TheStructureofIdempotentsinNeutrosophicRings andNeutrosophicQuadrupleRings
YingcangMa 1,*,XiaohongZhang 2,FlorentinSmarandache 3 andJuanjuanZhang 1
1 SchoolofScience,Xi’anPolytechnicUniversity,Xi’an710048,China;20080712@xpu.edu.cn
2 SchoolofArtsandSciences,ShaanxiUniversityofScience&Technology,Xi’an710021,China; zhangxiaohong@sust.edu.cn
3 DepartmentofMathematics,UniversityofNewMexico,Gallup,NM87301,USA;smarand@unm.edu
* Correspondence:mayingcang@xpu.edu.cn
Received:12September2019;Accepted:5October2019;Published:8October2019
Abstract: Thispaperaimstorevealthestructureofidempotentsinneutrosophicringsand neutrosophicquadruplerings.First,allidempotentsinneutrosophicrings R ∪ I aregiven when R is C, R, Q, Z or Zn.Secondly,theneutrosophicquadruplering R ∪ T ∪ I ∪ F is introducedandallidempotentsinneutrosophicquadruplerings C ∪ T ∪ I ∪ F , R ∪ T ∪ I ∪ F , Q ∪ T ∪ I ∪ F , Z ∪ T ∪ I ∪ F and Zn ∪ T ∪ I ∪ F arealsogiven.Furthermore,thealgorithms forsolvingtheidempotentsin Zn ∪ I and Zn ∪ T ∪ I ∪ F foreachnonnegativeinteger n are provided.Lastly,asageneralresult,ifallidempotentsinanyring R areknown,thenthestructure ofidempotentsinneutrosophicring R ∪ I andneutrosophicquadruplering R ∪ T ∪ I ∪ F can bedetermined.
Keywords: neutrosophicrings;neutrosophicquadruplerings;idempotents;neutrosophicextended tripletgroup;neutrosophicset
1.Introduction
ThenotionsofneutrosophicsetandneutrosophiclogicwereproposedbySmarandache[1]. Inneutrosophiclogic,everypropositionisconsideredbythetruthdegree T,theindeterminacy degree I,andthefalsitydegree F,where T, I and F aresubsetsofthenonstandardunitinterval ]0 ,1+ [= 0 ∪ [0,1] ∪ 1+
Usingtheideaofneutrosophicset,somerelatedalgebraicstructureshavebeenstudiedinrecent years.Amongthesealgebraicstructures,byextendingclassicalgroups,theneutrosophictripletgroup (NTG)andtheneutrosophicextendedtripletgroup(NETG)havebeenintroducedinrefs.[2–4].As anexample,paper[5]showsthat (Zp1 p2 pt , ·) isnotonlyasemigroup,butalsoaNETG,where · theclassicalmodmultiplicationand p1, p2, , pt aredistinctprimes.Afterthenotionswereput forward,NTGandNETGhavebeencarriedoutin-depthresearch.Forexample,theinclusion relationsofneutrosophicsets[6],neutrosophictripletcoset[7],neutrosophicdupletsemi-groups[8], AG-neutrosophicextendedtripletloops[9,10],theneutrosophicsettheorytopseudo-BCIalgebras[11], neutrosophictripletringandaneutrosophictripletfield[12,13],neutrosophictripletnormedspace[14], neutrosophicsoftsets[15],neutrosophicvectorspaces[16],andsoon.
Incontrasttotheneutrosophictripletring,theneutrosophicring R ∪ I ,whichisaringgenerated bythering R andtheindeterminateelement I (I2 = I),wasproposedbyVasanthaandSmarandache in[17].Theconceptofneutrosophicringwasfurtherdevelopedandstudiedin[18–20].
Asaspecialkindofelementinanalgebraicsystem,theidempotentelementplaysamajorrole indescribingthestructureandpropertiesofthealgebra.Forexample,Booleanringsrefertoringsin whichallelementsareidempotent,cleanrings[21]refertoringsinwhicheachelementisclean(an Symmetry 2019, 11,1254;doi:10.3390/sym11101254 www.mdpi.com/journal/symmetry
elementinaringisclean,ifitcanbewrittenasthesumofanidempotentelementandaninvertible element),andAlbelringisaringifeachelementintheringiscentral.Fromthesewecanseethatsome ringscanbecharacterizedbyidempotents.Thus,itisalsoquitemeaningfultofindallidempotentsin aring.Inthispaper,theidempotentsinneutrosophicringsandneutrosophicquadrupleringswill bestudiedindepth,andallidempotentsinthemcanbeobtainediftheidempotentsinRareknown. Inaddition,therelationshipbetweenidempotentsandneutralelementswillbegiven.Theelements ofeachNETGcanbepartitionedbyneutrals[10].Therefore,asanapplication,if R = F,where F isanyfield,wecandividetheelementsof R ∪ I (or R ∪ T ∪ I ∪ F )byidempotents.Asanother application,inpaper[22],theauthorsexploretheidempotentsandsemi-idempotentsinneutrosophic ring Zn ∪ I andsomeopenproblemsandconjecturesaregiven.Inthispaper,wewillanswerpartial openproblemsandconjecturesinpaper[22]andsomefurtherstudiesarediscussed.
Theoutlineofthispaperisorganizedasfollows.Section 2 givesthebasicconcepts.InSection 3, theidempotentsinneutrosophicring R ∪ I willbeexplored.Forneutrosophicrings Zn ∪ I , C ∪ I , R ∪ I , Q ∪ I and Z ∪ I ,allidempotentswillbegiven.Moreover,theopenproblemand conjecturesproposedinpaper[22]aboutidempotentsinneutrosophicring Zn ∪ I willbesolved. InSection 4,theneutrosophicquadruplering R ∪ T ∪ I ∪ F isintroducedandallidempotentsin neutrosophicquadruplerings C ∪ T ∪ I ∪ F , R ∪ T ∪ I ∪ F , Q ∪ T ∪ I ∪ F , Z ∪ T ∪ I ∪ F and Zn ∪ T ∪ I ∪ F willbegiven.Finally,thesummaryandfutureworkispresentedinSection 5.
2.BasicConcepts
Inthissection,therelatedbasicdefinitionsandpropertiesofneutrosophicring R ∪ I andNETG areprovided,thedetailscanbeseenin[3,4,17,18].
Definition1. ([17,18])Let (R, +, ) beanyring.Theset
R ∪ I = {a + bI : a, b ∈ R} iscalledaneutrosophicringgeneratedby R and I.Let a1 + b1 I, a2 + b2 I ∈ R ∪ I ,Theoperators ⊕ and ⊗ on R ∪ I aredefinedasfollows:
(a1 + b1 I) ⊕ (a2 + b2 I)=(a1 + a2)+(b1 + b2)I, (a1 + b1 I) ⊗ (a2 + b2 I)=(a1 a2)+(a1 b2 + b1 a2 + b1 b2)I.
Remark1. Itiseasytoverifythat ( R ∪ I , ⊕, ⊗) isaring,so R ∪ I isnamedbyaneutrosophicring isreasonable.
Remark2. Itshouldbenotedthattheoperators +, · aredefinedonring R and ⊕, ⊗ aredefinedonneutrosophic ring R ∪ I .Forsimplicityofnotation,wealsouse +, · toreplace ⊕, ⊗ onring R ∪ I .Thatis a + b also means a ⊕ b if a, b ∈ R ∪ I a · b alsomeans a ⊗ b if a, b ∈ R ∪ I .Forshort a · b denotedby ab and a · a denotedbya2
Example1. Z ∪ I , Q ∪ I , R ∪ I and C ∪ I areneutrosophicringsofinteger,rational,realandcomplex numbers,respectively. Zn ∪ I isneutrosophicringofmodulointegers.Ofcourse, Z, Q, R, C and Zn are neutrosophicringswhenb = 0.
Definition2. ([17,18])Let R ∪ I beaneutrosophicring. R ∪ I issaidtobecommutativeif ab = ba, ∀a, b ∈ R ∪ I
Inaddition,ifthereexists 1 ∈ R ∪ I suchthat 1 a = a 1 = a forall a ∈ R ∪ I thenwecall R ∪ I a commutativeneutrosophicringwithunity.
Definition3. ([17,18])Anelement a inaneutrosophicring R ∪ I iscalledanidempotentelementif a2 = a.
Definition4. ([3,4])Let N beanon-emptysettogetherwithabinaryoperation ∗.Then, N iscalleda neutrosophicextendedtripletsetifforany 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.
Definition5. ([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.
Proposition2. ([10])Let (N, ∗) isaNETG,denotethesetofalldifferentneutralelementin N by E(N).For anye ∈ E(N),denoteN(e)= {x|neut(x)= e, x ∈ N}.Then: (1)N(e) isaclassicalgroup,andtheunitelementise. (2)Foranye1, e2 ∈ E(N), e1 = e2 ⇒ N(e1) ∩ N(e2)= ∅ (3)N = e∈E(N) N(e).i.e., e∈E(N) N(e) isapartitionofN.
3.TheIdempotentsinNeutrosophicRings
Inthissection,wewillexploretheidempotentsinneutrosophicrings R ∪ I .If R is Z, Q, R, C or Zn,allidempotentsinneutrosophicrings Zn ∪ I , C ∪ I , R ∪ I , Q ∪ I or Z ∪ I willbegiven. Moreover,wecanalsoobtainallidempotentsinneutrosophicring R ∪ I ifallidempotentsinany ring R areknown.Asanapplication,theopenproblemandconjecturesabouttheidempotentsof neutrosophicring Zn ∪ I inpaper[22]willbesolved.Moreover,anexampleisgiventoshowhowto usetheidempotentstogetapartitionforaneutrosophicring.Thefollowingpropositionrevealthe relationofaneutralelementandanidempotentelement.
Proposition3. Let G beanon-emptyset, ∗ isabinaryoperationon G.Foreach a ∈ G, a isidempotentiffitis aneutralelement.
Proof. Necessity:If a isidempotent,i.e., a ∗ a = a,fromDefinition 4,whichshowsthat a hasneutral element a andoppositeelement a,so a isaneutralelement.
Sufficiency:If a isaneutralelement,fromProposition 1(2),wehave a ∗ a = a,thus a isidempotent.
Theorem1. Thesetofallidempotentsinneutrosophicring C ∪ I , R ∪ I , Q ∪ I or Z ∪ I is {0,1, I,1 I}.
Proof. Wejustgivetheprooffor R ∪ I ,andthesameresultcanbeobtainedfor C ∪ I , Q ∪ I or Z ∪ I
Let a + bI ∈ R ∪ I .If a + bI isidempotent,so (a + bI)2 = a + bI,whichmeans
a2 = a 2ab + b2 = b (1)
From a2 = a,wecanget a = 0 or a = 1.When a = 0,from 2ab + b2 = b,wecanget b = 0 or b = 1.Thatis 0 and I areidempotents.When a = 1,from 2ab + b2 = b,wecanget b = 0 or b = 1 Thatis 1 and 1 I areidempotents.Thus,thesetofallidempotentsofneutrosophicring R ∪ I is {0,1, I,1 I}
Theabovetheoremrevealsthatthesetofallidempotentsinneutrosophicring R ∪ I is {0,1, I,1 I} when R is C, R, Q or Z.Foranyring R,wehavethefollowingresults.
Proposition4. IfaisidempotentinanyringR,thenaIisalsoidempotentinneutrosophicring R ∪ I
Proof. If a ∈ R isidempotent,i.e., a2 = a,so (aI)2 =(0 + aI)(0 + aI)= a2 I = aI,thus, aI isalso idempotentinneutrosophicring R ∪ I
Proposition5. Inneutrosophicring R ∪ I ,thena aIisidempotentiffaisidempotent.
Proof. Necessity:If a aI isidempotent,i.e., (a aI)2 = a aI,so (a aI)2 =(a aI)(a aI)= a2 2aI + a2 I = a2 +(a2 2a)I = a aI,whichmeans a2 = a and a2 2a = a.Thus,wehave a2 = a,so a isidempotent.
Sufficiency:If a isidempotent,so (a aI)2 = a2 +(a2 2a)I = a aI,thus a aI isidempotent.
Theorem2. Inneutrosophicring R ∪ I ,let a + bI ∈ R ∪ I ,then a + bI isidempotentiff a isidempotent inRandb = c a,wherecisanyidempotentelementinR.
Proof. Necessity:If a + bI isidempotent,i.e., (a + bI)2 = a + bI,so (a + bI)2 = a2 +(2ab + b2)= a + bI,whichmeans a2 = a and 2ab + b2 = b.From a2 = a,wecanget a isidempotent.From 2ab + b2 = b and a2 = a,wecanget (b + a)2 = b2 + 2ab + a2 = b + a,so b + a isalsoidempotentin R, denotedby c,so b = c a.
Sufficiency:If a and c areanyidempotentsin R,let b = c a,so (a + bI)2 =(a +(c a)I)2 = a2 +(2a(c a)+(c a)2)I = a2 +(2ac 2a2 + c2 2ac + a2)= a +(c a)I = a + bI,thus a + bI isidempotent.
Theorem3. Ifthenumberofdifferentidempotentsinring R is t,thenthenumberofdifferentidempotentsin theneutrosophicring R ∪ I ist2
Proof. Ifthenumberofidempotentsin R is t andlet a + bI ∈ R ∪ I isidempotent,sofromTheorem 2, wecaninferthat a isidempotentin R,i.e., a has t differentselections.When a isfixed,set b = c a, where c isanyidempotentin R and c alsohas t differentselections,whichmeans b has t different selections.Thus, a + bI has t t = t2 differentselections,i.e.,thenumberofallidempotentsin R ∪ I is t2
Fromtheaboveanalysis,foranyring R,allidempotentsin R ∪ I canbedeterminedifall idempotentsin R areknown.Inthefollowing,wewillexploreallidempotentsinneutrosophicring Zn ∪ I ,i.e.,when R = Zn
Theorem4. ([5])Inthealgebrasystem (Zn, ) (seeAppendix A), istheclassicalmodmultiplication,foreach a ∈ Zn,ahasneut(a) andanti(a) iff gcd(gcd(a, n), n/gcd(a, n))= 1
Theorem5. ([5])Foranalgebrasystem (Zn, ) and n = p k1 1 p k2 2 p kt t ,whereeach pi (i = 1,2, , t) isa prime,thenthenumberofdifferentneutralelementsin Zn is 2t
Remark3. FromProposition 3 andTheorem 5,wecaninferthatthenumberofallidempotentsin Z p k1 1 p k2 2 pkt t isalso 2t
Example2. For (Z36, ), n = 36 = 2232.FromTheorem 5,thenumberofdifferentneutralelementsin Z36 is 22 = 4.Theyare:
(1) [0] hastheneutralelement [0] (2) [1], [5], [7], [11], [13], [17], [19], [23], [25], [29], [31] and [35] havethesameneutralelement [1].
(3) [9] and [27] havethesameneutralelement [9] being gcd(9,36)= gcd(27,36)= 9. (4) [4] and [8] havethesameneutralelementbeing gcd(4,36)= gcd(8,36)= 4.Infact, [4], [8], [16], [20], [28] and [32] havethesameneutralelement,whichis [28].
FromRemark 3,thenumberofidempotentsin Z36 isalso 4,whichare [0], [1], [9] and [28]
FromTheorems 2 and 3 andRemark 3,itfollowseasilythat:
Corollary1. Inneutrosophicring Zn ∪ I ,let a + bI ∈ Zn ∪ I ,then a + bI isidempotentiff a2 = a and b = c a,wherecisanyidempotentelementin Zn.
Corollary2. Foranalgebrasystem (Zn, ) and n = p k1 1 p k2 2 p kt t ,whereeach p1, p2, , and pk aredistinct primes.Thenthenumberofdifferentidempotentsin Zn ∪ I is 22t .
Thesolvingprocessfor Zn ∪ I isgivenbyAlgorithm 1.Justonlyinput n,thenwecangetall idempotentsin Zn ∪ I .TheMATLABcodeisprovidedintheAppendix B
Example3. Solveallidempotentsin Z600 ∪ I
Since n = 600 = 23 3 52,fromTheorem 5,wecangetthedifferentneutralelementsin Z600 are neut(1), neut(23), neut(3), neut(52), neut(23 3), neut(23 52), neut(3 52) and neut(0),i.e.,thedifferent idempotentsin Z600 are1,376,201,25,576,400,225,0.FromCorollary 2,thenumberofdifferentidempotents inneutrosophicring Z600 ∪ I is 22 3 = 64.
FromAlgorithm 1,thesetofall 64 idempotentsin Z600 ∪ I is: {0, I,25I,201I,225I,376I,400I,576I,1 + 599I,1,1 + 24I,1 + 200I,1 + 224I,1 + 375I,1 + 399I,1 + 575I, 25 + 575I,25 + 576I,25,25 + 176I,25 + 200I,25 + 351I,25 + 375I,25 + 551I,201 + 399I,201 + 400I,201 + 424I,201,201 + 24I,201 + 175I,201 + 199I,201 + 375I,225 + 375I,225 + 376I,225 + 400I,225 + 576I,225,225 + 151I,225 + 175I,225 + 351I,376 + 224I,376 + 225I,376 + 249I,376 + 425I,376 + 449I,376,376 + 24I,376 + 200I,400 + 200I,400 + 201I,400 + 225I,400 + 401I,400 + 425I,400 + 576I,400,400 + 176I,576 + 24I,576 + 25I,576 + 49I,576 + 225I,576 + 249I,576 + 400I,576 + 424I,576}
Algorithm1: Solvingthedifferentidempotentsin Zn ∪ I
Input: n
1:Factorizationofinteger n,wecanget n = p k1 1 p k2 2 ··· p kt t
2:Computingtheneutralelementof1, p k1 1 , p k2 2 , ··· , p kt t , p k1 1 p k2 2 , ··· p k1 1 p kt t , ··· , p k2 2 p k3 3 ··· p kt t and p k1 1 p k2 2 ··· p kt t .So,wecangetallidempotentsin Zn ,denotedby a1, a2, ··· , a2t 3:LetID=[]; 4:for i = 1:2t 5: a = ai 6:for j = 1:2t 7: b = mod(aj a, n); 8: ID =[ID; [a, b]]; 9:end 10:end Output:ID:alltheidempotentsin Zn ∪ I
Inpaper[22],theauthorsstudiedtheidempotentsandsemi-idempotentsin Zn ∪ I andproposed someopenproblemsandconjectures.Welistpartialopenproblemsandconjecturesaboutidempotents in Zn ∪ I asfollowsandanswerthem.
Problem1. ([22])Let S = Zpq, +, · ,where p and q aretwodistinctprimes,betheneutrosophicring.Can S havenon-trivialidempotentsotherthantheonesmentionedin(b)oftheTheorem 6?
Conjecture1. ([22])Let S = Zn, +, · betheneutrosophicring n = pqr,where p, q and r arethree distinctprimes.
1. Zn = Zpqr hasonlysixnon-trivialidempotentsassociatedwithit.
2. If m1, m2, m3, m4, m5 and m6 aretheidempotents,then,associatedwitheachrealidempotent mi,wehave sevennon-trivialneutrosophicidempotentsassociatedwithit,i.e., {mi + nj I, j = 1,2, ,7},suchthat mi + nj ≡ t,where tj takesthesevendistinctvaluesfromtheset {0,1, mk, k = i; k = 1,2,3, ,6}.i = 1,2, ,6.
Conjecture2. ([22])Given Zn ∪ I ,wheren = p1 p2 pt; t > 2 andpisarealldistinctprimes,find:
1. thenumberofidempotentsin Zn; 2. thenumberofidempotentsin Zn ∪ I \Zn;
Conjecture3. ([22])Proveif Zn ∪ I and Zm ∪ I aretwoneutrosophicringswhere n > m and n = pt q (t > 2,andpandqtwodistinctprimes)andm = p1 p2 ps wherepisaredistinctprimes. 1 ≤ i ≤ s,then 1. prove Zn hasagreaternumberofidempotentsthan Zm;and 2. prove Zn ∪ I hasagreaternumberofidempotentsthan Zn ∪ I
Theorem6. ([22])LetS = Zpq, +, · wherepandqaretwodistinctprimes:
(a) Therearetwoidempotentsin Zpq sayrands. (b) {r, s, rI, sI, I, r + tI, s + tI|t ∈{Zpq \0}} suchthat r + t = s,1 or 0 and s + t = 0,1 or r isthepartial collectionofidempotentsofS.
ForProblem1,fromRemark 3,therearefouridempotentsin Zpq,whichare {1, neut(p), neut(q), neut(pq)= 0}.Let r = neut(p), s = neut(q),sotherearetwonon-trivial idempotents r, s in Zpq.FromCorollary 1 and 2,thenumberofallidempotentsin Zpq ∪ I is 24 = 16,theyare {0 +(0 0)I = 0,0 +(1 0)I = I,0 +(r 0)I = rI,0 +(s 0)I = sI,1 +(0 1)I = 1 +(n 1)I,1 +(1 1)I = 1,1 +(r 1)I,1 +(s 1)I, r +(0 r)I = r +(n r)I, r +(1 r)I = r +
(n + 1 r)I, r +(r r)I = r, r +(s r)I, s +(0 s)I = s +(n r)s, s +(1 s)I = s +(n + 1 s)I, s + (r s)I, s +(s s)I = s}.Sothereare14non-trivialidempotentsin Zpq ∪ I ,butthereareonlyinclude 11non-trivialidempotentsin(b)oftheTheorem 6,missing {1 +(n 1)I,1 +(r 1)I,1 +(s 1)I}.
ForConjecture1,fromCorollary 1 and 2,thereareeightidempotentsin Zpqr,whichare {1 = m0, neut(p)= m1, neut(q)= m2, neut(r)= m3, neut(pq)= m4, neut(pr)= m5, neut(qr)= m6, neut(pqr)= 0 = m7}.Therearesixnon-trivialidempotentsin Zpqr.In Zn ∪ I ,allidempotents are {mi +(mj mi )I|i, j = 0,1,2, ··· ,7}
ForConjecture2,fromRemark 3,thenumberofidempotentsin Zp1 p2 pt is 2t,andthenumberof idempotentsin Zp1 p2 pt ∪ I \Zp1 p2 pt is22t 2t
ForConjecture3,fromRemark 3,thenumberofidempotentsin Zn is 22,andthenumberof idempotentsin Zm is 2s,where n = pt q, m = p1 p2 · ps.So,if s > 2, Zm ischaracterizedbyalarger numberofidempotentsthan Zn.Insimilarlyway,thenumberofidempotentsin Zn ∪ I is 24,and thenumberofidempotentsin Zm ∪ I is 22s.So,if s > 2,wecaninferthat Zm ∪ I ischaracterized byalargernumberofidempotentsthan Zn ∪ I
Asanotherapplication,wewillusetheidempotentstodividetheelementsoftheneutrosophic rings R ∪ I when R = F
ForeachNETG (N, ∗), a ∈ N,fromProposition 1,theneutralelementof a isuniquelydetermined. FromProposition 2, e∈E(N) N(e) isapartitionof N.Sincetheidempotentsandneutralelements aresame,wecanusetheidempotentstogetapartitionof N.Letusillustratethesewiththe followingexample.
Example4. Let R = Z3,whichisafield.Since n = 3,fromTheorem 5,wecangetthedifferentneutral elementsin Z3 are neut(1) and neut(0),i.e.,thedifferentidempotentsin Z3 are 1,0.FromCorollary 2,the numberofdifferentidempotentsinneutrosophicring Z3 ∪ I is 22 1 = 4
FromAlgorithm 1,thesetofall 4 idempotentsin Z3 ∪ I is: {0,1, I,1 + 2I} Wehave E(0)= {0}, E(1)= {1,2,1 + I,2 + 2I}, E(I)= {I,2I}, E(1 + 2I)= {1 + 2I,2 + I}.So Z3 ∪ I = E(0) ∪ E(1) ∪ E(I) ∪ E(1 + 2I)
4.TheIdempotentsinNeutrosophicQuadrupleRings
Intheabovesection,weexploredtheidempotentsin R ∪ I .Inneutrosophiclogic,each propositionisapproximatedtorepresentrespectivelythetruth(T),thefalsehood(F),andthe indeterminacy(I).Inthissection,accordingtheideaofneutrosophicring R ∪ I ,theneutrosophic quadruplering R ∪ T ∪ I ∪ F isproposedandtheidempotentsaregiveninthissection.
Let (R, +, ·) beanyring.Theset
Remark4. Itiseasytoverifythat ( R ∪ T ∪ I ∪ F , ⊕, ∗) isaring,moreover,italsohasthesamealgebra structurewithneutrosophicquadruplenumbers(see[23–25]),sothewecall R ∪ T ∪ I ∪ F isaneutrosophic quadrupleringisreasonable.
Remark5. SimilarlywithRemark 2,forsimplicityofnotation,weuse +, · toreplace ⊕, ∗ onneutrosophic quadruplering R ∪ T ∪ I ∪ F .Thatis a + b alsomeans a ⊕ b if a, b ∈ R ∪ T ∪ I ∪ F .and a · b alsomeans a ∗ bifa, b ∈ R ∪ T ∪ I ∪ F .Forshorta · bdenotedbyabanda · adenotedbya2
Example5. Z ∪ T ∪ I ∪ F , Q ∪ T ∪ I ∪ F , R ∪ T ∪ I ∪ F and C ∪ T ∪ I ∪ F areneutrosophicquadruple ringsofinteger,rational,realandcomplexnumbers,respectively. Zn ∪ T ∪ I ∪ F isneutrosophicquadruple ringofmodulointegers.Ofcourse, Z, Q, R, C and Zn areneutrosophicquadrupleringswhencoefficientsof T, I andFequalzero.
Definition7. Let R ∪ T ∪ I ∪ F beaneutrosophicquadruplering. R ∪ T ∪ I ∪ F iscommutativeif ab = ba, ∀a, b ∈ R ∪ T ∪ I ∪ F
Inaddition,ifthereexists 1 ∈ R ∪ T ∪ I ∪ F ,suchthat 1 a = a 1 = a forall a ∈ R ∪ T ∪ I ∪ F ,then R ∪ T ∪ I ∪ F iscalledacommutativeneutrosophicquadrupleringwithunity.
Definition8. Anelement a inaneutrosophicquadruplering R ∪ T ∪ I ∪ F iscalledanidempotentelement ifa2 = a.
Theorem7. Thesetofallidempotentsofneutrosophicquadruplerings C ∪ T ∪ I ∪ F , R ∪ T ∪ I ∪ F , Q ∪ T ∪ I ∪ F and Z ∪ T ∪ I ∪ F is
{(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)}
Proof. Weonlygivetheprooffor R ∪ T ∪ I ∪ F ,andthesameresultcanbeobtainedfor C ∪ T ∪ I ∪ F , Q ∪ T ∪ I ∪ F or Z ∪ T ∪ I ∪ F
Let a = a1 + a2T + a3 I + a4 F,if a isidempotentin R ∪ T ∪ I ∪ F ,so a2 = a,i.e., (a1 + a2T + a3 I + a4 F)2 =(a1 + a2T + a3 I + a4 F),whichmeans a2 1 = a1, 2a1 a2 + a2 2 = a2, 2(a1 + a2)a3 + a2 3 = a3, 2(a1 + a2 + a3)a4 + a2 4 = a4
Since a1 ∈ R,sofrom a2 1 = a1,wecanget a1 = 0or a1 = 1.
CaseA:if a1 = 0,thenfrom2a1 a2 + a2 2 = a2,wecaninfer a2 2 = a2,so a2 = 0or a2 = 1.
CaseA1:if a1 = 0 and a2 = 0,sofrom 2(a1 + a2)a3 + a2 3 = a3,wecaninfer a2 3 = a3,so a3 = 0 or a3 = 1.
CaseA11:if a1 = 0, a2 = 0 and a3 = 0,sofrom 2(a1 + a2 + a3)a4 + a2 4 = a4,wecaninfer a2 4 = a4, so a4 = 0or a4 = 1.
CaseA111:if a1 = a2 = a3 = a4 = 0,i.e., (0,0,0,0) isidempotentin R ∪ T ∪ I ∪ F .
CaseA112:if a1 = a2 = a3 = 0and a4 = 1,i.e., (0,0,0, F) isidempotentin R ∪ T ∪ I ∪ F
CaseA12:if a1 = a2 = 0 and a3 = 1,sofrom 2(a1 + a2 + a3)a4 + a2 4 = a4,wecaninfer 2a4 + a2 4 = a4,so a4 = 0or a4 = 1.
CaseA121:if a1 = a2 = 0, a3 = 1and a4 = 0,i.e., (0,0, I,0) isidempotentin R ∪ T ∪ I ∪ F .
CaseA122:if a1 = a2 = 0, a3 = 1and a4 = 1,i.e., (0,0, I, F) isidempotentin R ∪ T ∪ I ∪ F .
CaseA2:if a1 = 0 and a2 = 1,sofrom 2(a1 + a2)a3 + a2 3 = a3,wecaninfer 2a3 + a2 3 = a3,so a3 = 0or a3 = 1.
CaseA21:if a1 = 0, a2 = 1,and a3 = 0,sofrom 2(a1 + a2 + a3)a4 + a2 4 = a4,wecaninfer 2a4 + a2 4 = a4,so a4 = 0or a4 = 1.
CaseA121:if a1 = 0, a2 = 1, a3 = 0and a4 = 0,i.e., (0, T,0,0) isidempotentin R ∪ T ∪ I ∪ F
CaseA112:if a1 = 0, a2 = 1, a3 = 0 and a4 = 1,i.e., (0, T,0, F) isidempotentin R ∪ T ∪ I ∪ F
CaseA22:if a1 = 0, a2 = 1 and a3 = 1,sofrom 2(a1 + a2 + a3)a4 + a2 4 = a4,wecaninfer a2 4 = a4, so a4 = 0or a4 = 1.
CaseA121:if a1 = 0, a2 = 1, a3 = 1 and a4 = 0,i.e., (0, T, I,0) isidempotentin R ∪ T ∪ I ∪ F
CaseA112:if a1 = 0, a2 = 1, a3 = 1 and a4 = 1,i.e., (0, T, I, F) isidempotentin R ∪ T ∪ I ∪ F
CaseB:if a1 = 1,thenfrom2a1 a2 + a2 2 = a2,wecaninfer2a2 + a2 2 = a2,so a2 = 0or a2 = 1.
CaseB1:if a1 = 1 and a2 = 0,sofrom 2(a1 + a2)a3 + a2 3 = a3,wecaninfer 2a3 + a2 3 = a3,so a3 = 0or a3 = 1.
CaseB11:if a1 = 1, a2 = 0 and a3 = 0,sofrom 2(a1 + a2 + a3)a4 + a2 4 = a4,wecaninfer 2a4 + a2 4 = a4,so a4 = 0or a4 = 1.
CaseB111:if a1 = 1, a2 = 0, a3 = 0and a4 = 0,i.e., (1,0,0,0) isidempotentin R ∪ T ∪ I ∪ F .
CaseB112:if a1 = 1, a2 = 0, a3 = 0 and a4 = 1,i.e., (1,0,0, F) isidempotentin R ∪ T ∪ I ∪ F . CaseB12:if a1 = 1, a2 = 0 and a3 = 1,sofrom 2(a1 + a2 + a3)a4 + a2 4 = a4,wecaninfer a2 4 = a4, so a4 = 0or a4 = 1.
CaseB121:if a1 = 1, a2 = 0, a3 = 1 and a4 = 0,i.e., (1,0, I,0) isidempotentin R ∪ T ∪ I ∪ F . CaseB122:if a1 = 1, a2 = 0, a3 = 1 and a4 = 1,i.e., (1,0, I, F) isidempotentin R ∪ T ∪ I ∪ F . CaseB2:if a1 = 1 and a2 = 1,sofrom 2(a1 + a2)a3 + a2 3 = a3,wecaninfer a2 3 = a3,so a3 = 0 or a3 = 1.
CaseB21:if a1 = 1, a2 = 1,and a3 = 0,sofrom 2(a1 + a2 + a3)a4 + a2 4 = a4,wecaninfer a2 4 = a4, so a4 = 0or a4 = 1.
CaseB121:if a1 = 1, a2 = 1, a3 = 0 and a4 = 0,i.e., (1, T,0,0) isidempotentin R ∪ T ∪ I ∪ F CaseB112:if a1 = 1, a2 = 1, a3 = 0 and a4 = 1,i.e., (1, T,0, F) isidempotentin R ∪ T ∪ I ∪ F CaseB22:if a1 = 1, a2 = 1 and a3 = 1,sofrom 2(a1 + a2 + a3)a4 + a2 4 = a4,wecaninfer 2a4 + a2 4 = a4,so a4 = 0or a4 = 1.
CaseB121:if a1 = 1, a2 = 1, a3 = 1 and a4 = 0,i.e., (1, T, I,0) isidempotentin R ∪ T ∪ I ∪ F CaseB112:if a1 = 1, a2 = 1, a3 = 1 and a4 = 1,i.e., (1, T, I, F) isidempotentin R ∪ T ∪ I ∪ F
Fromtheaboveanalysis,wecangetthesetofallidempotentsinneutrosophicquadruplering R ∪ T ∪ I ∪ F are {(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)}
Theabovetheoremrevealsthattheidempotentsinneutrosophicquadruplering R ∪ T ∪ I ∪ F isfixedwhen R is C, R, Q or Z.Foranyring R,wehavethefollowingresults. Theorem8. Forneutrosophicquadruplering R ∪ T ∪ I ∪ F , a = a1 + a2T + a3 I + a4 F isidempotentin neutrosophicquadruplering R ∪ T ∪ I ∪ F iff a1 isidempotentin R, a2 = c a1, a3 = d (a1 + a2) and a4 = e (a1 + a2 + a3),wherec, dandeareanyidempotentsinR.
Proof. Necessity:If a = a1 + a2T + a3 I + a4 F isidempotent,i.e., (a1 + a2T + a3 I + a4 F)2 = a1 + a2T + a3 I + a4 F,whichmeans
a2 1 = a1, 2a1 a2 + a2 2 = a2, 2(a1 + a2)a3 + a2 3 = a3, 2(a1 + a2 + a3)a4 + a2 4 = a4
Since a1 ∈ R,from a2 1 = a1,wecanget a1 isidempotentin R.
From 2a1 a2 + a2 2 = a2 and a2 1 = a1,wecanget (a1 + a2)2 = a2 1 + 2a1 a2 + a2 2 = a1 + a2,so a1 + a2 isalsoidempotentin R,denotedby c,so a2 = c a1.
From 2(a1 + a2)a3 + a2 3 = a3,and (a1 + a2)2 = a1 + a2,wecanget (a1 + a2 + a3)2 =(a1 + a2)2 + 2(a1 + a2)a3 + a2 3 = a1 + a2 + a3,so a1 + a2 + a3 isalsoidempotentin R,denotedby d,so a3 = d a1 a2
From 2(a1 + a2 + a3)a4 + a2 4 = a4,and (a1 + a2 + a3)2 = a1 + a2 + a3,wecanget (a1 + a2 + a3 + a4)2 =(a1 + a2 + a3)2 + 2(a1 + a2 + a3)a3 + a2 4 = a1 + a2 + a3 + a4,so a1 + a2 + a3 + a4 isalso idempotentin R,denotedby e,so a4 = e a1 a2 a3
Sufficiency:If a1, c, d and e arearbitraryidempotentsin R,let a2 = c a1, a3 = d (a1 + a2) and a4 = e (a1 + a2 + a3).so (a1 + a2T + a3 I + a4 F)2 =(a1 +(c a1)T +(d a1 a2)I +(e a1 a2 a3)F)2 = a2 1 +(2(c a1)a1 +(c a1)2)T +(2c(d a1 a2)+(d a1 a2)2)I +(2d(e a1 a2 a3)+(e a1 a2 a3)2)F = a1 +(c a1)T +(d a1 a2)I +(e a1 a2 a3)F.Thus, a = a1 + a2T + a3 I + a4 F isidempotent.
Theorem9. Ifthenumberofdifferentidempotentsin R is t,thenthenumberofdifferentidempotentsin neutrosophicquadruplering R ∪ T ∪ I ∪ F ist4 .
Proof. Ifthenumberofdifferentidempotentsin R is t,let a1 + a2T + a3 I + a4 F ∈ Zn ∪ T ∪ I ∪ F is idempotent,so a1 isidempotentin R,i.e., a1 has t differentselections.When a1 isselected, a2 = c a1, where c isidempotent,whichalsohas t differentselections.When a1, a2 areselected, a3 = d a1 a2, where d isidempotent,whichalsohas t differentselections.When a1, a2, a3 isselected, a4 = e a1 a2 a3,where e isidempotent,whichalsohas t differentselections.Thus,thenumberofallselections is t · t · t · t = t4,i.e.,thenumberofdifferentidempotentsin R ∪ T ∪ I ∪ F is t4 FromTheorems 8 and 9 andRemark 3,itfollowseasilythat:
Corollary3. Inneutrosophicquadruplering Zn ∪ T ∪ I ∪ F , a = a1 + a2T + a3 I + a4 F isidempotentin neutrosophicquadruplering Zn ∪ T ∪ I ∪ F iff a1 isidempotentin Zn, a2 = c a1, a3 = d (a1 + a2) and a4 = e (a1 + a2 + a3),wherec, dandeareanyidempotentsin Zn
Algorithm2: Solvingthedifferentidempotentsin Zn ∪ T ∪ I ∪ F
Input: n
1:Factorizationofinteger n,wecanget n = p k1 1 p k2 2 ··· p kt t
2:Computingtheneutralelementof1, p k1 1 , p k2 2 , ··· , p kt t , p k1 1 p k2 2 , ··· p k1 1 p kt t , ··· , p k2 2 p k3 3 ··· p kt t and p k1 1 p k2 2 ··· p kt t .So,wecangetallidempotentsin Zn ,denotedby c1, c2, ··· , c2t
3:LetID=[];
4:for i = 1:2t 5: a1 = ci 6:for j = 1:2t
7: a2 = mod(cj a1, n);
8:for m = 1:2t
9: a3 = mod(cm a1 a2, n); 10:for q = 1:2t
11: a4 = mod(cq a1 a2 a3, n); 12: ID =[ID; [a1, a2, a3, a4]]; 13:end 14:end 15:end 16:end
Output:ID:alltheidempotentsin Zn ∪ T ∪ I ∪ F
Example6. Solveallidempotentsin Z12 ∪ T ∪ I ∪ F .
Since n = 12 = 22 3,fromTheorems 4 and 5,wecangetthedifferentneutralelementsin Z12 are neut(1), neut(22), neut(3), neut(23 3) and neut(0),i.e.,thedifferentidempotentsin Z12 are 1,4,9,0.From Corollary 4,thenumberofdifferentidempotentsinneutrosophicquadruplering Z12 ∪ T ∪ I ∪ F is 24 2 = 256.
FromAlgorithm 2,thesetofall 256 idempotentsin Z12 ∪ T ∪ I ∪ F is: {0,1F,4F,9F, I + 11F, I, I + 3F, I + 8F,4I + 8F,4I + 9F,4I,4I + 5F,9I + 3F,9I + 4F,9I + 7F,9I, T + 11I, T + 11I + F, T + 11I + 4F, T + 11I + 9F, T + 11F, T, T + 3F, T + 8F, T + 3I + 8F, T + 3I + 9F, T + 3I, T + 3I + 5F, T + 8I + 3F, T + 8I + 4F, T + 8I + 7F, T + 8,4T + 8I,4T + 8I + F,4T + 8I + 4F,4T + 8I + 9F,4T + 9I + 11F,4T + 9I,4T + 9I + 3F,4T + 9I + 8F,4T + 8F,4T + 9F,4T,4T + 5F,4T + 5I + 3F,4T + 5I + 4F,4T + 5I + 7F,4T + 5I,9T + 3I,9T + 3I + F,9T + 3I + 4F,9T + 3I + 9F,9T + 4I + 11F,9T + 4I,9T + 4I + 3F,9T + 4I + 8F,9T + 7I + 8F,9T + 7I + 9F,9T + 7I,9T + 7I + 5F,9T + 3F,9T + 4F,9T + 7F,9T,1 + 11T,1 + 11T + F,1 + 11T + 4F,1 + 11T + 9F,1 + 11T + I + 11F,1 + 11T + I,1 + 11T + I + 3F,1 + 11T + I + 8F,1 + 11T + 4I + 8F,1 + 11T + 4I + 9F,1 + 11T + 4I,1 + 11T + 4I + 5F,1 + 11T + 9I + 3F,1 + 11T + 9I + 4F,1 + 11T + 9I + 7F,1 + 11T + 9I,1 + 11I,1 + 11I + F,1 + 11I + 4F,1 + 11I + 9F,1 + 11F,1,1 + 3F,1 + 8F,1 + 3I + 8F,1 + 3I + 9F,1 + 3I,1 + 3I + 5F,1 + 8I + 3F,1 + 8I + 4F,1 + 8I + 7F,1 + 8I,1 + 3T + 8I,1 + 3T + 8I + F,1 + 3T + 8I + 4F,1 + 3T + 8I + 9F,1 + 3T + 9I + 11F,1 + 3T + 9I,1 + 3T + 9I + 3F,1 + 3T + 9I + 8F,1 + 3T + 8F,1 + 3T + 9F,1 + 3T,1 + 3T + 5F,1 + 3T + 5I + 3F,1 + 3T + 5I + 4F,1 + 3T + 5I + 7F,1 + 3T + 5I,1 + 8T + 3I,1 + 8T + 3I + F,1 + 8T + 3I + 4F,1 + 8T + 3I + 9F,1 + 8T + 4I + 11F,1 + 8T + 4I,1 + 8T + 4I + 3F,1 + 8T + 4I + 8F,1 + 8T + 7I + 8F,1 + 8T + 7I + 9F,1 + 8T + 7I,1 + 8T + 7I + 5F,1 + 8T + 3F,1 + 8T + 4F,1 + 8T + 7F,1 + 8T,4 + 8T,4 + 8T + F,4 + 8T + 4F,4 + 8T + 9F,4 + 8T + I + 11F,4 + 8T + I,4 + 8T + I + 3F,4 + 8T + I + 8F,4 + 8T + 4I + 8F,4 + 8T + 4I + 9F,4 + 8T + 4I,4 + 8T + 4I + 5F,4 + 8T + 9I + 3F,4 + 8T + 9I + 4F,4 + 8T + 9I + 7F,4 + 8T + 9I,4 + 9T + 11I,4 + 9T + 11I + F,4 + 9T + 11I + 4F,4 + 9T + 11I + 9F,4 + 9T + 11F,4 + 9T,4 + 9T + 3F,4 + 9T + 8F,4 + 9T + 3I + 8F,4 + 9T + 3I + 9F,4 + 9T + 3I,4 + 9T + 3I + 5F,4 + 9T + 8I + 3F,4 + 9T + 8I + 4F,4 + 9T + 8I + 7F,4 + 9T + 8I,4 + 8I,4 + 8I + F,4 + 8I + 4F,4 + 8I + 9F,4 + 9I + 11F,4 + 9I,4 + 9I + 3F,4 + 9I + 8F,4 + 8F,4 + 9F,4,4 + 5F,4 + 5I + 3F,4 + 5I + 4F,4 + 5I + 7F,4 + 5I,4 + 5T + 3I,4 + 5T + 3I + F,4 + 5T + 3I + 4F,4 + 5T + 3I + 9F,4 + 5T + 4I + 11F,4 + 5T + 4I,4 + 5T + 4I + 3F,4 + 5T + 4I + 8F,4 + 5T + 7I + 8F,4 + 5T + 7I + 9F,4 + 5T + 7I,4 + 5T + 7I + 5F,4 + 5T + 3F,4 +
5T + 4F,4 + 5T + 7F,4 + 5T,9 + 3T,9 + 3T + F,9 + 3T + 4F,9 + 3T + 9F,9 + 3T + I + 11F,9 + 3T + I,9 + 3T + I + 3F,9 + 3T + I + 8F,9 + 3T + 4I + 8F,9 + 3T + 4I + 9F,9 + 3T + 4I,9 + 3T + 4I + 5F,9 + 3T + 9I + 3F,9 + 3T + 9I + 4F,9 + 3T + 9I + 7F,9 + 3T + 9I,9 + 4T + 11I,9 + 4T + 11I + F,9 + 4T + 11I + 4F,9 + 4T + 11I + 9F,9 + 4T + 11F,9 + 4T,9 + 4T + 3F,9 + 4T + 8F,9 + 4T + 3I + 8F,9 + 4T + 3I + 9F,9 + 4T + 3I,9 + 4T + 3I + 5F,9 + 4T + 8I + 3F,9 + 4T + 8I + 4F,9 + 4T + 8I + 7F,9 + 4T + 8I,9 + 7T + 8I,9 + 7T + 8I + F,9 + 7T + 8I + 4F,9 + 7T + 8I + 9F,9 + 7T + 9I + 11F,9 + 7T + 9I,9 + 7T + 9I + 3F,9 + 7T + 9I + 8F,9 + 7T + 8F,9 + 7T + 9F,9 + 7T,9 + 7T + 5F,9 + 7T + 5I + 3F,9 + 7T + 5I + 4F,9 + 7T + 5I + 7F,9 + 7T + 5I,9 + 3I,9 + 3I + F,9 + 3I + 4F,9 + 3I + 9F,9 + 4I + 11F,9 + 4I,9 + 4I + 3F,9 + 4I + 8F,9 + 7I + 8F,9 + 7I + 9F,9 + 7I,9 + 7I + 5F,9 + 3F,9 + 4F,9 + 7F,9.}
Similarly,wewillusetheidempotentstodividetheelementsoftheneutrosophicrings R ∪ T ∪ I ∪ F when R = F.Letusillustratethesewiththefollowingexample. Example7. Let R = Z3,whichisafield.FromExample 4,thedifferentidempotentsin Z3 are 1,0.From Corollary 4,thenumberofdifferentidempotentsinneutrosophicquadruplering Z3 ∪ T ∪ I ∪ F is 24 = 16. FromAlgorithm 2,thesetofall 16 idempotentsin Z3 ∪ I is: E = {0, F, I + 2F, I, T + 2I, T + 2I + F, T + 2F, T,1 + 2T,1 + 2T + F,1 + 2T + I + 2F,1 + 2T + I,1 + 2I,1 + 2I + F,1 + 2F,1}. Wehave E(0)= {0}, E(F)= {F,2F}, E(I + 2F)= {I + 2F,2I + F}, E(I)= {I, I + F,2I,2I + 2F}, E(T + 2I)= {T + 2I,2T + I}, E(T + 2I + F)= {T + 2I + F, T + 2I + 2F,2T + I + F,2T + I + 2F}, E(T + 2F)= {T + 2F, T + I + F,2T + F,2T + 2I + 2F}, E(T)= {T + F, T, T + I, T + I + 2F,2T,2T + 2F,2T + 2I,2T + 2I + F}, E(1 + 2T)= {1 + 2T,2 + T}, E(1 + 2T + F)= {1 + 2T + F,1 + 2T + 2F,2 + T + F,2 + T + 2F}, E(1 + 2T + I + 2F)= {1 + 2T + I + 2F,1 + 2T + 2I + F,2 + T + I + 2F,2 + T + 2I + F}, E(1 + 2T + I)= {1 + 2T + I,1 + 2T + I + F,1 + 2T + 2I,1 + 2T + 2I + 2F,2 + T + I,2 + T + I + F,2 + T + 2I,2 + T + 2I + 2F}, E(1 + 2I)= {1 + 2I,1 + T + I,2 + I,2 + 2T + 2I}, E(1 + 2I + F)= {1 + 2I + F,1 + 2I + 2F,1 + T + I + F,1 + T + I + 2F,2 + I + F,2 + I + 2F,2 + 2T + 2I + F,2 + 2T + 2I + 2F}, E(1 + 2F)= {1 + 2F,1 + I + F,1 + T + F,1 + T + 2I + 2F,2 + F,2 + 2I + 2F,2 + 2T + 2F,2 + 2T + I + F}, E(1)= {1,1 + F,1 + I,1 + I + 2F,1 + T,1 + T + 2F,1 + T + 2I,1 + T + 2I + F,2,2 + 2F,2 + 2I,2 + 2I + F,2 + 2T,2 + 2T + F,2 + 2T + I,2 + 2T + I + 2F}.So Z3 ∪ T ∪ I ∪ F = e∈E E(e)
5.Conclusions
Inthispaper,westudytheidempotentsinneutrosophicring R ∪ I andneutrosophicquadruple ring R ∪ T ∪ I ∪ F .Wenotonlysolvetheopenproblemandconjecturesinpaper[22]about idempotentsinneutrosophicring Zn ∪ I ,butalsogivealgorithmstoobtainallidempotentsin Zn ∪ I and Zn ∪ T ∪ I ∪ F foreach n.Furthermore,if R = F,thentheneutrosophicrings(neutrosophic quadruplerings)canbeviewedasapartitiondividedbytheidempotents.Asageneralresult,ifall idempotentsinring R areknown,thenallidempotentsin R ∪ I and R ∪ T ∪ I ∪ F canbeobtained too.Moreover,ifthenumberofallidempotentsinring R is t,thenthenumbersofallidempotentsin R ∪ I and R ∪ T ∪ I ∪ F are t2 and t4 respectively.Inthefollowing,ontheonehand,wewillexplore semi-idempotentsinneutrosophicrings,ontheotherhand,wewillstudythealgebrapropertiesof neutrosophicringsandneutrosophicquadruplerings.
AuthorContributions: Allauthorshavecontributedequallytothispaper.
Funding: ThisresearchwasfundedbyNationalNaturalScienceFoundationofChina(GrantNo.61976130), DisciplineConstructionFundingofXi’anPolytechnicUniversity,InstructionalScienceandTechnologyPlan ProjectsofChinaNationalTextileandApparelCouncil(No.2016073)andScientificResearchProgramFundedby ShaanxiProvincialEducationDepartment(ProgramNo.18JS042).
Acknowledgments: Theauthorswouldliketothankthereviewersfortheirmanyinsightfulcomments andsuggestions.
ConflictsofInterest: Theauthorsdeclarenoconflictofinterest.
AppendixA.TheMATLABcodeforsolvingtheidempotentsin (Zn, ) function neut=solve_neut(n)
%n:nonnegativeinteger %neut:allidempotentsinZ_n
B=[]; digits(32); for i=1:n for j=1:n A1(i,j)=mod((i 1) * (j 1),n); end end a1=factor(n); a2=unique(a1); for i=1: length (a2) b=length ( find (a1==a2(i))); B(i)=a2(i)^b; end D=[1]; for i=1: length (a2) C=combnk(B,i); A=prod (C,2); D=[D;A]; end D=mod(D,n); for i=1: length (D) if D(i)==1 neut(i)=1; elseif D(i)==0 neut(i)=0; else for j=1:n if mod(D(i) * j,n)==D(i) for k=1:n if mod(D(i) * k,n)==j neut(i)=j; break end end end end end end neut= sort (neut);
AppendixB.TheMATLABcodeforsolvingtheidempotentsin Zn ∪ I function ID=Idempotents_ZR(n)
%n:nonnegativeinteger
%ID:allidempotentsininneutrosophicring<Z_n\cupI>
neut=solve_neut(n); neutall=[]; for i=1: length (neut) for j=1: length (neut) c1=mod(neut(j) neut(i),n); neutall=[neutall;[neut(i),c1]]; end end
%ID:allidempotentsininneutrosophicquadruplering<Z_n\cupT\cupI\cupF>
neut=solve_neut(n); neutall=[]; for i=1: length (neut) a1=neut(i); for j=1: length (neut) a2=mod(neut(j) a1,n); for m=1: length (neut) a3=mod(neut(m) a1 a2,n); for q=1: length (neut) a4=mod(neut(q) a1 a2 a3,n); neutall=[neutall;[a1a2a3a4]]; end end end end
ID=sortrows(neutall’,1)’;
References
1. Smarandache,F. Neutrosophy:NeutrosophicProbability,Set,andLogic:AnalyticSynthesisandSyntheticAnalysis; AmericanResearchPress:SantaFe,NM,USA,1998.
2. Smarandache,F.;Ali,M.Neutrosophictripletgroup. NeuralComput.Appl. 2018, 29,595–601.
3. Smarandache,F. NeutrosophicPerspectives:Triplets,Duplets,Multisets,HybridOperators,ModalLogic,Hedge Algebras.AndApplications;InfiniteStudy:Conshohocken,PA,USA,2017.
4. Zhang,X.;Hu,Q.;Smarandache,F.;An,X.Onneutrosophictripletgroups:basicproperties,NT-subgroups, andsomenotes. Symmetry 2018, 10,289.
5. Ma,Y.;Zhang,X.;Yang,X.;Zhou,X.Generalizedneutrosophicextendedtripletgroup. Symmetry 2019, 11,327.
Symmetry 2019, 11,1254 15of15
6.
Zhang,X.H.;Bo,C.X.;Smarandache,F.;Dai,J.H.Newinclusionrelationofneutrosophicsetswith applicationsandrelatedlatticestructure. Int.J.Mach.Learn.Cybern 2018, 9,1753–1763.
7. Bal,M.;Shalla,M.M.;Olgun,N.Neutrosophictripletcosetsandquotientgroups. Symmetry 2017, 10,126.
8.
Zhang,X.H.;Smarandache,F.;Liang,X.L.Neutrosophicdupletsemi-groupandcancellableneutrosophic tripletgroups. Symmetry 2017, 9,275.
9. Wu,X.Y.;Zhang,X.H.ThedecompositiontheoremsofAG-neutrosophicextendedtripletloopsandstrong AG-(l,l)-loops. Mathematics 2019, 7,268.
10.
11.
Zhang,X.H.;Wu,X.Y.;Mao,X.Y.;Smarandache,F.;Park,C.Onneutrosophicextendedtripletgroups(loops) andAbel-Grassmann sgroupoids(AG-groupoids). J.Intell.FuzzySyst. 2019,inpress.
Zhang,X.H.;Mao,X.Y.;Wu,Y.T.;Zhai,X.H.Neutrosophicfiltersinpseudo-BCIalgebras. Int.J. UncertaintyQuant. 2018, 8,511–526.
12. Smarandache,F.Hybridneutrosophictripletringinphysicalstructures. Bull.Am.Phys.Soc. 2017, 62,17.
13. Ali,M.;Smarandache,F.;Khan,M.Studyonthedevelopmentofneutrosophictripletringand neutrosophictripletfield. Mathematics 2018, 6,46.
14. Sahin,M.;Abdullah,K.Neutrosophictripletnormedspace. OpenPhys 2017, 15,697–704.
15. Zhang,X.H.;Bo,C.X.;Smarandache,F.;Park,C.Newoperationsoftotallydependent-neutrosophicsetsand totallydependent-neutrosophicsoftsets. Symmetry 2018, 10,187.
16. Agboola,A.;Akinleye,S.Neutrosophicvectorspaces. NeutrosophicSetsSyst. 2014, 4,9–18.
17. Vasantha,W.B.;Smaradache,F. NeutrosophicRings;Hexis:Phoenix,AZ,USA,2006.
18. Agboola,A.A.D.;Akinola,A.D.;Oyebola,O.Y.NeutrosophicringsI. Int.J.Math.Comb. 2011, 4,115.
19. Broumi,S.;Smarandache,F.;Maji,P.K.Intuitionisticneutrosphicsoftsetoverrings. Math.Stat. 2014, 2, 120–126.
20. Ali,M.;Shabir,M.;Smarandache,F.;Vladareanu,L.NeutrosophicLA-semigrouprings. NeutrosophicSets Syst. 2015, 7,81–88.
21. Nicholson,W.K.Liftingidempotentsandexchangerings. Trans.Am.Math.Soc. 1977, 229,269–278.
22. Vasantha,W.B.;Kandasamy,I.;Smarandache,F.Semi-idempotentsinneutrosophicrings. Mathematics 2019, 7,507.
23. SmarandacheF.Neutrosophicquadruplenumbers,refinedneutrosophicquadruplenumbers,absorbance law,andthemultiplicationofneutrosophicquadruplenumbers. NeutrosophicSetsSyst. 2015, 10,96–98.
24. Akinleye,S.A.;Smarandache,F.;Agboola,A.A.A.Onneutrosophicquadruplealgebraicstructures. NeutrosophicSetsSyst. 2016, 12,122–126.
25. Li,Q.;Ma,Y.;Zhang,X.;Zhang,J.Neutrosophicextendedtripletgroupbasedonneutrosophicquadruple numbers. Symmetry 2019, 11,187.
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