Neutrosophic Subset Semigroups

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2 MOD Nturl Neutrosophic Subset Semiroups W. B. Vsnth Kndsmy lnthenrl K Florentin Smrndche 216

3 MOD Nturl Neutrosophic Subset Semiroups 2

4 Copyriht 216 by EuropNov ASBL nd the Authors Peer reviewers: Prof. Hud E. Khlid University of Telfer Hed - Mthemtics Deprtment Collee of Bsic Eduction Telfer - Mosul rq. Dr. Ahmed Slm Port Sid University Eypt. Phm Hon Phon Ntionl University of Civil Enineerin Hnoi Vietnm. Prof. Bui Con Cuon nstitute of Mthemtics Hnoi Vietnm. Mny books cn be downloded from the followin Diitl Librry of Science: SBN-13: EAN: Printed in the United Sttes of Americ 3

5 This book cn be ordered from: EuropNov ASBL Clos du Prnsse 3E 1 Bruxelles Belium E-mil: info@europnov.be URL: 4

6 CONTENTS Prefce 4 Chpter One BASC CONCEPTS 9 Chpter Two MOD SUBSET ALGEBRAC STRUCTURES 11 Chpter Three MOD NATURAL NEUTROSOPHC SUBSET SEMGROUPS 85 5

7 Chpter Four SEMGROUPS BULT ON MOD SUBSET MATRCES AND MOD SUBSET POLYNOMALS 161 Chpter Five MOD SUBSET NATURAL NEUTROSOPHC SEMGROUPS 299 FURTHER READNG 347 NDEX 355 ABOUT THE AUTHORS 36 6

8 PREFACE n this book uthors for the first time introduce the notion of MOD subsets usin Z n (or C(Z n ) or Z n or Z n or Z n h or Z n k or Zn or C (Z n ) or Z n or Z n or Z n h or Z n k. On these MOD subsets the opertion + is defined S(Z n ) denotes the MOD subset nd {S(Z n ) +} hppens to be only Smrndche semiroup. These S-semiroups enjoy severl interestin properties. The notion of MOD universl subset nd MOD bsorbin subsets re defined nd developed. {S(Z n ) } is lso semiroup nd severl properties ssocited with them re derived. MOD nturl neutrosophic subsets forms only semiroup under +. n fct the min feture enjoyed by this structure is they hve subset idempotents with respect to +. They re S-semiroups under +. These MOD nturl neutrosophic subsets of ll 6 types re only semiroups under. 7

9 These enjoy distinct properties dependin on the subset tht is used. Further in these cses two types of products cn be used one zero dominted product nd the other MOD nturl neutrosophic zero dominted product. Both the product re different. Finlly usin these MOD subsets mtrices nd polynomils re defined nd developed. The new notion of MOD subset mtrices re defined nd these collections lso under + (or ) re only semiroups. On similr lines MOD subset polynomils re defined nd they lso form only semiroup under + (or ). These new notions re interestin nd reserchers cn find lots of pplictions where semiroups find their pplictions. Severl problems t reserch level re lso suested for the reders. We wish to cknowlede Dr. K Kndsmy for his sustined support nd encourement in the writin of this book. W.B.VASANTHA KANDASAMY LANTHENRAL K FLORENTN SMARANDACHE 8

10 Chpter One BASC CONCEPTS n this chpter we introduce the bsic references which is essentil for this study. Here we study the MOD subsets semiroups of vrious types usin the six types of MOD inteers s well s MOD nturl neutrosophic numbers [66]. Further for subset structures refer [51]. We used dul numbers [48] specil qusi dul numbers [5] nd specil dul like numbers [49]. For the notion of finite complex numbers refer [47]. For neutrosophic concept nd structures on them refer [6-7]. The study here is sinificnt for we re ble to provide finite order semiroups under + the ddition opertion in most of the cses they re monoids nd they re not roups but however ll of them under + opertion re Smrndche semiroups. These semiroups re of finite order. On the other hnd study of MOD subset mtrices nd MOD mtrix subsets re introduced. They lso re only finite semiroups under + which re Smrndche semiroups. Finlly we introduce the notion of MOD subset polynomils subsets. Both of them under + re only semiroups. However

11 1 MOD Nturl Neutrosophic Subset Semiroups if the polynomils re of the form infinite order. i x i i then they re of Thus these provide exmples of monoid of infinite order which re not roups. But if we use polynomils of type m i x i i 2 m < then we et clss of MOD polynomil subsets s well s MOD subset polynomils both of which re of finite order. Still these re lso only semiroups under +. We replce the + opertion by nd study the structure of the semiroup. The MOD semiroups under hve severl interestin properties. The book is endowed with severl illustrtive exmples for tht lone cn ive more clrity to the reder bout these bstrct concepts. Finlly we study the sme structure in cse of MOD nturl neutrosophic subsets. These behve in distinct wy for we re forced to define two types of product; one the usul zero dominnt product in which t m = for ll t {n c h k } nd m is zero divisor or nilpotent or n idempotent in Z n or C(Z n ) or Z n or Z n h or Z n k or Z n [6]. The MOD nturl neutrosophic number dominnt zero product is = where m {n c h k }. m m

12 Chpter Two MOD SUBSET ALGEBRAC STRUCTURES n this chpter for the first time uthors study the probble lebric structures on Z n C(Z n ) Z n Z n Z n h nd Z n k 2 n <. Althouh in the 19 th century set theory ws introduced to mke prdim shift in mthemtics modern lebr nd set topoloicl spces ws introduced. But unfortuntely set theoretic notions ws not exploited in lebric structures they were used only in topoloicl spces so it is deem fit to study lebric structures built usin Z n C(Z n ) nd so on. However we hve defined nd developed subset lebric structures in [51]. Here for the first time we introduced lebric structures not only on Z n C(Z n ) nd so on but usin Z n C (Z n ) Z n nd so on. The ltter theory ives mny types of properties ssocited with them different from the existin ones.

13 12 MOD Nturl Neutrosophic Subset Semiroups Before we o for systemtic definition first we will illustrte by some exmples. Exmple 2.1: Let Z 3 be the modulo inteers. S(Z 3 ) = {All subsets of Z 3 includin Z 3 } = {{} {1} {2} {1 2} {1 } {2 } {1 2 }}. Clerly o(s(z 3 )) = 7. Exmple 2.2: Let S(Z 9 ) = {{} {1} {8} { 1} { 2} {78} { 1 2} { 1 3} {6 7 8} { 1 2 3} { } { } { } { } { } { } { } nd so on { }} be the MOD subset collection nd o(s(z 9 )) = n view of ll these we prove the followin theorem. THEOREM 2.1: Let Z n be the modulo inteers. S(Z n ) be the collection of ll subsets of Z n. o(s(z n )) = 2 n 1. Proof is direct nd hence left s n exercise to the reder. Throuhout this book S(Z n ) will denote the collection of ll subsets of Z n. We will first define the opertion of ddition + on S(Z n ). DEFNTON 2.1: Let Z n be the rin of modulo inteers S(Z n ) be the collection of ll subsets of Z n. S(Z n ) is defined s the MOD subset. Let A B S(Z n ) A + B = { i + b j (mod n) for ech i A nd ech b j B} S(Z n ). This is the wy + opertion is defined on S(Z n ). We will describe with exmples before we proceed onto determine the properties enjoyed by this MOD subset semiroup {S(Z n ) +}.

14 MOD Subset Alebric Structures 13 Exmple 2.3: Let S(Z 4 ) = {{} {1} {2} {3} { 1} { 2} { 3} {1 2} {1 3} {2 3} { 1 2} { 1 3} { 2 3} {1 2 3} { } = Z 4 } be the MOD subsets of Z 4. Define + on S(Z 4 ) s follows. {} + {1 2 3} = {1 2 3} {} + { 3} = { 3}nd so on. t is left for the reder to prove {} cts s the dditive identity of S(Z 4 ). t is importnt to note for ny X S(Z 4 ) we do not in enerl hve Y S(Z 4 ) such tht X + Y = {}. For if X = { 2 3} Y = {2 3 1} S(Z 4 ). X + Y = { 2 3} + {2 3 1} = { } = { } = { 1 2 3} = Z 4. This is the wy + opertion is performed on S(Z 4 ). We see {2} + { 1 3} = {2 3 1} S(Z 4 ). {2 } + { 1 3} = {2 3 1 } = Z 4. We see Z 4 = {2 3 1 } S(Z 4 ) is such tht for every x S(Z 4 ) x + Z 4 = Z 4. We cll this Z 4 s the bsorbin element of S(Z 4 ). { 2} + {1 3} = { } = {1 3} S(Z 4 ).

15 14 MOD Nturl Neutrosophic Subset Semiroups We hve lso sets in S(Z 4 ) such tht X + Y = Y for Y in S(Z 4 ) nd for prticulr X in S(Z 4 ). This is not possible in usul cses tht in semiroups or roups or rins under +. This sitution does not follow the set theoretic results. For { 2} nd {1 3} hve no common fctor infct pir of disjoint sets but their sum ives the set {1 3}. This sort of behvior in lebric opertions is very rre. { 1} + {2 3 } = {2 3 1} = Z 4. Now consider { 1 2} + {3 1} = {3 1 2} = Z 4. However the sets { 3 1} nd { 1 2} s sets re not disjoint. Exmple 2.4: Let S(Z 6 ) = {(collection of ll subsets of Z 6 }. Define + opertion on S(Z 6 ). Let {3 2} S(Z 6 ). {3 2} + {3 2} = { 5 4} {3 2} + { 5 4} = {3 2 1} {3 2} + {3 2 1} = { } { } + {3 2} = { } { } + {3 2} = { } { } + {3 2} = { } = {Z 6 }. After ddin sets number of times we rrive t fixed set; such study is new nd innovtive. { 3} + { 3} = { 3}. { 2} + { 2} = { 2 4} { 2 4} + { 2} = { 2 4}. Thus these sets ttins fixed point.

16 MOD Subset Alebric Structures 15 Tht is X + X = Y Y + XY. The followin opertion re pertinent. f X = { t } S(Z n ) is subroup under + of Z n then we see X + X = X. nd Thus {S(Z n ) +} hs elements X such tht X + X = X so {S(Z n ) +} is not roup under + only semiroup under +. Let X = {5} S(Z 6 ) {5} + {5} = {4} {4} + {5} = {3} {3} + {5} = {2} {2} +{5} = {1} {5} + {1} = {}. Let {3 5 } = Y S(Z n ) Y + Y = { 3 5} + { 3 5} = { } Y + { } = { } { 3 5} + { } = { }. Thus we see Z 6 = { } dded with ny set X S(Z 6 ) ives only Z 6. X = { 2 4} nd Y = {1 5} S(Z 6 ); X + Y = { 2 4} + {1 5} = {1 5 3} = Z 1. X + X = { 2 4} + { 2 4} = { 2 4} = X. Y + Y = {1 5} + {1 5} = {2 4 } = V 1.

17 16 MOD Nturl Neutrosophic Subset Semiroups We see X + X = Y + Y but X Y. Further Z 1 + Z 1 = {1 5 3} + {1 5 3} = {2 4} = X X + Y + X + Y = X. (X + Y X) Y + Y = X (X + Y) + (X + Y) = X. X + Y + Y = {1 5 3} + {1 5} = {2 4} (X + Y) + X = {1 5 3} + { 2 4} = X + Y. Severl interestin fetures cn be nlysed usin the subsets. We see the structure of {S(Z 6 )+} is only semiroup of order However {} S(Z 6 ) serves s the identity for ny X S(Z 6 ); X + {} = X = {} + X. nfct {S(Z 6 ) +} is commuttive monoid of finite order. For ny X S(Z 6 ) we cnnot in enerl find Y in S(Z 6 ) such tht X + Y = {}. For tke {2 4} S(Z 6 ) we do not hve Y S(Z 6 ) such tht Y + {2 4} = {}. Thus {S(Z 6 ) +} is only monoid nd not roup. However {S(Z 6 ) +} is Smrndche MOD semiroup s {Z 6 +} is roup under +. n view of ll these we hve the followin theorem. THEOREM 2.2: Let S(Z n ) = {(collection of ll subsets of Z n )}; 2 n i) o(s(z n )) = 2 n 1. ii) {S(Z n ) +} is only commuttive monoid.

18 MOD Subset Alebric Structures 17 iii) The set Z n = { 1 2 n 1} S (Z n ) is such tht for every X S(Z n ); X + Z n = Z n + X = Z n. Proof is direct nd hence left s n exercise to the reder. DEFNTON 2.2: Let {S(Z n ) +} be the MOD subsets semiroup of Z n. The set { 1 2 n 1} = Z n S(Z n ) is the defined s the MOD universl subset of S(Z n ). We see S(Z n ) hs one nd only one MOD universl subset. The MOD universl subset of S(Z n ) will lso be known s the MOD lrest bsorbin subset of S(Z n ). But there re mny bsorbin subsets of S(Z n ). We will ive some more exmples before we proceed to define nd develop more properties bout these new MOD subset structures. Exmple 2.5: Let S(Z 12 ) = {Collection of ll subsets of Z 12 }; the MOD subset of Z 12 {S(Z 12 ) +} is the MOD subset semiroup. Let X = { 3} nd Y = { 6 9 3} S(Z 12 ) X + Y = { 3} + { 6 9 3} = { 3 6 9} = Y. Let X 1 = { 2 4} nd Y = { 6 9 3} we see X 1 + Y = { 2 4} + { 6 9 3} = { } = Z 12 the MOD universl subset of S(Z 12 ) or the lrest MOD subset of S(Z 12 ). Let X 2 = { 8} nd Y = { 6 9 3} S(Z 12 ). X 2 + Y = { }. Clerly X 2 + Y is not the MOD universl subset of S(Z 12 ).

19 18 MOD Nturl Neutrosophic Subset Semiroups Let X 3 = { 2 8} nd Y = { 6 3 9} S(Z 12 ); X 3 + Y = { 2 8} + { 6 3 9} = { } Z 12. So only X 1 = { 2 4} with Y ives the MOD lrest bsorbin subset of S(Z 12 ). Let X 4 = { 4 8} S(Z 12 ) X 4 + Y = { 4 8} + { 6 3 9} = { } = Z 12. Once in X 4 = { 4 8} with Y ives the MOD lrest bsorbin subset of S(Z 12 ). Let X 5 = { 5 7} S(Z 12 ) now we find X 5 + Y = { 5 7} + { 3 6 9} = { } = Z 12. We see X 5 = { 5 7} lso yields with Y the MOD lrest bsorbin subset of S(Z 12 ). Consider { 5 11} = X 6 S(Z 12 ). We find X 6 + Y = { 5 11} + { 3 6 9} = { } Z 12. So X 6 cnnot yield Z 12 when dded with Y. Let X 7 = { 7 11} S(Z 12 ). We find X 7 + Y = { 7 11} + { 3 6 9} = { } = Z 12. Y. However X 7 yields Z 12 the MOD lrest bsorbin set with Let Z = { } nd B 1 = { 1} S(Z 12 ). B 1 + Z = { 1} + { }

20 MOD Subset Alebric Structures 19 = { } = Z 12. So Z with B 1 or B 1 with Z ive the MOD universl subset of S(Z 12 ). Let B 2 = { 3} S(Z 12 ). B 2 + Z = { 3} + { } = { } = Z 12. Ain B 2 with Z yield the MOD universl subset of S(Z 12 ). Let B 3 = {1 3} S(Z 12 ) B 3 + Z = {1 3} + { } = { }. So B 3 cnnot yield with Z the MOD lrest bsorbin subset or the MOD universl subset of S(Z 12 ). Let B 4 = { 5} S(Z 12 ). B 4 + Z = { 5} + { } = { } = Z 12. So with Z yields the MOD universl subset of S(Z 12 ). Let B 5 = { 7} S(Z 12 ). B 5 + Z = { 7} + { } = { } = Z 12. B 5 lso yields with Z the MOD universl subset of S(Z 12 ). Let B 6 = {3 5} S(Z 12 ) B 6 + Z = {3 5} + { } = { } Z 12. Z. So B 6 cnnot yield the MOD universl subset of S(Z 12 ) with

21 2 MOD Nturl Neutrosophic Subset Semiroups Let B 7 = { 9} S(Z 12 ). B 7 + Z = { 9} + { } = { } = Z 12. So B 7 with Z yields the MOD universl subset of S(Z 12 ). Let B 8 = { 1} S(Z 12 ); B 8 + Z = { 11} + { } = { } = Z 11. Thus B 8 with Z yields the MOD universl subset of S(Z 12 ). B 9 = {1 11} S(Z 12 ); B 9 + Z = {1 11} + { } = { }. Thus some sets yield the MOD universl subset. Let W = { 6} S(Z 12 ) be the subset of S(Z 12 ). Let X = { 5 2 7} S(Z 12 ) W + X = { 6} + { 2 5 7} = { } Z 12. Thus for us to hve the resultnt sum to be the MOD universl subset we need tlest the product of the crdinlity to be 12 nd should be the common element between the two sets. Exmple 2.6: Let S(Z 7 ) = {(collection of ll subsets of Z 7 } be the MOD subset semiroup under +. Let X = { 2} nd Y = { 1 4} S(Z 7 ) X+ Y = { } Z 7. Let X = { 2} but Y = { 1 6} S(Z 7 ) X + Y = { 2} + { 1 6} = { } Z 7. Let X= { 1 3} nd Y = { 4 5} S(Z 7 ) X + Y = { } Z 7.

22 MOD Subset Alebric Structures 21 Let X = { 1 2 3} nd Y = { 4 5} S(Z 7 ) X+ Y = { 1 2 3} + { 4 5} = { } = Z 7. Let X = { 1 2 3} nd Y = {5 6} S(Z 7 ) X + Y = { 1 2 3} + {5 6 } = { } Z 7. Let X = { 1 4 5} nd Y = { 3} S(Z 7 ) X+ Y = ( 1 4 5} + { 3} = { } Z 7. So if in Z n n is prime S(Z n ) hppens to behve in very different wy evident from this exmple. Exmple 2.7: Let S(Z 4 ) = {collection of ll subsets of Z 4 } be the MOD subset semiroup under +. Let S = { 2} nd Y = {1 3} S + Y = { 2} + {1 3} = {1 3} So S = { 2} leves {1 3} s it is even fter ddition. Let X = {1 2} nd Y = { 3} S(Z 4 ) X+ Y = {1 2} + { 3} = {1 2 } Z 4. Let X = { 1 2} nd Y = { 3} S(Z 4 ) X + Y = { 1 2} + { 3} = { 1 2 3} = Z 4. Let X = {1 3 2} nd Y = { 1} S(Z 4 ) X + Y = { 1} + {1 3 2} = {1 3 2 } = Z 4 However it is seen Y X yet Y + X X but Z 4. This is the specilty of MOD subset semiroups. We see { 3} = Y nd X = {1 2 3} S(Z 4 ) is such tht X + Y = { 3} + {1 2 3} = {1 2 3 } = Z 4. So we see if Y = { 3} is replced by Y 1 = { 2} then Y 1 + X = { 2} + {1 2 3} = {1 2 3 } = Z 4. Let Y 2 = { 1} S(Z 4 ).

23 22 MOD Nturl Neutrosophic Subset Semiroups We find Y 2 + X = { 1} + {1 2 3} = { 1 2 3} = Z 4 thus for X = {1 2 3} ll the three sets { 1} { 2} nd { 3} re such tht their sum with X yield the MOD universl subset. Exmple 2.8: S(Z 1 ) = {(collection of ll subsets of Z 1 } be the MOD subset semiroup under +. Let X = { } nd Y = { 5} S(Z 1 ). X + Y = { } + { 5} = { } = Z 1. Thus X nd Y sum led to the MOD universl subset of S(Z 1 ). Let Y 1 = { 3} nd X = { } S(Z 1 ). We find Y 1 + X= { 3} + { } = { } = Z 1. Thus Y 1 with X lso yields the MOD universl subset of S(Z 1 ). Let Y 2 = { 1} S(Z 1 ); Y 2 + X = { 1} + { } = { } = Z 1. Thus Y 2 with X lso yields the MOD universl subset Z 1 of S(Z 1 ). Let Y 3 = { 7} S(Z 1 ) Y 3 + X = { 7} + { } = { } = Z 1. Thus Y 3 with X lso yields the MOD universl subset of S(Z 1 ). Let Y 4 = { 9} S(Z 1 ); we find Y 4 + X = { 9} + { } = { } = Z 1.

24 MOD Subset Alebric Structures 23 So Y 4 lso with X yields the MOD universl subset of S(Z 1 ). These subset Y Y 1 Y 2 Y 3 nd Y 4 re the smllest order subsets of S(Z 1 ) which when dded with X = { } yield the MOD universl subset of S(Z 1 ). Exmple 2.9: Let S(Z 15 ) = {Collection of ll subsets of Z 15 } be the MOD subset semiroup under +. Let X = { } nd Y = { 2} S(Z 15 ). X + Y = { } + { 2} = { }. Let Y = { 2 8} S(Z 15 ). We find X + Y = { 2 8} + { } = { } Z 15. So Y with X does not yield the MOD universl subset of S(Z 15 ). Let Y 1 = { 2 4} S(Z 15 ). X + Y 1 = { 2 4} + { } = { } = Z 15. We see Y 1 with X yields the MOD universl subset of S(Z 15 ). Let Y 2 = { 8 1} S(Z 12 ) Y 2 + X = { 8 1} + { } = { } = Z 15 so Y 2 lso yields when dded with X the MOD universl subset of S(Z 15 ). Let Y 3 = { 1 14} S(Z 15 ). We find

25 24 MOD Nturl Neutrosophic Subset Semiroups Y 3 + X = { 1 14} + { } = { } = Z 15. Thus Y 3 lso when dded with X yields the MOD universl subset of S(Z 15 ). Let Y 4 = { 8 14} S(Z 15 ). Y 4 + X = { 8 14} + { } = { } Z 15. So Y 4 cnnot serve s the subset of S(Z 15 ) which cn yield Z 15 ddin with the subset X. Let Y 5 = { 8 12} S(Z 15 ). Y 15 + X = { 8 12} + { } = { } Z 15. Thus it is left s open conjecture for S(Z n ). Now we propose the followin open problem. Problem 2.1: Let S(Z n ) be the MOD subset semiroup under +; n nonprime 2 n <. Find the number of subsets X in S(Z n ) which when dded with the subset P of S(Z n ) where the elements of P under + from roup nd the sum of X + P s MOD universl subset of S(Zn). Study this question for ll subsets of Z n. DEFNTON 2.3: Let S(Z n ) be the MOD subset semiroup. Let P S(Z n ) if the collection of elements in P forms roup under + then we define P to be subroup subset of S(Z n ). We will illustrte these situtions by some exmples.

26 MOD Subset Alebric Structures 25 Exmple 2.1: Let S(Z 24 ) be the MOD subset semiroup under +. P 1 = { 12} P 2 = { 8 16} P 3 = { } P 4 = { } P 5 = { } nd P 6 = { } re the only subset subroups of S(Z 24 ). Findin subsets in S(Z 24 ) so tht their sum with P 1 or P 2 or P 3 or P 4 or P 5 or P 6 yieldin the subset Z 24 hppens to be mtter of routine nd hence left s n exercise to the reder. We see in cse of P 3 = { } if we tke X = { } S(Z 24 ) nd we clculte P 3 + X = { } + { } = { } Z 24. Tke B = { } S(Z 24 ). We find P 3 + B = { } + { } = { } = Z 24. Thus for the MOD subroup subset P 3 B in S(Z 24 ) serves s the subset when summed up with P 3 yields Z 24 the MOD universl subset of S(Z 24 ). Now for the sme P 3 consider D = { } P 3 + D = { } + { }

27 26 MOD Nturl Neutrosophic Subset Semiroups = { }. Thus D is lso nother subset of S(Z 24 ) which yield the MOD universl subset of S(Z 24 ) when dded with the MOD subset subroup P 3 of S(Z 24 ). Consider P 2 = { 8 16} of S(Z 24 ). We see X = { } S(Z 24 ) is such tht P 2 + X = { 8 16} + { } = { } = Z 24 the MOD universl subset of S(Z 24 ). Let P 3 = { 8 16} nd Y = { } S(Z 24 ) is such tht P 3 + Y = { 8 16} + { } = { } = Z 24 S(Z 24 ) is such tht P 3 + Y yield the MOD universl subset of S(Z 24 ). Clerly Y is not MOD subroup subset of S(Z 24 ); Y is only just MOD subset of S(Z 24 ). Consider P 6 = { } S (Z 24 ) we see D = { 1 2 3} S(Z 24 ) is such tht P 6 + D = { } + { 1 2 3} = { } = Z 24. Thus D when dded with P 6 yields the MOD universl subset of S(Z 24 ).

28 MOD Subset Alebric Structures 27 Consider E = { 5 6 7} S(Z 24 ). We find P 6 + E = { } + { 5 6 7} = { } = Z 24. Thus we see for the MOD subset subroup P 6 we hve E S(Z 24 ) such tht the sum of P 6 with E yields Z 24 the MOD universl subset of S(Z 24 ). Consider F = { } S(Z 24 ). F + P 6 = { } + { } = { } = Z 24. Thus F dded with P 6 yield the MOD universl subset of S(Z 24 ). Let G = { } S(Z 24 ) we find P 6 + G = { } + { } = { } = Z 24 thus G is lso MOD subset of S(Z 24 ) such tht P 6 + G is the MOD universl subset of S(Z 24 ). Let H = { } S(Z 24 ) we find P 6 + H = { } + { } = { } = Z 24. Hence H summed up with P 6 ives the MOD universl subset of S(Z 24 ).

29 28 MOD Nturl Neutrosophic Subset Semiroups Let = { } S(Z 24 ) we find J + P 6 = { } + { } = { } = Z 24. We see J summed up with P 6 yields the MOD universl subset Z 24 of S(Z 24 ). Consider K = { } S(Z 24 ). We find P 6 + K = { } + { } = { } Z 24. Tke L = { } S(Z 24 ) we find L + P 6 = { } + { } = { } = Z 24. Thus L lso serves s MOD subset which yields the MOD universl subset of S(Z 24 ). Hence we hve severl such subsets in S(Z 24 ) when dded with P 6 cn yield Z 24. Wht is the problem is findin mens nd methods to et ll of them for iven MOD subroup subset in S(Z n ). Thus we leve the followin problem s open. Problem 2.2: Let {S(Z n ) +} be the MOD subset semiroup Z n the MOD universl subset of S(Z n ). Let P S(Z n ) be MOD subroup subset of S(Z n ). Find ll X S(Z n ) such tht P + X = Z n.

30 MOD Subset Alebric Structures 29 Next we proceed onto define the notion of product on MOD subsets of S(Z n ). We will first illustrte this sitution by some exmples. Exmple 2.11: Let S(Z 6 ) = {(collection of ll subsets of Z 6 } on S(Z 6 ) we define product. f X = { 3 2} nd Y = {1 4} S(Z 6 ) then X Y = { 3 2} {1 4} = { 3 2} S(Z 6 ). X X = { 3 2} { 3 2} = { 4 3}. Y Y = {1 4} {1 4} = {1 4} = Y. Let X = {3} nd Y = { 2 4} S(Z 6 ) ; X Y = {3} { 2 4} = {} X X = {3} {3} {3} = X. Y Y = { 2 4} { 2 4} = { 4 2} = Y. X X = {3} {3} = {3} = X. Y Y = { 2 4} { 2 4} = { 4 2} = Y. Thus both X nd Y re idempotents tht is they re defined s MOD subset idempotents. We cll {S(Z 6 ) } is the MOD subset semiroup. Let B = { 2 3} S(Z 6 ) B B = { 4 3} B. Let C = { 4 3} S(Z 6 ) C C = { 4 3} = C is MOD subset idempotent of S(Z 6 ). Now {1} S(Z 6 )} is such tht {1} A = A {1} = A for ll A S(Z 6 ). {1} is clled s the MOD subset unit of S(Z 6 ).

31 3 MOD Nturl Neutrosophic Subset Semiroups {} S(Z 6 ) is such tht {} A = A {} = {} for ll A S(Z 6 ) so {} is defined s the MOD subset zero of S(Z 6 ). Clerly S(Z 6 ) is only MOD commuttive semiroup of order = {2} nd b = {3} S(Z 6 ) is such tht b = {2} {3} = {}. Let P = { 3 2} nd Q = {4 5} S(Z 6 ). P P = { 3 2} { 3 2} = { 3 4} P. Q Q = {4 5} {4 5} = {4 2 1} Q. P Q = { 3 2} {4 5} = { 2 3 4}. Let T = { } S(Z 6 ). T T = { } { } = { } = Z 6. Thus the MOD set T in S(Z 6 ) is such tht T T is the MOD universl subset of S(Z 6 ). Let B = {1 5} S(Z 6 ). B B = {1 5} {1 5} = {1 5} is MOD idempotent subset of S(Z 6 ). Let M = {1 5 2} S(Z 6 ) M M = {1 5 2} {1 5 2} = { } M. Let Y = {1 3 2} nd B = {1 5} S(Z 6 ) B Y = {1 5} {1 3 2} = { } = { } Z 6. Consider C = {1 3 5} nd Y = {1 3 2} S(Z 6 ) we find C Y = {1 3 5} {1 3 2} = { } = Z 6.

32 MOD Subset Alebric Structures 31 Thus the MOD subsets C nd Y in S(Z 6 ) re such tht their product yields the MOD universl subset. Consider C + Y = { } = Z 6 Thus we see the pir of MOD subsets {C Y} of S(Z 6 ) re tht C Y = Z 6 nd C + Y = Z 6. Such MOD subsets re very unique nd the study relted with them will be crried out lter. Exmple 2.12: Let {S(Z 11 ) } be the MOD subset semiroup under. Let A = {2 3} S(Z 11 ) A A = {2 3} {2 3} = { 4 6 9} = A 2. A 2 A = { 4 6 9} { 2 3} = { } = A 3. A 3 A = { } { 2 3} = { } = A 4 A 4 A = { } { 2 3} = { } = A 5. A 5 A = { } { 2 3} = { } = A 6. A 6 A = { } { 2 3} = { } = A 7. A 7 A = { } { 2 3} = { } = A 8. A 8 A = { } { 2 3} = { } = A 9 = Z 11. So findin such A in S(Z 11 ) such tht A m for some m is the MOD universl subset of S(Z 11 ).

33 32 MOD Nturl Neutrosophic Subset Semiroups Let B = { 3 5} S(Z 11 ). B B = { 3 5} { 3 5} = { 9 4 3} = B 2. B 2 B = { 9 4 3} { 3 5} = { } { 3 5} = { } = B 3. B 3 B = { } { 3 5} = { } = B 4. B 4 B = { } { 3 5} = { } = B 5. B 5 B = B 6 = B 5. A = { } S(Z 11 ). A A = { } { } = { } = A 2 = Z 11. A 2 A = A 3 = A 2 = Z 11. Thus A ives fter product with A we et A 2 to be MOD universl subset of S(Z 11 ). Let P = { } S(Z 11 ). P P = { } { } = { } = Z 11. Thus P is MOD universl subset of S(Z 11 ) s P 2 = Z 11. B = { } S(Z 11 ). B B = { } { } = { } = Z 11. Similrly B 2 ives the MOD universl subset of S(Z 11 ). Let W = { } S(Z 11 ).

34 MOD Subset Alebric Structures 33 W W = { } { } = { } = W 2. W 2 W = { ) { } = { } = W 3. Clerly W 2 = W 3 so this MOD subset will not yield the MOD universl subset of S(Z 11 ). Let V = { 7 2 1} S(Z 11 ); V V = { 1 2 7} { 1 2 7} = { } = V 2. V 2 V = { } { 7 1 2} = { } = V 3. V 3 V = { } { 7 1 2} = { } = Z 11. Thus V is MOD subset of S(Z 11 ) which ives V 2 = Z 11. nfct if D = { 3} then D 2 = { 3} { 3} = { 9}. D 3 = { 3} { 9} = { 5} nd D 4 = { 5} { 3} = { 4}. D 4 D = { 4} { 3} = { 1} D 5 D = { 1} { 3} = { 3} = D. Thus D cn never be MOD subset of S(Z 11 ) which cn contribute Z 12 when producted ny number of times. D t Z 11 for ny t > 1. Let E = { 2 5} S(Z 11 ) E E = { 2 5} { 2 5} = { 4 1 3} = E 2 E 2 E = { 4 1 3} { 2 5} = { } = E 3

35 34 MOD Nturl Neutrosophic Subset Semiroups E 3 E = { } { 2 5} = { } = E 4 E 4 E = { } { 2 5} = { } = E 5. E 5 E = { } { 2 5} = { } = E 6. E 6 E = { } { 2 5} = { } = E 7. E 7 E = { } { 2 5} = { } = E 8. E 8 E = { } { 2 5} = { } = E 9 = Z 11. Tht is E 9 = Z 11. We see if A S (Z 11 ) with A 3 A we men the number of distinct elements in A nd one of elements in A is zero then there is m > 1 let m 1 such tht A m = Z 11 ; then A ives the MOD universl subset fter productin finite number of times. Exmple 2.13: Let {S(Z 7 ) } be the MOD subset semiroup under product. Let A = { 3} S(Z 7 ); A A = { 3} { 3} = { 2} = A 2 A 2 A = { 2} { 3} = { 6} = A 3 A 3 A = { 6} { 3} = { 4} = A 4 A 4 A = { 4} { 3} = { 5} = A 5 A 5 A = { 5} { 3} = { 1} = A 6 A 6 A = { 1} { 3} = { 3} = A. Thus A for no vlue of m cn rech Z 7 the MOD universl subset of S(Z 7 ).

36 MOD Subset Alebric Structures 35 Let B = { 1 4} S(Z 7 ); B B = { 1 4} { 1 4} = { 1 4 2} = B 2 B 2 B = { 1 4 2} { 1 4} = { 1 4 2} = B 3. Clerly B 2 = B 3 so we cnnot et for ny m B m = Z 7 Let B = { 2 5} S(Z 7 ) B B = { 2 5} { 2 5} = { 4 3} = B 2 B 2 B = { 2 5} { 4 3} = { 1 6} = B 3 B 3 B = { 1 6} { 2 5} = { 2 5} = B 4 = B. Thus B 4 = B so we do not hve positive inteer m such tht B m = Z 7. Let C = { 3 5} S(Z 7 ); C C = { 3 5} { 3 5} = { 2 1 4} = C 2 C 2 C = { 2 1 4} { 3 5} = { 6 3 5} = C 3 C 3 C = { 6 3 5} { 3 5} = { 4 2 1} = C 4 = C 2. So this C behves in n odd wy such tht C 4 = C 2 so C 3 = C 5 nd so on Let M = { 4 1 5} S(Z 7 ) M M = { 1 4 5} { 1 4 5} = { 1 4 5} = { } = M 2. M 2 M = { } { 1 4 5} = { } = M 3 = Z 7. Thus M 3 = Z 7 so M 3 ives the MOD universl subset of S(Z 7 ). Let N = { 2 3 6} S(Z 7 ). W find N N = { 2 3 6} { 2 3 6} = { } = N 2 = Z 7.

37 36 MOD Nturl Neutrosophic Subset Semiroups Thus N 2 ives the MOD universl subset of S(Z 7 ). Let L = { 1 2 3} S(Z 7 ); L L = { 1 2 3} { 2 1 3} = { } = L 2 L 2 L = { } { 1 2 3} = { } = L 3 = Z 7. Thus L 3 ives the MOD universl subset of S(Z 7 ) Let V = { 2 4 6} S(Z 7 ); we find V V = { 2 4 6} { 2 4 6} = { } = V 2 V 2 V = { } { } = { } = Z 7. Thus V 3 ives the MOD universl subset of S(Z 7 ). Let S = { 1 6} S(Z 7 ) S S = { 1 6} { 1 6} = { 1 6} = S. So no power of S cn ive Z 7 nd S 2 = s is MOD idempotent subset of S(Z 7 ). Let R = { 3 5} S(Z 7 ) R R = { 3 5} { 3 5} = { 2 1 4} = R 2 R 2 R = { 1 2 4} { 3 5} = { 3 6 5} = R 3 R 3 R = { 3 5 6} { 3 5} = { 2 1 4} = R 4 = R 2 So R 4 = R 2 cn never ive the MOD universl subset of S(Z 7 ). Let G = { 2 4} S(Z 7 ) G G = { 2 4} { 2 4} = { 4 1 2} = G 2 G 2 G = { 1 2 4} { 2 4} = { 2 4 1} = G 3 = G 2. So G 3 = G 2 hence for no m (m positive inteer);

38 MOD Subset Alebric Structures 37 G m = Z 7 the MOD universl subset of S(Z 7 ). Let T = { } S(Z 7 ) we find T T = { } { } = { } = Z 7. Thus T is such tht T T = T 2 = Z 7 ives the MOD universl subset of S(Z 7 ). So for S(Z 7 ) we see unlike S(Z 11 ) we do not see if A S(Z 7 ) with A = 3 nd A then in enerl A m Z 7. However there re subsets in S(Z 7 ) which re such tht their product tken t stipulted number of times yields Z 7. But {S(Z 11 ) } behves in different wy. So we re forced to suest the followin problem. Problem 2.3: Let {S(Z p ) } be the MOD subset semiroup under product (p prime). Chrcterize those MOD subset A of S(Z p ) so tht their power A m = Z p the MOD universl subset of S(Z p ); m > 1. Next we tke S(Z n ) where n is composite number nd ive exmples of the sme. Exmple 2.14: Let {S(Z 8 ) } be the MOD subset semiroup under product. A = { 2 4 6} S(Z 8 ) A A = { 2 4 6} { 2 4 6} = { 4} = A 2 A 2 A = { 2 4 6} { 4} = {}. A 3 = {}. We see A is MOD nilpotent subset of S(Z 8 ). Let P = { 2 5} S(Z 8 ) P P = { 2 5} { 2 5} = { 4 2 1} = P 2 P 2 P = { 4 2 1} { 2 5} = { } = P 3

39 38 MOD Nturl Neutrosophic Subset Semiroups P 3 P = { } { 2 5} = { } = P 4 = P 3. P 4 = P 3 so for no m (m > 1). P m = Z 8. S = {1 5 7} S(Z 8 ); S S = { 1 5 7} { 1 5 7} = { } = S 2 S 2 S = { } { } = { } = S 3 = S 2. Thus S 3 = S 2 so for no m (m positive inteer m > 1) we cn et S m = Z 8. Let T = { 2 7} S(Z 8 ); T T = { 2 7} { 2 7} = { 4 6 1} = T 2 T 2 T = { 1 4 6} { 2 7} = { } = T 3 T 3 T = { } { 2 7} = { } = T 4 T 4 T = { } { 2 7} = { } = T 5. T t T = { } { 2 7} = { } = T 6 = T 4. Thus this will not led to the MOD universl subset Z 8 S(Z 8 ) s T 6 = T 4 ; T 7 = T 5 nd so on. Let P = { 1 4} S(Z 8 ) P 2 = P P = { 1 4} { 1 4} = { 1 4} = P. So for no m > 1 P m universl subset Z 8 of S(Z 8 ). = Z 8 so P cnnot ive the MOD W = { 6 1} S(Z 8 );

40 MOD Subset Alebric Structures 39 W W = { 1 6} {1 6} = { 1 6 4} = W 2. W 2 W = { 1 4 6} { 1 6} = { 4 6 1} = W 3 = W 2 As W 3 = W 2 it is impossible to rrive t W m = Z 8 for ny m > 1. Exmple 2.15: Let {S(Z 12 ) } be the MOD subset semiroup under. Let B = { 3 5} S(Z 12 ); B B = { 3 5} { 3 5} = { 9 3 1} B 2 B = { 1 3 9} { 3 5} = { 3 5 9} = B 3 B 3 B = { 3 5 9} { 3 5} = { } = B 4 B 4 B = { } { 3 5} = { } = B 5 B 5 B = { } { 3 5} = { } = B 6 B 6 B = { } { 3 5} = { } = B 7 B 7 B 5 so for no m > 1 B m = Z 12 the MOD universl subset of S(Z 12 ). Let P = { 3 1} S(Z 12 ). P P = { 3 1} { 3 1} = { 9 6 4} = P 2 P 2 P = { 6 4 9} { 1 3} = { 4 6 3} = P 3 P 3 P = { 4 6 3} { 3 1} = { 6 9 4} = P 4 = P 2. So P 2 = P 4 so we see P is not MOD subset of S(Z 12 ) which cn contribute to the MOD universl subset of S(Z 12 ). Let V = { 1 2 5} S(Z 12 ); V V = { 1 2 5} { 1 2 5} = { } = V 2. V 2 V = { } { 1 2 5} = { } = V 3. V 3 V = { } { 1 2 5} = { } = V 4. But V 4 = V 3 so we see V cnnot for ny (m > 1) ive V m = Z 12 the MOD universl subset of S(Z 12 ).

41 4 MOD Nturl Neutrosophic Subset Semiroups Let W = { } S(Z 12 ); W W = { } { } = { } = W 2. W 2 W = { } { } = { } = W 3. W 3 = W 2 so this lso does not yield the MOD universl subset Z 12 of S(Z 12 ). Let M = { } S(Z 12 ) M M = { } { } = { } = M 2. M 2 M = { } { } = { } = M 3. But M 3 = M so M cnnot yield the MOD universl subset Z 12 of S(Z 12 ). Thus we see in cse of findin the MOD subsets which cn yield MOD universl subset hppens to be very difficult problem. n view of this we propose the followin problem. Problem 2.4: Let {S(Z n ) } ( n bi composite number) be the MOD semiroup under product. Find ll MOD subsets B of S(Z n ) such tht B m = Z n (m > 1). Next we proceed onto study MOD subsemiroups of S(Z n ) by some exmples. Exmple 2.16: Let {S(Z 6 ) } be the MOD subset semiroup of Z 6 under product.

42 MOD Subset Alebric Structures 41 Let P = { 2} S(Z 6 ) P P = { 2} { 2} = { 4} = P 2 P 2 P = { 2} { 4} = { 2} = P 3 = P. So M = {P P 2 P 3 = P} is MOD subset subsemiroup of order two. Let B = { 1 2} S(Z 6 ) B B = { 1 2} { 2 1} = { 1 2 4} = B 2 B 2 B = { 1 2 4} { 1 2} = { 1 2 4} = B 3 = B 2. L = {B B 2 B 3 = B 2 } is MOD subset subsemiroup of order two in S(Z 6 ). Clerly L nd M re not isomorphic s MOD subset subsemiroups. Let S = { 1 5 2} S(Z 6 ) S S = { 1 5 2} { 1 2 5} = { } = S 2 S 2 S = { } { 1 2 5} = { } = S 3 = S 2. Clerly N = {S S 2 S 3 = S 2 } is MOD subset subsemiroup of S(Z 6 ) nd N L s MOD subset subsemiroups of S(Z 6 ). Let W = { 2 3} S(Z 6 ) we find W W = { 2 3} { 2 3} = { 4 3} = W 2 W 2 W = { 2 3} { 3 4} = { 3 2} = W 3 = W. Now R = {W W 2 W 3 = W} is MOD subset subsemiroup of S(Z 6 ). Clerly R M s MOD subset subsemiroups of S(Z 6 ). Let A = { 3} S(Z 6 ) A A = { 3} { 3} = { 3} = A 2 = A.

43 42 MOD Nturl Neutrosophic Subset Semiroups Thus B = {A A 2 = A} is MOD subset subsemiroup of order one. Let C = {2} S(Z 6 ); C C = {2} {2} = {4} = C 2 C 2 C = {4} {2} = {2} = C 3 = C. Thus S = {C C 2 C 3 = C} is MOD subset subsemiroup of order two. Let T = { 1} nd S = { 2} S(Z 6 ). We find the MOD subset semiroup enerted by T nd S. T T = { 1} { 1} = { 1} = T 2 = T. S T = { 2} { 2} = { 4} = S 2 S 2 S = { 4} { 2} = { 2} = S 3 = S S T = { 1} { 2} = { 2}. So H = {{ 1} { 2} { 4} / { 1} { 1} = { 1} { 2} { 2} = { 4} { 2} { 4} = { 2} { 2} { 1} = { 2} nd { 4} { 1} = { 4}} is MOD subset subsemiroup of order three. t is to be noted S(Z 6 ) hs the elements {1} to be the identity but { 1} cn be defined s pseudo identity of the subsets Y of S(Z 6 ) where Y for in the cse { 1} Y = Y for ll such Y S(Z 6 ). So H cn be described s MOD subset subsemiroup with the pseudo identity or specificlly S is pseudo MOD subset monoid { 1} {1} = { 1}. However the pseudo identity is not the identity for if Z = {2 3 1} S(Z 6 ); { 1} {2 3 1} = { 1 2 3} Z. Hence our clim but {1} {2 3 1} = {1 3 2} = Z.

44 MOD Subset Alebric Structures 43 This is the mjor different between the identity {1} nd { 1}. nfct {1} serves s the identity for the pseudo identity { 1}. n view of ll these we ive followin definition. DEFNTON 2.4: Let {S(Z n ) } be the MOD subset semiroup under {1} S(Z n ) is defined s the identity nd { 1} is defined s the pseudo identity for ll subset Y S(Z n ) such tht is one of the element in Y. Further {1} { 1} = { 1} so {1} serves s the identity for the pseudo identity. DEFNTON 2.5: Let {S(Z n ) } be the MOD subset semiroup. P = {X 1 X t / 2 t < X i S(Z n )} S(Z n ) is defined s the pseudo MOD subset monoid if { 1} cts s the pseudo identity tht is { 1} X i = X i for 1 i t. We will ive more exmples of this sitution. Exmple 2.17: Let {S(Z 11 ) } be the MOD subset semiroup under. Let A = {6 3 2} S(Z 11 ) A A = {2 3 6} {2 3 6} = { } = A 2 A 2 A = { } {2 3 6} = { } = A 3 A 3 A = { } {2 3 6} = { } = A 4 A 4 A = { } = A 5 = A 4. Thus A 5 = A 4 so A hs no power to yield the MOD universl subset of S(Z 11 ) viz Z 11.

45 44 MOD Nturl Neutrosophic Subset Semiroups However if we hve tken {} A = { 2 3 6} = B then certinly B 4 = Z 11. t is pertinent to observe nd keep on record tht in cse we use in S(Z n ) n prime then if set A S(Z n ) such tht A m = Z n then necessrily A must contin. We consider now tke B = { 5 8 9} S(Z 11 ); B B = { 5 8 9} { 5 8 9} = { } = B B 2 B = { } { 5 8 9} = { } = B 3. B 3 B = { } { 5 8 9} = { } = B 4 = Z 11. Thus B yield the MOD universl subset of S(Z 11 ). Let M = { 1 5} S(Z 11 ); M M = { 1 5} { 1 5} = { 1 5 3} = M 2 M 2 M = { 1 5 3} { 1 5} = { } = M 3 M 3 M = { } { 1 5} = { } = M 4 M 4 M = { } { 1 5} = { } = M 5 M 5 = M 4 so M cnnot yield the MOD universl subset Z 11 for ny n n > 1. Now we ive the followin theorem. THEOREM 2.3: Let {S(Z p ) } (p prime) be the MOD subset semiroup under product. f A S(Z p ) A m = Z p (m > 1) only if A. Proof iven A m = Z p so A m. But s p is prime no product cn yield so A. However if A one cnnot sy tht A m = Z p.

46 MOD Subset Alebric Structures 45 This is will be illustrted by the followin exmple. Exmple 2.18: Let {S(Z 17 ) } be the MOD subset semiroup under product. Let A = {6 } S(Z 17 ). A A = { 6} {6} = { 2} = A 2 A 2 A = { 2} { 6} = { 12} = A 3 A 3 A = { 12} { 6} = { 4} = A 4 A 4 A = ( 4} { 6} = { 7} = A 5 A 5 A = { 7} { 6} = { 8} = A 6 A 6 A = { 8} { 6} = { 14} = A 7 A 7 A= { 14} { 6} = { 6} = A 8 = A. Since A 8 = A for no m m > 1 we cn hve A m = Z 17. But A. Hence the converse of the theorem is not true. Tht is if set A S(Z p ) p prime nd A then in enerl A m Z p for some m > 1. Wht we cn sy is if A m = Z p then A. Let P = { 5 2} S(Z 17 ) P P = { 2 5} { 2 5} = { 4 1 8} = P 2 P 2 P = { 4 8 1} { 2 5} = { } = P 3 P 3 P = { } { 2 5} = { } = P 4 P 4 P = { } { 2 5} = { } = P 5 P 5 P = { } { 2 5} = { } = P 6

47 46 MOD Nturl Neutrosophic Subset Semiroups P 6 P = { } { 2 5} = { } = P 7 P 7 P = { } { 2 5} = { } = P 8 P 8 P = { } { 2 5} = { } = P 9 P 9 P = { } { 2 5} = { } = P 1 P 1 P = { } { 2 5} = { } = P 11 P 11 P = { } { 2 5} = { } = P 12 P 12 P = { } { 2 5} = { } = P 13 P 13 P = { } { 2 5} = { } = P 14 P 14 P = { } { 2 5} = { } = P 15 = Z 17. Thus P 15 = Z 17 yields the MOD universl subset of S(Z 17 ). Let B = { 2 4} S(Z 17 ) B B = { 2 4} { 2 4} = { } = B 2 B 2 B = { } { 2 4} = { } = B 3 B 3 B = { } { 2 4}

48 MOD Subset Alebric Structures 47 = { } = B 4 B 4 B = { } { 2 4} = { } = B 5 B 5 B = { } { 2 4} nd so on. t is esily proved B 15 Z 17 ; this tsk is left s n exercise to the reder. Let S = { 4 5} S(Z 17 ) S S = { 4 5} { 4 5} = { } = S 2 S 2 S = { } { 4 5} = { } = S 3 S 3 S = { } { 4 5} = { } = S 4 S 4 S = { } { 4 5} = { } = S 5 nd so on. For this S lso we et S 15 universl subset of S(Z 17 ) = Z 17 tht S yield the MOD Let M = { 1 3} S(Z 17 ) M M = { 1 3} { 1 3} = { 1 9 3} = M 2 M 2 M = { 1 3 9} { 1 3} = { } = M 3 M 3 M = { } { 1 3} = { } = M 4 M 4 M = { } { 1 3} = { } = M 5 nd so on. Thus we see fter some steps sy 15 M 15 = Z 17 so M lso yield the MOD universl subset Z 17 of S(Z 17 ). Let T = { 1 3 5} S(Z 17 ) T T = { 1 3 5} { 1 3 5} = { } = T 2 T 2 T = { } { 1 3 5} = { } = T 3 T 3 T = { } { 1 3 5} = { } = T 4 T 4 T = { }

49 48 MOD Nturl Neutrosophic Subset Semiroups { 1 3 5} = { } = T 5. T 5 T = T 6 = Z 17. Thus T is MOD subset of S(Z 17 ) such tht T 6 = Z 17 the MOD universl subset of S(Z 17 ). Let V = { } S(Z 17 ) V V = { } { } = { } = V 2 V 2 V = { } { } = { } = V 3. V 3 V = { } = { } = { } = V 4 = Z 17. Thus V 4 S(Z 17 ). = Z 17 contributes the MOD universl subset of Let W = { } S(Z 17 ) W W = { } { } = { } = W 2. W 2 W = { } { } = { } = W 3. W 3 W = W 4 S(Z 17 ) = Z 17 ives the MOD universl subset of

50 MOD Subset Alebric Structures 49 Let A = { } nd B = { } S(Z 17 ) A B = { } { } = { } Z 17. So the product of these two sets A nd B does not yield the MOD universl subset of S(Z 17 ). Let P = { } nd Q = { } S(Z 17 ) P Q = { } { } = { } = Z 17. Thus this pir of sets P nd Q when their product P Q is tken ives the MOD universl subset Z 17 of S(Z 17 ). We cll the pir (P Q) s the MOD universl subset pir of S(Z 17 ). Let M = { } nd N = { } S(Z 17 ). M N = { } { } = { } Z 17. So we cnnot sy lwys pir of MOD subsets will contribute to the MOD universl subset of S(Z 17 ). We ive few more exmples before we put forth some results. Exmple 2.19: Let {S(Z 1 ) } be the MOD subset semiroup under. Let A = { 5 9 7} nd B = { } S(Z 14 )

51 5 MOD Nturl Neutrosophic Subset Semiroups A B = { 5 9 7} { } = { } Z 1. So {A B} is not MOD universl subset product. Let A = {1 4 6} nd B = {5 8 7} S(Z 1 ) A B = {1 4 6} {5 8 7} = { } Z 1. So this pir {A B} is not MOD universl subset pir of S(Z 1 ). Let S = { } nd P = {1 3 7} S(Z 1 ) S P = { } {1 3 7} = { } = Z 1. So the pir {S P} yields the MOD universl subset Z 1 of S(Z 1 ). However not ll MOD subsets pir cn yield the MOD universl subset Z 1 of S(Z 1 ). Consider A = { 5} nd B = { 2 4 6} S(Z 1 ) A B = { 5} { 2 4 6} = {}. Thus the MOD subset pir {A B} yields the MOD subset zero divisor of S(Z 1 ) Let A A = { 5} { 5} = { 5} so the MOD subset A of S(Z 1 ) is defined s the MOD subset idempotent of S(Z 1 ). Consider B B = { 2 4 6} { 2 4 6} = { } B. So it is esily observed tht in enerl ll elements of S(Z 1 ) re not MOD subset idempotents of S(Z 1 ). Let S = { } S(Z 1 ). S S = { } { }

52 MOD Subset Alebric Structures 51 = { } = S. Thus S in S(Z 1 ) is the MOD subset idempotent of S(Z 1 ). Similrly let S = { 3 9} S(Z 1 ); S S = { 3 9} { 3 9} = { 9 7 1} S. Now we define nd develop results relted with MOD subset universl pir in S(Z n ). DEFNTON 2.6: Let S(Z n ) be the MOD subset semiroup under product. A pir of MOD subsets {P Q} is MOD subset universl pir if P Q = Z n ; Z n P Q S(Z n ). P Z n nd Q Z n. We hve iven exmples of them. However the followin result is recorded. THEOREM 2.4: Let {S(Z n ) } be MOD subset semiroup under product; n prime. There exists MOD subset universl pirs {PQ} such tht P Q = Z n. Proof is left s n exercise to the reder. The followin problem is left open. Problem 2.5: Let {S(Z n ) } be the MOD subset semiroup; n composite number. Obtin conditions on n so tht S(Z n ) hs MOD subset universl pir P Q S(Z n ) \ Z n. DEFNTON 2.7: Let {S(Z n ) } be the MOD subset semiroup. A pir {P Q} P Q S(Z n ) \ {} is sid to be MOD zero divisor subset pir if P Q = {}.

53 52 MOD Nturl Neutrosophic Subset Semiroups A MOD subset P in S(Z n ) is sid to be MOD nilpotent subset if P m = {} for m > 1. A MOD subset P in S(Z n ) \ {{} { 1}} is sid to be MOD idempotent subset of S(Z n ) if P P = P. We hve iven exmples of them. We present only one more exmple before we proceed onto define S(C(Z n )). Exmple 2.2: Let {S(Z 24 ) } be the MOD subset semiroup under product. Let A = { 6 12} nd B = { } S(Z 24 ) A B = { 6 12} { } = {}. Thus {A B} is the MOD zero divisor pir of S(Z 24 ). Let M = { 12} S(Z 24 ) is such tht M M = {} so M is MOD subset nilpotent of order two in S(Z 24 ). Let N = { 6} S(Z 24 ); N N = { 6} { 6} = { 12} N 2 N 2 N = { 12} { 6} = {} = N 3. Thus N is MOD subset nilpotent of order three. Let S = { 9 1 6} S(Z 24 ) S S = { 9 16} { 9 16} = { 9 16} = S. Thus S in S(Z 24 ) is MOD subset idempotent of S(Z 24 ). Let T = { } nd R = { } (Z 24 ). R T = { } { } = {}. Thus {R T} is MOD subset zero divisor pir of S(Z 24 ). Let V = { 12} nd

54 MOD Subset Alebric Structures 53 W = { } S(Z 24 ). V W = { 12} { } = {}. Thus {V W} the MOD subset pir is MOD zero divisor. We hve shown exmples of MOD subset zero divisor pir MOD subset nilpotents nd MOD subset idempotents in S(Z 24 ). We now illustrte properties of MOD finite complex number subsets. Throuh out this book S(C(Z n ) denotes the collection of ll subsets of C(Z n ). Exmple 2.21: Let S(C(Z 3 )) = {{} {1} {2}{i F } {2i F } {1 + i F } {2 + i F } {1+2i F } {2+2i F } { 1} { 2} { i F } { 2i F } { 1 + i F } { 2 + 2i F } {1 i F } {1 i F + 2} {1 1 + i F } {1 2i F + 2} {2 i F } {1 2} {2 1 + i F } {2 2i F + 2} nd so on {C(Z 3 )}}. Let A = { 1 + i F 2 + i F 1} nd B = {1 2i F i F } S(C(Z 3 )). A + B = { 1 + i F 2 + i F 1} + {1 i F 2i F } = {1 2 + i F i F 1 + 2i F 2 + 2i F i F } is n element of S(C(Z 3 ). A = 4 tht is number of elements in A is 4. Number of elements in B is 3. But number of elements in A + B = 7. Thus is the wy + opertion is performd on S(C(Z 3 )). Exmple 2.22: Let S(C(Z 6 )) = {collection of ll subsets of the set C(Z 6 )}. Clerly C( Z 6 ) o(s(c(z 6 ))) = 2 1 where C(Z 6 )} denotes the number of elements in C(Z 6 ) S(C(Z 6 )) is defined s the MOD finite complex number subsets.

55 54 MOD Nturl Neutrosophic Subset Semiroups Let P = {3 + 2i F i F 5i F } nd Q = {2 3i F 1 + i F } S(C(Z 6 )). We find P + Q = {3 + 2i F i F 5i F + {2 3i F 1 + i F } = {5 + 2i F 4 + 5i F 2 + 5i F 3 + 5i F 4 + 3i F 2 + 2i F 2i F 5 + i F 3 1}. We see o(p) = 4 nd o(q) = 3 but o(p + Q) = 11 o(p) + o(q). Let B = { 3 3i F 2} nd C = {3i F 3 } S(C(Z 6 )). B + C = { 3 2 3i F } + { 3 3i F } = { 3 + 3i F i F }. Clerly o(b) = 4 o(c) = 3 but o(b+c) = 4 < o(b) + o(c). Thus we see in enerl o(b+c) is less thn or reter thn or equl o(b) + o(c). Let X = {2} nd Y = {4} X + Y = {2} + {4} = {}. o(x + Y) = 1 o(x) = 1 o(y) = 1. Thus o(a + B) = is n impossibility. Let T = { i F 5} nd R = {i F 2i F 3i F } S(C(Z 6 )). T + R = { i F 5} + {i F 2i F 3i F } = {i F 2i F 3i F 2 + i F 2 + 2i F 2 + 3i F 3 + i F 3 + 2i F 3 + 3i F 4 + 2i F 4 + 3i F 4 + 4i F 5 + i F 5 + 2i F 5 + 3i F }. (T + R) = 15 nd o(t) = 5 nd o(r) = 3. Clerly o(t + R) > o(t) + o(r). Next we proceed define ddition on S(C(Z n )).

56 MOD Subset Alebric Structures 55 DEFNTON 2.8: Let S(C(Z n )) be the MOD finite complex number subsets of C(Z n ). Define opertion + on S(C(Z n )) for ny A B S(C(Z n )) s follows. A + B = {every element of A is dded with every element of B}; A + B S(C(Z n )). Clerly A + B = B + A {} S(C(Z n )) is such tht A + {} = A for ll A S(C(Z n )). {S(C(Z n )) +} is defined s the MOD subset finite complex number semiroup under +. Exmple 2.23: Let {S(C(Z 1 )) +} be the MOD finite complex number subset semiroup. For A = { 5} nd B = {5} we et A + B = { 5} + {5} = { 5} = A. A + A = { 5} + { 5} = { 5}. So A is MOD finite complex modulo inteer subset idempotent of S((Z 1 )). We see if X S(C(Z 1 )) in enerl we cnnot find Y S(C(Z 1 )) such tht X + Y = {}. For tke X = { i F 9 + i F }; we hve no Y S(C(Z 1 )) such tht X + Y = {}. Let M = {5 + 2i F } nd N = {1 + 8i F } S(C(Z 1 )); M + N = {5 + 2i F } + {1 + 8i F } = {6}. Exmple 2.24: Let {S(C(Z 2 )) +} be the MOD finite complex number semiroup. Let X = { 1 + i F i F } nd Y = { 1} X + Y = { i F 1 + i F 1} = C(Z 2 ).

57 56 MOD Nturl Neutrosophic Subset Semiroups Thus X with Y yields the MOD universl finite complex number subset of S(C(Z 2 )). Exmple 2.25: Let {S(C(Z 4 )) +} be the MOD universl finite complex number subset semiroup under +. Let X = { i F 2 + i F 3 + i F i F 2 + 2i F 3 + 2i F } nd Y = { 1 + 3i F 2 + 3i F 3 + 3i F i F 2i F 3i F } S(C(Z 4 )). Clerly X + Y = C(Z 4 ). Thus X + Y yields the MOD finite complex number universl subset of S(C(Z 4 ). We ive the followin result. THEOREM 2.5: Let {S(C(Z n )) +} be the MOD finite complex number subset semiroup under +. i) There exists pir {P Q} such tht P + Q = C(Z n ). ii) f P + Q = C(Z n ) then P Q = {}. Proof is direct nd hence left s n exerise to the reder. We suest the followin problem. Problem 2.6: Let S(C(Z n )) be the MOD finite complex number subset semiroup under +. i) s it mndtory if {P Q} is MOD universl subset pir then P Q = {}? ii) Cn we find the number of such pirs? iii) s it possible for {P Q} nd {P R} to be MOD universl subset pir where R Q Q or R? We will ive some exmples where we find sum of P + P; P S(C(Z n )).

58 MOD Subset Alebric Structures 57 Exmple 2.26: Let {S(C(Z 13 ) +} be the MOD subset finite complex number semiroup under +. Let P = { i F 5} S(C(Z 12 )). We find P + P = { i F 5} + { i F 5} = { i F i F i F i F 8 + i F 1} = P + P. We see P + P P nd o(p) = 5 nd o(2p) = 14. We find P + P + P = { i F 4 + 4i F 5 + i F 6 + i F 8 + i F i F 9 + i F i F 1 + i F 6 + 2i F 7 + 2i F 8 + 2i F 9 + 2i F i F 11 + i F 2 + i F } = P + P + P; o(p + P + P) = 28 nd so on. The reder is left with the tsk of findin m such tht P P P = C(Z 12). mtimes So findin P P P = C(Z n) for m > 1 nd P mtimes S(C(Z n )) hppens to be chllenin nd difficult problem. However one is sure there re P S(C(Z n )) such tht P P P = C(Z n); but one is not in position to know the mtimes number of such MOD finite complex number subsets P in S(C(Z n )) whose sum result in the MOD finite complex number universl subset C(Z n ). Problem 2.7: Let {S(C(Z n ) +} be the MOD subset finite complex number semiroup. i) Find ll P S(C(Z n ) such tht P + P + + P m times = C(Z n ); m > 1. ii) Find ll MOD finite complex number subset universl pirs; {P Q} such tht P + Q = C(Z n ); P Q S(C(Z n ) \ C(Z n ).

59 58 MOD Nturl Neutrosophic Subset Semiroups One cn find MOD subset subsemiroups MOD subset idempotents. All these work is considered s mtter of routine so left s n exercise to the reder. Next we proceed onto describe the MOD subset neutrosophic set; S(Z n ) = {collection of ll subsets of Z n }. Exmple 2.27: Let S(Z 5 ) = {(collection of ll subsets of Z 5 } be the MOD subset neutrosophic set. Let A = { } nd B = { } be elements of S(Z 5 ). We cn define the opertion + on S(Z 5 ). Let P = { } nd Q = { } S(Z 5 ). P + Q = { } + { } = { } is in S(Z 5 ). Clerly o(p) = 4 nd o(q) = 4 o(p + Q) = 15. Let us now find P + P = { } + { } = { } S(Z 5 ). The followin observtion is mndtory. o(p) = 4 nd o(p + P) = 1. Clerly P + P 2P; for 2P = 2{( )} = {2 3 2(2 + 1) 2(4 + 3) 2 2}

60 MOD Subset Alebric Structures 59 = { } P + P. Tht is why we do not write P + P + P = 3P or more enerlly P P P mp. mtimes However mp is differently obtined. Tht is why we cnnot count s in cse of usul lebric structures. Let L = { } S(Z 5 ). L + L = { } + { } = { } 2L. Further L L + L. nfct L L + L if nd only if L. Let W = { } be in S(Z 5 ). W + W = { } + { } = { }. Clerly W W + W s subset. Hence in enerl if V S(Z 5 ) then V V + V. Let R = { } nd S = { } S(Z 5 ). We find R + S = { } + { } = { }. However R R + S S R + S but R S.

61 6 MOD Nturl Neutrosophic Subset Semiroups There re severl of the properties enjoyed by the MOD subset neutrosophic set under + which is not property of usul modulo inteers or rels or complex or neutrosophic numbers but true in cse of MOD subsets of S(Z n )) nd MOD finite complex number subsets of S(C(Z n )). Exmple 2.28: Let S(Z 12 ) be the collection of ll MOD subsets of neutrosophic inteers. Let M = { } S(Z 12 ). We find M + M = { } + { } = { }. o(m) = 5 but order M + M = 15. Let V = {6 6} nd W = {6} W + V = {6 + 6}. Let X = {6 + 6} nd Y = {6 6} X + Y = {6 6}. Clerly {6 + 6} = W S(Z 5 ) is such tht W + W = {6 + 6} + {6 + 6} = {}. But + V = V for ll V S (Z 12 ). Thus {S(Z 12 ) +} is MOD neutrosophic subset monoid which is commuttive nd is of finite order. We see there re elements in S(Z 12 ) such tht the pir {P Q} forms the MOD neutrosophic universl subset pir. Tht is P + Q = Z 12. Also there re MOD neutrosophic subsets P S (Z n ) such tht P P P = Z n m > 1. mtimes

62 MOD Subset Alebric Structures 61 All these study is considered s mtter of routine so is left s n exercise to the reder. We lso see severl properties derived in cse of S(Z n ) nd S(C(Z n )) cn lso be derived in cse of S(Z n ). Next we proceed onto describe by exmples the opertion of product on the MOD subsets of S(C(Z n )) nd S(Z n ). Exmple 2.29: Let S(C(Z 8 )) be the collection of ll MOD finite complex number subsets of C(Z 8 ). Let A = { i F 2 6} nd {5i F 2 + 4i F 7} = B S(C(Z 8 )). We find A B = { i} {7 5i F 2 + 4i F } = { i F 6i F 2i F 7i F 4i F i F 4i F } S(C(Z 8 )). o(a) = 4 o(b) = 3 o(a B) = 11. A A = { i F } { i F } = { i F 4i F 4 + 2i F } = o(a) = 4 but o(a A) = 7. This is the wy product opertion is performed on S(C(Z 8 )). Let M = {4i F + 4 2i F 6i F + 4} nd N = {4 + 4i F 4 4i F } S(C(Z 8 )). M N = {2i F 4+4i F 6i F +4} {4 + 4i F 4 4i F } = {}. Thus the MOD finite complex number subset pir {M N} yields the MOD zero divisor subsets. Let B = { i F 4i F } S(C(Z 8 )); B B = {4 4i F 4 + 4i F } {4 4i F 4 + 4i F } = {}.

63 62 MOD Nturl Neutrosophic Subset Semiroups Thus B is the MOD finite complex number nilpotent subset of S(C(Z 8 )). nfct it is left s n exercise for the reder to verify tht S(C(Z 8 )) is finite commuttive monoid. {1} in S(C(Z 8 )) is such tht A {1} = A for ll A S(C(Z 8 )) serves s the MOD subset identity of S(C(Z 8 )) under product. Let D = {2 2i F 4 + 4i F 4} S(C(Z 8 )). D D = {2 2i F 4 + 4i F 4} {2 2i F 4 + 4i F 4} = {4 4i F }; D D D = {4 4i F } {2 2i F 4 + 4i F + 4} = {}. Thus D is the MOD finite complex number nilpotent subset of S(C(Z 8 )). However we see S(C(Z 8 )) cnnot hve idempotents. Thus S(C(Z 8 )) hs MOD finite complex number zero divisor subset pir MOD finite complex number nilpotent subsets. However findin MOD finite complex number universl subset pir in S(C(Z 8 )) hppens to be chllenin one s C(Z 8 ). i = 7 in The reder is dvised to work in this direction. Further find P in S(C(Z 8 )) so tht P m = C(Z 8 ); m > 1. Exmple 2.3: Let {S(C(Z 3 )) } be the MOD finite complex number subset semiroup under product where; C(Z 3 ) = { 1 2 i F 2i F 1 + i F 1 + 2i F 2 + i F 2 + 2i F }. Tke A = { 1 2 i F 2 + i F } nd B = {1 1 + i F i F 1 + 2i F } S(C(Z 8 )). A B = { 1 2 i F 2 + i F } {1 i F 1 + i F 1 + 2i F } = { 1 2 i F 2 + i F 1 + i F 1 + 2i F 2 + 2i F 2i F } 2 F

64 MOD Subset Alebric Structures 63 = S(C(Z 3 )). Thus {A B} is the MOD finite complex number subset universl pir of S(C(Z 3 )). Tke A = { i F i F } S(C(Z 3 )). A A = { 1 i F 1 + i F } { 1 i F 1 + i F } = { 1 i F 1 + i F 2 i F + 2} = A 2 A 2 A = { 1 2 i F 1 + i F 2 + i F } { 1 i F 1 + i F } = { 1 2 i F 1 + i F 2 + i F 2i F 2 + i F 2 + 2i F 1+2i F } = C(Z 3 ). Thus A S(C(Z 3 )) is such tht A 3 = C(Z 3 ). Thus A is MOD finite complex subset which ives the MOD universl subset for m = 3. Exmple 2.31: Let {S(C(Z 6 )) } be the MOD finite complex number subset semiroup under product. Let A = { i F 3i F } nd B = { i F 4 + 2i F 2 + 4i F 4i F } S(C(Z 6 )). A B = { i F 3i F } { 2 4i F 2 + 2i F 4 + 2i F 2 + 4i F } = {}. So the pir {A B} is the MOD finite complex number subset zero divisor of S(C(Z 6 )). Further A A = { i F 3i F } { i F 3i F } = { i F 3i F } = A. Thus A S(C(Z 6 )) is such tht A A = A so A is the MOD finite complex number subset idempotent of S(C(Z 6 )). Let D = { i F i F 4i F 2i F 2 + 2i F 4 + 4i F } S(C(Z 6 )).

65 64 MOD Nturl Neutrosophic Subset Semiroups D D = { i F 4 + 2i F 4i F 2 2i F 2 + 2i F 4 + 4i F } { 4 2 4i F 2 + 2i F 4 + 4i F 2 + 4i F 4 + 2i F 2i F } = D so D is in MOD finite complex number of S(C(Z 6 )). Severl properties in this direction cn be obtined nd it is considered s mtter of routine nd is left s n exercise to the reder. Next few exmples of MOD neutrosophic subset semiroup under is defined. Exmple 2.32: Let {S(CZ 1 ) } be the MOD neutrosophic subset semiroup under product. Let A = { } nd B = { } S (Z 1 ). A B = { } { } = { }. This is the wy product is performed on S(CZ 1 ). Let A = { } nd B = { } S(CZ 1 ). Clerly A B = {}. A A = { } { } = { } = A. Thus A is MOD neutrosophic subset idempotent of S(CZ 1 ). Let B = {6 6} S(CZ 1 ) is such tht B B = {6 6} = B is in MOD neutrosophic subset idempotent.

66 MOD Subset Alebric Structures 65 Let { } = A nd B = { 6 2} S(Z 1 ) A B = { } { 6 2} = { 2 4 8} S(Z 1 ). Let M = { } nd N = {5}. M N = {}. Thus the {M N} MOD neutrosophic subset pir is zero divisor subset of S(Z 1 ). Let W = { } S(Z 1 ) W W = { } { } = { } = W. This is lso MOD neutrosophic idempotent subset of S(Z 1 ). Let V = { } S(Z 1 ); V V = { } { } = { } = V. This is lso MOD neutrosophic idempotent subset of S(Z 1 ). Let S = {collection of ll subsets from the set { }} S(Z 1 ). Clerly S is MOD neutrosophic subset subsemiroup of the MOD neutrosophic subset semiroup S(Z 1 ). B = {collection of ll subsets from the set { }} be the MOD neutrosophic subset subsemiroup of the MOD neutrosophic subset semiroup S(Z 1 ).

67 66 MOD Nturl Neutrosophic Subset Semiroups We see B is n MOD neutrosophic subset idel of S(Z 1 ) but S is not MOD neutrosophic subset idel of S(Z 1 ). Consider F = {collection of ll subsets from the set Z 1 }. F is only MOD neutrosophic subset subsemiroup nd not n idel of S(Z 1 ). Let P = {collection of ll subsets from the set Z 1 }; P is MOD neutrosophic subset subsemiroup nd lso MOD neutrosophic subset idel of S(Z 1 ). n view of ll these we hve the followin theorem. THEOREM 2.6: Let {SZ n ) } be the MOD neutrosophic subset semiroup under. i) {S(Z n ) } hs MOD neutrosophic subsemiroup which re not idels. ii) {SZ n } hs MOD neutrosophic subsemiroup which is lso idel. Proof is direct nd hence left s n exercise to the reder. Now consider the MOD dul number subset semiroup. S(Z n ) = {collection of ll subsets from the MOD dul number subset}; we will describe this by n exmple. Exmple 2.33: Let S(Z 7 ) = {collection of ll MOD subset dul numbers}. Let A = { } nd B = { } S(Z 7 ). A B = { } { } = { }.

68 MOD Subset Alebric Structures 67 M = { } S(Z 7 ). M M = { } { } = {}. Thus M is MOD dul number subset nilpotent of order two. Let B = { } S(Z 7 ). B B = { } { } = { } {}. Clerly B = {collection of ll MOD dul number subset from Z 7 } S(Z 7 ) is MOD dul number subsemiroup of B which is lso n idel. C = {collection of ll MOD subset from Z 7 } S(Z 7 ) is MOD subset subsemiroup which is not n idel. So findin idels other thn B hppens to be very chllenin one. Exmple 2.34: Let {S(Z 12 )} be the MOD dul number subset semiroup. We cn s in cse of S(Z n ) S(C(Z n )) S(Z n ) define multipliction. Let B = { } nd C = { } S(Z 12 ) B C = { } { } = { }. We now show how the opertion of ddition is performed on S(Z 12 ).

69 68 MOD Nturl Neutrosophic Subset Semiroups B + C = { } + { } = { }. We see o(b) = 6 o(c) = 3 o(b C) = 8 wheres o(b + C) = 14. B + B = { } + { } = { }. o(b) = 5 but o(b + B) = 15. B B = { } { } = { }. o(b B) = 1. Consider D = { } nd E = { } S(Z 12 ). D E = { } { } = {}. o(d) = 6 o(e) = 4 o(d E) = 1. D + E = { } + { } = { } o(d) = 6 o(e) = 4 but o(d + E) = 11. Let M = { } S(Z 12 ). M + M = { } + { } = { } = M. M M = { } { } = { 4 8} M.

70 MOD Subset Alebric Structures 69 Thus M under + is MOD dul number idempotent subset wheres M under is not n MOD dul number idempotent subset. Let G = { 6 6} S(Z 12 ) G G = { 6 6} { 6 6} = {}. Thus G is MOD dul number nilpotent subset of S(Z 12 ) of order two. nfct S(Z 12 ) hs severl MOD subset dul number nilpotents of order two. Findin properties bout MOD dul number subset semiroups under + nd is left s n exercise s it is mtter of routine. Next we proceed to describe MOD specil dul like number subsets by some exmples. Exmple 2.35: Let S(Z 15 h) be the MOD specil dul like number subset. Let A = { h h 2} nd B = {3 1 12h} S(Z 15 h). We cn define the opertion of ddition on A. A A = { h h 2 } { h h 2} = {1 5h 5 + 9h 8h 1 + 6h h 2h 4}. o(a) = 4 o(a A) = 7. A + A = {2 h h} + {2 h h} = {4 2 + h h 2h 5 + h 5 + 4h h 1 = 6h}. o(a + A) = 1.

71 7 MOD Nturl Neutrosophic Subset Semiroups Let B = {3h h 4h 8h} nd D = {6h 7h 5h 1h} S(Z 15 h). B D = {h 3h 4h 8h} {6h 7h 5h 1h} = {6h 3h 9h 7h 13h 11h 5h 1h}. o(b) = 4 o(d) = 4 o(b D) = 9. o(b + D) = {h 3h 4h 8h} + {6h 7h 5h 1h} = {7h 9h 1h 14h 8h 11h 6h 13h 3h}. o(b + D) = 1. This is the wy opertions of + nd re defined on S(Z 15 h). Let M = { 5h 1h} S(Z 15 h). M + M = { 5h 1h} + { 5h 1h} = { 5h 1h} = M. Thus the MOD specil dul like number is n idempotent subset under +. M M = { 5h 1h} { 5h 1h} = { 1h 5h} = M. Thus M is in MOD specil dul like number idempotent subset under. Clerly M + M 2M but M + M = M. P = { 3h 6h 9h 12h} S(Z 13 h). P + P = { 3h 6h h 12h} + { 3h 6h 9h 12h} = { 9h 3h 12h 6h} = P. Clerly P is in MOD specil dul like number idempotent subset under +.

72 MOD Subset Alebric Structures 71 P P = { 3h 6h 9h 12h} { 9h 3h 6h 12h} = P. Ain P is MOD specil dul like number subset idempotent under. Exmple 2.36: Let {S(Z 11 h) +} be the MOD specil dul like number subset semiroup under +. Let A = {3h 2 + h 6h } nd B = {8h 9+1h 5h 2} S(Z 11 h). We find A + B = {3h 2 + h 6h } + {2 5h 1h + 9 8h} = {3h h 2 + 6h 2 8h 2 + 6h 5h 2h + 9 4h + 9 1h + 9}. This is the wy + opertion is performed on S(Z 11 h). Let A = {3h + 7} nd B = {8h + 4} in S(Z 11 h) is such tht A + B = {3h + 7} + {8h + 4} = {}. Let B = { h 2h 3h 4h 5h 6h 7h 8h 9h 1h} S(Z 11 h). Clerly B + B = B; so B is MOD subset specil dul like number idempotent of S(Z 11 h). nfct B is MOD subset subsemiroup under + of order one. Next we proceed on to describe the opertion of product on MOD subset specil dul like number set S(Z n h). Exmple 2.37: Let {S(Z 2 h) } be the MOD specil dul like number subset semiroup under product.

73 72 MOD Nturl Neutrosophic Subset Semiroups Let A = {5 1h 15h 1 + 5h 5h h +1} nd B = {4 4h + 8 8h 8h h 12h h} S(Z 2 h). A B = {}. Thus the MOD specil dul like number pir subset is MOD zero divisor subset pir. Consider M = {5h 5 16h 16} S(Z 2 h). M M = { 5 5h 16h 16} { 5 5h 16 16h} = { 5 5h 16 16h} = M. Hence M is the MOD specil dul like number idempotent subset of S(Z 2 ). Let L = {1 1h 1 + 1h } S(Z 2 h). Clerly L L = { 1 1h 1 + 1h} {1 1h 1 + 1h} = {} so L is the MOD specil dul like number nilpotent subset of order two in S(Z 2 h). Let B = {Collection of ll subsets from the Z 2 } S(Z 2 h) B is MOD specil dul like number subset subsemiroup of S(Z 2 h) nd is not n idel. Consider C = {Collection of ll subsets from the set Z 2 h} S(Z 2 h). We see C is MOD specil dul like number subset subsemiroup which is lso n idel of S(Z 2 h). Findin MOD subset idels. MOD subset subsemiroups hppens to be mtter of routine nd hence left s n exercise to the reder.

74 MOD Subset Alebric Structures 73 Next we briefly describe the MOD subset specil qusi dul number set by some exmples. Exmple 2.38: Let S(Z 12 k) = {collection of ll MOD subsets from the set Z 12 k} be the MOD subset specil qusi dul number set. Define ddition on S(Z 2 k) s follows: Let A = {1k k 5 + 2k 9} nd B = {2k 5k 2 + 8k 3} S(Z 12 k). A + B = {1k k 5 + 2k 9} + {2k 5k 2 + 8k 3} = { 3k 2k 5 + 4k 6k 5k 5 + 7k 2 + 6k 2 + 9k 2 + 8k 7 +1k 3 + 1k 3 + k k 9 + 2k 9 + 5k k}. o(a) = 5 o(b) = 4 nd o(a + B) = 18. A + A = {1k k 5 + 2k 9} + { 1k k 5 + 2k 9} = {1k k 5 + 2k 9 8k 11k k 2k 3k k 5 + 3k 1 + 4k 2 + 2k 6}. o(a) = 5 but o(a + A) = 16. Clerly A + A 2A. Let A = { } S(Z 12 k). A + A = { } + { } = { } = A. Thus A is MOD specil qusi dul number idempotent subset of S(Z 12 k under +. Let B = { 3k 6k 9k} S{Z 12 k). B + B = { 3k 6k 9k} + { 3k 6k 9k} = B. Thus B is in MOD specil qusi dul number idempotent subset of S(Z 12 k).

75 74 MOD Nturl Neutrosophic Subset Semiroups We see S(Z 12 k) hs MOD subsets such tht A + A = A. We see if A = { k 4k 8k 1k + 1 4k + 4k 8 + 8k} nd B = {6 6k 6 + 6k} S(Z 12 k) then A + B = { k 4k 8k 1 + 1k 4 + 4k 8 + 8k} + {6 6k 6 + 6k} = { k 6 + 4k 6 + 8k 4 + 1k 1 + 4k 2 + 8k 4 + 6k 4 + 2k 1k 2k 1 + 4l k 8 + 2k 1 + 6k 1 + 2k 6 + 1k 6 + 2k 4 + 4k 1 + 1k 2 + 2k}. o(a) = 7 o(b) = 3 but o(a + B) = 21. B + B = {6 6k 6 + 6k} + {6 6k 6 + 6k} = { 6 + 6k 6k 6} B. f T = B {} = {6 6k 6 + 6k} then T + T = T so T is in MOD specil qusi dul number subset idempotent under + of S(Z 12 k). Thus {SZ 12 k) +} is only MOD specil qusi dul number subset semiroup of finite order infct finite commuttive monoid. Exmple 2.39: Let S(Z 6 k) = {collection of ll MOD specil qusi dul number subsets} be the MOD specil qusi dul number subset. Let A = {3 + 3k 3k 3 }. We find A A = {3 + 3k 3k 3 } {3 + 3k 3k 3 } = { 3 3k 3 + 3k} = A.

76 MOD Subset Alebric Structures 75 Thus we et A to be MOD specil qusi dul number idempotent subset of S(Z 6 k). Let T = {2 2k 4k 2 + 4k 2k k } nd S = {3 3k 3 + 3k } S (Z 6 k}. T S = {2 2k 4k 2 + 4k 2k k } {3 3k 3 + 3k } = {}. Thus we see the MOD specil qusi dul number subset pir of zero divisors of S(Z 6 k). However we do not find ny MOD specil qusi dul number subsets which re nilpotents in S(Z 6 k). So the followin problem is thrown open. Problem 2.8: Find condition on {SZ n k } the collection of ll MOD specil qusi dul number subsets semiroup under product to contin MOD specil qusi dul number nilpotents of order m m 2. We suest few problems for the reder. Problems: 1. Let S(Z 11 ) be the MOD subset semiroup under +. i) Find ll subset in S(Z 11 ) such tht their sum with X = { } ives Z 11 the MOD universl subset of S(Z 11 ). ii) Let Y = { } S(Z 11 ); does there exists P i subsets in S(Z 11 ) so tht P i + Y = Z 11? iii) Let X = { } S(Z 11 ) find ll Y in S(Z 11 ) so tht X + Y = Z 11.

77 76 MOD Nturl Neutrosophic Subset Semiroups iv) Let B = { } S(Z 11 ) does there exist sets C in S(Z 11 ) so tht B + C = Z Let {S(Z 16 ) +} be the MOD subset semiroup. i) Let X = { } S(Z 16 ) find ll subsets P of S(Z 16 ) so tht X + P = Z 16. ii) Find ll R S (Z 16 ) such tht R + R = Z 16 ; R Z 16 or R {}. iii) Find ll MOD subsemiroups of S(Z 16 ). iv) Find ll MOD subsemiroups of S(Z 16 ) which re MOD subroups. v) Find ll T S (Z 16 ) such tht T + T = T. vi) Cn we hve Y S (Z 16 ) such tht Y + Y = {}? 3. Let {S(Z 24 ) +} be the MOD subset semiroup under +. i) Find ll MOD subsets P in S(Z 24 ) so tht P + P = {}. ii) Find ll MOD subsets R in S(Z 24 ) so tht R + R = R. iii) Find ll MOD subsets T in S(Z 24 ) so tht T T T = mtimes Z 24 ; m 2. iv) Find ll MOD subsets subsemiroups which re not MOD subroups. v) Find ll MOD subsets subsemiroups which re MOD subroups. vi) Obtin ny other specil strikin fetures ssocited with S(Z 24 ). 4. Let {S(C(Z 1 )) +} be the MOD finite complex subset semiroup under +. i) Study questions (i) to (vi) of problem 3 for this S(C(Z 1 )).

78 MOD Subset Alebric Structures 77 ii) Obtin ny other specil feture ssocited with S(C(Z 1 )). 5. Let {S(C(Z 19 )) +} be the MOD finite complex subset semiroup under + i) Study questions (i) to (vi) of problem 3 for this S(C(Z 19 )). ii) Compre S(C(Z 19 )) in problem (4) with S(C(Z 19 )). 6. Let S(C(Z 24 )) be the MOD finite complex number subset semiroup under +. i) Study questions (i) to (vi) of problem (3) for this S(C(Z 24 )). ii) Compre S(C(Z 24 )) with S(C(Z 1 )) in problem 4 nd S((C(Z 19 )) in problem Let {S(Z 12 ) +} be the MOD neutrosophic subset semiroup under +. i) Study questions (i) to (vi) of problem (3) for this S(Z 12 ). ii) Compre the properties of S(Z 12 ) nd S(C(Z 12 )) with S(Z 12 ). 8. Let {S(Z 29 ) +} be the MOD neutrosophic subset semiroup. i) Study questions (i) to (vi) of problem (3) for this S(Z 29 ). ii) Compre S(Z 29 ) with S(Z 29 ) nd S(C(Z 29 ) s semiroups under +. iii) Let P 1 = {S(Z 24 ) +} nd P 2 = {S(Z 64 ) +} be MOD neutrosophic subsets with + opertion.

79 78 MOD Nturl Neutrosophic Subset Semiroups Compre P 1 nd P 2 with S{(Z 29 ) +}. 9. Let {S(Z 18 ) +} be the MOD subset dul number semiroup under +. i) Study questions (i) to (vi) of problem (3) for this S(Z 18 ). ii) Discuss the specil fetures ssocited with S(Z 18 ). iii) Compre {S(Z 18 ) +} with {S(Z 18 ) +} = P 1 P 2 = {S(C(Z 18 )) +} nd P 3 = {S(Z 18 ) +}. 1. Let {S( Z 1 ) +} be the MOD subset dul number 3 semiroup under +. i) Compre {S( Z 1 ) +} with {S( Z 1 ) +} 3 3 {S( Z 1 ) +} {S(C(Z 3 48 ) +} nd {S(Z 48 ) +}. ii) Study questions (i) to (vi) of problem (3) for this {S( Z 1 ) +} Let M = {S(Z 31 h) +} be the MOD specil dul like number subset semiroup under +. i) Study questions (i) to (vi) of problem (3) for this M. ii) Compre M with N = {S(Z 48 h) +}. iii) Compre M with P 1 = {S(Z 1 ) +} P 2 = {S(C(z 47 )) +} P 3 = {S(Z 48 h) +} P 4 = {S(Z 12 ) +} nd P 5 = {S(Z 6 ) +}. 12. Obtin ll specil nd distinct fetures ssocited with MOD specil dul like number subset semiroup {S(Z n h) +}.

80 MOD Subset Alebric Structures Let W = {S(Z 15 k +} be the MOD specil qusi dul number subset semiroup under +. i) Study questions (i) to (vi) of problem (3) for this W. ii) Compre this W with P 1 = {S(Z 15 ) +} P 2 = {S(C(Z 15 )) +} {P 3 = {S(Z 15 ) +} P 4 = {S(Z 15 h) +} nd P 5 = {S (Z 15 ) +}. iii) Compre W with B 1 = {S(Z 12 k) +} B 2 = {S(Z 11 k) +} B 3 = {S(Z 48 k) +} nd B 4 = {S(Z 243 k) +}. 14. Determine ny other specil feture ssocited with {S(Z n k) +} the MOD specil qusi dul number subset semiroup. 15. Let V = {S(Z 18 ) +} be the MOD subset semiroup under ddition. Chrcterize ll MOD subsets of V which contribute to MOD universl set Z Study problem (15) for W = {S(C(Z 42 )) +} nd M = {S(Z 4 ) +}. 17. Let W = {S(Z 2 ) } be the MOD subset semiroup under product. i) Find ll MOD subsets P of W such tht P Q = Z 2 the MOD universl subset of S(Z 2 ). ii) Find ll MOD subsets P Q such tht P Q = {}. iii) Cn S(Z 2 ) hve MOD subset nilpotents? iv) Chrcterize ll MOD zero divisor subset pirs of S(Z 2 ). v) Find ll MOD subset idempotents of S(Z 2 ). vi) Find ll MOD subset subsemiroups which re MOD subset subroups of S(Z 2 ).

81 8 MOD Nturl Neutrosophic Subset Semiroups vii) Find ll MOD subset subsemiroups which re not MOD subset subroups of S(Z 2 ). viii) Find ll MOD subset idels of S(Z 2 ). ix) Cn MOD subset idels be MOD subset subroup? 18. Let S = {S(Z 24 ) } be the MOD subset semiroup under. Study questions (i) to (ix) of problem (17) for this S(Z 48 ). 19. Let T = {S(Z 37 ) } be the MOD subset semiroup under. i) Study questions (i) to (ix) of problem (17) for this S(Z 37 ). ii) Compre S(Z 37 ) with S(Z 48 ) nd S(Z 2 ) in problems 18 nd 17 respectively. 2. Let Z = {S(Z 64 ) } be the MOD subset semiroup. i) Study questions (i) to (ix) of problem (17) for this S(Z 64 ). ii) Compre S( 64 ) with S(Z 2 ) of problem 18 S(Z 48 ) of problem 17 nd S(Z 37 ) of problem Let {S(C(Z 1 )) } = V be MOD finite complex number subset semiroup under. i) Study questions (i) to (ix) of problem (17) for this V. ii) Compre V with P = {S(Z 1 ) }. iii) Compre V with S = {S(Z 11 ) }. iv) Compre V with M = {S(Z 81 ) }. 22. Let W = {S(C(Z 43 )) } be the MOD finite complex number subset semiroup under. i) Study questions (i) to (ix) of problem (17) for this W. ii) Compre W with V of problem (2).

82 MOD Subset Alebric Structures 81 iii) Find those P W such tht there is Q with P Q C(Z 43 ) the MOD universl subset of S(C(Z 43 )). 23. Let Z = {S(Z 48 ) } be the MOD neutrosophic subset semiroup under product. i) Study questions (i) to (ix) of problem (16) for this Z. ii) Compre Z with {S(C(Z 48 )) } nd {S(Z 48 ) }. iii) Enumerte ll specil fetures enjoyed by Z. 24. Let M = {S(Z 19 ) be the MOD neutrosophic subset semiroup under product. Study question (i) to (ix) of problem (17) for this M. 25. Let P = {S(Z 12 ) } be the MOD dul number subset semiroup under product. i) Study questions (i) to (ix) of problem 17 for this P. ii) Prove P hs more number of MOD subset zero divisors thn S 1 = {S(Z 12 ) } S 2 = {S(C(Z 12 )) } S 3 = {S(Z 12 ) }. iii) Prove P hs more MOD subset idels thn S 1 S 2 nd S 3! 26. Let B = {S(Z 23 ) } be the MOD dul number subset semiroup under. i) Study questions (i) to (ix) of problem (17) for this B. ii) Obtin ll specil fetures ssocited with MOD dul number subset semiroups under. iii) Compre P in problem (24) with this B.

83 82 MOD Nturl Neutrosophic Subset Semiroups 27. Let E = {S( Z 8 ) } be the MOD dul number subset 2 semiroup under. i) Study questions (i) to (ix) of problem (17) for this E. ii) Compre E with P nd B of problem (24) nd (25) respectively. ii) All {SZ n ) } hs severl MOD nilpotent subsets of order two irrespective of n 2 n <. 28. Let W = {S(Z 15 h) } be the MOD specil dul like number subset semiroup under. i) Study questions (i) to (ix) of problem (17) for this W. ii) Compre W withv 1 = {S(Z 15 ) } V 2 = {S(C(Z 15 )) } V 3 = {S(Z 15 ) } nd V 4 = {S(Z 15 ) }. iii) Which of the MOD subset semiroups V 1 or V 2 or V 3 or V 4 or V 5 hs the mximum number of MOD subset zero divisor pirs? iv) Which of the MOD subset semiroups mentioned in (iii) hs mximum number MOD subset nilpotents of order two? v) Which of the MOD subsets semiroups mentioned in (iii) hs mximum number of MOD subset idempotents? vi) Which of the MOD subset semiroups mentioned in (iii) hs mximum number of MOD subset idels? 29. Let R = {S(Z 29 h) } be the MOD subset semiroup under product. i) Study questions (i) to (ix) of problem (17) for this R. ii) Cn R hve MOD subset nilpotents of order two? iii) Compre R with T 1 = {S(Z 29 ) } T 2 = {S(C(Z 29 )) } T 3 = {S(Z 29 ) } nd T 4 = {S(Z 29 ) }. iv) Obtin ny other specil feture ssocited with R.

84 MOD Subset Alebric Structures Let Y = {S(Z 42 k) } be the MOD subset specil qusi dul number semiroup under. i) Study questions (i) to (ix) of problem (17) for this Y. ii) Find o(y). iii) Compre Y with D 1 = {S(Z 42 ) } D 2 = {S(Z 42 ) } D 3 = {S(C(Z 42 )) } D 4 = {S(Z 42 ) } nd D 5 = {S(Z 42 ) }. iv) Which of the MOD subset semiroups iven in (iii) will hve mximum number of MOD subset idempotents? 31. Let Z = {S(Z 19 k) } be the MOD subset specil qusi dul number semiroup. i) Study questions (i) to (ix) of problem (16) for this Z. ii) Compre Z with R in problem Enumerte ll specil fetures in S(Z n k) i) n is prime ii) n is composite number iii) n is of the form p t t 2 p prime. 33. Prove MOD subsets semiroups re of finite order nd re commuttive. 34. Cn one sy ll MOD subsets semiroups under re S- MOD subset semiroups? 35. Cn these MOD subset semiroups under product hve i) S-MOD subset zero divisor pirs. ii) S MOD subset idempotents. 36. Let W = {S(Z 612 ) } be the MOD subset semiroup.

85 84 MOD Nturl Neutrosophic Subset Semiroups i) Does W contin S-MOD subset zero divisors? ii) Cn W hve MOD S-subset idempotents. 37. Chrcterize those MOD subset semiroups under which hs MOD subset S-zero divisors. 38. Chrcterize those MOD subset semiroups under which hs no MOD subset S-zero divisors. 39. Chrcterize those MOD subset semiroups under which hs MOD subset S-idempotents. 4. Study for Smrndche MOD zero divisors subsets nd MOD S-idempotents subsets in {SZ 48 k) }.

86 Chpter Three MOD NATURAL NEUTROSOPHC SUBSET SEMGROUPS n this chpter we define the new notion of MOD nturl neutrosophic elements subset semiroups under + nd. This study is very new however MOD nturl neutrosophic numbers hve been introduced in [6] nd lso complete study of semiroups constructed usin MOD nturl neutrosophic numbers hve been crried out in [64]. Here we study semiroups constructed usin subsets of MOD nturl neutrosophic numbers usin Zn or Z n or Z n or Z n h or Z n k or C (Z n ). As continuin the nottion S( Z n ) = {collection of ll subsets from the set Z n } which will lso be known s MOD nturl neutrosophic subset set. We will describe these by the followin exmples.

87 86 MOD Nturl Neutrosophic Subset Semiroups Exmple 3.1: Let S( Z ) = {Collection of ll subsets from the set 4 2 Z = { A = { neutrosophic subset of S( Z ) }}. ) S( Z ) is MOD nturl B = { 1 2 3} is MOD nturl neutrosophic subset of S( Z ). 4 We cn define opertions of + nd on S( Z ). Thus S( Z ) = {collection of ll subsets of the set n 4 4 Z n } 2 n <. Clerly o(s( Z n )) <. However even findin order of difficult tsk. We will ive exmples of them. Zn is Exmple 3.2: Let {S( Z 5 ) +} be the MOD nturl neutrosophic elements subset under + ddition nd it is defined s the MOD nturl neutrosophic subset semiroup under +. f A = {4 + 5 B = { } nd } S( Z ) then 5 5 A + B = {4 + = {4 + = Z } + { } } This is the wy the opertion + is performed on S( Z ). 5

88 MOD Nturl Neutrosophic Subset Semiroups 87 Thus the pir of MOD nturl neutrosophic elements subset is MOD universl nturl neutrosophic elements subset pir. Let M = { 1 2} nd N = { 1 + } S( Z ). 5 5 M + N = { 1 2} + { 1 + } = { }. We find M + M = { 1 2} + { 1 2} = { }. N + N = { } + { } = { }. Let P = { P + P = { } S( Z ) 5 5 } + { 5 } = { 5 5 } Thus P is the MOD nturl neutrospohi elements idempotent subset of S( Z ). 5 Let M = { } S( Z ). M + M = { }+ = { 1 + { } 5 } M. So in enerl we my not hve M + M = M but here M + M = Z. 5 Consider P = {collection of ll subsets from { }} S( Z ). 5 P is MOD nturl neutrosophic elements subset subsemiroup s if for ny X Y P then X + Y P.

89 88 MOD Nturl Neutrosophic Subset Semiroups Likewise M = {collection of ll subsets from { }} S( 5 elements subsets subsemiroup of S( Z ) is MOD nturl neutrosophic Z ). 5 Let M = { } nd N = { 1 3} S( Z ). 5 5 M + N = { 2 + = { 2 + } + { 1 3} }. 5 Next we provide one more exmple of MOD nturl neutrosophic subset semiroup. Exmple 3.3: Let {S( Z 12 ) +} be MOD nturl neutrosophic subset semiroup. Let A = { } S( Z ). A + A = { } + { } = { nd so on}. 6 1 This is the wy + opertion on S( Z ) is performed. Let M = { } nd N = { } S( M + N = { } + { } = { } S( Z ). 12 Z ). 12

90 MOD Nturl Neutrosophic Subset Semiroups 89 Clerly for iven A S( Z ) we my or my not be ble to find B such tht A + B = Z 12 tht the pir of MOD nturl neutrosophic elements subsets my or my not yield the MOD nturl neutrosophic elements subset universl pir. For we cn sy iven pir A B of MOD nturl neutrosophic elements subset S( Z ); sure we cnnot sy their sum will be Z but we cn estblish iven A S( Z ) we cn lwys find B S( Z ) such tht A + B = 12 Z 12. n view of this we propose the followin problem. Problem 3.1: Let {S( Z n ) +} be the MOD nturl neutrosophic elements subset semiroup under +. For ny iven A S( Z ) cn we find B such tht A + B = n Z n the MOD nturl neutrosophic elements subset universl pir. ii) Given two MOD nturl neutrosophic elements subsets A B prove the pir (A B) in enerl is not MOD nturl neutrosophic elements universl subset pir. We next study MOD nturl neutrosophic elements finite complex number subsets sets {S(C (Z n ))} by some exmples. Exmple 3.4: Let S(C (Z 6 )) = {collection of ll subsets of C (Z 6 )} be the MOD finite complex number nturl neutrosophic elements subset. A = { C 3 C 2iF C 2 4iF 12 C 4 + 3i F } S(C (Z 6 )) is MOD finite complex number nturl neutrosophic elements subset of S(C (Z 6 )). C Let M = { 2 + 3i F 1 + i F 4 5i F } S(C i (Z 6 )). 3 3iF

91 9 MOD Nturl Neutrosophic Subset Semiroups C C We find M + M = { 2 + 2i F i F 5i F } + { 2 3 3iF 3 3iF C 3 3iF C 3 3iF 3 3iF C + 2i F i F 5i f } = { 2 + 2i F i F + 5i F + C 3 3iF 2 + 2i F + C 3 3iF 4 + 4i F 2i F 3 + 3i F 2 + i F 2 C 3 3iF 5 + i F 4 + 5i F 1 4i F } S(C (Z 6 )). This is the wy + opertion is performed on S(C (Z 6 )). We see {S(C (Z 6 )) +} is only semiroup for iven ny A S(C (Z 6 )) we my not be lwys in position to find B S(C (Z 6 )) such tht A + B = {}. However A + {} = {} + A = A for ll A S(C (Z 6 )). Let A = { 2 + C 3 } we cnnot find B S(C (Z 6 )) such tht A + B = {}. nfct if M = { C 4 } nd N = { C 3 } then M + N = { C 4 + C 3 }. Further if S = { C } S(C (Z 6 )) then S + S = { C } + { C } = { C }. C C C Let T + T = { }+ { } = { } T S (C (Z 6 )). 2 2iF 2 2iF 2 2iF Thus T is MOD finite complex nturl neutrosophic subsets which re idempotents with respect to ddition. Hence {S(C (Z n )) +} is only finite commuttive semiroup with {} s the dditive identity. Exmple 3.5: Let {S(C (Z 8 )) +} be the MOD nturl neutrosophic finite complex number subset semiroup under +. Let P = { C 2 2i F + C 4iF C C Q = {6 2 } S(C (Z 8 )). 2iF 6iF } nd

92 MOD Nturl Neutrosophic Subset Semiroups We find P + Q = { C 2iF + C 2iF C C 6iF 2 + C 2iF 2i F + C 4iF C C 6iF C 2 2i F + C C 2i 2 + F 6i F C 6iF C } + {2 6 } = {2 C 4iF i F + C 4iF 4 + C 2iF C 4iF } S(C (Z 8 )). Clerly o(p) = 5 o(q) = 4 o(p + Q) = 18. 2iF C 6iF C 6iF i F + So o(p + Q) = o(p) + o(q) is not true in enerl. C 4iF C 2 C 2 2i F + Of course iven P one cn find Q such tht P + Q = C (Z 8 ); tht is (P Q) is MOD nturl neutrosophic finite complex number universl subset pir of S(C (Z 8 ). We hve R S(C (Z 8 )) such tht R + R =. We hve lso T S(C (Z 8 )) such tht T + T = T. n view of ll these we cn prove the followin theorem. THEOREM 3.1: Let {S( Z ) (or S(C (Z n )) +} be the MOD n nturl neutrosophic finite subset (or MOD nturl neutrosophic finite complex number subset) semiroup under +. i) For ny A S( Z ) (or S(C (Z n )) we cn lwys find n B in S( Z n ) (or S(C (Z n ))) so tht A + B = Z n (or (C (Z n )). Tht is the MOD nturl neutrosophic subset pir is universl pir. ii) Given pir of MOD nturl neutrosophic subsets then in enerl A + B Z (or C (Z n )). n Proof is direct nd hence left s n exercise to the reder. Exmple 3.6: Let {S(Z 12 )} be the MOD nturl neutrosophic-neutrosophic subsets with entries from Z 12. A = { } nd B = {

93 92 MOD Nturl Neutrosophic Subset Semiroups } re MOD nturl neutrosophicneutrosophic subsets of S(Z 12 ). We cn define the opertion + on S(Z 12 ) nd {S(Z 12 ) +} is MOD nturl neutrosophic-neutrosophic subset semiroup. Let A = { 4 2 A + A = { 4 2 = { { } (S(Z 12 ) } } = A } Thus A is MOD nturl neutrosophic-neutrosophic idempotents subset of (S(Z 12 ) Let A = { B = { } S(Z 12 ) 2 A + B = { = { } nd } + { } S((Z 12 ). Order of A = 6 o(b) = 3 nd o(a + B) = 18. Let A = { } S((Z 12 ) } 2 A + A = { = { } + { } } Let B = {6 + 6} then B + B = {6 +6} + {6 + 6} = {}.

94 MOD Nturl Neutrosophic Subset Semiroups 93 Consider P = {Collection of ll subsets of {Z 12 }} S(Z 12 ). Clerly P is MOD subset nturl neutrosophic - neutrosophic subsemiroup of S(Z 12 ). We cn find severl MOD subset nturl neutrosophicneutrosophic subsemiroups. Exmple 3.7: Let W = {{collection of ll subsets from Z 43 } = S(Z 43 ) +} be MOD nturl neutrosophic neutrosophic subsemiroup of finite order. nfct W monoid. There re severl MOD subset nturl neutrosophic neutrosophic subsemiroups iven by B 1 = {collection of ll subset of Z 43 } B 2 = {collection of ll subsets of Z 43 } B 3 = {collection of ll subsets of Z 43 } B 4 = { B 6 = { + } B 5 = { + / Z 43 }. / Z 43 } nd All these MOD subset nturl neutrosophic- neutrosophic subsemiroups re of finite order. Now for ny subset A S(Z 43 ); we cn lwys find B S(Z 43 ) such tht A + B = Z 43. Exmple 3.8: Let {V +} = {collection of ll subsets from Z 4 +} be the MOD subset nturl neutrosophicneutrosophic semiroup under +. Let B = { } nd

95 94 MOD Nturl Neutrosophic Subset Semiroups C = { } S(Z 4 ). Clerly B + C Z 4. However we cn lwys find D S (Z 4 ) such tht D + B =Z 4. Exmple 3.9: Let {S +} = {S(Z 2 ) +} be the MOD subset nturl neutrosophic - neutrosophic semiroup under {Z 2 = { }. So for ny A S we cn find B S such tht A + B = Z 2. n view of ll these we hve the followin theorem. THEOREM 3.2: Let G = {S(Z n ) +} be the MOD subset nturl neutrosophic - neutrosophic semiroup under +. i) o(g) <. ii) For every A G there is B G such tht A + B = Z n. Proof is direct nd we leve it s n exercise to the reder. Now consider ny A S (Z n ) we hv B such tht A + B = Z n We cll the pir (A B) to be the MOD subset nturl neutrosophic-neutrosophic universl pir if A + B = Z n nd B is of lest crdinlity for iven A.

96 MOD Nturl Neutrosophic Subset Semiroups 95 We leve it s open conjecture to find ll B such tht B = o(b) is the lest such tht for iven B A + B = Z n. i) s tht B only one set or we cn hve more thn one B? ii) For iven A find ll B in S(Z n ) so tht A + B = Z n. How mny such B exist for iven A? Further findin o(z n ) is itself very difficult problem. Now we proceed onto describe MOD subset nturl neutrosophic dul number semiroups under + by exmples. Exmple 3.1: Let S = {S(Z 1 ) +} be the MOD nturl neutrosophic subset dul number semiroup. 5} The reder is left with the tsk of findin o(z 1 ). Let A = { B = { A + B = { = { } nd 3 + 5} S; we find } + { } S(Z 1 ). This is the wy + opertion is performed on S

97 96 MOD Nturl Neutrosophic Subset Semiroups S. However {} S(Z 1 ) cts s the dditive identity of We see iven A S we in enerl cnnot find B in S such tht A + B = {}. We cn sy only for sinleton sets we cn et for A with o(a) = 1 nd A Z n lone is such tht there is A 1 such tht A + A 1 = {}. However if A = { 2 } S then we hve no B S such tht A + B = {}. { 2} = A S; B = {8} then A + B = { 2} + {8} = {8 } {}. Thus S is only MOD subset nturl neutrosophic semiroup under + s every A S hs no dditive inverse. Exmple 3.11: Let B = {S(Z 13 +} be the MOD subset nturl neutrosophic dul number semiroup under +. We see B hs MOD subset subsemiroup o(b) is finite. f A = { A + A = { = { } B. 4 } + { 4 } = A } Thus A is n MOD subset dul number idempotent of B. B hs lso MOD subset dul number idempotents. Thus in view of ll these we hve the followin result the proof of which is left s n exercise to the reder.

98 MOD Nturl Neutrosophic Subset Semiroups 97 THEOREM 3.3: Let M = {S(Z n ) +} be the MOD subset nturl neutrosophic dul number semiroup. i) M hs MOD subset dul number idempotents with respect to +. ii) M hs MOD subset nturl neutrosophic dul number subsemiroups. However it is difficult tsk to find the number of MOD subset dul number idempotents. Further findin order of Z n nd S(Z n ) is lso difficult tsk. Findin the number of MOD subset nturl neutrosophic dul number subsemiroups is lso chllenin problem. Next we proceed onto describe by exmples the notion of MOD subset nturl neutrosophic specil dul like number semiroups under +. Exmple 3.12: Let G = {S(Z 12 h ) +} be the MOD nturl neutrosophic specil dul like number subset semiroup under + o(g) <. G hs severl MOD subset nturl neutrosophic specil dul like number subset subsemiroup. Let A = { 6 B = {3 5 8h h 3 4h h 8h h 2 } nd h 9 2h h 2 h } G. We show how the opertion + is performed on G. A + B = { 6 h h {3 5 8h 2 8h 8h + 6 8h + h 3 4h + 6 h 9 2h h 2 + h 9 2h h 2 h 9 2h h 2 } + {3 5 8h h 3 4h h + h 3 4h + h 3 4h 3 + h h 9 2h h 2 h 8h } = h 8h h 3 4h h 9 2h h h 2 h 2 h 8h h h 2 h 8h + h 3 4h } G. h 3 4h

99 98 MOD Nturl Neutrosophic Subset Semiroups h 3h + This is the wy + opertion is performed on G. Let A = { h 3h h 4h + h 8 h 4h } G. h 3h + h 4h t is esily verified A + A = A. h 8 h 8 + h 3h h 4h + h 8 Thus A is the MOD subset nturl neutrosophic specil dul like number idempotent of G. G hs severl such MOD subset nturl neutrosophic specil dul like number subsemiroups. nterested reder cn find the number of such MOD subset nturl neutrosophic specil dul like number subsemiroups in G. Even findin o(z 12 h ) is difficult tsk. Findin the number of MOD nturl neutrosophic specil dul like number idempotents is yet chllenin problem. Exmple 3.13: Let W = {SZ 7 h ) +} be the MOD nturl neutrosophic specil dul like number subset semiroup under +. h + Let D = { h h 3h 3 4h + 1} nd E = { 2 h + 5} W. We show how D + E is clculted. 4h D + E = { h 3h } = { h 3h h + h 4h + h 2h 3 + h + h + h 3h + h h 2h h 3 4h + 1} + { h 3 4h h h 3h 4h h 3h 4h h 2h h 3h h 4h 5 + h 3h + + } W. h 2 + h 3h h 2h h h + + h 4h h 2h h h 3h h 2h + 4 h 3h + h h h + 3h h h + h + + h 3h 1 + h h 4h

100 MOD Nturl Neutrosophic Subset Semiroups 99 This is the wy + opertion is performed on W. Findin ll MOD subset nturl neutrosophic specil dul like number idempotents is left s n exercise to the reder. Findin o(w) is lso chllenin one. The reder is expected to find the number of MOD subset nturl neutrosophic specil dul like number subsemiroups of W. We hve the followin result. THEOREM 3.4: Let S = {S(Z n h +} be the MOD nturl neutrosophic subset specil dul like number semiroup under +. ) o(s) <. b) S hs MOD subset nturl neutrosophic specil dul like number subsemiroups. c) S hs MOD subset nturl neutrosophic specil dul like number idempotents. Proof is direct nd hence left s n exercise to the reder. Next we describe by exmples MOD nturl neutrosophic subset specil qusi dul number semiroup under +. Exmple 3.14: Let S = {S(Z 15 k ) +} be the MOD specil qusi dul number nturl neutrosophic subset semiroup under +. Let A = { 5k B = {3 + 6k A + B = { 5k k 3k k 9k 6 k + 3 4k k 3k k 9k 6 } nd k 6k } S. } + {3 + 6k k + 3 4k k 6k }

101 1 MOD Nturl Neutrosophic Subset Semiroups = {3 + 6k 11k k + k 3k 6k k k k k k 5k + + 3k k 4k k + + 4k } S. k 6k k 3k k 6k k 9k k 9k k 6k This is the wy + opertion is performed on S. Let A = { k k 6k k + k 6k } S; k 9k k 6k A + A = { k = { k k 6k k 6k + k k + } + { k } = A. k 6k k 6k k 6k k + k 6k } Thus A is the MOD subset specil qusi dul number nturl neutrosophic idempotent element of S. S hs severl such MOD subset idempotents. However the reder is left with the tsk of findin the number of such MOD nturl neutrosophic specil qusi dul number subset idempotents of S. Exmple 3.15: Let M = {S(Z 17 k ) +} be the MOD subset nturl neutrosophic specil qusi dul number semiroup under +. M hs MOD nturl neutrosophic subset specil qusi dul number subsemiroups. A = { k k k k + 6k 6k } M is such tht A + A = A is MOD nturl neutrosophic subset specil qusi dul number idempotent of M. The reder is left with the tsk of findin the totl number of MOD nturl neutrosophic subset specil qusi dul number idempotents of M. n view of this we hve the followin theorem the proof of which is left s n exercise to the reder.

102 MOD Nturl Neutrosophic Subset Semiroups 11 THEOREM 3.5: Let S = {(Z n k ) +} be the MOD subset nturl neutrosophic subset specil qusi dul number semiroup under +. i) o(s) <. ii) S hs MOD subset nturl neutrosophic specil qusi dul number subsemiroups. iii) S hs MOD subset nturl neutrosophic specil qusi dul number idempotents. iv) For every A S there exists B (o(b) the lest) in S such tht A + B = Z n k. However we propose the followin problems. Problem 3.2: Let {S(Z n k ) +} = S be the MOD subset nturl neutrosophic specil qusi dul number semiroup under +. i) For iven A S for fixed n; 2 n < how mny B exists such tht A + B = Z n k? ii) How mny B with o(b) lest exist for iven A in S so tht A + B = Z n k? iii) Find ll MOD nturl neutrosophic specil qusi dul number subsets B of fixed A such tht A + B = Z n k. iv) Find ll MOD nturl neutrosophic specil qusi dul number subsets of S which re MOD idempotent subsets nd study the three problems. When n is t prime ive lre vlue for n. b. When n = p t p prime. c. When n is odd composite number (lre). d. When n is n even composite number. Next we proceed onto describe the properties of S( Z n ) under product. Here we define two types of products.

103 12 MOD Nturl Neutrosophic Subset Semiroups n n i) t = t for ll pproprite t s the nturl neutrosophic dominnt MOD nturl neutrosophic semiroups. n ii) t = for ll pproprite t s the usul zero dominnt semiroup. However even if we do not mention explicitly the structure from the wy product is used we will know whether the MOD semiroup is zero dominnt or MOD nturl neutrosophic element dominnt. We hve lredy seen the concept of zero dominnt this ws used in [ ] for otherwise the MOD rphs edes would be very clumsy or mny levin no ede free. We will see when it is dominnt wht re the specil properties enjoyed by it nd when MOD nturl neutrosophic element is dominted wht re the specil fetures ssocited with it s semiroups under product. Exmple 3.16: Let S = {S( Z 1 ) } be the MOD nturl neutrosophic subset semiroup in which S is zero dominted. Tht is {} A = {} for ll A S. Let A = {3 5 7 We find A B; } nd B = {2 4 A B = { } { } = { }. Suppose A = { } nd B = {2 1 5 } S. A B = { } {2 1 5 } = { } S. 1 6 } S.

104 MOD Nturl Neutrosophic Subset Semiroups 13 Clerly S is MOD nturl neutrosophic subset semiroup. nfct S is monoid s {1} S is such tht {1} A = A {1} = A for ll A S. Further if A = {5 1 2 } nd B = { 2} then A B = {5 1 2 } { 2} = { 1 2 }. We see o(a) = 3 o(b) = 2 but o(a B) = 2. Thus these MOD subset nturl neutrosophic semiroup under product behve differently. Further if A = { } nd B = {5} S then A B = { } {5} = {}. Thus the pir {A B} is MOD nturl neutrosophic zero divisor pir subsets. However we cn find MOD zero divisor subset if 1 t A or B. Only when A = {} or B = {}. The zero is possible. So we see if A S such tht 1 t A then for no B S \ {} A B = {}. The nturl problem is to find for ny A S B S such tht A B = Z. 1 This tsk is relly difficult nd left s n exercise to the reder. Let A = {1 1 5 } nd B = { } S. A B = {1 1 5 } { = { } Z }

105 14 MOD Nturl Neutrosophic Subset Semiroups However if A = {1 A B = Z } then tke B Z \ { 1 } then 1 5 in t is n interestin problem to find the number of subsets B Z so tht A B = Z. 1 1 Findin those subsets B in S which hs the lest crdinlity which ives A B = Z 1. Such study is innovtive nd interestin so left s n exercise to the reder. Also nother nturl question is how mny MOD neutrosophic subsets in S re idempotents. nturl Let A = { 5 1 6} S; A A = { 5 6 1} { 6 5 1} = { 6 5 1} = A. So A is MOD nturl neutrosophic idempotent subset of S. Let B = { } S B B = { } { } = { } = B. Thus B is lso MOD nturl neutrosophic subset idempotent of S. Findin the totl number of such MOD nturl neutrosophic subset idempotents in S hppens to be chllenin problem. Let D ={ } S. D D = { } { } = { } = D.

106 MOD Nturl Neutrosophic Subset Semiroups 15 Thus D is lso MOD nturl neutrosophic subset idempotent of S. t is yet chllenin problem to find the MOD nturl neutrosophic subset idempotent M of such tht o(m) is the hihest crdinlity. However there re severl MOD subset nturl neutrosophic idempotents P of S such tht o(p) = 1. Exmple 3.17: Let M ={S( Z 11 ) } be the MOD nturl neutrosophic subset semiroup which is zero dominted tht is 11 =. Let A = { 2 6} nd B = { } M. To find A B A B = { 2 6} { } = { } o(a) = 3 o(b) = 4. o(a B) = 8. So we see o(a B) o(a) o(b) it my be less thn o(a) o(b) it my be less thn o(a) o(b). We cnnot ctully ttin the equlity. Let A = { } nd B = { } M. To find A } { } = { } M. Clerly o(a) = 5 o(b) = 5 o(a B) = 12. Now findin MOD subset nturl neutrosophic idempotents nd nilpotents in M is n impossibility. For if A = {} M

107 16 MOD Nturl Neutrosophic Subset Semiroups then A A = {}. f A = { 1} M then A A = { 1}. f B = { 1 11 } M then B B = { 1 11 }. D = {1 11 } M is such tht D D = {1 11 }. These re some of the MOD nturl neutrosophic subset idempotents of M. All these will be termed only s trivil MOD nturl neutrosophic subset idempotents of M. A = { 4 3 1} M is nontrivil MOD subset idempotent. Now the only MOD nturl neutrosophic subset nilpotents re A = {} nd B = { 11 }. So this M hs no nontrivil MOD nturl neutrosophic subset nilpotents. n view of ll these we hve the followin theorem. THEOREM 3.6: Let S = {S( Z ) } n prime be the MOD n nturl neutrosophic subset semiroup under which is zero dominted. i) S hs no nontrivil MOD nturl neutrosophic subset nilpotents. ii) This hs MOD nturl neutrosophic idempotents subset iven by { 1 b} where b = 1; b Z p or the set {Z p } = B or the set T = { Z }. Proof is direct nd hence left s n exercise to the reder. Exmple 3.18: Let S = {S( Z 8 ) } be the MOD nturl neutrosophic subset semiroup under zero dominnt product tht is 8 t = for ll t = p

108 MOD Nturl Neutrosophic Subset Semiroups 17 Let A = {2 4 6} S. A A = { 2 4 6} { 2 4 6} = { 4} = B A A A = B A = { 4} { 2 6 4} = {}. Thus A is MOD nilpotent subset nturl neutrosophic element of S of order three s A 3 = {} Let M = { } S; M M = { } { } = { } = M is the MOD nturl neutrosophic trivil idempotent subset of S. Let P = { } S; P P = { } { } = { } = P 2 ; P P P = P 2 P = { } { } = { 8 }. Thus P is the MOD nturl neutrosophic subset which is nturl neutrosophic nilpotent s P 3 = { 8 } the nturl neutrosophic zero subset of S. So interested reder cn study this sort of MOD nturl neutrosophic subset nilpotents P nd MOD nturl neutrosophic subset nilpotents R such tht P t = {} nd R s = { n }; t 2 nd s 2 with n = p m p prime; m 2. Next we proceed onto illustrte by exmples MOD nturl neutrosophic finite complex number subset semiroups under the zero dominnt product. Exmple 3.19: Let S = {S(C (Z 6 )) } be the MOD nturl neutrosophic finite complex number subset semiroup under C zero dominnt product tht is t = for ll relevnt t; t in C(Z 6 ).

109 18 MOD Nturl Neutrosophic Subset Semiroups Let A = { C 3 8 B = { 3 + 3iF A B = { = { C 2 4iF 4} nd C 2 3 i F } S; C 3 8 3iF 8 4} { 3 + C 2 4iF C C 3 + C 3iF + C 2 4iF C 2 3 i F } C 2 4i F }. This is the wy product opertion is performed on S. o(a) = 4 nd o(b) = 5 o(a B) = 7. Let A = { i F A A = { i F = { C 5 1} S; C 5 1} {5 1 i F 5 i F 1 5i F } A. C C } Thus findin MOD nturl neutrosophic finite complex subset idempotents hppens to be difficult problem. Let P = { 1 i F P P = { 1 i F = { 1 i F 5 5i F } S. We find C 5 5i F } { 1 i F C C 5 5i F } = P. C 5 5i F } Thus P is the MOD nturl neutrosophic finite complex number subset which is idempotent s P 2 = P. Consider D = { 3 4 C 1} S. We find D D = { = { } { C C } = D. C } This D is lso MOD nturl neutrosophic finite complex number idempotent subset of S.

110 MOD Nturl Neutrosophic Subset Semiroups 19 Findin the totl number of MOD nturl neutrosophic finite complex number idempotent subsets hppens to be chllenin problem. Exmple 3.2: Let S = {S(C (Z 3 )) } be the MOD nturl neutrosophic finite complex number subset semiroup under the zero dominnt product where C (Z 3 ) = { 1 2 i F 2i F 1 + i F 2 + i F 1 + 2i F 2 + 2i F C 1 + 2i F C 2 + C C 2 + 2i F + }. C i F + Let A = {1 i F 2 } nd B = { C 1 + A B = {1 i F 2 } { 1 = { 1 C 1 + C C o(a) = 4 o(b) = 4 but o(a B) = 8. C C C 2i F + C 1 + i F + C C 1} C (Z 3 ). + 1} i F i F + C } S. This is the wy product opertion is performed. Findin MOD nturl neutrosophic finite complex number zero divisors hppens to be difficult tsk. Let us find the product of A = { 1 2} S A A = { 1 2} { 1 2} = { 1 2} = A S. Thus A is the MOD nturl neutrosophic subset finite complex number idempotent. B = { 2 1 i F 1 + i F 2i F 2 + i F 2 + 2i F 1 + 2i F } S is such tht B B = B. This B is in the MOD nturl neutrosophic finite complex number subset idempotent of S. Thus even if in C (Z n ) n is prime we cn hve MOD nturl neutrosophic finite complex number C (Z 3 ) cn hve MOD nturl neutrosophic finite complex number subset idempotents. n view of this we hve the followin theorem.

111 11 MOD Nturl Neutrosophic Subset Semiroups THEOREM 3.7: Let S = {S(C (Z n )) } be the MOD nturl neutrosophic finite complex number subset semiroup under zero dominnt product. ) S hs certinly MOD nturl neutrosophic finite complex number subset idempotents. b) S hs MOD nturl neutrosophic finite complex number subset subsemiroups. c) S my or my not hve MOD nturl neutrosophic finite complex number subset nilpotents dependin on M. Proof is left s n exercise to the reder. We now ive exmples of MOD nturl neutrosophicneutrosophic subset semiroup under zero dominnt product. Exmple 3.21: Let M = {S(Z 4 } be the MOD nturl neutrosophic- neutrosophic subset semiroup. P = { } M such tht P P = { } { } = {}. So P is MOD nturl neutrosophic-neutrosophic subset nilpotent of M. Let N = { 1} M N N = { 1 } { 1 } = { 1 } = N. Thus N M is the MOD nturl neutrosophic- neutrosophic idempotent subset of M. Exmple 3.22: We consider M = {S(C (Z 4 ) }. Let D = { 1 2 3} M to find

112 MOD Nturl Neutrosophic Subset Semiroups 111 D D = { 1 2 3} { 1 2 3} = { 2 3 1} = D is in MOD nturl neutrosophic - neutrosophic idempotent subset of M. Let Z = {1 + i F i F 3 + i F } nd Y = {2i F 3i F } M; to find Z Y; Z Y = {i F 1 + i F 3 + i F } { 3i F 2i F } = { 1 3i F + 1 i F i F + 2}. Here o(y) = 3 o(z) = 3 but o(z Y) = 6. Let T = { 2 i F 1 + i F 2 + i F C } nd C C R = {1 2i F 2 + 2i F 1 + i F + } M. i F + C C C + T R = { 2 i F 1 + i F 2 + i F } = { 2 i F 1 + i F 2 + i F C 1 + i f + }. C 2 + i F + C 2i F + C } {1 2i F 2 + 2i F 1 + C C 2 2i F i F o(t) = 6 o(r) = 5 but o(t R) = 16. C 3 + C i F + C i F + 3 This is the wy product is obtined in M. However iven A M findin B M such tht A B = C (Z 4 ) hppens to be difficult tsk. Exmple 3.23: Let S = {S(Z 5 ) } be the MOD nturl neutrosophic-neutrosophic subset semiroup under zero dominnt product. Let A = { 4 + 4} nd B = { We find A B = { 4 + 4} { = { This is the wy is performed on S. C } S. } } S.

113 112 MOD Nturl Neutrosophic Subset Semiroups Let P = { } S P P = { } { } = { } S. Thus P is MOD nturl neutrosophic-neutrosophic subset idempotent of S. Let T = { 2 3 4} S. Consider T T = { 2 3 4} { 2 3 4} = { 2 3 4} = T. Hence once in T is MOD nturl neutrosophic - neutrosophic subset idempotent of S. Let W = { } S. We see W W = W so W is in MOD nturl neutrosophic - neutrosophic idempotent subset of S. However iven A S findin B S such tht A B =Z 5 hppens to be difficult tsk. Let V = { Z = { 1 V Z = { = { } S } nd } { This is the wy product opertion is performed on S. o(v) = 4 o(z) = 4 nd o(v Z) = 7. } } S.

114 MOD Nturl Neutrosophic Subset Semiroups 113 Let M = { 1} S M M = { 1 } { 1 } = { 1 } Let N = { } S N N = { } { } = { } = N Thus N is the MOD nturl neutrosophic- neutrosophic idempotent subset of S. Findin nilpotent subset of S is ner to n impossibility. n view of this we hve the followin conjecture 3. Conjecture 3.Let S = {S(Z n ) } n prime be zero dominnt product MOD nturl neutrosophic neutrosophic subset semiroup. Prove S hs no nontrivil MOD nturl neutrosophic - neutrosophic nilpotent subsets. Exmple 3.24: Let S = {S(Z 16 ) } be the MOD nturl neutrosophic - neutrosophic subset semiroup under zero dominnt product. Let B = { } S; B B = { } { } = {}. Thus B is the MOD nturl neutrosophic - neutrosophic nilpotent subset of order two. Let D = { } S. D D = { } {

115 114 MOD Nturl Neutrosophic Subset Semiroups 4 + 4} = { } = D 2. D 2 D = { } { } = { } = D 3. D 3 D = { } { } = {} = D 4. Thus D is MOD nturl neutrosophic neutrosophic nilpotent subset of order four. S. Let R = { } R R = R so R is MOD nturl neutrosophicneutrosophic idempotent subset of S. Let T = { } S. t is esily verified T T = T so once in T is MOD nturl neutrosophic - neutrosophic idempotent subset of S. f B = {Z 16 } S then lso B B = B is the MOD nturl neutrosophic - neutrosophic idempotent subset of S. Thus even if n = p t t 2 p prime still S(Z n ) hs MOD nturl neutrosophic - neutrosophic idempotent subsets. n view of ll these we hve the followin theorem. THEOREM 3.8: Let S = {S(Z n ) } be the MOD nturl neutrosophic - neutrosophic subset semiroup under the zero dominted product.

116 MOD Nturl Neutrosophic Subset Semiroups 115 ) S hs MOD nturl neutrosophic - neutrosophic idempotent subsets for ll n 2 n <. b) S hs MOD nturl neutrosophic - neutrosophic nilpotent subsets if n = p t t 2 nd p prime. c) S hs MOD nturl neutrosophic - neutrosophic subset subsemiroups. Proof is direct nd hence left s n exercise to the reder. However the tsk of providin iven A S (S s in theorem 3) to find B S such tht o(b) is the lest is left s n exercise to the reder. Next we describe by exmples the MOD nturl neutrosophic dul number subset semiroup under the product is zero dominted product. Exmple 3.25: Let B = {S(Z 9 } be the MOD nturl neutrosophic dul number subset semiroup under zero dominnt product. Let M = { } B. Clerly M M = {}. So M is the MOD nturl neutrosophic dul number nilpotent subset of B. Let D = {5 6 8 } M D D = {}. Thus B hs severl MOD nturl neutrosophic dul number subset nilpotents. Now we describe few MOD nturl neutrosophic dul number subset idempotents A = { } B is such tht A A = A is MOD nturl neutrosophic dul number subset idempotent. Let Z = { } B; Z Z = {} is the MOD nturl neutrosophic dul number nilpotent subset of B.

117 116 MOD Nturl Neutrosophic Subset Semiroups T = { 5} B is such tht T T = {}. t is importnt to keep on record tht MOD nturl neutrosophic dul number subset semiroups hve mny MOD nturl neutrosophic dul number subsets which re nilpotent of order two. Let W = { } B. Clerly W W = { } thus W is MOD nturl neutrosophic dul number nturl neutrosophic zero nilpotent subset of B. Thus in this sitution only we hve MOD nturl neutrosophic dul number neutrosophic zero nilpotent subsets N = { 2 4 } B is such tht N N = { }. E { Let A = { A D = { Thus { f E = {3 1 } nd D = { 3 } { 3 4 } B; 4 } = { }. } in B is defined s the mixed zero of B. } = { Thus cn we sy for ll x B. x { 4 + 3} B then 2 } { } = { } = { }? }. Exmple 3.26: Let S = {S(Z 7 ) } be the MOD nturl neutrosophic subset semiroup under the zero dominnt product. Let A = { } nd

118 MOD Nturl Neutrosophic Subset Semiroups 117 B = { } S. A B = { = { } {1 This is the wy product is obtined. We find A { } }. } = { } { So we see for every A S; A { } = { Further A {} = {} for ll A in S. }. } = { }. Finlly A { } = { } or {} or { } for ll A S. Thus we hve three zeros nd in this zero dominted MOD nturl neutrosophic dul number subsets {} is the zero nd { } nd { } re such tht { } {} = {} nd { } {} = {}. Consider M = { t + 2 t + t t t + 4 t + 6 t t t t t t + 6 t t t = 1 2 6} S is such tht if S(M) = {collection of ll MOD nturl neutrosophic dul number subsets from M} then S(M) is MOD nturl neutrosophic dul number subsets subsemiroup of S nd infct MOD nturl neutrosophic dul number subset idel of S. n view of ll these we hve the followin theorem.

119 118 MOD Nturl Neutrosophic Subset Semiroups THEOREM 3.9: Let S = {S(Z n ) } be the MOD nturl neutrosophic dul number subset semiroup under zero dominnt product. i) S hs severl MOD nturl neutrosophic dul number nilpotent subsets of order two. ii) S hs MOD nturl neutrosophic subset dul number idels. iii) S hs MOD nturl neutrosophic subset dul number idempotents. iv) S hs MOD nturl neutrosophic subset dul number subsemiroups which re not idels. Proof is direct nd hence left s n exercise to the reder. Next we proceed onto describe by exmples MOD nturl neutrosophic subset specil dul like number semiroups under zero dominnt product. Exmple 3.27: Let G = {S(Z 1 h ) } be the MOD nturl neutrosophic specil dul like number subset semiroup under the zero dominnt product. Let B = { h 1 5 5h 5 + 5h} G B B = { h 1 5 5h 5 + 5h} { h 1 5 5h 5 + 5h} = { 1 h 5 5h 5 + 5h} = B. Thus B is the MOD nturl neutrosophic specil dul like number idempotent subset of G. Let H = { 5 5h 5 + 5h K = { 2 2h h 2 H K = { 5 5h 5 + 5h = { h h 5 h h 2h } G. h 2 h h 2h h 5 } nd } G. h 5 } { 2 2h h 2 h 2h }

120 MOD Nturl Neutrosophic Subset Semiroups 119 This is the wy MOD nturl neutrosophic specil dul like number subsets product is obtined. Let T = { 5 5h 5 + 5h T S = { 5 5h 5 + 5h h } nd S = { 2 2h h } { 2 2h h } = { h } G. h }. Thus the product of the MOD nturl neutrosophic specil dul like number subsets yields the MOD nturl neutrosophic specil dul like number zero pir { }. Let M = { h 2h h 4h We find M N = { h h 5h + 5} nd N = { h 2h h 4h h 5h + 5} { Thus we cn prove for every Z G. Z { h } G. h } = { h } = { h }. h }. Let L = { } G. Clerly L L = L so L is MOD nturl neutrosophic specil dul like number idempotent subset of G. Let R = { h 2h 3h 4h 9h} G. R R = R so R is in MOD nturl neutrosophic specil dul like number idempotent subset of G. W = {Z 1 h} G is in MOD nturl neutrosophic specil dul like number subset idempotent of G s W W = W. We see G hs MOD nturl neutrosophic specil dul like number subset idempotents nd zero divisors. However findin MOD nturl neutrosophic specil dul like number nilpotents hppens to be difficult tsk for S(Z 1 h ) hs no nontrivil nilpotent subsets.

121 12 MOD Nturl Neutrosophic Subset Semiroups Exmple 3.28: Let W = {S(Z 13 h ) } be the MOD nturl neutrosophic specil dul like number subset semiroup under the zero dominnt product. Given A W the tsk of findin B such tht A B = Z 13 h is left s n exercise to the reder. T = {{Z 13 }} W is MOD nturl neutrosophic specil dul like number subset idempotent of W. V = {Z 13 h} W is lso MOD nturl neutrosophic specil dul like number subset idempotent of W. M = { h 2h 3h 12h} W is in MOD nturl neutrosophic specil dul like number subset idempotent of M s M M = M. Let P = {collection of ll subsets from {Z 13 h t } relevnt t Z 13 h} W is MOD subset nturl neutrosophic specil dul like number idel of W. n view of ll these we hve the followin theorem. THEOREM 3.1: Let S = {S(Z n h ) } be the MOD nturl neutrosophic specil dul like number subset semiroup with zero dominnt product. i) S hs MOD nturl neutrosophic specil dul like number subset idempotents 2 n <. ii) S hs MOD nturl neutrosophic specil dul like number nilpotents only for n = p t t > 2 p prime or n pproprite composite number. iii) S hs MOD nturl neutrosophic specil dul like number subset zero divisors tht is A B = {} or h h A B = { } or A B = { } for some A B S. iv) S hs MOD nturl neutrosophic specil dul like number subset subsemiroups which re not idels. v) S hs MOD nturl neutrosophic specil dul like number subset subsemiroups which re idels.

122 MOD Nturl Neutrosophic Subset Semiroups 121 Proof is direct nd hence left s n exercise to the reder. Next we propose the followin problem. Problem 3.3: Let S = {S(Z n h ) } be the MOD nturl neutrosophic specil dul like number subset semiroup with zero dominnt product. i) Find conditions on n so tht S hs MOD nturl neutrosophic specil dul like number subset nilpotents. ii) Given A S prove we cn find lwys B with miniml order (o(b) miniml) such tht A B = Z n h. iii) s tht B unique or there re more thn one B mentioned in (ii) which is such tht A B = Z n h? Next we proceed onto describe by exmples the MOD nturl neutrosophic specil qusi dul number subset semiroups under the zero dominnt product. Exmple 3.29: Let M = {S(Z 15 k } be the MOD nturl neutrosophic specil qusi dul number subset semiroup under the zero dominnt product. Let P = {Z 5 } M clerly P P = P so P is nontrivil MOD nturl neutrosophic specil qusi dul number subset idempotent of M. R = {Z 15 k} M is in MOD nturl neutrosophic specil qusi dul number subset idempotent of M s R R = R. nfct M hs severl such MOD nturl neutrosophic specil qusi dul number subset idempotents. Let A = {3 + k 5 k 3k k 1} nd

123 122 MOD Nturl Neutrosophic Subset Semiroups B = {5k + 1 1k k 3 5k } M. We find A B = {3 + k 5 1} {5k + 1 1k k 3k k k 3 5k } = { 5k + 1 1k k 3 5k k k 3k k 9k 1k + 5 5k}. This is the wy the product opertion is performed on M. Let A = {5 5k 1k 1} nd B = {3 6 9 } M. A B = {5 5k 1k 1} {3 6 9 } = {}. Thus {A B} is the MOD nturl neutrosophic specil qusi dul number subset zero divisor pir. Further A {} = {} for ll A M. A { k } = { k } or { k } or {} for ll A M Let A = { k k 3k 5k A B = { }. k Let A = { k 3k A B = { k }. k 5k k k 6k k 6k 6 k 6k 6 } nd B = { k } then k 9k } nd B = { k } then Let A = { k 5} nd B = {3k} A B = { k }. Thus we see we cn hve ny of these MOD nturl neutrosophic specil qusi dul number subset zero divisors. Let N = {collection of ll subsets from Z 15 } M N is MOD nturl neutrosophic specil qusi dul number subsemiroup of M which is not n idel of M.

124 MOD Nturl Neutrosophic Subset Semiroups 123 Let S = {collection of ll subsets from Z 15 k} M S is in MOD nturl neutrosophic specil qusi dul number subsemiroup of M which is not n idel of M. Severl interestin results cn be obtined in this direction. Exmple 3.3: Let S = {S(Z 16 k } be the MOD nturl neutrosophic specil qusi dul number subset semiroup under the zero dominnt product. S hs MOD nturl neutrosophic specil qusi dul number subset nilpotents. Tke A = {2 4k 8k 1} S A A = {2 4k 8k 12} {2 4k 8k 12} = {4 8k 8} = A 2 A 2 A = {4 8 8k} {2 4k 8k 12}= {8} = A 3 A 3 A = {8} {2 4k 8k 12} = {} = A 4. Thus A is the MOD nturl neutrosophic specil qusi dul number subset nilpotent of order four. Let B = { k 2 k 4 k 8 k 12k } S. B B = { k 2 = { k 4 k 4 k 8 k 8 k 8k k 12k } { k 2 } = B 2. k 4 k 8 k 12k } B 2 B = { k 4 k 8 B 3 B = { k 8 } {{ k 2 k 8k } { k k k k k k k } = { k } = B 4. k 12k } = { k 8 } = B 3 Thus B is MOD nturl neutrosophic specil qusi dul number MOD nturl neutrosophic zero nilpotent subset of order 4. Let D = { k 4 4k D D = { k k k 4 4k k 2 4k } S k 2 4k } { k k 4 4k k 2 4k }

125 124 MOD Nturl Neutrosophic Subset Semiroups = { k k 8 8k } = D 2. D 2 D = { = { k k k 8 8k } { } = D 3 k k 4 4k k 2 4k } Thus D in S is the MOD nturl neutrosophic specil qusi dul number subset mixed nilpotent of order three s D 3 = { k } which is MOD nturl neutrosophic specil qusi dul number mixed zero of S. Now we see S hs MOD nturl neutrosophic specil qusi dul number subset zero divisors. Let A = {8 + 4k 12} nd B = { k} S. A B = { k} { k} = { 8 8k} {}. Let A = {8 + 4k 12} nd B = {8 + 12k 8k + 4} S. A B = {8 + 4k 12} {8 + 12k 8k + 4} = {}. Thus S hs MOD nturl neutrosophic specil qusi dul number subset pir of zero divisors. nterested reder cn work in this direction for more types of MOD nturl neutrosophic specil qusi dul number zero divisors nd nilpotents. However we hve the followin theorem. THEOREM 3.11: Let S = {S(Z n k ) } be the MOD nturl neutrosophic specil qusi dul number subset semiroup under the zero dominnt product. i) S hs MOD nturl neutrosophic specil qusi dul number subset idempotents. ii) S hs MOD nturl neutrosophic specil qusi dul number subset nilpotents only for n not prime.

126 MOD Nturl Neutrosophic Subset Semiroups 125 iii) S hs MOD nturl neutrosophic specil qusi dul number subset subsemiroups which re not idels. iv) S hs MOD nturl neutrosophic specil qusi dul number subset subsemiroup which re idels. Proof is direct nd hence left s n exercise to the reder. Next we proceed on describe by one or two exmples MOD nturl neutrosophic subsets under dominnt MOD nturl n neutrosophic zero divisor ( k h C or or or or ). Tht is t = t for t {n k h c }. Exmple 3.31: Let S = {S( Z 1 ) } be the MOD nturl neutrosophic subsets semiroup in which MOD nturl neutrosophic zero divisor dominnt product is defined. Let P = { } nd Q = { } S P Q = { ) { } = { }. This is the wy MOD subset nturl neutrosophic semiroup where MOD nturl neutrosophic zero divisor product is dominnt is defined. Tht is 1 = 1. Let B = {5 } nd D = { } S B D = { 5} { } = {}. Thus this pir {B D} ives the MOD nturl neutrosophic zero divisor pir. Let A = { } nd B = { } S

127 126 MOD Nturl Neutrosophic Subset Semiroups A B = { } { } = { 1 }. Thus the MOD nturl neutrosophic subset pir ives the MOD nturl neutrosophic zero divisor. Let E = { } nd D = { } S; E D = { } { } = { 1 }. Thus this MOD nturl neutrosophic pir ives the MOD nturl neutrosophic zero divisor pir. We see 1 = 1 nd so on We see T = {Z 1 } S is such tht T T = T is MOD nturl neutrosophic subset idempotent of S. We see when from the cse when n t = the semiroup behves differently = n t n t. However from the context one cn esily find out how these products differ. 1 t f B = { } nd C = { 1 5 } S then B C = { 1 = 1. t } { But if we ssume 1 t = then } = { } where B C = { } { 5 } = { 1 } is the MOD nturl neutrosophic zero divisor pir.

128 MOD Nturl Neutrosophic Subset Semiroups 127 Thus we see the very difference when 1 t =. 1 t = 1 t nd Findin MOD nturl neutrosophic subsemiroups nd idels is considered s mtter of routine nd hence left s n exercise to the reder. Exmple 3.32: Let M = {S( Z 19 ) } be the MOD nturl neutrosophic subset semiroup under the MOD nturl neutrosophic zero dominnt product tht is 19 = 19. We see if A B M \ {} then A B = {} is n impossibility A {} {} in M. Let A = { } nd B = {} M then A B ={ } {} = { 19 } {}. Let A ={ } nd E = { 19 } S A E = { } { 19 } = { 19 }. Thus this is the cse of MOD nturl neutrosophic zero dominnt product semiroup. n cse of {} dominnt MOD nturl neutrosophic product semiroup. We see A E = { } { 19 } = { 19 }. But if E = { 19 } is replced by F = {} A F = { } {} = {} in cse {} A = {}.

129 128 MOD Nturl Neutrosophic Subset Semiroups n cse A { 19 } = 19 we see A F = { } {} = { 19 }. So there is difference in the results. As per need of the sitution one cn use either MOD nturl neutrosophic zero dominnt product or just zero dominnt product. Now we will provide exmples of MOD nturl neutrosophic finite complex number subset under MOD nturl neutrosophic finite complex number zero dominnt semiroup. Tht is { C } {} = { C }. Exmple 3.33: Let S = {S(C (Z 2 )) } be the MOD nturl neutrosophic finite complex number subset MOD nturl neutrosophic finite complex number zero dominnt product. We see A = { C } nd B = {5 + 3i F 2i + 12 C 12 C 18 + C 4iF A B = { C } {5 + 3i F 2i F + 12 C } S. C 12 C 18 + } = { C }. C 4iF Thus for ll A S we hve A { C }= { C } however A {} {} in this product. Let P = {3 + i F Q = { C C 4 C 15 5iF C 5iF } S C 4 + C 5 + 3i F + C } nd P Q = {3 + i F C 15 5iF C 4 + C 5 + 3i F + C } { C C 4 } C 5iF

130 MOD Nturl Neutrosophic Subset Semiroups 129 = { C C 4 C 5iF C 15iF 15 }. C 1iF This is the wy product opertion is performed on S. C Let A = { 1iF B = { C i F C A B = { 1iF C 5 + 5i F + } S C 8iF C 5 + 5i F C 2 C 2 C i F } nd C i F } { C 4 C 8iF 5 + 3i F } = { C C 4 + 6}. C 4 + C 8 C 16 + C 4 + C 8iF C 16iF + C 12iF C 8iF C 1iF C 2 This is the wy product opertion is performed on S. Clerly P = {Z 2 } is MOD nturl neutrosophic finite complex number subset which is n idempotent {C(Z 2 )} = B S is in MOD nturl neutrosophic finite complex number subset idempotent of S. Let M = {collection of ll subsets from the set Z 2 } S be MOD nturl neutrosophic finite complex number subset subsemiroup which is n idel. N = {collection of ll subsets of the set C(Z 2 )} S is in MOD nturl neutrosophic finite complex number subset subsemiroup which is not n idel of S. So S hs MOD nturl neutrosophic finite complex number subset subsemiroups which re not idels. Exmple 3.34: Let W = {collection of ll subsets from S(C (Z 11 )) } be the MOD nturl neutrosophic finite complex

131 13 MOD Nturl Neutrosophic Subset Semiroups number subset semiroup under where is MOD nturl neutrosophic zero dominted tht is =. We see C = {collection of ll subsets from Z 11 } W is MOD nturl neutrosophic finite complex number subset subsemiroup of W. D = {collection of ll subsets from C(Z 11 )} W is in MOD nturl neutrosophic finite complex number subset subsemiroup of W. Both C nd D re dels of W. Findin MOD nturl neutrosophic zero divisors nd nilpotents in W hppens to be difficult tsk. However M = {Z 11 } W is MOD subset nturl neutrosophic finite complex number idempotent of W. N = {C(Z 11 )} W is lso MOD subset nturl neutrosophic finite complex number idempotent of W. Thus if in {S(C (Z p )} p prime findin MOD nturl neutrosophic subset finite complex number zero divisors nd nilpotents is left s open conjecture. However for p = 2 we hve S(C (Z 2 )) hs MOD nturl neutrosophic finite complex number nilpotents of order two. Also S(C (Z 2 )) hs MOD nturl neutrosophic finite complex number subset zero divisors s A = { 1 +i F } nd B = {1 + i F } ives A B = {}. Further S(C (Z 2 )) hs MOD nturl neutrosophic finite complex number MOD nturl neutrosophic finite complex number mixed zero divisors. C For tke A = { 1 + i F } nd B = { C 1 + i F } {S(C (Z 2 )) }. C C

132 MOD Nturl Neutrosophic Subset Semiroups 131 A B = { C 1 + i F } { C 1 + i F } = { C }. f A = { C 1 + i F } {S(C (Z 2 )) } then A A = { C 1 + i F } { C 1 + i F } = { C } is the MOD nturl neutrosophic subset finite complex number mixed nilpotent of order two. n view of ll these we re forced to suest in {S(C (Z p )) } {p > 2 p prime} findin MOD nturl neutrosophic finite complex number rel zero divisors {} or MOD nturl neutrosophic finite complex number subset zero divisors { C } or MOD nturl neutrosophic finite complex number mixed zero divisors { C } hppens to be chllenin problem. However we re lwys urnteed of MOD nturl neutrosophic finite complex number subsets idempotents nd subsemiroups in cse of (S(C (Z n )) } whtever be n. n view of these we hve the followin theorem. THEOREM 3.12: Let M = {S(C (Z n )) } be the MOD nturl neutrosophic finite complex number under MOD nturl C neutrosophic finite complex number zero domintin. ) M hs MOD nturl neutrosophic finite complex number subset idempotents. b) M hs MOD nturl neutrosophic finite complex number subset subsemiroups which re not idels. c) M hs MOD nturl neutrosophic finite complex number subset nilpotents nd zero divisors only if n is suitble composite number. Proof is direct nd hence left s n exercise to the reder.

133 132 MOD Nturl Neutrosophic Subset Semiroups Next we proceed onto describe MOD nturl neutrosophic dul number subset semiroup under MOD nturl neutrosophic dul number zero product tht is = = by some exmples. Exmple 3.35: Let S = {S(Z 14 ) } by the MOD nturl neutrosophic dul number subset semiroup under MOD nturl neutrosophic dul number zero product tht is = 9 = A = { } nd B = { } S. Clerly A B = {} is MOD nturl neutrosophic dul number subset zero divisor. Let C = { 2 4 } nd D = { } S; C D = { } is the MOD nturl neutrosophic dul number zero divisor pir which does not ive the rel zero. Thus these MOD nturl neutrosophic dul number subset semiroups hs pirs of zero divisors which my yield either {} or { }. nfct S hs severl MOD nturl neutrosophic dul number subsets which re rel nilpotent of order two or MOD nturl neutrosophic nilpotents of order two. Let A = {5} S A A = {}. Let B = { 4 } S B B = { }. Let D = {3 8} S; D D = {}. Let M = { } S. Clerly M M = { }.

134 MOD Nturl Neutrosophic Subset Semiroups 133 Findin zero divisors of the other form hppens to be difficult tsk. Let P = {collection of ll subsets from Z 14 } S; P is only MOD nturl neutrosophic dul number subset subsemiroup which is not n idel of S. B = {collection of ll subsets from Z 14 } S; B is only MOD nturl neutrosophic dul number subset subsemiroup which is not n idel of S. Similrly W = {collection of ll subsets from the set Z 14 } S is only MOD nturl neutrosophic dul number subset subsemiroup of S which is not n idel but W is lso zero squre semiroup s W W = {}. nfct for every b W b =. Let us ive one more exmple of the sitution before we proceed onto enumerte their relted properties. Exmple 3.36: Let V = {collection of ll subsets from the set S(Z 19 }; } be the MOD nturl neutrosophic dul number subset semiroup. This lso hs MOD nturl neutrosophic dul number subset zero divisor pirs A B V A B = {}; lso hs MOD nturl neutrosophic dul number subset MOD nturl neutrosophic zero pirs tht is A B V; with A B = { }. Finlly mixed MOD nturl neutrosophic dul number subset pir iven by A B = { }; A B V. We mke it cler tht even thouh 19 is prime number we see V hs MOD nturl neutrosophic dul number zero divisors nd MOD nturl neutrosophic dul number nilpotents of order two.

135 134 MOD Nturl Neutrosophic Subset Semiroups n view of ll these we hve the followin theorem. THEOREM 3.13: Let S = {S(Z n ); } be the MOD nturl neutrosophic dul number subset semiroup in which the product is MOD nturl neutrosophic dul number zero dominted tht is = =. Then the followin re true. i) S hs MOD nturl neutrosophic dul number subset rel zero divisor pirs. ii) S hs MOD nturl neutrosophic dul number subset MOD nturl neutrosophic dul number zero divisor pir; tht is A B ={ } for A B S. iii) S hs MOD nturl neutrosophic dul number subset subsemiroups which re not idels. iv) S hs MOD nturl neutrosophic dul number subset idempotents. v) S hs MOD nturl neutrosophic dul number subset subsemiroups which re zero squre semiroups. The proof is direct nd hence left s n exercise to the reder. Next we proceed onto describe the MOD nturl neutrosophic specil dul like number subset semiroups by exmples. Exmple 3.37: Let Z = {S(Z 1 h ) } be the MOD nturl neutrosophic specil dul like number subset semiroup. Let A = {5 + 2h 5h 3h B = {3h 2h + 4 h 2h h h 4 } nd h 5} Z;

136 MOD Nturl Neutrosophic Subset Semiroups 135 A B = {5 + 2h 3h 5h } {3h 2h + 4 h h h 4 h 4 = {h 9h 5h 2h 8h h 2h }. h 2h 5 h } This is the wy product opertion is performed on Z. Here = =. h h h Let A = {5h h } nd B = {2 2h 6h 8h 4h 6h + 2 8h h} Z. A B = {}; tht is the pir A B Z ives the MOD nturl neutrosophic specil dul like number subset rel zero divisor. Z. Let P = { h h h h h 6h } nd Q = { h 5 h h 5h h 5 5h } P Q = { h } so tht pir P Q Z contributes to the MOD nturl neutrosophic specil dul like number subset pir the product of which yields MOD nturl neutrosophic specil dul like number zero. T = {Z 1 } S is such tht T T = {Z 1 }. Thus T is MOD nturl neutrosophic specil dul like number subset idempotent of Z. Similrly R = {Z 1 h} is lso MOD nturl neutrosophic specil dul like number subset idempotent of Z. M = {Z 1 h} is lso MOD nturl neutrosophic specil dul like number subset idempotent of Z. However findin nilpotents of MOD nturl neutrosophic specil dul like number subset hppens to be chllenin one. Let N = {collection of ll subsets from Z 1 } Z be the MOD nturl neutrosophic specil dul like number subset subsemiroups of Z which is not n idel of Z.

137 136 MOD Nturl Neutrosophic Subset Semiroups B = {collection of ll subsets from the set Z 1 h} Z be the MOD nturl neutrosophic specil dul like number subset subsemiroup of Z which is not n idel. Exmple 3.38: Let B = {S(Z 17 h ); } be the MOD nturl neutrosophic specil dul like number subset semiroup. This B lso hs MOD nturl neutrosophic specil dul like number zero divisors s well s MOD nturl neutrosophic specil dul like number idempotents. Thus if T = {Z 17 } B. T is such tht T T = T; tht is T is MOD nturl neutrosophic specil dul like number idempotent subset of B. R = {Z 17 h} B is in MOD nturl neutrosophic specil dul like number idempotent subset of B s R R = R. Similrly W = {Z 17 h} B is in MOD nturl neutrosophic specil dul like number idempotent subset of B s W W = W. D ={collection of ll subsets from the set Z 17 } B is in MOD nturl neutrosophic specil dul like number subset subsemiroup of B which is not n idel of B. We see findin MOD nturl neutrosophic specil dul like number subsets of zero divisors nd nilpotents hppens to be difficult problem. Problem 3.4: Let M = {S(Z n h ) } be the MOD nturl neutrosophic specil dul like number subset semiroup under the MOD nturl neutrosophic specil dul like number zero dominted product tht is = =. h h h

138 MOD Nturl Neutrosophic Subset Semiroups 137 ) f n is prime does M contin MOD nturl neutrosophic specil dul like number subset pirs A B M such tht i) A B = {}. ii) A B = { h }. h iii) A B = { }. b) f n is prime cn M contin A such tht A A = {} A A = { h h } or A A = { }. Next we proceed onto describe the MOD nturl neutrosophic specil qusi dul number subset semiroup under MOD nturl neutrosophic zero product tht is = = by exmples. k k k Exmple 3.39: Let S = {(Z 12 k ) } be the MOD nturl neutrosophic subset specil qusi dul number semiroup under MOD nturl neutrosophic zero dominnt product = =. k k k Let A = {6k + 6 3k 6k 3 6} nd B = {4 4k 8k 8 + 4k 4 + 8k} S A B = {}. Let A = { k k k k k 6 6k 9 9k 9 } nd B = { k k k k k 4 4 4k 8k 8k 8k 4 } S; A B = { k } is the MOD nturl neutrosophic specil qusi dul number MOD nturl neutrosophic specil qusi dul zero. A = { k k 6 k 6 6k } nd B = { k 4 k 4 4k k 8k } S. k Clerly A B = { } is the mixed MOD nturl neutrosophic specil qusi dul number zero.

139 138 MOD Nturl Neutrosophic Subset Semiroups Let D = {Z 12 } be the subset of S; D D = D is MOD nturl neutrosophic specil qusi dul number subset idempotent of S. Let E = { Z 12 } be the subset of S E E = E is MOD nturl neutrosophic specil qusi dul number subset idempotent of S. Let F = {Z 12 k} be the subset of S F F = F is MOD nturl neutrosophic specil qusi dul number subset idempotent of S. Let G = {collection of ll subsets of Z 12 k} S be the MOD nturl neutrosophic specil qusi dul number subset subsemiroup of S nd not n idel of S. H = {collection of ll subsets of Z 12 k} S be the MOD nturl neutrosophic specil qusi dul number subset subsemiroup of S nd not n idel of S. Let K = { 6k 6 + 6k 6} S; K K = {} is MOD nturl neutrosophic specil qusi dul number nilpotent subset of S. Let L = { k k k 6 6k } S L L = { k } is the MOD nturl neutrosophic specil qusi dul number subset MOD nturl neutrosophic qusi dul number nilpotent. Exmple 3.4: Let B = {S(Z 7 k ) be the collection of ll MOD nturl neutrosophic specil qusi dul number subset } be the MOD nturl neutrosophic specil qusi dul number subset under the MOD nturl neutrosophic specil qusi dul number zero divisor product dominted semiroup tht is = k = k. k Let P 1 = { } B.

140 MOD Nturl Neutrosophic Subset Semiroups 139 Clerly P 1 P 1 = P 1 so P 1 is MOD nturl neutrosophic specil qusi dul number idempotent subset of B. Let P 2 = { k 2k 3k 4k 5k 6k} B. We see P 2 P 2 = P 2 so P 2 is MOD nturl neutrosophic specil qusi dul number idempotent subset of B. P 3 = { Z 7 } B is such tht P 3 P 3 = P 3 is the MOD nturl neutrosophic specil qusi dul number idempotent subset of B. P 4 = {Z 7 k } B is MOD nturl neutrosophic specil qusi dul number idempotent subset of B s P 4 P 4 = P 4. Now findin MOD nturl neutrosophic specil qusi dul number nilpotent subsets or MOD nturl neutrosophic specil qusi dul number subset zero divisor pirs yieldin {} or{ k } or { k } hppens to be difficult tsk. Further M = {collection of ll subsets of Z 7 } B is MOD subset nturl neutrosophic specil qusi dul number subsemiroup of B. Let N = {collection of ll subsets of Z 7 k} B; N is MOD nturl neutrosophic specil qusi dul number subset subsemiroup of B. T = {collection of ll subsets from the set Z 7 k} is the MOD nturl neutrosophic specil qusi dul number subset subsemiroup of B. None of them re idels of B. n view of ll these we propose the followin problem. Problem 3.5: Let S = {{collection of ll subsets from Z p k } = S(Z p k ); } be the MOD nturl neutrosophic specil qusi dul number

141 14 MOD Nturl Neutrosophic Subset Semiroups subset semiroup under MOD nturl neutrosophic specil qusi dul number zero dominnt product opertion. Tht is = k = k. i) Cn S hve MOD nturl neutrosophic specil qusi dul number nilpotent subsets? ii) Cn S hve MOD nturl neutrosophic specil qusi dul number subset pirs A B such tht A B = {} or A B = { k } or A B = { k }? n view of ll these we hve the followin result. THEOREM 3.14: Let G = {collection of ll subsets from the set S(Z n k ) } be the MOD nturl neutrosophic specil qusi dul number subset under MOD nturl neutrosophic specil qusi dul number zero dominnt product semiroup. i) G hs MOD nturl neutrosophic specil qusi dul number idempotent subsets for ll 2 n <. ii) G hs MOD nturl neutrosophic specil qusi dul number subset subsemiroup which re not idels; 2 n <. iii) G hs MOD nturl neutrosophic specil qusi dul number subsets A in G such tht A A = {} or A A = k k { } or A A = { } nilpotents of order two; n n pproprite composite number. iv) G hs MOD nturl neutrosophic specil qusi dul number subset pirs A B G such tht they re MOD nturl neutrosophic specil qusi dul number zero k divisors tht is A B = {} or A B = { } or A B = k { }; for ll composite number n; n 2. Proof is direct nd hence left s n exercise to the reder. Next we proceed onto describe by exmples the MOD nturl neutrosophic neutrosophic subset semiroups under product in which = = or. k

142 MOD Nturl Neutrosophic Subset Semiroups 141 Exmple 3.41: Let M = {S(Z 16 ) be the collection of ll subsets of Z 16 } be the MOD nturl neutrosophicneutrosophic subset semiroup under the MOD nturl neutrosophic-neutrosophic subset zero dominnt product tht is = Let A = { B ={ } M. 4 4 A B = { = { } nd 8 } { } this the wy product opertion is performed on M. Let A = { A B = { } nd B = {4 4 8} M. } {4 4 8} = { } 8 } M is the MOD nturl neutrosophic - neutrosophic mixed zero divisor in M. Let T = { } M. T T = {}; thus T is MOD nturl neutrosophic neutrosophic nilpotent subset of order two. Let P = { }M; clerly P P = { }; tht is P is MOD nturl neutrosophic - neutrosophic subset nilpotent of MOD nturl neutrosophic - neutrosophic zero s P P = { }. Study in this direction is innovtive nd interestin.

143 142 MOD Nturl Neutrosophic Subset Semiroups B = {Z 16 } M is the MOD nturl neutrosophicneutrosophic subset idempotent of M s B B = B. Further A = {Z 16 } M is in MOD nturl neutrosophic - neutrosophic subset idempotent of M s A A = A. Thus M hs MOD nturl neutrosophic - neutrosophic subset nilpotents nd idempotents. Now W = {collection of ll subset of Z 16 } M is MOD nturl neutrosophic - neutrosophic subset subsemiroup of M but is not n idel of M. V = {collection of ll subsets from the set Z 16 } M is lso MOD nturl neutrosophic - neutrosophic subset subsemiroup of M which is not n idel. Let D = {collection of ll subsets of { t ; t Z 16 is either nilpotent or idempotent or zero divisor in Z 16 } M is MOD nturl neutrosophic subset subsemiroup which is n idel of M. nfct this idel D of M is defined s the pure MOD nturl neutrosophic - neutrosophic idel of M. Exmple 3.42: Let G = {{collection of ll subsets from the set Z 11 } } be the MOD nturl neutrosophic - neutrosophic subset semiroup under the MOD nturl neutrosophic neutrosophic zero dominnt product tht is = = Findin nontrivil MOD nturl neutrosophic - neutrosophic subset nilpotents is chllenin tsk. For in Z 2..

144 MOD Nturl Neutrosophic Subset Semiroups 143 P = {1 + } Z 2 is such tht P P = {}. However B 1 = {Z 11 } G is MOD nturl neutrosophic - neutrosophic idempotent subset of G. B 2 = {Z 11 } G is in MOD nturl neutrosophic - neutrosophic idempotent subset of G. B 3 = {Z 11 } G is in MOD nturl neutrosophic - neutrosophic idempotent of G. Further P 1 = {collection of ll subsets from the set Z 11 } G 1 is MOD nturl neutrosophic - neutrosophic subsemiroup of G which is not n idel of G. P 2 = {collection of ll subsets from the set Z 11 } G is lso MOD nturl neutrosophic - neutrosophic subsemiroup of G which is not n idel of G. P 3 = {collection of ll subsets from the set Z 11 } G is MOD nturl neutrosophic - neutrosophic subset subsemiroup of G which is not n idel of G. Thus we propose the followin problem. Problem 3.6: Let K = {collection of ll subsets from the set Z p ; p prime } be the MOD nturl neutrosophic - neutrosophic semiroup under product dominted by MOD nturl neutrosophic - neutrosophic zero ; ( = = ). i) Cn K contin non trivil MOD nturl neutrosophic - neutrosophic nilpotents subsets A in K such tht A A = {} or A A = { } or A A = { }? ii) Cn K contin nontrivil MOD nturl neutrosophicneutrosophic zero divisor subset pirs A B K

145 144 MOD Nturl Neutrosophic Subset Semiroups such tht A B = {} or A B = { } or A B = { }? iii) Cn P = { t / t Z p where t is either nilpotent or n idempotent or zero divisor of Z p } be n idel such tht o(p) > p + 1? Now we proceed onto express the followin result. THEOREM 3.15: Let S = {S(Z n ) } be the MOD nturl neutrosophic - neutrosophic subset semirin under the MOD nturl neutrosophic - neutrosophic zero dominted product (i.e. = = ). Then the followin re true. i) S hs MOD nturl neutrosophic - neutrosophic subset idempotents. ii) S hs MOD nturl neutrosophic neutrosophic subsets A S such tht A A = {} or A A = { } or A A = { } i.e. nilpotent subsets of order two (n n pproprite composite number). iii) S hs MOD nturl neutrosophic neutrosophic subset pir A B in S such tht A B = {} or A B = { } or A B = { } (For n - composite number). iv) S hs MOD nturl neutrosophic neutrosophic subset subsemiroups which re not idels. v) S hs MOD nturl neutrosophic neutrosophic subset idel iven by P = { / t Z n where t is n idempotent or nilpotent or zero divisor or elements of the form m = m positive inteer. Proof is direct nd hence left s n exercise to the reder. Next we proceed onto propose some problems for the interested reder. t

146 MOD Nturl Neutrosophic Subset Semiroups 145 Problems 1. Find ll MOD nturl neutrosophic subset subsemiroups of H = {S( Z 24 ) +} where H is the MOD nturl neutrosophic subset semiroup under +. i) Find o(h). ii) Find ll MOD nturl neutrosophic subset universl pirs of H. iii) Cn H hve MOD nturl neutrosophic subset idempotents? 2. Let G = {S( Z 16 ) +} be the MOD nturl neutrosophic subset semiroup under +. i) Find o(g). ii) How mny MOD nturl neutrosophic subset pirs A B G with A + B = Z 16? iii) Find ll MOD nturl neutrosophic subset idempotent of G. iv) Find ll MOD nturl neutrosophic subset subsemiroups of G. 3. Let K = {S( Z 19 ) +} be the MOD nturl neutrosophic subset semiroup under +. i) Study questions (i) to (iv) of problem (2) for this K. ii) For iven A is K how mny B K exists such tht A + B = Z Enumerte ll specil nd interestin fetures ssocited with S = {S( Z n ) +} the MOD nturl neutrosophic subset semiroup under Let W ={S(C (Z 1 )) +} be the MOD nturl neutrosophic finite complex number subset semiroup under +.

147 146 MOD Nturl Neutrosophic Subset Semiroups i) Study questions (i) to (iv) of problem (2) for this W. ii) Compre W with V = {S( Z 1 ) +} s MOD nturl neutrosophic subset semiroups. iii) Prove o(v) < o(w). 6. Let M = {S(C (Z 23 )) +} be the MOD nturl neutrosophic finite complex number subset semiroup under +. i) Study questions (i) to (iv) problem (2) for this M. ii) Study the specil nd distinct fetures enjoyed by M. 7. Let N = {S(C (Z 243 )) +} be the MOD nturl neutrosophic subset finite complex number semiroup under +. i) Study questions (i) to (iv) of problem (2) for this N. ii) Compre N with T = {S( Z 243 ) +} nd derive the specil fetures ssocited with N. 8. Let W = {S(Z 15 ) +} be the MOD nturl neutrosophic- neutrosophic subset semiroup under +. i) Study questions (i) to (iv) of problem (2) for this W. ii) Compre W with N = {S( (Z 15 )) +} nd M = {S( Z 15 ) +}. iii) Will o(n) > o(w) or o(w) > o(n)? 9. Let B = {S(Z 43 ) +} be the MOD nturl neutrosophic -neutrosophic subset semiroup under +. Study questions (i) to (iv) of problem (2) for this B. 1. Let D = {S(Z 64 ) +} be the MOD nturl neutrosophic - neutrosophic subset semiroup under +.

148 MOD Nturl Neutrosophic Subset Semiroups 147 Study questions i) to (iv) of problem (2) for this D. 11. Obtin ll specil nd distinct fetures enjoyed by S ={(Z n ) +}. 12. Let E = {S(Z 9 ) +} be the MOD nturl neutrosophic dul number subset semiroup under +. i) Study questions (i) to (iv) of problem (2) for this E. ii) f A = { } E find ll B in E 3 5 such tht A + B = Z Let Z = {S(Z 43 ) +} be the MOD nturl neutrosophic subset dul number semiroup under +. i) Study questions (i) to (iv) of problem (2) for this Z. ii) Compre Z with P = {S( Z 43 ) +} the MOD nturl neutrosophic subset semiroup. iii) Compre Z with T = {S(Z 43 ) +} the MOD nturl neutrosophic-neutrosophic subset semiroup iv) Compre Z with S = {S(C (Z 43 ) +} the MOD nturl neutrosophic finite complex number subset semiroup. v) Let A ={ } Z; how mny B in Z re such tht A + B = Z 43.? 14. Let F = {S(Z 48 ) +} be the MOD nturl neutrosophic dul number subset semiroup under +. i) Study questions (i) to (iv) problem (2) for this F. ii) Cn we sy if in Z n n is composite number S(Z n ) hs more number of MOD nturl neutrosophic dul number subset subsemiroups?

149 148 MOD Nturl Neutrosophic Subset Semiroups 15. Does H = {S(Z n ) +} enjoy ny other specil fetures thn S = {S(Z n +} or M = {S( Z n ) +} or N = {S(C (Z n )) +}? 16. Let T = {S(Z 8 h ) +} be the MOD nturl neutrosophic specil dul like number subset semiroup under +. i) Study questions (i) to (iv) of problem (2) for this T. ii) Obtin ll the specil nd distinct fetures enjoyed by T. h h iii) Let P ={7h 2 4h h 6h h + h 2 2 5h h} T. Find ll Q in T such tht P + Q = Z 8 h. 17. Let Y = {S(Z 11 h ) +} be the MOD nturl neutrosophic specil dul like number subset semiroup under +. i) Study questions (i) to (iv) of problem (2) for this Y. ii) Compre T in problem (16) with this Y. 18. Let R = {S(Z 12 h ) +} be the MOD nturl neutrosophic specil dul like number subset semiroup under +. i) Study questions (i) to (iv) of problem (2) for this R. ii) Compre this R with T nd Y of problems (16) nd (17) respectively. 19. Obtin ll specil nd distinct fetures ssocited with M = {S(Z n h ) +} the MOD nturl neutrosophic specil dul like number subset semiroup under Let S = {S(Z 4 k ) +} be the MOD nturl neutrosophic specil qusi dul number subset semiroup under +.

150 MOD Nturl Neutrosophic Subset Semiroups 149 i) Study questions (i) to (iv) of problem (2) for this S. ii) Compre S with T = {S( Z ) +}. 21. Let B = {S(Z 23 k ) +} be the MOD nturl neutrosophic specil qusi dul number subset semiroup under +. i) Study questions (i) to (iv) of problem (2) for this B. ii) Compre this B with S in problem (2). 22. Let M = {S(Z 1 k ) +} be the MOD nturl neutrosophic specil qusi dul number subset semiroup under +. i) Study questions (i) to (iv) of problem (2) for this M. k k ii) Let A = {5 + 5k 5 + 6k + 8 k k 2k 4 4 4k k 8k k} M how mny subsets B in M exist such tht A + B = Z 1 k. iii) Find the lest order of B in (ii) such tht A + B = Z 1 k. iv) Does there exist more thn one such B with lest order? 23. Enumerte ll specil fetures ssocited with S = {S(Z n k ) +}. 24. Let P = {S( Z 12 ) } be the MOD nturl neutrosophic subset semiroup under zero dominted product tht is 12 =. i) Find o(p). ii) How mny MOD nturl neutrosophic subset idempotents re in P? iii) How mny MOD nturl neutrosophic subset nilpotents re in P? iv) Find ll MOD nturl neutrosophic subsets zero divisor pirs of P. 14

151 15 MOD Nturl Neutrosophic Subset Semiroups v) Find ll MOD nturl neutrosophic subset subsemiroups of P which re not idels of P. vi) Find ll MOD nturl neutrosophic subset idels of P. vii) Find for ny A P the totl number of B in P such tht A B = Z 12 ; the MOD nturl neutrosophic universl subset of P. viii) Cn we sy for every Z P there exists t lest one Y P with Z Y = Z? 25. Let B = {S( Z 19 ) } be the MOD nturl neutrosophic subset semiroup which is rel zero product dominted tht is 19 = 19 =. 12 i) Study questions (i) to (viii) of problem (24) for this B. ii) Cn we hve for this B MOD nturl neutrosophic nilpotent subsets nd MOD nturl neutrosophic zero divisor subset pirs? 26. Let W = {S( Z 64 ) } be the MOD nturl neutrosophic subset semiroup under rel zero dominted product tht is =. 64 Study questions (i) to (viii) of problem (24) for this W. 27. Let V = {S(C (Z 6 )) } be the MOD nturl neutrosophic subset semiroup under zero dominted product =. C i) Study questions (i) to (viii) of problem 24 for this V. ii) Obtin ll specil nd distinct fetures ssocited with V. iii) Compre V with T = {S( Z 6 ) } the MOD nturl neutrosophic subset semiroup under in which

152 MOD Nturl Neutrosophic Subset Semiroups =. 28. Let N = {S(C (Z 43 )) } be the MOD nturl neutrosophic finite complex number subset semiroup under rel zero dominnt product tht is =. i) Study questions (i) to (viii) of problem (24) for this N. ii) Compre N with V of problem (27). 29. Enumerte ll specil nd interestin fetures ssocited with S = {S(C (Z n )) } the MOD nturl neutrosophic finite complex number subset semiroup under rel zero dominnt product tht is C =. 3. Let W = {S(Z 13 ) } be the MOD nturl neutrosophic neutrosophic subset semiroup under rel zero dominnt product tht is =. i) Study questions (i) to (viii) of problem (24) for this W. ii) Obtin ny other specil feture ssocited with W. iii) Will P = { C t where t Z 13 nd t is zero divisor or nilpotent or n idempotent } W be MOD nturl neutrosophic - neutrosophic subset idel of W. 31. Let D = {S(Z 24 ) } be the MOD nturl neutrosophic - neutrosophic subset semiroup under the rel zero dominted product. i) Study questions (i) to (viii) of problem (24) fo this D. ii) Compre this D with W in problem (3). 32. Let E = {S( 6 5 Z ) } be the MOD nturl neutrosophic - neutrosophic subset semiroup under the rel zero dominted product. C C

153 152 MOD Nturl Neutrosophic Subset Semiroups i) Study questions (i) to (viii) of problem (24) for this E. ii) Compre this E with D nd W of problem (31) nd (3) respectively. 33. Study ll specil nd distinct fetures enjoyed by MOD nturl neutrosophic - neutrosophic subset semiroup under rel zero dominted product. 34. Let S = {S((Z 1 ) } be the MOD nturl neutrosophic dul number subset semiroup under rel zero dominnt product tht is = =. i) Study questions (i) to (viii) of problem (25) for this S ii) Cn we sy S hs more number of MOD nturl neutrosophic dul number subset nilpotents A such tht A A = {} or A A = { } or A A = { }? iii) Cn we conclude S will hve more number of MOD nturl neutrosophic dul number subsets (A B) zero divisor pirs; tht is A B = {} or A B = { } or A B = { }? iv) Compre S with D = {S( nd F = {S(Z 1 }. 1 Z ) } E = {S(C (Z 1 )) } 35. Let R = {S(Z 43 } be the MOD nturl neutrosophic dul number subset semiroup under rel zero dominnt product. i) Study questions (i) to (viii) of problem (24) for this R. ii) Compre R with S of problem (34). 36. Enumerte ll specil nd distinct fetures ssocited with MOD nturl neutrosophic dul number subset semiroup P under rel zero dominnt product where P = {S(Z n ) }.

154 MOD Nturl Neutrosophic Subset Semiroups Let M = {S(Z 14 h ) } be the MOD nturl neutrosophic specil dul like number subset semiroup under rel zero dominnt product =. i) Study questions (i) to (viii) of problem (24) for this M. ii) Compre M with P ={S( Z 14 ) } the MOD nturl neutrosophic subset semiroup. iii) Compre M with V ={S(Z 14 ) } the MOD nturl neutrosophic dul number subset semiroup. iv) Compre M with W = {S(Z 14 ) } the MOD nturl neutrosophic - neutrosophic subset semiroup. 38. Let F = {S(Z 23 h ) } be the MOD nturl neutrosophic specil dul like number subset semiroup under rel zero dominted product. i) Study questions (i) to (viii) of problem (24) for this F. ii) Compre F with M of problem (37). h 39. Describe ll the distinct fetures enjoyed by MOD nturl neutrosophic specil dul like number subset semiroup S under rel zero dominnt product where S ={S(Z n h ) }. 4. Let T ={S(Z 4 k ) } be the MOD nturl neutrosophic specil qusi dul number subset semiroup under rel zero dominnt product i.e. =. k i) Study questions (i) to (viii) of problem (24) for this T. ii) Compre this T with P = {S(C (Z 4 )) } the MOD nturl neutrosophic finite complex number subset semiroup with rel zero dominted product.

155 154 MOD Nturl Neutrosophic Subset Semiroups iii) Compre T with W = {S(Z 4 ) } the MOD nturl neutrosophic dul number subset semiroup under the rel zero dominnt product. 41. Let G = {S(Z 17 k ) } be the MOD nturl neutrosophic specil qusi dul number subset semiroup under the rel zero dominted product. i) Study questions (i) to (viii) of problem (24) for this G. ii) Compre this G with T of problem (4). 42. Let H = {S(Z 28 k ) } be the MOD nturl neutrosophic specil qusi dul number subset semiroup under rel zero dominted product. i) Study questions (i) to (viii) of problem (24) for this H. ii) Compre this H with G nd T of problems (41) nd (4) respectively. 43. Let J = {S(Z n k ) } be the MOD nturl neutrosophic specil qusi dul number subset semiroup under rel zero product dominted semiroup. Enumerte ll specil nd distinct fetures enjoyed by J nd compre it with H = {S(Z n ) } K = {S(C (Z n )) } nd L = {S(Z n ) }. 44. Let G = {S( Z 2 ) } be the MOD nturl neutrosophic subset semiroup under MOD nturl neutrosophic zero dominted product. i) Study questions (i) to (viii) of problem (24) for this G. ii) Compre G with the MOD nturl neutrosophic product semiroup which is defined under rel zero product.

156 MOD Nturl Neutrosophic Subset Semiroups Let S = {S( Z 47 ) } be the MOD nturl neutrosophic subset semiroup under the MOD nturl neutrosophic zero dominted product tht is =. i) Study questions (i) to (viii) of problem (24) for this S. ii) f A = { } S find B in S so tht A B = Z 47 the MOD universl subset of S. ) How mny such B S? b) Wht is the lest order of such B? c) Wht is the retest order of such B? 46. Let P = {( Z 64 ) } be the MOD nturl neutrosophic subset semiroup which is MOD nturl neutrosophic zero dominted = i) Study questions (i) to (viii) of problem (24) for this P. ii) Find ll MOD nturl neutrosophic mixed zero divisors of P iii) s A { } = { } for A in P? 47. Let K = {S( Z n ) } be the MOD nturl neutrosophic subset semiroup under the MOD nturl neutrosophic zero dominted product. i) Study ll specil fetures enjoyed by K. ii) f the product is under rel zero dominted product compre both the semiroups for sme S( Z ). 48. Let W = {S(C (Z 48 )) } be the MOD nturl neutrosophic finite complex number subset semiroup under the MOD nturl neutrosophic zero dominted product n

157 156 MOD Nturl Neutrosophic Subset Semiroups i) Study questions (i) to (viii) of problem (24) for this W. C C C C ii) f P = { i F 24 4i 1i iii) C C C i + F 14 12i } W find Q W such F tht P Q = C (Z 48 ) the MOD nturl neutrosophic finite complex number universl subset of W. For tht iven P W how mny such Q W such tht P Q = C (Z 48 ) exist? 49. Obtin ll specil nd interestin fetures enjoyed by S = {S(C (Z n )) } the MOD nturl neutrosophic finite complex number subset semiroup under MOD nturl neutrosophic zero dominted product. 5. Let G = {S(Z 15 ) } be the MOD nturl neutrosophic - neutrosophic subset semiroup under the MOD nturl neutrosophic - neutrosophic zero dominted product =. i) Study questions (i) to (viii) of problem (24) for this G. ii) f W = { } G find V G such tht 9 W V = Z 15 the MOD nturl neutrosophic - neutrosophic universl subset of G. iii) How mny such V s exist for this iven W? 51. Let H = {S(Z 47 ) } be the MOD nturl neutrosophic - neutrosophic subset semiroup under the MOD nturl neutrosophic neutrosophic zero dominted product. Study questions (i) to (viii) of problem (25) for this H. 52. Let F = {S( Z 1 ) } be the MOD nturl 5 neutrosophic - neutrosophic subset semiroup under the F F

158 MOD Nturl Neutrosophic Subset Semiroups 157 MOD nturl neutrosophic - neutrosophic zero dominted product. i) Study questions (i) to (viii) of problem (24) for this F. ii) Compre F with H nd G of problems (51) nd (5) respectively. 53. Describe ll specil fetures ssocited with MOD nturl neutrosophic - neutrosophic subset semiroup under the MOD nturl neutrosophic zero dominted product M = {S(Z n ) }. Compre M with P = {S(Z n ) } where P enjoys the rel zero dominted product tht is = in M. 54. Let B = {S(Z 18 ) } be the MOD nturl neutrosophic dul number subset semiroup under the MOD nturl neutrosophic dul number zero dominted product tht is =. i) Study questions (i) to (viii) of problem (24) for this B. ii) Let T = { iii) 5 7 1} B. Find V in B so tht T B = (Z 18 the MOD universl nturl neutrosophic dul number subset of B. For this iven T how mny such B s exist. 55. Let W = {S(Z 128 ) } be the MOD nturl neutrosophic dul number subset semiroup under the MOD nturl neutrosophic dul number zero dominted product tht is =. i) Study questions (i) to (viii) of problem (24) for this W. ii) Compre the W with B in problem (54).

159 158 MOD Nturl Neutrosophic Subset Semiroups 56. Let C = {S(Z 53 ) } be the MOD nturl neutrosophic dul number subset semiroup under the MOD nturl neutrosophic dul number zero dominted product tht is =. i) Study questions (i) to (viii) of problem 24 for this C. ii) Compre C with W nd B in problems 54 nd 55 respectively. 57. Let S ={S(Z n ) } be the MOD nturl neutrosophic dul number subset semiroup under the MOD nturl neutrosophic dul zero number dominted product. Wht re the specil fetures enjoyed by S s 2 = nd = for ny r s Z r s 58. Let M = {S(Z 8 h ) } be the MOD nturl neutrosophic specil dul like number subset semiroup under the MOD nturl neutrosophic specil dul like number zero dominnt product tht is n = i) Study ll specil fetures ssocited with M. ii) Study questions (i) to (viii) of problem (24) for this M. h h iii) f H = {6 + 2h h + 2 4h } M find K in M such tht H K = Z 8 h ) the MOD nturl neutrosophic specil dul like number universl subset. How mny such K in M exists? 59. Let W ={S(Z 2 h ) } be the MOD nturl neutrosophic specil dul like number subset semiroup under the product =. h h h. Study questions (i) to (viii) of problem (24) for this W. h

160 MOD Nturl Neutrosophic Subset Semiroups Enumerte ll specil nd distinct fetures enjoyed by h h S = {S(Z n h ) } = the MOD nturl neutrosophic specil dul like number subset semiroup. Compre S with R = {S(Z n ) }; = the MOD nturl neutrosophic dul number subset semiroup. 61. Let S = {S(Z 27 k ) } be the MOD nturl neutrosophic qusi dul number subset semiroup under the MOD nturl neutrosophic specil qusi dul number zero dominted product tht is. i) Study questions (i) to (viii) of problem (24) for this S. ii) Compre S with P = {S(Z 27 h ) } nd R = {S(Z 27 ) }. 62. Let T = {S(Z 5 k } be the MOD nturl neutrosophic specil qusi dul number subset semiroup under the product of. k k i) Study questions (i) to (viii) of problem (24) for this T. ii) Compre S in problem (61) with this T. iii) f A = { k 2k k k k k k 2 3 1} T find ll B T such tht A B = Z 5 k the MOD nturl neutrosophic specil qusi dul number universl subset. 63. Let V = {S(Z 4 k } be the MOD nturl neutrosophic specil qusi dul number subset semiroup. i) Study questions (i) to (viii) of problem (24) for this V.

161 16 MOD Nturl Neutrosophic Subset Semiroups ii) iii) Compre V with W = {S(Z 4 ) } the MOD nturl neutrosophic specil qusi dul number subset semiroup. f D = {Z 4 } V does there exists E V such tht D E = Z 4 k the MOD nturl neutrosophic specil qusi dul number universl subset of V. 64. Let S = {S(Z n k ) } be the MOD nturl neutrosophic specil qusi dul number subset semiroup under the product =. k k i) f B = {Z n k} S find ll A in S with A B = Z n k. ii) f T = {Z n T} find ll Q S with Q T = Z n k. iii) f V = Z n k find ll W in S with V W = Z n k the MOD nturl neutrosophic specil qusi dul number universl subset of S.

162 Chpter Four SEMGROUPS BULT ON MOD SUBSET MATRCES AND MOD SUBSET POLYNOMALS n this chpter for the first time we introduce the new notion of MOD subset mtrices of two types. Likewise we introduce the notion of MOD subset polynomils. Let S(Z n ) = {collection of ll subsets of Z n }. M = ( ij ) st where ij S(Z n ); 1 i s 1 j t} denotes the collection of ll MOD subset s t mtrices ssocited with S(Z n ). We will illustrte this sitution by some exmples. Exmple 4.1: Let M = { A B C D / A B C D S(Z 6) be the collection of ll 2 2 MOD subset mtrices with entries from S(Z 6 ).

163 162 MOD Nturl Neutrosophic Subset Semiroups Let P = {3 2} {12} {1234} {5} M. R = {1} {2} {2} {53} M. Exmple 4.2: Let S = {( ) / i S(Z 1 ); 1 i 4} be the MOD subset row mtrix. x = ({ 1 3} { } { } {1 2}) S y = ({1} {} {Z 1 } {4 2}) S. Exmple 4.3: Let b1 b 2 T = { b 3 b4 b j S(Z 42 ); 1 j 4} be the MOD subset column mtrix with entries from S(Z 42 ). A = {4 284} {124 25} {3 462} {Z 42} nd B = {1} {3} {5} {41} T. Exmple 4.4: Let W = i S(Z 19 ); 1 i 8} be the MOD subset mtrix.

164 Semiroups built on MOD Subset Mtrices 163 B = C = {234} {14} {1237} {Z 19} {1818} {1} {1 215} {3165} {} {3} {Z 19} {2} {1} {4} {61} {6 4} W. W. Now hvin seen exmples of MOD subset mtrices we proceed onto show how opertions of + nd re performed on them. We will describe these situtions by exmples. Exmple 4.5: Let A = i S(Z 1 ); 1 i 3} be the MOD subset mtrix collection. Let x = {32} {5} {742} nd y = {157} {3} {92} A. x + y = {32} {5} {742} + {157} {3} {92} = { } {8} A. {96431}

165 164 MOD Nturl Neutrosophic Subset Semiroups This is the wy + opertion is performed on A. Let S = {5} {52} {Z 1} nd R = {5} {583} {175} A. S + R = {5} {52} {Z 1} + {5} {583} {175} = {5} {7385} A. {Z 1} Thus we see {A +} is only MOD subset mtrix semiroup nd is not roup. However we cn prove A is Smrndche semiroup. n view of ll these we prove the followin theorem. THEOREM 4.1: Let M = {(m ij ) ts / m ij S(Z n ); 1 i t 1 j s +} be MOD subset mtrix semiroup under +. i) o(m) < ii) M is Smrndche semiroup iii) M is monoid s () M nd () + A = A + () = A. Proof is direct nd hence left s n exercise to the reder. We now describe MOD finite complex number subset mtrix sets by some exmples. Exmple 4.6: Let B = {( ) / i S (C(Z 12 )); 1 i 6}

166 Semiroups built on MOD Subset Mtrices 165 be the MOD finite complex number subset mtrix set. M = ({ 1 3 5} {6 7 1} {11} { } {8 9 11} {2}) B is MOD finite complex number subset of B. We cn s in cse of S(Z n ) the MOD subset mtrices define the opertion + on B. Let = ({ 21} {78} {9} {2 4 3} {5} {3i F }) nd b = ({2i F + 8} {} {1} {3} {1 2 4i F } {6 + i F 1}) B. To find + b + b = ({ 1 2} {7 8} {9} {2 4 3} {5} {3i F }) + ({2i F + 8} {} {1} {3} {1 2 4i F } {6 + i F 1}) = ({2i F + 8 2i F + 9 2i F + 9 2i F + 1} {7 8} {1} {6 7 5} { i F } {6 + 4i F 1 + 3i F }). This is the wy + opertion is performed on B. Clerly {B +} is only finite commuttive monoid s ({} {} {} {} {} {}) cts s the dditive identity. Exmple 4.7: Let S = i S(C(Z 7 )); 1 i 5} be the MOD finite complex number subset column mtrix.

167 166 MOD Nturl Neutrosophic Subset Semiroups Let x = { if 62 6i F} {1 i F4} {235} {} {6 2i F55i F} nd y = {23} {4 i F3} {3}. {{2 i F3 4i F3} {2 5i F2 2i F} We now define + on S s follows: x + y = { if 62 6i F} {1 i F4} {235} {} {6 2i F55i F} + {23} {4 i F3} {3} {{2 i F3 4i F3} {2 5i F2 2i F} = {23i F 1i F 24 6i F5 6i F} {1 i F4 i F5 2i F} {561} {{2 i F33 4i F} {15i F2 3i F1 4i F2i F2} S. This is the wy + opertion is performed on S. Thus {S +} is finite commuttive monoid. However findin the order of S is chllenin tsk for enerl C(Z n ); 2 n <. Next we proceed onto describe the MOD neutrosophic subset mtrix sets under the + opertion on them by some exmples. Exmple 4.8: Let

168 Semiroups built on MOD Subset Mtrices 167 P = i S(Z 4 ); 1 i 8} be the MOD neutrosophic subset mtrix set. Let A = { 23} {1 2} {} { 3} { 2 3} {1 22} {3} {1 3 2} nd B = {1 2} {3} {1 23} { 2} {} {2} {1 312} {2 2} A + B = { 23} {1 2} {} { 3} { 2 3} {1 22} {3} {1 3 2} + {1 2} {3} {1 23} { 2} {} {2} {1 312} {2 2} {21 3 1} {1 21 } {1 23} { } = {2 3} {2 23 2} {13} { } P. This is the wy the sum of two MOD subset neutrosophic mtrices re obtined.

169 168 MOD Nturl Neutrosophic Subset Semiroups A + A = { 23} {1 2} {} { 3} { 2 3} {1 22} {3} {1 3 2} + { 23} {1 2} {} { 3} { 2 3} {1 22} {3} {1 3 2} = { 231} {21 32} {} {2 2} {22} {21} {2} { } This is the wy + opertion is performed on P. We cn prove in enerl if S(M) = {Collection of ll m t mtrices with entries from S(Z n )} then S(M) under the opertion of + is MOD neutrosophic subset mtrix semiroup of finite order. Clerly S(M) is not roup. Exmple 4.9: Let S(N) = i S(Z 6 ) 1 i 4 +} be the MOD neutrosophic subset mtrix semiroup under +. Let A = {32 4} {125 1} {3 3} {4 23 5} nd B = {42} {3} S(N). {2} {4 4}

170 Semiroups built on MOD Subset Mtrices 169 A + B = {32 4} {125 1} {3 3} {4 23 5} + {42} {3} {2} {4 4} = { } { } {25 3} {21 3} S(N). This is the wy + opertion is performed on S(N). Let {} = {} {} {} {} S(N) clerly A + {} = {} + A = A for ll A S(N). Given AS(N) it is not esy to find the inverse infct inverse of A under + opertion does not exist in enerl for every A S(N). Let T = {3} {2} {5} {} S(N) clerly P = {3} {4} {} {5} S(N) is such tht T + P = {3} {2} {5} {} + {3} {4} {} {5} = {} {} {} {}

171 17 MOD Nturl Neutrosophic Subset Semiroups Thus P is the MOD neutrosophic subset mtrix inverse of T in S(N). All Y S(N) my not hve inverse ssocited with it. For the Z = {3 2} {4 2 1} {2 44} { } S(N) we cnnot find Y in S(N) such tht Z + Y = {} {}. {} {} Hence we my not be ble to find inverse for every Y S(N). Thus {S(N) +} is only MOD neutrosophic subset semiroup of finite order nd is commuttive. Exmple 4.1: Let S(T) = {( ) / i S(Z 11 ); 1 i 3 +} be the MOD neutrosophic subset semiroup under +. ({}) = ({} {} {}) is the dditive identity in S(T) for A + ({}) = ({}) + A = A for ll A S(T). Thus S(T) is finite commuttive monoid. Let P = ({ } {} {2}) nd Q = ({ } { } { }) S(T)

172 Semiroups built on MOD Subset Mtrices 171 P + Q = ({ } {} {2}) + ({ } { } {3+4 7}) = ({ } { } { }) S(T). This is the wy + opertion is performed on S(T). Next we proceed onto ive exmples of MOD dul number subset mtrices nd describe their properties. Exmple 4.11: Let S(W) = i S(Z 1 ); 1 i 4} be the MOD dul number subset squre mtrices. B = { } { } { } {5137 2} S(W). ({}) = {} {} {} {} is defined s the MOD dul number subset identity mtrix with respect to ddition opertion +. Let M = {3 89 2} {4 86 1} {3 4 1} {5 42 3} nd N = { } {3 15} {123} {7 26 1} S(W)

173 172 MOD Nturl Neutrosophic Subset Semiroups M + N = {3 89 2} {4 86 1} {3 4 1} {5 42 3} + { } {3 15} {123} {7 26 1} = { { } 3 4 1} { { } 58 4} This is the wy + opertion is performed on S(W). Thus {S(W) +} is MOD dul number subset semiroup under +. Exmple 4.12: Let S(V) = i S(Z 3 ) 1 i +} be the MOD dul number subset semiroup under +; S(V) is of finite order. The tsk of findin the order of S(V) is left s n exercise to the reder. ({}) = {} {} {} {} S(V) {} {}

174 Semiroups built on MOD Subset Mtrices 173 cts s the dditive identity of S(V). For ll G S(V). G + {} = {} = {} + G = G. Let A = {1 22 } { 22 12} { 2 2} {1 1} {11 2} {2 2} nd B = { 21 2} {1 2 } {12} { 2 1} {1 2} {1} S(V). Consider A + B = {1 22 } { 22 12} { 2 2} {1 1} {11 2} {2 2} + { 21 2} {1 2 } {12} { 2 1} {1 2} {1} = { { } 2 11 } {122 { } } { { } 1 2} S(V). This is the wy + opertion is performed on S(V).

175 174 MOD Nturl Neutrosophic Subset Semiroups S(V) is only MOD dul number semiroup s every Z S(V) in enerl does not hve n inverse. B = { } {} {} {} {} {} S(V). We cnnot find A such tht A + B = {} {} {} {} {} {} = ({}). We next describe the MOD specil dul like number subset mtrix by some exmples. Exmple 4.13: Let S(M) = { / i S(Z 12 h); 1 i 8 +} be the MOD specil dul like number subset mtrix semiroup under +. Clerly ({} = {} {} {} {} {} {} {} {} S(M) is the dditive identity of S(M). Let A = {34h} {5 h} {13h} {h 4} {2 h} {14 3h} {2} {1 h3} nd

176 Semiroups built on MOD Subset Mtrices 175 B = {7} {2 4 3h} {1 h1} {7} {13h} {5} {7 3h} {92h} S(M). A + B = {34h} {5 h} {13h} {h 4} {2 h} {14 3h} {2} {1 h3} + {7} {2 4 3h} {1 h1} {7} {13h} {5} {7 3h} {92h} = {1 7 4h {2 4 3h 7 h {2 h1 21 4h {11 7 h} 3 4h} 9 4h} 1 3h} {1 3h h {1 h1 3h {6 9 3h} {7 3h 9 3h} 2 4h} 3 2h} S(M). This is the wy + opertion is performed on S(M). The reder is expected to find o(s(m)). Clerly if B = {} {} {} b {} {} {} b S(Z 12 h) +} S(M) is MOD specil dul like number subsemiroup of S(M). S(M) hs severl MOD specil dul mtrix subsemiroups. like number subset Exmple 4.14: Let

177 176 MOD Nturl Neutrosophic Subset Semiroups S(P) = i S(Z 7 h) 1 i 5 +} be the MOD specil dul like number subset mtrix semiroup under +. ({}) = {} {} {} {} {} S(P) cts s the identity element for S(P). P 1 = S(Z 7 h) +} S(P) is MOD specil dul like number subset mtrix subsemiroup. P 2 = S(Z 7 h) +} S(P)

178 Semiroups built on MOD Subset Mtrices 177 is in MOD specil dul like number subset mtrix subsemiroup. P 3 = S(Z 7 ) +} S(P) is MOD specil dul like number subset mtrix semiroup. P 4 = b b S(Z 7 ) +} is lso MOD specil dul like number subset mtrix subsemiroup of S(P). P 5 = b b S(Z 7 h) +} S(P) is lso MOD specil dul like number subset mtrix subsemiroup of S(P). S(P) hs severl MOD specil dul like number subset mtrix subsemiroups.

179 178 MOD Nturl Neutrosophic Subset Semiroups The reder is left with the tsk of findin the number of MOD specil dul like number subset mtrix subsemiroups of S(P). Next we ive some exmples of MOD specil qusi dul number subset mtrix semiroups. Exmple 4.15: Let S(B) = i S(Z 9 k); 1 i 6 +} be the MOD specil qusi dul number subset mtrix semiroup under +. ({}) = {} {} {} {} {} {} is the MOD specil qusi dul number subset identity mtrix of S(B); for every A in S(B) A + ({}) = A. P 1 = b b S (Z 9 k) +} S(B) is MOD specil qusi dul number subset mtrix subsemiroup of S(B). P 2 = b c b S(Z 9 k) nd C S(Z 9 k) +} S(B) is lso MOD specil qusi dul number subset mtrix subsemiroup of S(B).

180 Semiroups built on MOD Subset Mtrices 179 The reder is left with the tsk of findin the order of P 1 nd P 2. Let P 3 = b c d e f b c S(Z 9 k) d e S((Z 9 ) nd f S(Z 9 k) +} S(B) is MOD specil qusi dul number mtrix subset subsemiroup. Exmple 4.16: Let S(M) = i S(Z 19 k); 1 i 12 +} be the MOD specil qusi dul number subset mtrix semiroup under +. nfct S(M) is monoid s ({}) = {} {} {} {} {} {} ; {} {} {} {} {} {}

181 18 MOD Nturl Neutrosophic Subset Semiroups in S(M) is A + ({}) = A for ll A S(M). The reder is left with the tsk of findin the order of S(M). We see P 1 = {} {} {} {} {} {} {} {} i S(Z 19 ) 1 i 4 +} S(M) is MOD specil qusi dul number subset mtrix subsemiroup of S(M). P 2 = {} {} {} {} {} {} {} {} i S(Z 19 k); 1 i 3 1} S(M) is in MOD specil qusi dul number subset mtrix subsemiroup of S(M). P 3 = {} {} {} {} {} {} {} {} {} 1 {} {} i S(Z 19 k) +} S(M)

182 Semiroups built on MOD Subset Mtrices 181 is in MOD specil qusi dul number subset mtrix subsemiroup of S(M). nfct ll these three MOD subset mtrix subsemiroups re infct MOD subset mtrix submonoids of S(M). The reder is left with the tsk of findin ll MOD specil qusi dul number subset mtrix subsemiroups of S(M). nfct cn there be MOD mtrix subset mtrix subsemiroups in S(M) which re not MOD mtrix subset submonoids of S(M)? n view of ll these we hve the followin theorem. THEOREM 4.2: Let S(M) = {collection of ll t s mtrices with entries from S(Z n ) (or S(C(Z n )) or S(Z n ) or S(Z n ) or S(Z n h) or S(Z n k)); +} be the MOD subset mtrix monoid (or MOD subset mtrix finite complex number monoid or MOD neutrosophic subset mtrix monoid or MOD dul number subset mtrix monoid or MOD specil dul like number subset mtrix monoid or MOD specil qusi dul number subset mtrix monoid respectively). i) o(s(m)) <. ii) S(M) hs severl MOD mtrix subset subsemiroups. iii) All MOD mtrix subset subsemiroups re MOD subset mtrix submonoids. Proof is direct nd hence left s n exercise to the reder. Next we proceed on to describe MOD subset mtrix semiroups under nturl product n of mtrices by some exmples. Exmple 4.17: Let

183 182 MOD Nturl Neutrosophic Subset Semiroups S(M) = i S(Z 6 ); 1 i 5 n } be the MOD subset column mtrix subset semiroup. We see S(Z 6 ) the collection of ll subsets of Z 6 under product is semiroup. Let A = {2} {34} {5} {1} {215} nd B = {31} {52} {34} {2} {1} S(M). = {2} {31} {34} {52} A n B = {5} n {34} {1} {2} {215} {1} {2} {31} {2} {34} {25} { 23} {5} {34} = {32} S(M). {1} {2} {2} {215} {1} {215} This is the wy product n opertion is performed on S(M).

184 Semiroups built on MOD Subset Mtrices 183 ({}) = {} {} {} {} {} S(M) nd for ll A S(M) we hve A n ({}) = ({}) n A = ({}). Further ({1}) = {1} {1} {1} {1} {1} S(M) is defined s the MOD identity element of S(M) with respect to n. Clerly ({1}) n A = A n ({1}) = A for ll A S(M). Let A = {3 4} {25} {} {1 23} {345} nd B {} {} {123} S(M) {} {}

185 184 MOD Nturl Neutrosophic Subset Semiroups clerly A n B = {3 4} {25} {} {1 23} {345} n {} {} {123} {} {} = {} {} {}. {} {} Thus we sy A is MOD subset mtrix zero divisor B nd vice vers. Let N = {13} {3} {13} {4} {43} S(M). N n N = {13} {3} {13} {4} {43} n {13} {3} {13} {4} {43} = {13} {3} {3} {4} {43} = N. Thus we see N in S(M) is the MOD subset idempotent mtrix with respect to the product opertion. Thus S(M) is MOD subset mtrix commuttive monoid of finite order which hs MOD subset mtrix zero divisors nd MOD subset mtrix idempotents in it. However S(M) hs no MOD subset mtrix nilpotent elements.

186 Semiroups built on MOD Subset Mtrices 185 Exmple 4.18: Let S(P) = {( ) where i S(Z 8 ); 1 i 4 } be the MOD subset mtrix semiroup under product. Let A = ({3 5 6} {2 7} {1} {}) nd B = ({4 3} {5 } { } {5 6 7 }) S(P). A B = ({3 5 6} {2 7} {1} {}) ({4 3} {5 } { } {5 6 7 }) = ({3 5 6} {4 3} {2 7} {5 } {1} { } {} {5 6 7 }) = ({ } {2 3 } { } {}) S(P). This is the wy product opertion is performed on the MOD mtrix subset semiroup S(P). ({1}) = ({1} {1} {1} {1}) S(P) cts s the multiplictive identity. For A ({1}) = ({1}) A = A for ll A S(P). ({}) = ({} {} {} {}) S(P) is such tht A ({}) = ({} A = ({}); for ll A S(P). Let A = ({4 2} {} {4} {}) nd B = ({4} {3 4 5} {2 4} { }) S(P); A B = ({} {} {} {}) = ({}). So A is MOD subset mtrix zero divisor of B nd vice vers. Let P 1 = {( 1 ) / 1 S(Z 8 ) } S(P); clerly P 1 is MOD subset mtrix subsemiroup of S(P) which is lso MOD subset mtrix idel of S(P).

187 186 MOD Nturl Neutrosophic Subset Semiroups Let P 2 = {( 1 2 ) / 1 2 S(Z 8 ) } S(P) clerly P 2 is lso MOD subset mtrix idel of S(P). Further P 1 P 2 = ({}). Let M = {( ) / i S({ 2 4 6}); 1 i 4 } S(P) we see M is MOD subset mtrix idel of S(P). Now let T = ({4} {4} {} {2 4}) S(P); we find T T = ({4} {4} {} {2 4}) ({4} {4} {} {2 4}) = ({4} {4} {4} {4} {} {} {2 4} {2 4}) = ({} {} {} {4 }) = T 2 T 2 T = ({} {} {} {4 }) ({4} {4} {} {2 4}) = ({} {} {} {}) = T 3 = ({}) Thus T S(P) is MOD subset nilpotent mtrix of order three. Hence this S(P) hs MOD mtrix subset nilpotents. However findin nontrivil MOD subset mtrix idempotents hppens to be chllenin problem. n view of this we hve the followin theorem. THEOREM 4.3: Let S(B) = {collection of ll m p subset mtrices with entries from S(Z n ); n = p t ( t 2); n } be the MOD subset mtrix semiroup under nturl product n. i) S(B) hs MOD subset mtrices A such tht A s = ({}) for some s 2. ii) S(B) hs only trivil MOD subset mtrix idempotents.

188 Semiroups built on MOD Subset Mtrices 187 Proof is direct nd hence left s n exercise to the reder. Exmple 4.19: Let S(W) = i S(Z 12 ); 1 i 6 n } be the MOD subset mtrix semiroup under the nturl product n. Let A = {6} {6} {42} {53} {132} {} nd B = {37} {931} {15} {7} {3} {8913} S(W). A n B = {6} {6} {42} {53} {132} {} n {37} {931} {15} {7} {3} {8913} = {6} {37} {6} {391} {42} {15} {53} {7} {1 23} {3} {} {3981} = {6} {6} {428} {35} {369} {} S(W). This is the wy product opertion n is performed on S(W).

189 188 MOD Nturl Neutrosophic Subset Semiroups ({}) = {} {} {} {} S(W) is the zero of S(W) nd {} {} ({1}) = {1} {1} {1} {1} {1} {1} S(W) is the identity element of S(W). S(W) is commuttive MOD subset mtrix monoid of finite order. The tsk of findin o(s(w)) is left s n exercise to the reder. P = {6} {6} {} {6} S(W) is such tht {6} {6} P n P = {} {} {} {} {} {} = ({}). Thus P is MOD subset mtrix nilpotent of order two. Left us find some MOD subset mtrix idempotents of S(W). Let M = {49} {4} {9} {19} S(W) {419} {1}

190 Semiroups built on MOD Subset Mtrices 189 M n M = {49} {4} {9} {19} {419} {1} = M. Thus M is MOD subset mtrix idempotent of S(W). Let B = {482} {48} {369} {69} {48} {39} nd C = {6} {63} {48} {48} {369} {48} S(W). Clerly B n C = {} {} {} {} {} {} = ({}). Thus B is zero divisor of C nd vice vers. We hve seen this S(W) hs MOD subset mtrix zero divisors MOD subset mtrix idempotents nd MOD mtrix subset mtrix nilpotents. So if n = 12 = we see it hs ll such MOD specil subset mtrices. n view of ll these we put forth the followin result. THEOREM 4.4: Let S(M) = {collection of ll s t subset mtrices with entries from S(Z n ) n } be the MOD subset mtrix semiroup under the nturl product n where n = p r q where p is prime nd r 2 q ny composite or prime number.

191 19 MOD Nturl Neutrosophic Subset Semiroups i) Then S(M) hs nontrivil MOD mtrix subset idempotent. ii) S(M) hs nontrivil MOD mtrix subset nilpotents (order of the nilpotents depends on r). iii) S(M) hs MOD subset mtrix zero divisors. Proof is left s n exercise to the reder. Exmple 4.2: Let S(V) = i S(Z 5 ); 1 i 8 n } be the MOD subset mtrix semiroup under the nturl product n. Let A = {2} {1} {3} {} {14} {} {12} {2} nd B = A n B = {31} {1 2} {4} {234} {13} {43} {} {13} S(V). {2} {1} {3} {} {14} {} {12} {2} n {31} {1 2} {4} {234} {13} {43} {} {13} = {2} {31} {1} {12} {3} {4} {} {234} {14} {13} {} {43} {12} {} {2} {13} = {1 2} {12} {2} {} {1432} {} {} {21} S(V).

192 Semiroups built on MOD Subset Mtrices 191 This is the wy product opertion n is performed on S(V). We see S(V) hs no nontrivil MOD subset mtrix nilpotents non trivil MOD subset mtrix idempotent nd nontrivil MOD mtrix subset zero divisors. n view of ll these we hve the followin result. THEOREM 4.5: Let S(D) = {collection of ll m n subset mtrices with entries from S(Z p ); p prime n } be the MOD subset mtrix semiroup under nturl product n. i) S(D) hs no nontrivil MOD subset mtrix idempotents. ii) S(D) hs no nontrivil MOD subset mtrix nilpotents. iii) S(D) hs only MOD subset mtrix zero divisors of the form A = ( ij ) nd B = (b ij ); ij b ij S (Z p ). Then A B = ({}) only if either ij = {} or b ij = {}. Proof is direct nd hence left s n exercise to the reder. Exmple 4.21: Let S(P) = i S(Z 1 ); 1 i 4} be the MOD subset mtrix. We cn define two types of products s the collection is squre mtrix one the usul product nd the other is the nturl product n. {S(P) } is non commuttive MOD subset mtrix semiroup of finite order; infct mtrix subset monoid with

193 192 MOD Nturl Neutrosophic Subset Semiroups ({1}) = {1} {} {} {1}. But {S(P) n } is commuttive MOD mtrix subset semiroup under the nturl product n ; infct MOD subset mtrix monoid with ({1}) = {1} {1} {1} {1} s the identity with respect to the nturl product n. Both nd n re two distinct opertions on S(P). Let A = B = {524} {31} {1} {5} nd {12} {5} {7} {42} S(P). A B = {524} {31} {1} {5} {12} {5} {7} {42} = {52 4} {12} {52 4} {5} {13} {7} {31} {42} {1} {12} {1} {5} {5} {7} {5} {42}

194 Semiroups built on MOD Subset Mtrices 193 = {52 48} {5} {71} { 426} {12} {5} {} {5} = {63519 { } 71} {567} {5} This is the wy usul product opertion is performed on S(P). Now we find B A = {12} {5} {7} {42} {245} {31} {1} {5} = {1 2} {2 45} {1 2} {31} {5} {1} {5} {5} {7} {245} {7} {31} { 42} {1} {4 2} {5} = {2485} {3612} {5} {5} {485} {17} {42} {}

195 194 MOD Nturl Neutrosophic Subset Semiroups = {793} {8167} {4852 {17} 967} Clerly nd re not equl so in enerl A B = B A for A B S(P). Thus {S(P) } is MOD subset mtrix non commuttive semiroup. Now we find A n B = {524} {31} {1} {5} n {12} {5} {7} {42} = {524} {12} {31} {5} {1} {7} {5} {42} = {52 48} {5} {7} {} Clerly A n B = B n A but A B A n B nd B A A n B evident from equtions () () nd (). Findin MOD subset mtrix zero divisors in {S(P) } is left s mtter of routine. Further {S(P) } my contin MOD subset mtrix left zero divisor which my not be MOD subset mtrix riht zero divisor.

196 Semiroups built on MOD Subset Mtrices 195 We see {S(P) } my hve riht unit MOD subset mtrix which my not be MOD subset left unit nd vice vers. Now in view of ll these we hve the followin result. THEOREM 4.6: Let S(M) = {collection of ll m n subset mtrices with entries from S(Z n ); } be the MOD subset squre mtrix semiroup under the usul product. i) o(s(m)) < nd is non commuttive monoid. ii) S(M) my contin MOD subset mtrix riht zero divisors which my not in enerl be MOD subset mtrix left zero divisors. iii) S(M) will hve MOD subset mtrix riht idels which re not left idels. Proof is direct nd hence left s n exercise to the reder. Next we just describe MOD finite complex number mtrix subset semiroups under n the nturl product by some exmples. Exmple 4.22: Let S(M) = {( 1 6 ) be the collection of ll MOD subset finite complex number mtrix with entries from S(C(Z 1 )); } be the MOD subset finite complex number mtrix semiroup. We see ({}) = ({} {} {} {} {} {}) S(M) is such tht A ({}) = ({}) for ll A S(M). Further ({1}) = ({1} {1} {1}) S(M) is such tht A ({1}) = A is the identity or unit MOD subset mtrix of S(M). Thus S(M) is MOD subset finite complex number mtrix monoid of finite order. Reder is expected to find o(s(m)).

197 196 MOD Nturl Neutrosophic Subset Semiroups Consider P 1 = {( 1 2 ) / 1 2 S(C(Z 1 )) } S(M) is MOD finite complex number subset mtrix subsemiroup or submonoid of S(M). Let A = ({3 i F 2 + 5i F } {2i F } { i F } {1} {i F } { i F }) nd B = ({ 1 i F + 2} {1 + 2i F 3} {3 4i F 2 + i F } {3} {4i F } { 1 3}) S(M). A B = ({3 i F 2 + 5i F } {2i F } { i F } {1} {i F } { i F }) ({ 1 i F + 2} {1 + 2i F 3} {3 4i F 2 + i F } {3} {4i F } { 1 3}) = ({3 i F 2 + 5i F } { 1 i F + 2} {2i F } {1 + 2i F 3} { i F } {3 4i F 2 + i F } {1} {3} {i F } {4i F } { i F } { 1 3}) = ({ 3 i F 2 + 5i F 3i F i F 9 + 2i F } {6i F 2i F + 8} { i F 4i F 8i F 12i F i F 4 + 2i F 5 + 5i F } {3} {6} { i F i F }). This is the wy product opertion is performed on MOD finite complex number subset mtrices. Let A = ({} {24 6i F + 3} {2} {4 + 5i F 3i F } {} {4 2i F + 9 3}) B = ({3 + 5i F 2 4 3i F + 7} {} {5 + 5i F 5}{} {3 + 2i F 7i F i F } {}) S(M). A B = ({} {} {} {} {} {}) = ({}). A = ({5} {5i F } {5 + 5i F } {1 6} {5 6} { 1}) S(M). Thus A A = A is the MOD subset finite complex number mtrix idempotent of S(M).

198 Semiroups built on MOD Subset Mtrices 197 A = ({5} {5 + 5i F 5 5i F } {4 + 4i F 8 8i F 2} {} {4 + 4i F 6 + 2i F 6i F + 8} {5 + 5i F }) nd B = ({2 4 6} { i F 2i F } {5 5i F 5 + 5i F } { i F 5 + 6i F 3 + 8i F 1 + 9i F } {5 + 5i F 5} { i F 8i F i F }) S(M) is such tht A B = ({} {} {} {} {} {}) = ({}) is MOD subset mtrix zero divisor of S(M). The reder is left with the tsk of findin the number of MOD subset mtrix zero divisors of S(M). Exmple 4.23: Let S(V) = v1 v 2 v3 v4 v5 v i S(C(Z 7 )); 1 i 5 n } be the MOD subset finite complex number mtrix semiroup under the nturl product n. Let A = {3i F4 2i F1} {25i F} {i F} {26i F} {15} nd B = {34i F} {6 3i F} {2 4i F3} S(V). {1 i F} {3 5i F3}

199 198 MOD Nturl Neutrosophic Subset Semiroups We find A n B = {3i F4 2i F1} {25i F} {i F} {26i F} {15} n {34i F} {6 3i F} {2 4i F3} {1 i F} {33 5i F} = {3i F14 2i F} {34i F} {25i F} {6 3i F} {i F} {32 4i F} {26i F} {1 i F} {15} {33 5i F} = {2i F35 3i F4iF 5i F6 2i F} {5 6i F2iF 6} {3i F2iF 3} {2 2i F6iF 1} {33 5i F1 1 4i F} This is the wy the nturl product n opertion is performed on S(V). Clerly ({1}) = {1} {1} {1} {1} {1} cts s the multiplictive identity; for A ({1}) = A for ll A S(V). ({}) = {} {} {} S(V) {} {}

200 Semiroups built on MOD Subset Mtrices 199 is such tht A ({}) = ({}) for ll A S(V). We see A = {} {234 i F} {} {5 2i F3i F} {} nd B = {1 i F2i F3iF 56 2i F} {} {4 i F2 3i F} {} {1 i F2 5i F4i F} S(V) is such tht A n B = {} {} {} {} {} = S(V) hs only MOD finite complex number subset mtrix zeros of the followin form. t is left s n open problem to find other types of rel zero divisors. For these type of zero divisors cn be relized s the trivil type of MOD finite complex number subset mtrix zero divisors.

201 2 MOD Nturl Neutrosophic Subset Semiroups However if S(B) = i S(Z 7 ); 1 i 5 n } S(V) is MOD finite complex number subset mtrix subsemiroup of S(V) which is not n idel of S(V). n view of ll these the followin result is put forth. THEOREM 4.7: Let S(B) = {collection of ll subset mtrices with entries from S(C(Z n )); n } be the MOD finite complex number subset mtrix semiroup. i) S(B) hs nontrivil MOD finite complex number subset mtrix zero divisor which re nontrivil (n not prime). ii) S(B) hs MOD finite complex number subset mtrix subsemiroups which re not idels. iii) S(B) hs MOD finite complex number subset mtrix idels. Proof is direct hence left s n exercise to the reder. We leve the followin s open conjecture. Conjecture 4.1: Let S(W) = {collection of ll s t MOD finite complex number subset mtrix with entries from S(C(Z p )) p nd prime n }be the MOD finite complex number subset mtrix semiroup under nturl product. Cn S(W) hve MOD finite complex number subset mtrix nontrivil zero divisors?

202 Semiroups built on MOD Subset Mtrices 21 Next we proceed onto describe MOD subset neutrosophic mtrix semiroup under nturl product n by some exmples. Exmple 4.24: Let S(W) = i S(Z 12 ; 1 i 6 n } be the MOD neutrosophic subset mtrix semiroup under nturl product n. Let A = {3 4 26} {2 } {4 6 2} {} {125} {3 72 3} nd B = {2} {4 5} {1 3} {2 5 4} S(W). {3 8} {2 4} {34 26} {2} {2} {4 5} An B {4 62} {1 3} {} {42 5} {1 25} {3 8} { } {2 4} = {68 436} {8 19} {84 6} {} { } {6 66 2} S(W). This is the wy the product opertion n is performed on S(W).

203 22 MOD Nturl Neutrosophic Subset Semiroups Let A = { } {4 4 48} {1 11 1} {6 63} {4 66 8} {68 8} nd B = is such tht { } { } {66 66} {4 488} {666 6} {66 6} S(W) A n B = {} {} {} {} {} {} = ({}) is the MOD neutrosophic subset mtrix zero divisor of S(W). Let B = {44} {99} {44} {} {9} {9} S(W) is such tht B n B = {44} {99} {44} {} {9} {9} = B. Thus S(W) hs MOD neutrosophic mtrix subset idempotents. nfct S(W) lso hs nontrivil MOD neutrosophic mtrix subset nilpotents iven by the followin mtrix.

204 Semiroups built on MOD Subset Mtrices 23 F = {66} {6} {6 6} {6 6} {} {666 6} S(W) is such tht F n F = {} {} {} {} {} {} = ({}) is the MOD neutrosophic subset mtrix nilpotents of order two. n view of ll these we hve the followin result. THEOREM 4.8: Let S(M) = {collection of ll s t subset mtrices with entries from S(Z n ); n } be the MOD neutrosophic subset mtrix semiroup under nturl product n. i) f n = p t q (t 2 p prime of ny composite number) then S(M) hs MOD neutrosophic subset mtrix nilpotents the order of which is dependent on t. ii) S(M) hs MOD neutrosophic subset mtrix idempotents. iii) S(M) hs MOD neutrosophic subset mtrix zero divisors which re nontrivil. iv) S(M) hs MOD neutrosophic subset mtrix idels. Proof is direct nd hence left s n exercise to the reder. Next we proceed onto describe MOD dul number subset mtrix semiroups under nturl product by exmples. Exmple 4.25: Let S(P) = i S(Z 6 );

205 24 MOD Nturl Neutrosophic Subset Semiroups 1 i 1; n } be the MOD dul number subset mtrix semiroup under nturl product. Let A = {5} {3} {2} {2} { 2} {3} P4} {2} {4} {5} S(P); clerly A n A = {} {} {} {} {} {} {} {} {} {} = ({}). Thus A is MOD nilpotent dul number subset mtrix of S(P). nfct S(P) hs severl MOD nilpotent subset mtrices of order two. nfct S(M) = i {S(Z 6 )}; 1 i 1} S(P) is MOD dul number subset mtrix subsemiroup which is lso n idel of S(P). nfct S(M) is zero squre MOD dul number subset idel of S(P) s S(M) S(M) = ({}). We see S(P) hs MOD dul number subset mtrix subsemiroups which re not idels. For let

206 Semiroups built on MOD Subset Mtrices 25 S(T) = i S(Z 6 ); 1 i 1 n } S(P) is MOD dul number subset mtrix subsemiroup of S(P) which is clerly not n idel of S(P). L = {4} {3} {3} {4} {} {} {3} {4} {4} {3} S(P) we see L n L = {4} {3} {3} {4} {} {} {3} {4} {4} {3} = L. Thus L is MOD dul number mtrix subset idempotent of S(P). Hence S(P) hs MOD dul number mtrix subset nilpotents MOD dul numbers mtrix subset idempotents nd MOD dul number mtrix subset zero divisors. S(P) hs MOD dul number subset idels which re squre zero divisors nd MOD dul number subset subsemiroups which re not idels. Exmple 4.26: Let S(B) = i S(Z 11 ); 1 i 4 } be the MOD subset dul number mtrix semiroup under usul product. Clerly S(B) is non commuttive MOD dul number subset mtrix semiroup finite order.

207 26 MOD Nturl Neutrosophic Subset Semiroups Let A = {3 41} {25} {986} {27} S(B); clerly A A = {} {} {} {} = ({}) so is MOD dul number subset mtrix nilpotent of order two. nfct S(B) hs severl MOD dul number subset mtrix nilpotents of order two under usul product. The tsk of findin MOD dul number subset mtrix idempotents of S(B). Thus findin o(sb)) nd ettin MOD dul number subset mtrix left zero divisors which re not MOD dul number subset mtrix riht zero divisors re left for the reder. n view of ll these we hve the followin result. THEOREM 4.9: Let S(V) = {collection of ll s t subset mtrices with entries from S(Z n ) } be the MOD dul number subset mtrix semiroup under usul product. i) S(V) is non commuttive MOD subset dul number mtrix semiroup of finite order. ii) S(V) hs severl MOD subset dul number mtrix nilpotents of order two. iii) S(V) hs MOD dul number subset mtrix idel which is zero squre subsemiroup. iv) S(V) hs MOD dul number subset mtrix idempotents dependin on Z n. Proof is direct nd hence left s n exercise to the reder. We leve it s n exercise to the reder to show MOD dul number subset mtrix semiroup under nturl product n is different from the usul product.

208 Semiroups built on MOD Subset Mtrices 27 Next we proceed onto describe MOD specil dul like number subset mtrix semiroup under both the nturl product n nd tht of usul product by some exmples. Exmple 4.27: Let S(D) = i S(Z 1 h); 1 i 8 n } be the MOD specil dul like number subset mtrix semiroup under the nturl product n. Let A = {3h2h 15} {4h} {} {2} {h} { 2h 18h} {5h 55} {8h 32h} nd We find B = {h3h} {4h 23} {7h 28h 3 {4 7h8h3h 5h 83h4} 7h 3 2} {3 4h5h 1} {2} {3} {h2 3h} S(D).

209 28 MOD Nturl Neutrosophic Subset Semiroups A n B = {3h2h 15} {4h} {} {2} {h} { 2h 18h} {5h 55} {8h 32h} n = {h3h} {4h 23} {7h 28h 3 {4 7h8h3h 5h 83h4} 7h 3 2} {3 4h5h 1} {2} {3} {h2 3h} {3h 2h 15} {4h} {h3h} {34 h 2} {} {7h 28h 3 {2} {4 7h8h 5h 83h4} 3h7h 2} {h} {3 4h {8h2h 1} 5h 1} {2} {55h 5} {3} {2h8h 3} {h2 3h} = {3h7h5h9h} {2h4h} {} {8 4h6h 44h 6}. {7h6h} {6h 4h 2} {55h 5} {2hh6 9h} This is the wy nturl product is performed on S(D). Let A = {6} {6h} {5 5h} {5h} S(D). {56} {5h6h} {66h} {55h}

210 Semiroups built on MOD Subset Mtrices 29 A n A = A so A is the MOD specil dul like number subset mtrix idempotent of S(D). nd Let M = {24h6h} {55h5h 5} {5} {2 4h8h 2} {6h 68 6h {5h} 4 6h4h} {2 2h8 6h} {8h 42h 8} B = {2 2 4h 4h8h6h {55h5 5h} 6 2h} {22h4h 68h 2 {5} 4 6h48h} {22 8h6h 4 {45h} 8 2h} {55h5 5h} {5} M n B = {} {} {} {} = ({}). {} {} {} {} Thus M is MOD specil dul like number mtrix subset zero divisor. Let S(P) = i S(Z 1 h); 1 i 8 n } S(D)

211 21 MOD Nturl Neutrosophic Subset Semiroups be the MOD specil dul like number mtrix subset subsemiroup of S(D). Clerly S(P) is lso MOD specil dul like number mtrix subset idel of S(D). We ive one more exmple. Exmple 4.28: Let S(W) = i S(Z 11 h); 1 i 4 n } be the MOD specil dul like number subset mtrix semiroup under the nturl product n. Let A = {3h 53h2h1} {1 h h4 3h 2 h} {1 h5h6h 6} {h2h 16h9} nd B = { h4h 2 3h 6} {17h5h1h1} {6h 41} {115h5} S(W); A n B = {3h 53h2h1} {1 h h4 3h 2 h} {1 h5h6h 6} {h2h 16h9} n { h4h 2 {6h 41} 3h 6} {17h5h1h1} {115h5}

212 Semiroups built on MOD Subset Mtrices 211 = {3h 53h2h1} {h1 h 4 3h 2 h} {4h 23h 6} {16h 4} {1 h5h6 6h} {h2h 16h9} {17h5h1h1} {1155h} = {8h3h2h1h7h 9 {h1 h4 3h 2 h 5 8hh7h1 5h8 9h} 1h 5h5 1h8} {1h5h6 6h7h 2h7h {h2h 16h95h1h 5 3h1h6h15 5h} 8h11h1 9h 4hh} This is the wy nturl product n is obtined in S(W). Let S(V) = i S(Z 11 h); 1 i 4 n } S(W) is the MOD subset specil dul like number mtrix subsemiroup which is lso n idel of S(W). Now if we replce in S(W) the nturl product n by then S(W) will be non commuttive MOD subset specil dul like number mtrix non commuttive semiroup under the usul product. This two opertions re distinct we will just show by some n illustrtion. Let A = {h 2h1 h} {1h15h} {5 1h1 5h {1 2h1 nd h5h1} h 2}

213 212 MOD Nturl Neutrosophic Subset Semiroups B = {3h1} {4h2 6h8h} {1 2h6h 4h48} {875h1h1} S(W); we now define the usul product in S(W). A B = {h 2h1 h} {1h15h} {5 1h1 5h {1 2h1 h5h1} h 2} {3h1} {4h2 6h8h} {1 2h6h 4h48} {875h1h1} = {1h9h1 1h3h 6h {8h 4h5h 2 3h} {1h1 9h5h8h {33h 4 4h 7h 6h 9h7h 73h} 2h h15h11h} {h 4h 3h6 h1 6h {5h9h 4h1h 7h 8h 2 6h 6h1h } {11 2h9 3h 9 h1} {5h 388h 5 4 3h 6h7h4hh7 8h 7 7h 3 h3h1 2h1 h 21 9h1 4 4h 8 3 5h88h 5} 2h1h 95h} = =

214 Semiroups built on MOD Subset Mtrices 213 {1h9h1 1h3h 6h8h1 9h 2h5h 7h1 8h1 7h1 h1 4 h 4h1 4h6h1 8h h1 6h 1 h 79 h 71h 6 7 3h 6h 7 1 2h1 h 3 9h 3 2 1h3 6h} {h4h3h6 h1 6h6h1h8h 1 4 h 13h 1 7 h 2 6h5h51 h11 1 3h8h 51 6h1 5h5 3h 8h 9h 7 h 51 8h1 h1 2h 9 4h3h 59 7h 9 6h 4 4h1 9h 6 3h8 h9 9h9 3h 2h8 4 h 1h9h6 7 h7h8 3h6 8h1 2h 6 2 h1 7 h3 h5h 88h 87 h h9 6 h91h 83h 84h 8 8 6h3 6h3 9h3 8h 9h 5 h h3 5h} {33h44h7h6h2hh15h11h 8h 34 8h9h1 8h1 8h 8h3 4h4 4h1 4h5h 15h 2 2 4h6 3h2 7h2 9h3 3h 2 8h1 3h2 h} {1h 33h 39h 34h 3h 3 2h 3h51 6h9h1h1 7h 11 5h7 8h2 5h8 5h 8 9h67h9 6h8 4h8 1h 8 7h6h5 h 2h4 8h8 8h 8 6h6 h1h 4 h8h7 2h 56h 55h4 7 hh 57h 5 4h 54 2h7 3h3 9h4 1h 6 9h8h 47 4h7 1h4h 3 3 3h7 5h7 9hh 35h 3 36h 33h 34h 3 5 2h 2 5h2 7h2 7 h 9 2h1 h} This is the wy the usul product opertion is performed on S(D).

215 214 MOD Nturl Neutrosophic Subset Semiroups f we replce the usul product opertion by nturl product n certinly S(D) is commuttive nd both opertions re distinct. set Further we see under both the opertions s well s n the S(V) = i S(Z 11 h); 1 i 4; or n } is MOD specil dul like number subset mtrix subsemiroup which is n idel. This is the only common feture enjoyed by S(D) under both nd n. However MOD specil dul like number zero divisor subset mtrices sy A B S(D) need not in enerl imply A B = ({}) then A n B = ({}) nd vice vers. Next we supply yet nother exmple of this set up. Exmple 4.29: Let S(W) = i S(Z 4 h); 1 i 4 n } be the MOD specil dul like number subset mtrix semiroup under the nturl product n.

216 Semiroups built on MOD Subset Mtrices 215 Let A = {32h} {2 2h2} {31 h} {2h 2} nd B = {3h1} {22h} {1 2h} {3 2h} S(W); A n B = {32h} {22 2h} {31 h} {2 2h} n {13h} {2 2h} {1 2h} {3 2h} = {32 h} {13h} {22 2h} {22h} {31 h} {1 2h} {22h} {3 2h} = {3 2 h2h3h} {} {3 2h1 2h} {2 2h} S(W). This is the wy nturl product opertion is performed on S(W). P 1 = {} {} {} S(Z 4 h); n } S(W) is MOD specil dul like number subset mtrix idel of S(W). Let P 2 = {} {} {} S(Z 4 ) n } S(W)

217 216 MOD Nturl Neutrosophic Subset Semiroups is MOD specil dul like number subset mtrix subsemiroup of S(W) which is not n idel of S(W) Let P 3 = {} {} {} S(Z 4 h); n } S(W) is in MOD subset specil dul like number mtrix idel of S(W). nfct S(W) hs severl MOD specil dul like number subset mtrices subsemiroup which re not idel. n view of ll these we hve the followin theorem. THEOREM 4.1: Let S(M) = {collection of ll s t mtrices with entries from subsets of S(Z n h); n } be the MOD specil dul like number subset mtrix semiroup under the nturl product n. i) S(M) hs MOD specil dul like number nilpotents idempotents nd zero divisors if n is n pproprite composite number of the form n = p m q m 2 p prime nd q composite or nother prime. ii) S(M) hs lwys MOD specil dul like number subset mtrix idels if the entries re from S(Z n h) no mtter whtever be n. Proof is direct nd hence left s n exercise for the reder. Next we proceed onto describe the MOD specil qusi dul number subset mtrix semiroups under n nd lso under usul product provided the mtrices under considertion re squre mtrices by exmples.

218 Semiroups built on MOD Subset Mtrices 217 Exmple 4.3: Let S(B) = i S(Z 6 k); 1 i 9 n } be the MOD specil qusi dul number subset mtrix semiroup under the nturl product n. Let A = {34} {14} {1} {3} {3k3} {3 3k} S(B). {43k} {43k} {13k} We see A n A = A tht is A is the MOD specil qusi dul number subset mtrix idempotent. Suppose we impose the usul product tht is we find A A then A A = {34} {14} {1} {3} {33k} {3 3k} {43k} {43k} {13k} {34} {14} {1} {3} {33k} {3 3k} {43k} {43k} {13k}

219 218 MOD Nturl Neutrosophic Subset Semiroups = {3 4} {3 4} {3 4} {1 4} {3 4} {1} {3} {1 4} {1 4} {33k} {1 4} {3 3k} {1} {43k} {1} {43k} {1} {13k} {3} {34} {3} {14} {3} { 1} {33k} {3} {33k} {33k} {33k} {3 3k} {3 3k} {43k} {3 3k} { 43k} {3 3k} {13k} {43k} {3 4} {43k} {1 4} {43k} {1} { 43k} {3} {33k} { 43k} { 43k} {3 3k} {13k} {43k} {13k} { 43k} {1 3k} {13k} {3 4} {3 4} {3} {3 4} {33k} {3 3k} {43k} { 43k} {13k} {3} {3} {33k} {3} {33k} = {3 3k} {} {} {3 3k} {3k 4} {3k 4} {43k} {} {3k} {3k} {13k} {43k} { 43k} = {41253k {133 k3k {4 3k1 3k 3 3k3k 4 3k 451 3k 2 3k 1 3k} 3 3k2 3k4} 13} {33k3 3k} {3 3k33k} {33k3 3k} {43k { 43k2 {53k 1 2 3k4 3k} 2 3k 4 3k} 4 3k} A. Thus we see the two opertions behve in very distinct wy.

220 Semiroups built on MOD Subset Mtrices 219 Let A = {3k} {3 3k} {33k} {2} {4} {2 4k} {4k 4} {2 2k4} {3 3k} nd B = {4} {44k} {2} {33k} {3 3k} {33k} {3} {33 3k} {4 2} be elements of S(B). Clerly A n B = {33k} {3 3k} {33k} {2} {4} {2 4k} {4 4k} {4 2 2k} {3 3k} n = {4} {44k} {2} {33k} {3 3k} {33k} {3k} {33 3k} {42} {33k} {4} {3 3k} {44k} {{33k} {2} {2} {33k} {4} {3 3k} {2 4k} {33k} {3 3k} {4 4k} {3} {4 2 2k} {33 3k} {42} = {} {} {} {} {} {} {} {} {} = ({}) Thus under the nturl product n. A B is zero divisor pir. Now we find A B

221 22 MOD Nturl Neutrosophic Subset Semiroups {33k} {3 3k} {33k} = {2} {4} {2 4k} {4k 4} {4 2 2k} {3 3k} {4} {44k} {2} {33k} {3 3k} {33k} {3} {33 3k} {2 4} = {33k} {4} {33k} {44k} {33k} {2} {3 3k} {33k} {3 3k} {3 3k} {3 3k} {33k} {33k} {3} {33k} {33 3k} {33k} {2 4} {2} {44k} {4} {2} {2} {4} {2} {4} {4} {33k} {3 3k} {2 4k} {33k} {2 4k} {2 4k} {3} {33 3k} {24} {4k 4} {4} {44k} {4 4k} {4k 4} {2} {4 2 2k} {33k} {3 3k} {4 2 2k} {4 2 2 k} {33k} {3 3k} {3} {3 3k} {33 3k} {3 3k} {2 4} {3 3k33k} {3 3k3k3} {3 3k} = {2} {22k} {2k 2k4k}. {4 4k k 1} {3 3k1 k} {2 2k} Clerly the pir A B in S(B) does not yield MOD zero divisor of specil qusi dul numbers under the usul product. However S(P) = i S(Z 6 k); 1 i 9 n }

222 Semiroups built on MOD Subset Mtrices 221 s well {S(P) } re both MOD specil qusi dul number subset mtrix idels of S(D). This property lone is common to both the product nd n. n view of ll these we ive the followin result. THEOREM 4.11: Let S(M) = {collection of ll s t mtrices with entries from subsets S(Z n k); n } be the MOD specil qusi dul number subset mtrix semiroup under nturl product. i) S(M) hs MOD specil qusi dul number subset mtrix idempotents nilpotents nd zero divisors for n = p m q m 2 p is prime q my be prime or non prime. ii) S(W) = {collection of ll s t mtrices with subset entries from S(Z n k) under nturl product n } is MOD specil qusi dul number subset mtrix idel of S(M). Proof is direct nd hence left s n exercise to the reder. Next we proceed onto study MOD subset coefficient polynomil semiroup under product by some exmple. Exmple 4.31: Let S[x] = i ix i S(Z 12 ); } i be the MOD subset coefficient polynomil semiroup under product. Clerly o(s[x]) =. Let p(x) = { 3 6} + {2 6}x + {5 1 8}x 2 nd

223 222 MOD Nturl Neutrosophic Subset Semiroups q(x) = {4 8 1} + {4 6 3} x 3 S[x] p(x) q(x) = ({ 3 6} +{2 6}x + {5 1 8}x 2 ) x x + ({5 1 8} {1 4 8})x 2 + ({ 3 6} {4 6 3})x 3 + ({2 6} x 4 + ({5 1 8} {4 6 3})x 5 = { 3 6} + { }x + { }x 2 + { 96}x 3 + {8 6 }x 4 + {2 4 6 }x 5 S[x]. This is the wy product of two polynomils is mde in S[x]. () = + x + x x n S[x] is such tht p(x) () = () S[x] is commuttive semiroup of infinite order. Let p(x) = { 4} + { 8} x 2 + { 8 4} x 4 nd q(x) = { 3 6} + {6 3 9}x 3 S[x]. p(x) q(x) = ({ 4} + { 8}x 2 + { 4 8}x 4 ) ({ 3 6} + {3 6 9}x 3 ) ={} + {}x 2 + {}x 4 + {}x 3 + {}x 5 + {}x 7 = (). This we see S[x] cn hve pirs of polynomils such tht p(x) q(x) = ({}) is zero divisor in S[x]. We cn lso find MOD subset coefficient polynomils subsemiroups under s well s idels of S[x]. Let i P[x] = ix i S(Z 12 ) } S[x] i2z {}

224 Semiroups built on MOD Subset Mtrices 223 is MOD subset coefficient polynomil subsemiroup of S[x] of infinite order which is clerly not n idel of S[x]. For p(x) = { 5 1 2} x + {8 4 1} S[x] nd q(x) = { 5 8} + { 9 8 2} x 6 P[x] Consider p(x) q(x) = ({8 4 1} + { 2 5 1} x) ({ 5 8} + { 2 8 9})x 6 = ({8 4 1} { 5 8}) + ({ 2 5 1} { 5 8})x + ({8 4 1} { 2 8 9})x 6 + ({ 2 5 1} { 2 8 9})x 7 = { } + { }x + { }x 6 + { }x 7 P[x]. Hence our clim P[x] is only MOD subset polynomil coefficient subsemiroup. i R[x] = ix i S ({ }) } S[x] i is MOD subset coefficient polynomil subsemiroup which is lso n idel of S[x]. The reder is left with the tsk of findin MOD subset coefficient polynomil subsemiroups idels nd zero divisors. However it is kept on record tht S[x] no idempotents. Exmple 4.32: Let T[x] = i ix i S(Z 19 ); } i be the MOD subset coefficient polynomil semiroup under.

225 224 MOD Nturl Neutrosophic Subset Semiroups Clerly T[x] hs no zero divisors however T[x] hs no idels but hs subsemiorups of infinite order. n view of ll these we hve the followin theorem. THEOREM 4.12: Let W[x] = i i x i S(Z n ) } be the i MOD subset coefficient polynomil semiroup. i) W[x] hs no idempotent MOD subset polynomils. ii) W[x] hs MOD subset coefficient polynomils which re nilpotents nd zero divisors only for some pproprite n n = p t q (t 2 p prime nd q ny number other thn p). iii) W[x] hs MOD subset coefficient polynomil subsemiroup for ll n. iv) W[x] hs no MOD subset coefficient polynomil idels for n prime. Proof is left s n exercise to the reder. Next we describe by exmple MOD finite complex number subset coefficient polynomil semiroups under. Exmple 4.33: Let D[x] = i ix i S(C(Z 6 )); } i be the MOD subset finite complex number coefficient polynomil semiroup under product. P[x] = i ix i S(Z 6 ) } D[x] i is MOD subset finite complex coefficient polynomil subsemiroup under of D[x] however P[x] is not MOD

226 Semiroups built on MOD Subset Mtrices 225 subset finite complex number coefficient polynomil idel of D[x]. Find MOD subset finite complex number coefficient polynomil idels of D[x]. Clerly D[x] hs MOD subset finite complex number coefficient zero divisor polynomils. However D[x] hs no nontrivil MOD subset finite complex number coefficient nilpotent or idempotent polynomils. t is left for the reder to find under wht conditions MOD finite complex number subset coefficient polynomils hs idels. For findin idels hppens to be yet nother chllenin problem. Exmple 4.34: Let S[x] = i ix i S(C(Z 13 )) } i be the MOD finite complex number subset coefficient polynomil semiroup under. S[x] hs no MOD finite complex number coefficient subset zero divisors or idempotents or nilpotents. However S[x] hs MOD finite complex number coefficient subset polynomil subsemiroups but no idels. n view of this we propose the followin problem. Problem 4.1: Let S[x] = i ix i S(C(Z p )) p prime } i be the MOD finite complex number subset coefficient polynomil semiroup under.

227 226 MOD Nturl Neutrosophic Subset Semiroups Cn S[x] hve MOD finite complex number subset coefficient polynomil idel? THEOREM 4.13: Let S[x] = i i x j S(C(Z n )); } i be the MOD finite complex number subset coefficient polynomil semiroup under. i) For pproprite n S[x] hs zero divisors nilpotents nd idels. ii) For n prime S[x] hs no zero divisors nd nilpotents. iii) S[x] hs no nontrivil MOD finite complex number subset coefficient polynomil idempotent. The reder is left with the tsk of provin this theorem. Next we proceed onto describe MOD dul number subset coefficient polynomil semiroups under by exmple. Exmple 4.35: Let S[x] = i ix i S(Z 1 ); } i be the MOD dul number subset coefficient polynomil semiroup under product. S[x] hs infinite number of MOD subset dul number coefficient polynomils such tht they re zero divisors. S[x] hs infinite number of MOD subset dul number coefficient nilpotent polynomils of order two.

228 Semiroups built on MOD Subset Mtrices 227 Let p(x) = { } x 3 + { 2 9 8}x + { 8 7} S[x]. Consider p(x) p(x) = ({ }x 3 + { 2 9 8}x + { 8 7}) ({ }x 3 + { 2 9 8}x + { 7 8}) = ({ } { })x 6 + ({ } { 2 9 8})x 4 + ({ 7 8} { })x 3 + ({ 2 8 9} { })x 4 + ({ 2 8 9} { 2 8 9})x 2 + ({ 2 8 9} { 7 8})x + ({ 7 8} { }) x 3 + ({ 7 8} { 2 9 8}) x + ({ 7 8} { 7 8}) = {} x 6 + {}x 4 + {}x 3 + {}x 4 + {}x 2 + {}x + {}x 3 + {}x + {} = ({}). Thus p(x) is MOD subset dul number coefficient polynomil which is nilpotent of order two in S[x]. nfct it is pertinent to keep on record tht S[x] hs infinite number of MOD dul number subset coefficient nilpotent polynomils of order two. Let p(x) = {3 } x 3 + {2 4 7} x + { } nd q(x) = { } x 2 + { } S[x]. Clerly p(x) q(x) = ({}). Thus p(x) is MOD dul number subset coefficient polynomil zero divisor of S[x]. We see S[x] hs infinite number of MOD subset dul number coefficient polynomil pirs which re zero divisors. This is the mrked difference between the MOD subset coefficient polynomil semiroup nd MOD dul number coefficient subset polynomil semiroup under.

229 228 MOD Nturl Neutrosophic Subset Semiroups Further B[x] = ix i i i S(Z 1 ); } S[x] is MOD dul number subset coefficient polynomil subsemiroup which is lso MOD dul number subset coefficient polynomil idel of S[x]. nfct B[x] is zero divisor subsemiroup s for ny p(x) q(x) B[x]; p(x) q(x) = ({}) tht is B[x] B[x] = ([}). However in S[x] we do not hve MOD dul number subset coefficient polynomils which re idempotents. Exmple 4.36: Let S[x] = i ix i S(Z 7 ); } i be the MOD dul number subset coefficient polynomil semiroup under product. Let p(x) = {3 2 } + {4 5 6}x 2 + {6 }x 3 nd q(x) = {4 } + {2 3 }x + {5 6 }x 5 S[x]. p(x) q(x) = ({}). Thus we see even thouh Z 7 is prime field yet we hve MOD dul number subset coefficient polynomil pirs which re zero divisors. Also p(x) p(x) = ({}) nd q(x) q(x) = ({}) re both MOD dul number subset coefficient polynomils which re nilpotents of order two.

230 Semiroups built on MOD Subset Mtrices 229 D[x] = i ix i S(Z 7 ); } S[x] i is MOD subset dul number coefficient polynomil subsemiroup which hs no nontrivil MOD dul number subset coefficient polynomil zero divisors or nilpotents. D[x] is not n idel of S[x]. B[x] = ix i i i S(Z 7 ); } S[x] is MOD dul number subset coefficient polynomil subsemiroup which is lso n idel. This B[x] hs nontrivil MOD subset dul number coefficient polynomil pirs which re zero divisors s well s nilpotents of order two. However B[x] D[x] nd S[x] hs no MOD subset dul number coefficient polynomil idempotents which re nontrivil. For = { 1} S[x] then we hve = { 1} { 1} = { 1} = is trivil MOD subset dul number coefficient constnt polynomil idempotent. nview of ll these we hve the followin theorem. THEOREM 4.14: Let S[x] = { i x i i S((Z n ); } be MOD dul number subset coefficient polynomil semiroup under. i) S[x] hs no nontrivil MOD dul number subset coefficient polynomil idempotents.

231 23 MOD Nturl Neutrosophic Subset Semiroups ii) S[x] hs nontrivil MOD dul number subset coefficient polynomil zero divisors nd nilpotents whtever be n. iii) S[x] hs nontrivil MOD dul number coefficient polynomil subsemiroups which re not idels. iv) S[x] hs nontrivil MOD dul number coefficient polynomils subsemiroups D[x] which re not idels but D[x] D[x] = ({}) v) S[x] hs nontrivil MOD dul number coefficient i polynomil subsemiroups T[x] = i x i i S(Z n ); } S[x] to be n idel with T[x] T[x] = ({}). Proof is direct nd hence left s n exercise to the reder. Next we proceed onto describe MOD neutrosophic subset coefficient polynomil semiroups by exmples. Exmple 4.37: Let S[x] = i ix i S(Z 17 ); } i be the MOD neutrosophic subset coefficient polynomil semiroup under. Clerly S[x] hs no nontrivil neutrosophic subset coefficient idempotent polynomils or nilpotent polynomils or zero divisor polynomils s in Z is prime. However if p(x) = { } + { }x 2 nd q(x) = { 4 3} + { } x 2 S[x]; then p(x) q(x) = ({ } + { }x 2 ) ({ 4 3} { }x 2 )

232 Semiroups built on MOD Subset Mtrices 231 = ({ } { 4 3}) + ({ } { 4 3})x 2 + ({ } { })x 2 + ({ } { })x 4 = { } + { }x 2 + { }x 2 + { }x 4 = { } +{ i } x 2 + { }x 4. This is the wy product opertion is performed on S[x]. B[x] = i ix i S{Z 17 ); } S[x] i is MOD neutrosophic subset coefficient polynomil subsemiroup which is lso n idel of S[x]. This S[x] hs no nontrivil idempotents. The trivil idempotent MOD neutrosophic subsets re {} {1} {} { 1} { } {1 } nd { 1 }. They cn only yield to constnt MOD subset neutrosophic polynomils.

233 232 MOD Nturl Neutrosophic Subset Semiroups Exmple 4.38: Let S[x] = i ix i (Z 12 ); } i be the MOD neutrosophic subset coefficient polynomil semiroup under product opertion. Let p(x) = { 4 2} x 3 + { 8 8 4} nd q(x) = { } + { }x S[x]. p(x) q(x) = ({ 2 4}x 3 + { 8 8 4}) ({ } + { }x) = ({}). Thus S[x] hs MOD neutrosophic subset coefficient polynomil pir which re zero divisors. However S[x] hs no idempotents. Let p(x) = { 6} + { }x 3 S[x]; clerly p(x) p(x) = ({}) so p(x) is MOD neutrosophic subset coefficient polynomil nilpotent of S[x]. S[x] hs trivil idempotents iven by { 4 9 1} { 4 9} { 4 1} { 9 1} { 4} {1 4} {1 9} nd { 9}. Clerly M[x] = ix i i i S(Z 12 ) } S[x] is MOD neutrosophic subset coefficient polynomil subsemiroup which is lso n idel of S[x]. But

234 Semiroups built on MOD Subset Mtrices 233 T[x] = i ix i S(Z 12 ) } S[x] i is MOD neutrosophic subset coefficient polynomil subsemiroup which is not n idel. n view of ll these we hve the followin result. THEOREM 4.15: Let S[x] = i i x i S(Z n ); } i be the MOD neutrosophic subset coefficient polynomil semiroup under product. i) S[x] hs no nontrivil MOD neutrosophic subset coefficient polynomil idempotents. ii) S[x] hs MOD neutrosophic subset coefficient polynomil pirs which cn contribute to zero divisors provided n is n pproprite composite number. iii) S[x] hs MOD neutrosophic subset coefficient polynomil nilpotents if n = p t q t 2 p prime nd q composite number i iv) D[x] = ix i S(Z n ) } is the MOD i neutrosophic subset coefficient polynomil subsemiroup which is n idel of S[x]. v) S[x] hs MOD neutrosophic subset coefficient polynomil subsemiroups which re not idels. Proof is direct nd hence left s n exercise to the reder. Next we proceed onto describe by exmples MOD subset specil dul like number coefficient polynomil semiroups under product.

235 234 MOD Nturl Neutrosophic Subset Semiroups Exmple 4.39: Let B[x] = i ix i S(Z 13 h) } i be the MOD subset specil dul like number coefficient polynomil semiroup under product. We see B[x] hs no nontrivil idempotents. B[x] hs no nontrivil nilpotents s well s zero divisor polynomils. However B[x] hs MOD subset specil dul like number coefficient polynomil subsemiroups which re idels. Let p(x) = {3 5h 2 + h 1} + { 4h 3}x + { 1 3h}x 2 nd q(x) = { 4h 2}x 4 + {1 2 3 } B[x]. We find p(x) q(x) = ({3 5h 2 + h 1} + { 4h 3}x + { 1 3 h})x 2 ({ 1 2 3} + { 2 4h}x 4 ) = {3 5h h} { 1 2 3} + ({3 5h h} { 2 4h})x 4 + ({ 4h 3} { 1 2 3})x + ({ 3 4h} { 2 4h}) x 5 + ({ 1 3h} { 1 2 3})x 2 + ({ 1 3h} { 2 4h})x 6 = { 3 5h h 6 1h h 9 2h 6 + 3h} + { 6 1h h 4 + 2h 12 7h 4h}x 4 + { 4h 3 8h 6 12h 9}x + { 6 8h 12h 3h}x 5 + { 1 3h 2 6h 3 9h}x 2 + { 2 6h 4h 12h}x 6 B[x]. This is the wy product opertion is performed in B[x]. B[x] hs no MOD idempotents or zero divisors or nilpotents. Exmple 4.4: Let i M[x] = ix i S(Z 15 h); } i

236 Semiroups built on MOD Subset Mtrices 235 be the MOD subset specil dul like number coefficient polynomil semiroup under the product. M[x] hs zero divisors hs no nontrivil nilpotents or idempotents. p(x) = {5 5h 1} + { h 5 + 1h } x 2 nd q(x) = {3 3h} + {3 + 3h 6}x 2 + {3h 6h 6 + 3h}x 4 M[x]. p(x) q(x) = ({5 5h 1} + { h 5 + 1h })x 2 ({ 3 3h} + {3 + 3h 6h})x 2 + {6h 3h 6 + 3h} x 4 ) = {5 5h 1} { 3 3h} + ({5 5h 1} {3 + 3h 6h})x 2 + ({5 5h 1} {6h 3h 6 + 3h})x 4 + ({ h 5 + 1h } { 3 3h}) x 2 + ({ h 5 + 1h } {6 + 3h 3h 6h}) x 6 + ({ h 5 + 1h } {6h 3+ 3h})x 4 = ({}). Thus this pir of MOD specil dul like number subset coefficient polynomils is zero divisor subset polynomil pir. We cn find infinite number of MOD specil dul like number subset coefficient polynomil zero divisor pirs. However M[x] hs no nilpotents only trivil idempotents of the form {} {1} {6} {1} {6 1 } {1 6} { 6} {1 1} { 1} { 1 6 1} s 6 2 = 6 (mod 15) nd 1 1 = 1 (mod 15). This hs no nontrivil nilpotent subset only {} is the trivil nilpotent subset. n view of ll these we put forth the followin theorem. i THEOREM 4.16: Let S[x] = i x i S(Z n h); } be i the MOD specil dul like number subset coefficient polynomil semiroup under product.

237 236 MOD Nturl Neutrosophic Subset Semiroups Then i) S[x] hs MOD specil dul like number subset coefficient polynomil subsemiroup which is not n idel (wht ever be n). ii) S[x] hs MOD specil dul like number subset coefficient polynomil semiroups which is lso n idel. iii) S[x] hs MOD specil dul like number subset coefficient polynomil nilpotents nd zero divisors only for pproprite n; n = p t s (t 2 p prime s composite or prime different from p). iv) S[x] hs no MOD specil dul like number subset coefficient polynomil which is n idempotent polynomil. Proof is direct nd hence left s n exercise to the reder. Next we proceed onto describe MOD specil qusi dul number subset coefficient polynomil semiroup under product opertion by some exmples. Exmple 4.41: Let S[x] = i ix i S (Z 2 k); } i be the MOD specil qusi dul number subset coefficient polynomil semiroup under product. Let p(x) = { k 2k + 1 4k} + {1 5k 2 + 3k} x + {1 2k 8k 4}x 2 nd q(x) = { 4k 1k} + { k 1 k}x 4 S[x]. p(x) q(x) = ({ k 4k 2k + 1} + {1 5k 2+ 2k}x + {1 4 8k 2k}x 2 ) ({ 4k 1k} + { k 1 k}x 4 )

238 Semiroups built on MOD Subset Mtrices 237 = { k 4k 2k + 1} { 4k 1k} + ({ k 4k 2k + 1} {1 1k k})x 4 + ({1 5k 2 + 3k} { 4k 1l})x + ({1 5k 2 + 3k} { k 1 k})x 5 + ({ 1 4 2k 8k} { 4k 1k})x 6 = { 16k 4k 16k 1k} + { 1k 1 1k + 1 k 4k 2k k 16k 19k}x 4 + {1 1k 1 + 1k 1 5k 2 + 3k k 15k 19k}x 5 + { 4k 6k 1k} x + { 4k 16k 12k18k 1k}x 2 + { k 1 4 8k 2k k 4k 12k 18k} x 6. This is the wy product opertion is performed on S[x]. Consider p(x) = {1 1k} + { 1 1k + 1} x 2 + {1 + 1k 1k}x 4 S[x]. Clerly p(x) p(x) = ({}). Thus p(x) is nilpotent MOD subset coefficient polynomil. Thus S[x] hs both MOD specil qusi dul number subset coefficient polynomil zero divisors nd nilpotents however hs no idempotents. Exmple 4.42: Let S[x] = i ix i S (Z 5 k); } i be the MOD specil qusi dul number subset coefficient polynomil semiroup under product. Let p(x) = {k 1 } + {1 + k 4 2}x + {1 3k 3}x 2 nd q(x) = { k} + {1 2 3 k 4} x 4 S[x]. p(x) q(x) = ({ 1 k} + { k} x + {1 3k 3}x 2 ) ({ k} + {1 2 3 k 4}x 4 )

239 238 MOD Nturl Neutrosophic Subset Semiroups = { 1 k} { k} + ({ k} { k}) x + ({1 3k 3} { k}) x 2 + ({ 1 k} {1 2 3 k 4})x 4 + ({ k}{ k}) x 5 + ({1 3 3k} { k})x 6 = { k 4k} + { 2k 4k}x + { k 3k 2k}x 5 { 1 k 2 2k 3 3k k 4k 4} x 4 + { k k k 4 + 4k 2k 4k }x 5 + { k 3k 4k 2k}x 6. This is the wy product opertion is performed on S[x]. S[x] hs no MOD specil qusi dul number subset coefficient polynomil idempotents or zero divisors or nilpotents. n view of ll these we hve the followin result. THEOREM 4.17: Let S[x] be the MOD specil qusi dul number subset coefficient polynomil semiroup under product S[x] = i i x i i S (Z n k); }. i) S[x] hs no MOD nilpotents or idempotents or zero divisors if n = p prime. ii) S[x] for ll n hs MOD specil qusi dul number subset coefficient subsemiroups which re not idels. i iii) P[x] = x i i S (Z n k); } S[x] is MOD i specil qusi dul number subset coefficient polynomil idel of S[x]. iv) S[x] hs MOD nilpotents nd zero divisors if n = p t q t 2 p prime nd q composite number of prime different from p. Proof is direct nd hence left s n exercise to the reder.

240 Semiroups built on MOD Subset Mtrices 239 Next we proceed onto ive by simple illustrtions MOD subset coefficient polynomil semiroup under product of finite order very briefly. Exmple 4.43: Let S[x] 8 = i ix i S (Z 6 ); x 9 = 1 } i be the MOD subset coefficient polynomil semiroup under product. S[x] 8 is of finite order nd is commuttive. Let p(x) = {3 2 4} + { 5 1}x 5 nd q(x) = {1 3} + { } x 4 S[x] 8. p(x) q(x) = ({3 2 4} + { 5 1} x 5 ) ({1 3} + { } x 4 ) = {3 2 4} {1 3} + {3 2 4} { } x 4 ({ 7 5} {1 3}) x 5 + ({ 1 5} { })x 9 = {3 2 4 } + {3 2 4 }x 4 + { 1 5 3} x 5 + { } (s x 9 = 1} = ({3 2 4 } + { }) + {3 2 4 } x 4 + { 1 5 3} x 5 = { } + {3 2 4 }x 4 + { 1 5 3} x 5. This is the wy product opertion is performed on S[x] 8. The reder is left with the tsk of findin the order of S[x] 8. We see if p(x) = { 3} x 2 + {3} nd q(x) = { 2} + { 2 4}x 6 S[x] 8 then p(x) q(x) = [{3} + { 3}x 2 ] [{ 2} + { 2 4} x 6 } = {} + {} x 2 + {}x 6 + {}x 8 = ({}). Thus S[x] 8 hs zero divisors.

241 24 MOD Nturl Neutrosophic Subset Semiroups However S[x] 8 hs no idempotents or nilpotents. Exmple 4.44: Let S[x] 4 = i ix i S (Z 7 ); x 5 = 1 } i be the MOD subset coefficient polynomil semiroup under product opertion. Let p(x) = { 3 2} + {6 5 1} x 4 nd q(x) = {1 2} + {3 4 2} x 3 S[x] 4 x 3 ) p(x) q(x) = ({ 3 2} + {6 5 1} x 4 ) ({1 2} + {3 4 2} = { 3 2} {1 2} + { 3 2} {3 4 2}x 3 + {6 5 1} {1 2} x 4 + {6 5 1} {3 4 2}x 3 = { } + { }x 3 + { } x 4 + { }x 2 S[x] 4. This is the wy the product opertion is performed on S[x] 4. Clerly S[x] 4 hs no zero divisors or nilpotents or idempotents. But S[x] 4 is of finite order nd the reder is expected with the tsk of findin the order of S[x] 4. n view of ll these we hve the followin result. THEOREM 4.18: Let S[x] m = m i i x i S (Z n ); x m+1 = 1 } i be the MOD subset coefficient polynomil semiroup under product. i) o(s[x] m ) <.

242 Semiroups built on MOD Subset Mtrices 241 ii) S[x] m hs zero divisors nd nilpotents for pproprite n. iii) S[x] m hs no zero divisors nd nilpotents when n is prime. Proof is direct nd hence left s n exercise to the reder. Now we proceed onto describe by exmple the MOD finite complex number subset coefficient polynomil semiroup of finite order by exmples. Exmple 4.45: Let M[x] 6 = 6 i ix i S (C(Z 1 ); x 7 = 1 } i be the MOD finite complex number subset coefficient polynomil semiroup. The reder is left with the tsk of findin the order of M[x] 6. M[x] 6 hs MOD zero divisors but no nilpotents zero divisors but no nilpotents. nfct M[x] 6 hs idels. P[x] 6 = 6 i ix i S ({ }); x 2 = 1 } M[x] 6 i is MOD subset finite complex number coefficient polynomil subsemiroup which is not n idel of M[x] 6 is MOD subset finite complex number coefficient polynomil subsemiroup which is not n idel of M[x] 6. Let p(x) = { 5} + { 5 5i F } x 3 nd q(x) = { 2} + { 2i F 4} x + { 4 + 4i F 8 8i F } x 4 M[x] 6.

243 242 MOD Nturl Neutrosophic Subset Semiroups Clerly p(x) q(x) = ({}). Consider T[x] 6 = 6 i ix i S ({ 5 5i F 5 + 5i F }); x 7 = 1 } i M[x] 6 is in MOD finite complex number subset subsemiroup which is lso n idel of M[x] 6. However M[x] 6 hs no nilpotents. Exmple 4.46: Let B[x] 7 = 7 i ix i S (C(Z 7 )); x 8 = 1 } i be the MOD finite complex number subset coefficient polynomil semiroup under product of finite order B[x] 7 hs no idempotents zero divisors or nilpotents. Even findin MOD finite complex number coefficient subsemiroups nd idels hppens to be difficult tsk. nview of ll these we propose the followin problem. Problem 4.2: Let S[x] m = m i ix i S (C(Z p )); x m+1 = 1 } i be the MOD finite complex number subset coefficient polynomil semiroup (p prime). i) Cn S[x] m hve MOD finite complex number subset coefficient polynomil idels?

244 Semiroups built on MOD Subset Mtrices 243 ii) Cn S[x] m hve MOD zero divisor subset coefficient polynomil pirs? iii) Cn S[x] m hve MOD nilpotent subset coefficient polynomils? Next we describe one or two exmples of MOD neutrosophic subset coefficient polynomil semiroup of finite order. Exmple 4.47: Let W[x] 1 = 1 i ix i S{(Z 12 ) x 11 = 1 } i be the MOD neutrosophic subset coefficient polynomil semiroup under. W[x] 1 hs MOD neutrosophic subset coefficient polynomil zero divisors nd nilpotents. Also W[x] 1 hs idels. Exmple 4.48: Let 12 i V[x] 12 = { ix x 13 =1 i S{(Z 13 ) } i be the MOD neutrosophic subset coefficient polynomil semiroup under. V[x] 12 hs no MOD neutrosophic subset coefficient polynomil idempotent or nilpotent or zero divisors. V[x] 12 hs nontrivil MOD neutrosophic subset coefficient polynomils idels. The tsk of findin the order of V[x] 12 is left s n exercise to the reder.

245 244 MOD Nturl Neutrosophic Subset Semiroups Next we proceed onto describe MOD dul number subset coefficient polynomil semiroups under product by some exmples. Exmple 4.49: Let P[x] 15 = 15 i ix i S{(Z 14 ) x 16 = 1 } i be the MOD dul number subset coefficient polynomil semiroup under. Clerly P[x] 15 hs MOD dul number subset coefficient polynomils which re nilpotents of order two. Also P[x] 15 hs MOD dul number subset coefficient polynomils which re zero divisors. Let p(x) = { } + {4 8}x 3 + { }x 5 P[x] 15. Clerly p(x) p(x) = ({}). Let q(x) = {3 1} + {5 12 }x 3 + { }x 5 + { } x 8 P[x] 15. We see p(x) q(x) = ({}). This P[x] 5 hs MOD nilpotent polynomils s well s MOD zero divisor polynomil pirs. Further W[x] 15 = 15 i ix x 16 =1 i S(Z 14 ) } P[x] 15 i is MOD subset dul number coefficient subset polynomil idel of P[x] 15.

246 Semiroups built on MOD Subset Mtrices 245 Let B[x] 15 = 15 i ix i S(Z 14 ) } P[x] 15 ; i B[x] 15 is only MOD dul number coefficient subset polynomil subsemiroup of P[x] 15 which is not n idel of P[x] 5. n view of ll these we hve the followin theorem. THEOREM 4.19: Let S[x] m = m i i x i S{(Z n ) x 11 = i 1 } be the MOD dul number coefficient subset polynomil semiroup under. i) S[x] m hs severl MOD dul number coefficient subset polynomil zero divisors nd nilpotents even if n is prime. ii) S[x] m hs MOD dul number subset coefficient polynomil subsemiroups which re not idels even if n is prime. iii) S[x] m hs MOD dul number subset coefficient polynomil idel even if n is prime. iv) S[x] m hs no nontrivil MOD dul number subset coefficient polynomil idempotent. Proof is direct nd hence left s n exercise to the reder. Next we proceed onto describe MOD specil dul like number subset coefficient polynomil semiroups of finite order by some exmples. Exmple 4.5: Let 8 i W[x] 8 = ix i S{Z 12 h); x 9 = 1} i

247 246 MOD Nturl Neutrosophic Subset Semiroups be the MOD specil dul like number subset coefficient polynomil semiroup under. Clerly if p(x) = { 4 8} + { 8h 4h+4}x 2 + {4 + 8h 8h + 8 }x 4 nd q(x) = { 3h} + { 3h + 3 6h}x 2 + { 6 + 6h 6h 6 + 3h}x 5 W[x] 8 then clerly p(x) q(x) = ({}); tht is W[x] 8 hs non trivil MOD specil dul like number coefficient subset zero divisor pirs. Let p(x) = { 6} + { 6h}x +{ 6 + 6h}x 2 W[x] 8. We see p(x) p(x) = ({}) is MOD specil dul like number subset coefficient polynomil nilpotent of order two. However W[x] 8 hs no nontrivil MOD subset specil qusi dul number coefficient polynomil idempotents. We see V[x] 8 = 8 i ix i S(Z 12 h); x 9 = 1 } W[x] 8 i is MOD specil dul like number subset coefficient polynomil subsemiroup of W[x] 8 whicih is lso n idel of w[x] 8. Exmple 4.51: Let S[x] 1 = 1 i ix i S{(Z 13 h) x 11 = 1 } i be the MOD specil dul like number subset coefficient polynomil semiroup under. 1 i Clerly P[x] 1 = ix i S(Z 13 ); x 11 = 1 } S[x] 1 i

248 Semiroups built on MOD Subset Mtrices 247 is the MOD specil dul like number subset coefficient polynomil subsemiroup of S[x] 1 which is not n idel of S[x] 1. Consider R[x] 1 = 1 i ix i S(Z 13 h) x 11 = 1 } S[x] 1 i is MOD specil dul like number subset coefficient polynomil subsemiroup of S[x] 1 which is lso n idel of S[x] 1. However findin nontrivil MOD specil dul like number subset coefficient polynomil zero divisor pirs or nilpotents or idempotents in S[x] 1 is different tsk. n view of ll these we hve the followin theorem. THEOREM 4.2: Let Q[x] m = m i i x i S{(Z n h); x m+1 i = 1; } be the MOD specil dul like number subset coefficient polynomil semiroup under. i) o(q[x]) <. ii) Q[x] m hs MOD specil dul like number subset coefficient polynomil subsemiroups which re not idels (2 n < ) iii) Q[x] m hs MOD specil dul like number subset coefficient polynomil subsemiroups which re idels of Q[x] m. iv) f n is composite number of the form n = p t q; t 2 p prime nd q distinct prime from p or composite number then Q[x] m hs nontrivil MOD specil dul like number subset coefficient polynomils which re nilpotents or zero divisor pirs. v) Wht ever be n Q[x] m hs no nontrivil idempotent.

249 248 MOD Nturl Neutrosophic Subset Semiroups Proof is direct nd hence left s n exercise to the reder. Next we proceed onto describe MOD specil qusi dul number subset coefficient polynomils semiroups under by some exmples. Exmple 4.52: Let P[x] 6 = 6 i ix i S{(Z 2 k) x 7 = 1 } i be the MOD specil qusi dul number subset coefficient polynomil semiroup under. Let p(x) = { k 1} + {5 + 1k 1k 1 + 5l} x 2 + { 15k k} x 4 nd q(x) = {4 8k} + { k 12}x + { k k k} x 3 P[x] 6. Clerly p(x) q(x) = ({}). Thus there re MOD specil qusi dul number subset coefficient polynomil zero divisor pirs. We cn hve only nilpotents of the form { 1} + {1 1k }x 2 + { 1 + 1k}x 3 + { 1k}x 4 = p(x) such tht p(x) p(x) = ({}). However we do not hve nontrivil idempotents. Only very few nilpotents risin from subset coefficient { 1 1k 1 + 1}. But

250 Semiroups built on MOD Subset Mtrices 249 D[x] 6 = 6 i ix i S(Z 2 ) x 7 = 1 } P[x] 6 i is MOD specil qusi dul number subset coefficient polynomil subsemiroup which is not n idel of P[x] 6. Consider E[x] 6 = 6 i ix i S(Z 2 k); x 7 = 1 } P[x] 6 ; i clerly E[x] 6 is MOD specil qusi dul number subset coefficient polynomil subsemiroup which is lso n idel of P[x] 6. Exmple 4.53: Let G[x] 9 = 9 i ix i S{(Z 5 k) x 1 = 1 } i be the MOD specil qusi dul number subset coefficient polynomil semiroup under product. G[x] 9 hs no nontrivil nilpotents zero divisors or idempotents. However G[x] 9 hs both MOD specil qusi dul number subset coefficient polynomil idels nd subsemiroups which re not idels. n view of ll these we hv the followin theorem. THEOREM 4.21: Let H[x] m = m i i x i S{(Z n k) x m+1 i = 1 } be the MOD specil qusi dul number subset coefficient polynomil semiroup under.

251 25 MOD Nturl Neutrosophic Subset Semiroups i) H[x] m hs both MOD specil qusi dul number subset coefficient polynomil subsemiroup which is not n idel nd MOD specil qusi dul number subset polynomil subsemiroup which is n idel for ll 2 n <. ii) H[x] m hs no nontrivil MOD specil qusi dul number subset coefficient polynomil idempotents for 2 n <. iii) H[x] m hs nontrivil MOD specil qusi dul number subset coefficient polynomil zero divisors nd nilpotents in n if n is of the type n = p t q (t 2 p prime nd p q nd composite or some other prime). Prime is direct nd hence left s n exercise to the reder. n lmost ll cses these MOD subset coefficient polynomil semiroups re Smrndche semiroups. n view of this we hve the followin result. THEOREM 4.22: Let S[x] (or S[x] m ) be the MOD subset (finite complex or dul number or specil qusi dul number of specil dul like number or specil neutrosophic) coefficient polynomil semiroup under infinite semiroup (or finite semiroup) respectively. S[x] (or S[x] m ) is Smrndche MOD semiroup if nd only if Z n in S(Z n ) or Z n in S(C(Z n )) or S(Z n ) or S(Z n k) or S(Z n h) or S(Z n ) is Smrndche semiroup. Proof : Let S[x] = i ix i S(Z n ) (or S(C(Z n )) or S(Z n i ) or S(Z n h) or S (Z n k) S(Z n )) } (S[x] m = m { i x i / i S(Z n ) (or S(C(Z n ) or S(Z n ) or S(Z n ) i or S(Z n h) or S(Z n k)) x m+1 = 1 } be the MOD subset coefficient polynomil semiroup under product.

252 Semiroups built on MOD Subset Mtrices 251 f S(Z n ) is Smrndche semiroup then n is prime or Z n hs subsemiroup H Z n such tht H is roup under product. Hence B = {{1} {2} {3} {4} {5} {6} {p}} (or if h = {h 1 h t } is roup under product H Z n then M = {{h 1 } {h 2 } {h t }}) under product is subset roup under if nd only if S(Z n ) is S-semiroup under product. Thus the result. We will illustrte this sitution by n exmple or two. Exmple 4.54: Let S[x] = i ix i S(Z 13 ); } i be the MOD subset coefficient polynomil semiroup under product. B = {{1} {2} {3} {4} {5} {6} {12}} S[x] is MOD subset subsemiroup under which is in fct roup. Hence S[x] is Smrndche MOD - subset coefficient polynomil semiroup under. Exmple 4.55: Let S[x] = i ix i S(C(Z 12 )); } i be the MOD subset finite complex coefficient polynomil semiroup P 1 = {{1} {5}} P 2 = {{1} {11}} nd P 3 = {{1} {7}} re ll subroups under nd they re cyclic roups of order two. Hence S[x] is MOD subset finite complex number polynomil coefficient Smrndche semiroup.

253 252 MOD Nturl Neutrosophic Subset Semiroups Exmple 4.56: Let 9 i B[x] 9 = ix x 1 = 1 i S(Z 17 h) } i be the MOD subset specil dul like number coefficient polynomil semiroup under product. B[x] 9 is Smrndche MOD subset specil dul like number coefficient polynomil semiroup s T = {{1} {2} {3} {4} {16} B[x] 9 is roup under. nterested reder cn et more properties bout these Smrndche notions. nfct we cn hve the followin exmple. Exmple 4.57: Let S[x] 1 = 1 i ix i S(Z 24 ); x 11 = 1 } i be the MOD dul number coefficient subset polynomil semiroup under product. S[x] 1 is S-semiroup s {{1} {7}} = B 1 nd B 2 = {{1} {23}} re subset roups under product. 1 x {1} {7} {1} {1} {7} {7{ {7} {1} Clerly B 1 is cyclic subset roup of order two. Tble for B 2

254 Semiroups built on MOD Subset Mtrices 253 x {1} {23} {1} {1} {23} {23{ {23} {1} s MOD subset subroup of S[x] 1. Hence our clim S[x] 1 is Smrndche MOD subset dul number coefficient polynomil semiroup of finite order. nfct we cn sy under ll MOD subset semiroups re Smrndche MOD subset semiroups wht ever be n of Z n. Next we wnt to briefly describe MOD subset mtrices tht is collection of subset mtrices where mtrices tke entries from Z n or C(Z n ) or Z n or Z n or Z n k or Z n h. We describe them by exmples. Secondly we lso build MOD subset polynomils with coefficients from Z n or C(Z n ) or Z n nd so on. We will describe both the situtions by some exmples. We ive definition of both these notions. DEFNTON 4.1: Let M = {collection of ll m t mtrices with entries from Z n or C(Z n ) or Z n or Z n h or Z n or Z n k under n nturl product opertion}. S(M) = {collection of ll subset mtrices with elements from M}. We cn define n on S(M) nd {S(M) n } is defined s the MOD subset mtrix semiroup. i DEFNTON 4.2: Let S[x] = i x i Z n or Z n or i Z n k or C(Z n ); } be the MOD polynomils under. Let S(S[x]) = {collection of ll subsets from S[x]}.

255 254 MOD Nturl Neutrosophic Subset Semiroups {S(S[x]) } is defined s the MOD subset polynomils semiroup under of infinite order. f S[x] is replced by S[x] m then {S(S[x] m ) } is defined s the MOD subset polynomil semiroup under. Clerly S(S[x] m ) is of finite order nd is infct commuttive monoid. Next we proceed onto describe the MOD subset mtrix sets by some exmples. Exmple 4.58: Let M = {collection of ll 1 5 row mtrices with entries from Z 9 } = {{ } {b 1 b 2 b 3 b 4 b 5 } {t 1 t 2 t p }} where i b i nd t j re 1 5 mtrices with entries from Z 9. Let P = {( 1 2) (2 3 4 ) ( 8 )} nd Q = {( ) ( ) ( 1 1) (2 5 6)} M. P Q = {( 1 2) (2 3 4 ) ( 8 )} {(21 1 2) ( ) {( 1 1) (2 5 6)} = {( 1 4) (4 3 ) ( 8 ) ( 1 ) (2 3 4 ) ( 8 ) ( 2) ( ) ( 5 3) (4 6 ) ( 4 )}. This is the wy product opertion is performed on M. Clerly {( )} M is such tht A {( )} = {( )} A = {( )} for ll A M. B = {( )} M is such tht A B = B A = A for ll A M.

256 Semiroups built on MOD Subset Mtrices 255 Exmple 4.59: Let S(P) = {collection of ll subset mtrices from the set of mtrices P = i Z 12 ; 1 i 3 n } n } be the MOD subset mtrix semiroup under nturl product n. 1 A = { } nd B = { } S(P) 1 1 A n B = { } n { } 1 1 = { 2 1 n n n n n n n 1 3 } 1 1 = { } S(P). 3 This is the wy product opertion is performed on S(P). Clerly S(P) hs MOD subset mtrix zero divisors idempotents nd nilpotents iven by

257 256 MOD Nturl Neutrosophic Subset Semiroups A = { 9 4 is such tht A n A = { } S(P) 4 }. 9 Thus A is MOD subset idempotent mtrix of S(P). Let B = { } nd 8 6 A = { } S(P). 6 Clerly A n B = { }. Thus A B is MOD zero divisor subset pir of S(P). 6 Let B = { } S(P). We see B n B = { }.

258 Semiroups built on MOD Subset Mtrices 257 Thus B is MOD subset nilpotent mtrix of order two in S(P). Let S(M) = {collection of ll subsets of mtrices from M = i { } 1 i 3} n } be the MOD subset mtrix subsemiroup of S(P) which is lso MOD subset mtrix idel of S(P). Let S(W) = {collection of ll subsets from W = Z 12 n } n } S(P) be MOD subset mtrix idel of S(P). Thus S(P) in this sitution hs MOD subset subsemiroups idels idempotents zero divisors nd nilpotents. Exmple 4.6: Let S(W) = {collection of ll subsets from W = i C(Z 5 ); n ; 1 i 6} n } be the MOD subset finite complex mtrix semiroup under the nturl product. S(W) hs MOD finite complex number subset subsemiroups s well s idels. However findin nontrivil zero divisors is chllenin one.

259 258 MOD Nturl Neutrosophic Subset Semiroups For we sy if A = { ) nd 2 3 b1 b2 B = { } S(W) then b3 A n B = { }; such type of MOD mtrix zero divisors will only be termed s trivil MOD subset mtrix zero divisors. S(W) hs no nontrivil MOD subset finite complex number idempotents or nilpotents or zero divisors. S(M) = {collection of ll subset mtrices from M = i Z 5 1 i 6 n } n } S(W) is only MOD subset finite complex number mtrix subsemiroup which is not n idel of S(W). Consider S(N) = {collection of ll subset finite complex number mtrices from 1 N = { 2 / 1 2 C(Z 5 ) n } n }} S(W) is MOD finite complex number subset mtrix subsemiroup which is lso n idel of S(W). Exmple 4.61: Let S(Y) = {collection of ll MOD subsets dul number mtrices from

260 Semiroups built on MOD Subset Mtrices 259 Y = { / i Z 6 ; 1 i 6 n } n } be the MOD subset dul number mtrix semiroup under the nturl product n. 3 4 Let A = { } nd B = { } S(Y); 2 5 clerly A n B = { } nd A n A = { } nd B n B = { }.

261 26 MOD Nturl Neutrosophic Subset Semiroups Thus S(Y) hs MOD subset dul number mtrix zero divisors nd nilpotents. 4 M = { } S(Y) is such tht M n M = M so M is MOD subset dul number mtrix idempotent of S(Y). This S(Y) hs MOD subset dul number mtrix idels s well s subsemiroups which re not idels. This tsk is left s n exercise to the reder s it is considered s mtter of routine. Here we proceed onto ive exmples of MOD subset mtrix sets nd their properties under nturl product n. Exmple 4.62: Let S(P) = {collection of ll subsets from the set of MOD specil qusi dul number 6 1 mtrices i Z 12 k 1 i 6 n }} be the MOD subset mtrices set.

262 Semiroups built on MOD Subset Mtrices 261 Let A = { 1 k 8 9 k k } B ={ 9k 3 6 k 6k 2 } nd C = { k 1 5 k 1k k k k }; clerly it is left s n exercise to verify tht (A n B) n C = A n (B n C).

263 262 MOD Nturl Neutrosophic Subset Semiroups Clerly { } = {()} cts s {()} n A = A n {()} = {()} for ll A S(P). Further {(1)} = S(P) is such tht A {(1)} = {(1)} A = A for ll A S(P). t is esily verified S(P) hs MOD subset specil qusi dul number mtrix zero divisors nilpotents nd idempotents. Further S(P) hs MOD subset specil qusi dul number mtrix subsemiroups which re not idels s well s subsemiroups which re idels. Exmple 4.63: Let B = i Z 11 ; 1 i 4 (or n )};

264 Semiroups built on MOD Subset Mtrices 263 S(B) = {collection of ll subsets from B or ( n )} be the MOD subset neutrosophic mtrix semiroup under (or n ) {S(B) } be non commuttive MOD subset neutrosophic mtrix semiroup; wheres {S(B) n } is MOD subset neutrosophic commuttive mtrix semiroup. S(T) = {collection of ll subset mtrices from T = i Z 11 1 i 4} n (or )} S(B) is MOD subset neutrosophic mtrix idel of S(B). S(V) = {collection of ll subset mtrices from V = i Z 11 ; 1 i 4} (or n )} S(B) is MOD subset mtrix neutrosophic subsemiroup of S(B) which is not n idel of S(B). Now hvin seen MOD subset mtrix semiroups under product we ive few relted results. THEOREM 4.23: Let S(M) = {collection of ll subsets from M = {m t mtrices with entries from Z n (or C(Z n ) or Z n or Z n h or Z n or Z n k; n (m t)} n } be the MOD subset mtrix semiroup. i) o(s(m)) <. ii) S(M) is MOD subset mtrix commuttive monoid. iii) S(M) hs MOD subset subsemiroups which re not idels. iv) S(M) hs MOD subset subsemiroups which re idels. v) S(M) hs MOD zero divisors nilpotents nd idempotents for pproprite n.

265 264 MOD Nturl Neutrosophic Subset Semiroups Proof is direct nd left s n exercise to the reder. All properties (i) to (v) hold ood if Z n is replced by C(Z n ) or Z n or Z n h or Z n k or Z n. Corollry 4.2: f in the bove theorem m = t then ll results of theorem hold ood under the usul product n. Corollry 4.3: f in the bove theorem m = t nd the nturl product n is replced by the usul product the MOD subset semiroup hs riht idels which re not left idels subsemiroups which re not idels riht zero divisors which re not left zero divisors. The proof of the bove two corollries is left s n exercise to the reder. Next we proceed onto describe by exmples MOD subset polynomil semiroups. Exmple 4.64: Let S(M[x]) = {collection of ll subset from the MOD semiroups of polynomils M[x] = i ix i Z 1 } } i be the MOD subset polynomil semiroup. Clerly o(s(m[x])) =. Let A = {x x + 4 5} nd B = {2x x 4 + 8x 2 + 2} S(M[x]). A B = {x x + 4 5} {2x x 4 + 8x 2 + 2} = {5 2x x 4 + 5x 3 2x x 4 + 8x x 7 + 8x 5 + 2x 3 6x x 2 + 6x 4 + 6x + 2x 5 + 6x 5 } S(M[x]). This is the wy product opertion is performed on S(M[x]). Let {()} = {( + x + x x n )} be the MOD subset polynomil.

266 Semiroups built on MOD Subset Mtrices 265 Clerly {()} p(x) = {()} for ll {p(x)} S(M[x]); P[x] = {collection of some polynomils from M[x]}. Likewise if {(1)} = {(1 + x + + x n )} S(M[x]); then P[x] {(1)} = P[x]. Let T[x] = {5x + 5 5x 3 5x 9 + 5x 12 5} nd S[x] = {2x + 4 4x 2 + 8x + 2 8x 6 + 4x + 2} S(M[x]); T[x] S[x] = {()}. However S(M[x]) hs no MOD subset polynomil idempotents whtever be n. i Consider (P[x]) = ix i { } } let i S(P[x]) = {collection of ll MOD subset polynomils from P[x] } S(M[x]) is MOD subset polynomil subsemiroup which is lso MOD subset polynomil idel of S(M[x]). The tsk of findin MOD subset polynomil structures is left s n exercise to the reder. Exmple 4.65: Let S(W[x]) = {collection of ll MOD subset polynomils from the set W[x] = i ix i (Z 17 ) } i be the MOD dul number subset polynomil semiroup. Let p(x) = {3x 4 + 4x x x} nd q(x) = {14 x 4 + 1x x x x 18 1x } S(W[x]). Clerly p(x) q(x) = {()} p(x) q(x) = {()} nd q(x) q(x) = {()} thus S(W[x]) hs MOD subset dul number polynomils which re MOD subset dul number zero divisor

267 266 MOD Nturl Neutrosophic Subset Semiroups polynomil pirs nd MOD subset dul number nilpotent polynomils of order two. nfct S(W[x]) hs MOD subset dul number polynomil subsemiroups which re not idels s well s subsemiroups which re idels. This tsk is left s n exercise to the reder. Exmple 4.66: Let V[x] = i ix i C(Z 12 ) } i be the MOD finite complex number coefficient polynomil semiroup under. S(V[x]) = {collection of ll subset polynomils from V[x] } be the MOD complex coefficient polynomil subsets semiroup. t is left s n exercise for the reder to verify tht S(V[x]) hs MOD subset zero divisors nd MOD subset nilpotents. nfct S(V[x]) hs MOD subset subsemiroups s well s idels. Both idels nd subsemiroups of S(V[x]) re of infinite order. Exmple 4.67: Let S(M[x]) = {collection of ll subsets from the MOD neutrosophic coefficient polynomils from M[x] = i ix i Z 19 } } i be the MOD subset neutrosophic coefficient polynomils semiroup under product opertion. Reder is left with the tsk of findin MOD subset neutrosophic coefficient idels nd subsemiroups.

268 Semiroups built on MOD Subset Mtrices 267 Findin zero divisors nd nilpotents hppens to be difficult job hs no nontrivil idempotent. The reder is left with the tsk of studyin MOD semiroups by replcin Z 19 in exmple 4.68 by Z n h nd lter by Z n k nd obtin similr results. As this study is mtter of routine the reder is left t this tsk. However we ive the followin theorem. THEOREM 4.24: Let S(P[x]) = {collection of ll subsets from the MOD polynomils P[x] = i i x i Z n } (or i P[x] = i i x i C(Z n ) } or i P[x] = i i x i Z n } or i P[x] = i i x i i Z n } or P[x] = i i x i Z n h } or i P[x] = i i x i i Z n k }) nd so on. i) S(P[x]) hs zero divisors nd nilpotents for pproprite n. ii) S([x]) hs MOD subset polynomil subsemiroups nd idels for ll n in cse of Z n h nd Z n k.

269 268 MOD Nturl Neutrosophic Subset Semiroups iii) S(P[x]) hs MOD subset polynomil subsemiroups nd idels for pproprite n when C(Z n ) or Z n is used. Proof is direct nd hence left s n exercise to the reder. We now briefly illustrte S(P[x] m ) by some exmples. Exmple 4.68: Let S(P[x] 9 ) = {collection of ll subsets from P[x] 9 = 9 i ix i Z 12 x 1 = 1 } } i be the MOD neutrosophic polynomil subset semiroup under product. S(P[x] 9 ) is of finite order. Let A = {1x 3 + 3x + 4 (2+4) x 2 + 2} nd B = {5x 3 + 7} S[p[x] 9 ). We cn clculte A B s mtter of routine. nfct S(P[x] 9 ) hs MOD zero divisors nd nilpotents of course only trivil idempotents. Exmple 4.69: Let S(M[x] 3 ) = {collection of ll subsets from M[x] 3 = 3 i ix x 4 = 1 i C(Z 7 ) } } i be the MOD finite complex number coefficient subset polynomil semiroup. The reder is expected to find substructures nd specil elements like nilpotents nd idempotents if it exists.

270 Semiroups built on MOD Subset Mtrices 269 n view of ll these we cn hve the followin result. THEOREM 4.25: Let S(M[x] m ) = {collection of ll subsets from MOD polynomil semiroup M[x] m = m i i x i i Z n x m+1 = 1 } (or M[x] m = m i i x i C(Z n ) x m+1 = 1 } or i M[x] m = m i i x i i Z n x m+1 = 1 } or M[x] m = m i i x i Z n x m+1 = 1 } or i M[x] m = m i i x i i Z n h x m+1 = 1 } or M[x] m = m i i x i Z n k x m+1 = 1x } i be the MOD subset polynomil semiroup (or MOD) subset finite complex number polynomil semiroup or MOD subset neutrosophic polynomil semiroup or MOD subset dul number polynomil semiroup or MOD subset specil dul like number subset polynomil semiroup nd MOD subset specil qusi dul number polynomil semiroup respectively. i) S(M[x] m ) hs MOD subset polynomil subsemirins nd idels for C(Z n ) Z n Z n Z n h nd Z n k. ii) o(s(m[x])) < for ll sets Z n C(Z n ) nd so on. iii) Clerly S(M[x] m ) hs zero divisors nd nilpotents.

271 27 MOD Nturl Neutrosophic Subset Semiroups iv) The MOD dul number polynomil subset S(M[x] m ) usin Z n behves distinctly different from other MOD semiroups. Proof is direct nd hence left s n exercise to the reder. Next we proceed onto describe MOD mtrix subset collection under + opertion by exmples. Exmple 4.7: Let P(M) = {collection of ll subsets from the MOD mtrix semiroup 1 2 M = 3 4 i Z 8 ; 1 i 4 +} +} be the MOD mtrix subset semiroup under ddition. Let A ={ } nd 2 3 B = { } P(M). 1 1 A + B = { } + { } 1 1

272 Semiroups built on MOD Subset Mtrices 271 = { } = { } 3 1 This is the wy + opertion is performed on S(M). S(M). S(M) is MOD subset mtrix commuttive monoid of finite order s {()} = { } S(M) is such tht A + {()} = A for ll A S(M). Clerly S(M) hs MOD subsemiroup. Exmple 4.71: Let M = i Z 13 k 1 i 6 +} be the MOD specil qusi dul number mtrix semiroup.

273 272 MOD Nturl Neutrosophic Subset Semiroups S(M) = {collection of ll subsets mtrices from M under +} be the MOD specil qusi dul number subset mtrix semiroup under +. Let A = nd 5 B = S(M) A + B k k 2k k = k k k k k k k k k k k k S(M). This is the wy + opertion is performed on S(M). The reder is left with the tsk of findin MOD specil dul like number subset mtrix subsemiroup.

274 Semiroups built on MOD Subset Mtrices 273 Next we describe + opertion on S(P[x]) the MOD polynomil subsets by some exmples. Exmple 4.72: Let P[x] = i ix i Z 9 ; +} i be the MOD dul number coefficient polynomil semiroup under +. Let S(P[x]) = {collection of ll subsets from P[x] +} be the MOD dul number coefficient polynomil subset semiroup under +. Let A = {(3 + ) x 8 + (4 + 8) 5x 3 + 8} nd B = {(4 + 3) x 2 + 7x + (8 + 2)} S(P[x]). A + B = {(3 + ) x 8 + (4 + 8) 5x 3 + 8} + {(4 + 3) x 2 + 7x } = (3 + ) x 8 + (4 + 3)x 2 + 7x + (3 + ) 5x 3 + (4 + 3)x 2 + 7x } S(P[x]). This is the wy + opertion is performed on S(P[x]). The reder is left with the tsk of provin (S(P[x]) +) is MOD dul number coefficient polynomil monoid under + of infinite order. Clerly {()} = { +.x +.x x n } S(P[x]) is such tht for every A S(P[x]); A + {()} = A. Hence S(P[x]) is n infinite order monoid. The reder is left with the tsk of findin subsemiroups nd submonoids of S(P[x]). We ive yet nother exmple.

275 274 MOD Nturl Neutrosophic Subset Semiroups Exmple 4.73: Let S(P[x]) = {collection of ll MOD finite complex number coefficient polynomil subsets with entries from P[x] = i ix i C(Z 15 ); } i be the MOD finite complex number coefficient polynomils subsets semiroup under + opertion. P[x] hs subsemiroups of both finite nd infinite order. The reder is left with the tsk of findin such subsemiroups. Exmple 4.74: Let S(M[x]) = {collection of ll subsets from the MOD neutrosophic coefficient polynomil semiroup M[x] = i ix i Z 11 } i be the MOD neutrosophic coefficient polynomil subset semiroup under +. o(s(m[x]) = nd S(M[x]) is nd S(M[x]) is n infinite monoid. Let S(P[x]) = {collection of ll MOD neutrosophic polynomil coefficient from the semiroup; P[x] = i ix i Z 11 } S(M[x]) be the MOD i neutrosophic polynomil coefficient subset semiroup. 1 i Let M = ix i < Z 11 > } nd i

276 Semiroups built on MOD Subset Mtrices 275 S(M) = {collection of ll subsets from M +} is MOD neutrosophic polynomil coefficient subset subsemiroup of finite order. Thus S(M[x]) cn hve severl MOD neutrosophic coefficient polynomil subset subsemiroup of finite order under +. Let S(W) = {collection of ll MOD neutrosophic polynomil coefficient subsets from W = 7 i ix i Z 11 ; +} +} i be the MOD neutrosophic polynomil coefficient subset subsemiroup of finite. The reder is left with the tsk of findin o(s(m)) nd o(s(w)). Exmple 4.75: Let S(M[x] 9 ) = {collection of ll MOD dul number coefficient polynomil subsets from M[x] 9 = 9 i ix i Z 12 x 1 = 1 } i be the MOD dul number coefficient polynomil subset semiroup under +. o(s(m[x] 9 ) is finite nd hs finite order MOD dul number coefficient polynomil subsemiroups. Exmple 4.76: Let S(V[x] 12 ) = {collection of ll subsets from V[x] 12 = 12 i ix i Z 1 x 13 = 1 } i be the MOD neutrosophic coefficient polynomil subset semiroup of finite order.

277 276 MOD Nturl Neutrosophic Subset Semiroups This hs subsemiroups ll of which re finite order. S(P[x] 12 ) = {collection of ll subsets from P[x] 12 = 12 i ix i i Z 12 ; x 13 = 1 +} is MOD neutrosophic coefficient polynomil subset subsemiroup of finite order. Let A = {5x 3 + 6x x 6 } nd B = {5 + 6x 3 1x 7 + 3} S(P[x] 12 ). A + B = {5x 3 + 6x x 6 } + {5 + 6x 3 1x 3} = {5x 3 6x x x x 3 5x 3 + 6x x x 6 + 1x} = {11x 3 + 6x + 9x 6 + 6x x 6 + 5x 3 + 6x + 6 9x 6 + 1x 7 + } S(P[x] 12 ). This is the wy the + opertion is performed on S(P[x] 12 ). nterested reder cn nlyse the relted properties of S(P[x] 12 ). Exmple 4.77: Let S(W[x] 18 ) = {collection of ll MOD polynomils subsets from the MOD polynomil semiroup. W[x] 18 = 18 i ix i Z 17 x 19 = 1 } i be the MOD polynomil subset semiroup under +. Let A = {x 1 16x x 5 + 2} nd B = {3x x } S(W[x] 18 ). 8} A + B = {x 1 6x x 5 + 2} + {3x x 4 + 1

278 Semiroups built on MOD Subset Mtrices 277 = {x 1 + 3x x 7 + 6x x x 7 + 3x x x 1 6x x 4 1x x 5 + 1x x 1 6x x x x x 4 + 1} S(W[x] 18 ). The reder is left with the tsk of findin MOD polynomil subset subsemiroups. Thus we see one cn hve two opertions MOD polynomil subsets. Under both of them they behve differently. Further we cn hve likewise two types of opertions on MOD subset mtrices they re lso different nd distinct. t is left s n exercise to work with these structures. nfct we suest set of problems some of which re chllenin some of them esy just routine exercises. Problems: 1. Let M = {( ) / i S(Z 9 ) 1 i 5 +} be the MOD subset mtrix semiroup under +. i) Find order of M. ii) How mny MOD subset subsemiroups of M re there? iii) s M Smrndche MOD subset mtrix semiroup? 2. Let S = { / i S(Z 43 ) 1 i 8 +} be the MOD subset mtrix semiroup under +.

279 278 MOD Nturl Neutrosophic Subset Semiroups i) Study questions (i) to (iii) of problem (1) for this S. ii) Compre S with M is problems (1). 3. Let A = { i S(Z 48 ) 1 i 15 +} be the MOD subset mtrix semiroup. i) Study questions (i) to (iii) of problem (1) for this A. ii) Compre A with S nd M of problems (2) nd (1) respectively. 4. Obtin ny other specil nd interestin fetures ssocited with M = {( ij ) st / ij S(Z n ); 1 i s nd 1 j t +}. 5. Let B = { / i S(C(Z 1 )); 1 i 15 +} be the MOD semiroup. subset finite complex number mtrix 6. Let Study questions (i) to (iii) of problem (1) for this B.

280 Semiroups built on MOD Subset Mtrices T = { / i S(Z 41 ); 1 i 9 +} 7. Let be the MOD finite complex number mtrix semiroup. Study questions (i) to (iii) of problem (1) for this T W= i S(Z 7 ; 1 i 15 +} be the MOD neutrosophic subset mtrix semiroup. i) Study questions (i) to (iii) of problem (1) for this W. ii) Compre W with T of problem Let V = { / i S(Z 12 ; 1 i 15 +} be the MOD subset dul number mtrix semiroup. i) Study questions (i) to (iii) of problem (1) for this V. ii) Compre V with W in problem Obtin ll specil nd strikin fetures enjoyed by MOD subset dul number mtrix semiroups.

281 28 MOD Nturl Neutrosophic Subset Semiroups 1. Let S = { / i S(Z 12 h; 1 i 1 +} 11. Let be the MOD specil dul like number mtrix semiroup. Study questions (i) to (iii) of problem (1) for this S. D = { / i S(Z 15 k; 1 i 16 +} be the MOD specil qusi dul number mtrix semiroup. Study questions (i) to (iii) problem (1) for this D. 12. Enumerte ll specil nd distinct fetures ssocites with MOD subset specil qusi dul number mtrix semiroup. 13. Let M = {( ) / i S(Z 6 ); 1 i 6 } be the MOD subset mtrix semiroup. i) Find o(m). ii) Find ll MOD subset mtrix subsemiroups which re not idels. iii) Find ll MOD subset mtrix subsemiroups which re idels. iv) Find ll MOD subset zero divisors nd S-zero divisors if ny. v) Find ll MOD subset nilpotent nd idempotents of M. vi) Obtin ny other specil feture enjoyed by M nd compre some M under +.

282 Semiroups built on MOD Subset Mtrices Let S = { / i S(C(Z 12 )); 1 i 15 n } be the MOD subset finite complex number semiroup under. 15. Let Study questions (i) to (v) of problem (13) for this S. P = { / i S(Z 1 ; 1 i 12 n } 16. Let be the MOD subset neutrosophic semiroup under. Study questions (i) to (v) of problem (13) for this P. V = { / i S(Z 23 ; 1 i 16 n }

283 282 MOD Nturl Neutrosophic Subset Semiroups be the MOD subset dul number mtrix semiroup under product. Study questions (i) to (v) of problem (13) for this V. 17. Obtin ll specil fetures enjoyed by MOD subset dul number mtrix semiroups nd compre it with MOD subset neutrosophic mtrix semiroups both under. 18. Let N = { / i S(Z 15 k; 1 i 15 n } be the MOD subset specil qusi dul number semiroup. 19. Let Study questions (i) to (v) of problem (13) for this N x = { / i S(Z 19 ; 1 i 9 n } be the MOD subset neutrosophic semiroup under nturl product n. i) Study questions (i) to (v) of problem (13) for this n. ii) Replce the opertion n by nd study nd distinuish ll the properties under n nd under.

284 Semiroups built on MOD Subset Mtrices Let S[x] = i ix i S(Z 12 ); +} i be the MOD subset polynomil coefficient semiroup under +. i) Find ll MOD subset subsemiroups of S[x] ii) Show S[x] cn hve finite order subset subsemiroups of S[x]. iii) Obtin ny other specil feture ssocited by S[x]. 21. Let P[x] = i ix i S(C(Z 17 )) +} i be the MOD subset finite complex number polynomil coefficient semiroup. Study questions (i) to (iii) of problem (2) for this P[x]. 22. Let D[x] = i ix i S(Z 18 ) be the MOD i neutrosophic subset coefficient polynomil semiroup. Study questions (i) to (iii) of problem (2) for this D[x]. 23. Let W[x] = i ix i S(Z 23 ) } be the MOD dul i number subset coefficient polynomil semiroup. Study questions (i) to (iii) of problem (2) for this W[x].

285 284 MOD Nturl Neutrosophic Subset Semiroups 24. Let B[x] = i ix i S(Z 15 h) } i be the MOD specil dul like number subset coefficient polynomil semiroup. Study questions (i) to (iii) of problem (2) for this B[x]. 25. Let L[x] = i ix i S(Z 18 k) } i be the MOD specil qusi dul number subset coefficient semiroup. Study questions (i) to (iii) of problem (2) for this L[x]. 26. Obtin ll specil nd distinct fetures enjoyed by the 6 different MOD subset coefficient polynomils semiroup under +. Compre nd construct them with ech other by enumertin ll specil fetures. 27. Let T[x] 9 = 9 i ix i S(Z 48 ); x 1 = 1} i be the MOD subset coefficient polynomil subsemiroups of finite order under + i) Find o(t[x] 9 ). ii) Find ll MOD subset coefficient polynomil subsemiroups. iii) Cn we sy usin S (Z 48 ) will yield more MOD subset coefficient polynomil subsemiroups thn by usin

286 Semiroups built on MOD Subset Mtrices 285 S(Z 47 )? 28. Let M[x] 2 = 2 i ix i S(C(Z 28 )) x 21 = 1 +} i be the MOD subset finite complex number coefficient polynomil semiroup under +. i) Study questions (i) to (ii) of problem (27) for this M[x] 2. ii) Enumerte ll specil fetures ssocited with M[x] 2 nd compre it with T[x] 9 in problem (27) 29. Let 5 W[x] 5 = { i x i / i SZ 1 x 6 = 1 } i be the MOD subset finite neutrosophic coefficient polynomil semiroup. i) Study questions (i) to (ii) of problem (27) for this W[x] 5. ii) Compre W[x] 5 with M[x] 2 nd T[x] 9 of problem 28 nd 27 respectively. 3. Let V[x] 8 = { i x i / i S(Z 2 k x 9 = 1 } i be the MOD subset specil qusi dul number coefficient polynomil semiroup under +. Study questions (i) to (ii) of problem (27) for this V[x] 8.

287 286 MOD Nturl Neutrosophic Subset Semiroups 31. Wht re the specil fetures ssocited with MOD subset specil qusi dul number subset coefficient polynomil semiroups under Let 18 F[x] 8 = { i x i / i S(Z 17 x 19 = 1 } i be the MOD subset specil dul number coefficient polynomil semiroup under +. Study questions (i) to (ii) of problem (27) for this F[x] Let 8 P[x] 8 = { i x i / i S(Z 27 h); x 9 = 1 +} i be the MOD subset specil dul like number coefficient polynomil semiroup under +. i) Study questions (i) to (ii) of problem (27) for this P[x] 8. ii) Study ll specil nd distinct fetures ssocited with MOD subset specil dul like number coefficient polynomil semiroups. 34. Let S[x] = { i x i / i S(Z 12 )} i be the MOD subset coefficient polynomil semiroup under. i) Find ll MOD subset coefficient polynomil subsemiroup of under. ii) Find ll MOD subset coefficient polynomil idels of S[x] under. iii) Find ll MOD zero divisors of S[x]. iv) Find ll MOD nilpotents of S[x].

288 Semiroups built on MOD Subset Mtrices 287 v) Prove S[x] cnnot hve MOD idempotents. vi) Cn S[x] of finite order MOD subsemiroup? 35. Let P[x] = { i x i / i S(Z 18 } be the MOD dul i number subset coefficient polynomil semiroup under. Study questions (i) to (vi) of problem (34) for this P[x]. 36. Let i T[x] = ix i SC(Z 19 ); } i be the MOD subset finite complex number coefficient polynomil semiroup under. i) Study questions (i) to (vi) of problem (34) for this T[x]. ii) Find ll specil feture enjoyed by MOD subset finite complex number coefficient polynomil under. 37. Let W[x] = i ix i S(Z 92 k; } i be the MOD specil qusi dul number subset coefficient polynomil semiroup. i) Study questions (i) to (vi) of problem (33) for this M[x] ii) Find ll specil fetures enjoyed by the MOD subset specil dul like number coefficient polynomil semiroup Let P[x] = { i x i / i S(Z 24 ); x 1 = 1 } be the MOD i subset polynomil semiroup under. i) Find o(p[x])

289 288 MOD Nturl Neutrosophic Subset Semiroups ii) Find ll MOD subset polynomil subsemiroups which re not idels. iii) Find ll MOD subset polynomil subsemiroups which re idels. iv) Find ll MOD zero divisors nd MOD nilpotents of P[x]. v) Give ll specil fetures ssocited P[x] Let M[x] = { i x i / i S(Z 12 ; x 7 = 1 } be the i MOD subset dul number polynomil semiroup. i) Study questions (i) to (v) of problem 38. ii) Distinuish M[x] nd P[x] of problem 38. iii) Prove M[x] enjoys specil fetures entirely different from other MOD subset coefficient polynomil semiroup. 4. Let V[x] ={ i x i / i S(C(Z 18 )); } i be the MOD specil qusi dul number subset coefficient polynomil semiroup. 41. Let Study questions (i) to (vi) of problem (34) for this V[x]. G[x] = i ix i S(Z 1 k) } i be the MOD subset specil qusi dul like number coefficient polynomil semiroup. i) Study questions (i) to (vi) of problem (34) for this G[x]. ii) Compre G[x] with V[x] of problem (4).

290 Semiroups built on MOD Subset Mtrices Let 12 H[x] 12 = ={ i x i / i S(Z 1 h); x 13 = 1 } i be the MOD subset specil dul like number coefficient polynomil semiroup. i) Study questions (i) to (v) of problem (38) for this H[x] Let S(M) = {collection of mtrix subsets from M = { i Z 18 1 i 8 +} +} 8 be the MOD mtrix subset semiroup under +. i) Find o(m). ii) Find ll MOD subsemiroups of M. iii) Cn M hve MOD idempotents under +. iv) Enumerte ll specil fetures ssocited with S(M). 44. Let S(T) = {collection of ll subsets from 1 2 T ={ 3 i C(Z 8 ); 1 i 5 +} 4 5 be MOD finite complex number subset mtrices semiroup under +. Study questions (i) to (iv) of problem (43).

291 29 MOD Nturl Neutrosophic Subset Semiroups 45. Let W(S) = {collection of ll subsets from the mtrix set with entries from Z 1. S = { i Z 1 1 i 4 +} _} be the MOD neutrosophic mtrix subset semiroup under +. Study questions (i) to (iv) of problem (43) for this W(S). 46. Let H(M) = {collection of ll subset mtrix from M = { i Z 23 1 i 1 +} +} be the MOD dul number subset mtrix semiroup under +. Study questions (i) to (iv) of problem (43). 47. Let V(N) = {collection of ll subset mtrices from N = { / i Z 48 ; 1 i 12 n } n } be the MOD subset mtrix semiroup under. i) Find (V(N)) ii) Find ll MOD subset mtrix zero divisors nd nilpotents of V(N). iii) Find ll MOD subset mtrix subsemiroups which re not idels of V(N). iv) Find ll MOD subset mtrix subsemiroups which re

292 Semiroups built on MOD Subset Mtrices 291 idels of V(N). v) Prove / disprove V(N) is Smrndche MOD mtrix subset semiroup. vi) Obtin ll specil fetures ssocited with V(N). 48. Let W(M) = {collection of ll subset mtrix from M = { / i Z 13 1 i 6 n } n } be the MOD subset dul number mtrix semiroup under n the nturl product. Study questions (i) to (vi) of problem (47) for this W(M). 49. Let G(P) = {collection of mtrix subsets from P = { / i C(Z 49 ); 1 i 9 + (or n ) (or n )} be the MOD subset mtrix finite complex number semiroup under (or n ). i) Study questions (i) to (vi) of problem (47) for this G(P). ii) f x is used prove G(P) is non commuttive. iii) Find in G(P) MOD left zero divisor which is not riht zero divisor nd vice vers under the usul product. iv) Find ll MOD left idels which re not MOD riht idels in G(P) under the usul product.

293 292 MOD Nturl Neutrosophic Subset Semiroups 5. Let T(H) = {collection of ll of mtrix subsets from H = { / i Z 24 n 1 i 1} n } be the MOD dul number subset mtrix semiroup. i) Study questions (i) to (vi) of problem (47) for this T(H). ii) Prove T(H) hs more number of MOD zero divisors nd nilpotents even if n is prime. 51. Let S(W) = {collection of ll subsets from the MOD specil qusi dul number mtrix semiroup W = { / i Z 9 1 i 21 n } n } be the MOD neutrosophic subset mtrix semiroup under nturl product n i) Study questions (i) to (vi) of problem (47) for this S(W). ii) Distinuish T(H) of problem (5) from S(W). 52. Let S(M) = {collection of ll subsets from the MOD mtrix specil dul like number semiroup.

294 Semiroups built on MOD Subset Mtrices 293 M = { / i Z 13 h 1 i 15 n } n } be the MOD subset specil dul like number semiroup under n. Study questions (i) to (vi) of problem (47) for this S(M). 53. Let S(P[x]) = {collection of ll subsets from the set P[x] where i P[x] ={ d i x i / d i Z 48 } } i be the MOD subset polynomil semiroup under product. i) Prove S(P[x]) is commuttive monoid of infinite order. ii) Find ll MOD idels of S(P[x]). iii) Find ll MOD subsemiroups of S(P[x]) which re not idels. iv) Find ll MOD subset zero divisors nd nilpotents of S(P[x]). v) Prove S(P[x]) cnnot hve idempotents. 54. Let S(W[x]) = {collection of ll subsets from the MOD semiroup. W[x] ={ d i x i / d i (Z 47 ) } } i be the MOD subset of finite complex number polynomils semiroup.

295 294 MOD Nturl Neutrosophic Subset Semiroups Study questions (i) to (v) of problem (53) for this W[x]. 55. Let S(Z[x]) = {collection of ll subsets from the MOD specil dul like number polynomil semiroup z[x] ={ i x i / i Z 4 h } } i be the MOD specil dul like number polynomils subsets semiroup. Study questions (i) to (v) of problem (53) for this S(Z[x]). 56. Let S(B[x]) = {collection of ll subsets from the MOD dul number coefficient polynomil semiroup B[x] = { i x i / i Z 2 ; } } i be the MOD subset polynomil dul number semiroup. i) Study questions (i) to (v) of problem (53) for this S(B[x]). ii) Prove S(B[x]) hs more number of MOD zero divisors nd MOD nilpotents. 57. Let S(P[x] 8 ) = {collection of ll subsets polynomils from MOD polynomil semiroup. P[x] 8 ={ i x i / i Z 26 x 9 = 1 } } i be the MOD subset polynomil semiroup under. i) Find o(sp[x] 8 ). ii) Find ll MOD zero divisors nd nilpotents. iii) Find ll MOD subset idels. iv) Find ll MOD subset subsemiroups which re not

296 Semiroups built on MOD Subset Mtrices 295 idels. v) Prove S(P[x] 8 ) hs no idempotents. 58. Let S(Q[x] 6 ) = {collection of ll subset polynomils from 6 Q[x] 6 ={ i x i / i Z 1 x 7 = 1 } } i be the MOD neutrosophic subset polynomils semiroup. Study questions (i) to (v) of problem (57) for this S(Q[x] 6 ). 59. Let S(T[x] 1 ) = {collection of ll subset polynomils from 1 T[x] 1 = { i x i / x 11 = 1 i Z 19 ; } } i be the MOD dul number subset polynomil semiroup. i) Study questions (i) to (v) of problem (59) for this T[x] Let S(V[x] 18 ) = {collection of ll polynomil subsets from 18 V[x] 18 ={ i x i / x 19 = 1 i Z 12 k } } i be the MOD polynomil specil qusi dul number subset semiroup. i) Study questions (i) to (v) of problem (57) for this S(V[x] 18 ). ii) Compre S(V[x] 18 ) when Z 12 k is replced by Z 12 h or Z 12 or Z 12 or C(Z 12 ) nd Z Let S(P[x]) = {collection of ll MOD subset polynomils from

297 296 MOD Nturl Neutrosophic Subset Semiroups 18 P[x] ={ i x i / i Z 483 +} } be the MOD subset polynomil semiroup. i i) Prove S(P[x]) hs MOD subset semiroups of finite order. ii) Find ll MOD subset subsemiroups of infinite order. iii) Prove o(s(p[x]) =. 62. Let S (G[x]) = {collection of ll subset polynomils from 18 G[x] ={ i x i / i C(Z 11 ) } } i be the MOD finite number complex subset polynomil semiroup under +. Study questions (i) to (iii) of problem (61) for this S(G[x]). 63. Let S(V[x]) = {collection of ll subset polynomil from 18 V[x] ={ i x i / i Z 12 } } } i be the MOD finite dul number subset polynomil semiroup under +. i) Study questions (i) to (iii) of problem (61) for this S(V[x]). ii) Compre S(V[x]) with S(G[x]) of problem (62). 64. Let S(B[x] 1 ) = {collection of ll subset polynomils from 1 B[x] 1 = { i x i / x 11 = 1 i Z 48 +} } i

298 Semiroups built on MOD Subset Mtrices 297 be the MOD subset polynomil semiroup under +. i) Find o(s(b[x] 1 )). ii) Find ll MOD subset polynomil subsemiroups. iii) Cn S(B[x] 1 ) hve nontrivil idempotents? 65. Let S(V[x] 6 ) = {collection of ll polynomil subset from 6 V[x] 6 ={ i x i / x 7 = 1 i C(Z 27 ) +} +} i be the MOD finite complex number polynomil subset semiroup under +. Study questions (i) to (iii) of problem (64) for this S(V[x] 6 ). 66. Let S(G[x] 12 ) = {collection of ll polynomil subsets from 12 G[x] 12 ={ i x i / x 13 = 1 i Z 43 } } i be the MOD dul number coefficient polynomil subset semiroup under +. Study questions (i) to (iii) of problem (64) for this S(G[x] 12 ). 67. Let S(M[x] 19 ) = {collection of ll polynomil subsets from 19 M[x] 19 ={ i x i / x 2 = 1 i Z 1 k } } i be the MOD specil qusi dul number polynomil subset semiroup under +. Study questions (i) to (iii) of problem (64) for this S(M[x] 14 ).

299 298 MOD Nturl Neutrosophic Subset Semiroups 68. Differentite ll specil fetures ssocited with S(M[x] 19 ) in problem 67 with S(M[x] 19 ) under product opertion. 69. Compre S(M[x] 19 ) on problem 67 with S(G[x] 12 ) in problem Enumerte ll specil fetures enjoyed by the followin MOD semiroups. S(Q[x] m ) with entries from Z n or C(Z n ) or Z n Z n h Z n k nd Z n ; 2 m < x m+1 = 1. Compre ech of them.

300 Chpter Five MOD SUBSET NATURAL NEUTROSOPHC SEMGROUPS n this chpter we just describe MOD nturl neutrosophic number semiroups of different types built usin subsets from Z or C (Z n ) or Z n or Z n h or Z n or Z n k. n We describe by exmples nd derive severl importnt nd interestin properties ssocited with them. For properties of MOD nturl neutrosophic numbers plese refer [6]. We first describe the notions by exmples. S( Z ) = {Collection of ll subsets from n Z n }. Exmple 5.1: S( Z ) = {Collection of ll subsets from 6 Z 6 }. A = { } nd B = { subset nturl neutrosophic numbers in S( Z ) } re MOD

301 3 MOD Nturl Neutrosophic Subset Semiroups Exmple 5.2: S (C (Z 8 )) = {Collection of ll subsets from MOD nturl neutrosophic finite complex numbers of C (Z 8 )}. Let A = { C 4 C 2 + C +5 C 6 + C 4 C + + 5i F + 6} nd 2i F C B = { + 4iF C C C 6iF 4 2 4iF 2 + 6i F + } S(C (Z 8 )). C 4 2iF S(Z n ) = {Collection of ll subsets from the MOD nturl neutrosophic - neutrosophic set Z n }. Exmple 5.3: S(Z 1 ) = {Collection of ll MOD nturl neutrosophic-neutrosophic subsets from Z 1 }. Let B = { } nd C = { } S(Z 1 ) re MOD nturl neutrosophic - neutrosophic subsets of S(Z 1 ). Exmple 5.4: S(Z 18 ) = {Collection of ll MOD nturl neutrosophic dul number subsets from S(Z 18 }. Let S = { } nd T = { S(Z 18 ) } Thus S(Z n ) = {Collection of ll subsets from the MOD nturl neutrosophic dul number set Z n }. Let S(Z n h ) = {Collection of ll subsets from the MOD nturl neutrosophic specil dul like number set with entries from Z n h }.

302 MOD Subset Nturl Neutrosophic Semiroups 31 Exmple 5.5: Let S(Z 12 h ) be the MOD nturl neutrosophic specil dul like number subset. Let A = { h h + h 6h 2 h + 9} nd B = { h 3 + h 4 + 3h 4 + h 1h + h h h h 1 11+h h 1h + + h + h 1h 2 h 2h h 8 } S(Z 12 h ) re MOD nturl neutrosophic specil dul like number subset of S(Z 12 h ). S(Z n k ) = {Collection of ll MOD nturl neutrosophic specil qusi dul number subsets of Z n k }. Exmple 5.6: Let S(Z 16 k ) = {Collection of ll MOD nturl neutrosophic qusi dul number subsets of Z 16 k }. k 8 8k Let P = { k 4 k 2k k 8 2k 3 +9k + k 1k } nd Q = { k k k k k k 1 8k k + 2k k k } S(Z 16 k ) re MOD nturl neutrosophic specil qusi dul number subsets of S(Z 16 k ). We see on S( Z ) or S(C (Z n )) or S(Z n ) or S(Z n n h ) or S(Z n ) nd S(Z n k ) we cn define lebric opertions + or nd under ech of these opertions ll these six MOD sets re only MOD semiroups. We will illustrte this sitution by some exmples. Exmple 5.7: Let H = {S( Z 6 ) +} be the MOD nturl neutrosophic subset semiroup under +. Let A = { B = { } nd 6 4 3} H.

303 32 MOD Nturl Neutrosophic Subset Semiroups A + B = { } + { } = { }. Thus is the wy + opertion is performed on H. H is infct monoid s {} H is such tht {} + A = A for ll A H. Exmple 5.8: Let M = {S(Z 1 ) +} be the MOD nturl neutrosophic dul number subset monoid under +. Let A = { B = { 4 2 } M. 6 8 A + B = { } nd } + { 4 2 } = { } M. The reder is expected to find MOD nturl neutrosophic dul number subset monoid of finite order. Exmple 5.9: Let V = {S(Z 15 ) +} be the MOD nturl neutrosophic - neutrosophic number subset monoid of finite order. Let A = { 3 B = { } nd } V. The reder is left with the tsk of findin A + B.

304 MOD Subset Nturl Neutrosophic Semiroups 33 Exmple 5.1: Let G = S(C (Z 9 ) +} be the MOD nturl neutrosophic finite complex number subset monoid of finite order. Let P = { C C 3 6i F C i F } nd C Q = { C 3iF 6 3 C 6iF 3 + C i F } G. C 3iF The reder is expected to find P + Q. Likewise we cn find {S(Z 19 k ) +} nd {S(Z 2 k ) +} nd this is left s n exercise to the reder s it is considered s mtter of routine. THEOREM 5.1: Let {H +} = {S( Z ) (or S(C (Z n )) or S(Z n ) or S(Z n h ) or S(Z n k ) or S(Z n )) +} be the MOD nturl neutrosophic subset (or finite complex number subset or dul number subset or specil dul like number subset or specil qusi dul number subset or neutrosophic subset) semiroup under +. (i) o(h) <. (ii) H hs MOD nturl neutrosophic subsemiroups. The proof is left s n exercise s it is considered s mtter of routine. We now ive exmples of MOD nturl neutrosophic subset monoid under by some exmples. Exmple 5.11: Let {S } = {S( Z neutrosophic subset semiroup. Let A = { } nd 12 B = { } {S }. n ) } be the MOD nturl

305 34 MOD Nturl Neutrosophic Subset Semiroups A B = { } { } = { } This is the wy product opertion is performed on B. Thus B hs MOD nturl neutrosophic zeros lso. We cn on B define nother type of opertion in which = nd =. By the context we cn understnd wht product is used f = then we sy the product is nturl neutrosophic zero dominted product. Tht is wht we hve used in A B. 12 The other product = then we sy the product is nturl zero dominted product. Now we find A B under nturl zero dominted product. A B = { } The reder is left with the tsk of findin both types of products. However under both types of product they ttin only monoid structure nd in one the dominted usul zero will contribute to zero divisors in nother type the MOD nturl neutrosophic dominnt zero will contribute to more MOD nturl neutrosophic zero divisors. Further the reder is left with the tsk of findin MOD nturl neutrosophic subset idels nd subsemiroups which re not idels.

306 MOD Subset Nturl Neutrosophic Semiroups 35 Further subset nilpotents of two types nd subset idempotents. We will ive some more exmples. Exmple 5.12: Let {V } = {S(Z 18 ) } be the MOD nturl neutrosophic - neutrosophic subset semiroup. A = { } nd B = { } V. We find the MOD nturl neutrosophic - neutrosophic subsets of V. We find the usul zero dominted product first. + A B = { } Now we find the MOD nturl neutrosophic zero dominted product tht is =. A B = { } Clerly nd re distinct thus the product defined two distinct MOD nturl neutrosophic - neutrosophic subset monoid which is commuttive nd is of finite order. There re severl MOD nturl neutrosophic subset zero divisors s well s usul zero divisors. Study in this direction is mtter of routine with only chne which is to be observed is whether usul zero dominnt product is tken or MOD nturl neutrosophic dominnt zero is tken.

307 36 MOD Nturl Neutrosophic Subset Semiroups Such study is similr to one done in chpter. Exmple 5.13: Let W = {S(Z 2 ) } be the MOD nturl neutrosophic dul number subset semiroup under product. This hs mny subset nilpotents nd zero divisors. P 1 = {S(Z 2 ) } W is subset subsemiroup which is not n idel. P 2 = {S(Z 2 ) } W is in subset subsemiroup which is not n idel. P 3 = {S(Z 2 ) } W is in subset subsemiroup which is not n idel. Study in this direction is interestin nd innovtive. P 2 is zero squre semiroup. Exmple 5.14: Let M = {S(Z 13 k ) } be the MOD nturl neutrosophic subset specil qusi dul number semiroup. P 1 = {S(Z 13 ) } is MOD subset subsemiroup which is not n idel. P 2 = {S(Z 13 k) } is in MOD subset subsemiroup which is not subset idel. P 3 = {S(Z 13 k) } is in MOD subset subsemiroup which is not subset idel. We see findin MOD subset zero divisors nd nilpotents hppens to be problem. Further findin MOD subset idempotents is lso chllenin one s Z 13 is used in this exmple. We see P 4 = {S( k k t t is n idempotent or zero divisor in Z 13 ) } is in MOD nturl neutrosophic subset

308 MOD Subset Nturl Neutrosophic Semiroups 37 subsemiroups which is lso n idel in the MOD nturl neutrosophic zero dominted product. n view of ll these we hve the followin theorem. THEOREM 5.2: Let P 1 = {S( Z ) } (or P 2 = {S(C (Z n ) } or n {P 3 = S(Z n ) } or {P 4 = S(Z n ) } or {P 5 = S(Z n h ) } or {P 6 = S(Z n k ) }) be MOD nturl neutrosophic subset semiroup under MOD nturl zero dominted product. Then the followin re true. (i) o(s( Z n )) (nd ll other such subset MOD nturl neutrosophic semiroups under ) is of finite order. (ii) {S( Z n ) } hs MOD nturl neutrosophic subset zero divisors nd subset nilpotent if n is n pproprite number. (iii) Whtever be n prime or otherwise P 1 P 2 P 6 hs MOD nturl neutrosophic subset subsemiroups which re not idels. t t (iv) T = S({ m m is zero divisor or idempotent or nilpotent in Z n (or C(Z n ) or Z n or Z n or Z n h or Z n k)} P i i = re MOD subset idels of P i i = (v) P 3 P 4 P 5 nd P 6 hs idels of the form {S(Z n t } {S(Z n t } {S(Z n h h t } {S(Z n k k t } in P 3 P 4 P 5 nd P 6 respectively. Proof is direct nd hence left s n exercise to the reder. n cse of product in theorem 5 is replced by n MOD nturl neutrosophic usul zero dominted product then with pproprite chnes;

309 38 MOD Nturl Neutrosophic Subset Semiroups this theorem cn be proved. Next we briefly describe MOD nturl neutrosophic mtrix subset semiroups under + nd two types of products by exmples. Exmple 5.15: Let P = {( ) i S( Z 6 ) 1 i 3 +} be the MOD nturl neutrosophic subset mtrix semiroup under +. Let A = ({ B = ({ } { } { } { 6 4 We find A + B = ({ ({ } { } { }) = ({ }) { } + { = ({ } + { } { } { performed on P } { }) nd }) P } { } { } { }) } + { } }). This is the wy + opertion is 6 4 The reder is left with the tsk of findin MOD nturl neutrosophic subset mtrix subsemiroups. Exmple 5.16: Let 6 4 W = i Z 4 ); 1 i 4 +}

310 MOD Subset Nturl Neutrosophic Semiroups 39 be the MOD nturl neutrosophic - neutrosophic subset mtrix semiroup under +. This hs MOD nturl neutrosophic - neutrosophic subset mtrix subsemiroup under +. As this is considered s mtter of routine the reder is left with the tsk of findin them. Also if A B W then one cn find A + B. Exmple 5.17: Let P = i S(Z 12 k ); 1 i 9 +} be the MOD nturl neutrosophic specil qusi dul number subset mtrix semiroup under +. P hs MOD nturl neutrosophic subset mtrix subsemiroup this is left s n exercise to the reder. n view of ll these we ive the followin result. THEOREM 5.3: Let P = {Collection of ll m t subset mtrices with entries from S( Z ) or S(C (Z n )) or S(Z n ) or (S(Z n n h ) or S(Z n k ) or S(Z n )) +} be the MOD nturl neutrosophic subset mtrix semiroup under +. (i) o(p) <. (ii) P hs MOD nturl neutrosophic subset mtrix subsemiroups under +. (iii) P hs MOD nturl neutrosophic mtrix subsets which re idempotents under +.

311 31 MOD Nturl Neutrosophic Subset Semiroups The reder is left with the tsk of ivin the proof. Next we proceed onto describe both types of MOD nturl neutrosophic subset mtrix semiroups under the product by some more exmples. Exmple 5.18: Let P = i S( Z ); 1 i 6 n } 12 be the MOD nturl neutrosophic subset mtrix semiroup under usul zero dominted product tht is 12 t = {3 4 3 } { 6 } { 6 6} Let A = { 6 8 } { 9 } { } 12 {} {1} { 2} nd B = {6} { 1 4 } {9 8 } S. We find A n B = {3 4 3 } { 6 } { 6 6} { 6 8 } { 9 } { } n 12 {} {1} { 2} {6} { 1 4 } {9 8 } = {} { 6 } { 6 } { 6 8 } { 6 } { } This is the wy product opertion is performed on S.

312 MOD Subset Nturl Neutrosophic Semiroups 311 We see S hs MOD nturl neutrosophic mtrix subset zero divisors nilpotents nd idempotents. This tsk is left s n exercise to the reder. Further this S hs MOD nturl neutrosophic mtrix subset subsemiroups which re not idels s well s subsemiroups which re idels. Exmple 5.19: Let M = i S (Z 1 ); 1 i 14 n } be the MOD nturl neutrosophic dul number subset mtrix semiroup under the MOD nturl neutrosophic zero dominnt product =. The tsk of findin MOD nturl neutrosophic dul number subset mtrix zero divisors nilpotents idempotents idels nd subsemiroups re left s n exercise to the reder. n view of ll these we hve the followin result. THEOREM 5.4: Let D = {Collection of ll m t subset mtrices with entries from S( Z ) or (S(C (Z n )) or S(Z n ) or n (S(Z n ) or S(Z n k ) or S(Z n h )) n } be the MOD nturl neutrosophic subset mtrix semiroup under the nturl zero dominted product or MOD nturl neutrosophic zero dominted product.

313 312 MOD Nturl Neutrosophic Subset Semiroups Then (i) D hs MOD nturl neutrosophic subset mtrix zero divisors nilpotents nd idempotents only if n = p s q (s 2 p prime) nd for ll n is cse of S(Z n ) brrin only MOD nturl neutrosophic mtrix subset idempotents. (ii) D hs MOD nturl neutrosophic subset mtrix subsemiroups which re not idels. (iii) D hs MOD nturl neutrosophic subset mtrix subsemiroups which re idels. Proof is direct nd hence left s n exercise to the reder. Next we proceed onto describe with exmples MOD nturl neutrosophic mtrix subset semiroups under + nd. Exmple 5.2: Let S(M) = {Collection of ll mtrix subsets from M = i ( Z ); 1 i 3 +} +} 1 be the MOD nturl neutrosophic mtrix subset semiroup under +. Let A = nd B = S(M). 1 8

314 MOD Subset Nturl Neutrosophic Semiroups 313 A + B = = =

315 314 MOD Nturl Neutrosophic Subset Semiroups This is the wy + opertion is performed on S(M). The reder is left with the tsk of findin MOD nturl neutrosophic mtrix subset subsemiroups. Exmple 5.21: Let S(P) = {Collection of ll mtrix subsets from P = i S(Z 9 ); 1 i 6 +} +} be the MOD nturl neutrosophic dul number mtrix subsets semiroup under +. Let A = nd B S(P). The reder is expected to find A +B. Further prove; {()} =

316 MOD Subset Nturl Neutrosophic Semiroups 315 cts s the dditive identity of S(P); tht is A + {()} = A for ll A S(P). Find tlest 3 MOD nturl neutrosophic subset mtrix subsemiroups of S(P). Exmple 5.22: Let S(V) = {Collection of ll mtrix subsets from V = i C (Z 15 ); 1 i 1 +} +} be the MOD nturl neutrosophic subset mtrix finite complex number semiroup under +. Find MOD nturl neutrosophic finite complex number subset mtrix subsemiroups. Find o(s(v)). n view of ll these we hve the followin theorem. THEOREM 5.5: Let S(W) = {Collection of ll mtrix subsets from W = {M = (m ij ) st m ij Z or (C (Z n )) or Z n or Z n n ) or Z n k or Z n h 1 i s 1 j t; +} be the MOD nturl neutrosophic subset mtrix semiroup under +. (i) o(s(w)) <. (ii) S(W) hs severl MOD nturl neutrosophic mtrix subset subsemiroups. Next we describe by n exmple or two the notion of MOD mtrix subset nturl neutrosophic semiroups under zero

317 316 MOD Nturl Neutrosophic Subset Semiroups dominnt product s well s MOD nturl neutrosophic zero dominnt product. Exmple 5.23: Let S(B) = {Collection of ll mtrix subsets from B = i Z 14 ; 1 i 8 n } n } be the MOD nturl neutrosophic subset MOD nturl neutrosophic mtrix subset semiroup under zero dominnt product. Let P = nd R = S(B). 6 The reder is ssined the tsk of findin A n A nd B n A. Further find MOD nturl neutrosphic mtrix subset zero divisors nd idempotents. Find lso MOD nturl neutrosophic mtrix subset idels nd subsemiroups.

318 MOD Subset Nturl Neutrosophic Semiroups 317 Exmple 5.24: Let S(M) = {Collection of ll subset mtrix from M = i Z 1 h 1 i 6 n } n } be the MOD nturl neutrosophic specil dul like number subset mtrix semiroup under the MOD nturl neutrosphic zero dominnt product n. Find MOD subset mtrix nturl zero divisors nd idempotents of S(M). n view of ll these we hve the followin result. THEOREM 5.6: Let S(W) = {Collection of ll mtrix subsets from W = {Collection of ll s t mtrices with entries from Z or C (Z n ) or Z n or Z n or Z n k or Z n h ; n n } n } be the MOD nturl neutrosophic subset mtrix or MOD nturl neutrosophic finite complex number subset mtrix or MOD nturl neutrosophic - neutrosophic subset mtrix or MOD nturl neutrosophic dul number subset mtrix or MOD nturl neutrosophic specil dul like number subset mtrix or MOD nturl neutrosophic specil qusi dul number subset mtrix semiroup under zero dominted product or MOD dominted semiroup respectively. (i) Then o(s(w)) <. (ii) S(W) hs MOD subset mtrix subsemiroups which re not idels s well s subsemiroups which re idels.

319 318 MOD Nturl Neutrosophic Subset Semiroups (iii) S(W) hs MOD subset mtrix zero divisor nilpotents nd idempotents for pproprite n. The proof is direct nd hence left s n exercise to the reder. Next we proceed onto describe briefly the concept of MOD nturl neutrosophic number subset coefficient polynomil semiroups under + nd two types of products by exmples. Exmple 5.25: Let S[x] = i ix i S( i Z ) +} 1 be the MOD nturl neutrosophic subset coefficient polynomil semiroup under +. Let p(x) = { }x 3 + { } nd q (x) = { }x 2 + { } S[x]. p(x) + q(x) = { }x 3 + { } + { }x 2 + { } 1 6 = { }x 3 +{ }x 2 + { } S[x] This is the wy + opertion is performed on S[x]. The reder is left with the tsk of findin MOD nturl neutrosophic subset subsemiroups of finite order. Further the reder is expected to prove ll idels of S[x] hve infinite crdinlity.

320 MOD Subset Nturl Neutrosophic Semiroups 319 We prove S[x] hs no nilpotents or zero divisors or non trivil idempotents. Let p(x) = { 5}x 3 + {5} nd q(x) = {2 4 6}x 4 + {6 8 2}x 2 + { } S[x]. Clerly p(x) + q(x) S[x]. Exmple 5.26: Let W[x] = i ix i S(Z 2 ); +} be i the MOD nturl neutrosophic dul number subset coefficient polynomil semiroup under +. Enumerte ll properties enjoyed by W[x]. We ive the followin result. i THEOREM 5.7: Let S[x] = i x i S( Z n ) or S(C (Z n )) or i S(Z n ) or S(Z n ) or S(Z n h ) or S(Z n k ); +} be the MOD nturl neutrosophic subset coefficient polynomil semiroup or MOD nturl neutrosophic finite complex number subset coefficient polynomil semiroup or MOD nturl neutrosophic - neutrosophic subset coefficient polynomil semiroup or MOD nturl neutrosophic dul number coefficient polynomil semiroup or MOD nturl neutrosophic specil dul like number subset coefficient polynomil semiroup or MOD nturl neutrosophic specil qusi dul number subset coefficient polynomil semiroup respectively. (i) o(s[x]) =. (ii) S[x] hs MOD nturl neutrosophic subset coefficient polynomil subsemiroups of finite order. Next we proceed onto describe MOD nturl neutrosophic subset polynomil semiroups under both the products zero

321 32 MOD Nturl Neutrosophic Subset Semiroups dominted product s well s MOD nturl neutrosophic zero dominted product by some exmples. Exmple 5.27: Let S[x] = i ix i S(C (Z 12 )); } i be the MOD nturl neutrosophic finite complex number coefficient polynomil subset semiroup under zero dominted product. S[x] hs MOD nturl neutrosophic finite complex number subset coefficient polynomil zero divisors nd nilpotents but hs no nontrivil idempotents. However ll MOD nturl neutrosophic subset polynomil subsemiroups nd idels re infinite order only. All these re left s n exercise for the reder to verify. Exmple 5.28: Let P[x] = i ix i S(Z 13 ); } be i the MOD nturl neutrosophic dul number subset coefficient polynomil semiroup under MOD nturl neutrosophic zero dominnt product. P[x] hs infinite number of MOD nturl neutrosophic subset polynomil zero divisors nd nilpotents. However P[x] hs non trivil idempotents. n view of this we prove the followin result. THEOREM 5.8: Let B[x] = i i x i S( i Z ) or S(C (Z n )) or S(Z n ) or S(Z n ) or S(Z n h ) or S(Z n k ); } be the MOD nturl neutrosophic subset coefficient polynomil semiroup of ny one of the coefficient subsets semiroup under n

322 MOD Subset Nturl Neutrosophic Semiroups 321 the zero dominnt product or MOD nturl neutrosophic dominnt product. (i) B[x] is of infinite order. (ii) B[x] hs no nontrivil idempotents. (iii) B[x] hs MOD nturl neutrosophic subset coefficient nilpotents or zero divisors which minly depend on n. (iv) B[x] hs ll MOD nturl neutrosophic subset coefficient subsemiroup or idels re only of infinite order. Proof is direct nd hence left s n exercise to the reder. Corollry 5.1: f in theorem the set S(Z n ) is tken then the MOD nturl neutrosophic dul number coefficient polynomils hs more number of nilpotents nd zero divisors. The proof is lso left s n exercise to the reder. Next we will briefly describe MOD nturl neutrosophic subset polynomils nd define the + nd two types of product opertions by exmples. Exmple 5.29: Let S(P[x]) = {Collection of ll subsets from i P[x] = ix i Z 16 +} +} be the MOD nturl i neutrosophic polynomil subset semiroup. Let A = {3x 3 + ( )x x 2 5x 2 + 1x + 3} nd B = {5x 7 + ( )x x x 2 } S(P[x]). We cn find A + B which is mtter of routine nd this tsk is left s n exercise to the reder. nfct S(P[x]) is of infinite order.

323 322 MOD Nturl Neutrosophic Subset Semiroups This hs subsemiroup both finite nd infinite order. Exmple 5.3: Let S(R[x]) = {Collection of ll subsets from R[x] = i ix i Z 12 h ) +}; +} i be the MOD nturl neutrosophic specil dul like number polynomil subset semiroup under +. All properties cn be derived bout S(R[x]) s mtter of routine. n view of ll these we hve the followin result. THEOREM 5.9: polynomil from Let S(B[x]) = {Collection of ll subset B[x] = i i x i i Z or C (Z n ) or Z n or Z n or n Z n h ) or Z n k ; +} +} be the MOD nturl neutrosophic subset polynomil semiroup of ny one of the six types. S(B[x]) hs MOD nturl neutrosophic subset polynomil subsemiroups of both finite nd infinite order. Proof is direct nd hence left s n exercise to the reder. Now we proceed onto ive exmples of MOD nturl neutrosophic polynomils subset semiroups under the product in which is dominted nd nother product is which the MOD nturl neutrosophic zero is dominted by the followin exmple.

324 MOD Subset Nturl Neutrosophic Semiroups 323 Exmple 5.31: Let S(P) = {Collection of ll polynomil subsets from P[x] = i ix i Z 1 k ; } } i be the MOD nturl neutrosophic specil qusi dul number polynomil subset semiorup under the usul zero dominted product. o(s(p[x])) =. Further ll MOD nturl neutrosophic specil qusi dul number polynomil subset subsemiroups or idels re of infinite order. This S(P[x]) hs MOD nturl neutrosophic specil qusi dul number polynomil subset zero divisors. However S(P[x]) hs no idempotents. Let A = {5x 3 + ( k k )x (6 + 5 B = {( k k + 5k )x 3 ( k 5k 2 S(P[x]). k ) + (4 + k 2 )x 2 +3) + ( k 2 5k + k 4 8k + k + k 6 k x 3 } nd )x 2 ] t is mtter of routine to find A B so is left s n exercise for the reder. Exmple 5.32: Let S(T[x]) = {Collection of ll polynomil subsets from T[x] = i ix i C (Z 27 ) } } i be the MOD nturl neutrosophic finite complex number subset polynomil semiroup under with usul zero dominted product. The reder is left with the tsk of findin ll relted properties of S(T[x]).

325 324 MOD Nturl Neutrosophic Subset Semiroups n view of ll these we hve the followin theorem the proof of which is left s n exercise to the reder. THEOREM 5.1: Let S(P[x]) = {Collection of ll polynomil i subsets from P[x] = i x i Z n or C (Z n ) Z n or Z n i or Z n h or Z n k under zero dominted product or the MOD nturl neutrosophic zero dominted product} } be the MOD nturl neutrosophic subset polynomil semiroup under defined over ny of the six sets ( Z n or C (Z n ) nd so on) then (i) o(s(p[x])) =. (ii) S(P[x]) is commuttive monoid. (iii) S(P[x]) hs MOD nturl neutrosophic subset polynomil subsemiroups s well s idels ll of them re only of infinite order. (iv) S(P[x]) hs MOD nturl neutrosophic zero divisors nd nilpotents for pproprite n. Next we proceed onto describe MOD nturl neutrosophic subset polynomil semiroup of finite order by exmples. Exmple 5.33: Let S(P[x] 8 ) = {Collection of ll polynomil 8 i subsets from P[x] 8 = ix i Z 1 ); x 9 = 1 +} +} i be the MOD nturl neutrosophic dul number polynomil subset semiroup under + of finite order. o(s(p[x] 8 )) <. S(P[x] 8 ) hs MOD nturl neutrosophic dul number polynomil subset subsemiroup. nfct S(P[x] 8 ) is commuttive finite monoid. Exmple 5.34: Let S(R[x] 3 ) = {Collection of ll polynomil 3 i subset from R[x] 3 = ix i C (Z 12 ); x 4 = 1 +} +} be the i

326 MOD Subset Nturl Neutrosophic Semiroups 325 MOD nturl neutrosophic finite complex number polynomil subset semiroup under + of finite order. Find ll subsemiroups of S(R[x] 3 ). n view of ll these we hve the followin result. THEOREM 5.11: Let S(P[x] m ) = {Collection of ll polynomil subsets from P[x] m = m i x i i i Z or C (Z n ) or Z n or Z n or n Z n h or Z n k x m+1 = 1 +} +} be the MOD nturl neutrosophic polynomil subsets semiroup of ny one (tkin entries from Zn or C Z n or so on Z n k ) under + (i) o(s(p[x] m ) < nd is finite commuttive monoid. (ii) S(P[x] m ) hs severl MOD nturl neutrosophic polynomil subset subsemiroups. Proof is direct nd hence left s n exercise to the reder. Next we describe MOD nturl neutrosophic polynomil subset semiroup of finite order under both the types of opertions by exmples. Exmple 5.35: Let S(P[x] 8 ) = {Collection of ll polynomil 8 i subset from the set P[x] 8 = ix i Z 18 x 9 = 1 } } be i the MOD nturl neutrosophic polynomil subset semiroup of finite order under zero dominnt product. S(P[x] 8 ) hs MOD nturl neutrosophic polynomil subset zero divisors nd nilpotents.

327 326 MOD Nturl Neutrosophic Subset Semiroups This hs idels nd subsemiroups which re not idels. The reder is left with the tsk of findin them. Exmple 5.36: Let S(B[x] 5 ) = {Collection of ll polynomil subsets from B[x] 5 = 6 i ix i Z 1 x 6 = 1 } } i be the MOD nturl neutrosophic dul number polynomil subset semiroup under MOD nturl neutrosophic dul number zero product. Find ll MOD nturl neutrosophic dul number subsemiroups nd idels. This hs lots of MOD subset zero divisors. nview of this we hve the followin results. THEOREM 5.12: Let S(M[x] m ) = {Collection of ll polynomil subsets from M[x] m = m i i x i i n Z x m+1 = 1 or C (Z n ) or Z n or Z n h or Z n k or Z n } } be the MOD nturl neutrosophic polynomil subsets semiroup usin ny of Z n (or C (Z n ) or Z n nd so on) under zero dominnt product or under MOD nturl neutrosophic zero dominnt product. (i) o(s(m[x] m )) <. (ii) This S(M[x] m ) hs MOD idels nd subsemiroups.

328 MOD Subset Nturl Neutrosophic Semiroups 327 (iii) S(M[x] m ) hs MOD nturl neutrosophic polynomil subset zero divisors nd nilpotents. Proof is direct nd hence left s n exercise to the reder. Thus we hve iven only few exmples nd results however the reder is left with the tsk of findin more properties of them. All these re suested by the followin problems some of which re difficult some re t reserch level nd some of them re simple exercises. Problems: 1. Let B = {S( Z 6 ) +} be the MOD nturl neutrosophic subset semiroup under +. (i) Find o(s( Z 6 )). (ii) Find ll MOD nturl neutrosophic subset idempotents under + of B. (iii) Find ll MOD nturl neutrosophic subset subsemiroups of B. 2. Let A = {S( Z 47 ) +} be the MOD nturl neutrosophic subset semiroup under +. Study questions (i) to (iii) of problem (1) for this A. 3. Let D = {S( Z 128 ) +} be the MOD nturl neutrosophic subset semiroup under +. Study questions (i) to (iii) of problem (1) for this D. 4. Let M = {S(C (Z 12 )) +} be the MOD nturl neutrosophic finite complex number subset semiroup under +.

329 328 MOD Nturl Neutrosophic Subset Semiroups (i) Study questions (i) to (iii) of problem (1) for this M. (ii) Compre M with P = {S( Z ) +}. 5. Let N = {S(C (Z 43 )) +} be the MOD nturl neutrosophic finite complex number subset semiroup under +. (i) Study questions (i) to (iii) of problem (1) for this N. (ii) Compre this N with M of problem Let P = S(Z 1 ) +} be the MOD nturl neutrosophic - neutrosophic subset semiroup under +. Study questions (i) to (iii) of problem (1) for this P. 7. Let V = S(Z 23 ) +} be the MOD nturl neutrosophic - neutrosophic subset semiroup under +. (i) Study questions (i) to (iii) of problem (1) for this V. (ii) Compre V with this P of problem Let E = S(Z 19 ) +} be the MOD nturl neutrosophic dul number subset semiroup under +. Study questions (i) to (iii) of problem (1) for this E. 9. Let F = S(Z 48 ) +} be the MOD nturl neutrosophic dul number subset semiroup. (i) Study questions (i) to (iii) of problem (1) for this F. (ii) Compre E of problem 8 for this F. 1. Let G = S(Z 244 k ) +} be the MOD nturl neutrosophic qusi dul number subset semiroup under +. Study questions (i) to (iii) of problem (1) for this G. 11. Let H = S(Z 53 k ) +} be the MOD nturl neutrosophic specil qusi dul number subset semiroup under +. 12

330 MOD Subset Nturl Neutrosophic Semiroups 329 (i) Study questions (i) to (iii) of problem (1) for this H. (ii) Compre H with G of problem Let J = S(Z 113 h ) +} be the MOD subset nturl neutrosophic specil dul like number subset semiroup under +. Study questions (i) to (iii) of problem (1) for this J. 13. Let K = {S(Z 48 h ) +} be the MOD subset nturl neutrosophic specil dul like number subset semiroup under +. (i) Study questions (i) to (iii) of problem (1) for this K. (ii) Compre J of problem (12) with this K. 14. Enumerte ll specil fetures ssocited with MOD nturl neutrosophic subset semiroups under Let T = {S( Z 24 ) } where is zero dominted product MOD nturl neutrosophic subset semiroup. (i) Find o(t). (ii) Enumerte ll MOD subset zero divisors. (iii) Prove T hve idels find ll MOD subset nturl neutrosophic idels of T. (iv) Find ll MOD nturl neutrosophic subsemiroup of T which re not idels of T. (v) List out ll MOD nturl neutrosophic nilpotents subsets of T. (vi) List out ll MOD nturl idempotent subsets of T (if ny find out). 16. Let P = {S( Z 29 ) } be the MOD nturl neutrosophic zero dominted product MOD nturl neutrosophic semiroup. Study questions (i) to (vi) of problem (15) for this P.

331 33 MOD Nturl Neutrosophic Subset Semiroups 17. Let M = {S(C (Z 4 )) } be the MOD nturl neutrosophic finite complex number zero dominted product semiroup. Study questions (i) to (vi) of problem (15) for this M. 18. Let W = {S(C (Z 59 )) } be the MOD nturl neutrosophic finite complex number zero dominted product semiroup. Study questions (i) to (vi) of problem (15) for this W. 19. Let S = {S(C (Z 42 )) } be the MOD nturl neutrosophic finite complex number zero dominted product semiroup. (i) Study questions (i) to (vi) of problem (15) for this S. (ii) Compre S with M nd W of problem 17 nd 18 respectively. 2. Let W = {S(Z 42 ) } be the MOD nturl neutrosophic neutrosophic subset semiroup under usul zero dominted product. Study questions (i) to (vi) of problem (15) for this W. 21. Let T = {S(Z 193 ) } be the MOD nturl neutrosophic - neutrosophic subset semiroup under usul zero dominted product. (i) Study questions (i) to (vi) of problem (15) for this T. (ii) Compre T with W of problem Let E = {S(Z 26 ) } be the MOD nturl neutrosophic dul number subset semiroup which is usul zero dominted. Study questions (i) to (vi) of problem (15) for this E. 23. Let F = {S(Z 53 ) } be the MOD nturl neutrosophic dul number MOD nturl neutrosophic zero dominted subset semiroup.

332 MOD Subset Nturl Neutrosophic Semiroups 331 (i) Study questions (i) to (vi) of problem (15) for this F. (ii) Compre F with E in problem Let Z = {S(Z 24 h ) } be the MOD nturl neutrosophic specil dul like number subset semiroup under the usul zero dominted product. (i) Study questions (i) to (vi) of problem (15) for this Z. (ii) f in Z the zero dominted product is replced by MOD nturl neutrosophic zero dominted product compre them. 25. Let K = {S(Z 19 k ) } be the MOD nturl neutrosophic specil qusi dul number subset semiroup under usul zero divisor dominted product. Study questions (i) to (vi) of problem (15) for this K. 26. Let V = {S(Z 424 k ) } be the MOD nturl neutrosophic specil qusi dul number subset semiroup under MOD nturl neutrosophic zero dominted product. (i) Study questions (i) to (vi) of problem (15) for this V. (ii) Compre K of problem 25 with this V. 27. Let N = {S(Z n ) } be the MOD nturl neutrosophic - neutrosophic subset semiroup. Enumerte ll properties ssocited with N. 28. Let M = i S( Z ); 1 i 12; +} 24 be the MOD nturl neutrosophic subset mtrix semiroup under +.

333 332 MOD Nturl Neutrosophic Subset Semiroups (i) Study the specil fetures of this semiroup. (ii) Find ll subsemiroup of M. (iii) Find o(m). 29. Let Y = i S(C (Z 47 )). +} be the MOD nturl neutrosophic finite complex number subset semiroup under +. Study questions (i) to (iii) of problem (28) for this Y. 3. Let Y = i S(Z 16 ) 1 i 1; +} be the MOD nturl neutrosophic - neutrosophic mtrix subset semiroup under +. Study questions (i) to (iii) of problem (28) for this C. 31. Enumerte ll specil properties enjoyed by MOD nturl neutrosophic - neutrosophic mtrix subset semiroups under Let M = i S(Z 1 k ) 1 i 12; +}

334 MOD Subset Nturl Neutrosophic Semiroups 333 be the MOD nturl neutrosophic specil qusi dul number subset mtrix semiroup under +. Study questions (i) to (iii) of problem (28) for this M. 33. Let M = i S(Z 42 ) 1 i 12; +} be the MOD nturl neutrosophic dul number mtrix subset semiroup under +. Study questions (i) to (iii) of problem (28) for this T. 34. Let M = i S(Z 28 h ) 1 i 2; +} be the MOD nturl neutrosophic specil dul like number mtrix semiroup. Study questions (i) to (iii) of problem (28) for this B. 35. Let D = {( ) i S( Z ); 1 i 5; } be the MOD nturl neutrosophic subset mtrix semiroup under zero dominted product. 48 (i) (ii) Find o(p). Find ll MOD nturl neutrosophic subset mtrix subsemiroups which re not idels.

335 334 MOD Nturl Neutrosophic Subset Semiroups (iii) Find ll MOD nturl neutrosophic subset mtrix subsemiroup which re idels. (iv) Find ll MOD zero divisors of D. (v) Find ll nilpotents of D (if ny). (vi) Find ll idempotents of D (if ny). (vii) Find ny other specil fetures ssocited with D Let E = i S(C (Z 47 )); 1 i 6; n } be the MOD nturl neutrosophic subset mtrix MOD nturl neutrosophic zero dominted finite complex number semiroup under nturl product n. Study questions (i) to (vii) of problem (35) for this E. 37. Let M = i S(Z 216 ) 1 i 16; n } be the MOD nturl neutrosophic dul number mtrix subset semiroup under ny of the product. (i) Study questions (i) to (vii) of problem (35) for this B. (ii) Compre B with both the products. 38. Let P = i S(Z 12 h ) 1 i 15; n }

336 MOD Subset Nturl Neutrosophic Semiroups 335 be the MOD nturl neutrosophic specil dul like number subset mtrix semiroup. Study questions (i) to (vii) of problem (35) for this P. 39. Let W = i S(Z 144 k ); 1 i 14; n } be the MOD nturl neutrosophic specil qusi dul number subset mtrix semiroup under nturl product n. Study questions (i) to (vii) of problem (35) for this W. 4. Let S(M) = {Collection of ll subsets from M = i Z 24 ; 1 i 5; +} +} be the MOD nturl neutrosophic mtrix subsets (from M) semiroup under +. (i) Find o(s(m)). (ii) Find ll MOD nturl neutrosophic subset subsemiroup. (iii) Find ll MOD nturl neutrosophic subset elements which re idempotent subsets under +.

337 336 MOD Nturl Neutrosophic Subset Semiroups 41. Let V = i (C (Z 2 )); 1 i 12; +} be the MOD nturl neutrosophic semiroup under +. Define S(V) = {Collection of ll mtrix subsets from V; +} the MOD nturl neutrosophic finite complex number subset mtrix semiroup under +. Study questions (i) to (iii) of problem (4) for this S(V). 42. Let S(W) = {Collection of ll subsets from W = i SZ 2 ; 1 i 4; +} +} be the MOD nturl neutrosophic - neutrosophic mtrix subset semiroup under +. Study questions (i) to (iii) of problem (4) for this S(W). 43. Let S(B) = {Collection of ll subsets from B = i Z 13 ; 1 i 8; +} +} be the MOD nturl neutrosophic - neutrosophic mtrix subset semiroup under +. Study questions (i) to (iii) of problem (4) for this S(B).

338 MOD Subset Nturl Neutrosophic Semiroups Let T = i Z 24 k ; 1 i 21; +} be the MOD nturl neutrosophic specil qusi dul number semiroup under +. S(T) = {Collection of ll subset mtrices from T +} be the MOD nturl neutrosophic specil qusi dul number subset mtrix semiroup under +. Study questions (i) to (iii) of problem (4) for this T. 45. Let S(M) = {Collection of ll mtrix subsets from M = i Z 12 ; 1 i 12; n }; n } be the MOD nturl neutrosophic mtrix subsets semiroup under n. (i) Find o(s(m)). (ii) Find ll MOD nturl neutrosophic zero divisors. (iii) Find ll MOD nturl neutrosophic nilpotents. (iv) Prove this hs MOD nturl neutrosophic idempotents. (v) How mny MOD nturl neutrosophic subsemiroup which re not idels?

339 338 MOD Nturl Neutrosophic Subset Semiroups (vi) How mny MOD nturl neutrosophic idels exist in S(M)? (vii) Obtin ny other specil property ssocited with S(M). 46. Let S(D) = {Collection of mtrix subset from D = i Z 27 ; 1 i 24; n }; n } be the MOD nturl neutrosophic dul number mtrix subset semiroup under n. Study questions (i) to (vii) of problem (45) for this S(D). 47. Let S(D) = {Collection of ll mtrix subsets from P = i Z 12 ; 1 i 6; n }; n } be the MOD nturl neutrosophic - neutrosophic subset mtrix semiroup. Study questions (i) to (vii) of problem (45) for this S(P).

340 MOD Subset Nturl Neutrosophic Semiroups Let S(Z) = {Collection of ll mtrix subsets from Z = i Z 5 k ; 1 i 4; n (or )}; n (or )} be the MOD nturl neutrosophic specil qusi dul number mtrix subsets semiroup under nturl product or usul product. (i) Study questions (i) to (vii) of problem (45) for this S(Z) under both the nturl product or usul product. (ii) Prove {S(Z) } is non commuttive finite monoid. (iii) Show {S(Z) } hs rihts zero divisors which re not left zero divisors nd vice vers. (iv) Find in {S(Z) } riht idels which re not left idels nd vice vers. i 49. Let S[x] = ix i S (Z 2 ) ; +} i be the MOD nturl neutrosophic polynomil semiroup under +. subset coefficient (i) Prove S[x] hs MOD nturl neutrosophic subset coefficient polynomil subsemiroups of finite order. (ii) Find idempotents under + in S[x]. i 5. Let P[x] = ix i S(C (Z 44 )); +} i be the MOD nturl neutrosophic finite complex number subset coefficient polynomil semiroup under +. Study questions (i) to (ii) of problem (49) for this P[x].

341 34 MOD Nturl Neutrosophic Subset Semiroups i 51. Let M[x] = ix i S(Z 9 ); +} i be the MOD nturl neutrosophic coefficient polynomil semiroup. dul number subset Study questions (i) to (ii) of problem (49) for this M[x]. i 52. Let V[x] = ix i S(Z 43 ); +} i be the MOD nturl neutrosophic - neutrosophic subset coefficient polynomil semiroup under +. Study questions (i) to (ii) of problem (49) for this V[x]. i 53. Let P[x] = ix i S (Z 48 ) ; } i be the MOD nturl neutrosophic subset coefficient polynomil semiroup under zero dominnt product of MOD nturl neutrosophic zero dominnt product. (i) Prove ll MOD nturl neutrosophic subset idels nd subsemiroups of P[x] re of infinite order. (ii) Does P[x] hve MOD nturl neutrosophic subset coefficient polynomil zero divisors? (iii) Does P[x] hve MOD nturl neutrosophic subset coefficient polynomil idempotent? (iv) Cn P[x] hve MOD nturl neutrosophic subset polynomil coefficient nilpotents? (v) Enumerte ll specil fetures enjoyed by P[x]. i 54. Let M[x] = ix i S(Z 93 ); } i be the MOD nturl neutrosophic dul number subset coefficient polynomil semiroup.

342 MOD Subset Nturl Neutrosophic Semiroups 341 Study questions (i) to (v) of problem (53) for this M[x]. i 55. Let B[x] = ix i S(Z 14 ); } i be the MOD nturl neutrosophic - neutrosophic subset coefficient polynomil semiroup. Study questions (i) to (v) of problem (53) for this B[x]. i 56. Let T[x] = ix i S(Z 43 k ); } i be the MOD nturl neutrosophic subset coefficient polynomil semiroup of specil qusi dul numbers. Study questions (i) to (v) of problem (53) for this T[x]. 57. Let S(P[x]) = {Collection of ll subset polynomils from i P[x] = ix i S(C (Z 284 ) +}; +} i be the MOD nturl neutrosophic finite complex number subset polynomils semiroup under +. (i) Find ll MOD nturl neutrosophic finite complex number subset polynomil subsemiroup of S(P[x]). (ii) Cn S(P[x]) hve MOD nturl neutrosophic idempotents under +?

343 342 MOD Nturl Neutrosophic Subset Semiroups 58. Let S(R[x]) = {Collection of ll subset polynomils from i R[x] = ix i Z 24 k ; +}; +} i be the MOD nturl neutrosophic specil qusi dul number coefficient polynomils subsets semiroup under +. Study questions (i) to (ii) of problem (57) for this R[x]. 59. Let S(W[x]) = {Collection of ll subsets from i W[x] = ix i Z 9 ; +}; +} i be the MOD nturl neutrosophic - neutrosophic polynomils subsets semiroup under +. Study questions (i) to (ii) of problem (57) for this W[x]. i 6. Let P[x] 8 = ix i S( Z 87 ); x 9 = 1; +} i be the MOD nturl neutrosophic subset coefficient polynomils under + of finite order. (i) Find o(p[x] 8 ). (ii) Find ll idempotents in P[x] 8. (iii) Find ll MOD nturl neutrosophic subset coefficient polynomil subsemiroup of P[x] Let S[x] 11 = 11 i ix i S(C (Z 24 )); x 12 = 1; +} i be the MOD nturl neutrosophic finite complex number subset coefficient polynomil semiroup.

344 MOD Subset Nturl Neutrosophic Semiroups 343 Study questions (i) to (iii) of problem (6) for this S[x] Let W[x] 9 = 9 i ix i Z 12 ;; x 1 = 1; +} i be the MOD nturl neutrosophic dul number subset coefficient polynomil semiroup under +. Study questions (i) to (iii) of problem (6) for this W[x] Let M[x] 3 = 3 i ix i S(Z 17 k ; x 4 = 1; +} i be the MOD nturl neutrosophic specil qusi dul number subset coefficient polynomil semiroup under +. Study questions (i) to (iii) of problem (6) for this M[x] Let B[x] 8 = 8 i ix i S(C (Z 42 ); x 9 = 1; } i be the MOD nturl neutrosophic finite complex number subset coefficient semiroup under product (cn be usul zero dominted product or MOD nturl neutrosophic zero dominted product). (i) Find o(b[x] 8 ). (ii) Find ll zero divisors in B[x] 8. (iii) Prove B[x] 8 cnnot hve idempotents. (iv) Prove B[x] 8 hs nilpotents. (v) Find ll MOD nturl neutrosophic subset subsemiroups which re not idels of B[x] 8. (vi) Find ll MOD nturl neutrosophic idels of B[x] 8. (vii) Enumerte ny of the specil fetures ssocited with it.

345 344 MOD Nturl Neutrosophic Subset Semiroups 65. Let C[x] 5 = 5 i ix i S(Z 24 ; x 6 = 1; } i be the MOD nturl neutrosophic - neutrosophic subset coefficient polynomil semiroup. Study questions (i) to (vii) of problem (64) for this C[x] Let V[x] 15 = 15 i ix i S(Z 12 ; x 16 = 1; } i be the MOD nturl neutrosophic dul number subset coefficient polynomil semiroup. Study questions (i) to (vii) of problem (64) for this V[x] Let S(P[x] 7 ) = {Collection of ll subsets from P[x] 7 = 7 i ix i i Z 43 ; x 8 = 1; +} be the MOD nturl neutrosophic polynomil subsets semiroup under +. (i) Find o(s(p[x]) 7. (ii) Find ll MOD polynomil subset subsemiroups of S(P[x] 7 ). (iii) Find idempotents if ny in S(P[x] 7 ). 68. Let S(B[x] 18 ) = {Collection of ll polynomil subsets from B[x] 18 = 18 i ix i Z 12 ; x 19 = 1; +}; +} i be the MOD nturl neutrosophic - neutrosophic coefficient polynomil subsets semiroup under +.

346 MOD Subset Nturl Neutrosophic Semiroups 345 Study questions (i) to (iii) of problem (67) for this S(B[x] 18 ). 69. Let S(S[x] 1 ) = {Collection of ll polynomil subsets from S[x] 1 = 11 i ix i Z 9 k ; x 12 = 1; +}; +} i be the MOD nturl neutrosophic specil qusi dul number coefficient polynomil subsets semiroup under +. Study questions S(S[x] 1 ). (i) to (iii) of problem (67) for this 7. Let S(W[x] 1 ) = {Collection of ll subsets from W[x] 1 = 1 i ix i i Z 25 ; x 11 = 1; }; } be the MOD nturl neutrosophic polynomil neutrosophic subsets (from W[x] 1 ) semiroup under usul zero dominnt product or MOD nturl neutrosophic zero dominnt product. (i) Find o(s(w[x] 1 )). (ii) Cn S(W[x] 1 ) divisors? (iii) Does S(W[x] 1 ) contin non trivil idempotents? (iv) Cn S(W[x] 1 ) hve non trivil nilpotents? (v) Find ll MOD subset subsemiroups in S(W[x] 1 ) which re not idels of S(W[x] 1 ). (vi) Find ll idels of S(W[x] 1 ). (vii) Obtin ny other specil feture enjoyed by S(W[x] 1 ). 71. Let S(P[x] 18 ) = {Collection of ll subsets from

347 346 MOD Nturl Neutrosophic Subset Semiroups P[x] 18 = 18 i ix i C (Z 12 ); x 19 = 1; }; } i be the MOD nturl neutrosophic finite complex number coefficient polynomils subsets semiroup under. Study questions (i) to (vii) of problem (7) for this S(P[x] 18 ). 72. Study problem (7) when C (Z 12 ) is replced by Z Let S(B[x] 27 ) = {Collection of ll subsets from 27 i B[x] 27 = ix i Z 18 ; x 28 = 1; }; } i be the MOD nturl neutrosophic specil dul like number coefficient polynomil subset semiroup under. Study questions (i) to (vii) of problem (7) for this S(B[x] 27 ).

348 FURTHER READNG 1. Hrry F. Grph Theory Nros Publictions (reprint ndin edition) New Delhi Kosko B. Fuzzy Conitive Mps nt. J. of Mn-Mchine Studies 24 (1986) Kosko B. Fuzzy Thinkin Hyperion Kosko B. Hidden Ptterns in Combined nd Adptive Knowlede Networks Proc. of the First EEE nterntionl Conference on Neurl Networks (CNN-86) 2 (1988) Kosko B. Neurl Networks nd Fuzzy Systems: A Dynmicl Systems Approch to Mchine ntellience Prentice Hll of ndi Smrndche F. (editor) Proceedins of the First nterntionl Conference on Neutrosophy Neutrosophic Set Neutrosophic Probbility nd Sttistics Univ. of New Mexico Gllup dins.pdf 7. Smrndche F. Neutrosophic Loic - Generliztion of the ntuitionistic Fuzzy Loic presented t the Specil Session on

349 348 MOD Nturl Neutrosophic Semirins ntuitionistic Fuzzy Sets nd Relted Concepts of nterntionl EUSFLAT Conference Zittu Germny 1-12 September Smrndche F. Collected Ppers Editur Abddb Orde Vsnth Kndsmy W.B. Love. Life. Lust. Loss: 11 Rel Life Stories of Mirtion nd ADS A Fuzzy Anlysis Tmil Ndu Stte ADS Control Society Chenni Vsnth Kndsmy W.B. Women. Work. Worth. Womb. 11 rel life stories of rurl women Tmil Ndu Stte ADS Control Society Chenni Vsnth Kndsmy W.B. Public. Pity. Ptients. Pece. Public opinion on HV/ADS Tmil Ndu Stte ADS Control Society Chenni Vsnth Kndsmy W.B. nd Minor A. Estimtion of Production nd Loss or Gin to ndustries Usin Mtrices Proc. of the Ntionl Conf. on Chllenes of the 21 st century in Mthemtics nd its llied topics Feb Univ. of Mysore Vsnth Kndsmy W.B. nd Blu M. S. Use of Weihted Multi-Expert Neurl Network System to Study the ndin Politics Vrhimir J. of Mth. Sci. 2 (22) Vsnth Kndsmy W.B. nd ndr V. Mximizin the psseners comfort in the mdrs trnsport corportion usin fuzzy prormmin Proress of Mt. Bnrs Hindu Univ. 32 (1998) Vsnth Kndsmy W.B. nd Kruppusmy Environmentl pollution by dyein industries: A FAM nlysis Ultr Sci 18(3) (26) Vsnth Kndsmy W.B. nd Mry John M. Fuzzy Anlysis to Study the Pollution nd the Disese Cused by

350 Further Redin 349 Hzrdous Wste From Textile ndustries Ultr Sci 14 (22) Vsnth Kndsmy W.B. nd Rm Kishore M. Symptom- Disese Model in Children usin FCM Ultr Sci. 11 (1999) Vsnth Kndsmy W.B. nd Prmod P. Prent Children Model usin FCM to Study Dropouts in Primry Eduction Ultr Sci. 13 (2) Vsnth Kndsmy W.B. nd Prseeth R. New Fuzzy Reltion Equtions to Estimte the Pek Hours of the Dy for Trnsport Systems J. of Bihr Mth. Soc. 2 (2) Vsnth Kndsmy W.B. nd Um S. Combined Fuzzy Conitive Mp of Socio-Economic Model Appl. Sci. Periodicl 2 (2) Vsnth Kndsmy W.B. nd Um S. Fuzzy Conitive Mp of Socio-Economic Model Appl. Sci. Periodicl 1 (1999) Vsnth Kndsmy W.B. nd Smrndche F. Anlysis of socil spects of mirnt lbourers livin with HV/ADS usin fuzzy theory nd neutrosophic conitive mps Xiqun Phoenix Vsnth Kndsmy W.B. nd Smrndche F. Bsic Neutrosophic lebric structures nd their pplictions to fuzzy nd Neutrosophic models Hexis Church Rock Vsnth Kndsmy W.B. nd Smrndche F. Fuzzy nd Neutrosophic nlysis of Periyr s views on untouchbility Hexis Phoenix Vsnth Kndsmy W.B. nd Smrndche F. Fuzzy Conitive Mps nd Neutrosophic Conitive Mps Xiqun Phoenix 23.

351 35 MOD Nturl Neutrosophic Semirins 26. Vsnth Kndsmy W.B. nd Smrndche F. Fuzzy Reltionl Equtions nd Neutrosophic Reltionl Equtions Neutrosophic Book Series 3 Hexis Church Rock USA Vsnth Kndsmy W.B. nd Smrndche F. ntroduction to n-dptive fuzzy models to nlyse Public opinion on ADS Hexis Phoenix Vsnth Kndsmy W.B. nd Smrndche F. Vedic Mthemtics: Vedic or mthemtics A Fuzzy nd Neutrosophic Anlysis utomton Los Aneles Vsnth Kndsmy W.B. Smrndche F. nd lnthenrl K. Elementry Fuzzy mtrix theory nd fuzzy models for socil scientists utomton Los Aneles Vsnth Kndsmy W.B. nd Anith V. Studies on Femle nfnticide Problem usin Neurl Networks BAM-model Ultr Sci. 13 (21) Vsnth Kndsmy W.B. nd ndr V. Applictions of Fuzzy Conitive Mps to Determine the Mximum Utility of Route J. of Fuzzy Mths publ. by the nt. fuzzy Mt. nst. 8 (2) Vsnth Kndsmy W.B. nd Victor Devdoss dentifiction of the mximum e roup in which the riculturl lbourers suffer helth hzrds due to chemicl Pollution usin fuzzy mtrix Dimension of pollution 3 (25) Vsnth Kndsmy W.B. nd Ysmin Sultn FRM to Anlyse the Employee-Employer Reltionship Model J. Bihr Mth. Soc. 21 (21) Vsnth Kndsmy W.B. nd Ysmin Sultn Knowlede Processin Usin Fuzzy Reltionl Mps Ultr Sci. 12 (2)

352 Further Redin Vsnth Kndsmy W.B. Mry John M. nd Knmuthu T. Study of Socil nterction nd Womn Empowerment Reltive to HV/ADS Mths Tier 1(4) (22) Vsnth Kndsmy W.B. Neelkntn N.R. nd Rmthilm S. Mximize the Production of Cement ndustries by the Mximum Stisfction of Employees usin Fuzzy Mtrix Ultr Science 15 (23) Vsnth Kndsmy W.B. Neelkntn N.R. nd Knnn S.R. Operbility Study on Decision Tbles in Chemicl Plnt usin Hierrchicl Genetic Fuzzy Control Alorithms Vikrm Mthemticl Journl 19 (1999) Vsnth Kndsmy W.B. Neelkntn N.R. nd Knnn S.R. Replcement of Alebric Liner Equtions by Fuzzy Reltion Equtions in Chemicl Enineerin n Recent Trends in Mthemticl Sciences Proc. of nt. Conf. on Recent Advnces in Mthemticl Sciences held t T Khrpur on Dec published by Nros Publishin House (21) Vsnth Kndsmy W.B. Neelkntn N.R. nd Rmthilm S. Use of Fuzzy Neurl Networks to Study the Proper Proportions of Rw Mteril Mix in Cement Plnts Vrhmihir J. Mth. Sci. 2 (22) Vsnth Kndsmy W.B. Pthinthn nd Nryn Murthy. Child Lbour Problem usin Bi-directionl Associtive Memories (BAM) Model Proc. of the 9 th Ntionl Conf. of the Vijnn Prishd of ndi on Applied nd ndustril Mthemtics held t Netji Subhs nst. of Tech. on Feb Vsnth Kndsmy W.B. Rmthilm S. nd Neelkntn N.R. Fuzzy Optimistion Techniques in Kiln Process Proc. of the Ntionl Conf. on Chllenes of the 21 st century in Mthemtics nd its llied topics Feb. 3-4 (21) Univ. of Mysore (21)

353 352 MOD Nturl Neutrosophic Semirins 42. Vsnth Kndsmy W. B. nd Smrndche F. Neutrosophic lebric structures nd neutrosophic N-lebric structures Hexis Phoenix Arizon (26). 43. Vsnth Kndsmy W. B. nd Smrndche F. Smrndche Neutrosophic lebric structures Hexis Phoenix Arizon (26). 44. Vsnth Kndsmy W.B. nd Smrndche F. Fuzzy ntervl Mtrices Neutrosophic ntervl Mtrices nd their Applictions Hexis Phoenix (26). 45. Vsnth Kndsmy W.B. Smrndche F. nd lnthenrl K. Specil fuzzy mtrices for socil scientists nfolernquest Ann Arbor (27). 46. Vsnth Kndsmy W.B. nd Smrndche F. ntervl Semiroups Kpp nd Ome Glendle (211). 47. Vsnth Kndsmy W.B. nd Smrndche F. Finite Neutrosophic Complex Numbers Zip Publishin Ohio (211). 48. Vsnth Kndsmy W.B. nd Smrndche F. Dul Numbers Zip Publishin Ohio (212). 49. Vsnth Kndsmy W.B. nd Smrndche F. Specil dul like numbers nd lttices Zip Publishin Ohio (212). 5. Vsnth Kndsmy W.B. nd Smrndche F. Specil qusi dul numbers nd Groupoids Zip Publishin Ohio (212). 51. Vsnth Kndsmy W.B. nd Smrndche F. Alebric Structures usin Subsets Eductionl Publisher nc Ohio (213). 52. Vsnth Kndsmy W.B. nd Smrndche F. Alebric Structures usin [ n) Eductionl Publisher nc Ohio (213).

354 Further Redin Vsnth Kndsmy W.B. nd Smrndche F. Alebric Structures on the fuzzy intervl [ 1) Eductionl Publisher nc Ohio (214). 54. Vsnth Kndsmy W.B. nd Smrndche F. Alebric Structures on Fuzzy Unit squres nd Neutrosophic unit squre Eductionl Publisher nc Ohio (214). 55. Vsnth Kndsmy W.B. nd Smrndche F. Nturl Product on Mtrices Zip Publishin nc Ohio (212). 56. Vsnth Kndsmy W.B. nd Smrndche F. Alebric Structures on Rel nd Neutrosophic squre Eductionl Publisher nc Ohio (214). 57. Vsnth Kndsmy W.B. lnthenrl K. nd Smrndche F. MOD plnes EuropNov (215). 58. Vsnth Kndsmy W.B. lnthenrl K. nd Smrndche F. MOD Functions EuropNov (215). 59. Vsnth Kndsmy W.B. lnthenrl K. nd Smrndche F. Multidimensionl MOD plnes EuropNov (215). 6. Vsnth Kndsmy W.B. lnthenrl K. nd Smrndche F. Nturl Neutrosophic numbers nd MOD Neutrosophic numbers EuropNov (215). 61. Vsnth Kndsmy W.B. lnthenrl K. nd Smrndche F. Alebric Structures on MOD plnes EuropNov (215). 62. Vsnth Kndsmy W.B. lnthenrl K. nd Smrndche F. MOD Pseudo Liner Alebrs EuropNov (215). 63. Vsnth Kndsmy W.B. lnthenrl K. nd Smrndche F. Problems in MOD Structures EuropNov (216). 64. Vsnth Kndsmy W.B. lnthenrl K. nd Smrndche F. Semiroups on MOD Nturl Neutrosophic Elements EuropNov (216).

355 354 MOD Nturl Neutrosophic Semirins 65. Vsnth Kndsmy W.B. lnthenrl K. nd Smrndche F. Specil Type of Fixed Points of MOD Mtrix Opertors EuropNov (216). 66. Vsnth Kndsmy W.B. lnthenrl K. nd Smrndche F. Specil Type of Fixed Points pirs of MOD Rectnulr Mtrix Opertors EuropNov (216). 67. Vsnth Kndsmy W.B. Smrndche F. nd lnthenrl K. Distnce in mtrices nd their pplictions to fuzzy models nd neutrosophic models EuropNov (214). 68. Vsnth Kndsmy W.B. Smrndche F. nd lnthenrl K. MOD Conitive Mps models nd MOD nturl neutrosophic Conitive Mps models EuropNov (216). 69. Vsnth Kndsmy W.B. Smrndche F. nd lnthenrl K. MOD Reltionl Mps models nd MOD nturl neutrosophic reltionl Mps models EuropNov (216). 7. Zdeh L.A. A Theory of Approximte Resonin Mchine ntellience 9 (1979) Zhn W.R. nd Chen S. A Loicl Architecture for Conitive Mps Proceedins of the 2 nd EEE Conference on Neurl Networks (CNN-88) 1 (1988) Zimmermnn H.J. Fuzzy Set Theory nd its Applictions Kluwer Boston 1988.

356 NDEX MOD dul number nilpotent subsets 67-9 MOD dul number subset mtrix semiroup under MOD dul number subset semiroup 66-8 MOD dul number subset subsemiroup67-9 MOD finite complex number idempotent subset 63-5 MOD finite complex number nturl neutrosophic subset set MOD finite complex number subset mtrix semiroup under MOD finite complex number subset semiroup under 61-3 MOD finite complex number subset semiroups under MOD finite complex number subset subsemiroups 55-7 MOD finite complex number universl subset pir 63-4 MOD lrest bsorbin subset 17-9 MOD mtrix subset idempotents under n MOD mtrix subset neutrosophic semiroup under n (or ) MOD mtrix subset nilpotents MOD mtrix subset semiroup under (or ) MOD mtrix subset specil dul like number semiroup under n (or ) 26-9

357 348 MOD Nturl Neutrosophic Subset Semiroups MOD mtrix subset specil qusi dul number semiroup under n (or ) MOD mtrix subset subsemiroup under MOD mtrix subset zero divisors MOD nturl neutrosophic dul number mtrix subset semiroup under MOD nturl neutrosophic dul number subset mtrix semiroup under n (or ) MOD nturl neutrosophic dul number subset polynomil semiroup under MOD nturl neutrosophic dul number subset semiroup 95-7 MOD nturl neutrosophic dul number subset subsemiroup 98-9 MOD nturl neutrosophic elements subset universl pir MOD nturl neutrosophic finite complex number subset polynomil semiroup under MOD nturl neutrosophic finite complex number subset semiroup under MOD nturl neutrosophic mtrix subset semiroup under MOD nturl neutrosophic numbers 9-15 MOD nturl neutrosophic specil dul like number subset semiroup 97-9 MOD nturl neutrosophic specil qusi dul number subset idempotent under MOD nturl neutrosophic specil qusi dul number subset mtrix semiroup under n MOD nturl neutrosophic specil qusi dul number subset polynomil semiroup under MOD nturl neutrosophic specil qusi dul number subset semiroup under

358 ndex 349 MOD nturl neutrosophic specil qusi dul number under MOD nturl neutrosophic specil qusi dul number under zero dominnt product semiroup MOD nturl neutrosophic subset coefficient polynomil semiroup under 32-1 MOD nturl neutrosophic subset idempotents under MOD nturl neutrosophic subset mtrix semiroup under (or ) MOD nturl neutrosophic subset semiroup under MOD nturl neutrosophic subset semiroups 32-5 MOD nturl neutrosophic subset set 85-7 MOD nturl neutrosophic subset subsemiroups 87-9 MOD nturl neutrosophic zero dominnt product semiroup MOD nturl neutrosophic-neutrosophic idempotent subset under MOD nturl neutrosophic-neutrosophic semiroup 35-7 MOD nturl neutrosophic-neutrosophic subset mtrix under MOD nturl neutrosophic-neutrosophic subset polynomil semiroup under MOD nturl neutrosophic-neutrosophic subset semiroup under MOD nturl neutrosophic-neutrosophic subset set 91-3 MOD nturl neutrosophic-neutrosophic subset subsemiroup under MOD nturl neutrosophic-neutrosophic subset universl pir under MOD nturl specil qusi dul number subset mtrix semiroup under MOD neutrosophic subset coefficient polynomil semiroup MOD neutrosophic subset idel 65-7 under 231-4

359 35 MOD Nturl Neutrosophic Subset Semiroups MOD neutrosophic subset mtrix semiroup under MOD pseudo identity 42-9 MOD specil dul like number mtrix subset semiroup under MOD specil dul like number subset coefficient polynomil semiroup under MOD specil dul like number subset idempotents 7-3 MOD specil dul like number subset semiroup MOD specil qusi dul number mtrix subset semiroup under MOD specil qusi dul number subset coefficient polynomil semiroup under MOD specil qusi dul number subset idempotents 72-4 MOD specil qusi dul number subset mtrices semiroup under n MOD subset coefficient polynomil zero divisors MOD subset coefficients polynomil semiroup under MOD subset dul number coefficient polynomil semiroup under MOD subset finite complex number coefficient polynomil MOD subset idempotent MOD subset mtrices 161 semiroup under MOD subset mtrix dul number semiroup under n (or ) 24-6 MOD subset mtrix finite complex number idel under n MOD subset mtrix finite complex number semiroup under n (or ) MOD subset mtrix finite complex number semiroup under n 257-9

360 ndex 351 MOD subset mtrix finite complex number subsemiroup under MOD subset mtrix semiroup under n 254 MOD subset mtrix semiroup under MOD subset neutrosophic semiroup 58 MOD subset nilpotents 53-5 MOD subset polynomils 161 MOD subset pseudo monoid 43-9 MOD subset semiroup bsorbin element 13-5 MOD subset semiroup nd MOD subset semiroup with pseudo identity 42-9 MOD subset semiroup under MOD subset specil qusi dul number semiroup 72-4 MOD subset subsemiroups 41-9 MOD subset universl pir 51-4 MOD universl finite complex number subset 55-6 MOD universl subset under product 3-6 MOD universl subset 17-9 MOD zero divisor subset pir 51-4 n 258-9

361 ABOUT THE AUTHORS Dr.W.B.Vsnth Kndsmy is Professor in the Deprtment of Mthemtics ndin nstitute of Technoloy Mdrs Chenni. n the pst decde she hs uided 13 Ph.D. scholrs in the different fields of non-ssocitive lebrs lebric codin theory trnsporttion theory fuzzy roups nd pplictions of fuzzy theory of the problems fced in chemicl industries nd cement industries. She hs to her credit 694 reserch ppers. She hs uided over 1 M.Sc. nd M.Tech. projects. She hs worked in collbortion projects with the ndin Spce Reserch Orniztion nd with the Tmil Ndu Stte ADS Control Society. She is presently workin on reserch project funded by the Bord of Reserch in Nucler Sciences Government of ndi. This is her 119 th book. On ndi's 6th ndependence Dy Dr.Vsnth ws conferred the Klpn Chwl Awrd for Coure nd Drin Enterprise by the Stte Government of Tmil Ndu in reconition of her sustined fiht for socil justice in the ndin nstitute of Technoloy (T) Mdrs nd for her contribution to mthemtics. The wrd instituted in the memory of ndin-americn stronut Klpn Chwl who died bord Spce Shuttle Columbi crried csh prize of five lkh rupees (the hihest prize-money for ny ndin wrd) nd old medl. She cn be contcted t vsnthkndsmy@mil.com Web Site: or Dr. K. lnthenrl is Assistnt Professor in the School of Computer Science nd En VT University ndi. She cn be contcted t ilnthenrl@mil.com Dr. Florentin Smrndche is Professor of Mthemtics t the University of New Mexico in USA. He published over 75 books nd 2 rticles nd notes in mthemtics physics philosophy psycholoy rebus literture. n mthemtics his reserch is in number theory non-eucliden eometry synthetic eometry lebric structures sttistics neutrosophic loic nd set (enerliztions of fuzzy loic nd set respectively) neutrosophic probbility (enerliztion of clssicl nd imprecise probbility). Also smll contributions to nucler nd prticle physics informtion fusion neutrosophy ( enerliztion of dilectics) lw of senstions nd stimuli etc. He ot the 21 Telesio-Glilei Acdemy of Science Gold Medl Adjunct Professor (equivlent to Doctor Honoris Cus) of Beijin Jioton University in 211 nd 211 Romnin Acdemy Awrd for Technicl Science (the hihest in the country). Dr. W. B. Vsnth Kndsmy nd Dr. Florentin Smrndche ot the 212 New Mexico-Arizon nd 211 New Mexico Book Awrd for Alebric Structures. He cn be contcted t smrnd@unm.edu

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