (NQ, ∗) isacommutativemonoid. (2) (NQ, ) isacommutativemonoid.
Theorem1. [17]Forthealgebrasystem (NQ, ∗)(or (NQ, )),foreveryelement a ∈ NQ,thereexists theneutralelement neut(a) andoppositeelement anti(a),whichmeansthatthealgebrasystem (NQ, ∗) (or (NQ, ))isaNETG.
3.TheAlgebraStructureofNeutrosophicQuadrupleNumbersonGeneralField
Fromtheabovesection,wecanseethattheneutrosophicquadruplenumbersaredefinedon numberfield R or C.Inthissection,thenotionsoftheneutrosophicquadruplenumbersonageneral fieldareintroducedandthealgebrastructureoftheneutrosophicquadruplenumbersongeneralfield isexplored.
Let (F, +, ·) beafield,and 0 and 1 aretheunitelementsforoperator + and ·,respectively.For every a ∈ F, a istheinverseelementof a foroperator +,and a 1 istheinverseelementof a for operator ·.Inthefollowing,field (F, +, ·) isdenotedby F forshortand a · b isdenotedby ab
Definition8. Let F beafield;aneutrosophicquadruplenumberisanumberoftheform (a, bT, cI, dF), where T, I, F havetheirusualneutrosophiclogicmeanings,i.e.,truth,indeterminacyandfalse,respectively, anda, b, c, d ∈ F.ThesetNQFdefinedby
NQF = {(a, bT, cI, dF)|a, b, c, d ∈ F}.(10) iscalledaneutrosophicsetofquadruplenumbersonfield F.
Definition9. Let a =(a1, a2T, a3 I, a4 F), b =(b1, b2T, b3 I, b4 F) ∈ NQF,thentheadditionoperatoris definedasfollows: a ⊕ b =(a1 + b1, (a2 + b2)T, (a3 + b3)I, (a4 + b4)F).(11)
Definition10. Let a =(a1, a2T, a3 I, a4 F), b =(b1, b2T, b3 I, b4 F) ∈ NQF.Suppose,inanpessimisticway, theneutrosophicexpertconsiderstheprevalenceorder T ≺ I ≺ F.Then,themultiplicationoperationisdefined asfollows:
a ∗ b =(a1, a2T, a3 I, a4 F) ∗ (b1, b2T, b3 I, b4 F) =(a1b1, (a1b2 + a2b1 + a2b2)T, (a1b3 + a2b3 + a3b1 + a3b2 + a3b3)I, (a1b4 + a2b4 + a3b4 + a4b1 + a4b2 + a4b3 + a4b4)F)
(12)
SupposeinanoptimisticwaytheneutrosophicexpertconsiderstheprevalenceorderT I F.Then, a b =(a1, a2T, a3 I, a4 F) (b1, b2T, b3 I, b4 F) =(a1b1, (a1b2 + a2b1 + a2b2 + a3b2 + a4b2 + a2b3 + a2b4)T, (a1b3 + a3b1 + a3b3 + a3b4 + a4b3)I, (a1b4 + a4b1 + a4b4)F).
(13)
Theorem2. (NQF, ⊕) isanabeliangroup. Proof. Itisobvious.
Theorem3. Forthealgebrasystem (NQF, ∗) (or (NQF, )),foreveryelement a ∈ NQF,thereexists theneutralelement neut(a) andoppositeelement anti(a),thusthealgebrasystem (NQF, ∗) (or (NQF, )) isaNETG.
Theproof’smethodissimilartotheproofwhen F = R in[17].Thedetailedproofisomitted. Foralgebrasystem (NQF, ∗),Table 1 givesallthesubsetswhichhavethesameneutralelement,and thecorrespondingneutralelementandoppositeelements.Inthefollowing,fromtwoexamples,we showthathowtosolvethetheneutralelementandoppositeelementsofeachelementforalgebra system (NQF, ∗) ondifferentfields.
Example1. Let F = Z5 = {[0], [1], [2], [3], [4]},then (Z5, +, ·) isafield,where + and · aretheclassical modadditionandmultiplication,respectively.Foralgebrasystem (NQF, ∗),if a =(a1, a2T, a3 I, a4 F)= ([2], [4]T, [3]I, [1]F),i.e., a1 =[0], a1 + a2 =[0], a1 + a2 + a3 =[0], a1 + a2 + a3 + a4 =[0], then,fromTable 1,wecanget neut(a)=([1], [0], [0], [4]F).Let anti(a)=(c1, c2T, c3 I, c4 F), so c1 = a 1 1 =[3], c2 =[3], c3 =[3], c4 ∈ Z5,thus anti(a)=([3], [3]T, [3]I, c4 F),where c4 ∈ Z5.Thus,we caneasilygettheneutralelementandoppositeelementsofeachneutrosophicquadruplenumberongeneralfield. Formoreexamples,seethefollowing:
1. Letb =([1], [2]T, [1]I, [3]F),thenneut(b)=([1], [0], [0], [0]) andanti(b)=([1], [1]T, [2]I, [4]F).
2. Let c =([0], [0], [1]I, [4]F),then neut(c)=([0], [0], [1], [4]) and anti(c)=(c1, c2T, c3 I, c4 F), wherec1 ⊕ c2 ⊕ c3 =[1], c4 ∈ Z5.
3. Let d =([0], [1]T, [1]I, [1]F),then neut(d)=([0], [1]T, [0], [0]) and anti(d)=(c1, c2T, [2]I, [4]F), wherec1 ⊕ c2 =[1]
Table1. Thecorrespondingneutralelementandoppositeelementsfor (NQF, ∗)
TheSubsetof NQF Neutral Element OppositeElements (c1, c2T, c3 I, c4 F)
{(0,0,0,0)} (0,0,0,0) ci ∈ F
{(0,0,0, a4 F)|a4 = 0} (0,0,0, F) c1 + c2 + c3 + c4 = a 1 4 {(0,0, a3 I, a3 F)|a3 = 0} (0,0, I, F) c1 + c2 + c3 = a 1 3 , c4 ∈ F {(0,0, a3 I, a4 F)|a3 = 0, a3 + a4 = 0} (0,0, I,0) c1 + c2 + c3 = a 1 3 , c4 = (a4 a 1 3 (a3 + a4) 1) {(0, a2T, a2 I,0)|a2 = 0} (0, T, I,0)} c1 + c2 = a 1 2 , c3, c4 ∈ F {(0, a2T, a2 I, a4 F)|a2 = 0, a4 = 0} (0, T, I, F) c1 + c2 = a 1 2 , c3 + c4 = a 1 4 +( a 1 2 ) {(0, a2T, a3 I, a4 F)|a2 = 0, a2 + a3 = 0, a2 + a3 + a4 = 0} (0, T,0, F) c1 + c2 = a 1 2 , c3 = (a3 a 1 2 (a2 + a3) 1), c4 ∈ F
{(0, a2T, a3 I, a4 F)|a2 = 0, a2 + a3 = 0, a2 + a3 + a4 = 0} (0, T,0,0) c1 + c2 = a 1 2 , c3 = (a3 a 1 2 (a2 + a3) 1), c4 = (a4(a2 + a3) 1(a2 + a3 + a4) 1)
{(a1, a1T,0,0)|a1 = 0} (1, T,0,0)} c1 = a 1 1 , c2, c3, c4 ∈ F {(a1, a1T,0, a4 F)|a1 = 0, a4 = 0} (1, T,0, F) c1 = a 1 1 , c2 + c3 + c4 = a 1 4 +( a 1 1 )
{(a1, a1T, a3 I, a3 F)|a1 = 0, a3 = 0} (1, T, I, F) c1 = a 1 1 , c2 + c3 = a 1 3 a 1 1 , c4 ∈ F
{(a1, a1T, a3 I, a4 F)|a1 = 0, a3 = 0, a3 + a4 = 0} (1, T, I,0) c1 = a 1 1 , c2 + c3 = a 1 3 +( a 1 1 ), c4 = (a4 a 1 3 (a3 + a4) 1)
{(a1, a2T, a3 I,0)|a1 = 0, a1 + a2 = 0, a1 + a2 + a3 = 0}
{(a1, a2T, a3 I, a4 F)|a1 = 0, a1 + a2 = 0, a1 + a2 + a3 = 0, a4 = 0}
{(a1, a2T, a3 I, a4 F)|a1 = 0, a1 + a2 = 0, a1 + a2 + a3 = 0, a1 + a2 + a3 + a4 = 0}
{(a1, a2T, a3 I, a4 F)|a1 = 0, a1 + a2 = 0, a1 + a2 + a3 = 0, a1 + a2 + a3 + a4 = 0}
(1,0, I,0) c1 = a 1 1 , c2 = (a2 a 1 1 (a1 + a2) 1), c3, c4 ∈ F
(1,0, I, F) c1 = a 1 1 , c2 = (a2 a 1 1 (a1 + a2) 1), c3 + c4 = a 1 4 +( (a1 + a2) 1)
(1,0,0, F) c1 = a 1 1 , c2 = (a2 a 1 1 (a1 + a2) 1), c3 = (a3(a1 + a2) 1(a1 + a2 + a3) 1), c4 ∈ F
(1,0,0,0) c1 = a 1 1 , c2 = (a2 a 1 1 (a1 + a2) 1), c3 = (a3(a1 + a2) 1(a1 + a2 + a3) 1), c4 = (a4(a1 + a2 + a3) 1(a1 + a2 + a3 + a4) 1)
Example2. Let F4 = {0,1, x, y},theoperators + and on F4 isdefinedbyTable 2.
Table2. Theoperators + and on F4 + 01 xy 01 xy 001 xy 00000 110 yx 101 xy xxy 01 x 0 xy 1 yyx 10 y 0 y 1 x
Then, (F4, +, ·) isafield.SetNQF = {(a, bT, cI, dF)|a, b, c, d ∈ F4}.Wehave:
1. Let a =(0,0, xI, xF),then neut(a)=(0,0, I, F) and anti(a)=(c1, c2T, c3 I, c4 F),where c1 + c2 + c3 = y, c4 ∈ F4
2. Let b =(0, xT, xI, yF),then neut(b)=(0, T, I, F) and anti(b)=(c1, c2T, c3 I, c4 F),where c1 + c2 = y, c3 + c4 = 1
3. Letc =(x, xT,0,0),thenneut(c)=(1, T,0,0) andanti(c)=(y, c2T, c3 I, c4 F),wherec2, c3, c4 ∈ F4.
Inthesameway,foralgebrasystem (NQF, ),Table 3 givesallthesubsetswhichhavethesame neutralelement,andthecorrespondingneutralelementandoppositeelements.
Table3. Thecorrespondingneutralelementandoppositeelementsfor (NQF, )
TheSubsetof NQF Neutral Element OppositeElements (c1, c2T, c3 I, c4 F)
{(0,0,0,0)}
(0,0,0,0) ci ∈ F {(0, a2T,0,0)|a2 = 0} (0, T,0,0) c1 + c2 + c3 + c4 = a 1 2
{(0, a3T, a3 I,0)|a3 = 0} (0, T, I,0) c1 + c3 + c4 = a 1 3 , c2 ∈ F {(0, a2T, a3 I, a4 F)|a3 = 0, a2 + a3 = 0} (0,0, I,0) c1 + c3 + c4 = a 1 3 , c2 = (a2 a 1 3 (a2 + a3) 1)
{(0,0, a4 I, a4 F)|a4 = 0} (0,0, I, F)} c1 + c4 = a 1 4 , c2, c3 ∈ F {(0, a2T, a4 I, a4 F)|a2 = 0, a4 = 0} (0, T, I, F) c1 + c4 = a 1 4 , c2 + c3 = a 1 2 +( a 1 4 )
{(0, a2T, a3 I, a4 F)|a4 = 0, a3 + a4 = 0, a2 + a3 + a4 = 0} (0, T,0, F) c1 + c4 = a 1 4 , c3 = (a3 a 1 4 (a3 + a4) 1), c2 ∈ F
{(0, a2T, a3 I, a4 F)|a4 = 0, a3 + a4 = 0, a2 + a3 + a4 = 0} (0,0,0, F) c1 + c4 = a 1 4 , c3 = (a3 a 1 4 (a3 + a4) 1), c2 = (a2(a3 + a4) 1(a2 + a3 + a4) 1)
{(a1,0,0, a1 F)|a1 = 0} (1,0,0, F)} c1 = a 1 1 , c2, c3, c4 ∈ F
{(a1, a2T,0, a1 F)|a1 = 0, a2 = 0} (1, T,0, F) c1 = a 1 1 , c2 + c3 + c4 = a 1 2 +( a 1 1 )
{(a1, a3T, a3 I, a1 F)|a1 = 0, a3 = 0} (1, T, I, F) c1 = a 1 1 , c3 + c4 = a 1 3 +( a 1 1 ), c4 ∈ F
{(a1, a2T, a3 I, a1 F)|a1 = 0, a3 = 0, a2 + a3 = 0}
{(a1,0, a3 I, a4 F)|a1 = 0, a1 + a4 = 0, a1 + a3 + a4 = 0}
{(a1, a2T, a3 I, a4 F)|a1 = 0, a1 + a4 = 0, a1 + a3 + a4 = 0, a2 = 0}
{(a1, a2T, a3 I, a4 F)|a1 = 0, a1 + a4 = 0, a1 + a3 + a4 = 0, a1 + a2 + a3 + a4 = 0}
{(a1, a2T, a3 I, a4 F)|a1 = 0, a1 + a4 = 0, a1 + a3 + a4 = 0, a1 + a2 + a3 + a4 = 0}
(1,0, I, F) c1 = a 1 1 , c3 + c4 = a 1 3 +( a 1 1 ), c2 = (a2 a 1 3 (a2 + a3) 1)
(1,0, I,0) c1 = a 1 1 , c4 = (a4 a 1 1 (a1 + a4) 1), c2, c3 ∈ F
(1, T, I,0) c1 = a 1 1 , c4 = (a4 a 1 1 (a1 + a4) 1), c2 + c3 = a 1 2 +( (a1 + a4) 1)
(1, T,0,0) c1 = a 1 1 , c4 = (a4 a 1 1 (a1 + a4) 1), c3 = (a3(a1 + a4) 1(a1 + a3 + a4) 1), c2 ∈ F
(1,0,0,0) c1 = a 1 1 , c4 = (a4 a 1 1 (a1 + a4) 1), c3 = (a3(a1 + a4) 1(a1 + a3 + a4) 1), c2 = (a2(a1 + a3 + a4) 1(a1 + a2 + a3 + a4) 1)
4.TheAlgebraStructureofRefinedNeutrosophicNumbersonGeneralField
Intheabovesection,werevealthatthealgebrastructureoftheneutrosophicquadruplenumbers ongeneralfield.Inthissection,weexplorethethealgebrastructureoftherefinedneutrosophic numbers(refinedneutrosophicquadruplenumbers)ongeneralfield.
Definition11. Let F beafield;arefined n-aryneutrosophicnumberisanumberofthe form (a0, a1 I1, a2 I2, ··· , an In ),where I1, I2, ··· ,and In aredifferenttypesofindeterminacies, anda0, a1, a2, , an ∈ F.ThesetRNFn definedby
RNFn = {(a0, a1 I1, a2 I2, , an In )|a0, a1, a2, , an ∈ F}.(14) iscalledarefinedneutrosophicsetonfield F
Definition12. Let x =(a0, a1 I1, a2 I2, , an In ), y =(b0, b1 I1, b2 I2, , bn In ) ∈ RNFn,thentheaddition operatoronRNFn isdefinedasfollows:
x ⊕ y =(a0 + b0, (a1 + b1)I1, (a2 + b2)I2, ··· , (an + bn )In ).(15) Definition13. Let x =(a0, a1 I1, a2 I2, , an In ), y =(b0, b1 I1, b2 I2, , bn In ) ∈ RNFn,theneutrosophic expertconsiderstheprevalenceorder I1 ≺ I2 ≺···≺ In.Then,themultiplicationoperationisdefined asfollows:
x ∗ y =(a0, a1 I1, a2 I2, , an In ) ∗ (b0, b1 I1, b2 I2, , bn In ) =(a0b0, (a0b1 + a1b1 + a1b2)I1, (a0b2 + a1b2 + a2b0 + a2b1 + a2b2)I2, , (a0bn + a1bn + a2bn + + an 1bn + an b0 + an b1 + + an bn )In ). (16)
TheneutrosophicexpertconsiderstheprevalenceorderI1 I2 ··· In.Then,
x y =(a0, a1 I1, a2 I2, , an In ) (b0, b1 I1, b2 I2, , bn In ) =(a0b0, (a0b1 + a1b1 + + an b1 + a1b0 + a1b2 + a1b3 + + a1bn )I1, , (a0bn 1 + an 1b0 + an 1bn 1 + an 1bn + an bn 1)In 1, (a0bn + an b0 + an bn )In ).
Theorem4. (RNFn, ⊕) isanabeliangroup.
Proof. Theproofisobvious.
(17)
Theorem5. Forthealgebrasystem (RNFn, ∗) (or (RNFn, )),foreveryelement a ∈ RNFn,thereexists theneutralelement neut(a) andoppositeelement anti(a),thusthealgebrasystem (RNFn, ∗) (or (RNFn, )) isaNETG.
Proof. Weuseappliedmathematicalinductionfor n andonlydiscussthealgebrasystem (RNFn, ∗) Thealgebrasystem (RNFn, ) hasasimilarproof.
If n = 2,forrefined2-aryneutrosophicset,whichissameasneutrosophicbinarynumbersset in[17],fromTheorem7in[17],wecanseethatforeveryelement a ∈ RNF2,thereexiststheneutral element neut(a) andoppositeelement anti(a),thusthealgebrasystem (RNF2, ∗) isaNETG.
Assumethattherefined n-aryneutrosophicset RNFn isaNETG.Thatis,foreveryelement a ∈ RNFn,thereexiststheneutralelement neut(a) andoppositeelement anti(a).Inthefollowing,we provethatfortherefined (n + 1)-aryneutrosophicset,whichisaNETG.
Foreach a =(a0, a1 I1, a2 I2, , an+1 In+1) ∈ RNFn+1,let a =(a0, a1 I1, a2 I2, , an In ), being a ∈ RNFn,then,fromtheaboveassumptioncondition, neut(a ) and anti(a ) exist andlet neut(a )=(b0, b1 I1, b2 I2, , bn In ), anti(a )=(c0, c1 I1, c2 I2, , cn In ).Weprovefor a =(a0, a1 I1, a2 I2, , an+1 In+1), neut(a) and anti(a) exist.Wediscussfromthedifferent casesof an+1.
CaseA:If an+1 = 0,Being (a0, a1 I1, a2 I2, , an In,0) ∗ (b0, b1 I1, b2 I2, , bn In, bn+1 In+1)= (a0, a1 I1, a2 I2, , an In,0),thatis (a0 + a1 + + an )bn+1 = 0,thuswediscussfrom a0 + a1 + + an = 0or a0 + a1 + + an = 0.
CaseA1:If an+1 = 0, a0 + a1 + + an = 0,so (a0, a1 I1, a2 I2, , an In,0) ∗ (c0, c1 I1, c2 I2, , cn In, cn+1 In+1)=(b0, b1 I1, b2 I2, , bn In, bn+1 In+1),thatis bn+1 = 0 and cn+1 can bechosenarbitrarilyin F
CaseA2:If an+1 = 0, a0 + a1 + ··· + an =
0,sofrom (a0 + a1 + ··· + an )bn+1 = 0,wehave bn+1 = 0, from (a0, a1 I1, a2 I2, ··· , an In,0) ∗ (c0, c1 I1, c2 I2, ··· , cn In, cn+1 In+1)=(b0, b1 I1, b2 I2, ··· , bn In,0), thus cn+1 = 0.
CaseB:If an+1 = 0,being (a0, a1 I1, a2 I2, ··· , an In, an+1 In+1) ∗ (b0, b1 I1, b2 I2, ··· , bn In, bn+1 In+1)= (a0, a1 I1, a2 I2, ··· , an In, an+1 In+1),thatis (a0 + a1 + ··· + an+1)bn+1 + an+1(b0 + b1 + ··· + bn )= an+1, thatis (a0 + a1 + ··· + an+1)bn+1 = an+1(1 b0 b1 −···− bn),wediscussfrom a0 + a1 + ··· + an+1 = 0 or a0 + a1 + ··· + an+1 = 0.
CaseB1:If an+1 = 0, a0 + a1 + + an+1 = 0,wehave b0 + b1 + + bn = 1,sofrom (a0, a1 I1, a2 I2, , an In, an+1 In+1) ∗ (c0, c1 I1, c2 I2, , cn In, cn+1 In+1)= (b0, b1 I1, b2 I2, , bn In, bn+1 In+1),thatis bn+1 =(c0 + c1 + + cn )an+1 = bn+1 and cn+1 canbe chosenarbitrarilyin F
CaseB2:If an+1 = 0, a0 + a1 + + an+1 = 0,wehave bn+1 = an+1(1 b0 b1 bn )(a0 + a1 + a2 an ) 1,sofrom (a0, a1 I1, a2 I2, , an In, an+1 In+1) ∗ (c0, c1 I1, c2 I2, , cn In, cn+1 In+1)= (b0, b1 I1, b2 I2, , bn In, bn+1 In+1),thatisand cn+1 =(bn+1 an+1(c0 + c1 + + cn ))(a0 + a1 + + an+1) 1 .
Fromtheaboveanalysis,wecanseethat,foreach a ∈ RNFn,whichhastheneutralelement neut(a) andoppositeelement anti(a),fromthemathematicalinductionmethod,wecanobtainthat thealgebrasystem (RNF, ∗) isaNETG.
Foralgebrasystem (RNFn+1, ∗),if a =(a0, a1 I1, a2 I2, , an+1 In+1),let a = (a0, a1 I1, a2 I2, , an In ) ∈ RNFn,ifwehave neut(a )=(b0, b1 I1, b2 I2, , bn In ), anti(a )= (c0, c1 I1, c2 I2, , cn In ).Then,thecorrespondingneutralelementandoppositeelementsof a aregiven inTable 4 accordingtothedifferentcasesof an+1 Table4. Thecorrespondingneutralelementandoppositeelementsfor (RNFn+1, ∗)
TheSubset {(a0, a1 I1, ··· , an+1 In+1)} NeutralElement (b0, b1 I1, ··· , bn In , bn+1 In+1)
OppositeElements (c0, c1 I1, ··· , cn In , cn+1 In+1) an+1 = 0, a0 + a1 + ··· + an = 0 bn+1 = 0 cn+1 ∈ F an+1 = 0, a0 + a1 + ··· + an = 0 bn+1 = 0 cn+1 = 0 an+1 = 0, a0 + a1 + ··· + an + an+1 = 0 bn+1 =(c0 + c1 + ··· + cn )an+1 cn+1 ∈ F an+1 = 0, a0 + a1 + ··· + an + an+1 = 0 bn+1 = an+1(1 b0 b1 −···− bn )(a0 + a1 + a2 + ··· + an ) 1) cn+1 =(bn+1 an+1(c0 + c1 + ··· + cn ))(a0 + a1 + ··· + an+1) 1
Foralgebrasystem (RNFn, ∗),accordingtotheresultsinTables3and4in[17],wecaneasily obtaintheneutralelementandoppositeelementswhen n = 1,2 ongeneralfields.InTable 1,wecanget theneutralelementandoppositeelementswhen n = 3 ongeneralfield.Thus,fromTheorem 5,wecan gettheneutralelementandoppositeelementsofeachelementin RNFn step-by-step.Thesolving methodisgivenbyAlgorithm 1 andthefollowingexampleisusedtoexplainthealgorithm.
Algorithm1
Solvingtheneutralelementandoppositeelementsofeachelementin (RNFn, ∗)
Input: a =(a0, a1 I1, a2 I2, ··· , an In ), n ≥ 3
1: i = 3; 2:For i = 3: n
3: Ai =(a0, a1 I1, a2 I2, , ai Ii ); 4:If i == 3
5:Obtain neut(Ai ), anti(Ai ) byTable 1; 6:else
7:Obtain neut(Ai ), anti(Ai ) byTable 4 combiningthevaluesof neut(Ai 1) and anti(Ai 1); 8:end 9:Save neut(Ai ), anti(Ai ); 10:end
Output: neut(a)= neut(An ), anti(a)= anti(An )
Example3. Foralgebrasystem (RNF2, ∗),andset F = R.If a =(a0, a1 I1)=(0, I1),fromTable3in[17], wecangetneut(a)=(0, I1) andanti(a)=(c0, c1 I1),wherec0 + c1 = 1. Inthefollowing,weusetwomethodstosolvethetheneutralelementandoppositeelementsof b = (a0, a1 I1, a 2I2)=(0, I1, I2) ∈ RNF3 andwegetthesameresults.
1. Algorithm 1:FromTable 2,being a3 = 0 and a0 + a1 + a2 = 0,thus b2 =(c0 + c1) 1 = 1, thatis neut(b)=(0, I1, I2) and anti(a)=(c0, c1 I1, c2 I2),where c0 + c1 = 1 and c2 canbechosen arbitrarilyin R.
2. RsultsfromTable4in[17]:Being a2 = 0 and a0 + a1 + a2 = 0,thus neut(b)=(0, I1, I2) and anti(a)=(c0, c1 I1, c2 I2),wherec0 + c1 = 1 andc2 canbechosenarbitrarilyin R
Example4. Foralgebrasystem (RNF3, ∗),andset F = R.If a =(a0, a1 I1, a2 I2)=(1, I1, I2) ∈ RNF3, fromTable4in[17],wecangetneut(a)=(1, I1, I2) andanti(a)=(1, c1 I1, c2 I2),wherec1 + c2 = 0 Inthesameway,weusetwomethodstosolvethetheneutralelementandoppositeelementsof b = (a0, a1 I1, a2 I2, a3 I3)=(1, I1, I2, I3) ∈ RNF4.
1. Algorithm 1:FromTable 2,being a3 = 0 and a0 + a1 + a2 + a3 = 0,thus b3 = 0,thatis neut(b)= (0, I1, I2,0) andc3 =(0 1 1 1 2 )= 1 2 ,thusanti(a)=(1, c1 I1, c2 I2, 1 2 I3),wherec1 + c2 = 0
2. ResultsfromTable1in[17]:Being a0 = 0, a2 = 0 and a2 + a3 = 0,thus neut(b)=(1, I1, I2,0) and c0 = 1,c1 + c2 = 0, c3 = 1 1 (1+1) = 1 2 ,thusanti(a)=(1, c1 I1, c2 I2, 1 2 I3),wherec1 + c2 = 0.
Foralgebrasystem (RNFn+1, ),set a =(a0, a1 I1, a2 I2, , an+1 In+1),beingtheorder I1 I2 ··· In,thusweshouldobtaintheneutralelementandoppositeelements of a =(a0, a2 I2, , an+1 In+1).Knowingthat neut(a )=(b0, b2 I2, b3 I3, , bn In ), anti(a )= (c0, c2 I2, c3 I3, , cn In ),thenthecorrespondingneutralelementandoppositeelementsof a aregiven inTable 5 accordingtothedifferentcasesof a1.
Table5. Thecorrespondingneutralelementandoppositeelementsfor (RNF, ).
TheSubset {(a0, a1 I1, ··· , an+1 In+1)} NeutralElement (b0, b1 I1, ··· , bn In , bn+1 In+1)
OppositeElements (c0, c1 I1, ··· , cn In , cn+1 In+1)
b1 = 0 c1 ∈ F a1 = 0, a0 + a2 + + an+1 = 0 b1 = 0 c1 = 0 a1 = 0, a0 + a1 + + an + an+1 = 0
a1 = 0, a0 + a2 + + an+1 = 0
b1 =(c0 + c2 + + cn+1)a1 c1 ∈ F a1 = 0, a0 + a1 + + an + an+1 = 0
b1 = a1(1 (b0 + b2 + + bn+1))(a0 + a1 + + an+1) 1 c1 =(b1 a1(c0 + c2 + + cn+1))(a0 + a1 + + an+1) 1
Similarly,wealsocangettheneutralelementandoppositeelementsofeachelementin (RNFn, ) step-by-step.ThesolvingmethodisgivenbyAlgorithm 2 andthefollowingexampleisusedtoexplain thealgorithm.
Algorithm2 Solvingtheneutralelementandoppositeelementsofeachelementin (RNFn, )
Input: a =(a0, a1 I1, a2 I2, ··· , an In ), n ≥ 3 1: i = n 2; 2:While i ≥ 1 3: Ai =(a0, ai Ii, , an 1 In 1, an In ); 4:If i == n 2 5:Obtain neut(Ai ), anti(Ai ) byTable 3; 6:else 7:Obtain neut(Ai ), anti(Ai ) byTable 5 combiningthevaluesof neut(Ai+1) and anti(Ai+1); 8:end 9:Save neut(Ai ), anti(Ai ); 10: i = i 1; 11:end Output: neut(a)= neut(A1), anti(a)= anti(A1)
Example5. Foralgebrasystem (RNF6, ),andset F = R, a =(0,0, 2I2, I3, I4,0),solvetheneutral elementandoppositeelementsofa.
AccordingAlgorithm2foralgebrasystem (RNF6, ):Firstly,wesolvetheneutralelementandopposite elementsof a =(0, I3, I4,0) fromTable 3,andthensolvetheneutralelementandoppositeelements of a =(0, 2I2, I3, I4,0) fromTable 5,lastly,wesolvetheneutralelementandoppositeelementsof a fromTable 5.
1. FromTable 3: neut(a )=(0, I3, I4,0) and anti(a )=(c0, c3 I3, c4 I4, c5 I5),where c0 + c4 + c5 = 1, c3 ∈ R
2. FromTable 5 andcombiningtheresultsoftheabovestep:Being 2 = 0 and 0 +( 2)+( 1)+ 1 + 0 = 0,thus neut(a )=(0, I2, I3, I4,0) and anti(a )=(c0, c2 I2, c3 I3, c4 I4, c5 I5),where c0 + c4 + c5 = 1, c2 + c3 = 3 2
3. FromTable 5 andcombiningtheresultsoftheabovestep:Being 0 = 0 and 0 + 0 +( 2)+( 1)+ 1 + 0 = 0,thus neut(a)=(0,0, I2, I3, I4,0) and anti(a)=(c0,0, c2 I2, c3 I3, c4 I4, c5 I5),where c0 + c4 + c5 = 1,c2 + c3 = 3 2 .
Similarly,weexplorethethealgebrastructureoftherefinedneutrosophicquadruplenumberson generalfieldinthefollowing.
Definition14. Let F beafield;arefinedneutrosophicquadruplenumberisa numberoftheform (a, b1T1, b2T2, , bp Tp, c1 I1, c2 I2, , cq Iq, d1 F1, d2 F2, , dr Fr ), wherea, b1, b2, , bp, c1, c2, , cq, d1, d2, , dr ∈ F.ThesetRNQFpqr definedby RNQFpqr = {(a, b1T1, b2T2, ··· , bp Tp, c1 I1, c2 I2, ··· , cq Iq, d1 F1, d2 F2, ··· , dr Fr )| a, b1, b2, , bp, c1, c2, , cq, d1, d2, , dr ∈ F}
iscalledarefinedneutrosophicquadruplesetonfield F
Definition15. Let x =(a, b1T1, b2T2, ··· , bp Tp, c1 I1, c2 I2, ··· , cq Iq, d1 F1, d2 F2, ··· , dr Fr ), y =(e, f1T1, f2T2, ··· , f p Tp, g1 I1, g2 I2, ··· , gq Iq, h1 F1, h2 F2, ··· , hr Fr ) ∈ RNQFpqr,thentheaddition operatorisdefinedasfollows: x ⊕ y =(a + e, (b1 + f1)T1, (b2 + f2)T2, ··· , (bp + f p )Tp, (c1 + g1)I1, (c2 + g2)I2, ··· , (cq + gq )Iq, (d1 + h1)F1, (d2 + h2)F2, , (dr + hr )Fr ) (19)
Definition16. Let x =(a, b1T1, b2T2, , bp Tp, c1 I1, c2 I2, , cq Iq, d1 F1, d2 F2, , dr Fr ), y =(e, f1T1, f2T2, , f p Tp, g1 I1, g2 I2, , gq Iq, h1 F1, h2 F2, , hr Fr ) ∈ RNQFpqr;theneutrosophic expertconsiderstheprevalenceorder T1 ≺ T2 ≺···≺ Tp ≺ I1 ≺ I2 ≺···≺ Iq ≺ F1 ≺ F2 ≺···≺ Fr. Then,themultiplicationoperationisdefinedasfollows:
x ∗ y =(a, b1T1, b2T2, , bp Tp, c1 I1, c2 I2, , cq Iq, d1 F1, d2 F2, , dr Fr ) ∗(e, f1T1, f2T2, ··· , f p Tp, g1 I1, g2 I2, ··· , gq Iq, h1 F1, h2 F2, ··· , hr Fr ) =(ae, (af1 + b1e + b1 f1)T1, (af2 + b1 f2 + b2e + b2 f1 + b2 f2)T2, ··· , (ahr + b1hr + b2hr + ··· + dr 1hr + dr e + dr f1 + ··· + dr hr )Fr ) (20)
Theneutrosophicexpertconsiderstheprevalenceorder T1 T2 ··· Tp I1 I2 ··· Iq F1 F2 ··· Fr.Then,
x y =(a, b1T1, b2T2, ··· , bp Tp, c1 I1, c2 I2, ··· , cq Iq, d1 F1, d2 F2, ··· , dr Fr ) (e, f1T1, f2T2, ··· , f p Tp, g1 I1, g2 I2, ··· , gq Iq, h1 F1, h2 F2, ··· , hr Fr ) =(ae, (af1 + b1 f1 + ··· + dr f1 + b1e + b1 f2 + b1 f3 + ··· + b1hr )T1, ··· , (ahr 1 + dr 1e + dr 1hr 1 + dr 1hr + dr hr 1)Fr 1, (ahr + dr e + dr hr )Fr ) (21)
Similarly,wealsohavethefollowingresults.
Theorem6. (RNQFpqr, ⊕) isanabeliangroup.
Theorem7. Forthealgebrasystem (RNQFpqr, ∗) (or (RNQFpqr, )),foreveryelement a ∈ RNQFpqr, thereexiststheneutralelement neut(a) andoppositeelement anti(a),thusthealgebrasystem (RNQFpqr, ∗) (or (RNQFpqr, ))isaNETG.
Example6. Foralgebrasystem (RNQF213, ∗),andset F = R, a =(1,0,2T2, 3I1,2F1,0, 2F3),solvethe neutralelementandoppositeelementsofa.
AccordingAlgorithm1foralgebrasystem (RNF213, ∗),firstly,wesolvetheneutralelementandopposite elementsof a =(1,0,2T2, 3I1) fromTable 1.Wethensolvetheneutralelementandoppositeelements of a =(1,0,2T2, 3I1,2F1) fromTable 4.Nextwesolvetheneutralelementandoppositeelementsof a =(1,0,2T2, 3I1,2F1,0) fromTable 4.Finally,wesolvetheneutralelementandoppositeelementsof a fromTable 4
1. FromTable 1,neut(a )=(1,0,0, I1) andanti(a )=(1,0, 2 3 T2, c4 I1),wherec4 ∈ R
2. FromTable 4 andcombiningtheresultsoftheabovestep:Being 2 = 0 and 1 + 0 + 2 +( 3)+ 2 = 0, thusneut(a )=(1,0,0, I1, F1) andanti(a )=(1,0, 2 3 T2, c4 I1, c5 F1),wherec4 + c5 = 1 6
3. FromTable 4 andcombiningtheresultsoftheabovestep:Being 0 = 0 and 1 + 0 + 2 +( 3)+ 2 + 0 = 0, thusneut(a )=(1,0,0, I2, F1,0) andanti(a )=(1,0, 2 3 T2, c4 I1, c5 F1,0),wherec4 + c5 = 1 6 .
4. FromTable 4 andcombiningtheresultsoftheabovestep:Being 2 = 0 and 1 + 0 + 2 +( 3)+ 2 + 0 +( 2)= 0,thus neut(a)=(1,0,0, I1, F1,0, F3) and anti(a)=(1,0, 2 3 T2, c4 I1, c5 F1,0, c7 F3), wherec4 + c5 = 1 6 , c7 ∈ R
Example7. Foralgebrasystem (RNQF213, ∗),andset F = Z5 = {[0], [1], [2], [3], [4]}, a =([1], [0], [2]T2, [2]I1, [2]F1,0, [3]F3),solvetheneutralelementandoppositeelementsofa. SimilartoExample 6,accordingAlgorithm1foralgebrasystem (RNF213, ∗),firstly,wesolvetheneutral elementandoppositeelementsof a =([1], [0], [2]T2, [2]I1) fromTable 1.Wethensolvetheneutralelement andoppositeelementsof a =([1], [0], [2]T2, [2]I1, [2]F1) fromTable 4.Next,wesolvetheneutralelementand oppositeelementsof a =([1], [0], [2]T2, [2]I1, [2]F1, [0]) fromTable 4.Finally,wesolvetheneutralelement andoppositeelementsofafromTable 4.
1. FromTable 1,neut(a )=([1], [0], [0], [4]I1) andanti(a )=([1], [0], [1]T2, c4 I1),wherec4 ∈ F
2. FromTable 4 andcombiningtheresultsoftheabovestep:Being [2] =[0] and [1]+[0]+[2]+[2]+ [2] =[0],thus neut(a )=([1], [0], [0], [4]I1, [1]F1) and anti(a )=([1], [0], [1]T2, c4 I1, c5 F1),where c4 + c5 =[1].
3. FromTable 4 andcombiningtheresultsoftheabovestep:Being [0]=[0] and [1]+[0]+[2]+[2]+[2]+ [0] = 0,thus neut(a )=([1], [0], [0], [4]I1, [1]F1, [0]) and anti(a )=([1], [0], [1]T2, c4 I1, c5 F1, [0]), wherec4 + c5 =[1]
4. FromTable 4 andcombiningtheresultsoftheabovestep:Being [3] =[0] and [1]+[0]+[2]+ [2]+[2]+[0]+[3]= 0,thus neut(a)=([1], [0], [0], [4]I1, [1]F1, [0], [4]F3) and anti(a)= ([1], [0], [1]T2, c4 I1, c5 F1, [0], c7 F3),wherec4 + c5 =[1], c7 ∈ F
5.Conclusions
Inthispaper,westudythealgebrastructureof (NQF(RNFn, RNQFn), ⊕), (NQF(RNFn, RNQFn), ∗) and (NQF(RNFn, RNQFn ), ),andweprovethat (NQF(RNFn, RNQFn ), ∗) (or (NQF(RNFn, RNQFn ), ))isaneutrosophicextendedtripletgroup,andprovidenewexamplesofneutrosophic extendedtripletgroupandtheneutralelementandoppositeelementsofeachrefined n-ary neutrosophicnumber(refinedneutrosophicquadruplenumber)canbeobtainedbygiven algorithms.Inthefollowing,wecanexplorethealgebrastructureof (NQF(RNFn, RNQFn ), ⊕, ∗) or (NQF(RNFn, RNQFn ), ⊕, ).Wecanalsoexploretherelationofneutrosophicquadruplenumbers andotheralgebrasystemsinpapers[21–23].Moreover,ontheonehand,wewilldiscussthe neutrosophicquadruplenumbersbasedonsomeparticularringwhichcanformaneutrosophic extendedtripletgroup,while,ontheotherhand,wewillintroduceanewoperation ◦ inorderto guarantee (NQF(RNFn, RNQFn ), ∗, ◦) isaneutrosophictripletring.
AuthorContributions:
Acknowledgments: Theauthorswouldliketothankthereviewersfortheirmanyinsightfulcomments andsuggestions. ConflictsofInterest: Theauthorsdeclarenoconflictsofinterest.
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