Problems on MOD Structures. W. B. Vasantha Kandasamy Ilanthenral K Florentin Smarandache

Size: px
Start display at page:

Download "Problems on MOD Structures. W. B. Vasantha Kandasamy Ilanthenral K Florentin Smarandache"

Transcription

1

2 Problems o MOD Structures W. B. Vasatha Kadasamy latheral K Floreti Smaradache 2016

3 This book ca be ordered from: EuropaNova ASBL Clos du Parasse, 3E 1000, Bruxelles Belgium URL: Copyright 2016 by EuropaNova ASBL ad the Authors Peer reviewers: Professor Paul P. Wag, Ph D, Departmet of Electrical & Computer Egieerig, Pratt School of Egieerig, Duke Uiversity, Durham, NC 27708, USA Dr.S.Osma, Meofia Uiversity, Shebi Elkom, Egypt. Dr. Sebastia Nicolaescu, 2 Terrace Ave., West Orage, NJ 07052, USA. May books ca be dowloaded from the followig Digital Library of Sciece: SBN-13: EAN:

4 CONTENTS Preface 5 Chapter Oe PROBLEMS ON MOD NTERVALS 7 Chapter Two PROBLEMS ASSOCATED WTH MOD PLANES 31 Chapter Three PROBLEMS ON SPECAL ELEMENTS N MOD STRUCTURES 59 3

5 Chapter Four PROBLEMS ON MOD NATURAL NEUTROSOPHC ELEMENTS 75 Chapter Five PROBLEMS ON MOD NTERVAL POLYNOMALS AND MOD PLANE POLYNOMALS 99 Chapter Six PROBLEMS ON DFFERENT TYPES OF MOD FUNCTONS 117 Chapter Seve PROBLEMS ON MOD DFFERENTATON AND MOD NTEGRATON OF FUNCTONS N THE MOD PLANES 121 FURTHER READNG 139 NDEX 143 ABOUT THE AUTHORS 145 4

6 PREFACE this book authors for the first time give several types of problems o MOD structures happes to be a iterestig field of study as it makes the whole 4 quadrat plae ito a sigle quadrat plae ad the ifiite lie ito a half closed ope iterval. So study i this directio will certaily yield several iterestig results. The law of distributivity is ot true. Further the MOD fuctio i geeral do ot obey all the laws of itegratio or differetiatio. Likewise MOD polyomials i geeral do ot satisfy the basic properties of polyomials like its roots etc. Thus over all this study is ot oly iovative ad iterestig but challegig. So this book which is oly full of problems based o MOD structures will be a boo to researchers. Further the MOD series books of the authors will certaily be a appropriate guide to solve these problems. 5

7 We wish to ackowledge Dr. K Kadasamy for his sustaied support ad ecouragemet i the writig of this book. W.B.VASANTHA KANDASAMY LANTHENRAL K FLORENTN SMARANDACHE 6

8 Chapter Oe PROBLEMS ON MOD NTERVALS this chapter we just recall all types of MOD itervals ad the algebraic structure ejoyed by them. All problems related with these MOD itervals ad the algebraic structures ejoyed by them are proposed here. DEFNTON 1.1: S = {[0, ); Z + \ {1} is the MOD real iterval mod. We see the ifiite real lie (, ) ca be mapped oto the MOD iterval [0, ); Z + \ {1} i the followig way. Let η: (, ) [0, ) x if 0 < x < 0 if x = 0 or t ( t (, )) x if < x < 0 η(x) = x t t if x > > 0 ad = s + x t t if < x < 0 ad = s + η is the MOD real trasformatio.

9 8 Problems o MOD Structures We will illustrate this by some examples. Example 1.1: Let S = {[0, 6)} be the real MOD iterval modulo 6. η: (, ) [0, 6) η(4.332) = 4.332, η(9.82) = 3.82, η( 7.68) = = η(48) = 0 η(0) = 0 ad η( 24) = 0 η( ) = η( ) = η is a MOD real trasformatio from (, ) to [0, 6). Example 1.2: Let S = {[0, 13)} be the real MOD iterval. η: (, ) [0, 13) is defied by η(0.84) = 0.84 η(12.993) = , η( 1.234) = , η(65) = 0, η( 26) = 0, η(0) = 0. η( 13) = 0 ad so o. This the way MOD real trasformatio η from (, ) to [0, 13) is made. We see η is maps may poits (or ifiitely may poits) oto a sigle poit. f we wat to study the reverse the MOD trasformatio say η 1 : [0, ) (, ).

10 Problems o MOD tervals 9 We see η 1 (0) = 0, η 1 (x) = tx where t Z + ; so a sigle x ca be or should be mapped oto ifiitely may. This is true for every x [0, ). We will illustrate this situatio by some examples. Example 1.3: Let S = {[0, 10)} be the MOD real iterval. Let η 1 : [0, 10) (, ) defied by η 1 (3) = {10 + 3; 10Z + ; Z } Thus η : (, ) [0, 10) η(3) = 3 η( 7) = 3 η(13) = 3 η( 17) = 3 ad so o. Thus η 1 the iverse map is a pseudo map (as a sigle elemet is map oto) ifiitely may elemets, that is why the term pseudo map is used. η 1 (0.75) = {0.75, 10.75, 20.75, 30.75, 40.75,, 9.25, 19.25, 29.25, }. Thus the sigle poit 0.75 i [0, 10) is mapped oto ifiitely may poits i (, ). That is why we ca η 1 as the pseudo MOD real trasformatio. Likewise the mappigs are carried out i a very systematic way. We give oe more example of this situatio. Example 1.4: Let η 1 : [0, 12) (, ) be the pseudo MOD map. η 1 (0) = {0, 12, Z + } η 1 (1) = {1, 13, 25, 37, ad 11, 23, 35, }

11 10 Problems o MOD Structures η 1 (2) = {2, 14, 26, 38, 50,, 10, 22, 34, 46, } η 1 (3) = {3, 15, 27, 39, 51, 63,, 9, 21, 33, } Thus ifiitely may poits of (, ) are associated with a sigle poit of [0, 12). Cosider [0, 12). η 1 (0.3189) = {0.3189, , , ,, , , , ad so o}. Thus the sigle poit is mapped o a ifiite umber of poits by the pseudo MOD map η 1 : [0, 12) (, ). Further both MOD trasformatio map η ad pseudo iverse MOD maps are uique. We have ifiitely may real MOD itervals, however oly oe real iterval (, ). Next we ca defie operatios of + ad o [0, ). {[0, ), +} is a group of ifiite order. η : (, ) [0, ) is compatible with respect to additio. For if we take the MOD iterval [0, 4). Let x = ad y = (, ). x + y = η(x) = η(y) = η(x) = η(x) + η(y) η(5.925) = η(5.732) + η(0.193) = = Hece we see η ca be compatible with respect to +.

12 Problems o MOD tervals 11 We have oly oe group uder + usig (, ) but there are ifiitely may MOD iterval real groups {[0, m) m Z + \ {1}, +}. This has subgroups of both fiite ad ifiite order. Example 1.5: Let G = {[0, 24), +} be the MOD iterval real group of ifiite order. {Z 24, +} = H 1 is a subgroup of fiite order. H 2 = H 3 = H 4 = {0, 6, 12, 18, +} is a subgroup of fiite order. {0, 8, 16} is a subgroup of fiite order. {0.1, 0.2,, 23, 23.1, 23.2,, 23.9, 0} G is a subgroup of G. Example 1.6: Let G = {[0, 17), +} be a MOD iterval group of ifiite order. Ca we say G is a torsio group? We ca as i case of usual groups defie MOD homomorphism from G = {[0, m), +} to H = {[0, t), +}. This will be illustrated by examples. Example 1.7: Let G = {[0, 6), +} ad H = {[0, 12), +} be two MOD groups of ifiite order. Defie φ : G H φ(0) = 0 φ(x) = 2x if x G φ(0.76) = 1.52 ad so o. Clearly φ(x + y) = 2x + 2y Let x = 3.84 ad y = 4.57 G.

13 12 Problems o MOD Structures φ(x + y) = φ( ) = = 4.82 φ(x) + φ(y) = φ(3.84) + φ(4.57) = = 4.82 ad are idetical hece the claim. This is the way operatios or MOD homomorphisms are made. Clearly oe ca defie MOD isomorphisms of MOD iterval group. Next we proceed oto defie the otio of MOD iterval semigroups. DEFNTON 1.2: Let S = {[0, m), } be the MOD iterval semigroup for a, b S. a b (mod m) S. We will illustrate this situatio by some examples. Example 1.8: Let M = {[0, 12), } be the MOD iterval semigroup. Let x = 3.2 ad y = 6.9 M x y = = (mod 12). Let x = 6 ad y = 2 so x y = (mod 12). Thus M has zero divisors. Clearly 1 M is such that 1 x = x for all x M.

14 Problems o MOD tervals 13 Thus M is a mooid. M has zero divisors. 11 M is such that = 1 (mod 12). Let x = 2.4 ad y = 5 [0, 12). Clearly x y = = 0 (mod 12) so is a zero divisor. But 5 [0, 12) is a uit as (mod 12). Thus a uit cotributes to a zero divisor. This gives oe a atural problem amely why i the MOD iterval semigroups we have uits cotributig to zero divisors. This is the oe of the marked differece betwee the usual semigroups ad MOD iterval semigroups. Thus MOD iterval semigroup behave i a odd way so we defie such zero divisors cotributed by uits of the MOD iterval semigroup as MOD pseudo zero divisors or just pseudo zero divisors. Example 1.9: Let S = {[0, 7), } be the MOD iterval semigroup of ifiite order. This has oly pseudo zero divisors. For x = 3.5 ad y = 2 S is such that x y = 0 (mod 7) is a pseudo zero divisor however y = 2 is a uit i S as (mod 7). Cosider x = 1.75 ad y = 4 S. x y = 0 (mod 7) is agai pseudo zero divisor; however y is a uit of S. Thus S has pseudo zero divisors which are ot zero divisors for oe of them is a uit. Next we proceed oto study about the substructures i MOD iterval mooids. MOD iterval mooids have fiite order subsemigroups but ca it have ideals of fiite order is a questio mark. Example 1.10: Let S = {[0, 18), } be the MOD iterval mooid.

15 14 Problems o MOD Structures P 1 = {0, 9}, P 2 = {0, 2, 4,, 16}, P 3 = {0, 3, 6,, 15}, P 4 = {0, 6, 12} ad P 5 = {Z 18, } are fiite order MOD subsemigroups of S. H 1 = { 0.01, } is a ifiite order subsemigroup of S. H 2 = { 0.1 }, H 3 = { 0.2 } are all some of the ifiite order MOD subsemigroup of S. Noe of them are ideals. Fidig ideals happes to be a very difficult problem. Example 1.11: Let S = {[0, 23), } be the MOD iterval semigroup (mooid). S is of ifiite order. P 1 = {Z 23 } is a subsemigroup of fiite order. P 2 = { 0.1 } is a subsemigroup of ifiite order. P 3 = { 0.2 } is a subsemigroup of ifiite order ad so o. Fidig ideals of S happes to be a challegig problem. Now MOD iterval groups ad semigroups are also built usig matrices with etries from MOD itervals [21-2, 24]. Study i this directio is iterestig. Next we proceed oto describe the otio of MOD real iterval pseudo rigs for more [21].

16 Problems o MOD tervals 15 Example 1.12: Let R = {[0, 10), +, } be the MOD iterval pseudo rig. R has zero divisors, uits ad idempotets. R has subrigs as well as pseudo subrigs. However fidig pseudo ideals of R happes to be a challegig problem. Example 1.13: Let W = {[0, 29), +, } be the MOD iterval pseudo rig. This MOD pseudo rig has pseudo zero divisors, uits ad subrigs of fiite order. Ca P = { 0.1, +, } be the MOD pseudo subrig of W? Fidig MOD pseudo subrigs of ifiite order happes to be a very difficult problem. Z 29 = P is a subrig of order 29 which is ot pseudo. fact P is a field so W is a S-real MOD pseudo rig. Studyig properties associated with them happes to be a very difficult problem. Example 1.14: Let S = {[0, 48), +, } be the pseudo MOD real iterval rig of ifiite order. This has several subrigs of fiite order which are ot pseudo subrigs. Oce agai fidig ideals of S happes to be a very difficult problem. Next we itroduce the MOD eutrosophic iterval. Let (, ); 2 = be the real eutrosophic iterval. [0, m) = {a a [0, m), 2 = }, 1 < m < is defied as the MOD eutrosophic iterval.

17 16 Problems o MOD Structures Clearly N = {[0, m)} uder + is a abelia group uder product is a semigroup ad uder + ad. N is oly a MOD pseudo eutrosophic iterval rig. All properties associated with MOD real iterval ca be derived for N also. For more refer [24]. Next we proceed oto study the otio of MOD fiite 2 complex modulo itegers [0, m)i F ; i F = (m 1), got from the complex ifiite iterval ( i, i). We ca have MOD trasformatios both for MOD iterval eutrosophic set [0, m) as well as MOD iterval fiite complex modulo itegers [0, m)i F ; i 2 F = m 1. This is left as a matter routie for the reader however for more iformatio refer [18, 21, 24]. Clearly o [0, m)i F for o value of m; 1 < m < we ca built MOD iterval fiite complex modulo iteger semigroup uder or a MOD iterval fiite complex modulo iteger pseudo rig as is ot a closed operatio o [0, m)i F as i F 3i F = 3 i 2 F = 3(m 1) [0, m)i F. This is the marked differece betwee the other types of MOD itervals ad the MOD fiite complex modulo iteger iterval. 2 Example 1.15: Let M = {[0, 8)i F, i F = 7, +} be a MOD iterval fiite complex modulo iteger group of ifiite order. However M is ot eve closed uder as 2i F 2.4i F = 4.8 i 2 F = = 1.6 M. Hece {M, +, } is also ot a MOD complex modulo iteger iterval pseudo rig. Oly the algebraic structure viz. group uder + ca be defied o [0, m)i F ; 1 < m < as i 2 F = m 1.

18 Problems o MOD tervals 17 Now we proceed oto defie the MOD dual umber iterval got from the real dual umber iterval ( g, g) g 2 = 0; the trasformatio from ( g, g) to [0, m)g is the usual oe discussed i [17]. We will illustrate this by some examples. Example 1.16: Let M = {[0, 18)g g 2 = 0, +} be the MOD iterval dual umber group. M is of ifiite order. M has subgroups of fiite order also. Example 1.17: Let P = {[0, 29)g g 2 = 0, +} be the MOD dual umber group. M has subgroups of fiite order. Now we proceed oto build semigroup usig [0, m)g; g 2 = 0. [17, 21, 24]. Example 1.18: Let B = {[0, 12)g g 2 = 0, } be the MOD iterval dual umber semigroup. Clearly B is ot a mooid. Further every proper subset of B is a subsemigroup of B. fact B is a MOD iterval dual umber zero square semigroup of ifiite order. Further every subsemigroup is also a ideal of B. These happe for ay 1 < m <. Next we proceed to give examples of MOD iterval dual umber rig. Example 1.19: Let M = {[0, 25)g; +, } be the MOD iterval dual umber zero square rig. Sice distributive law is true it is ot a pseudo rig oly a rig. Every subgroup of M uder + is a ideal of M. M has both fiite order as well as ifiite order ideals. For istace P 1 = {0, 5g, 10g, 15g, 20g} M is a ideal. P 2 = {Z 25 g g 2 = 0} is a ideal of M of order 25.

19 18 Problems o MOD Structures P 3 = {0, 0.1g, 0.2g,, g,, 24g, 24.1g, 24.2g,, 24.9g} M is a ideal of fiite order. Study of dual umber MOD iterval rigs is importat ad iterestig for two reasos it ca cotribute to zero square semigroups as well as subrigs. Next we cosider the MOD iterval special dual like umber set [0, m)h, h 2 = h. O [0, m)h we proceed to give operatios as + or or both + ad. [18, 21, 24]. All these situatios will be represeted by some examples. Example 1.20: Let S = {[0, 15)h, h 2 = h, +} be a ifiite MOD iterval special dual like umber group. S has subgroups of both fiite ad ifiite order. P 1 = Z 15 h is a subgroup of order 15. P 2 = {0, 3h, 6h, 9h, 12h} is a subgroup of order 5 ad so o. Example 1.21: Let P 1 = {[0, 19)h, h 2 = h, +} be the MOD iterval special dual like umber group of ifiite order. P 1 has subgroups of both fiite ad ifiite order. For more refer [18, 21, 24]. Now we proceed oto give examples of MOD iterval special dual like umber semigroup. Example 1.22: Let M = {[0, 24)h, h 2 = h, } be the MOD iterval special dual like umber semigroup. M has a subsemigroup of fiite order. M has zero divisors, idempotets but o uits. M has subsemigroups of fiite order. But M has o subgroups. That is M is ot a S-subsemigroup.

20 Problems o MOD tervals 19 Example 1.23: Let B = {[0, 47)h h 2 iterval special dual like semigroup. = h, } be the MOD P 1 = Z 47 h is a fiite subsemigroup. However B is ot a S-semigroup. Example 1.24: Let M = {[0, 16)h, h 2 = h; } be the MOD iterval special dual like semigroup which has zero divisors ad idempotets. Thus we have several properties associated with MOD iterval special dual like umber semigroup of ifiite order. Next we proceed oto study the properties of MOD iterval special dual like umber pseudo rig. All these rigs are pseudo as distributive law is ot true i geeral. Example 1.25: Let S = {[0, 82)h, h 2 = h,, +} be the MOD iterval pseudo special dual like umber rig of ifiite order. S has subrigs of fiite order like P 1 = {0, 41h}, P 2 = {Z 82 h} ad P 3 = {0, 2h, 4h,, 80h} are subrigs of fiite order which are ot pseudo. S has also subrigs which are pseudo. Fidig subrigs of ifiite order ad ideals of S happes to be a challegig problem. For more about these structures please refer [19]. Example 1.26: Let P = {[0, 31)h, h 2 = h, +, } be the MOD iterval special dual like umber pseudo rig. P has fiite subrigs which are ot pseudo. P has subrigs of ifiite order. Fid ideals of P.

21 20 Problems o MOD Structures Thus P has pseudo zero divisors ad so o. Next we proceed oto study the otios associated with MOD iterval special quasi dual umbers [0, m)k, k 2 = (m 1)k, 1 < m <. For more about these refer [19]. We will illustrate this by some example. Example 1.27: Let M = {[0, 24)k, k 2 = 23k, +} be the MOD iterval special quasi dual umber group. M is a ifiite group. M has subgroups of fiite order as well as subgroups of ifiite order. Example 1.28: Let P = {[0, 43)k, k 2 = 42k, +} be the MOD iterval special quasi dual umber group. P has oly few umber of subgroups of fiite order. Next we proceed oto study MOD iterval special quasi dual umber semigroups. Example 1.29: Let B = {[0, 48)k, k 2 = 47k, } be the MOD iterval special quasi dual umber semigroup. B has zero divisors. Fidig idempotets happe to be a difficult problem. For more about these refer [19, 24]. Example 1.30: Let M = {[0, 29)k, k 2 = 28k, } be the MOD iterval special quasi dual umber semigroup. M has pseudo zero divisors. Fidig S-uits or uits is a challegig problem. We propose several problems associated with this structure.

22 Problems o MOD tervals 21 Next we proceed oto describe MOD iterval specific quasi dual umber pseudo rigs. Example 1.31: Let B = {[0, 12)k, k 2 = 11k, +, } be the MOD iterval special quasi dual umber pseudo rigs. B is of ifiite order. B has zero divisors. Fidig idempotets is a difficult task. However B has subrigs of fiite order which are ot i geeral ot pseudo. Fidig ideals is yet aother difficult task. Example 1.32: Let S = {[0, 19)k, k 2 = 18k, +, } be the MOD iterval special quasi dual umber pseudo rig. P = {Z 19 k} S is a subrig of fiite order. s S a S-rig? This questio i geeral is difficult as we have to fid a uit, but k 2 = 18k. [19]. So fidig ideals, subrigs etc; happes to be a challegig task. This pseudo rig has pseudo zero divisors. fact all MOD iterval special quasi dual umber pseudo rigs i which m is a prime always cotais pseudo zero divisors. However eve fidig zero divisors is a challegig problem. Eve if p is a prime fidig uits i those rigs are very difficult. But we have just recalled these cocepts with illustrative examples maily to make this book a self cotaied oe. Further these cocepts are elaborately aalysed i books [19-24] but the problems i this book are maily give as the cocept of MOD mathematics or more to be i a layma s laguage the otio of small mathematics is very ew ad advetures i the mathematical world. They behave i may places i a chaotic way as the real world.

23 22 Problems o MOD Structures They are ot well orgaized as the real mathematical world. So this property of chaos has attracted the authors, so authors wish to share these cocepts with those iterested comathematicias. The best way to do it is give or suggest a series of problems. Some problems are simple ad direct. Some of them are difficult problems. Some ca be cosidered as ope cojectures. The mai ad sustaied work o MOD mathematics is surprisig us for whe they are coverted to small scale they do ot possess all the properties whe they are large. This difficulty is ot yet completely overcome by us. the followig we suggest some problems for this chapter. Problems 1. Ca we say G = {[0, m), +, m Z + \ {1}} have ifiite umber of subgroups fiite order? 2. How may subgroups of ifiite order ca G = {[0, m), +}, (m a prime) cotai? 3. Ca G = {[0, m), +}, m Z + \ {1} be a torsio free group? 4. Defie φ : G H where G = {[0, 43), +} ad H = {[0, 24), +} be two MOD iterval groups be a MOD homomorphism. i) How is φ defied? ii) s it ever possible to be have a oe to oe MOD homomorphism?

24 Problems o MOD tervals 23 iii) Ca we have kerφ to be a fiite subgroups of G? iv) Ca we defie a φ : G H so that kerφ is a ifiite subgroup? v) Let S = {Collectio of all MOD homomorphism from G to H}. What is the algebraic structure ejoyed by S? vi) Fid δ : H G. Let R = {Collectio of all MOD homomorphism from H to G}. What is the algebraic structure ejoyed by R? vii) What is relatio exist betwee R ad S? 5. Let G = {[0, 48), +} ad H = {[0, 35), +} be ay two MOD iterval group. Study questios (i) to (vii) of problem 4 for this G ad H. 6. Characterize the property of MOD iterval pseudo zero divisors i MOD iterval semigroups (mooids) S = {[0, m), }. 7. Fid special ad distict features ejoyed by MOD iterval mooids S = {[0, ), }. 8. Ca S = {[0, 19), } have oly fiite umber of pseudo zero divisors?

25 24 Problems o MOD Structures 9. Ca a MOD iterval mooid S = {[0, m), } be free from MOD pseudo zero divisors? 10. Prove all MOD iterval mooids S = {[0, m), } are Smaradache semigroups. 11. Prove or disprove MOD iterval mooids S = {[0, 24), } has more umber of zero divisors tha the MOD iterval mooid R = {[0, 43), }. 12. Ca P = {[0, p), } (p a prime) the MOD iterval mooid have ifiite umber of pseudo zero divisor? 13. Ca P i problem (12) have zero divisors which are ot pseudo zero divisors? 14. s it ever possible to fid MOD iterval real semigroups (mooids) which ca have MOD iterval ideals which is of fiite order? Justify your claim. 15. Let S = {[0, 23), +, } be the MOD iterval pseudo real rig. i) Fid all zero divisors of S. ii) Ca S have ifiite umber of zero divisors? iii) Ca S have idempotets? iv) Ca S have pseudo zero divisors? v) Ca S have S-uits, S-idempotets ad S-zero divisors? vi) Ca S have ideals of fiite order? vii) How may of the subrigs of S are of fiite order? viii) Fid all pseudo ideals of S. ix) Ca S have S-pseudo ideals?

26 Problems o MOD tervals 25 x) Ca S have S-subrigs of ifiite order? xi) s S a S-pseudo iterval rig? 16. Let P = {[0, 145), +, } be the MOD real iterval pseudo rig. Study questios (i) to (xi) of problem (15) for this P. 17. Let S = {[0, m), +,, m a composite umber} be the MOD real iterval pseudo rig. Study questios (i) to (xi) of problem (15) for this S. 18. Let W = {[0, p), +, ; p a prime} be the MOD real iterval pseudo rig. Study questios (i) to (xi) of problem (15) for this W. 19. What are the distict ad special features ejoyed by N = {[0, m); 2 = }? i) s N a group uder +? ii) Ca{N, } be a semigroup? iii) Ca{N, +, } be a MOD pseudo iterval eutrosophic rig? Study all properties of N, 1 < m <. 20. What are the special features ejoyed by MOD iterval fiite complex modulo itegers [0, m)i F, i 2 F = m 1? 21. Prove this set [0, m)i F is ot closed uder product operatio. 22. Show the MOD iterval dual umbers [0, m)g, g 2 = 0 behave i a very differet way from the other MOD itervals.

27 26 Problems o MOD Structures 23. Obtai all the special features ejoyed by MOD special dual like umbers iterval [0, m)h, h 2 = h; 1 < m <. 24. Prove W = {[0, 23)h, h 2 = h, +} is a MOD iterval special dual like umber group of ifiite order. P 1 = Z 23 h is a subgroup of W of order 23. P 2 = {0, 0.1h, 0.2h,, h,, 20h, 20.1h,, 22h, 22.4h, 22.5h,, 22.9h, +, h 2 = h} is also subgroup of fiite order i the MOD iterval dual like umber group W = {[0, 23)h, h 2 = h, +}. Fid all fiite umber of MOD iterval special dual like umber subgroups. 25. Let N = {[0, 42)h, h 2 = h, +} be the MOD iterval special dual like umber group. i) Fid the umber of subgroups of fiite order. ii) Fid the umber of subgroups of ifiite order. 26. Let T = {[0, 31)h, h 2 = h, +} be the MOD iterval special dual like umber group. Study questios (i) ad (ii) of problem (25) for this T. 27. Fid all special properties satisfied by the MOD iterval special dual like umber semigroup {[0, m)h, h 2 = h, } (1 < m < ). 28. Let B = {[0, 20)h, h 2 = h, } be the MOD iterval special dual like umber semigroup.

28 Problems o MOD tervals 27 i) Fid the umber of fiite order subsemigroups of B. ii) Fid all ifiite order subsemigroups of B. iii) Ca B ever be a S-semigroup? iv) Fid all ideals of B. v) Ca ideals of B be of fiite order? Justify your claim. vi) Fid ay S-ideal of B. vii) Fid all idempotets ad S-idempotets of B. viii) Fid all zero divisors ad S-zero divisors of B. ix) Fid ay other special feature ejoyed by B. 29. Let S = {[0, 23)h, h 2 = h, } be the MOD iterval special dual like umber semigroup. Study questios (i) to (ix) of problem (28) for this S. 30. Let M = {[0, 45)h, h 2 = h, } be the MOD iterval special dual like umber semigroup. Study questios (i) to (ix) of problem (28) for this M. 31. Let A = {[0, 26)h, h 2 = h, +, } be the MOD iterval special dual like umber pseudo rig. i) Fid the umber of subrigs which are ot pseudo i A. ii) Ca A be a S-rig? iii) Ca A be a pseudo rig which has zero divisors as well as S-zero divisors? iv) Ca A have S-idempotets? v) Ca A have S-ideals? vi) Ca A have ideals of fiite order? vii) Ca A have a subfield of ifiite order? viii) Obtai ay other special feature ejoyed by A.

29 28 Problems o MOD Structures 32. Let M = {[0, 24)h, h 2 = h, +, } be the MOD iterval pseudo special dual like umber rig. Study questios (i) to (viii) of problem (31) for this M. 33. Let N = {[0, 47)h, h 2 = h, +, } be the MOD iterval pseudo special dual like umber rig. Study questios (i) to (viii) of problem (31) for this N. 34. Let Z = {[0, 248)h, h 2 = h, +, } be the MOD iterval special dual like umber pseudo rig. Study questios (i) to (viii) of problem (31) for this Z. 35. Obtai all special features associated with MOD iterval special dual like umber algebraic structures. 36. Let S = {[0, 28)k, k 2 = 27k, +} be the MOD iterval special quasi dual umber group. i) Prove o(s) =. ii) iii) How may subgroups of fiite order does S have? Fid the umber of subgroups of S of ifiite order. iv) Obtai a trasformatio from group ( k, k) S. (k 2 = k i ( k, k)). 37. Let M = {[0, 15)k, k 2 = 14k, +} be the MOD iterval special quasi dual umber group. Study questios (i) to (iv) of problem (36) for this M.

30 Problems o MOD tervals Let S = {[0, 47)k, k 2 = 46k, +} be the MOD iterval special quasi dual umber group. Study questios (i) to (iv) of problem (36) for this S. 39. Let B = {[0, 45)k, k 2 = 44k, } be the MOD iterval special quasi dual umber semigroup. i) Fid all fiite order subsemigroups of B. ii) Ca B have ideals of fiite order? iii) Fid all ifiite subsemigroups of B which are ot ideals. iv) Ca B be a S-semigroup? v) Fid all pseudo zero divisors of B. vi) Fid all zero divisors which are ot pseudo zero divisors of B. vii) Ca B have uits? viii) Fid all idempotets ad S-idempotets of B. ix) Ca B have S-zero divisors? x) Ca B have S-ideals? xi) Obtai ay other special or iterestig property associated with B. 40. Let M = {[0, 23)k, k 2 = 22k, } be the MOD iterval special quasi dual umber semigroup. Study questios (i) to (xi) of problem (39) for this M. 41. Let W = {[0, 24)k, k 2 = 23k, } be the MOD iterval special quasi dual umber semigroup. Study questios (i) to (xi) of problem (39) for this W.

31 30 Problems o MOD Structures 42. Let T = {[0, 42)k, k 2 = 41k, +, } be the MOD iterval special quasi dual umber pseudo rig. Study all questios (i) to (xi) of problem (39) by appropriately chagig from semigroups to pseudo rigs. 43. Let S = {[0, 53)k, k 2 = 52k, +, } be the MOD iterval special quasi dual umber pseudo rig. Study questios (i) to (xi) of problem (39) by appropriately chagig from semigroup to pseudo rig. 44. Let P = {[0, m)k, k 2 = (m 1)k, +, } be the MOD iterval special quasi dual umber pseudo rig. Study questios (i) to (xi) of problem (39) by appropriately chagig from semigroup to pseudo rig whe i. m is a composite umber. ii. m is a prime umber. 45. Prove there exists a MOD trasformatio from the rig R = {( k, k); k 2 = k, +, } to S = {[0, m)k, k 2 = (m 1)k, +, }. 46. Obtai ay other special features of MOD iterval pseudo rigs.

32 Chapter Two PROBLEMS ASSOCATED WTH MOD PLANES this chapter we propose problems simple, difficult as well as ope cojectures related to the seve types of MOD plaes. To make this chapter a self cotaied oe we just recall or describe these otios very briefly. However for more about these otios please refer [26-7]. Throughout this book the real plae y Figure 2.1 x

33 32 Problems o MOD Structures is a four quadrat plae as show above. The MOD real plae R (m); 1 < m < is a plae which has oly oe quadrat. As it is the small size plae built o modulo itegers it will be kow as MOD-plae. We will give some examples of MOD real plaes. Figure 2.2 is a four quadrat plae as show above. The above diagram is the MOD real plae R (m) (1 < m < ) ad elemets of R (m) = {(a, b) where a,b [0, m); that is a ad b are elemets of the MOD iterval [0,m)} Thus R (7), R (148), R (251), R (10706), R (20451) ad so o are all MOD real plaes ad the mai advatage of usig MOD real plaes is that MOD real plaes are ifiite i umber whereas the real plae is oly oe. Several problems i this directio are suggested at the ed of this chapter. Next we recall the descriptio of the MOD complex modulo iteger plae C (m) ; 2 m <. C (m) = {a + b if a, b [0, m), i 2 F = (m 1)}. Just we give the diagram of this MOD complex modulo iteger plae i the followig.

34 Problems associated with MOD Plaes 33 Figure 2.3 Ay poit a + bi F is represeted as (a, b) i the MOD complex modulo iteger plae C (m). Thus i reality we have oe ad oly oe complex plae give b y the followig figure i Figure 2.4 x

35 34 Problems o MOD Structures However there are ifiitely may MOD complex modulo iteger plaes give by C (m), C (5), C (2051), C (999991), C (7024) ad so o. Thus we ca choose ay appropriate complex modulo iteger plae ad work with the problem. fact we have ifiite choice i cotrast with oe ad oly oe complex plae which is a advatage. We have suggested problems at the ed of the chapter some of which are ormal ad a few ca be take for future research. Next we proceed oto describe the MOD eutrosophic plaes R (m) ; 2 m < i the followig. For a fixed positive iteger m the MOD eutrosophic plae R (m) = {a + b a, b [0, m) 2 = } here the is a idetermiacy which caot be give a real value. This caot be misuderstood for idempotet, whe we see = it meas, it is a idempotet but a idetermiate idempotet which caot be described as trasformatio or as a matrix; however may times we may multiply a idetermiate with itself it will oly cotiue to be a idetermiate oly. This metio is made for oe should ot get cofused with special dual like umber plaes. R(g) = {a + bg g 2 = g where g is a special dual like umber}. This cocept is abudat i liear operators which are idempotets. Lattices which are abudat i idempotets. We will first describe the eutrosophic plae before we illustrate the MOD eutrosophic plae R (m); 2 < m < by the followig figure.

36 Problems associated with MOD Plaes Figure 2.5 The above figure is a eutrosophic ifiite plae. We have oe ad oly oe eutrosophic ifiite plae. But we have ifiitely may MOD eutrosophic plaes R (m); 2 m < ; we just give oe illustrative example. x Figure 2.6

37 36 Problems o MOD Structures Ay poit a + b i eutrosophic plae by (a,b). R (m) is deoted i the MOD Thus a + b = (a, b) is the represetatio of its elemets i the MOD eutrosophic plaes. Oce agai we have ifiitely may MOD eutrosophic plaes whereas there is oe ad oly oe eutrosophic plae. This is advatageous while workig with real world problem for istead of workig with oe ifiite structure we ca appropriately use ay oe of the compact oe quadrat MOD eutrosophic plae which has also the capacity to have ifiite values but they ca be realized as bouded ifiities. For a precise trasformatio exists from the four quadrat plae to this oe quadrat plae. Several problems ca be proposed i this directio which is doe i at the ed of this chapter. Next we proceed oto describe the dual umber plae ad the MOD dual umber plae i the followig. A dual umber plae R(g) is a ifiite four quadrat ifiite plae where R(g) = {a + bg a, b R ad g 2 = 0} is as follows

38 Problems associated with MOD Plaes g Figure 2.7 Ay poit a + bg is deoted by (a, b) i R(g). This four quadrat dual umber plae is uique ad there is oe ad oly oe such plae. However the MOD dual umber plaes are ifiite i umber ad for ay positive iteger m it is deoted by R (m) = {a + bg a, b [0, m); g 2 = 0} ; 2 m <. g So there are ifiitely may MOD dual umber plaes. x The dual umber plae followig figure: g R (m) is described by the

39 38 Problems o MOD Structures Figure 2.8 Thus as per eed of the problem which eeds the otio of dual umbers these plaes ca be used. g g g g g For R (2), R (3), R (12), R (18), R (204899) ad so o are some of the MOD dual umber plaes. Problems related with them is give at the ed of this chapter. fact oe ca built all algebraic structures, groups, semigroups ad pseudo rigs; but this is the oly algebraic structure which ca give ifiite order zero square semigroups ad rigs (the rigs are ot pseudo). Next we proceed oto describe special dual like umber plae ad MOD special dual like umber plaes i the followig. g R = {(a + bg g 2 = g, a, b R} where a + bg = (a, b) is the otatio used.

40 Problems associated with MOD Plaes g Figure 2.9 However i some places we have used R (h) istead of R (g) but from the very cotest oe ca easily uderstad whether we are usig dual umber or special dual like umber. Ay poit a + bg i R(g) is deoted i the special dual like umber plae by (a, b). Next we proceed oto describe the otio of MOD special dual like umber plaes i the follows. g R (m) = {a + bg a, b [0, m), g 2 = g} deotes the MOD special dual like umber plae. The diagrammatic represetatio of g R (m) is as follows. x

41 40 Problems o MOD Structures Figure 2.10 Ay poit a + bg i the MOD special dual like umber plae is deoted by (a, b). We have ifiitely may special dual like umber plaes for m ca vary i the ifiite iterval 2 m <. g g g g R (10), R (18), R (18999), R (250) ad so o are some of the examples of MOD special dual like umber plaes. As per eed of the problem oe ca use ay appropriate m; 2 m <, these plaes for storage space is comparatively less whe compared with R(g). Problems for the reader are give at the ed of this chapter. Some examples of special dual like umbers i the real world problems are square matrices A with A A = A, ay p q matrices B with B B = B where is the usual product ad is the atural product of matrices.

42 Problems associated with MOD Plaes 41 Also we have liear operators which ca cotribute to special dual like umbers. Fially oe ca get several special dual like umbers from Z t ; the modulo itegers where t is a composite umber. Also semi lattices ad lattices are abudat with special dual like umbers. Next we proceed oto describe the ifiite plae of special quasi umbers ad MOD special quasi dual umbers. Let R(g) = {a + bg g 2 = g, a, b R} be the ifiite special quasi dual umber plae. We have oly oe such plae. However there are ifiitely may MOD special quasi dual umber plaes which will be described. The special quasi dual umber plae R(g) is as follows g Figure 2.11 x

43 42 Problems o MOD Structures Ay poit a + bg R(g) is oly represeted as (a, b) i R(g), the real special quasi dual umber plae. Now the MOD special quasi dual umber plae. g Now the MOD special quasi dual umber plae R (m) = {a + bg a, b [0, m) ad g 2 = (m 1) g}, 2 m <. Here also ay a + bg is represeted as (a, b) i the MOD special quasi dual umber plae R (m). g The followig diagram gives the MOD special quasi dual umber plae R (m). g Figure 2.12 Now usig six types of MOD plaes we propose the followig problems. The authors wish to keep o record that all these defiitios ad otios are recalled to make this book self cotaied. As the book is problems o MOD structures all the more we wated to make it as self cotaied as possible.

44 Problems associated with MOD Plaes 43 Problems 1. What are the advatages of usig MOD real plaes istead of the real plae? 2. Ca oe say use of MOD real plaes saves both time ad ecoomy? 3. What are the ways of defig the cocept of distace betwee two poits i the MOD real plae; R (m), 2 m <? 4. llustrate the situatio i problem (3) by some examples (that is used a fixed m). 5. Ca the cocept of aalytical geometry be employed o MOD real plaes? 6. Ca the otio of Euclidea geometry be used o MOD real plaes? 7. Give some ice problems i which MOD real plaes would be better tha usual real plae. 8. Ca the cocept of cotiuity be defied i MOD real plaes? 9. Does the otio of cotiuity depeds o the value of the m of the MOD real plae R (m) ad the fuctio? 10. Give examples of fuctios which are cotiuous i the real plae ad are ot cotiuous i the MOD real plae. 11. Ca a o cotiuous fuctio i the real plae be cotiuous i the MOD real plae R (m) depedig o m? 12. Ca you give examples of fuctios i the MOD real plae which are defied but ot defied i the real plae?

45 44 Problems o MOD Structures 13. Clearly x = 3. 1 x 3 = f(x) is ot defied i the real plae at 1 i) Prove or disprove f(x) = is defied i the MOD x 3 real plae R (m) (for some suitable m). ii) which of the other MOD real plaes f(x) = defied? 1 x 3 is 14. Clearly 1 (x 2) (x 7) (x + 3) the real plai at x = 2, x = 7 ad x = 3. i) Give all MOD real plaes i which f(x) = 1 (x 2) (x 7) (x + 3) the MOD real plae. = f(x) is ot defied i is defied at all poits of ii) Ca we have more tha oe real MOD plae for which the above statemet (i) is true? 15. Ca the otio of a circle be defied i the MOD real plae R (m)? 16. Prove i polyomial real plaes R (m)[x] as the polyomials i R (m)[x] do ot satisfy distributive law the otio of solvig equatios is a impossibility i R (m)[x]. 17. llustrate the situatio of problem i (16) for m = 5 ad m = 12.

46 Problems associated with MOD Plaes Prove G = {R (m), +} is a ifiite real MOD plae group uder Ca G i problem 18 have decimals (that is elemets x R (m) \ {(a,b) / a, b Z m } such that x = 0 for some t? 20. Prove S = {R (m), } is a abelia MOD real plae semigroup. 21. Prove or disprove S is ot always a Smaradache semigroup. 22. Prove S = {R (m), } have ifiite umber of zero divisors. 23. Prove or disprove S = {R (m), } has oly fiite umber of idempotets. 24. Ca S = {R (m), } have ilpotet elemets of order greater tha or equal to 2? 25. Ca S = {R (m), } have torsio free elemets? 26. Ca S = {R (m), } have torsio elemets which are ifiite i umber? 27. Will S = {R (m), } have ideals which are of fiite order? 28. Study problem (27) for S = {R (28), }. 29. Fid all subsemigroups all of which are ifiite order but are ot ideals of S = {R (47), }. 30. Ca S = {R (48), } have ifiite umber of subsemigroups of fiite order? 31. Let M = {R (11), } be the semigroup of MOD real plae R (11). t

47 46 Problems o MOD Structures i) Fid all zero divisors of M. ii) Fid all idempotets of M. iii) Fid all the ideals of M. iv) Ca M have ideals of fiite order? v) Ca M have torsio free elemets? vi) vii) Ca M have ifiite umber of torsio elemets? Prove M ca have ilpotet elemets of order greater tha or equal to two. viii) Ca M have ifiite umber of uits? ix) Obtai all special properties associated with M. 32. Let R = {R (m), +} be the MOD pseudo rig. i) Ca we have elemets x, y, z R (m) \ {(a, b) a, b Z m } which satisfy the distributive law x (y + z) = x y + x z? ii) s the o distributive ature i R a advatage to problems i geeral or disadvatage to those problems? iii) Ca oe always accept that all atural problems for research really satisfy the distributive laws? Justify! iv) Ca we prove all practical problems i the world satisfy distributive law? v) Ca oe just ackowledge that the o acceptace of distributive law does ot drastically affect the solutio of the problem. vi) Uder ay of these circumstaces ca oe claim that use of MOD real plae pseudo rigs will give more accurate aswer if appropriately adopted to certai problems. vii) Fid all idempotets of {R (m),+ } the pseudo rig.

48 Problems associated with MOD Plaes 47 viii) Fid all uits of {R (m), +, }. ix) Ca oe say the umber of uits i both {R (m), } ad that of {R (m), +, } have same umber of uits? x) Ca we prove both {R (m), } ad {R (m), +, } have the same umber of idempotets? xi) xii) Ca we say both {R (m), } ad {R (m), + } have same umber of zero divisors? Obtai ay of the special features ejoyed by {R (m), +, }. 33. Let B = {R (23), +, } be the MOD real plae pseudo rig. Study questios (i) to (xii) of problem (32) for this B. 34. Study questios (i) to (xii) of problem (32) for this M = {R (48), +, }. 35. Let B = {R (m)[x]} be the MOD real plae polyomials. i) f p(x) B is of degree prove dp(x) dx polyomial of degree ( 1). is ot a MOD ii) Prove p(x) dx may or may ot be defied i B. iii) Prove if p(x) is a MOD polyomial of degree the p(x) eed ot have ad oly roots. iv) Prove p(x) ca have fiite umber of uits less tha.

49 48 Problems o MOD Structures v) Prove p(x) B ca have more tha roots eve though p(x) is a polyomial of degree i B. vi) Ca oe say as distributive law is ot true we have p(x) = (x α 1 ) (x α 2 )... (x α ) x (α 1 + α α ) x 1 + α iα j x 2 + α 1 α = q(x) i geeral. i j vii) Prove p(α 1 ) = 0 the q(α 1 ) 0 i geeral. viii) Does there exist a p(x) of degree with coefficiets from R (m) \ {(a,b) / a, b Z } satisfy the equality. (x α 1 ) (x α 2 ) (x α ) = x (α 1 + α α ) α i α j x 2 + α 1 α? x 1 + i j ix) Let p(x) = (x ) (x ) (x + 4.2) R (5); ca equatio i (viii) be true for this p(x). 36. Ca we have polyomials i R (m)[x] so that the derivatives are as i case of R[x]? 37. Ca we have p(x) R (m)[x] so that the itegral exists as i case of p(x) R[x]? 38. Give examples of polyomials i R (m)[x] i which both itegral ad derivatives does ot exists. 39. Give a MOD polyomial p(x) i R (m)[x] for which itegral exist but derivatives does ot exist. 40. Give a example of a MOD polyomial p(x) i R [m][x] for which derivative exist but itegral does ot exist. 41. What are the special features ejoyed by C (m)? 42. Ca C (m) have uits?

50 Problems associated with MOD Plaes Ca C (m) have idempotets? 44. Prove {C (m), } is oly a semigroup of ifiite order. 45. Obtai all the special features ejoyed by {C (m), +}. 46. Prove C (45) have subgroups of fiite order. 47. What are the special features ejoyed by {C (m), }? 48. Fid all zero divisors of {C (m), }. 49. Will {C (m), } be always a S-semigroup? 50. Ca ideals of {C (m), } be of fiite order? 51. Obtai ay other differeces betwee {C (m), } ad {R (m), }. 52. Prove {C (m), +, } is a MOD complex modulo iteger pseudo rig. 53. Let P = {C (45), } be the MOD complex modulo iteger semigroup. Study questios (i) to (ix) of problem (31) for this P. 54. Let M = {C (47), } be the MOD complex modulo iteger semigroup. Study questios (i) to (ix) of problem (31) for this M. 55. Let W = {C (15),, +} be the MOD complex modulo iteger pseudo rig. i) Study questios (i) to (xii) of problem (32) for this W.

51 50 Problems o MOD Structures ii) Obtai ay other special ad iterestig feature ejoyed by W. 56. Let V = {C (37),, +} be the MOD complex modulo iteger pseudo rig. Study questios (i) to (xii) of problem (32) for this V. 57. Let M = {C (48), + } be the MOD complex modulo iteger pseudo rig. Study questios (i) to (xii) of problem (32) for this M. 58. Characterize all properties that ca be associated with the MOD complex modulo iteger plae set {C (p)}, p a prime or a composite umber. 59. Let = { R (m)} be the MOD eutrosophic plae. Fid all the special features ejoyed by the MOD eutrosophic plae. 60. Let = { R (20), } be the MOD eutrosophic semigroup. i) Fid all ideals of M. ii) Prove ideals of M are of ifiite order. iii) Fid all subsemigroups of fiite order. iv) Fid all subsemigroups of ifiite order which are ot ideals. v) Fid all zero divisors ad S-zero divisors of M. vi) Fid all idempotets of M. vii) Fid ilpotets of order higher tha two i M. viii) What are the special features ejoyed by M? ix) Ca M be a S-semigroup? x) Ca M have S-ideals? 61. Let P = { R (37), } be the MOD eutrosophic semigroup.

52 Problems associated with MOD Plaes 51 Study questios (i) to (x) of problem (60) for this P. 62. Let Q = { R (42), +, } be the MOD eutrosophic pseudo rig. i) Fid zero divisors of Q. ii) iii) iv) Ca Q have S-zero divisors? Prove Q has idempotets. Ca Q have S-idempotets? v) Prove Q has ideals of oly ifiite order. vi) vii) Prove Q has S-ideals. Prove Q has subrigs of fiite order. viii) Ca Q have pseudo subrigs of ifiite order which are ot ideals of Q? ix) Obtai ay other special feature ejoyed by Q. x) Compare MOD eutrosophic pseudo rigs with the MOD complex modulo iteger pseudo rig ad MOD real pseudo rig. 63. Let Q = { R (53), +, } be the MOD eutrosophic pseudo rig. Study questios (i) to (x) of problem (62) for this Q. 64. Let T = { R (50), +, } be the MOD eutrosophic pseudo rig. Study questios (i) to (x) of problem (62) for this T.

53 52 Problems o MOD Structures 65. Let W = { R (m); g 2 = 0; q m < } be the MOD dual umber plae. i) Obtai all the special features ejoyed by W. ii) iii) iv) Ca the distace cocept be defied o this plae? Ca a circle be defied o this plae? Prove {W, } = S is the MOD special dual umber semigroup of ifiite order. v) Ca S have ifiite umber of idempotets? vi) vii) Prove S has ifiite umber of zero divisors? Prove {W, } = S has zero square subsemigroups. viii) S = {W, } has oly ideals of ifiite order prove. ix) Prove S = {W, } have subsemigroups of ifiite order which are ot ideals. x) Obtai ay other special feature ejoyed by S = {W, }. xi) xii) Ca S have ifiite umber of uits? Ca these MOD dual umber semigroup fid ay applicatios to real world problems? xiii) What are the advatages of usig these MOD dual umber plaes i the place of dual umber plae? xiv) Study all the special features ejoyed by B = {W, +}, the MOD dual umber group. xv) Ca B have subgroups of fiite order?

54 Problems associated with MOD Plaes 53 xvi) s every elemet i B a torsio elemet? Justify your claim! xvii) Study the algebraic structure ejoyed by the MOD dual umber pseudo rig where M = {W, +, }. xviii) Prove M has subsemigroups of ifiite order which are ot pseudo. xix) Prove all ideals of M are of ifiite order. xx) Prove or disprove all zero divisors of M ad that of the MOD semigroup S are the same. xxi) Prove or disprove all uits i S ad M are same. xxii) Prove or disprove all idempotets of S ad M are the same. xxiii) Fid all subrigs of M of fiite order. xxiv) Obtai ay other special feature ejoyed by M. 66. Let V = { R g (58), g 2 = 0} be the MOD dual umber plae. i) Study questios (i) to (xxiv) of problem (65) for this V. ii) Obtai ay other feature which is uique about V. 67. Let T = { R h (m), h 2 = h} be the MOD special dual like umber plae. i) Derive all the special features ejoyed by T. ii) iii) Ca T fid ice applicatios of the real world problem? {T, } = W be the MOD special dual like umber

55 54 Problems o MOD Structures plae semirig, prove W is a commutative semigroups of ifiite order. iv) Prove all ideals of W are of ifiite order. v) Fid the umber of subsemigroups of fiite order. vi) Show W ca have subsemigroups of ifiite order which are ot ideals. vii) Fid all zero divisors of W. viii) Ca W have S-zero divisors? ix) Characterize all idempotets of W. x) Ca W have S-uits? xi) xii) Prove or disprove W ca have oly fiite umber of uits. Prove {T, +} = A be the MOD special dual like umber group, A has ifiite order subsemigroups. xiii) How may subgroups of A are of fiite order for a give M? xiv) Prove S = {T, +, } be the MOD special dual like umber pseudo rig does ot satisfy the distributive law. xv) How may subrigs of S are of fiite order? xvi) Fid all zero divisors which are ot S-zero divisors of S (at least characterize). xvii) Ca S have ifiite umber of uits? xviii) Does S cotai ifiite umber of idempotets?

56 Problems associated with MOD Plaes 55 xix) Fid all idempotets of S which are ot S-idempotets. xx) Give ay other special feature ejoyed by these MOD special dual like umber of plaes? 68. Let W = { R h (53), h 2 = h} be the MOD special dual like umber plae. Study questios (i) to (xx) of problem (67) for this W. 69. Let M = { R h (48), h 2 = h} be the MOD special dual like umber plae. Study questios (i) to (xx) of problem (67) for this M. h 70. Let P = { R (256), h 2 = h} be the MOD special dual like umber plae. Study questios (i) to (xx) of problem (67) for this M. h 71. Let P = { R (49) h 2 = h} be the MOD special dual like umber plae. Study questios (i) to (xx) of problem (67) for this P. 72. Let S = { R k (m), k 2 = (m 1)k} be the MOD special quasi dual umber plae. i) Study the special ad distict features associated with S. ii) What are advatages of usig the MOD special quasi dual umber plaes?

57 56 Problems o MOD Structures iii) iv) Give some applicatios of this MOD special quasi dual umber plaes to real world problems. Prove the MOD special quasi dual umber plae semigroup (S, } is commutative ad is of ifiite order. v) Prove all ideals of {S, } are of ifiite order. vi) vii) Fid all subsemigroups of {S, } which are of fiite order. What are the advatages of usig this semigroup {S, } istead of the MOD real semigroup {R (m), } or {C (m), } or { R (m), } or { R (m), }? g viii) Fid all uits ad S-uits of {S, }. ix) Ca we claim {S, } has ifiite umber of zero divisors? x) Fid all idempotets of {S, } which are ot S- idempotets of {S, }. xi) Study the special properties ejoyed by the MOD special quasi dual umber group {S, +}. xii) Fid all subgroups of fiite order i {S, +}. xiii) Ca {S, +} have subgroups of ifiite order, ifiite i umber or oly fiite i umber? xiv) Prove {S, +, } is oly a pseudo MOD special quasi dual umber rig of ifiite order. xv) Fid all subrigs of {S, +, } of fiite order. xvi) Ca {S, +, } have ifiite umber of uits?

58 Problems associated with MOD Plaes 57 xvii) Does {S, +, } cotai S-uits? xviii) Ca {S, +, } have ifiite umber of idempotets? xix) Fid all S-idempotets of {S, +, }. xx) Ca {S, +, } have ifiite umber of zero divisors? xxi) Characterize all S-zero divisors of {S, +, }. xxii) Prove all ideals of {S, +, } are oly of ifiite order. xxiii) Obtai ay other iterestig special feature ejoyed by {S, +, }. 73. Let S 1 = { R k (19), k 2 = 18k} be the MOD special quasi dual umber plae. i) Study questios (i) to (xxii) of problem (72) for this S 1. ii) Does the product reflect o of the properties as m = 19 is a prime? 74. Let B = { R k (24), k 2 = 23k} be the MOD special quasi dual umber plae. i) Study questios (i) to (xx) of problem (72) for this B. ii) Does the fact m = 24 a composite umber have ay impact o umber of subrigs, umber of zero divisors ad umber of uits? iii) Compare S 1 of problem (73) with this B.

59 58 Problems o MOD Structures 75. Let M = { R k (3 7 ), k 2 = (3 7 1)k} be the MOD special quasi dual umber plae. i) Study questios (i) to (xx) of problem (72) for this M. ii) iii) iv) As m = 3 7 does it have ay impact o the umber of idempotets of M or o S-idempotets (if ay) o M. Compare M of this problem with B ad S 1 of problem (74) ad (73) respectively. Ca we say if m is a power of a prime umber some of the properties ejoyed by these MOD special quasi dual umber plae is distictly differet from whe m is a composite umber? 76. Compare the algebraic structures ejoyed by the plaes g h k R (m), R (m), C (m), R (m), R (m) ad R (m) whe they are defied o them. i) Which of the 6 plaes is a very powerful as a group? ii) iii) Which of the 6 plaes will fid more applicatios? Which of the six plaes behaves more like the real plae?

60 Chapter Three PROBLEMS ON SPECAL ELEMENTS N MOD STRUCTURES this chapter for the first time we propose the followig problems some of which are difficult ad others are at research level. We have already recalled the otio of MOD iterval ad MOD plaes of differet types [26, 30]. the first place the cocept of special pseudo zero divisors was defied that is i [0,5); 4 is a uit but 2.5 [0, 5) is such that (mod 5) as well as (mod 5) [21,24]. Thus 2 ad 4 are uits as (mod 5) ad (mod 5); but they also lead to zero divisors so we defie them as special pseudo divisors. For i usual algebraic structures this sort of situatio will ever occur.

61 60 Problems o MOD Structures Oly due to MOD itervals such sort of special pseudo divisors i possible. Let [0, 8) be the MOD iterval. Let 5 [0,8); clearly 5 5 = 1 (mod 8). Now for 1.6 [0,8) we have = 8 (mod 8) = 0. Thus 1.6 is a special pseudo MOD zero divisor but 5 is a uit. Will 1.6 [0,8) be a uit is aother ope cojecture? That is ca for some t; (1.6) t = 1 for some t > 2. Yet = 0. Ca this sort of mathematical magic occur? This is a ope cojecture which is give i the problems suggested at the ed of the chapter. Now barig the study of these MOD iterval special pseudo zero divisors we ca have these types of special MOD pseudo zero divisors i the plae. They happe to be differet from the special MOD iterval pseudo zero divisors. We will illustrate these by the followig examples. Let R = R (7) = {(x, y) x, y [0.7)}. Take (3.5, 1.75) R; we have (2, 4) R. (3.5, 1.75) (2, 4) = (0, 0) is agai a MOD pseudo zero divisors of the plae. Cosider (1.75, 0), (4, 3.116) R.

62 Problems o Special Elemets i MOD Structures 61 Clearly (1.75, 0) (4, 3.116) = (0, 0) is a MOD special pseudo zero divisor. So fidig all pseudo zero divisors, fidig idempotets ad S-idempotets of R is a challegig problem. Similarly fidig MOD ilpotets ad uits from (x, y) R (7) \ (Z 7 Z 7 ) happes to be a ope cojecture. Eve fidig the umber of elemets which are uits ad ilpotets happes to be a very difficult problem. We suggest the followig problems. Problems 1. Let B = {[0, m), } be the MOD real iterval uder product. i) Ca [0,m) have ifiite umber of uits? ii) Ca [0, m) have MOD pseudo special zero divisors? iii) Ca the elemets of [0, m) which is a special pseudo zero divisor be both the uits? iv) Characterize all MOD pseudo special pseudo zero divisors? v) Characterize all torsio elemets x [0, m) \ Z m. vi) Ca x [0,m) \ Z m be idempotets? vii) Ca x [0,m) \ Z m be S-uits? viii) Ca x [0,m) \ Z m be S-zero divisors? ix) Obtai ay other special feature associated with it. 2. Let P = {[0, m)k, } be the MOD special quasi dual umber iterval semigroup (k 2 = (m 1) k). i) Ca P have the idetity with respect to. ii) Study questios (i) to (ix) of problem (1) for this P.

63 62 Problems o MOD Structures 3. Let M = {[0,m) h, } be the MOD special dual like umber (h 2 = h) semigroup. i) What is the multiplicative idetity of this semigroup? ii) Study questios (i) to (ix) of problem (1) for this M. 4. Let W = {[0, m), ; 2 = } be the MOD eutrosophic iterval semigroup. i) Fid the multiplicative idetity of W. ii) Study questios (i) to (ix) of problem (1) for this W. iii) Obtai ay applicatios of W as a MOD eutrosophic set. 5. Let S = {[0, 25), 2 =, } be the MOD eutrosophic semigroup. Study questios (i) to (ix) of problem (1) for this S. 6. Let T = {[0, 47), 2 =, } be the MOD eutrosophic semigroup. Study questios (i) to (ix) of problem (1) for this T. 7. Let V = {[0, 24), 2 =, } be the MOD eutrosophic semigroup. Study questios (i) to (ix) of problem (1) for this V. 8. Compare the MOD eutrosophic semigroups i problems 5, 6 ad Let N = {[0,27), } be the MOD real iterval semigroup. Study questios (i) to (ix) of problem (1) for this N. 10. Let Z = {[0, 48), } be the MOD iterval real semigroup. Study questios (i) to (ix) of problem (1) for this Z.

64 Problems o Special Elemets i MOD Structures Let W = {[0, 53), } be the MOD iterval real semigroup. Study questios (i) to (ix) of problem (1) for this W. 12. Compare the MOD iterval real semigroups N, Z ad W i problems 9, 10 ad 11 respectively. 13. Let S = {[0, 19) k, k 2 = 18k, } be the MOD iterval special quasi dual umber semigroup. Study questios (i) to (ix) of problem (1) for this S. 14. Let D = {[0, 28)k, k 2 = 27k, } be the MOD iterval special quasi dual umber semigroup. Study questios (i) to (ix) of problem (1) for this D. 15. Let F = {[0, 243)k, k 2 = 242k, } be the MOD iterval special quasi dual umber semigroup. Study questios (i) to (ix) of problem (1) for this F. 16. Compare the MOD iterval special quasi dual umber semigroups S, D ad F i problems 13, 14 ad 15 respectively. 17. Let Y = {[0, 96)h, h 2 = h, } be the MOD iterval special dual like umber semigroup. i) Study questios (i) to (ix) of problem (1) for this Y. ii) Derive ay other distict feature ejoyed by Y. 18. Let R = {[0, 47)h, h 2 = h, } be the MOD iterval special dual like umber semigroup. Study questios (i) to (ix) of problem (1) for this R.

65 64 Problems o MOD Structures 19. Let P = {[0, 625) h, h 2 = h, } be the MOD iterval special dual like umber semigroup. Study questios (i) to (ix) of problem (1) for this P. 20. Compare the MOD iterval special dual like umber semigroups; Y, R ad P of problems 17, 18 ad 19 respectively. 21. Let N = {[0,m)g g 2 = 0, } be the MOD iterval special dual umber semigroup. i) N is a zero square semigroup. ii) Prove N has o idetity. iii) Prove N has o idempotets. iv) Prove N has all order subsemigroups {2, 3, 4,, }. v) Ca N have S-zero divisor? vi) Prove N caot be a S-semigroup. 22. Let M = {[0, 42)g, g 2 = 0, } be the MOD iterval dual umber semigroup. Study questios (i) to (vi) of problem (21) for this M. 23. Let R = {[0, m), + } be the MOD real iterval pseudo rig. i) Prove or disprove the umber of zero divisors of R is the same as the umber of zero divisors of S = {[0, m), }, the MOD real iterval semigroup. ii) Prove all S-zero divisors of S ad R are the same. iii) Will the umber of uits i S ad R be the same? iv) Prove or disprove all S-uits of S ad R are the same. v) Prove S ad R have the same S-idempotets ad idempotets. vi) vii) Prove ideals of R ad S are ot the same i geeral. Ca R have MOD pseudo zero divisors which are idetical with that of S ad vice versa?

66 Problems o Special Elemets i MOD Structures Let R = {[0, 48), +, } be the MOD real iterval pseudo rig. Study questios (i) to (vii) of problem (23) for this R. 25. Let M = {[0, 37), +, } be the MOD real iterval pseudo rig. Study questios (i) to (vii) of problem (23) for this M. 26. Let Z = {[0, 256), +, } be the MOD real iterval pseudo rig. Study questios (i) to (vii) of problem (23) for this Z. 27. Let W = {[0, 47) g, g 2 = 0, +, } be the MOD iterval dual umber pseudo rig. Study questios (i) to (vii) of problem (23) for this W. 28. Let T = {0, 49), k, k 2 = 48k, +, } be the MOD iterval special quasi dual umber pseudo rig. Study questios (i) to (vii) of problem (23) for this T. 29. Let N = {[0, 120) h, h 2 = h, +, } be the MOD iterval special dual like umber pseudo rig. Study questios (i) to (vii) of problem (23) for this N. 30. Let S = {[0, 55), 2 =, +, } be the MOD iterval eutrosophic pseudo rig. Study questios (i) to (vii) of problem (23) for this S Ca W = {[0, 120)i F, i F = 119; +, } be the MOD iterval complex modulo iteger pseudo rig. Justify?

67 66 Problems o MOD Structures 32. Fid all special ad distict features ejoyed by 2 R = {[0, m)i F ; i F = (m 1); +} the MOD iterval complex modulo iteger group. 33. Let R = {R (m), } be the MOD real plae semigroup. i) Fid all MOD pseudo zero divisors of R. ii) s the umber of such MOD pseudo zero divisors of R ifiite or fiite? iii) Ca we say the umber of uits of R \ {(Z m Z m )}? iv) s the collectio fiite or ifiite? v) Does T = R \ {(Z m Z m )} have idempotets? vi) Fid all otrivial idempotets of T. vii) Ca (x, y) R \ {(x, y) / x, y Z m } be pseudo zero divisor as well as zero divisor? viii) Ca the umber of ilpotets i R \ {(x, y) / x, y Z m } be ifiite? 34. Let P = {R (20), } be the MOD real plae semigroup. Study questios (i) to (viii) of problem 33 for this P. 35. Let M = {R (47), } be the MOD real plae semigroup. Study questios (i) to (viii) of problem (33) for this M. 36. Let T = {R (2 14 ), } be the MOD real plae semigroup. Study questios (i) to (viii) of problem (33) for this T. 37. Let S = { R (m); 2 =, } be the MOD eutrosophic plae semigroup. Study questios (i) to (viii) of problem (33) for this S.

68 Problems o Special Elemets i MOD Structures Let L = { R (42); 2 =, } be the MOD eutrosophic plae semigroup. Study questios (i) to (viii) of problem (33) for this L. 39. Let Z = { R (23); 2 =, } be the MOD eutrosophic plae semigroup. Study questios (i) to (viii) of problem (33) for this Z. 40. Let S = { R (3 20 ), 2 =, } be the MOD eutrosophic plae semigroup. Study questios (i) to (viii) of problem (33) for this S. 41. Let V = { R g (14), g 2 = 0, } be the MOD special dual umber plae semigroup. Study questios (i) to (viii) of problem 33 for this V. 42. Let Z = { R g (53), g 2 = 0, } be the MOD special dual umber plae semigroup. Study questios (i) to (viii) of problem (33) for this Z. 43. Let N = { R g (7 12 ), g 2 = 0, } be the MOD special dual umber plae semigroup. Study questios (i) to (viii) of problem (33) for this N. 44. Let M = { R g (m), g 2 = 0, } be the MOD special dual umber plae semigroup. i) Study questios (i) to (viii) of problem (33) for this M. ii) Obtai all special features ejoyed by M.

69 68 Problems o MOD Structures Let V = {C (m), ; i F = (m 1)} be the MOD special complex modulo iteger plae semigroup. i) Study questios (i) to (viii) of problem (33) for this V. ii) Obtai all special features ejoyed by V. iii) Fid the differece betwee V ad R (m), R (m) ad g R (m) Let S = {C (29); i F = 28, } be the MOD special complex modulo iteger semigroup. Study questios (i) to (viii) of problem (33) for this S Let Z = {C (48), i F = 47, } be the MOD complex modulo iteger semigroup. Study questios (i) to (viii) of problem (33) for this Z. 48. Let T = {C ( ), i F = , } be the MOD complex modulo iteger semigroup. Study questios (i) to (viii) of problem (33) for this T. 49. Let D = { R h (m), h 2 = h, } be the MOD special dual like umber semigroup. i) Study questios (i) to (viii) of problem 33 for this D. ii) Fid the special features ejoyed by D. iii) Compare D with R (m), R (m), R (m) ad C (m). g 50. Let P = { R h (128), h 2 = h, } be the MOD special dual like umber semigroup. Study questios (i) to (viii) of problem (33) for this P.

70 Problems o Special Elemets i MOD Structures Let B = { R h (47), h 2 = h, } be the MOD special dual like umber semigroup. Study questios (i) to (viii) of problem (33) for this B. 52. Let V = { R h (11 8 ), h 2 = h, } be the MOD special dual like umber semigroup. Study questios (i) to (viii) of problem (33) for this V. 53. Let S = { R k (m), k 2 = (m 1)k, } be the MOD special quasi dual umber plae semigroup. Study questios (i) to (viii) of problem (33) for this S. 54. Let Z = { R k (24); k 2 = (m 1)k, } be the MOD special quasi dual umber plae semigroup. Study questios (i) to (viii) of problem (33) for this Z. 55. Let R = { R k (59), k 2 = 58k, } be the MOD quasi dual umber plae semigroup. Study questios (i) to (viii) of problem (33) for this R. 56. Let Z = { R k (13 7 ), k 2 = (13 7 1)k, } be the MOD special quasi dual umber plae semigroup. Study questios (i) to (viii) of problem (33) for this Z. 57. Let M = {R (m), +, } be the MOD real plae pseudo rig. i) Fid all zero divisors of M. ii) Ca we say both M ad S = {R (m), } have same umber of zero divisors? iii) Fid all zero divisors of R (m) \ {(x, y) / x, y Z m }. iv) Fid all special MOD pseudo zero divisors of R (m) \ {(x, y) / x, y Z m }.

71 70 Problems o MOD Structures v) s the umber of zero divisors ad MOD special pseudo divisors differet? vi) Ca we say all MOD special pseudo zero divisors i R (m) \ {(x, y) / x, y Z m } are oly uits of R (m) \ {(x, y) / x, y Z m }? vii) Does R (m) \ {(x, y) / x, y Z m } have torsio elemets? viii) Ca R (m) \ {(x, y) / x, y Z m } have torsio free elemets? ix) Fid all ilpotets of R (m) \ {(x, y) / x, y Z m }. x) Obtai ay other special features associated with the pseudo rig R (m). xi) Fid all idempotets of R (m) \ {(x, y) / x, y Z m }. xii) Fid all ideals of R (m). xiii) Prove all ideals of R (m) are of ifiite order. xiv) Prove R (m) has subrigs of fiite order. xv) Ca there be fiite order subrig of order greater tha m i R (m)? xvi) Obtai all special features ejoyed by {R (m), +, }. 58. Let S = {R (47), +, } be the MOD real plae pseudo rig. Study questios (i) to (xvi) of problem (57) for this S. 59. Let M = {R (12), + } be the MOD real plae pseudo rig. Study questios (i) to (xvi) of problem (57) for this M. 60. Let N = {R (3 8 ), + } be the MOD real plae pseudo rig. Study questios (i) to (xvi) of problem (57) for this N. 61. Let M = { R g (m); g 2 = 0, +, } be the MOD dual umber plae pseudo rig. i) Study questios (i) to (xvi) of problem (57) for this M. ii) Prove M has subrigs which are zero square subrigs.

72 Problems o Special Elemets i MOD Structures 71 iii) Prove M has ifiite umber of zero divisors tha the real plae pseudo rig R (m). 62. Let V = { R g (24 3 ), g 2 = 0, +, } be the MOD dual umber plae pseudo rig. i) Study questios (i) to (xvi) of problem (57) for this V. ii) Study questios (ii) ad (iii) of problem (61) for this V. 63. Let S = { R g (2 21 ), g 2 = 0, +, } be the MOD dual umber plae pseudo rig. i) Study questios (i) to (xvi) of problem (57) for this S. ii) Study questios (ii) ad (iii) of problem (61) for this S. 64. Let A = { R g (41), g 2 = 0, +, } be the MOD dual umber plae pseudo rig. i) Study questios (i) to (xvi) of problem (57) for this A. ii) Study questios (ii) ad (iii) of problem (61) for this A. 65. Let B = { R (m), 2 =, +, } be the MOD eutrosophic plae pseudo rig. i) Study questios (i) to (xvi) of problem (57) for this B. ii) Obtai ay other special features ejoyed by B. 66. Let W = { R (97), 2 =, +, } be the MOD eutrosophic plae pseudo rig. i) Study questios (i) to (xvi) of problem 57 for this W. 67. Let F = { R (420), 2 =, +, } be the MOD eutrosophic plae pseudo rig. i) Study questios (i) to (xvi) of problem (57) for this F.

73 72 Problems o MOD Structures 68. Let G = { R (7 10 ), 2 =, +, } be the MOD eutrosophic plae pseudo rig. i) Study questios (i) to (xvi) of problem (67) for this G. 69. Compare the MOD eutrosophic pseudo rigs W, F ad G i problems (66), (67) ad (68) respectively Let T = {C (m), i F = m 1, +, } be the MOD complex modulo iteger pseudo rig. i) Study questios (i) to (xvi) of problem (57) for this T. ii) Obtai ay other ice feature ejoyed by T. 71. Let R = {C (96), i 2 F = 95, +, } be the MOD complex modulo iteger pseudo rig. Study questios (i) to (xvi) of problem (57) for this R. 72. Let S = {C (47), i 2 F = 46, +, } be the MOD complex modulo iteger plae pseudo rig. Study questios (i) to (xvi) of problem (57) for this S. 73. Let V = {C ( ); i F = , +, } be the MOD complex modulo iteger plae pseudo rig. Study questios (i) to (xvi) of problem (57) for this V. 74. Compare the MOD complex modulo iteger plae pseudo rigs R, S ad V i problems 71, 72 ad 73 respectively. 75. Let P = { R k (m), k 2 (m 1)k, +, } be the MOD special quasi dual umber pseudo rig. i) Study questios (i) to (xvi) of problem (57) for this P.

74 Problems o Special Elemets i MOD Structures 73 ii) Obtai ay other special feature associated with these pseudo rig. 76. Let B = { R k (42), k 2 = 41k, +, } be the MOD special quasi dual umber pseudo rig. Study questios (i) to (xvi) of problem (57) for this B. 77. Let D = { R k (37), k 2 = 36k, +, } be the MOD special quasi dual umber pseudo rig. Study questios (i) to (xvi) of problem (57) for this D. 78. Let E = { R k (7 21 ), k 2 = (7 21 1)k, +, } be the MOD special quasi dual umber plae pseudo rig. Study questios (i) to (xvi) of problem (57) for this E. 79. Compare the MOD special quasi dual umber plaes pseudo rigs i problems (76), (77) ad (78) with each other. 80. Let H = { R h (m); h 2 = h, +, } be the MOD special dual like umber plae pseudo rig. i) Study questios (i) to (xvi) of problem (57) for this H. ii) Obtai ay other special strikig feature associated with H. 81. Let J = { R h (23), h 2 = h, +, } be the MOD special dual like umber plae pseudo rig. Study questios (i) to (xvi) of problem (57) for this J.

75 74 Problems o MOD Structures 82. Let K = { R h m (48), h 2 = h, +, } be the MOD special dual like umber plae pseudo rig. Study questios (i) to (xvi) of problem (57) for this K. 83. Let L = { R h m (13 4 ), h 2 = h, +, } be the MOD special dual like umber plae pseudo rig. Study questios (i) to (xvi) of problem (57) for this L. 84. Compare the MOD special dual like umber plae pseudo rigs, J, K ad L i problems 81, 82 ad 83 respectively with each other. 85. Obtai ay special features ejoyed by these pseudo rigs built usig MOD dual umber plaes. 86. Ca S-zero divisors be preset i { R (m); g 2 = 0, +, }? 87. Prove the otio of special pseudo zero divisors is impossible i {[0, m)g, g 2 = 0, +, }. 88. s it possible to defie special pseudo zero divisors i case of { R (m), g 2 = 0, +, }? g g m

76 Chapter Four PROBLEMS ON MOD NATURAL NEUTROSOPHC ELEMENTS this chapter we propose some problems o MOD eutrosophic umbers i [0, m), [0, m), [0, m)g, [0, m)h, g h k [0, m)k, R (m), R (m), R (m), R (m) ad R (m); the MOD itervals ad MOD plaes respectively. For the first time the atural eutrosophic umbers were itroduced i [29]. The problem of fidig atural eutrosophic umbers was proposed by Floreti Smaradache o several occasios. The MOD mathematics series [29] has aswered this questio completely. Here a brief descriptio of this cocept is give for more oe ca refer [29]. Let Z 9 be the set of modulo itegers. Now we try to itroduce the operatio of divisio / o Z 9 by / (divisio). {Z 9, /} leads to atural eutrosophic umbers {Z 9, /} = {0, 1, 2,

77 76 Problems o MOD Structures, 8, 0/0, 1/0, 2/0,, 8/0, 1/3, 2/3,, 8/3, 0/6, 1/6, 2/6,, 8/6}. These are deoted by = {1/0, 2/0,, 8/0, 0/0}, = {1/3, 2/3, 0/3,, 8/3}, 6 = {1/6, 2/6, 0/6,, 8/6} that is divisio by 0 or 3 or 6 are idetermiates called as the atural eutrosophic umbers. Further these umbers behave i a differet way for; =, = ad = Thus there are atural eutrosophic umbers which are eutrosophic zero divisors. Cosider the modulo itegers Z 12 = {0, 1, 2,, 11} , 0,3 6 4, 8, 10, 9 are the collectio of all atural eutrosophic umbers of Z 12 or related with Z 12 ; where 12 2 = {0/2, 1/2, 2/2,, 12/2}, = {0/0, 1/0, 2/0,, 12/0} ad so o = {0/0, 1/9, 2/9,, 12/9} =, =, = = = = = = = = = 8, ad so o. Thus atural eutrosophic zero divisors ad atural eutrosophic ilpotets are got by this method. Further there are some eutrosophic idempotets for istace 9 ad 4 are eutrosophic idempotets related with Z 12.

78 Problems o MOD Natural Neutrosophic Elemets 77 Next we study the atural eutrosophic elemets of Z 16.,,,,,, ad 16, =, = = 0 is a ilpotet eutrosophic elemet of order four. = is a atural eutrosophic ilpotet of order two = 0 is agai a eutrosophic ilpotet of order four = 0 is agai a eutrosophic ilpotet of order four. = is a eutrosophic ilpotet of order two = 0 is agai a eutrosophic ilpotet of order four = 4 so of order four is agai a atural eutrosophic ilpotet We see the atural eutrosophic elemets associated with Z 16 are either eutrosophic ilpotets of order two or of order four oly. Clearly there are o eutrosophic atural idempotets associated with Z Let Z 15 be the modulo iteger. 0, 3, 6, 9, 5,10 are the associated atural eutrosophic elemets of Z ad =,

79 78 Problems o MOD Structures =, =, = So 3 is ot a eutrosophic ilpotet of order two or a idempotet. However = is a eutrosophic atural idempotet = ; is ot a eutrosophic ilpotet or eutrosophic idempotet. = = is a atural eutrosophic idempotet But =, =, =, =, = ad = All elemets are atural eutrosophic zero divisors.

80 Problems o MOD Natural Neutrosophic Elemets 79 Next we study the complex fiite modulo itegers C(Z 10 ) ad its associated atural eutrosophic elemets got by divisio which are as follows.,.,,,,,,,...,,,,, C C C C C C C C C C C C C C 0 5 5iF 2 2iF 4 4iF 8 8iF 8+ 2iF 8+ 5iF 4iF iF 8+ 8iF where C 0 = {0/0, 1/0, 2/0,, 9+9i F /0} = {0/5, 1/5, 2/5,, 9+8i F /5, 9+9i F /5} C 5 C 2iF = {0/2i F, 1/2i F, 2/2i F,, 9+9i F /2i F } ad so o. t is left as a ope cojecture to fid elemets which are atural eutrosophic elemets i C(Z m ) for a give m. C 0. Let C(Z 7 ) be the fiite complex modulo itegers. The atural eutrosophic elemets associated with C(Z 7 ) is fact is a difficult problem to fid other atural eutrosophic elemets associated with C(Z 7 ). Next we recall the otio of atural eutrosophic elemets of the eutrosophic set Z m = {a + b / 2 = ad a, b Z m }. For more about these otios refer [13,14, 29]. Now we study the atural eutrosophic elemets of Z m for m = 4. Clearly the atural eutrosophic elemets associated with Z 4 are,,, , 1 + 3,, ad so o where 0 = {0/0, 1/0, 2/0, 3/0, 4/0, /0, 1+/0,, 3+3/0},

81 80 Problems o MOD Structures 2 = {0/2, 1/2, 2/2,, /2,, 3+3/2} ad so o. = {0/, 1/, 2/,., 3+3/}. 3 + is also a atural eutrosophic zero divisor as 3 = + 2 ad = 0. Study i this directio is both iterestig ad iovative. Cosider Z 3. To fid the atural eutrosophic elemets associated with Z 3. The atural eutrosophic elemets are ( ) Z 4. = so ad = ad + we see is a atural eutrosophic idempotet of are such that 0 is a atural eutrosophic zero divisor. = is agai a atural eutrosophic zero divisor = 0 ad = 0 are all atural eutrosophic zero divisors. Thus some set of associated atural eutrosophic elemets of Z 3 are {,,,, } The problem of fidig all atural eutrosophic elemets is a difficult problem. However Z 3 has oly oe atural eutrosophic elemet viz. ad othig more. 0 Further Z 3 has more umber of atural eutrosophic elemets tha Z 3 which has oly oe eutrosophic elemet. 3 0

82 Problems o MOD Natural Neutrosophic Elemets 81 Thus Z 5 has more tha oe atural eutrosophic elemet. For,,,,,,,, are some of 0 2 3, the atural eutrosophic elemets associated with Z 5. Hece fidig the umber of atural eutrosophic elemets i Z m is a challegig problem. Cosider Z 6, the umber of atural eutrosophic elemets are give by,,,,,,,,,,,, , 2+ 2,4+ 4 ad so o = 3, 4 4 = 4 are some of the atural eutrosophic idempotets of Z 6. We see =, =, = 0, = 0. = ad so o are some of the atural eutrosophic zero divisors of Z 6. Thus Z 6 has several atural eutrosophic elemets. However it is difficult to fid atural eutrosophic ilpotets i Z m for m a prime or a composite umber ad m ot of the from p t where p is a prime ad t 2. Further we have several problems some of which are ope cojectures that are proposed i the ed of this chapter about Z m, 2 < m <. Next we proceed oto aalyse the atural eutrosophic elemets of Z m g the modulo dual umbers. Z m g = {a + bg / a, b Z m, g 2 = 0}. 9 6, The atural eutrosophic elemets of Z 9 g are { 9 9 0, 3, g, 2g, 3g, 4g, 5g, 6g, 7g, 8g, 3+ 3g, 6+ 6g, 3+ 6g,

83 82 Problems o MOD Structures, g, g 9,, g, g 6+ 4g g, 9 6+ g 9,, 3+ 8g, 9 3+ g 1+g/g,, 8+8g/g} ad so o. 9 9,,, g g }, where 6+ 2g 3+ 3g, g, g 9 g = {0/g, 1/g, 2/g,, For g 3g = 0, + 9 = 9 ad so o g 6g 0 9 3g divisors. 9 6g = 9 0 ad =. This has ilpotets ad zero Thus the atural eutrosophic elemets of the modulo dual umbers Z m g behave i a distict way [17]. Next the otio of atural eutrosophic umbers i Z m h ; h 2 = h. The special dual like umber modulo itegers is aalysed. Z m h = {a + bh / a, b Z m, h 2 = h} [18]. We ow describe the atural eutrosophic elemets of Z 12 h i the followig., h 2+ 10h h h 2h 4h h 4+ 8h 12 2 h The atural eutrosophic elemets of Z 12 h are, h 2h + 10 h h,,,, 5h 6h 7h h h 75+ 7h, 7+ 5h, h h h h h h 2, 4 6 8, 10, 3, h h h,,, h 8h 9h h h 6+ 6h 3+ 6h,,, h h 2 4,, h 10h 11h h 6+ 9h, +, h 6 3h,,, h h h h h h 1+ 11h h 11,, h 9 6h h 9, h h, h 01,, h 6 h h,, h 3h, h 3+ 9h + ad so o where + = {0/2+h, 1/2+h,, h/2+h,, 11h+11/2+h} ad so o. h h h h h h h h h h h h h 9+ 6h = 0, 3 4h = 0, 4h 4h = 4h ad 9 9 = 9. Thus the atural eutrosophic elemets associated with Z 12 h ca be atural eutrosophic idempotets, atural eutrosophic ilpotets ad atural eutrosophic zero divisors. = is a atural eutrosophic zero divisors. h h 7h 5+ 7h 0 Next we study the special dual like umber Z 7 h.

84 Problems o MOD Natural Neutrosophic Elemets 83 The atural eutrosophic elemets associated with Z 7 h h h h h h h h h are 0,,, 3h,, 5h, 6h, 1+ 6h,, 2+ 5h, 5+ 2h, 3+ 4h,, h 4+ 3h h h h 2h h 4h h 6+ h h h h h h h We see 3+ 4h 2h = 0, 1 + 6h 6h = 0 thus there are several atural eutrosophic zero divisors. Further Z 7 h. = is a atural eutrosophic idempotet of h h h h h h t is importat to ote all Z m h has atleast oe h idempotet give by h ad Z m h has atleast m umber of zero divisors if m is eve ad (m 1) umber of zero divisors if m is odd. Next we study the atural eutrosophic elemets associated with Z m k, k 2 = (m 1)k the special quasi dual umber modulo itegers. Clearly Z m k = {a + bk a, b Z m, k 2 = (m 1)k}. For more refer [19]. Now we give the atural eutrosophic elemets associated k k k with Z 5 k are,,,,, + ad so o, k 0 k k 1+ k 2+ 2k 3+ 3k where = {0/0, 1/0,, 4/0, k/0,, 4+4k/0}, k 0 k 4 4k k 2 + 2k = {0/2+2k, ½+2k,, 4+4k/2+2k} ad so o. We see = k k k k 1+ k 0 = k k k k 2+ 2k 2+ 2k k 3 + 3k, k 4 + 4k ad so o. We see + = zero divisors. k k k k 1 k 0 = are atural eutrosophic k k k k 2+ 2k 0 Next we study the atural eutrosophic elemets of Z 4 k.

85 84 Problems o MOD Structures k k, The atural eutrosophic elemets of Z 4 k are k k k k k 1 + k 2+ 2k 2 2k 3+ 3k,,,. k 0, Clearly Z 4 k has atural eutrosophic idempotets atural eutrosophic ilpotets ad atural eutrosophic zero divisors. k k k k k k k k 1 + k = 0, k 3+ 3k = 0 ad k k k 2+ 2k = 0 are atural eutrosophic zero divisors associated with atural eutrosophic elemets of Z 4 k. k 2 k k k k k k 2k = 0, 2k 2k = 0 ad 2 + 2k k k 2+ 2k = 0 are the atural eutrosophic ilpotets of order two. k k k Cosider 1 + k 1 + k = 1 + k are atural eutrosophic idempotet. For more refer [29]. view of all these we have the followig result. There is always a atural eutrosophic idempotet k k k k associated with Zm k give by 1+ 2k; for 1 + 2k 1 + k = 1 + k as (1 + k) (1 + k) = 1 + 2k + k 2 = 1 + 2k + (m 1) k = 1 + (m + 1)k = 1 + k (mod m). Hece the claim. Further all atural eutrosophic elemets associated with Z m k has atleast (m 1) umber of eutrosophic atural zero divisors. We see =, t = 1, 2,, m 1. k k k k t+ tk 0 Hece the claim.

86 Problems o MOD Natural Neutrosophic Elemets 85 Now havig see examples of all atural eutrosophic elemets of modulo itegers, we ow proceed oto study MOD real itervals, MOD complex modulo iterval itegers, MOD dual umber itervals ad so o. We see [0, m) the real MOD iterval has atural eutrosophic elemets which are pseudo zero divisors also. Cosider the MOD real iterval [0, 5). [0,5) [0,5) [0,5) [0,5) Clearly 0, 1.25, 2.5, where [0,5) 0 = {0/0, 0.0 5/0,, 4.999/0 } ad so o are all pseudo MOD zero divisors of [0,5). Thus it is left as a ope cojecture to fid the pseudo MOD eutrosophic umbers. [0,5) [0,5) [0.5) For = 0 so a uit acts as a zero divisor also i the MOD real eutrosophic iterval [0, 5). That is why we call these type of zero divisors as special pseudo zero divisors ad their associated eutrosophic elemet as pseudo eutrosophic zero divisors. Cosider the MOD real iterval [0, 6); = 0 (mod 6). But 5 is a uit i [0,6) so 1.2 is a special pseudo zero divisor ad is a atural eutrosophic special pseudo zero divisor [0,6) 1.2 [0,6) ad we are forced to iclude 5 as the atural eutrosophic uit as it is the factor which cotributes to the special pseudo eutrosophic zero divisors. Cosider the MOD dual umber iterval [0,m)g, g 2 = 0. [0,m)g Every elemet i [0,m)g is a zero divisor so x is a eutrosophic zero divisor for all x [0,m)g. So they are ifiite i umber so ot much of developmet is possible for the MOD dual umber itervals.

87 86 Problems o MOD Structures Cosider the MOD eutrosophic iterval [0,m); this behaves more like a real MOD iterval. Next cosider the complex MOD iterval [0,m)i F, 2 i F = (m 1) we see uder product [0, m)i F is ot eve closed so othig ca be said about MOD complex itervals [0,m)i F ; 2 m <. Now cosider the MOD special dual like umber iterval [0, m)h, h 2 = h. This MOD iterval also behaves more like the MOD real iterval. Next cosider the MOD special quasi dual umber iterval [0,m)k; k 2 = (m 1) k, 2 m <. Clearly this MOD iterval also behaves like the MOD real iterval. So the study of the MOD itervals [0,m), [0, m)h ad [0,m)k is aki to the study of MOD real iterval. Hece oly a few problems are proposed at the ed of this chapter. Next we proceed oto study about MOD eutrosophic elemets of the MOD real plae, MOD fiite complex umber plae, MOD eutrosophic plae ad so o. Let R (m) be the real MOD plae. This has several MOD atural eutrosophic elemets which are MOD eutrosophic idempotets, MOD eutrosophic ilpotets ad MOD eutrosophic zero divisors [29]. We will illustrate this situatio by some examples. Let R (10) be the MOD real plae.

88 Problems o MOD Natural Neutrosophic Elemets 87,,,,,,,,, R R R (0,0) (5,5) (0,5) R R R (2,0),(0,2),(2,2), R R R (0,0) (6,5) (5,6) eutrosophic elemets of Z m Z m. Here a, b Z m }. R R R (6,6) (6,0) (0,6) R R R (4,0),(0,4),(4,4) are some of the atural R (0,0) = {(0,0)/(0,0), (0,a)/(0,0),, (a,b)/(0,0) where t is pertiet to ote Z m Z m is oly the cross product ad ot the MOD plae. fact,, R R (2.5,0) (0,2.5),, R R (1.25,0) (0,1.25), ad so o R R (2.5,0) (1.25,2.5) are some of the MOD eutrosophic elemets of the MOD plae. For of R (10) ad = is a MOD zero eutrosophic divisor R R R (1,25,2.5) (8,8) (0,0) idempotets ad so o. = is the MOD eutrosophic R R R (6,0) (6,0) (6,0) fact all MOD eutrosophic elemets of the MOD iterval [0, 10) will cotribute to MOD eutrosophic elemets i the MOD real plae R (10). Thus all MOD eutrosophic elemets of [0, m) will cotribute to MOD eutrosophic elemets of the MOD plae R (m). Hece we oly propose some problems i the directio. Similarly R (m) = {a + b / a, b [0, m), 2 = } will cotribute to MOD eutrosophic of R (m). These MOD eutrosophic elemets are cotributed by the itervals [0, m), [0, m) ad more. Apart from other types of MOD eutrosophic elemets got from a + b.

89 88 Problems o MOD Structures So study of MOD eutrosophic elemets will remai mostly aki to MOD eutrosophic elemets from [0, m) ad [0, m); hece left as a exercise to the reader. Now we study R (m) = {a + bg g 2 = 0, a, b [0, m)}. The MOD eutrosophic elemets of the MOD dual umber plae will be MOD eutrosophic elemets from [0, m), [0,m)g ad more. However this study is ot that similar to the earlier case but i a way oly a little similar to R (m). We ca fid the MOD eutrosophic elemets from the MOD complex plae C (m) = {a + bi 2 F i F = (m 1); a, b [0,m)}. Oe part of MOD eutrosophic elemets will be cotributed by [0, m) however for other part oe has to work thoroughly. fact it ca be left as a ope cojecture. Next for the special dual like umber MOD plae R (m) = {a + bh a, b [0, m), h 2 = h} also works i a h similar fashio as that of R (m). Fially the MOD special quasi dual umber plae works etirely differet from all the other MOD plaes for k R (m) = {a + bk k 2 = (m 1)k, a, b [0, m)} 2 m <. Study is iovative ad iterestig but oe has work a lot i the directio. this regards several problems are suggest. Cosider R (6), g R (6), h R (6), k R (6) C (6) ad R (6). We take same sets of elemets ad show how differetly the product behaves.

90 Problems o MOD Natural Neutrosophic Elemets 89 Let x = (2.35, 1.5) ad y = (3.5, 2.7) R (6). x y = (2.225, 4.05). Let x = ad y = R (6). x y = Let x = g ad g y = g R (6) x y = ( g) h Let x = i F ad y = i F C (6). x y = i F = i F C (6). V Let x = ad h y = h R (6) x y = h V Lastly x = k ad k y = k R (6) x y = k V We see all the six values are differet except R (m) ad R (m) but is a idetermiate where as h is a determiate. We propose the followig problems some of which are at research level ad a few are ope cojectures.

91 90 Problems o MOD Structures Problems 1. For a give m fid all the atural eutrosophic elemets associated with Z m. 2. Fid all the atural eutrosophic elemets associated with Z Fid all atural eutrosophic elemets associated with Z 256. i) Ca we say that oe of the atural eutrosophic elemets associated with Z 256 will be eutrosophic idempotets except 0 0 = 0? ii) Ca we claim every atural eutrosophic elemet of Z 256 will be atural ilpotets? 4. Obtai ay other special features associated with the atural eutrosophic elemets related with Z m ; 2 m <. 5. Let C(Z m ) be the fiite complex modulo itegers. i) Fid the umber of atural eutrosophic elemets associated with C(Z m ) for a fixed m. ii) Ca we say whe m is a prime umber the umber of atural eutrosophic elemets would be the least? iii) Ca we claim the more m has divisors; C(Z m ) will have more atural eutrosophic elemets? iv) Compare the umber of atural eutrosophic elemets m C(Z 210 ) ad C(Z 505 ). 6. Let C(Z 101 ) be the fiite complex modulo itegers. i) Fid the umber of atural eutrosophic elemets associated with C(Z 101 ). ii) How may atural eutrosophic idempotets are related with C(Z 101 )?

92 Problems o MOD Natural Neutrosophic Elemets 91 iii) Ca there be atural eutrosophic ilpotets related with C(Z 101 )? 7. Let C(Z 96 ) be the fiite complex modulo itegers. Study questios (i) to (iii) of problem (6) for this C(Z 96 ). 8. Ca we say C(Z 96 ) will have more umber of atural eutrosophic elemets associated with it tha C(Z 101 )? 9. Compare the umber of atural eutrosophic elemets associated with C(Z m ) ad Z for a fixed m. 10. Characterize all m which are such that both C(Z m ) ad Z m have same umber of atural eutrosophic elemets. 11. Ca we say i geeral C(Z m ) will have more umber of atural eutrosophic elemets tha Z m whe m is a composite umber? 12. Obtai ay other special feature associated with the atural eutrosophic elemets of C(Z m ). 13. Fid the umber of atural eutrosophic umbers associated with Z m ; 2 m <. 14. Let Z 15 be the modulo eutrosophic itegers. i) Fid all the associated atural eutrosophic elemets of Z 15. ii) Prove there are atural eutrosophic zero divisors. iii) Prove there are atural eutrosophic elemets. iv) Does the associated atural eutrosophic elemets has ilpotets? v) Fid ay other special features ejoyed by these atural eutrosophic elemets. vi) Prove the collectio of all atural eutrosophic elemets is a semigroup uder product.

93 92 Problems o MOD Structures 15. Let Z 19 be the modulo eutrosophic itegers. Study questios (i) to (vi) of problem (14) for this Z Let Z 12 3 = S be the eutrosophic modulo itegers. Study questios (i) to (vi) of problem (14) for this S. 17. Let T = Z m g be the dual umber modulo itegers. i) Fid all the atural eutrosophic elemets associated with Z m g for a fixed m. ii) Prove there are may atural eutrosophic zero divisors. iii) Study questios (i) to (vi) of problem (14) for this T. 18. Let W = Z 19 g be the dual umber modulo itegers. Study questios (i) to (vi) of problem (14) for this W. 19. Let M = Z 18 g be the dual umber modulo itegers. Study questios (i) to (vi) of problem (14) for this M. 20. Let P = Z 20 2 g be the dual umber modulo itegers. Study questios (i) to (vi) of problem (14) for this P. 21. Let T = Z m h ; h 2 = h be the special dual like umber modulo itegers. Study questios (i) to (vi) of problem (14) for this T. 22. Let S = Z 10 h ; h 2 = h be the special dual like umber modulo itegers. Study questios (i) to (vi) of problem (14) for this S.

94 Problems o MOD Natural Neutrosophic Elemets Let B = Z 6 h 5 modulo iteger. ; h 2 = h be the special dual like umber Study questios (i) to (vi) of problem (14) for this B. 24. Let D = Z 47 h ; h 2 = h be the special dual like umber modulo itegers. Study questios (i) to (vi) of problem (14) for this D. 25. Obtai all the special features of the atural eutrosophic elemets associated with Z m k ; the special quasi dual umber modulo itegers. 26. Let W = Z m k, k 2 = (m 1)k the special quasi dual umber modulo itegers. i) Fid all the atural eutrosophic elemets associated with Z m k. ii) How may are atural eutrosophic zero divisors? iii) Fid all atural eutrosophic ilpotets. iv) How may are atural eutrosophic idempotets? (Study for a fixed m, m a prime ad m a composite umber). 27. Let T = Z 23 k ; k 2 = 22k be the special quasi dual umber modulo itegers. Study questios (i) to (iv) of problem (26) for this T. 28. Let P = Z 6 7 k umber modulo itegers. ; k 2 = (7 6 1)k be the special quasi dual Study questios (i) to (iv) of problem (26) for this P. 29. Let B = Z 24 k ; k 2 = 23k be the special quasi dual umber modulo itegers.

95 94 Problems o MOD Structures Study questios (i) to (iv) of problem (26) for this B. 30. Compare the atural eutrosophic elemets of T, P ad B give i problems 27, 28 ad 29 respectively. 31. Ca there be a real MOD iterval [0, m) without special pseudo zero divisors? 32. Fid all special pseudo zero divisors of [0, 16). 33. Fid all special pseudo zero divisors of the MOD real iterval [0,43). 34. Obtai all special pseudo zero divisors of the MOD real iterval [0,45). 35. Fidig the total umber of MOD atural eutrosophic elemets of the MOD real iterval [0,m) 2 m < happes to be a ope cojecture. 36. t is a ope cojecture to fid the total umber of MOD special pseudo eutrosophic zero divisors of the MOD real iterval [0,m). 37. Fid all MOD eutrosophic elemets of P = [0, 13), the MOD real iterval. 38. Fid all MOD eutrosophic special pseudo zero divisors of P i problem Let M = [0, 48) be the MOD real iterval; i) Fid all MOD eutrosophic elemets of M. ii) Fid all MOD special pseudo zero divisors of [0,48). 40. Let W = [0, 3 24 ) be the MOD real iterval. Study questios (i) ad (ii) of problem (39) for this W.

96 Problems o MOD Natural Neutrosophic Elemets Compare the ature of the MOD eutrosophic elemets ad MOD special pseudo eutrosophic zero divisors i problems (37), (39) ad (40) for P, M ad W respectively. 42. Prove [0,m)g, g 2 = 0 the MOD dual umber iterval has ifiite umber of MOD eutrosophic zero divisors ad MOD eutrosophic ilpotets of order two ad has o MOD atural eutrosophic idempotets. 43. Prove or disprove [0,m) ad [0,m) have the same type of MOD eutrosophic zero divisors, idempotets, ad special pseudo zero divisors. 44. Compare the MOD eutrosophic elemets of [0,m) ad [0,m). 45. Study for P = [0,24) the MOD eutrosophic zero divisors, MOD eutrosophic idempotets ad MOD eutrosophic pseudo zero divisors. 46. Study M = [0, 3 10 ), the MOD eutrosophic iterval; for MOD eutrosophic elemets. 47. Compare P ad M of problems (45) ad (46) with W = [0,29). 48. Study the MOD special dual like iterval [0,m)h h 2 = h for atural MOD eutrosophic elemets. i) Ca we say [0,m) ad [0,m)h have same type of atural MOD eutrosophic elemets? ii) Fid the differece betwee [0,m)h ad [0,m). iii) Fid the differece betwee [0,m) ad [0,m). 49. Let P = [0,m),k k 2 = (m 1)k be the MOD special quasi dual umber iterval. i) Study questios (i) ad (ii) of problem (39) for this P.

97 96 Problems o MOD Structures 50. Let S = [0,96)k, k 2 = 985k be the MOD special quasi dual umber iterval. Study questios (i) to (ii) of problem (39) for this S. 51. Compare the MOD special quasi dual umber iterval [0,m)k, MOD special dual like umber iterval, [0,m)h MOD special quasi dual umber iterval; [0,m)g, ad the eutrosophic MOD iterval; [0,m) ad so o. 52. Fid the MOD eutrosophic elemets of the real MOD plae R (m). 53. Fid the collectio of all MOD eutrosophic elemets of the MOD real plae R (27). 54. Fid the collectio of all MOD eutrosophic elemets of the MOD real plae R (47). 55. Fid the collectio of all MOD eutrosophic elemets of the MOD real plae R (24). (i) Does R (24) cotai pseudo zero divisors? (ii) Does R (24) cotai pseudo atural MOD eutrosophic zero divisors? 56. Fid the collectio of all MOD eutrosophic elemets of the MOD complex modulo iteger plae C (m); i 2 F = m Fid the collectio of all MOD eutrosophic elemets of the MOD eutrosophic plae C (45); i 2 = 44. F 58. Let S = C (29), i 2 F = 28 be the MOD complex plae. Fid all the MOD eutrosophic elemets of S Let M = C (48), i F = 47 be the MOD complex modulo iteger plae. Fid all MOD eutrosophic elemets of M.

98 Problems o MOD Natural Neutrosophic Elemets Compare the MOD eutrosophic elemets of S ad M i problems 58 ad 59 respectively. 61. Fid the collectio of all MOD eutrosophic elemets of R (m) the MOD eutrosophic plae. 62. Let M = R (48) be the MOD eutrosophic plae fid the MOD eutrosophic elemets of M. 63. Let M 1 = R (13) be the MOD eutrosophic plae fid the MOD eutrosophic elemets of M Compare M ad M 1 i problems (62) ad (63) respectively. 65. Let N = g R (m) be the MOD dual umber plae. Fid all MOD eutrosophic elemets of N. 66. Let P = g R (28) be the MOD dual umber plae. i) Compare P with N of (65). ii) Obtai all MOD eutrosophic elemets of P. 67. Let S = g R (47) be the MOD dual umber plae. i) Study questios (i) ad (ii) of problem (66) for this S. 68. Let R = h R (m) be the MOD special dual like umber plae. (i) Study questios (i) ad (ii) of problem (66) for this R. (ii) Compare R of this problem with S of problem (67)

99 98 Problems o MOD Structures k 69. Let T = R (m), k 2 = (m 1)k be the MOD special quasi dual umber plae. i) Study questios (i) to (ii) of problem (66) for this T. ii) Compare T with R of problem i (67). 69. Let M = { R k (24), k 2 = 23k} be the MOD special quasi dual umber plae. i) Study questios (i) to (ii) of problem (66) for this M. ii) Compare M with P = { R (42), h 2 = h}. 70. Let Y = { R k (426), k 2 = 425k} be the MOD special quasi dual umber plae. i) Fid all MOD eutrosophic elemets of Y. ii) Prove Y has pseudo zero divisors. h

100 Chapter Five PROBLEMS ON MOD NTERVAL POLYNOMALS AND MOD PLANE POLYNOMALS this chapter for the first time we itroduce the otio of MOD iterval polyomials ad MOD plae polyomials. For more refer [26-30]. Clearly [0, m)[x] = iterval polyomials. i aix a i [0, m)} is the MOD real i= 1 Fidig roots of [0, m)[x] is a impossibly. As [0, m)[x] is oly a pseudo rig we see it is difficult to get roots. Several difficult ope problems are suggested i this which are ope cojecture. i Study [0,96)[x] = aix a i [0, 96)}, the MOD real i= 1 iterval polyomials.

101 100 Problems o MOD Structures Fid all roots of p(x) = x [0, 96)[x]. Let [0,m)[x] be the pseudo rig. eutrosophic polyomial iterval We see p(x) R[x] may be cotiuous i the plae R; however these polyomials i [0,m)[x] behave very differetly. Next we ca have MOD iterval dual umber polyomials, i [0, m)g[x] = aix a i [0, m)g}. i= 1 These polyomials p(x) [0,m)g[x] has o solutio i may cases as this polyomial rig is a zero square rig ad is ot pseudo. We see [0, m)[x] is the MOD eutrosophic iterval pseudo polyomial rig. The problem of solvig the equatios arises from the fact is a idetermiate ad ot a ivertible elemet that is has o iverse. So solvig these equatios is also impossible for we ca oly say or fid roots of x ad ot x. Similar situatio arise i case of [0,m)h[x] ad [0, m)k[x], h 2 = h ad k 2 = (m 1)k respectively. Thus we get oly values of x, gx, hx or kx ad ot for x. The case where x gets the value is [0, m)[x] the MOD real iterval pseudo rig. Next we study the polyomials i MOD plaes. i R (m)[x] = aix a i R (m)}. i= 1

102 Problems o MOD terval Polyomials 101 We see eve the simple equatio y = x 2 is ot cotiuous has several zeros i the MOD (real plae R (m)). Such illustratios ad study is carried out i [Books 1 & 2]. However here we propose some problems. Similarly the equatio y = sx + t R (m); 1 s, t < m behaves i a very differet way has several zeros. So a liear equatio i the MOD plae ca have several zeros. Next the study of a simple equatio y = x, 1 < < m i R (m) has may zeros. Hece we have give several ope problems i this chapter regardig roots of a polyomial. Next if we cosider MOD polyomials i the MOD complex 2 modulo umber plae C (m), m 1 = i F we see the similar situatio arises. For more iformatio refer (Books 1, 2). Thus for istace eve simple polyomials like p(x) = (a + bi F )x 2 + c + di F, a, b, d, c [0,10) behave i a very uatural way for varyig values of a, b, c, d [0, 10). Study i this directio is also left ope for iterested reader. Cosider 5x i F = 0 C (10). Solvig this equatio is ear to a impossibility as 5 2 = 5 (mod 10), so 5x = 2 + 8i F is the solutio. Cosider (2 + 4i F ) x + 5 = 0 C (10). Clearly (2 + 4i F )x = 5 is the solutio for 2 ad 4 are zero divisors i C (10). (2 + 4i F ) 5 0 (mod 10).

103 102 Problems o MOD Structures So oe caot lesse the coefficiet. fact (2 + 4i F ) x = 5 multiply by 3 oe gets (6 + 2i F ) x = 5 so are both same oe is left to woder. Next we study MOD eutrosophic polyomials i R (m)[x]. Sice is a idetermiate if the coefficiets of x or its power cotais it is ot solvable for x. Cosider 3x x = 3 + x = C (5)[x] So we solve this ot for x oly for x. Cosider 2x i F C (5)[x] 2x = i F. So x = i F is the value of x. Let 2.5 x i F C (5)[x] 2.5 x 2 = i F sice (mod 5); we see value of x caot be got oly the value of x 2 is got. Oe ca fid 2 2.5x = i F + so that x( ) = i F we do ot kow whether is a uit or ot i [0, 5). So solvig eve a quadratic equatio i C (m)[x] is ot that easy. Further i C (m)[x] or for that matter i all types of MOD plaes we see

104 Problems o MOD terval Polyomials 103 (x α 1 ) (x α 2 ) (x α ) x (α 1 + α α ) x αiα jx + α 1 α 2 α as the distributive law is ot true, i j so after solvig the p(x) oe faces lots of difficulty all these are left as ope cojecture to the reader. Next we study the polyomial i the MOD eutrosophic pseudo polyomial rigs. R (m)[x] = i aix a i = d i + c i i= 1 eutrosophic polyomial pseudo rig. R (m)} is the MOD As 2 m < we have ifiitely may such rigs for each value of m, however there is oly oe eutrosophic rig, R i [x] = aix a i R }. i= 1 Solvig MOD polyomial equatios i R (m)[x] is as difficult as i MOD complex polyomials ad MOD real polyomials. Cosider 0.5x R (12). Clearly 0.5x = However value of x caot be foud oly the value of 0.5x is got. Thus eve liear equatios i geeral are ot solvable i R (12). So solvig quadratic equatios ad equatios of higher degree is a very difficult problem, hece left as a ope cojecture or for future study. Let 5 x x 2 = 8.3 R (12)

105 104 Problems o MOD Structures x 2 = 5.5 so x so for this quadratic equatio has a solutio. However 5x if x = but its value is 7.2 oly. We see x 2 = 5.5 is true but 5x 2 = 8.3 is ot true. There is a fallacy i solvig so left as a ope cojecture. case of modulo additio or multiplicatio how to solve equatios. Thus there are several itricate problems ivolved while solvig equatios i MOD eutrosophic polyomials i R (m)[x]. Next we proceed oto describe ad discuss about the MOD g i dual umber polyomials R (m)[x] = aix a i R g (m) i= 1 where a i = x + yg, x, y [0,m), g 2 = 0}. Solvig these equatios is a very difficult task for 0.7g x = 0 is i R (10)[x]. g Clearly as g 2 = 0 we caot cacel g usig ay iverse. Now 0.7g x 3 = We caot take the cube root as g is a dual umber ad is it possible to take cube root of a dual umber is still a ope problem for the dual umber ca be matrix or a trasformatio or from Z the rig of modulo itegers a composite umber. Uder these circumstaces oly 0.7gx 3 value is got is ot further solvable. Cosider

106 Problems o MOD terval Polyomials 105 3x 3 g R (10)[x]. 3x 3 = (multiplyig by 7 takig mod10 we get) x 3 = So x = is oe of the roots of the give MOD polyomial. Thus oe caot always say all the polyomials i MOD plaes are ot solvable. Some of them may be solvable partially ad a few fully ad may of the polyomials ever solvable. i Eve solvig p(x) = ( g) x 2 + ( g)x + ( g) R (10)[x] is ear to a impossibility. g Readers ca try this as a recreatio or pursue this as a hobby. Several ope cojectures are laid before them. Next the otio of MOD special quasi dual like umber polyomials i h i R (m)[x] = aix a i = x + yh R h (m), h 2 = h ad x, y i= 1 [0, m), 2 m < } are discussed i the followig. Let p(x) = hx h x is a difficult problem. h R (7)[x]. Solvig this for For if hx 3 = h the what is the cube root of h. This is ot a easy work for square roots ca be doe as i case of complex umbers but fidig cube root or th root happes to be a challegig problem.

107 106 Problems o MOD Structures However if hx 3 = 6.4h the we ca solve for x as follows hx = h is oe of the roots. However as i case of other MOD plae some are solvable but certaily all polyomial equatios are ot solvable. Next we proceed oto study the solvig of MOD polyomials i k R (m)[x] = k 2 = ( m 1) k, 2 m < }. Let p(x) = 8x k i aix a i = x + yh i= 1 k R (11)[x]. R (m), x, y [0, m), 8x 3 = 4.75k, so multiplyig the equatio by 7 ad takig mod we get x 3 = 0.25 k x = is oe of the values of x for this p(x). Solvig equatios is ot a easy task but this research is iterestig ad iovative. Further each MOD plae differetly ad ifact oe has ifiite umber of such MOD plaes, be it real MOD plae or complex MOD plae or dual umber MOD plae ad so o. the followig we suggest some problems which are ope cojectures, some of them at research level ad some of them are simple. k Problems: 1. Prove or disprove the fudametal theorem of algebra for roots for the MOD real iterval polyomials [0,m)[x].

108 Problems o MOD terval Polyomials Ca a polyomial of degree have less tha umber of roots? Justify your claim. 3. Ca a polyomial of degree i [0,m)[x] have more tha roots? Justify your claim. 4. As distributive law is ot true ca we say fidig roots of [0,m)[x] is a impossibility? 5. Let [0,m)[x] be the MOD eutrosophic iterval polyomials. i) Does there exist p(x) [0,m)[x] of degree which have ad oly roots? ii) iii) iv) Give a polyomial p(x) [0,m)[x] of degree which has oly less tha roots. Prove a polyomial p(x) of degree i [0, m)[x] has more tha roots. Show the distributive law is ot true i [0,m)[x] which affects the roots ad the umber of roots of p(x) i [0,m)[x]. 6. Let R = [0,45)[x] be the MOD eutrosophic polyomial pseudo rig. Study questios (i) to (iv) of problem (5) for this R. 7. Let S = [0,37)[x] be the MOD eutrosophic polyomial pseudo rig.

109 108 Problems o MOD Structures Study questios (i) to (iv) of problem (5) for this S. 8. Let W = [0,m)g[x] be the MOD dual umber iterval pseudo rig. i) Solve p(x) = 0.8g x 3 = 4.8gx gx g W. a) Ca p(x) have more tha 3 roots? b) Ca p(x) have less tha 3 roots? ii) iii) Show roots of degree oe aloe is solvable provided coefficiets of x is 1 so it is ot possible. Ca 5.03gx g = 0 have two roots or o root? 9. Prove or disprove dual umber MOD iterval polyomials are ot solvable if the highest coefficiet is ot 1 or ot possible. 10. Let P = [0,m)h[x], h 2 = h be the MOD iterval special dual like umber polyomial pseudo rig. i) Show p(x) P caot be solved i geeral. ii) Eve liear polyomial i P caot be solved prove or disprove. iii) Obtai ay other special feature ejoyed by P. iv) Prove p(x) P behaves i a odd way i all MOD iterval polyomials barig [0,m)[x]. 11. Let M = [0,45)h[x], h 2 = h be the MOD iterval special dual like umber plae. Study questios (i) to (iv) of problem (10) for this M.

110 Problems o MOD terval Polyomials Study the MOD iterval special quasi dual umber polyomial pseudo rig B = [0,m)l[x]; k 2 = (m 1)k their roots. 13. Study questios (i) to (iv) of problem (10) for this B 14. Let M = [0,24)l[x], k 2 = 23k be the MOD iterval special quasi dual umber pseudo rig. Study questios (i) to (iv) of problem (10) for this M. 15. For a polyomial p(x) fixed compare the roots i [0,10)[x], [0,10)[x], [0,10)g[x]; g 2 = 0, [0,10) h[x], h 2 = h ad [0,10)k[x], k 2 = 9 k. 16. Fid ay special feature associated with polyomials i R (m)[x]. 17. Prove the equatio y = x for varyig m i R (m)[x] has differet sets of roots. 18. Give a polyomial of degree 5 i R (6)[x] which has more tha roots. 19. Let p(x) = x h of p(x). h R (5)[x]. Fid all roots 20. Does there exist a polyomial p(x) of degree 7 i R (5)[x] which has less tha seve roots? 21. Give the graph of the polyomial y = 4x + 2 R (8)[x]. i) s the curve a cotiuous oe? ii) Fid all the zeros of y = 4x + 2. iii) Compare y = 4x + 2 R (8)[x] with y = (4x + 2) R (6)[x].

111 110 Problems o MOD Structures 22. Let y = x 2 R (4)[x] be the fuctio i the MOD plae. i) Fid all zeros of y = x 2 i R (4)[x]. ii) iii) iv) Ca this fuctio be cotiuous? Trace the graph of y = x 2 i R (4)[x]. Ca this fuctio have more tha two zeros? v) s the fuctio cotiuous i the real plae? vi) Ca the fuctio be cotiuous i ay other MOD plae? 23. Let y = 3x R (5)[x]. i) Study questios (i) to (vi) of problem (22) for this fuctio. ii) Study this fuctio i the plaes R (6), R (7), R (10), R (12) ad R (23). 24. Let y = x R (3)[x]. i) Study questios (i) to (vi) of problem (22) for this fuctio. ii) Study the fuctio i the MOD plae R (8), R (91), R (24) ad R (19). 25. Let y = 4x 2 + x + 1 R (6)[x]. i) Study questios (i) to (vi) of problem (22) for this fuctio. ii) Study this questios i the plaes R (9), R (23), R (48) ad R (121). 26. Solvig polyomial equatios be it liear or otherwise is a difficult task.

112 Problems o MOD terval Polyomials 111 So this problem is left as a ope cojecture. 27. Solve p(x) R (20)[x] = i aix a i x i / a i R (20) where i= 1 a i = (t i, v i ), t i, u i [0,20)} where p(x) = (0,2)x + (10.31, 3.2). 28. Solve p(x) = (4,9) x 3 + (2,1.09)x + (6.67, 4) R (12)[x]. i) How may roots exist for p(x)? ii) Ca p(x) have more tha 3 roots? 29. Solve p(x) = (4,0.7)x , 0.8)x 4 + (5.2, 0)x 3 + (0, 4.3) R (4)[x]. i) Study questios (i) to (vi) of problem (21) for this p(x). ii) Study for this p(x) i R (19)[x], R (12)[x], R (43)[x]; R (4)[x] ad R (10)[x]. 30. Fid polyomials other tha y = x which is cotiuous i R (m) [x], 2 m <. 31. Let p(x) C (m) [x] be a MOD polyomial with complex coefficiets. Study for the same p(x) but for varyig m; 2 m <. 32. Compare polyomials i C (m)[x] ad R (m)[x]. 33. which MOD plae a polyomial p(x) will have more zeros i C (m)[x] or R (m)[x]? 34. Study the roots of the polyomial f(x) = ax 3 + bx + 1 C (7)[x] where a = i F, b = i F ad 1 = 1 + 0i F C (7).

113 112 Problems o MOD Structures i) Ca f(x) have more tha three roots? Justify your claim. ii) iii) f a = 3 + 4i F ad b = 4 + 3i F will f(x) has oly three roots? f a = 3.5, b = 1.75 ad 1 = 1 + 0i F, ca f(x) have oly three roots? 35. Let p(x) = 5 x 3 + 7x + 34 C (10)[x] ad also this p(x) R (10)[x]. i) Will p(x) have differet sets of roots i R (10)[x] ad C (10)[x]? ii) Ca we say p(x) will have 3 roots? iii) Ca p(x) have more tha three roots? iv) Will p(x) have less tha three roots? v) s p(x) solvable i R (10)[x]? 36. Ca we write program for solvig roots of a polyomial equatio i R [x] ad C [x]? (m) 37. What will be the probable applicatios of these polyomials i [0,m)[x]? 38. Ca we say this paradigm shift will give ew dimesio to solvig polyomial equatios? (m) 39. Let p(x) = 0.03 x i F C (2)[x]. i) Does the roots of p(x) exist? ii) Will p(x) have atleast oe root? iii) Ca p(x) have more tha 5 roots? iv) Derive a method to solve p(x) i C (2)[x]. v) s 0.03 [0,2) a ivertible elemet? 40. t is left as a ope cojecture to fid

114 Problems o MOD terval Polyomials 113 i) All zero divisors ad ilpotets i [0,2) ([0,m) i geeral, 2 m < ). ii) iii) Fidig idempotets i [0,2) ([0,m) i geeral, 2 m < ). Fidig uits of [0,2) ([0,m) i geeral, 2 m < ). 41 t is yet aother ope cojecture of fidig the total umber of MOD atural eutrosophic elemets i [0,m), [0, m)h ad [0,m)k for 2 m <. 42. Prove [0,m)g ad [0,m) has ifiite umber of MOD eutrosophic elemets. i) Does they cotai as may as the cardiality of [0,m) itself? 43. Prove oce [0,m) is solved oe ca solve all the MOD plae R (m), 2 m <. 44. Solve the equatio p(x) = 2.7 x R (8)[x]. i) Does p(x) have more tha oe root? ii) Ca x be got oly i terms of x; be got as a solutio? iii) Ca p(x) have more tha 2 roots? iv) Study this problem i R (11)[x] ad R (12)[x]. 45. Let p(x) = x R (3)[x]. Study questios (i) to (iv) of problem (44) for this p(x). 46. Let q(x) = x x + ( ) R (2)[x]. Study questios (i) to (iv) of problem (44) for this p(x).

115 114 Problems o MOD Structures 47. Let q(x) = 6x 3 + (3 + 4) R (7)[x]. Study questios (i) to (iv) of problem (44) for this q(x). 48. Let q(x) = 0.78x R (5)[x]. i) Study for roots i R (5)[x]. ii) Ca q(x) have more tha oe root? iii) s 0.78 a uit or a zero divisor i R (5)? 49. Let p(x) = ( g) x 2 + ( ) g R (4)[x]. i) Study questios (i) to (iv) of problem (44) for this p(x). ii) Ca we say g is a zero divisor or a idempotet i R (4)? 50. Let p(x) = 2x 2 + (4 + 6g) g g R (10)[x]. i) Study questios (i) to (iv) of problem (44) for this p(x). ii) Does a root exist as 2 is a zero divisor i g R (10)? 51. Let p(x) = (3 + 2g) x 2 + (4 + 3g) g R (6)[x]. i) Study questios (i) to (iv) of problem (44) for this p(x). ii) Ca we say (3 + 2g) is a zero divisor i g R (6)?

116 Problems o MOD terval Polyomials 115 iii) Ca we say polyomials of the form p(x) is ot solvable eve i Z 6 g [x] = { (a i + b i g) x i / a i i= 1 + b i g Z 6 g where g 2 = 0 ad a i b i Z 6 }? 52. Let p(x) = ( g) x g g R (6)[x]. i) Study questios (i) to (iv) of problem (44) for this p(x). 53. Fidig uits, idempotets, zero divisors ad ilpotets of g the MOD dual umber plae R (m), g 2 = 0, 2 m < happes to be a very challegig ope problem. 54. Fid all uits, zero divisors ad idempotets of g R (12) ad g R (625), g 2 = Let p(x) = (7 + h)x 2 + (4 + 3h) h R (14)[x], h 2 = h. g R (53), Fid the roots of p(x) ad study questios (i) to (iv) of problem (44) for this p(x). 56. Let f(x) = ( h) x 2 + (2 + 7h) h R (14)[x]. Study questios (i) to (iv) of problem (44) for this f(x). 57. Fidig all the uits, zero divisors ad idempotets of h R (m), h 2 = h, 2 m < happes to be a ope cojecture. 58. Fid all uits, zero divisors, idempotets ad ilpotet elemets of the MOD special dual like umber MOD plaes R (43), R (24) ad R (81). h h h

117 116 Problems o MOD Structures 59. Fid all iovative techiques that ca be used i solvig polyomials i R (m), h 2 = h ad 2 m <. 60. Let p(x) = 3kx 3 (9 + 5k) h k R (15)[x] k 2 = 14k. i) Solve for x or kx. ii) Study questios (i) to (iv) of problem (44) for this p(x). k iii) Will the solutio p(x) be the same i R (15) ad Z 15 g? 61. Let p 1 (x) 0.3kx 3 + ( k) k R (15)[x], k 2 = 14k. Study questios (i) to (iv) of problem (44) for this p 1 (x). Compare p(x) i problem (60) with this p 1 (x). 62. t is left as a ope cojecture to fid the uits, idempotets, zero divisors of the MOD special quasi dual umber plae R (m), k 2 = (m 1)k; 2 m <. k 63. Fid the zero divisors, uits ad idempotets of the MOD special quasi dual umber plae R (10), R (47) ad k R (16). 64. Solve the equatio q(x) = ( k) x kx 2 + (0.04 k + 4.4) x R (5)[x]; k 2 = 4k. i) Study questios (i) to (iv) of problem (44) for this q(x). h ii) Study q(x) i R (5)[x], h 2 = h, iii) Study q(x) i R (5)[x] iv) Study q(x) i C (5)[x]. g v) Study q(x) i R (5)[x]. vi) Compare the solutios i all the 5 MOD plaes. k k k

118 Chapter Six PROBLEMS ON DFFERENT TYPES OF MOD FUNCTONS this chapter we just recall differet types of MOD fuctios viz. MOD trigoometric, MOD logarithmic, MOD expoetial ad their geeralizatios. However such study is carried out i [27]. Here our mai aim is to suggest problems some of which are ope cojectures ad some of them are simple. For more about these MOD fuctios refer [27]. Problems: 1. Defie MODsie; six fuctio. 2. Does values / graph of six vary for varyig R (m); 2 m <. 3. Study problem (2) for tax, cosx ad secx.

119 118 Problems o MOD Structures 4. Describe the trigoometric fuctio MOD cosecx (cosecx) i R (7). 5. Ca we say MOD trigoometric fuctios are i o way differet from usual trigoometric fuctios? Justify your claim. 6. Give some special features about ta3x i the MOD real plae R (9). 7. What are the advatages of usig MOD trigoometric fuctio? 8. Obtai ay special properties ejoyed by MOD trigoometric fuctios. 9. Ca we see simx behaves i the same way i all MOD plaes m a fixed umber? 10. For problem (9), study the situatios i the MOD plaes R (t). i) if t / m ii) f (t 1 m) = 1 iii) f (t 1 m) = d iv) m / t. 11. Compare the MOD trigoometric fuctios six ad cosx i the MOD real plae R (15). 12. Study tax i the MOD plaes R (t) where t is as i problem Compare tax with cot x. 14. Ca ay of the MODtrigometric fuctios be cotiuous i R (t); for ay value of t; 2 t m.

120 Problems o Differet types of MOD Fuctios Let si5θ be the MOD trigoometric fuctio i R (5). i) s this fuctio cotiuous? ii) f R (5) is replaced by R (2) or R (3) or R 5 (4), R (10) ad R (6) study the ature of the curve. iii) Study cosec7θ i the MOD plae R (7). 16. Study sec9θ i the MOD plaes R (3), R (9) ad R (27). 17. Study sec(mθ); 2 t < i the MOD plaes R (t) where (t, θ) = 1, t / θ ad (t, θ) = d. 18. Obtai ay special feature associated with MOD trigoometric fuctios. 19. Study MOD expoetial fuctios i the MOD plaes. 20. How does the study of MOD trigoometric fuctios i MOD plaes R (m) differet from the trigoometric fuctios i the real plae? 21. Study the MOD trigoometric fuctio si -1 x i the MOD plae R (15). 22. Study cot -1 5x i the MOD plaes R (5), R (7), R (4), R (15) ad R (10). Are these fuctios graph differet i differet plaes or is it the same? 23. Study the MOD iverse trigoometric fuctio sec -1 x i the MOD plaes R (2), R (4), R (3), R (5), R (6), R (13) ad R (18). Draw the curves i these seve MOD plaes ad compare them.

121 120 Problems o MOD Structures 24. s this equatio siθ + ta2θ = 0 solvable i R (5). Fid the values of θ. 25. Solve for θ the followig fuctios. i) cos3θ 1+ ta θ i R (13) ii) cot 2θ si θ + cosθ (10) iii) sec3θ 2 1+ ta 3θ (5). 26. Solve for θ the followig MOD trigoometric iverse fuctios. i) ii) iii) iv) 1 si 2θ 1 + θ i R (17). 1 ta 1 cot 5θ 1 cis 2θ + 1 i R (5) 1 ta 5θ 1 1+ cot 5θ i R (15) 1 cos 3θ 1 1+ sec 3θ i R (7) 27. Trace the curve e x i the MOD plae R ( ). 28. Obtai all the special features ejoyed by the MOD logarithmic fuctios. 29. Derive all special features i studyig MOD expoetial fuctio. 30. Distiguish the logarithmic fuctios from the MOD logarithmic fuctios.

122 Chapter Seve PROBLEMS ON MOD DFFERENTATON AND MOD NTEGRATON OF FUNCTONS N THE MOD-PLANES this chapter we just give some relevat problems related with differetiatio ad itegratio of MOD fuctios defied o MOD plaes. For more about these please refer [27]. t is importat to keep o record that differetiatio ad itegratio of MOD fuctios i geeral behaves i a very odd way. They do ot obey most rules associated with differetiatio or itegratio. Further a fuctio which is cotiuous i the real plae or a complex plae eed ot i geeral be cotiuous i the MOD plaes. So at this stage itself most of the properties associated with MOD fuctios do ot behave like usual fuctios fact we see polyomial fuctios behave very haphazardly.

123 122 Problems o MOD Structures For a th degree polyomial i the MOD iterval eed ot have ay of the derivatives to exist i [0,6)[x]. Thus if f(x) = x x x [0,6)[x]. The df dx = 0. There is a vast differece betwee MOD polyomials ad usual polyomials i the study of differetiatio. This has bee elaborately discussed i the books [26-30]. Likewise if f(x) = 0.7x x x [0,6)[x] the we see f(x) dx = 0.74x 23 dx x 17 dx x 5 dx dx x 4.02x x 0.33 = x + c is ot defied for i [0,6) the values of 1 24, 1 18 ad 1 are ot defied. 6 Here we proceed oto give some more examples. Cosider the MOD iterval polyomial pseudo rig. [0,12)x = i aix a i [0,12)}. i= 1 p(x) = 3.7 x x x 4 + 4x 3 + 6x [0,12)x. Clearly dp(x) dx = 0.

124 Problems o Differet types of MOD Fuctios 123 That is the derivative is zero though p(x) is of degree 24. The itegral of p(x) is also ot defied for; x p(x) dx = + 1.2x c, where c is a costat. 5 3x x 4 + 6x x + Sice 1 4 ad 1 3 divisors i [0,12). are ot defied i [0,12) as they are zero We see the itegral of p(x) is also ot defied. Thus fidig polyomials i p(x) [0,m)x = aix a i [0,m)} whose derivative ad i= 1 itegrals are defied properly ad those which satisfy the usual or classical rules of differetiatio ad itegratio also true happes to be a very challegig problem. Cosider p(x) = 2x x x x 7 [0,24)x = [0,14)}. i aix a i i= 1 We see dp(x) dx = 0. Further p(x) dx = defied as 1 8, x x x 29 are udefied i [0,14) x 8 + c is ot

125 124 Problems o MOD Structures So i [0,m)x there exists several polyomials whose derivatives are zero ad the itegrals does ot exists or are udefied [26-30], Characterizig such polyomials happes to be a ope cojecture. Next we study polyomials i R (m)[x] = R (m) = {(a, b) a, b [0,m)}. i aix a i i= 1 t is importat to ote that eve i case of polyomial i the MOD plae R (m) [x] they i geeral do ot satisfy the classical properties of differetiatio or itegratio. Cosider p(x) R (4)[x] = a, b [0, 4)} where i aix a i R (4) = {(a, b) i= 1 p(x) = (2,1.2)x 8 + (3.1, 0.315)x 4 + (2, 2)x 2 + (0.1, 0). Clearly dp(x) = 0. dx However p(x) dx = (2,1.2)x (3.1, 0.315)x (2, 2)x (0.1,0)x + c. As 1 3 = 3 the itegral exists i this case. However the derivative is zero.

126 Problems o Differet types of MOD Fuctios 125 Thus eve the polyomials i the MOD plae do ot satisfy all the classical properties of derivatives ad itegrals i geeral. Suppose we have the polyomials from the MOD plae R (24)[x] the all polyomials whose power or degree is multiples of 24. Also if the polyomials have coefficiets from {(a,b a, b {0, 2, 4, 6,, 22}} the there are possibilities where the derivatives are zero. p(x) = (2,4)x 12 + (16,8)x 6 + (12,6)x 4 +(12,12) x 2 + (8,6) R (24)[x]. The the derivative is zero ad the itegrals do ot exist. Also all MOD polyomials whose powers are say p 1, p 2, p 3,, p ad these p i s take values from {1, 5, 7, 11, 13, 17, 19, 23} are such that their itegrals are ot defied. Let p(x) = (0, (5.33) + (2, 7.21) x + (6.022, 5.001)x 2 + (5, 0.331)x 5 R (24)[x] be such that dp(x) exists but its itegral dx is ot defied. Let p(x) R (m)[x], we see eve if p(x) is of degree, the p(x) eed ot have all the derivatives to exist. Secodly eve the secod derivative of p(x) eed ot be a polyomial of degree ( 1). For istace if p(x) = (3, 6)x 4 + (2.01, 3.2) x 2 + (0.77, 6.2) R (12)[x]. p(x) is of degree four. Cosider dp(x) dx = 0 + (4.02, 6.4) x + 0.

127 126 Problems o MOD Structures Thus the first derivative results i a polyomial of degree oe. 2 d p(x) dx 2 = (4.02, 6.4). That is the secod derivative is a costat polyomial ad 3 d p(x) the third derivative = (0,0). 3 dx Hece our claim that a th degree polyomial i MOD plaes eed ot be derivable times. Sometimes the first derivative itself ca be zero. Likewise oe ca defie polyomials usig the complex MOD iteger plae C (m). C (m)[x] = i aix a i = c + i= 1 di F C (m); c 1 d [0,m)}. The degree of the polyomial roots of the polyomials are defied i a aalogous way as that of R (m)[x]. Suppose p(x) = (5 + 3i F )x 10 ( i F ) x 4 = 15i F x i F C (30)[x]. We see eve the first derivative is zero. Further the itegral of p(x) does ot exist as the terms 1 1, 5 3 are ot defied i C (3). Hece oe caot as i case of polyomials i C[x] defie the derivatives or itegrals of polyomials i C (m)[x]; 2 m <.

128 Problems o Differet types of MOD Fuctios 127 Thus it is difficult to fid coditios o polyomials i C (m)[x] to have both the derivative ad itegrals to exist. Let p(x) = ( i F ) x i F x i F C (5)[x]. dp(x) dx = 0 ad p(x) dx is ot defied. Thus fidig itegrals ad derivatives that follows all classical coditios happes to be a challegig problem. Next we study polyomials i g i R (m)[x] = aix a i R g (m) = {a + bg a, b [0, m), i= 1 g 2 = 0}, kow as the dual umber coefficiet polyomials. f p(x) R g (m)[x] we caot say that p(x) is itegrable or derivable i geeral. Most of the polyomial do ot satisfy the properties of classical derivatives. For let p(x) = ( g)x + ( g)x g R (5)[x]. Clearly dp(x) dx ot satisfied. = 0 which shows the classical property is [0,5). Further p(x) dx is ot eve defied as 1 5 is ot defied i Thus i geeral a polyomial p(x) R (5)[x] is ot itegrable.

129 128 Problems o MOD Structures Thus characterizig those polyomials which are itegrable g (or (ad) differetiable) i R (m)[x] happes to be a ope cojecture. Next we ca study MOD special dual like umber polyomials i h R (m)[x] = h 2 = h, 2 m < }. i aix a i = c + bh i= 1 R (m), c, b [0,m), h Clearly polyomials p(x) i R (m)[x] will be kow as MOD special dual like umber polyomials. geeral as i case of other MOD polyomials these polyomials also do ot i geeral ad further do ot satisfy the classical properties of differetiatio or itegratio. Apart from this these polyomials also do ot satisfy the classical properties of roots of polyomials. Study i these directios are just left as ope cojectures. We will preset a example or two i this directio. Let p(x) = ( h)x 12 + ( h)x h R (6)[x]. h h Clearly dp(x) dx = 0 ad p(x) (dx) is ot defied. Let p(x) = hx 2 + ( h) h R (6)[x]. Clearly this is a quadratic MOD polyomial equatio i R (6)[x] but is ot solvable i R (6)[x]. h h

130 Problems o Differet types of MOD Fuctios 129 Similarly 3.2 hx + ( h) = 0 is ot solvable i R (6)[x]. h Next we study MOD polyomials with coefficiets from the special quasi dual umbers or equivaletly the MOD special quasi dual umber polyomials i R (m) [x]. k Let k R (m)[x] = i aix a i i= 1 [0, m) k 2 = (m 1)k, 2 m < }. k R (m) = {a + bk a, b Let p(x) = ( k) x 12 + (4 + 3k) x k R (6) [x]. k Clearly dp(x) dx = 0. However the itegral exist. Let (0.3k ) x 2 + ( k) Solve for x. k R (6)[x]. Further it is importat to keep o record that it is ot possible to solve for x eve whe it is a liear equatio i x. Cosider ( k)x + ( k) Solve for x. k R (6)[x]. Let p(x) = ( k)x 5 + ( )x 4 + (2.5k)x 2 + ( k) R (5). k Prove the derivative is zero ad the itegral does ot exist.

131 130 Problems o MOD Structures k Characterizig those polyomials i R (m) which are both itegrable as well as derivable happes to be a challegig problem. Next we cosider about solvig for roots for polyomials i R (m). Cosider p(x) = kx k R (10). k Clearly p(x) caot be solved for x however oe may be at times i a positio to get the value of kx ad x. k Solve 3.52 kx k caot solve for x oly for kx. k R (15). Here also we Let x 2 k + 4 R (20) solvig for x is a easy task for x 2 = 16 so x = ± 4. But if x 2 + 4k is cosidered istead the x 2 = 16k. x = 4 k. Fidig value of k is a very difficult task. Now i the followig we suggest some problems which are ope cojectures some of them are difficult problems other easy exercise. Problems 1. Fid coditios o the MOD iterval real polyomials p(x) [0, m)[x] to have derivatives ad to satisfy the classical properties. 2. Study the itegrals of polyomials p(x) [0,m)[x]. 3. Characterize those p(x) [0,12)[x] which follows the classical properties of derivatives. 4. Study questio three for p(x) [0,19)[x]. 5. Let p(x) R (m)[x]. Fid coditio o p(x) to have derivatives.

132 Problems o Differet types of MOD Fuctios Prove p(x) i geeral i R (15)[x] are ot itegrable. 7. Fid coditios o all those polyomials i R (4)[x] which are both itegrable ad derivable. 8. Characterize those polyomials p(x) R (5)[x] which are derivable but ot itegrable ad vice versa. 9. Let p(x) R (11)[x]. i) Fid all roots of p(x). ii) f p(x) R (11)[x] is such that p(x) is both derivable ad itegrable ca we say p(x) has same umber of roots as its degree? iii) Characterize all polyomials p(x) i R (11)[x] of degree, but has less tha roots. iv) Characterize all polyomial p(x) i R (11)[x] of degree but has roots greater tha. 10. What are special ad distict features associated with the MOD polyomial iterval [0,m)[x]? (2 m < ) 11. Eumerate all the special features ejoyed by the MOD polyomials i R (m)[x] (2 m < ). 12. Prove both [0, m)[x] ad R (m)[x] are oly pseudo rigs ad ot rigs. 13. Ca we say as the distributive property is ot true, is the oly reaso for polyomials ot satisfyig the classical properties? 14. What is the reaso for the polyomials i [0,m)[x] ad R (m)[x] to behave i a chaotic maer? 15. Study the MOD complex modulo iteger polyomials i C (m)[x].

133 132 Problems o MOD Structures 16. Characterize all those polyomials i C (m)[x] which follow both the classical laws of differetiatio ad itegratio. 17. Characterize all those polyomials i C (m)[x], 2 m < which satisfy all the classical laws of differetiatio. 18. Characterize all these polyomials i C (m)[x], 2 m < which satisfy all the classical laws of itegratio. 19. Characterize all those polyomials i C (m)[x] which satisfy the classical properties of roots. 20. Study questios (16) to (19) for the polyomials i C (15)[x]. 21. Study questios (16) to (19) for polyomials i C (43)[x]. 22. Study questios (16) to (19) for polyomials i C (3 12 )[x]. 23. Solve p(x) = ( i F ) x 2 + ( i F ) C (10)[x]. 24. Show as the distributive laws are ot true i C (m)[x] fidig roots of a polyomial i C (m)[x] is a difficult task by some examples. 25. Let p(x) = 27 x 3 + (5 + 6 i F ) C (7)[x]. i) Solve for x. ii) Will p(x) have three roots i C (7)[x]? iii) f α, β ad γ are the roots of p(x) ca we prove (3x α) (3x γ) (3x β) = p(x). iv) f p 1 (x) = x 3 + ( i F ) C (7)[x] will (iii) be true for the roots α, β, γ of p 1 (x). 26. Prove i C (m)[x]; (x α 1 ) (x α 2 ) (x α ) x (α 1 + α α ) x 1 + i j α i α j x 2 ± α 1 α where

134 Problems o Differet types of MOD Fuctios 133 α 1, α 2,, α are roots of a polyomial p(x) ad are i C (m)[x]. 27. Whe will problem (25) be true i R (m)[x]? 28. How will this problem i (25) be over come? 29. Ca we have this sort of problem i real world problem? (if so give some examples). 30. s it advatageous to use MOD polyomials of MOD plaes tha usual real plae ad complex plae? 31. Let p(x) g R (m)[x] = i aix a i i= 1 g R (m) = {a + bg = (a, b); g 2 = 0; 2 m < } be the MOD dual umber polyomial with coefficiets from the MOD dual umber plae R (m). g i) Whe is p(x) totally solvable? ii) f p(x) is of degree ca we say p(x) will have ad oly roots. iii) Whe will p(x) satisfy all the classical properties of differetiatio? iv) Whe will p(x) satisfy all the laws of differetiatio? v) Obtai ay other special or distict feature associated with p(x). vi) Ca we say because p(x) is a MOD dual umber coefficiet polyomial they ejoy some special features? g vii) Characterize all p(x) i R (m)[x] for which both the itegral or derivative does ot exist. g viii) Characterize all those polyomials i R (m)[x] which satisfies all properties of derivatives but are ot itegrable.

135 134 Problems o MOD Structures ix) g Characterize those polyomials i R (m)[x] which satisfy all properties of itegrals but does ot satisfy the properties of derivatives. g 32. Let R (12)[x] be the dual umber polyomials with coefficiets from the MOD dual umber plae R (12) = {a + bg = (a, b) g 2 = 0, a, b [0, 12)}. g 33. Let 34. Let Study questios (i) to (ix) of problem (31) for this polyomials i R (12)[x]. g g R (43)[x] be the dual umber polyomials with coefficiets from the MOD dual umber plae g R (43). Study questios (i) to (ix) of problem (31) of polyomials i R (43)[x]. g R (5 7 )[x] be the collectio of all polyomials with g coefficiets from the MOD dual umber plae g R (5 7 ). Study questios (i) to (ix) of problem (31) for polyomials i R (5 7 )[x]. g 35. Let p(x) = ( g)x g g R (12)[x]. i) Fid all roots of p(x). ii) s p(x) itegrable? iii) Does the derivative of p(x) satisfy the usual laws of derivatio? 36. Let p(x) = 4.3g x 4 + ( g)x g R (13)[x]. g Study questios (i) to (iii) of problem (35) for this p(x).

136 Problems o Differet types of MOD Fuctios Let g x 3 + (5 + 2g)x gx g R (2 5 )[x]. g Study questios (i) to (iii) of problem (35) for this p(x). 38. Obtai all special features associated with the MOD h special dual like umber polyomials i R (m)[x] with coefficiets from R (m) = {a + bh = (a,b) h 2 = h, a,b [0, m)}. h 39. Let p(x) = ( k) ( k)x + ( k)x 2 h R (6)[x]; k 2 = 5k special quasi dual MOD umber polyomial. i) Fid the roots of p(x). ii) Ca p(x) have more tha two roots? iii) s x uiquely solvable for this p(x)? 40. Let w(x) = ( h) x 3 + ( h) h R (7)[x] where h 2 = h be the MOD special dual like umber polyomial. Study questios (i) to (iii) of problem (39) for this w(x). 41. Let t(x) = ( h) x 6 + ( h)x 3 + (17 + g h) R (19)[x], h 2 = h, be the MOD special dual like umber polyomial. Study questios (i) to (iii) of problem (38) for this t(x). 42. Let m(x) = ( h) x 4 + ( h) x 3 + ( h) x 2 h + ( h) x h R (8)[x], h 2 = h be the MOD special dual like umber polyomial. Study questios (i) to (iii) of problem (38) for this m(x).

137 136 Problems o MOD Structures 43. Let p(x) = ( h)x 16 + ( h)x 8 + (4.32 h h)x + ( h) R (8)[x] be the MOD special dual like umber polyomial. i) Fid dp(x) does the derivative satisfy the usual dx laws of derivatio? ii) s p(x) dx defied? Justify your claim. h iii) Characterize those MOD polyomials i R (8)[x] for which the itegrals follow the usual laws of itegratio. iv) Characterize all those MOD polyomials which follows all the classical rules of derivatives. h v) Characterize those MOD polyomials i R (8)[x] which satisfy both the classical law of itegratio as well as that of differetiatio. h 44. Let R (47)[x] be the collectio of all MOD special dual like umber polyomials. Study questios (i) to (v) of problem 42 for this R (47)[x]. h h 45. Let R (24)[x] be the collectio of all MOD special dual like umber polyomials. Study questios (i) to (v) of problem (42) for this R (24)[x]. h 46. Eumerate all the special features ejoyed by the MOD k special quasi dual umber polyomials R (m)[x], m 2 = (m 1)k. 47. Characterize all MOD special quasi dual umber k polyomials i R (m)[x] which are itegrable but whose derivative is zero.

138 Problems o Differet types of MOD Fuctios Characterize all those MOD special quasi dual umber k polyomials i R (m)[x] which follows the usual laws of differetiatio but is ot itegrable. 49. Characterize all those MOD quasi dual umber polyomials which does ot satisfy both classical properties of differetiatio ad itegratio. 50. Let p(x) =kx 4 k + ( k) R (4)[x] be the MOD special quasi dual umber polyomial. i) Does p(x) satisfy the classical laws of derivatio? ii) Prove or disprove p(x) satisfy the classical laws of itegratio. k iii) Ca p(x) have four roots i R (4), k 2 = 3k? iv) s p(x) solvable? v) Let p 1 (x) = 5.32k x 3 k + ( k) R (10)[x], k 2 = 9k be the MOD special quasi dual umber polyomial. Study questios (i) to (iv) of problem (49) for this p 1 (x)? 51. Let p(x) = ( k) x 4 + ( k) x 3 + ( k)x 2 k + ( k)x + ( k) R (23)[x] be the MOD special quasi dual umber polyomial. Study questios (i) to (iv) of problem (49) for this p(x). k 52. Let q(x) R (140)[x] where coefficiets of the MOD special quasi dual umber polyomials are from Z 140 k = {a + bk a, b Z 140, k 2 = 139 k}. i) Ca we say all classical laws of derivatives will be true for these polyomials i Z 140 k [x]?

139 138 Problems o MOD Structures ii) Ca the classical properties of itegratio be true for these polyomials i Z 140 k [x]? iii) Ca we say a th degree polyomial i Z 140 k [x] will have ad oly roots? iv) Study questios (i) to (iii) of problems for p(x) = 14k x x k Z 140 k [x]. 53. Compare polyomials i Z m k [x] with the polyomials i Z m h [x] ad Z m g [x]. 54. Study MOD eutrosophic polyomials i 55. Derive all the special features ejoyed by R (m)[x]. R (43)[x]. 56. f p(x) = 2.3 x 7 + ( ) eutrosophic polyomial. R (10) be the MOD i) Ca we say p(x) has seve ad oly seve roots? ii) s p(x) solvable i R (10)? 57. Let q(x) = (5 + 4) x 3 + (2 + 7) x + 4 i) Will q(x) have oly 3 roots or more? R (8)[x]. ii) iii) Fid dq(x) dx Fid q(x) dx Eumerate all the special features ejoyed by R (m)[x]. 59. Characterize those MOD eutrosophic polyomials which has both derivatives ad itegrals which satisfy all the classical properties.

140 FURTHER READNG 1. Herstei.,.N., Topics i Algebra, Wiley Easter Limited, (1975). 2. Lag, S., Algebra, Addiso Wesley, (1967). 3. Smaradache, Floreti, Defiitios Derived from Neutrosophics, Proceedigs of the First teratioal Coferece o Neutrosophy, Neutrosophic Logic, Neutrosophic Set, Neutrosophic Probability ad Statistics, Uiversity of New Mexico, Gallup, 1-3 December (2001). 4. Smaradache, Floreti, Neutrosophic Logic Geeralizatio of the tuitioistic Fuzzy Logic, Special Sessio o tuitioistic Fuzzy Sets ad Related Cocepts, teratioal EUSFLAT Coferece, Zittau, Germay, September Vasatha Kadasamy, W.B., O Muti differetiatio of decimal ad Fractioal Calculus, The Mathematics Educatio, 38 (1994) Vasatha Kadasamy, W.B., O decimal ad Fractioal Calculus, Ultra Sci. of Phy. Sci., 6 (1994)

141 140 Problems o MOD Structures 7. Vasatha Kadasamy, W.B., Muti Fuzzy differetiatio i Fuzzy Polyomial rigs, J. of st. Math ad Comp. Sci., 9 (1996) Vasatha Kadasamy, W.B., Some Applicatios of the Decimal Derivatives, The Mathematics Educatio, 34, (2000) Vasatha Kadasamy, W.B., O Partial Differetiatio of Decimal ad Fractioal calculus, J. of st. Math ad Comp. Sci., 13 (2000) Vasatha Kadasamy, W.B., O Fuzzy Calculus over fuzzy Polyomial rigs, Vikram Mathematical Joural, 22 (2002) Vasatha Kadasamy, W.B., Liear Algebra ad Smaradache Liear Algebra, Bookma Publishig, (2003). 12. Vasatha Kadasamy, W. B. ad Smaradache, F., N-algebraic structures ad S-N-algebraic structures, Hexis, Phoeix, Arizoa, (2005). 13. Vasatha Kadasamy, W. B. ad Smaradache, F., Neutrosophic algebraic structures ad eutrosophic N-algebraic structures, Hexis, Phoeix, Arizoa, (2006). 14. Vasatha Kadasamy, W. B. ad Smaradache, F., Smaradache Neutrosophic algebraic structures, Hexis, Phoeix, Arizoa, (2006). 15. Vasatha Kadasamy, W.B., ad Smaradache, F., Fuzzy terval Matrices, Neutrosophic terval Matrices ad their Applicatios, Hexis, Phoeix, (2006). 16. Vasatha Kadasamy, W.B. ad Smaradache, F., Fiite Neutrosophic Complex Numbers, Zip Publishig, Ohio, (2011). 17. Vasatha Kadasamy, W.B. ad Smaradache, F., Dual Numbers, Zip Publishig, Ohio, (2012).

142 Further Readig Vasatha Kadasamy, W.B. ad Smaradache, F., Special dual like umbers ad lattices, Zip Publishig, Ohio, (2012). 19. Vasatha Kadasamy, W.B. ad Smaradache, F., Special quasi dual umbers ad Groupoids, Zip Publishig, Ohio, (2012). 20. Vasatha Kadasamy, W.B. ad Smaradache, F., Algebraic Structures usig Subsets, Educatioal Publisher c, Ohio, (2013). 21. Vasatha Kadasamy, W.B. ad Smaradache, F., Algebraic Structures usig [0, ), Educatioal Publisher c, Ohio, (2013). 22. Vasatha Kadasamy, W.B. ad Smaradache, F., Algebraic Structures o the fuzzy iterval [0, 1), Educatioal Publisher c, Ohio, (2014). 23. Vasatha Kadasamy, W.B. ad Smaradache, F., Algebraic Structures o Fuzzy Uit squares ad Neutrosophic uit square, Educatioal Publisher c, Ohio, (2014). 24. Vasatha Kadasamy, W.B. ad Smaradache, F., Special pseudo liear algebras usig [0, ), Educatioal Publisher c, Ohio, (2014). 25. Vasatha Kadasamy, W.B. ad Smaradache, F., Algebraic Structures o Real ad Neutrosophic square, Educatioal Publisher c, Ohio, (2014). 26. Vasatha Kadasamy, W.B., latheral, K., ad Smaradache, F., MOD plaes, EuropaNova, (2015). 27. Vasatha Kadasamy, W.B., latheral, K., ad Smaradache, F., MOD Fuctios, EuropaNova, (2015). 28. Vasatha Kadasamy, W.B., latheral, K., ad Smaradache, F., Multidimesioal MOD plaes, EuropaNova, (2015).

143 142 Problems o MOD Structures 29. Vasatha Kadasamy, W.B., latheral, K., ad Smaradache, F., Natural Neutrosophic umbers ad MOD Neutrosophic umbers, EuropaNova, (2015). 30. Vasatha Kadasamy, W.B., latheral, K., ad Smaradache, F., Algebraic Structures o MOD plaes, EuropaNova, (2015). 31. Vasatha Kadasamy, W.B., latheral, K., ad Smaradache, F., MOD Pseudo Liear Algebras, EuropaNova, (2015).

144 NDEX D Dual umber plae, 34-8 M MOD complex modulo iteger plae, 31-7 MOD dual umber plae, 34-6 MOD iterval complex modulo iteger pseudo rig, MOD iterval dual umber group, 16-9 MOD iterval fiite complex umber group, 16-9 MOD iterval polyomial pseudo rig, MOD iterval real groups, 10-9 MOD iterval semigroup, 12-9 MOD iterval special dual like umber group, MOD iterval special dual like umber pseudo rig, MOD iterval special dual like umber semigroup, MOD iterval special quasi dual umber group, MOD iterval special quasi dual umber pseudo rig, MOD iterval special quasi dual umber semigroup, MOD iterval special quasi dual umber subgroup, MOD logarithmic fuctios, MOD atural elemets, MOD eutrosophic plaes, 31-5 MOD real iterval, 7-12

145 144 Problems o MOD Structures MOD real plae, MOD real trasformatio, 7-13 MOD special complex modulo iteger pseudo subrig, MOD special dual like umber plae, 37-9 MOD special pseudo rig, MOD special quasi dual umber plae, MOD special quasi dual umber pseudo subrig, MOD special real pseudo subrig, MOD trigoometric fuctios, N Natural eutrosophic ilpotets, Neutrosophic zero divisors, Neutrosophic idempotets, P Pseudo MOD real trasformatio, 8-15 Pseudo zero divisors, S Special dual umber like plae, 36-9 Special quasi dual umber plae, 38-40

146 ABOUT THE AUTHORS Dr.W.B.Vasatha Kadasamy is a Professor i the Departmet of Mathematics, dia stitute of Techology Madras, Cheai. the past decade she has guided 13 Ph.D. scholars i the differet fields of o-associative algebras, algebraic codig theory, trasportatio theory, fuzzy groups, ad applicatios of fuzzy theory of the problems faced i chemical idustries ad cemet idustries. She has to her credit 694 research papers. She has guided over 100 M.Sc. ad M.Tech. projects. She has worked i collaboratio projects with the dia Space Research Orgaizatio ad with the Tamil Nadu State ADS Cotrol Society. She is presetly workig o a research project fuded by the Board of Research i Nuclear Scieces, Govermet of dia. This is her 111 th book. O dia's 60th depedece Day, Dr.Vasatha was coferred the Kalpaa Chawla Award for Courage ad Darig Eterprise by the State Govermet of Tamil Nadu i recogitio of her sustaied fight for social justice i the dia stitute of Techology (T) Madras ad for her cotributio to mathematics. The award, istituted i the memory of dia-america astroaut Kalpaa Chawla who died aboard Space Shuttle Columbia, carried a cash prize of five lakh rupees (the highest prize-moey for ay dia award) ad a gold medal. She ca be cotacted at vasathakadasamy@gmail.com Web Site: or Dr. K. latheral is the editor of The Maths Tiger, Quarterly Joural of Maths. She ca be cotacted at ilatheral@gmail.com Dr. Floreti Smaradache is a Professor of Mathematics at the Uiversity of New Mexico i USA. He published over 75 books ad 200 articles ad otes i mathematics, physics, philosophy, psychology, rebus, literature. mathematics his research is i umber theory, o-euclidea geometry, sythetic geometry, algebraic structures, statistics, eutrosophic logic ad set (geeralizatios of fuzzy logic ad set respectively), eutrosophic probability (geeralizatio of classical ad imprecise probability). Also, small cotributios to uclear ad particle physics, iformatio fusio, eutrosophy (a geeralizatio of dialectics), law of sesatios ad stimuli, etc. He got the 2010 Telesio-Galilei Academy of Sciece Gold Medal, Adjuct Professor (equivalet to Doctor Hooris Causa) of Beijig Jiaotog Uiversity i 2011, ad 2011 Romaia Academy Award for Techical Sciece (the highest i the coutry). Dr. W. B. Vasatha Kadasamy ad Dr. Floreti Smaradache got the 2012 New Mexico-Arizoa ad 2011 New Mexico Book Award for Algebraic Structures. He ca be cotacted at smarad@um.edu

147

MOD Planes: A New Dimension to Modulo Theory. W. B. Vasantha Kandasamy Ilanthenral K Florentin Smarandache

MOD Planes: A New Dimension to Modulo Theory. W. B. Vasantha Kandasamy Ilanthenral K Florentin Smarandache MOD Plaes: A New Dimesio to Modulo Theory W. B. Vasatha Kadasamy latheral K Floreti Smaradache 2015 This book ca be ordered from: EuropaNova ASBL Clos du Parasse, 3E 1000, Bruxelles Belgium E-mail: ifo@europaova.be

More information

CHAPTER I: Vector Spaces

CHAPTER I: Vector Spaces CHAPTER I: Vector Spaces Sectio 1: Itroductio ad Examples This first chapter is largely a review of topics you probably saw i your liear algebra course. So why cover it? (1) Not everyoe remembers everythig

More information

Modern Algebra. Previous year Questions from 2017 to Ramanasri

Modern Algebra. Previous year Questions from 2017 to Ramanasri Moder Algebra Previous year Questios from 017 to 199 Ramaasri 017 S H O P NO- 4, 1 S T F L O O R, N E A R R A P I D F L O U R M I L L S, O L D R A J E N D E R N A G A R, N E W D E L H I. W E B S I T E

More information

Exercises 1 Sets and functions

Exercises 1 Sets and functions Exercises 1 Sets ad fuctios HU Wei September 6, 018 1 Basics Set theory ca be made much more rigorous ad built upo a set of Axioms. But we will cover oly some heuristic ideas. For those iterested studets,

More information

6 Integers Modulo n. integer k can be written as k = qn + r, with q,r, 0 r b. So any integer.

6 Integers Modulo n. integer k can be written as k = qn + r, with q,r, 0 r b. So any integer. 6 Itegers Modulo I Example 2.3(e), we have defied the cogruece of two itegers a,b with respect to a modulus. Let us recall that a b (mod ) meas a b. We have proved that cogruece is a equivalece relatio

More information

Homework 2 January 19, 2006 Math 522. Direction: This homework is due on January 26, In order to receive full credit, answer

Homework 2 January 19, 2006 Math 522. Direction: This homework is due on January 26, In order to receive full credit, answer Homework 2 Jauary 9, 26 Math 522 Directio: This homework is due o Jauary 26, 26. I order to receive full credit, aswer each problem completely ad must show all work.. What is the set of the uits (that

More information

11. FINITE FIELDS. Example 1: The following tables define addition and multiplication for a field of order 4.

11. FINITE FIELDS. Example 1: The following tables define addition and multiplication for a field of order 4. 11. FINITE FIELDS 11.1. A Field With 4 Elemets Probably the oly fiite fields which you ll kow about at this stage are the fields of itegers modulo a prime p, deoted by Z p. But there are others. Now although

More information

Infinite Sequences and Series

Infinite Sequences and Series Chapter 6 Ifiite Sequeces ad Series 6.1 Ifiite Sequeces 6.1.1 Elemetary Cocepts Simply speakig, a sequece is a ordered list of umbers writte: {a 1, a 2, a 3,...a, a +1,...} where the elemets a i represet

More information

6.3 Testing Series With Positive Terms

6.3 Testing Series With Positive Terms 6.3. TESTING SERIES WITH POSITIVE TERMS 307 6.3 Testig Series With Positive Terms 6.3. Review of what is kow up to ow I theory, testig a series a i for covergece amouts to fidig the i= sequece of partial

More information

TEACHER CERTIFICATION STUDY GUIDE

TEACHER CERTIFICATION STUDY GUIDE COMPETENCY 1. ALGEBRA SKILL 1.1 1.1a. ALGEBRAIC STRUCTURES Kow why the real ad complex umbers are each a field, ad that particular rigs are ot fields (e.g., itegers, polyomial rigs, matrix rigs) Algebra

More information

62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 +

62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + 62. Power series Defiitio 16. (Power series) Give a sequece {c }, the series c x = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + is called a power series i the variable x. The umbers c are called the coefficiets of

More information

Chapter 2. Periodic points of toral. automorphisms. 2.1 General introduction

Chapter 2. Periodic points of toral. automorphisms. 2.1 General introduction Chapter 2 Periodic poits of toral automorphisms 2.1 Geeral itroductio The automorphisms of the two-dimesioal torus are rich mathematical objects possessig iterestig geometric, algebraic, topological ad

More information

PAPER : IIT-JAM 2010

PAPER : IIT-JAM 2010 MATHEMATICS-MA (CODE A) Q.-Q.5: Oly oe optio is correct for each questio. Each questio carries (+6) marks for correct aswer ad ( ) marks for icorrect aswer.. Which of the followig coditios does NOT esure

More information

CSE 1400 Applied Discrete Mathematics Number Theory and Proofs

CSE 1400 Applied Discrete Mathematics Number Theory and Proofs CSE 1400 Applied Discrete Mathematics Number Theory ad Proofs Departmet of Computer Scieces College of Egieerig Florida Tech Sprig 01 Problems for Number Theory Backgroud Number theory is the brach of

More information

Chapter 4. Fourier Series

Chapter 4. Fourier Series Chapter 4. Fourier Series At this poit we are ready to ow cosider the caoical equatios. Cosider, for eample the heat equatio u t = u, < (4.) subject to u(, ) = si, u(, t) = u(, t) =. (4.) Here,

More information

Lesson 10: Limits and Continuity

Lesson 10: Limits and Continuity www.scimsacademy.com Lesso 10: Limits ad Cotiuity SCIMS Academy 1 Limit of a fuctio The cocept of limit of a fuctio is cetral to all other cocepts i calculus (like cotiuity, derivative, defiite itegrals

More information

Basic Sets. Functions. MTH299 - Examples. Example 1. Let S = {1, {2, 3}, 4}. Indicate whether each statement is true or false. (a) S = 4. (e) 2 S.

Basic Sets. Functions. MTH299 - Examples. Example 1. Let S = {1, {2, 3}, 4}. Indicate whether each statement is true or false. (a) S = 4. (e) 2 S. Basic Sets Example 1. Let S = {1, {2, 3}, 4}. Idicate whether each statemet is true or false. (a) S = 4 (b) {1} S (c) {2, 3} S (d) {1, 4} S (e) 2 S. (f) S = {1, 4, {2, 3}} (g) S Example 2. Compute the

More information

PRELIM PROBLEM SOLUTIONS

PRELIM PROBLEM SOLUTIONS PRELIM PROBLEM SOLUTIONS THE GRAD STUDENTS + KEN Cotets. Complex Aalysis Practice Problems 2. 2. Real Aalysis Practice Problems 2. 4 3. Algebra Practice Problems 2. 8. Complex Aalysis Practice Problems

More information

MAT1026 Calculus II Basic Convergence Tests for Series

MAT1026 Calculus II Basic Convergence Tests for Series MAT026 Calculus II Basic Covergece Tests for Series Egi MERMUT 202.03.08 Dokuz Eylül Uiversity Faculty of Sciece Departmet of Mathematics İzmir/TURKEY Cotets Mootoe Covergece Theorem 2 2 Series of Real

More information

Math 155 (Lecture 3)

Math 155 (Lecture 3) Math 55 (Lecture 3) September 8, I this lecture, we ll cosider the aswer to oe of the most basic coutig problems i combiatorics Questio How may ways are there to choose a -elemet subset of the set {,,,

More information

(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3

(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3 MATH 337 Sequeces Dr. Neal, WKU Let X be a metric space with distace fuctio d. We shall defie the geeral cocept of sequece ad limit i a metric space, the apply the results i particular to some special

More information

Chapter 1. Complex Numbers. Dr. Pulak Sahoo

Chapter 1. Complex Numbers. Dr. Pulak Sahoo Chapter 1 Complex Numbers BY Dr. Pulak Sahoo Assistat Professor Departmet of Mathematics Uiversity Of Kalyai West Begal, Idia E-mail : sahoopulak1@gmail.com 1 Module-2: Stereographic Projectio 1 Euler

More information

4 The Sperner property.

4 The Sperner property. 4 The Sperer property. I this sectio we cosider a surprisig applicatio of certai adjacecy matrices to some problems i extremal set theory. A importat role will also be played by fiite groups. I geeral,

More information

In number theory we will generally be working with integers, though occasionally fractions and irrationals will come into play.

In number theory we will generally be working with integers, though occasionally fractions and irrationals will come into play. Number Theory Math 5840 otes. Sectio 1: Axioms. I umber theory we will geerally be workig with itegers, though occasioally fractios ad irratioals will come ito play. Notatio: Z deotes the set of all itegers

More information

ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 5: SINGULARITIES.

ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 5: SINGULARITIES. ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 5: SINGULARITIES. ANDREW SALCH 1. The Jacobia criterio for osigularity. You have probably oticed by ow that some poits o varieties are smooth i a sese somethig

More information

CHAPTER 5. Theory and Solution Using Matrix Techniques

CHAPTER 5. Theory and Solution Using Matrix Techniques A SERIES OF CLASS NOTES FOR 2005-2006 TO INTRODUCE LINEAR AND NONLINEAR PROBLEMS TO ENGINEERS, SCIENTISTS, AND APPLIED MATHEMATICIANS DE CLASS NOTES 3 A COLLECTION OF HANDOUTS ON SYSTEMS OF ORDINARY DIFFERENTIAL

More information

NBHM QUESTION 2007 Section 1 : Algebra Q1. Let G be a group of order n. Which of the following conditions imply that G is abelian?

NBHM QUESTION 2007 Section 1 : Algebra Q1. Let G be a group of order n. Which of the following conditions imply that G is abelian? NBHM QUESTION 7 NBHM QUESTION 7 NBHM QUESTION 7 Sectio : Algebra Q Let G be a group of order Which of the followig coditios imply that G is abelia? 5 36 Q Which of the followig subgroups are ecesarily

More information

The picture in figure 1.1 helps us to see that the area represents the distance traveled. Figure 1: Area represents distance travelled

The picture in figure 1.1 helps us to see that the area represents the distance traveled. Figure 1: Area represents distance travelled 1 Lecture : Area Area ad distace traveled Approximatig area by rectagles Summatio The area uder a parabola 1.1 Area ad distace Suppose we have the followig iformatio about the velocity of a particle, how

More information

Chapter 0. Review of set theory. 0.1 Sets

Chapter 0. Review of set theory. 0.1 Sets Chapter 0 Review of set theory Set theory plays a cetral role i the theory of probability. Thus, we will ope this course with a quick review of those otios of set theory which will be used repeatedly.

More information

Apply change-of-basis formula to rewrite x as a linear combination of eigenvectors v j.

Apply change-of-basis formula to rewrite x as a linear combination of eigenvectors v j. Eigevalue-Eigevector Istructor: Nam Su Wag eigemcd Ay vector i real Euclidea space of dimesio ca be uiquely epressed as a liear combiatio of liearly idepedet vectors (ie, basis) g j, j,,, α g α g α g α

More information

Chapter 10: Power Series

Chapter 10: Power Series Chapter : Power Series 57 Chapter Overview: Power Series The reaso series are part of a Calculus course is that there are fuctios which caot be itegrated. All power series, though, ca be itegrated because

More information

We are mainly going to be concerned with power series in x, such as. (x)} converges - that is, lims N n

We are mainly going to be concerned with power series in x, such as. (x)} converges - that is, lims N n Review of Power Series, Power Series Solutios A power series i x - a is a ifiite series of the form c (x a) =c +c (x a)+(x a) +... We also call this a power series cetered at a. Ex. (x+) is cetered at

More information

If a subset E of R contains no open interval, is it of zero measure? For instance, is the set of irrationals in [0, 1] is of measure zero?

If a subset E of R contains no open interval, is it of zero measure? For instance, is the set of irrationals in [0, 1] is of measure zero? 2 Lebesgue Measure I Chapter 1 we defied the cocept of a set of measure zero, ad we have observed that every coutable set is of measure zero. Here are some atural questios: If a subset E of R cotais a

More information

DIVISIBILITY PROPERTIES OF GENERALIZED FIBONACCI POLYNOMIALS

DIVISIBILITY PROPERTIES OF GENERALIZED FIBONACCI POLYNOMIALS DIVISIBILITY PROPERTIES OF GENERALIZED FIBONACCI POLYNOMIALS VERNER E. HOGGATT, JR. Sa Jose State Uiversity, Sa Jose, Califoria 95192 ad CALVIN T. LONG Washigto State Uiversity, Pullma, Washigto 99163

More information

CALCULUS BASIC SUMMER REVIEW

CALCULUS BASIC SUMMER REVIEW CALCULUS BASIC SUMMER REVIEW NAME rise y y y Slope of a o vertical lie: m ru Poit Slope Equatio: y y m( ) The slope is m ad a poit o your lie is, ). ( y Slope-Itercept Equatio: y m b slope= m y-itercept=

More information

Polynomial Functions and Their Graphs

Polynomial Functions and Their Graphs Polyomial Fuctios ad Their Graphs I this sectio we begi the study of fuctios defied by polyomial expressios. Polyomial ad ratioal fuctios are the most commo fuctios used to model data, ad are used extesively

More information

Some remarks for codes and lattices over imaginary quadratic

Some remarks for codes and lattices over imaginary quadratic Some remarks for codes ad lattices over imagiary quadratic fields Toy Shaska Oaklad Uiversity, Rochester, MI, USA. Caleb Shor Wester New Eglad Uiversity, Sprigfield, MA, USA. shaska@oaklad.edu Abstract

More information

MATH spring 2008 lecture 3 Answers to selected problems. 0 sin14 xdx = x dx. ; (iv) x +

MATH spring 2008 lecture 3 Answers to selected problems. 0 sin14 xdx = x dx. ; (iv) x + MATH - sprig 008 lecture Aswers to selected problems INTEGRALS. f =? For atiderivatives i geeral see the itegrals website at http://itegrals.wolfram.com. (5-vi (0 i ( ( i ( π ; (v π a. This is example

More information

APPENDIX F Complex Numbers

APPENDIX F Complex Numbers APPENDIX F Complex Numbers Operatios with Complex Numbers Complex Solutios of Quadratic Equatios Polar Form of a Complex Number Powers ad Roots of Complex Numbers Operatios with Complex Numbers Some equatios

More information

MIDTERM 3 CALCULUS 2. Monday, December 3, :15 PM to 6:45 PM. Name PRACTICE EXAM SOLUTIONS

MIDTERM 3 CALCULUS 2. Monday, December 3, :15 PM to 6:45 PM. Name PRACTICE EXAM SOLUTIONS MIDTERM 3 CALCULUS MATH 300 FALL 08 Moday, December 3, 08 5:5 PM to 6:45 PM Name PRACTICE EXAM S Please aswer all of the questios, ad show your work. You must explai your aswers to get credit. You will

More information

Brief Review of Functions of Several Variables

Brief Review of Functions of Several Variables Brief Review of Fuctios of Several Variables Differetiatio Differetiatio Recall, a fuctio f : R R is differetiable at x R if ( ) ( ) lim f x f x 0 exists df ( x) Whe this limit exists we call it or f(

More information

SOLVED EXAMPLES

SOLVED EXAMPLES Prelimiaries Chapter PELIMINAIES Cocept of Divisibility: A o-zero iteger t is said to be a divisor of a iteger s if there is a iteger u such that s tu I this case we write t s (i) 6 as ca be writte as

More information

Synopsis of Euler s paper. E Memoire sur la plus grande equation des planetes. (Memoir on the Maximum value of an Equation of the Planets)

Synopsis of Euler s paper. E Memoire sur la plus grande equation des planetes. (Memoir on the Maximum value of an Equation of the Planets) 1 Syopsis of Euler s paper E105 -- Memoire sur la plus grade equatio des plaetes (Memoir o the Maximum value of a Equatio of the Plaets) Compiled by Thomas J Osler ad Jase Adrew Scaramazza Mathematics

More information

Zeros of Polynomials

Zeros of Polynomials Math 160 www.timetodare.com 4.5 4.6 Zeros of Polyomials I these sectios we will study polyomials algebraically. Most of our work will be cocered with fidig the solutios of polyomial equatios of ay degree

More information

Sets. Sets. Operations on Sets Laws of Algebra of Sets Cardinal Number of a Finite and Infinite Set. Representation of Sets Power Set Venn Diagram

Sets. Sets. Operations on Sets Laws of Algebra of Sets Cardinal Number of a Finite and Infinite Set. Representation of Sets Power Set Venn Diagram Sets MILESTONE Sets Represetatio of Sets Power Set Ve Diagram Operatios o Sets Laws of lgebra of Sets ardial Number of a Fiite ad Ifiite Set I Mathematical laguage all livig ad o-livig thigs i uiverse

More information

September 2012 C1 Note. C1 Notes (Edexcel) Copyright - For AS, A2 notes and IGCSE / GCSE worksheets 1

September 2012 C1 Note. C1 Notes (Edexcel) Copyright   - For AS, A2 notes and IGCSE / GCSE worksheets 1 September 0 s (Edecel) Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets September 0 Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets September 0 Copyright

More information

1 Approximating Integrals using Taylor Polynomials

1 Approximating Integrals using Taylor Polynomials Seughee Ye Ma 8: Week 7 Nov Week 7 Summary This week, we will lear how we ca approximate itegrals usig Taylor series ad umerical methods. Topics Page Approximatig Itegrals usig Taylor Polyomials. Defiitios................................................

More information

Recursive Algorithms. Recurrences. Recursive Algorithms Analysis

Recursive Algorithms. Recurrences. Recursive Algorithms Analysis Recursive Algorithms Recurreces Computer Sciece & Egieerig 35: Discrete Mathematics Christopher M Bourke cbourke@cseuledu A recursive algorithm is oe i which objects are defied i terms of other objects

More information

Continuous Functions

Continuous Functions Cotiuous Fuctios Q What does it mea for a fuctio to be cotiuous at a poit? Aswer- I mathematics, we have a defiitio that cosists of three cocepts that are liked i a special way Cosider the followig defiitio

More information

Section 1.1. Calculus: Areas And Tangents. Difference Equations to Differential Equations

Section 1.1. Calculus: Areas And Tangents. Difference Equations to Differential Equations Differece Equatios to Differetial Equatios Sectio. Calculus: Areas Ad Tagets The study of calculus begis with questios about chage. What happes to the velocity of a swigig pedulum as its positio chages?

More information

lim za n n = z lim a n n.

lim za n n = z lim a n n. Lecture 6 Sequeces ad Series Defiitio 1 By a sequece i a set A, we mea a mappig f : N A. It is customary to deote a sequece f by {s } where, s := f(). A sequece {z } of (complex) umbers is said to be coverget

More information

It is always the case that unions, intersections, complements, and set differences are preserved by the inverse image of a function.

It is always the case that unions, intersections, complements, and set differences are preserved by the inverse image of a function. MATH 532 Measurable Fuctios Dr. Neal, WKU Throughout, let ( X, F, µ) be a measure space ad let (!, F, P ) deote the special case of a probability space. We shall ow begi to study real-valued fuctios defied

More information

Addition: Property Name Property Description Examples. a+b = b+a. a+(b+c) = (a+b)+c

Addition: Property Name Property Description Examples. a+b = b+a. a+(b+c) = (a+b)+c Notes for March 31 Fields: A field is a set of umbers with two (biary) operatios (usually called additio [+] ad multiplicatio [ ]) such that the followig properties hold: Additio: Name Descriptio Commutativity

More information

7 Sequences of real numbers

7 Sequences of real numbers 40 7 Sequeces of real umbers 7. Defiitios ad examples Defiitio 7... A sequece of real umbers is a real fuctio whose domai is the set N of atural umbers. Let s : N R be a sequece. The the values of s are

More information

Bertrand s Postulate

Bertrand s Postulate Bertrad s Postulate Lola Thompso Ross Program July 3, 2009 Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 1 / 33 Bertrad s Postulate I ve said it oce ad I ll say it agai: There s always a

More information

Solutions to Math 347 Practice Problems for the final

Solutions to Math 347 Practice Problems for the final Solutios to Math 347 Practice Problems for the fial 1) True or False: a) There exist itegers x,y such that 50x + 76y = 6. True: the gcd of 50 ad 76 is, ad 6 is a multiple of. b) The ifiimum of a set is

More information

Properties and Tests of Zeros of Polynomial Functions

Properties and Tests of Zeros of Polynomial Functions Properties ad Tests of Zeros of Polyomial Fuctios The Remaider ad Factor Theorems: Sythetic divisio ca be used to fid the values of polyomials i a sometimes easier way tha substitutio. This is show by

More information

Chapter 6 Infinite Series

Chapter 6 Infinite Series Chapter 6 Ifiite Series I the previous chapter we cosidered itegrals which were improper i the sese that the iterval of itegratio was ubouded. I this chapter we are goig to discuss a topic which is somewhat

More information

3. Z Transform. Recall that the Fourier transform (FT) of a DT signal xn [ ] is ( ) [ ] = In order for the FT to exist in the finite magnitude sense,

3. Z Transform. Recall that the Fourier transform (FT) of a DT signal xn [ ] is ( ) [ ] = In order for the FT to exist in the finite magnitude sense, 3. Z Trasform Referece: Etire Chapter 3 of text. Recall that the Fourier trasform (FT) of a DT sigal x [ ] is ω ( ) [ ] X e = j jω k = xe I order for the FT to exist i the fiite magitude sese, S = x [

More information

PAijpam.eu ON DERIVATION OF RATIONAL SOLUTIONS OF BABBAGE S FUNCTIONAL EQUATION

PAijpam.eu ON DERIVATION OF RATIONAL SOLUTIONS OF BABBAGE S FUNCTIONAL EQUATION Iteratioal Joural of Pure ad Applied Mathematics Volume 94 No. 204, 9-20 ISSN: 3-8080 (prited versio); ISSN: 34-3395 (o-lie versio) url: http://www.ijpam.eu doi: http://dx.doi.org/0.2732/ijpam.v94i.2 PAijpam.eu

More information

Vector Spaces and Vector Subspaces. Remarks. Euclidean Space

Vector Spaces and Vector Subspaces. Remarks. Euclidean Space Vector Spaces ad Vector Subspaces Remarks Let be a iteger. A -dimesioal vector is a colum of umbers eclosed i brackets. The umbers are called the compoets of the vector. u u u u Euclidea Space I Euclidea

More information

Sequences of Definite Integrals, Factorials and Double Factorials

Sequences of Definite Integrals, Factorials and Double Factorials 47 6 Joural of Iteger Sequeces, Vol. 8 (5), Article 5.4.6 Sequeces of Defiite Itegrals, Factorials ad Double Factorials Thierry Daa-Picard Departmet of Applied Mathematics Jerusalem College of Techology

More information

ACO Comprehensive Exam 9 October 2007 Student code A. 1. Graph Theory

ACO Comprehensive Exam 9 October 2007 Student code A. 1. Graph Theory 1. Graph Theory Prove that there exist o simple plaar triagulatio T ad two distict adjacet vertices x, y V (T ) such that x ad y are the oly vertices of T of odd degree. Do ot use the Four-Color Theorem.

More information

b i u x i U a i j u x i u x j

b i u x i U a i j u x i u x j M ath 5 2 7 Fall 2 0 0 9 L ecture 1 9 N ov. 1 6, 2 0 0 9 ) S ecod- Order Elliptic Equatios: Weak S olutios 1. Defiitios. I this ad the followig two lectures we will study the boudary value problem Here

More information

A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence

A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence Sequeces A sequece of umbers is a fuctio whose domai is the positive itegers. We ca see that the sequece,, 2, 2, 3, 3,... is a fuctio from the positive itegers whe we write the first sequece elemet as

More information

Mathematical Methods for Physics and Engineering

Mathematical Methods for Physics and Engineering Mathematical Methods for Physics ad Egieerig Lecture otes Sergei V. Shabaov Departmet of Mathematics, Uiversity of Florida, Gaiesville, FL 326 USA CHAPTER The theory of covergece. Numerical sequeces..

More information

Riemann Sums y = f (x)

Riemann Sums y = f (x) Riema Sums Recall that we have previously discussed the area problem I its simplest form we ca state it this way: The Area Problem Let f be a cotiuous, o-egative fuctio o the closed iterval [a, b] Fid

More information

Subject: Differential Equations & Mathematical Modeling-III

Subject: Differential Equations & Mathematical Modeling-III Power Series Solutios of Differetial Equatios about Sigular poits Subject: Differetial Equatios & Mathematical Modelig-III Lesso: Power series solutios of differetial equatios about Sigular poits Lesso

More information

MATH 205 HOMEWORK #2 OFFICIAL SOLUTION. (f + g)(x) = f(x) + g(x) = f( x) g( x) = (f + g)( x)

MATH 205 HOMEWORK #2 OFFICIAL SOLUTION. (f + g)(x) = f(x) + g(x) = f( x) g( x) = (f + g)( x) MATH 205 HOMEWORK #2 OFFICIAL SOLUTION Problem 2: Do problems 7-9 o page 40 of Hoffma & Kuze. (7) We will prove this by cotradictio. Suppose that W 1 is ot cotaied i W 2 ad W 2 is ot cotaied i W 1. The

More information

Dirichlet s Theorem on Arithmetic Progressions

Dirichlet s Theorem on Arithmetic Progressions Dirichlet s Theorem o Arithmetic Progressios Athoy Várilly Harvard Uiversity, Cambridge, MA 0238 Itroductio Dirichlet s theorem o arithmetic progressios is a gem of umber theory. A great part of its beauty

More information

Inverse Matrix. A meaning that matrix B is an inverse of matrix A.

Inverse Matrix. A meaning that matrix B is an inverse of matrix A. Iverse Matrix Two square matrices A ad B of dimesios are called iverses to oe aother if the followig holds, AB BA I (11) The otio is dual but we ofte write 1 B A meaig that matrix B is a iverse of matrix

More information

Ma 530 Introduction to Power Series

Ma 530 Introduction to Power Series Ma 530 Itroductio to Power Series Please ote that there is material o power series at Visual Calculus. Some of this material was used as part of the presetatio of the topics that follow. What is a Power

More information

Recitation 4: Lagrange Multipliers and Integration

Recitation 4: Lagrange Multipliers and Integration Math 1c TA: Padraic Bartlett Recitatio 4: Lagrage Multipliers ad Itegratio Week 4 Caltech 211 1 Radom Questio Hey! So, this radom questio is pretty tightly tied to today s lecture ad the cocept of cotet

More information

Complex Numbers Solutions

Complex Numbers Solutions Complex Numbers Solutios Joseph Zoller February 7, 06 Solutios. (009 AIME I Problem ) There is a complex umber with imagiary part 64 ad a positive iteger such that Fid. [Solutio: 697] 4i + + 4i. 4i 4i

More information

MATH 304: MIDTERM EXAM SOLUTIONS

MATH 304: MIDTERM EXAM SOLUTIONS MATH 304: MIDTERM EXAM SOLUTIONS [The problems are each worth five poits, except for problem 8, which is worth 8 poits. Thus there are 43 possible poits.] 1. Use the Euclidea algorithm to fid the greatest

More information

[ 11 ] z of degree 2 as both degree 2 each. The degree of a polynomial in n variables is the maximum of the degrees of its terms.

[ 11 ] z of degree 2 as both degree 2 each. The degree of a polynomial in n variables is the maximum of the degrees of its terms. [ 11 ] 1 1.1 Polyomial Fuctios 1 Algebra Ay fuctio f ( x) ax a1x... a1x a0 is a polyomial fuctio if ai ( i 0,1,,,..., ) is a costat which belogs to the set of real umbers ad the idices,, 1,...,1 are atural

More information

LESSON 2: SIMPLIFYING RADICALS

LESSON 2: SIMPLIFYING RADICALS High School: Workig with Epressios LESSON : SIMPLIFYING RADICALS N.RN.. C N.RN.. B 5 5 C t t t t t E a b a a b N.RN.. 4 6 N.RN. 4. N.RN. 5. N.RN. 6. 7 8 N.RN. 7. A 7 N.RN. 8. 6 80 448 4 5 6 48 00 6 6 6

More information

Patterns in Complex Numbers An analytical paper on the roots of a complex numbers and its geometry

Patterns in Complex Numbers An analytical paper on the roots of a complex numbers and its geometry IB MATHS HL POTFOLIO TYPE Patters i Complex Numbers A aalytical paper o the roots of a complex umbers ad its geometry i Syed Tousif Ahmed Cadidate Sessio Number: 0066-009 School Code: 0066 Sessio: May

More information

SEQUENCES AND SERIES

SEQUENCES AND SERIES Sequeces ad 6 Sequeces Ad SEQUENCES AND SERIES Successio of umbers of which oe umber is desigated as the first, other as the secod, aother as the third ad so o gives rise to what is called a sequece. Sequeces

More information

, then cv V. Differential Equations Elements of Lineaer Algebra Name: Consider the differential equation. and y2 cos( kx)

, then cv V. Differential Equations Elements of Lineaer Algebra Name: Consider the differential equation. and y2 cos( kx) Cosider the differetial equatio y '' k y 0 has particular solutios y1 si( kx) ad y cos( kx) I geeral, ay liear combiatio of y1 ad y, cy 1 1 cy where c1, c is also a solutio to the equatio above The reaso

More information

Regn. No. North Delhi : 33-35, Mall Road, G.T.B. Nagar (Opp. Metro Gate No. 3), Delhi-09, Ph: ,

Regn. No. North Delhi : 33-35, Mall Road, G.T.B. Nagar (Opp. Metro Gate No. 3), Delhi-09, Ph: , . Sectio-A cotais 30 Multiple Choice Questios (MCQ). Each questio has 4 choices (a), (b), (c) ad (d), for its aswer, out of which ONLY ONE is correct. From Q. to Q.0 carries Marks ad Q. to Q.30 carries

More information

Sequences and Series of Functions

Sequences and Series of Functions Chapter 6 Sequeces ad Series of Fuctios 6.1. Covergece of a Sequece of Fuctios Poitwise Covergece. Defiitio 6.1. Let, for each N, fuctio f : A R be defied. If, for each x A, the sequece (f (x)) coverges

More information

Seunghee Ye Ma 8: Week 5 Oct 28

Seunghee Ye Ma 8: Week 5 Oct 28 Week 5 Summary I Sectio, we go over the Mea Value Theorem ad its applicatios. I Sectio 2, we will recap what we have covered so far this term. Topics Page Mea Value Theorem. Applicatios of the Mea Value

More information

Lecture 4: Grassmannians, Finite and Affine Morphisms

Lecture 4: Grassmannians, Finite and Affine Morphisms 18.725 Algebraic Geometry I Lecture 4 Lecture 4: Grassmaias, Fiite ad Affie Morphisms Remarks o last time 1. Last time, we proved the Noether ormalizatio lemma: If A is a fiitely geerated k-algebra, the,

More information

Lecture 7: Fourier Series and Complex Power Series

Lecture 7: Fourier Series and Complex Power Series Math 1d Istructor: Padraic Bartlett Lecture 7: Fourier Series ad Complex Power Series Week 7 Caltech 013 1 Fourier Series 1.1 Defiitios ad Motivatio Defiitio 1.1. A Fourier series is a series of fuctios

More information

4x 2. (n+1) x 3 n+1. = lim. 4x 2 n+1 n3 n. n 4x 2 = lim = 3

4x 2. (n+1) x 3 n+1. = lim. 4x 2 n+1 n3 n. n 4x 2 = lim = 3 Exam Problems (x. Give the series (, fid the values of x for which this power series coverges. Also =0 state clearly what the radius of covergece is. We start by settig up the Ratio Test: x ( x x ( x x

More information

Sequences A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence

Sequences A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence Sequeces A sequece of umbers is a fuctio whose domai is the positive itegers. We ca see that the sequece 1, 1, 2, 2, 3, 3,... is a fuctio from the positive itegers whe we write the first sequece elemet

More information

Math 299 Supplement: Real Analysis Nov 2013

Math 299 Supplement: Real Analysis Nov 2013 Math 299 Supplemet: Real Aalysis Nov 203 Algebra Axioms. I Real Aalysis, we work withi the axiomatic system of real umbers: the set R alog with the additio ad multiplicatio operatios +,, ad the iequality

More information

The Borel hierarchy classifies subsets of the reals by their topological complexity. Another approach is to classify them by size.

The Borel hierarchy classifies subsets of the reals by their topological complexity. Another approach is to classify them by size. Lecture 7: Measure ad Category The Borel hierarchy classifies subsets of the reals by their topological complexity. Aother approach is to classify them by size. Filters ad Ideals The most commo measure

More information

Math 113, Calculus II Winter 2007 Final Exam Solutions

Math 113, Calculus II Winter 2007 Final Exam Solutions Math, Calculus II Witer 7 Fial Exam Solutios (5 poits) Use the limit defiitio of the defiite itegral ad the sum formulas to compute x x + dx The check your aswer usig the Evaluatio Theorem Solutio: I this

More information

2.1. The Algebraic and Order Properties of R Definition. A binary operation on a set F is a function B : F F! F.

2.1. The Algebraic and Order Properties of R Definition. A binary operation on a set F is a function B : F F! F. CHAPTER 2 The Real Numbers 2.. The Algebraic ad Order Properties of R Defiitio. A biary operatio o a set F is a fuctio B : F F! F. For the biary operatios of + ad, we replace B(a, b) by a + b ad a b, respectively.

More information

The value of Banach limits on a certain sequence of all rational numbers in the interval (0,1) Bao Qi Feng

The value of Banach limits on a certain sequence of all rational numbers in the interval (0,1) Bao Qi Feng The value of Baach limits o a certai sequece of all ratioal umbers i the iterval 0, Bao Qi Feg Departmet of Mathematical Scieces, Ket State Uiversity, Tuscarawas, 330 Uiversity Dr. NE, New Philadelphia,

More information

Lecture Chapter 6: Convergence of Random Sequences

Lecture Chapter 6: Convergence of Random Sequences ECE5: Aalysis of Radom Sigals Fall 6 Lecture Chapter 6: Covergece of Radom Sequeces Dr Salim El Rouayheb Scribe: Abhay Ashutosh Doel, Qibo Zhag, Peiwe Tia, Pegzhe Wag, Lu Liu Radom sequece Defiitio A ifiite

More information

Enumerative & Asymptotic Combinatorics

Enumerative & Asymptotic Combinatorics C50 Eumerative & Asymptotic Combiatorics Stirlig ad Lagrage Sprig 2003 This sectio of the otes cotais proofs of Stirlig s formula ad the Lagrage Iversio Formula. Stirlig s formula Theorem 1 (Stirlig s

More information

An Introduction to Randomized Algorithms

An Introduction to Randomized Algorithms A Itroductio to Radomized Algorithms The focus of this lecture is to study a radomized algorithm for quick sort, aalyze it usig probabilistic recurrece relatios, ad also provide more geeral tools for aalysis

More information

Fourier Series and the Wave Equation

Fourier Series and the Wave Equation Fourier Series ad the Wave Equatio We start with the oe-dimesioal wave equatio u u =, x u(, t) = u(, t) =, ux (,) = f( x), u ( x,) = This represets a vibratig strig, where u is the displacemet of the strig

More information

Polynomials with Rational Roots that Differ by a Non-zero Constant. Generalities

Polynomials with Rational Roots that Differ by a Non-zero Constant. Generalities Polyomials with Ratioal Roots that Differ by a No-zero Costat Philip Gibbs The problem of fidig two polyomials P(x) ad Q(x) of a give degree i a sigle variable x that have all ratioal roots ad differ by

More information

ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER / Statistics

ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER / Statistics ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER 1 018/019 DR. ANTHONY BROWN 8. Statistics 8.1. Measures of Cetre: Mea, Media ad Mode. If we have a series of umbers the

More information

Measure and Measurable Functions

Measure and Measurable Functions 3 Measure ad Measurable Fuctios 3.1 Measure o a Arbitrary σ-algebra Recall from Chapter 2 that the set M of all Lebesgue measurable sets has the followig properties: R M, E M implies E c M, E M for N implies

More information

Math 61CM - Solutions to homework 3

Math 61CM - Solutions to homework 3 Math 6CM - Solutios to homework 3 Cédric De Groote October 2 th, 208 Problem : Let F be a field, m 0 a fixed oegative iteger ad let V = {a 0 + a x + + a m x m a 0,, a m F} be the vector space cosistig

More information