Contributions to the studyof the connections between algebraic

hyperstructures and fuzzy sets

Irina Elena Cristea

Introduction

In 1934, at the 8th Congress of Scandinavian Mathematicians, F. Marty has intro-duced, for the first time, the notion of hypergroup, using it in different contexts:algebraic functions, rational fractions, non-commutative groups. This momentwas the first step in the history of the development of the Algebraic Hyper-structures Theory all over the world, especially in Europe (France, Italy, Greece,Romania, Czeck and Slovak Republics, Montenegro, Serbia, Finland, Holland,Germany), Australia, S.U.A. and Canada, and later also in Iran, China, Thai-land, Japan, Korea, Jordan. At the beginning, the general aspects of the theory,the connections with groups and various applications in geometry (redefining aspherical, a descriptive or a projective geometry using hypergroups) have beenstudied. The theory knew an important progress starting with the 70’s, when itsresearch area has enlarged. In France, M.Krasner, M.Koskas and Y.Sureau haveinvestigated the theory of subhypergroups and the relations defined on hyperstruc-tures; in Greece, J.Mittas, D.Stratigopoulos, M.Konstantinidou, K.Serafimidis,Ch.G.Massouros have studied the canonical hypergroups, the hyper-rings andthe hyper-modules, the hyper-lattices, the hyper-fields with applications in Auto-mata Theory. T.Vougiouklis has analyzed especially the cyclic hypergroups, theP-hyperstructures and the representation of the Hϑ-structures. Significant con-tributions in the study of the regular hypergroups, the complete hypergroups, ofthe heart and of the homomorphisms of hypergroups in general or with applica-tions in Combinatorics and Geometry have been brought by the group of Italianmathematicians represented by P.Corsini, M.De Salvo, R.Migliorato, F.De Maria,G.Romeo, P.Bonansinga, D.Freni.

Beginning with the ’90s, Hyperstructures Theory represents a constant con-cern also for the Romanian mathematicians, the decisive moment being the 5thInternational Congress of Hyperstructures and Applications, organized in 1993, atthe University ”Al.I.Cuza” of Iasi. This domain of the modern algebra is a topic ofa great interest also for the Romanian researchers, who published a lot of paperson hyperstructures in national and international journals, gave communications inconferences and congresses, wrote PhD theses in this field. We quote the paperswritten by M.Stefanescu, I.Tofan, Gh.Radu, V.Leoreanu, C.Volf, M.Gontineac,C.Pelea, M. and C.Gutan, M. Tarnauceanu, I.Cristea.

Till now applications of the hypergroups in various domains, as: geometry,topology, analysis of the convex systems, finite groups’ character theory, crypto-graphy and codes theory, automata theory, graphs and hypergraphs, theory offuzzy and rough sets, theory of binary relations, probability theory, ethnology,

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economy, etc, have been founded.The first chapter of this thesis contains the presentation of some notions and

results in the hypergroup theory, necessary later on. It is now well-known thatthe hypergroup theory is a natural generalization of the group theory, applied ininformatics. In hypergroups, identities and the inverses, as well as the notions ofsubhypergroups, homomorphism of hypergroups, relations on hypergroups, factorhypergroup are defined naturally. The properties of the hyperproduct allow us toconsider different types of subhypergroups: closed, ultraclosed, invertible, com-plete parts. We recall the definition of the fundamental relation β, which helpsus to determine the heart of a hypergroup, the subhypergroup which offers manyinformation about the structure of a given hypergroup. Our presentation goes onby introducing some particular hypergroups, the studying their properties, theirheart and determining connections among these classes of hypergroups. So weconsider the regular hypergroups, the reversible hypergroups, the canonical, thecomplete ones, the join spaces.

In the end of the first chapter we recall some results from the join spacestheory, established by Prenowitz and Jantosciak in [43], used prioritarily in thesecond chapter, where we show how to reconstruct, from an algebraic pointof view, three branches of the geometry: the projective, the descriptive and thespherical one. Each of them can be characterized by a set S of points and a ternaryrelation (abc), read ”b is between a and c”, which verifies some axioms. Using, ifnecessarily, an ideal point e /∈ S, Prenowitz and Jantosciak have defined on theset S, in the descriptive geometry, and respectively on the new set S ′ = S∪{e}, inthe other two geometries, a join hyperoperation (Theorem 2.2 and 2.3, Theorem2.5 and 2.6, Theorem 2.13), succeeding in characterizing the given geometries asparticular join spaces.

In the Chapter 3 we describe three types of connections between the hyper-groups and the fuzzy sets, which lead us to obtain new join spaces. The firstrelationship, considered by Rosenfeld in 1971, has been extended by Ameri andZahedi [1] in 1997. Given a group G and a fuzzy subset µ of it, one defines thehyperoperation induced by µ:

◦µ : G×G −→ P∗(G), a ◦µ b = µaµb.

The Theorem 3.1.6 determines conditions under which the new hyperstructure isa hypergroup, a canonical hypergroup or a join space.

In the subsection 3.2 some results obtained by Corsini and Leoreanu in [9],[10], [11], [13], [14] about the join space obtained when one endows a nonemptyset H with the following hyperproduct:

x ◦ y = {z ∈ H | min{µ(x), µ(y)} ≤ µ(z) ≤ max{µ(x), µ(y)}}

are recalled; here µ is a fuzzy subset of H. Necessary and sufficient conditionsthat two such join spaces associated with two different fuzzy subsets defined onthe same universe H are isomorphic are presented (Theorem 3.2.3 and Theorem3.2.4).

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The third connection between the hypergroups and the fuzzy sets has been de-fined by Corsini in [17] and studied then by P.Corsini, M.Stefanescu, V.Leoreanu,I.Cristea. With a hypergroupoid 〈H, ◦〉, one associates the fuzzy set µ by therelations (ω) from § 3.3.1 and then the join space 〈1H, ◦1〉, as in the previousassociation; using again the construction (ω) for the new hypergroupoid 〈1H, ◦1〉,one determines the fuzzy set µ1 and the corresponding join space 〈2H, ◦2〉 andso on. The process stops when one obtains two successively equal or isomorphicjoin spaces. In this way one constructs a sequence of join spaces and fuzzy sets(〈iH, ◦i〉, µi)i≥1 associated with H; its length is called the fuzzy grade of the hy-pergroupoid H.For any i ≥ 1, there exist r ∈ IN∗ and a partition π = {iCj}r

j=1 of iH such thatx, y ∈ iCj ⇐⇒ µi−1(x) = µi−1(y). Denoting kjl

=| iCjl|, 1 ≤ l ≤ r, with every join

space iH one associates an ordered r-tuple (kj1 , kj2 , ..., kjr). Particularly, with anyhypergroupoid H we associate an ordered r-tuple (k1, k2, ..., kr). Studying cer-tain types of ordered r-tuples, we obtain some properties of the fuzzy grade of ahypergroupoid (see § 3.3.2.).

If with the finite hypergroupoid H = {a1, a2, ..., an} we associate the n-tuple(1, 1, ..., 1), we can easily construct, depending on the parity of n and withoutcalculating the effective values µ1(ai) using the relations (ω), the associated joinspace 1H. Moreover, if n + 1 = 2s, we prove that s.f.g.(H) = s + 1 (see theRemark 3.3.11 ). An immediate consequence of this theorem is given by theCorollary 3.3.14: Given a natural number n ∈ IN∗ \ {1} one can construct ajoin space H with the fuzzy grade equal to n. The Theorem 3.3.15 expresses asufficient condition, but not a necessary one, that two successively join spacesof the sequence are not isomorphic: if with the join space iH we associate ther-tuple (k1, k2, ..., kr) with the property (k1, k2, ..., kr) = (kr, kr−1, ..., k1), then iHand i+1H are not isomorphic.

One of the class of hypergroups studied in the following is that of the finitecomplete hypergroups, that is the one of the hypergroups H which can be repre-

sented as H =⋃g∈G

Ag, where

1) (G, ·) is a group;

2) for any (g1, g2) ∈ G2, g1 6= g2, we have Ag1 ∩ Ag2 = ∅;

3) if (a, b) ∈ Ag1 × Ag2 , then a ◦ b = Ag1g2 .

We have shown that, for a complete hypergroup H of order n, its fuzzy gradedepends only on the m-decomposition of n. The Theorem 3.3.20 determines allthe complete hypergroups of order smaller than 7 and their fuzzy grade. Moreover,any complete hypergroup of order n is characterized by a m-tuple [k1, k2, ..., km],where m = |G|, 2 ≤ m ≤ n− 1, G = {g1, g2, ..., gm}, and for any i ∈ {1, 2, ..., m},ki = |Agi

|.If H is a complete hypergroup of the type [p, p, ..., p︸ ︷︷ ︸

k times

, kp], where n =|H|= 2kp,

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then s.f.g.(H)=2 (Proposition 3.3.21); if it is of the type [p, p, ..., p︸ ︷︷ ︸s times

, k, k, ..., k︸ ︷︷ ︸t times

, ps],

with 2 ≤ p < k < ps, n = |H| = 2ps + kt, then for n = 4ps, s.f.g.(H) = 3,and for kt 6= 2ps, f.g.(H) = 2 (Proposition 3.3.22); more general, if H is ofthe type [k, k, ..., k︸ ︷︷ ︸

l times

, p, p, ..., p︸ ︷︷ ︸s times

, s, s, ..., s︸ ︷︷ ︸p times

, l, l, ..., l︸ ︷︷ ︸k times

], with 2 ≤ k < p < s < l, n =

|H| = 2(ps + kl), we obtain that, for kl = ps, s.f.g.(H) = 3 and for kl 6= ps,f.g.(H) = 2 (Remark 3.3.25).

Next we gave an example of a non-complete 1-hypergroup of order n and wecalculated its fuzzy grade depending on the value of n.

Another class of hypergroups is that of the hypergroups with partial scalaridentities ( i.p.s.hypergroups). We determine the fuzzy grade of all the i.p.s.hy-pergroups of order smaller than 8 (Theorem 3.3.26).

Given an equivalence relation R on a universe H, one defines a rough set(R(A), R(A)), where A is a nonempty subset of H, and

R(A) =⋃

R(x)⊂A

R(x) and R(A) =⋃

R(x)∩A6=∅R(x).

It is well known that every rough set is a particular fuzzy set, but converselynot ([2]).

Defining on H the hyperoperation:

∀(x, y) ∈ H2, x ◦ y = R({x, y}) \R({x, y}),the hypergroupoid 〈H, ◦〉 is a join space if and only if, for any x ∈ H, |R(x)| ≥ 3.Attaching with this join space the fuzzy set µ defined by the relations (ω), we haveobtained that, for any two distinct elements u, v ∈ H, the following implicationshold:

(i) µ(u) = µ(v) ⇐⇒ |R(u)|=|R(v)|;(ii) µ(u) < µ(v) ⇐⇒ |R(u)|>|R(v)|.

If R is an equivalence relation on H such that | R(x) |= k = const, for anyx ∈ H, then s.f.g.(H) = 1.

In the last part of the thesis we give some properties of the join spaces iHassociated with a hypergroupoid H, in connection with the fundamental relationsdefined on H. Besides the fundamental relation β defined on a hypergroupoid H,there exist another three types of equivalence relations, called also fundamental byJantosciak in [29]: operational equivalence, inseparability and essential indistin-guishability. By using them, J.Jantosciak has introduced the notion of reducedhypergroup. We show that, for any hypergroupoid H and any i ≥ 1, the quotienthypergroup iH/Reµi

is reduced (Theorem 3.3.33). This property is not valid forthe associated join spaces iH, more clearly, the join space 1H is the unique joinspace from the sequence (〈iH, ◦i〉, µi)i≥1 corresponding to H which can be a re-duced hypergroup, this happens if and only if the n-tuple associated with H is ofthe form (1, 1, ..., 1).

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In the last section we find some properties of the three fundamental relationsand we give examples of remarkable hypergroups which are reduced or not.

In conclusion, this thesis is focused on the three types of connections esta-blished till now between the hypergroups and the fuzzy sets, studying especiallythe sequence of join spaces obtained with this method. We consider some openresearch directions and we present some problems, a few of them remaining stillnon-resolved; they are enumerated in the Chapter 4 of the thesis.

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Acknowledgements

Besides the efforts I put in writing up this thesis, many other people are involvedin this project and have contributed to its success in various direct and indirectways. I appreciate their help and I am glad I had the opportunity to learn fromthem and work with them.

First of all, I would like to express my deep and sincere gratitude to ProfessorMirela Stefanescu, for her helpful suggestions and discussions along the years ofmy PhD courses. She encourage me all the time and she took care that I stay onthe right track.

I am deeply grateful to Professor Piergiulio Corsini, for his detailed and con-structive comments, for his important support throughout this work, during mystaying to Udine’s University. He is the person who introduced me to the field ofhyperstructures and in the world of research.

I wish to express my thanks to Professor Ioan Tofan who encouraged me tostart the PhD programme and who gave me the opportunity to earn a scholarshipfor six months to the University of Udine. That was the moment when I startedto understand what means to do research.

My sincere thanks are due to Professor Violeta Leoreanu-Fotea, for her advicesand the discussions with her, so important for me in that difficult moments.

I am grateful to Mrs. Elena Mocanu, for helping me to study the LaTexprogramme.

I owe my loving thanks for my parents, my husband Renato and my step-parents; without their understanding and support it would have been impossiblefor me to finish this work.

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Contents

1 Fundamental definitions and results on Hypergroups 3

2 Geometries and Join spaces 11

3 Hypergroups and fuzzy sets 193.1 The fuzzy subgroup of a group . . . . . . . . . . . . . . . . . . . . 193.2 The classic association between fuzzy sets and hypergroups . . . . 223.3 The sequence of join spaces associated with a hypergroupoid . . . 31

3.3.1 The construction of the sequence of join spaces and fuzzy sets 313.3.2 Properties of the membership function µi . . . . . . . . . . 333.3.3 Fuzzy grade of the hypergroups . . . . . . . . . . . . . . . 353.3.4 Fuzzy grade of the complete hypergroups . . . . . . . . . . 493.3.5 An example of a non-complete 1-hypergroup . . . . . . . . 523.3.6 Fuzzy grade of i.p.s. hypergroups . . . . . . . . . . . . . . 733.3.7 The sequence of join spaces associated with a rough set . . 743.3.8 Properties of the join spaces iH associated with the hyper-

groupoid H . . . . . . . . . . . . . . . . . . . . . . . . . . 773.4 Other examples of reduced hypergroups . . . . . . . . . . . . . . . 79

4 Concluding remarks and future work 85

A The sequence of fuzzy sets and of join spaces determined by allthe complete hypergroups of order less than or equal to 6 87

B The sequence of fuzzy sets and of join spaces determined by allthe i.p.s. hypergroups of order less than or equal to 7 97

Bibliography 127

1

2

Chapter 1

Fundamental definitions andresults on Hypergroups

In this chapter are presented notions and results obtained on hypergroups (see[4]) which will be used during this thesis. More exactly, we define the notionof hypergroup, subhypergroup, the heart of a hypergroup, the homomorphismof hypergroups; we introduce different types of relations on hypergroups, diffe-rent types of hypergroups, indicating the connections between them; finally arepresented some results from the theory of join spaces, useful in the second chapter.

LetH be a nonempty set and P∗(H) the set of all nonempty subsets ofH.

Definition 1.1. A set H endowed with a hyperoperation ◦ : H2 −→ P∗(H) iscalled a hypergroupoid . The image of the pair (a, b) ∈ H2 is denoted by a ◦ b andcalled the hyperproduct of a and b.

If A and B are nonempty subsets of H, then A ◦B =⋃a∈Ab∈B

a ◦ b.

Definition 1.2. The hypergroupoid 〈H, ◦〉 is called semihypergroup if the hype-roperation ” ◦ ” is associative. If H satisfies the reproducibility law :

∀ a ∈ H, H ◦ a = a ◦H = H,

then H is called quasihypergroup.A hypergroup is a semihypergroup which is also a quasihypergroup.

Definition 1.3. Let 〈H, ◦〉 be a hypergroupoid .

(i) An element e ∈ H is called an identity if, for all x ∈ H, x ∈ x ◦ e ∩ e ◦ x.An identity e is called scalar identity if, for all x ∈ H, x ◦ e = e ◦ x = x.An identity e is called partial identity if, for any x ∈ H, x ∈ x◦e or x ∈ e◦x.

(ii) An element x′ ∈ H is called an inverse of x ∈ H if there is an identitye ∈ H, such that e ∈ x ◦ x′ ∩ x′ ◦ x.

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Notation. For a, b ∈ H, we denote:a/b = {x | a ∈ x ◦ b} and b\a = {y | a ∈ b ◦ y}.If A and B are subsets of a hypergroup H, we denote A/B =

⋃a∈Ab∈B

a/b.

The homomorpfisms of hypergroups represent an important domain of thetheory of hypergroups; they have been studied from D.Ore, M.Dresher, M.Krasner,M.Koskas and later from P.Corsini, S.D.Comer, J.Jantosciak, F.Rossi, P.Bonansigna,V.Leoreanu.

Definition 1.4. Let 〈H, ◦〉 and 〈K, ?〉 be hypergroupoids and f : H −→ K anapplication from H in K. We say:

(i) f is a homomorphism if for all (a, b) ∈ H2, f(a ◦ b) ⊂ f(a) ? f(b);

(ii) f is a good homomorphism if for all (a, b) ∈ H2, f(a ◦ b) = f(a) ? f(b).

The equivalence relations defined on a hypergroupoid H form a significantpart of theory of hypergroups. We present some results in this direction, insistingon the fundamental relation β, with which is defined the heart of a hypergroup,notion studied by P.Corsini, A.Orsatti and especially by Leoreanu in [35], whereshe determined the structure of the heart of a join space.

Definition 1.5. Let 〈H, ◦〉 be a hypergroupoid and let ρ be an equivalencerelation on H.

(i) We say that ρ is regular to the right if the following implication holds:

aρb =⇒ ∀ u ∈ H, ∀x ∈ a ◦ u, ∃y ∈ b ◦ u : xρy and∀ y ∈ b ◦ u,∃x ∈ a ◦ u : xρy.

Similarly, the regularity to the left can be defined.We say that ρ is regular if it is regular to the right and to the left.

(ii) We say that ρ is strongly regular to the right if the following implicationholds:

aρb =⇒ ∀ u ∈ H, ∀x ∈ a ◦ u,∀ y ∈ b ◦ u : xρy.

Similarly, the strong regularity to the left can be defined.We say that ρ is strongly regular if it is strongly regular to the right and tothe left.

Definition 1.6. Let 〈H, ◦〉 be a hypergroupoid. We define the relation β on Has it follows:

aβb ⇐⇒ ∃n ∈ IN∗,∃(x1, x2, ..., xn) ∈ Hn : a ∈n∏

i=1

xi 3 b.

Notice that β is a reflexive and a symmetric relation on H, but generally, nota transitive one. Let us denote by β∗ the transitive closure of β.

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Theorem 1.7. If 〈H, ◦〉 is a hypergroupoid, then β∗ is the smallest strongly regularequivalence on H, with respect to the inclusion. Moreover, if H is a hypergroup,then β∗ = β.

Definition 1.8. For any equivalence relation ρ defined on a hypergroupoid Hand for any element x ∈ H, we define the equivalence class of x related to ρ byx = {y ∈ H | xρy}; we denote by H/ρ the set of all equivalence classes of elementsof H, called the factor set.

Theorem 1.9. Let 〈H, ◦〉 be a semihypergroup and ρ an equivalence relation onH.

(i) If ρ is regular, then H/ρ is a semihypergroup, with respect to the followinghyperoperation:

∀ (x, y) ∈ (H/ρ)2, x⊗ y = {z | z ∈ x ◦ y}.

(ii) Conversely, if the hyperoperation ”⊗” is well-defined on H/ρ, then ρ isregular.

Theorem 1.10. Let 〈H, ◦〉 be a hypergroup and ρ a strongly regular equivalencerelation on H. Then H/ρ is a group.In particular, H/β is a group.

Definition 1.11. Let H be a hypergroup, 1 the identity element of the groupH/β and ϕH :−→ H/β the canonical projection: for any x ∈ H, ϕH(x) = x.The heart of a hypergroup H is the set ωH ={x ∈H | ϕH(x)= 1}.

The complete parts, introduced and studied for the first time by M.Koskas,further have been analyzed by P.Corsini and Y.Sureau, especially in the generalcontext of the theory, but M.De Salvo has studied them from a combinatorialpoint of view.

Definition 1.12. Let 〈H, ◦〉 be a semihypergroup and A a non empty subset ofH. We say that A is a complete part of H if

∀n ∈ IN∗, ∀ (x1, ..., xn) ∈ Hn,

n∏i=1

xi ∩ A 6= ∅ =⇒n∏

i=1

xi ⊂ A.

Proposition 1.13. Any intersection of complete parts of a hypergroup H is acomplete part of H.

Definition 1.14. If 〈H, ◦〉 is a semihypergroup and A ⊂ H, A 6= ∅, then thecomplete closure of A in H is the intersection of all the complete parts of H,which contain A. It will be denoted by C(A).

Theorem 1.15. Let 〈H, ◦〉 be a semihypergroup.

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(i) The relation ρ defined by aρb ⇐⇒ x ∈ C({y}) is an equivalence relation.

(ii) For any (a, b) ∈ H2 we have aρb ⇐⇒ aβ∗b

(iii) If A is a non empty subset of 〈H, ◦〉 , then C(A) =⋃a∈A

C(a).

Definition 1.16. A semihypergroup 〈H, ◦〉 is called complete if, for any (x, y) ∈H2, C(x ◦ y) = x ◦ y.

Theorem 1.17. A hypergroup H is complete if H =⋃g∈G

Ag, where

1) (G, ·) is a group;

2) for any (g1, g2) ∈ G2, g1 6= g2, we have Ag1 ∩ Ag2 = ∅;3) if (a, b) ∈ Ag1 × Ag2, then a ◦ b = Ag1g2 .

Definition 1.18. Let 〈H, ◦〉 be a hypergroupoid and K a non empty subset ofH. K is called a subhypergroupoid of H if K ◦K ⊂ K. A subhypergroupoid K ofH is called a subhypergroup of H if 〈K, ◦〉 is a hypergroup.

Definition 1.19. Let 〈H, ◦〉 be a hypergroup and K a subhypergroup of it. Wesay that:

(i) K is closed to the left in H if, for any a ∈ H and any (x, y) ∈ K2, fromx ∈ a ◦ y it follows a ∈ K.Similarly, we can define the notion of subhypergroup closed to the right.K is closed in H if it is closed to the left and to the right.

(ii) K is ultraclosed to the left in H if, for any x ∈ H,K ◦ x ∩ (H\K) ◦ x 6=∅.Similarly, we can define the notion of subhypergroup ultraclosed to the right.K is ultraclosed in H if it is ultraclosed to the left and to the right.

The regular hypergroups have been introduced by Dresher and Ore in 1938(see [25]), later they have been investigated by P.Corsini and A.Orsatti, whichdetermined the heart of a regular reversible hypergroup.

Definition 1.20. A hypergroup H is regular if it has at least one identity and eachelement has at least one inverse. A regular hypergroup 〈H, ◦〉 is called reversibleif, for any (x, y, z) ∈ H3, it satisfies the following conditions:

1) if y ∈ a ◦ x, then there exists an inverse a′ of a, such that x ∈ a′ ◦ y;

2) if y ∈ x ◦ a, then there exists an inverse a′′ of a, such that x ∈ y ◦ a′′.

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Theorem 1.21.

(i) Any complete hypergroup is regular and reversible.

(ii) The heart of a complete hypergroup H is formed by all its identities.

In 1975, R.Roth utilized the canonical hypergroups to prove some theoremsfrom the theory of the characters of finite groups (see [49]). J.R.McMullen andJ.F.Price used generalizations of the canonical hypergroups in the harmonic ana-lysis and in the particle physics.

Definition 1.22. We say that a hypergroup H is canonical if

(i) it is commutative;

(ii) it has a scalar identity;

(iii) every element has a unique inverse;

(iv) it is reversible.

Definition 1.23. A canonical hypergroup 〈H, +〉 is called strongly canonical if itsatisfies the following conditions:

(i) ∀ (x, a) ∈ H2, x ∈ x + a =⇒ x = x + a;

(ii) (x + y) ∩ (z + w) 6= ∅ =⇒ x + y ⊂ z + w or z + w ⊂ x + y.

An i.p.s. hypergroup (a hypergroup with partial identities) 〈H, +〉 is a canonicalhypergroup such that

∀ (x, a) ∈ H2, x ∈ x + a =⇒ x = x + a.

The notion of join space has been introduced by W.Prenowitz and togetherwith J.Jantosciak he reconstructed, from the algebraic point of view, the projec-tive, the spherical and the descriptive geometry. In the second chapter of thethesis we deal with this subject.

Definition 1.24. A commutative hypergroup 〈H, ◦〉 is called a join space if, forany (a, b, c, d) ∈ H4, the following implication holds:

a/b ∩ c/d 6= ∅ =⇒ a ◦ d ∩ b ◦ c 6= ∅.

If A and B are subsets of a hypergroup H, we denote A/B =⋃a∈Ab∈B

a/b.

Theorem 1.25.

(i) A commutative hypergroup is canonical if and only if it is a join space witha scalar identity.

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(ii) Any complete commutative hypergroup is a join space.

Examples of hypergroups:

1) Let G be a group; for any (x, y) ∈ G2, we define x ◦ y = 〈x, y〉 the subgroupof G generated by x and y. Then 〈G; ◦〉 is a hypergroup.

2) Let A be a non-empty set and for any x, y ∈ A, x ◦ y = A. The hypergroup〈A; ◦〉 is called the total hypergroup on A.

3) Let G be a group and H be a normal subgroup of G. We define on G thehyperoperation x ◦ y = Hxy and we obtain that 〈H, ◦〉 is a hypergroup.

4) Let 〈G, +〉 be a commutative group, α an element of second order of G. Forany x, y ∈ G we define x ◦ y = {x + y, x + y + α}. We obtain that 〈G; ◦〉 isa complete hypergroup.

5) Set Hn = A⋃

B⋃{e}, where |A| ≥ 2 ≤ |B|, A ∩B = ∅, e /∈ A

⋃B. We fix

an element b1 ∈ B and we define the hyperoperation in Hn: ∀a ∈ A, a ◦ a =b1; ∀(a1, a2) ∈ A2, a1 6= a2, a1 ◦ a2 = B; ∀(a, b) ∈ A × B, a ◦ b = b ◦ a = e;∀(b, b′) ∈ B2, b ◦ b′ = A; ∀a ∈ A, a ◦ e = e ◦ a = A; ∀b ∈ B, b ◦ e = e ◦ b = B;e ◦ e = e. Hn is a non complete 1-hypergroup.

6) Any commutative hypergroup endowed with an identity is regular.

Further on we present some results from the theory of join spaces (see [43]).

Proposition 1.26. Any intersection of closed subhypergroups of a hypergroup His a closed subhypergroup of H.

Definition 1.27. The closed subhypergroup generated by N , denoted 〈N〉, is thesmallest closed subhypergroup of H containing N , that is the intersection of allclosed subhypergroups of H containing N .

One introduces a relation of congruence with respect to a closed subhypergroupN of H in order to construct a new join space, the factor space of H and N .

Definition 1.28. Let N be a nonempty closed subhypergroup of a join spaceH. Let a, b ∈ H. Then a ≡ b(modN), read a is congruent to b modulo N , ifaN ∩ bN 6= ∅. The relation ≡ (modN) is an equivalence relation in H and (a)N

denotes the equivalence class of a inH, the coset of N determined by a.The relation ≡ (modN) can be extended to a relation on the subsets of H as itfollows: A ≡ B(modN) means that for each a ∈ A there exists b ∈ B such thata ≡ b(modN) and conversely.

On the factor set H :N = {(x)N | x ∈ H} formed by all cosets of N determinedby elements of H, we define the join of the cosets (a)N , (b)N by (a)N ¯ (b)N ={(x)N | x ∈ a ◦ b}. Then (H : N,¯) is a join space with identity N , called thefactor space H modulo N . The order of H:N is the cardinality of the set H:N .

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Definition 1.29. If A and B are closed subhypergroups of the join space 〈H, ◦〉such that B ⊂ A and the order of A : B is 3, then we say that B separates A.

Definition 1.30. If the join space H has a scalar identity e, we set E = {e},otherwise E = ∅. We define C∅B = E and if A ∈ P∗(H),CAB = 〈A〉.

A join space H is called an exchange space if it satisfies the following conditions:

(I) if a ∈ CbB, a /∈ E, then CaB = CbB;

(II) if c ∈ Ca, bB and c /∈ CbB, then Cc, bB = Ca, bB.

Theorem 1.31. Let H be a join space with a scalar identity e. Then H satisfies(I) if and only if satisfies (II).

Definition 1.32. Let B be a subset of a join space H. B is called independent iffor any b ∈ B we have b /∈ 〈B \ {b}〉.

Definition 1.33. A subset B of a closed subhypergroup N of an exchange spaceH is called a basis of N if it is independent and furthermore 〈B〉 = N.

Any closed subhypergroup of an exchange space has a basis.Any two bases of N have the same cardinal number called the dimension of

N , denoted by d(N).If N and M are finite dimensional closed subhypergroups of 〈H, ◦〉 such that

N ∩M 6= ∅ then the dimensional equality holds:

d(〈N,M〉) + d(N ∩M) = d(N) + d(M).

If N covers M , then d(N) = d(M) + 1.If d(N)=n is a finite number, any independent set ofn elements ofN represents

a basis ofN.

9

10

Chapter 2

Geometries and Join spaces

One of the major purpose of the hypergroups theory is that to construct newhyperstructures and to find necessary and sufficient conditions for them to becomejoin spaces. A such construction is studied in this thesis and, for this reason, inthe Chapter 2 we present the context where the concept of join space appeared.

The notion of join space has been introduced and studied for the first timeby W.Prenowitz. Later, together with J.Jantosciak, he has reconstructed, fromthe algebraic point of view, several branches of geometry: the projective, thedescriptive and the spherical geometry (see [28], [43], [44], [45], [46]). All thesethree geometries are join theoretic in character, in the sense that a central role isplayed by an operation ”join”, which assigns to two distinct points an appropriateconnective: in descriptive geometry, a segment; in spherical geometry, minor arc ofgreat circle; in projective geometry, a line. Although the three treatments employsimilar methods and contain many similar results, the formulations do not seemto permit a uniform development of the three theories.

For the detailed proofs of the following theorems see [44], [45], [46].

1. The projective geometry is a non-Euclidian geometry which has been de-veloped at he beginning of the 19th century, based on a fundamental principle:the parallel lines meet to the infinite. It is the geometry of the properties whichremain invariant on the projections.

Definition 2.1. A projective geometry is a system (S, T ) formed by a set S ofpoints and a set T of lines which verifies the following postulates:

(P1) A line is a set of points and contains at least three points.

(P2) Two distinct points a, b are contained in a unique line, denoted L(ab).

(P3) If a, b, c, d are distinct and L(ab)⋂

L(cd) 6= ∅, then L(ac)⋂

L(bd) 6= ∅.Given a projective geometry (S, T ), we can associate to it a join space. One

constructs a combinatory system (S ′, ·) in the following way.Let S ′ = S

⋃{e}, where e /∈ S, called an ideal point.First, we suppose T 6= ∅. A join operation ” · ” is defined in S ′ as it follows:

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1) If a, b ∈ S, a 6= b, then a · b = L(ab) \ {a, b}.2) Let a ∈ S. If some line of T contains exactly three points, then a · a = e;

otherwise a · a = {a, e}.3) If a ∈ S ′, then e · a = a · e = a.

For T = ∅ and S = {a}, it can be possible to define two operations on S ′ (forboth of them e is an identity): a · a = {e} or a · a = {e, a}.Finally, if T = ∅ and S = ∅ we define e · e = e.

Note that (S ′, ·) has the properties:

(i) e ∈ a · b if and only if a = b

(ii) a/b = a · b.Theorem 2.2. The system (S ′, ·) is a join space with identity e for which theclosed subhypergroup 〈a〉, a 6= e, generated by a has the cardinal number equal to2. (S ′, ·) is called the associated join space of the projective geometry (S, T ).

Proof. We have to verify that (S ′, ·) has the properties of a join space:

(JS1) a · b 6= ∅(JS2) a · b = b · a(JS3) (a · b) · c = a · (b · c)(JS4) a/b

⋂c/d 6= ∅

(JS5) a/b 6= ∅.It is obviously that (S ′, ·) satisfies (JS1), (JS2), (JS5).We proceed to verify (JS4). Suppose in S ′, a/b

⋂c/d 6= ∅. By the previous

property (ii), there exists p ∈ a · b ⋂c · d. We have to consider the situations:

a) the points a, b, c, d ∈ S are distinct and non collinear;

b) the points a, b, c, d ∈ S are distinct and collinear;

c) the points a, b, c, d ∈ S are non distinct;

d) one of the points a, b, c, d is the ideal point e.

We show the case a), the other possibilities are treated similarly. We have p ∈a · b ⋂

c · d ⇐⇒ p ∈ (L(ab) \ {a, b}) ⋂(L(cd) \ {c, d}), so p ∈ L(ab)

⋂L(cd). By

(P3), L(ad)⋂

L(bc) 6= ∅. Therefore (a · d ⋃{a, d}) ⋂(b · c ⋃{b, c}) 6= ∅.

Suppose b ∈ a · d ⇐⇒ d ∈ a/b, that is p 6= d ∈ L(ab)⋂

L(cd). By (P2), L(ab) =L(cd) contrary to hypothesis. Thus b /∈ a ·d. Similarly c /∈ a ·d, a /∈ b · c, d /∈ b · c.Hence the only possibility is a · d ⋂

b · c 6= ∅ and (JS4) holds.

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Now we verify (JS3). Let x ∈ (a · b) · c, with a, b, c ∈ S ′. Then x/c⋂

a · b 6= ∅,x/c

⋂b/a 6= ∅. By (JS4), x · a⋂

b · c 6= ∅, x/a⋂

b · c 6= ∅, then x ∈ a · (b · c),therefore (a · b) · c ⊂ a · (b · c). The reverse inclusion can be proved similarly.We may assert that (S ′, ·) is a join space. Clearly e is an identity of (S ′, ·).Since {a, e} is a closed subhypergroup and {a, e} ⊂ 〈a〉, it follows 〈a〉 = {a, e}.Thus 〈a〉 has the cardinal number equal to 2, if a 6= e.

Theorem 2.3. Let 〈J, ·〉 be a join space with identity e in which, for any a ∈J \{e}, the closed subhypergroup 〈a〉 generated by a has the cardinal number equalto 2. Then 〈J, ·〉 is an associated join space of a projective geometry.

Proof. Let S = J \ {e} and let L(ab) = 〈a, b〉 \ {e}, if a, b ∈ S, a 6= b. EachL(ab) is a line and the set of all L(ab) is denoted by T . One proves that (S, T )is a projective geometry, that is, it verifies the axioms (P1)-(P3) and finally that〈J, ·〉 is an associated join space. For this, since J = S

⋃{e}, we take e as theideal point to be adjoined to S and S ′ = J.If J = {e}, then S = ∅, T = ∅ and 〈J, ·〉 is an associated join space of (S, T ), bydefinition.If J = {a, e}, a 6= e, then S = {a}, T = ∅. We have a · e = e · a = a, e · e = e,e ∈ a · a. Thus a · a = e or a · a = {a, e} and, in the both cases, 〈J, ·〉 is anassociated join space of (S, T ), by definition.Suppose the cardinal number of J is at least 3. Then S has at least two elementsand T 6= ∅. Thus the join operation in S ′, denoted ” ◦ ”, is defined by:

(i) if a, b ∈ S, a 6= b, then a ◦ b = L(ab) \ {a, b};(ii) let a ∈ S; if some line of T contains exactly three points, then a ◦ a = e;

otherwise a ◦ a = {a, e};(ii) if a ∈ S ′, e ◦ a = a ◦ e = a.

To complete the proof, one shows that the hyperoperations ” · ” and ” ◦ ” areidentical.

The previous two theorems characterize projective geometries in terms of joinspaces in this way:A join space 〈J, ·〉 is an associated join space of a projective geometry if and onlyif it has an identity e and for any a ∈ J \ {e}, the closed subhypergroup 〈a〉 gene-rated by a has the cardinal number equal to 2.

2. The spherical geometry is the geometry of the two-dimensional surface ofa sphere. It is an example of a non-Euclidian geometry, the simplest model for anelliptic geometry, where, given a line L and a point p outside L, there exists noline parallel to L passing through p.

Definition 2.4. A spherical geometry is a system (S, (abc)) involving a set S ofpoints and a ternary betweenness relation (abc), which may be read b lies betweena and c, which satisfies the following postulates.

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(S1) If (abc) then a, b, c are distinct.

(S2) If (abc) then (cba).

(S3) For each a there exists a unique a′ (called the opposite of a) such that a′ 6= aand (axy) always implies (xya′).

(S4) If b 6= a, b 6= a′, there exists x such that (axb).

If b 6= a, b 6= a′, we define a · b = {x | (axb)}, a · a = a. The hyperoperation ” · ”is a partial join operation in S; it is extended to subsets A,B of S in the familiar

way A ·B =⋃a∈Ab∈B

a · b.

(S5) (a · b) · c = a · (b · c) provided both terms are defined.

Example. If we take S to be a Euclidean n-sphere and (abc) to mean b is aninterior point of the minor arc of a great circle which joins a and c, then theaxioms (S1)-(S5) are satisfied.

It remains the problem how to extend the definition of join to pairs of oppositepoints a, a′ in such a way as to conserve the associative law. W.Prenowitz hasproved that it is impossible to do this except in trivial cases (see [46]). Thus, withtrivial exceptions, it is impossible to convert a spherical geometry (S, (abc)) intoa join space (S, ·). However, the conversion can easily be made by adjoining to San ”ideal point” which will play the role of an identity.

Let S ′ = S⋃{e}, e /∈ S. The hyperoperation ” · ” is extended to S ′ as follows

a · a′ = {a, a′, e}, ∀ a ∈ S, a · e = e · a = a, ∀ a ∈ S ′.Thus, in S ′ take place the commutative and the associative laws.

Theorem 2.5. If S is a spherical geometry then S ′ is a join space with theidentity element e, satisfies the idempotent law aa = a and has the property thatthe closed subhypergroup 〈a〉 generated by a 6= e has the cardinal number equal to3.S ′ is called the associated join space of the spherical geometry S.

Proof. The axiom (S4) implies the axiom (JS1): a · b 6= ∅. The axioms (JS2) and(JS3) are evident. We show (S ′, ·) is a join space by verifying (JS4) and (JS5).Suppose a, b, c, d ∈ S ′ such that a/b

⋂c/d 6= ∅.

Using that a/b = a · b′, we have p ∈ a/b = a · b′, p ∈ c/d = c · d′ and thusa ∈ p/b′ = p · b = b · p ∈ b · (c · d′) = (b · c) · d′. Then a/d′

⋂b · c 6= ∅, so that

a · d ⋂b · c 6= ∅.

(JS5) is immediate from a/b = a · b′. Then one proves that 〈a〉 = {a, a′, e}.Clearly, a/a = {a, a′, e} is a closed subhypergroup contained by 〈a〉 and it is theleast closed subhypergroup which contains a, therefore 〈a〉 = {a, a′, e}.

Theorem 2.6. Let G be an idempotent join space with the identity e (that is, anidempotent regular hypergroup) with the property that for each a ∈ G \ {e}, the

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closed subhypergroup 〈a〉generated by a has the cardinal number equal to 3. ThenG is the join space associated to a spherical geometry.

Proof. Let S = G \ {e} and we define (abc) to mean b ∈ a · c, where c /∈ {a, a′}(in a regular hypergroup each element has an opposite). We show that S, withthe order so defined, is a spherical geometry and that its associated join space S ′

coincides with G. First we show for a 6= e, that a · a′ = {a, a′, e}.Adding a to the both members of the relation e ∈ a · a′ we havea = a · e ∈ a · (a · a′) = (a · a) · a′ = a · a′ and similarlya′ = e · a′ ∈ (a · a′) · a′ = a · (a′ · a′) = a · a′. Since card〈a〉 = 3, it resultsa · a′ = {a, a′, e}.To prove (S1), let us consider (abc), so that b ∈ a · c, c 6= a, c 6= a′. Supposea = b then a ∈ a · c = c · a and c ∈ a/a ⊂ 〈a〉 = {a, a′, e} false. Similarly for b = c;so a, b, c are distinct.(S2) follows from the commutative law of a join space.(S3) can be verified by taking p = a′ = e/a.(S4) follows from a · b 6= ∅, so that there exists x ∈ S ′ such that x ∈ a · b. Thus(axb).To verify (S5) consider the operation ” ¯ ” defined in S by a ¯ a = a anda¯ b = {x | (axb)} for b 6= a, b 6= a′. We see immediately that a¯ b = a · b, fora, b ∈ S provided b 6= a′. Thus, since ” · ” is associative, the associative law for”¯ ” certainly holds for those triples a, b, c ∈ S for which it is significant. Hence(S5) is verified and S is a spherical geometry.Now we prove that S ′ = S

⋃{e} is the associated join space of S.We extend ”¯ ” to S ′ as it follows:a¯a′ = {a, a′, e},∀ a ∈ S, a¯e = e¯a = a,∀ a ∈ S ′. As we have already proved,a · b = a¯ b, ∀ a, b ∈ S and therefore S ′ = G.

3. The descriptive geometry, introduced by Gaspard Monge in 1975, is abranch of the geometry which deals with the study of the two-dimensional repre-sentation of the three-dimensional objects.

Definition 2.7. A descriptive geometry is a system (S, (abc)), where S is a set ofelements called points and (abc) a ternary relation in S called betweenness, whichsatisfies the following postulates:

(D1) If a, b, c are points and (abc) then a, b, c are distinct.

(D2) If a, b, c are points and (abc) then (bca) is false.

If a, b (a 6= b) are points, the set consisting of a, b and all points x for which (xab)or (axb) or (abx) is called the line ab. The set of points x for which (axb) is calledthe segment ab. The set of points x satisfying (xab) is called a ray and it is saidto emanate from a.

(D3) If c, d (c 6= d) are points of the line ab then a is a point of the line cd.

(D4) If a, b (a 6= b) are points there is at least one point c such that (abc).

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(D5) There exist three points not in the same line.

(D6) (Transversal postulate) If a, b, c are distinct points and a is not in the linebc and if d, e are points such that (bcd) and (cea) then there is a point f inthe line de such that (afb).

A join operation is defined in S as follows:if a 6= b, a · b = {x | (axb)} and a · a = a and the extension of a from b isa/b = {x | a ∈ x · b} and clearly these operations satisfy the properties:

(i) a · b 6= ∅.

(ii) a · b = b · a.

(iii) (a · b) · c = a · (b · c).

(iv) a/b 6= ∅.

(v) a/a = a = a · a.

(vi) a/b⋂

c/d 6= ∅ =⇒ a · d ⋂b · c 6= ∅.

The last property is a formulation of a triangle postulate employed by Peano:Segments which join two vertices of a triangle to respective point of their oppositesides intersect.

Thus (S, ·) is a join space called the associated join space of the descriptivegeometry (S, (abc)).

Notice that the line ab is the set a · b ⋃a/b

⋃b/a

⋃{a, b}.

Let us consider a join space 〈J, ·〉 for which ∀ a ∈ J, a · a = a/a = {a}.Define on J the following ternary relation: (abc) means a 6= c and b ∈ a · c.

Proposition 2.8.

(i) If (abc) then a · b ⋂b · c = ∅.

(ii) If (abc) then a, b, c are distinct.

(iii) If (abc) and (bcd) then (abd) and (acd).

(iv) If (abc) and (acd) then (abd) and (bcd).

(v) (abx) and (aby) imply the falsity of (xay) and (xby). Similarly, (axb)and(ayb) imply the falsity of (xay) and (xby).

The following theorem establishes a first connection between the conditions ofa join space and the postulates of a descriptive geometry.

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Theorem 2.9. The ternary relation in J satisfies the postulates (D1), (D2), (D4),(D6).

Proof. (D1) is the assertion (ii) of the previous proposition.By (iii), (abc) and (bca) imply (aba), a contradiction with (ii). Thus (D2) is valid.(D4) is a restatement of the fact that in a join space a/b 6= ∅.We verify (D6). Suppose a, b, c distinct, a does not belong to the line bc and(bcd), (cea). Then c ∈ b · d and e ∈ c · a. We obtain a · b ⋂

e/d 6= ∅.Let f ∈ a · b ⋂

e/d 6= ∅. Then (afb).If d 6= e, then f belongs to the line de.If d = e then (bcd) and (cea) imply (bca), so that a belongs to the line bc, contrarywith the hypothesis. Thus d 6= e and the proof is complete.

Remark 2.10. Since the direct sum of two join spaces is a join space, W.Prenowitzhas shown that the postulate (D3) is independent of the conditions of a join spacedefinition.

In the following, let 〈J, ·〉 be a join space, satisfying

(J0) ∀ a ∈ J, a · a = a/a = a.

We introduce three new postulates for J in order to characterize a descriptivegeometry.

(J1) If (a, b) ∈ J2, a 6= b, then 〈a, b〉 covers a.

This is a consequence of the postulate ”two points belong to a unique line”.

Remark 2.11.

(i) A join space satisfying (J1) is an exchange space.

(ii) (J1) is independent of conditions of join space definition and of condition(J0).

(J2) There exist A and B, closed hypergroups of (J, ·) such that B ⊂ A and thefactor space (A : B) has order 3.

It means that J contains two closed hypergroups A,B such that B separates A.(J2) is verified in a descriptive geometry, since we can take A to be a line and Bone of its points.

The postulate(J3) d(J) > 2means that J contains a set of three independent elements.

Proposition 2.12. For any (a, b) ∈ J2 we have

〈a, b〉 = a · b ∪ a/b ∪ b/a ∪ {a, b}.

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Theorem 2.13. Descriptive geometries are characterized as join spaces satisfying(J0), (J1), (J2), (J3).

Proof. Let (S, (abc) be a descriptive geometry. For any (x, y) ∈ S2 we definex · x = x and x · y = {t | (xty)}. Then 〈S, ·〉 is a join space, which verifies (J0),(J1), (J2), (J3), as we have seen before.

Conversely, if 〈J, ·〉 is a join space satisfying (J0), then (D1), (D2), (D4), (D6)hold, by the Theorem 2.9.It remains to show (J1), (J2), (J3) imply (D3) and (D5).

The line ab (where a 6= b) is the set a · b ⋃a/b

⋃b/a

⋃{a, b} which coincideswith 〈a, b〉. We verify that if a 6= b and {c, d} ⊂ 〈a, b〉,where c 6= d, then 〈a, b〉 =〈c, d〉 and this obviously implies (D3). Suppose c 6= a, then 〈a, b〉 ⊃ 〈c, d〉 3 a andby (J1) it follows 〈a, b〉 = 〈a, c〉. Thus d ∈ 〈a, c〉. Since c 6= d, we can similarlyshow 〈a, c〉 = 〈c, d〉 and then 〈a, b〉 = 〈c, d〉.

By (J3) there are three distinct elements a, b, c of J , which form an independentset. Suppose a, b, c are contained in the same line pq = 〈p, q〉. But then 〈a, b〉 =〈p, q〉 3 c, contrary to the independence of a, b, c. Therefore a, b, c are not in thesame line and (D5) is verified.

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Chapter 3

Hypergroups and fuzzy sets

In the last decades, many connections between hypergroups and fuzzy sets havebeen established in order to obtain new algebraic structures, connections studiedfrom a theoretical point of view, but also for their applications obtained in variousdomains: the graphs theory, the probabilistic spaces theory, for example.

The ”fuzzy” concepts derive from the various fuzzy phenomena presented inthe nature and in our life. For example, the rain is a commune natural pheno-menon, which can not be described with precision, since it can rain with varyingintensity anywhere from a light sprinkle to a torrential downpour. But the phe-nomenon can be characterized by a function which associates a real number fromthe closed interval [0, 1] to any kind of rain: if it doesn’t rain the function takesthe value 0 and the greater is the intensity of the rain, the nearer to 1 is the valueof the function.

The concept of fuzzy set has been introduced by L.A.Zadeh in 1965 (see [58]),when he proposed the idea of a multi-valued logic, which extends the traditionalconcept of a bivalent logic, which becomes a particular case of the new theory. Thefuzzy set theory is based on the principle called by L.A.Zadeh ”the principle ofincompatibility”, that is ”the closer a phenomenon is studied, the more indistinctits definition becomes”.

In general, a fuzzy set of an arbitrary universe X is defined by a functionµA : X −→ [0, 1], where A is a non-empty subset of X. For any element x ∈ X,the value µA(x) is called the membership degree of x to A. When A is a set inthe classical sense, then µA(x) = 0, if x /∈ A and µA(x) = 1, if x ∈ A, that is, µA

is the characteristic function of the set A.

3.1 The fuzzy subgroup of a group

Since has been introduced the concept of fuzzy set, some notions have been ex-tended and generalized, substituting the classical sets with the fuzzy sets. Thefirst connection between fuzzy sets and algebraic structures has been consideredby A.Rosenfeld in 1971 (see [48]), when he defined the notion of fuzzy subgroupof a group (G, ·), with applications in the economical mathematics (see [3]). La-ter, R.Ameri and M.M.Zahedi have extended this association for constructing a

19

new hyperstructure 〈G, ◦µ〉, which under suitable conditions is a hypergroup, ormoreover, a join space (see [1]).

In the following, we present some results obtained in these directions.

Definition 3.1.1. Let (G, ·) be a group and µ be a fuzzy set of G. Then µ iscalled fuzzy subgroup of G if

(i) ∀x, y ∈ G,µ(xy) ≥ min(µ(x), µ(y));

(ii) µ(x−1) ≥ µ(x),∀ x ∈ G.

For a non-empty set X, we denote

FS(X) = {all non zero function µ:X −→ [0, 1]},named the set of all (non zero) fuzzy subsets of X. From now by G we mean agroup with the identity e.

Proposition 3.1.2. We have that µ is a fuzzy subgroup of G if and only if

µ(xy−1) ≥ min(µ(x), µ(y)), ∀x, y ∈ G.

Proof. If µ is a fuzzy subgroup we have µ(x) = µ((x−1)−1) ≤ µ(x−1) ≤ µ(x), soµ(x−1) = µ(x). Then µ(xy−1) ≥ min(µ(x), µ(y−1)) = min(µ(x), µ(y)).

Conversely, if µ(xy−1) ≥ min(µ(x), µ(y)),∀ x, y ∈ G, let y = x to obtainµ(e) ≥ µ(x),∀x ∈ G; hence µ(y−1) = µ(ey−1) ≥ min(µ(e), µ(y)) = µ(y) andµ(xy) = µ(x(y−1)−1) ≥ min(µ(x), µ(y−1)) ≥ min(µ(x), µ(y)).

Definition 3.1.3. µ ∈ FS(G) is called

(i) symmetric if µ(x) = µ(x−1), ∀x ∈ G;

(ii) invariant if µ(xy) = µ(yx), ∀x, y ∈ G;

(iii) subnormal if it is both symmetric and invariant.

Definition 3.1.4. Let µ ∈ FS(G) and x ∈ G. Then the left fuzzy coset of µ,denoted xµ ∈ FS(G), is defined by (xµ)(g) = µ(x−1g), for any g ∈ G.Similarly, the right fuzzy coset of µ, denoted µx ∈ FS(G), is defined by (µx)(g) =µ(gx−1), for any g ∈ G.

Notations. Let µ ∈ FS(G) and a, b ∈ G. Denoteaµ = {x ∈ G | xµ = aµ}µa = {x ∈ G | µx = µa}aµe = {ax | x ∈ µe}µaµb = {xy | x ∈ µa, y ∈ µb}.

If µ is invariant then it is easy to see that aµ = µa,∀a ∈ G.

Lemma 3.1.5. Let µ ∈ FS(G) be subnormal. Then

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(i) ∀x, y ∈ G, xµ = yµ ⇐⇒ y−1x ∈ µe;

(ii) µe is a normal subgroup of G;

(iii) µa = aµe,∀a ∈ G;

(iv) µaµb = µab,∀a, b ∈ G;

(v) xµ = yµ ⇐⇒ zxµ = zyµ, ∀x, y, z ∈ G;

(vi) xµ = yµ ⇐⇒ xzµ = yzµ, ∀x, y, z ∈ G.

Proof. (i) Let x, y ∈ G such that xµ = yµ =⇒ ∀g ∈ G,µ(x−1g) = µ(y−1g). Forg 7→ xg we have µ(g) = µ(y−1xg) =⇒ y−1x ∈ µe.Similarly the other implication.

(ii) We know µe ≤ G ⇐⇒ ∀x, y ∈ µe =⇒ xy−1 ∈ µe.Let x, y ∈ µe =⇒ µx = µy = µe and by (i) µe is a subgroup of G.Moreover, µe E G ⇐⇒ ∀x ∈ G,∀h ∈ µe =⇒ xhx−1 ∈ µe.Let x ∈ G and h ∈ µe =⇒ µh = µe =⇒ ∀g ∈ G,µ(h−1g) = µ(g).It is obvious that ∀g ∈ G,µ(xh−1x−1g) = µ(g).

(iii) Let a ∈ G and x ∈ µa =⇒ µx = µa =⇒ a−1x ∈ µe =⇒ a−1x = y withy ∈ µe =⇒ x = ay, y ∈ µe =⇒ x ∈ aµe =⇒ µa ⊂ aµe.Similarly the other inclusion.

(iv)We show µaµb ⊂ µab.Let xy ∈ µaµb =⇒ x ∈ µa = aµe, x = ax′, x′ ∈ µe and y ∈ µb = bµe, y = by′, y′ ∈µe. Since µe is a normal subgroup of G, aµe = µea, ∀a ∈ G. So xy = ax′by′ =abx′y′ ∈ abµe = µab.Similarly the other inclusion.

(v) If xµ = yµ then xy−1 ∈ µe; we know µe E G and thus zxµ = zyµ ⇐⇒zx(zy)−1 = zxy−1z−1 ∈ µe.We suppose now zxµ = zyµ, ∀z ∈ G. For z = e we have xµ = yµ.

(vi)xµ = yµ ⇐⇒ xy−1 ∈ µe ⇐⇒ xzz−1y−1 = xz(yz)−1 ∈ µe ⇐⇒ xzµ = yzµ.

Let µ ∈ FS(G). We define the hyperoperation

◦µ : G×G −→ P∗(G), a ◦µ b = µaµb

named the hyperoperation induced by µ.The next result determines the conditions under which the hypergroupoid

〈G, ◦µ〉 is a hypergroup, a canonical hypergroup, a join space.

Theorem 3.1.6. Let µ ∈ FS(G).

(i) Then 〈G, ◦µ〉 is a cuasihypergroup.

(ii) If µ ∈ FS(G) is subnormal, then 〈G, ◦µ〉 is a hypergroup.

(iii) If µ ∈ FS(G) is subnormal, then 〈G, ◦µ〉 is a polygroup. Moreover, thereexists a good homomorphism between 〈G, ·〉 and 〈G, ◦µ〉.

21

(iv) The hypergroup 〈G, ◦µ〉 is a join space if and only if the commutator subgroup[G,G] ⊆ µe, where e is the identity of the group G.

Proof. (i) We have to prove the reproductive law: a ◦µ G = G = G ◦µ a, ∀a ∈ G.This is clear because for x ∈ G, x ∈ a ◦µ (a−1x) = µaµa−1x = µx andx ∈ (xa−1) ◦µ a = µxa−1

µa = µx.

(ii) It is enough to show that ” ◦µ ” is associative. Let a, b, c ∈ G. Then

(a ◦µ b) ◦µ c =⋃

x∈µaµb

µxµc =⋃

x∈µab

µxc,

since µ is subnormal.But x ∈ µab implies µx = µab and therefore (a ◦µ b) ◦µ c = µ(ab)c.Similarly, a ◦µ (b ◦µ c) = µa(bc) and by consequence (a ◦µ b) ◦µ c = a ◦µ (b ◦µ c).

(iii) Let a ∈ G. Then a ∈ µa = µea = µeµa = e ◦µ a and a ∈ µa = µae == µaµe = a ◦µ e. So a ∈ e ◦µ a ∩ a ◦µ e. Similarly, e ∈ a ◦µ a−1 ∩ a−1 ◦µ a.Now suppose c ∈ a ◦µ b = µab. Thus µc = µab and so µa = µcb−1, that isa ∈ c ◦µ b−1. Also, b ∈ a−1 ◦µ c. Therefore (G, ◦µ ) is a polygroup.

Now we define ϕ : (G, ·) −→ (G, ◦µ ), ϕ(a) = µa. Then ϕ(ab) = µab = µaµb

and ϕ(a) ◦µ ϕ(b) = µa ◦µ µb =⋃

x∈µa,y∈µb

x ◦µ y =⋃

xy∈µab

µxy = µab = ϕ(ab).

Thus ϕ is a good homomorphism.

(iv) Now we check the commutativity and the transposition properties of〈G, ◦µ〉. Let a, b ∈ G. Thena ◦µ b = b ◦µ a⇐⇒µab = µba⇐⇒abµ= baµ⇐⇒ a−1b−1ab ∈µe ⇐⇒ [G,G]⊆ µe.Thus 〈G, ◦µ〉 is a commutative hypergroup if and only if [G,G] ⊆ µe.

Let a, b, c, d ∈ G and a/b ∩ c/d 6= ∅. Then ∃x ∈ a/b ∩ c/d such that a ∈ x ◦µ band c ∈ x ◦µ d. Thus µa = µxb and µc = µxd. But

1) µa = µxb =⇒ µad = µxbd

2) µc = µxd =⇒ µbc = µbxd.

From 1) and 2) we haveµad = µbc ⇐⇒ (xbd)(bxd)−1 ∈ µe ⇐⇒ xbx−1b−1 ∈ µe ⇐⇒ [G,G] ⊆ µe.Therefore 〈G, ◦µ〉 is a join space ⇐⇒ [G,G] ⊆ µe.

3.2 The classic association between fuzzy sets

and hypergroups

In [9], Corsini established a new connection between hypergroups and fuzzy sets;afterwards, P.Corsini and V.Leoreanu have obtained more results concerning thisassociation (see [11], [13], [14]); some of them are presented in the following.

22

Let µ : H −→ I be a function from a non-empty set H to the closed intervalI = [0, 1], that is 〈H; µ〉 is a fuzzy subset. Let us define on H the hyperoperation:for all (x, y) ∈ H2,

x ◦ y = y ◦ x = {z ∈ H | min{µ(x), µ(y)} ≤ µ(z) ≤ max{µ(x), µ(y)}}.

Theorem 3.2.1. The hypergroupoid 〈H, ◦〉 is a join-space.

Proof. It is clear that the hyperoperation ”◦” is associative and reproducible, thatis 〈H, ◦〉 is a commutative hypergroup. It remains to prove that 〈H, ◦〉 satisfiesthe condition

for all (a, b, c, d) ∈ H4, a/b ∩ c/d 6= ∅ =⇒ a ◦ d ∩ b ◦ c 6= ∅.Let us suppose x ∈ a/b ∩ c/d; then µ(a) ∈ [µ(x), µ(b)] and µ(c) ∈ [µ(x), µ(d)].We distinguish four cases:

1) µ(x) ≤ µ(b), µ(x) ≥ µ(d).Then µ(x) ≤ µ(a) ≤ µ(b), µ(d) ≤ µ(c) ≤ µ(x), from which it resultsµ(d) ≤ µ(c) ≤ µ(x) ≤ µ(a) ≤ µ(b), whence[µ(c), µ(x)] ⊂ [µ(d), µ(a)] ∩ [µ(c), µ(b)] and thus a ◦ d ∩ b ◦ c 6= ∅.

2) µ(x) ≥ µ(b), µ(x) ≤ µ(d).It follows µ(b) ≤ µ(a) ≤ µ(x) and µ(x) ≤ µ(c) ≤ µ(d); thenµ(b) ≤ µ(a) ≤ µ(x) ≤ µ(c) ≤ µ(d), which implies[µ(a), µ(x)] ⊂ [µ(b), µ(c)] ∩ [µ(a), µ(d)], therefore a ◦ d ∩ b ◦ c 6= ∅.

3) µ(x) ≤ µ(b), µ(x) ≤ µ(d).Then µ(a) ≤ µ(b) and µ(c) ≤ µ(d). We consider

(i) µ(b) ≤ µ(d) and then µ(a) ≤ µ(b) ≤ µ(d), thus a ◦ d ∩ b ◦ c 6= ∅.(ii) µ(b) ≥ µ(d) and therefore µ(c) ≤ µ(d) ≤ µ(b), so again b◦ c∩a◦d 6= ∅.

4) µ(x) ≥ µ(b), µ(x) ≥ µ(d).In this case we have µ(b) ≤ µ(a) and µ(d) ≤ µ(c). There are two possibilities:

(i) µ(a) ≤ µ(c), with µ(b) ≤ µ(a) ≤ µ(c)

(ii) µ(a) ≥ µ(c), with µ(d) ≤ µ(c) ≤ µ(a)

In the both cases a ◦ d ∩ b ◦ c 6= ∅.

Theorem 3.2.2.

1) We have, for all n ∈ IN∗, for all (z1, z2, ..., zn) ∈ Hn,

n∏i=1

zi =

{u ∈ H |

n∧i=1

µ(zi) ≤ µ(u) ≤n∨

i=1

µ(zi)

}.

Moreover, if R is the equivalence relation defined on H:

xRy ⇐⇒ µ(x) = µ(y), then R ⊂ β2;

23

2) H/R is a hypergroup with respect to the hyperoperation

x⊗ y = {z | z ∈ x ◦ y}.

Proof. 1) It follows inductively from the definition. Indeed we have

n∏i=1

zi =n−1∏i=1

zi◦zn =⋃

v∈n−1Qi=1

zi

v◦zn =⋃

v∈n−1Qi=1

zi

{λ | µ(v) ∧ µ(zn) ≤ µ(λ) ≤ µ(v) ∨ µ(zn)} .

By induction suppose

n−1∏i=1

zi =

{δ |

n−1∧i=1

µ(zi) ≤ µ(δ) ≤n−1∨i=1

µ(zi)

}.

Then we obtain

n∏i=1

zi =⋃

n−1Vi=1

µ(zi)≤µ(v)≤n−1Wi=1

µ(zi)

{µ(v) ∧ µ(zn) ≤ µ(λ) ≤ µ(v) ∨ µ(zn)} =

=

{λ |

n∧i=1

µ(zi) ≤ µ(λ) ≤n∨

i=1

µ(zi)

}.

Since 〈H, ◦〉 is a join space, it is a hypergroup, whence, for all (a,b) ∈ H2,there is q ∈ H such that a ∈ b ◦ q. So, if aRa′, it follows a′ ∈ µ−1µ(a) ⊂ b ◦ q,then R(a) ⊂ b ◦ q and therefore R ⊂ β2.

2) It is enough to prove that R is regular. Indeed, aRa′, bRb′ implies a ◦ b =a′ ◦ b′. Then, by the Theorem 3.2.1, 〈H/R,⊗〉 is a hypergroup.

In the following, we shall give a necessary and sufficient condition for theisomorphism of two join spaces, associated with two different fuzzy subsets, onthe same universe H.

First we study the case of a finite universe.Let us consider H = I(n) = {1, 2, ..., n} and µA a fuzzy subset on H. We

define on H the equivalence relation

u ∼A v if and only if µA(u) = µA(v).

We denote H ′ = H/ ∼A, H ′ = {h1, h2, ..., hs} and we order H ′ such that

I) ∀ (h, k) ∈ H2, h < k if and only if µA(h) < µA(k).

Let λ(µA) be the ordered partition of n into s parts defined as follows:

24

II) ∀ (h1, h2, ..., hs) ∈ H ′s, λ(µA) = (a1, a2, ..., as) if and only if ∀ i,ai =| µ−1

A (µA(hi)) | and ∀ (i, j) ∈ I(s)× I(s) such that i 6= j,i < j =⇒ hi < hj.

Clearly, we haves∑

i=1

ai = n and ∀ i, ai ≥ 1.

Let (a1, a2, ..., as) be an ordered partition of n into s parts. Let us set

III) ψ(a1, a2, ..., as) = (b1, b2, ..., bs) where for each i, 1 ≤ i ≤ s, bi = as−i+1.

Theorem 3.2.3. If µA, µB are fuzzy subsets on a finite universe H, then the joinspaces 〈H; ◦A〉 and 〈H; ◦B〉 are isomorphic if and only if either λ(µA) = λ(µB) orλ(µB) = ψ(λ(µA)).

Proof. First we prove the implication ”⇐=”.Let us suppose λ(µA) = λ(µB) = (a1, a2, ..., as). We can set

H =s⋃

j=1

Hj =s⋃

j=1

H ′j, where ∀ j∈I(s), Hj = µ−1

A (µA(hj)) and H ′j = µ−1

B (µB(hj)).

We take Hj = {x1,j, x2,j, ..., xaj ,j}, H ′j = {x′1,j, x

′2,j, ..., x

′aj ,j} and we order H in the

following manner

∀ (j, j′) ∈ I(s)× I(s),∀ (h, h′) ∈ I(aj)× I(aj)

xh,j < xh′,j ⇐⇒ h < h′.

If j 6= j′, for all (h, h′) ∈ I(aj)× I(a′j),

xh,j < xh′,j′ ⇐⇒ j < j′.

Moreover, for all (i, j) ∈ I(s)× I(s) we denote i∨ j = max{i, j}, i∧ j = min{i, j}.Then, for all (i, j) ∈ I(s)× I(s) and for all (h, k) ∈ I(ai)× I(aj), we have

xh,i ◦A xk,j =⋃

i∧j≤r≤i∨j

Hr,

x′h,i ◦B x′k,j =⋃

i∧j≤r≤i∨j

H ′r.

Therefore, if f : 〈H; ◦A〉 −→ 〈H; ◦B〉 is the function defined as follows: for each(u, t) ∈ I(at)× I(s), f(xu,t) = x′u,t, then we have

f(xh,i ◦A xk,j) = x′h,i ◦B x′k,j = f(xh,i) ◦B f(xk,j),

whence 〈H; ◦A〉 and 〈H; ◦B〉 are isomorphic hypergroups.Let us suppose now λ(µB) = ψ(λ(µA)).

We consider H =⋃

1≤j≤s

Hj, where, for any j ∈ I(s), we denote

Hj = {x1,j, x2,j, ..., xaj ,j} = µ−1A (µA(hj)) and similarly, H =

⋃

1≤j′≤s

H ′j′ , where

25

j′ = ψ(j) = s − j + 1 and H ′j′ = {x′1,j′ , x

′2,j′ , ..., x

′a′

j′ ,j′} = µ−1

B (µB(hj)), with

a′j′ = aj.Let us define for each (h, j) ∈ I(aj)× I(s), f(xh,j) = x′h,j′ . We have

f(xh,i ◦A xk,j) = f

( ⋃i∧j≤r≤i∨j

Hr

)=

⋃

ψ(i∨j)≤ψ(r)≤ψ(i∧j)

H ′ψ(r)

=⋃

i′∧j′≤r′≤i′∨j′H ′

r′ = x′h,i′ ◦B x′k,j′ = f(xh,i) ◦B f(xk,j),

whence f is an isomorphism.Let us prove now the implication ” =⇒ ”.Let p : H −→ I(s) be the function defined as follows:

for any x ∈ H, p(x) = j, where j is that unique element of I(s), such that it existsk ∈ I(aj) so that x = xk,j ∈ Hj. Similarly, we define p′ : H −→ I(s′), wheres′ =| H/ ∼B| .

Let f : 〈H; ◦A〉 −→ 〈H; ◦B〉 be the isomorphism of these two join spaces.Then, for any (x, y) ∈ H2 we have

f(x ◦A y) = f

⋃

p(x)∧p(y)≤j≤p(x)∨p(y)

Hj

=

⋃

p(x)∧p(y)≤j≤p(x)∨p(y)

f(Hj).

But, setting µ−1B µB(x′k,r) = H ′

r, we also have

f(x ◦A y) = f(x) ◦B f(y) =⋃

p′(f(x))∧p′(f(y))≤r≤p′(f(x))∨p′(f(y))

H ′r.

For any (u, v) ∈ I(n)× I(n), we shall denote by I(u, v) the set

{z ∈ I(n) | u ∧ v ≤ z ≤ u ∨ v}.

We remark now that

(1) if r1 6= r2, then H ′r1∩H ′

r2= ∅;

(2) if {x, y} ⊂ Hj, then x ◦A y = Hj = x ◦A x = y ◦A y, whence

f(x) ◦B f(y) = f(x ◦A y) = f(x ◦A x) = f(Hj) = f(x) ◦B f(x).

(3) by (1) and (2) there exists a unique t = p′(f(x)) such that

f(Hj) = f(x) ◦B f(y) = H ′t.

We set t = φ(j), whence f(Hj) = H ′φ(j).

So we have a function φ : I(p(x), p(y)) −→ I(p′(f(x)), p′(f(y))) and clearly, weobtain that, for any x ∈ H,φ(p(x)) = p′(f(x)).

26

We shall prove that φ is a bijection.φ is injective. Indeed, if there are j1, j2 such that j1 6=j2 and φ(j1)=φ(j2),

it follows f(Hj1) = f(Hj2), whence, for any k ∈ I(aj1), there exists h ∈ I(aj2)such that f(xk,j1) = f(xh,j2), which is absurd since Hj1 ∩ Hj2 = ∅ and f is aninjective function.

φ is also surjective. Indeed, since

f(xk,i ◦A xh,j) = f(xk,i) ◦B f(xh,j) =⋃

r∈I(p′(f(xk,i)),p′(f(xh,j)))

H ′r

and p′(f(xk,i)) = φ(i), p′(f(xh,j)) = φ(j), we have that, for any r ∈ I(φ(i), φ(j)),there exists t ∈ I(i, j) such that φ(t) = r.

Therefore, φ is a bijection from the interval I(i, j) to the interval I(φ(i), φ(j)).On the other hand, as f is a one-to-one function, it follows that for any y ∈ H,there exists an unique pair (k, j) ∈ I(aj)× I(s) such that f(xk,j) = y, whence

f(H) = H =⋃

j∈I(s)

f(Hj) =⋃

j∈I(s)

H ′φ(j) =

⋃

r∈I(φ(1),φ(s))

H ′r.

Moreover, φ is a function from I(s) to I(s′) and since s′ =| H/ ∼B| and f is anisomorphism, we have

| f(H)/ ∼B|=| H/ ∼B|=| p′(H) |=| p′(f(H)) |=| φ(p(H)) |,

whence s′ =| H/ ∼B|=| φ(p(H)) |=| p(H) |= s.Moreover, for any j ∈ I(s), we have

aj =| Hj |=| f(Hj) |=| Hφ(j) |= a′φ(j).

On the other hand, for any k ∈ I(a1) and for any h ∈ I(as),

f(H) = f(xk,1 ◦A xh,s) =⋃

t∈I(1,s)

H ′t = H.

Therefore, the interval I(φ(1), φ(s)) coincides with the interval I(1, s) = I(s). Itfollows {φ(1), φ(s)} = {1, s}. Hence

I(2, s−1) = I(1, s)\ {1, s}=I(φ(1), φ(s))\{φ(1), φ(s)}=φ(I(s))\{φ(1), φ(s)}== φ(I((2, s− 1)) = I(φ(2), φ(s− 1)).

From I(2, s− 1) = I(φ(2), φ(s− 1)), one obtains {φ(2), φ(s− 1)} = {2, s− 1}.In general, we have

(ε) for any k, φ(k) ∈ {k, s− k + 1}.Let V be the set of the permutations of I(s) which satisfy (ε).(η) We shall prove now that either φ is the identity function I of I(s) or it is

the permutation Ψ of I(s) defined:

for any k ∈ I(s), Ψ(k) = s− k + 1.

27

If s ≤ 3 we have V = {II(s), Ψ}. If s > 3 and one supposes φ(1) = 1, φ(2) == Ψ(2) = s − 2 + 1 = s − 1, it follows φ(I(1, 2)) = I(φ(1), φ(2)) = I(1, s − 1),whence 2 =| I(1, 2) |=| φ(I(1, 2)) |6=| I(1, s− 1) |≥ 3, absurd.

Analogously, if φ(1) = s, then φ(2) = 2 and we obtain a contradiction.Therefore either φI(2) = II(2) or φI(2) = ΨI(2).Let k be in I(s) and let us suppose

φI(k) = II(k), φ(k + 1) = Ψ(k + 1).

Then we havek+1 =|I(k + 1) |=|φ(I(k + 1)) |=| I(φ(1), φ(k)) |+ | I(φ(k), Ψ(k + 1) |−1=

= k+ | I(k, s− k) | −1 =k + s− 2k, from which s = 2k + 1,hence φ(k + 1) = s− (k + 1) + 1 = k + 1 and then φI(k+1) = II(k+1).

Similarly, if one supposes

φI(k) = ΨI(k), φ(k + 1) = k + 1

we have k +1 =|φ(I(k +1)) |=|φ(I(k)) | + | φ(I(k, s− k)) |−1 = k + s− 2k, fromwhich s = 2k + 1. Then Ψ(k + 1) = s − (k + 1) + 1 = k + 1 = I(k + 1) hence,φI(k+1) = ΨI(k+1).

Therefore, by induction, (η) is proved and consequently the implication ” =⇒ ”follows and the theorem is completed.

Now, it is interesting to study how two isomorphism problem of two join spacesassociated with different fuzzy subsets on a finite universe can be generalized forthe case of an arbitrary universe.

For this, we use the same notations: if µA and µB are two different fuzzysubsets on H, then H/ ∼A= {Hi | i ∈ I} and H/ ∼B= {H ′

i′ | i′ ∈ I ′}, where

∀ i ∈ I, Hi = {xk,i | k ∈ Ki}; | Ki |=| Hi |= ai and∀ i′ ∈ I ′, H ′

i′ = {x′k′,i′ | k′ ∈ K ′i′}; | K ′

i′ |=| H ′i′ |= a′i′ .

We order I (I ′, respectively) such that:i < j ⇐⇒ for any (x, y) ∈ Hi ×Hj, µA(x) < µA(y)(i′ < j′ ⇐⇒ for any (x′, y′) ∈ H ′

i′ ×H ′j′ , µB(x′) < µB(y′), respectively)

and we order also the quotient set H/ ∼A (H/ ∼B, respectively) such that:Hi < Hj ⇐⇒ i < j (H ′

i′ < H ′j′ ⇐⇒ i′ < j′, respectively).

We obtain the following result.

Theorem 3.2.4. If µA, µB are fuzzy subsets on a universe H, then the join spaces〈H; ◦A〉 and 〈H; ◦B〉 are isomorphic if and only if a strictly monotone and bijectivefunction ϕ : I −→ I ′ exists, such that, for any i ∈ I, ai = a′ϕ(i).

Proof. First, let us prove the implication ” =⇒ ”.We denote by f the isomorphism between 〈H; ◦A〉 and 〈H; ◦B〉.As in the finite case, we shall consider the function p : H −→ I, p(x) = j, where jis the unique element of I such that x∈Hj and, similarly, p′ : H−→I ′, p′(x) = j′,where j′ is the unique element of I ′ such that x ∈ H ′

j′ .

28

For {x, y} ⊂ Hi, we have x ◦A x = x ◦A y = y ◦A y = Hi, so we obtainf(x) ◦B f(x) = f(x ◦A x) = f(y ◦A y) = f(y) ◦B f(y), relation which meansH ′

p′(f(x)) = f(Hi) = H ′p′(f(y′)), whence p′(f(x)) = p′(f(y)), since for {r1, r2} ⊂ I ′,

r1 6= r2, H′r1∩H ′

r2= ∅.

We define the function ϕ : I −→ I ′ as it follows:

for i = p(x), x ∈ H,ϕ(i) = p′(f(x)).

We prove that ϕ is an injective function. We suppose that there exist j1, j2 ∈ I,j1 6= j2, such that ϕ(j1) = ϕ(j2). It follows f(Hj1) = f(Hj2), which is absurd sinceHj1 ∩Hj2 = ∅ and f is a bijective function.

ϕ is also surjective. Indeed, since f(H) = H, we have:

H =⋃

i′∈I′H ′

i′ = f

(⋃i∈I

Hi

)=

⋃i∈I

f(Hi) =⋃i∈I

H ′ϕ(i) =

⋃

i′∈Imϕ

H ′i′ ,

so I ′ = Imϕ.Therefore ϕ is a bijection and | I |=| I ′ | .f is an isomorphism, so for any i ∈ I, | f(Hi) |=| Hi |, that is, for any

i ∈ I, a′ϕ(i) =| H ′ϕ(i) |=| Hi |= ai.

We prove now the strict monotony of ϕ. We shall use the following notations:u ∧ v = min{u, v}, u ∨ v = max{u, v};∀ (i, j) ∈ I2, [i ∧ j, i ∨ j] = {t ∈ I | i ∧ j ≤ t ≤ i ∨ j}.

Let (i, j) ∈ I2, i < j and let us consider x ∈ Hi and y ∈ Hj. From f(x ◦A y) =f(x) ◦B f(y) it follows

f

( ⋃i≤r≤j

Hr

)=

⋃

ϕ(i)∧ϕ(j)≤k≤ϕ(i)∨ϕ(j)

H ′k,

whence ⋃i≤r≤j

H ′ϕ(r) =

⋃

ϕ(i)∧ϕ(j)≤k≤ϕ(i)∨ϕ(j)

H ′k;

therefore, {ϕ(r) | i ≤r≤ j}= {k ∈ I | ϕ(i) ∧ ϕ(j) ≤ k ≤ ϕ(i) ∨ ϕ(j)}, that is

(∗) ∀ (i, j) ∈ I2, i < j, ϕ([i, j]) = [ϕ(i) ∧ ϕ(j), ϕ(i) ∨ ϕ(j)].

We have ϕ(i) 6= ϕ(j), since i < j and ϕ is a bijection; so, there are two possibilities:ϕ(i) < ϕ(j) or ϕ(i) > ϕ(j).

For ϕ(i) < ϕ(j), ϕ is strictly increasing on [i, j]. Indeed, let us observe that:

1) if [i, j] = {i, j}, that is obviously;

2) if there exists r ∈ I such that i < r < j, thenϕ(i) = ϕ(i) ∧ ϕ(j) < ϕ(r) < ϕ(i) ∨ ϕ(j) = ϕ(j), since (∗).

3) if there exists (r, s) ∈ I2, such that i < r < s < j we obtain:ϕ([i, r]) = [ϕ(i) ∧ ϕ(r), ϕ(i) ∨ ϕ(r)]; so, using 2), ϕ(i) = ϕ(i) ∧ ϕ(r) << ϕ(s) < ϕ(i) ∨ ϕ(r) = ϕ(r) < ϕ(j).

29

For ϕ(j) < ϕ(i), we can prove that ϕ is strictly decreasing in a similar manner.Moreover, we shall prove that for any (i, j) ∈ I2, such that i < j, we have two

situations:

1◦. If ϕ(i) < ϕ(j), then ϕ is strictly increasing on I;

2◦. If ϕ(j) < ϕ(i), then ϕ is strict decreasing on I.

Indeed, in the first situation, we have already seen that ϕ is strictly increasingon [i, j]. Let l be an arbitrary element of I, such that j < l. So, ϕ(j) 6= ϕ(l); ifϕ(l) < ϕ(j), then ϕ(l) < ϕ(i).

Indeed, ϕ(l) /∈ [ϕ(i), ϕ(j)] = ϕ([i, j]), since ϕ is one-to-one.But, from ϕ(l) < ϕ(i), we obtain that ϕ is strictly decreasing on [i, l], as it

follows from the second case. So that, ϕ is strictly decreasing on [i, j] too, whichis not true. Therefore, for any l ∈ I, j < l, ϕ(j) < ϕ(l), that is, ϕ is strictlyincreasing on [j, l], ∀ l > j.

Similarly, we can prove that ϕ is strictly increasing on [l, i], for any l ∈ I, l < i.In conclusion, ϕ is strictly increasing on I.

Analogously it follows 2◦.

Now we prove the implication ” =⇒ ”.For any i ∈ I, we have | Ki |=| Hi |= ai = a′ϕ(i) =| H ′

ϕ(i) |=| K ′ϕ(i) |, so we can

suppose Ki = K ′ϕ(i).

Let us define f : (H, ◦A) −→ (H, ◦B) in this way: for any i ∈ I, for anyk ∈ Ki, f(xk,i) = x′k,ϕ(i). Hence, f is a bijection.

We verify that f is a homomorphism, too.For any (i, j) ∈ I2 and for any (x, y) ∈ Hi × Hj, there exists k ∈ Ki and thereexists h ∈ Kj such that x = xk,i and y = yh,j. We have:

f(x) ◦B f(y) = f(xk,i) ◦B f(yh,j) = x′k,ϕ(i) ◦B y′h,ϕ(j) =⋃

ϕ(i)∧ϕ(j)≤t≤ϕ(i)∨ϕ(j)

H ′t;

f(x ◦A y) = f

( ⋃

i∧j≤k≤i∨j

Hk

)=

⋃

i∧j≤k≤i∨j

H ′ϕ(k).

If ϕ is strictly increasing, then

ϕ(i ∧ j) = ϕ(i) ∧ ϕ(j) and ϕ(i ∨ j) = ϕ(i) ∨ ϕ(j);

ϕ is also a bijection, so

f(x ◦A y) =⋃

ϕ(i∧j)≤ϕ(k)≤ϕ(i∨j)

H ′ϕ(k) =

⋃

ϕ(i)∧ϕ(j)≤s≤ϕ(i)∨ϕ(j)

H ′s,

whence f(x ◦A y) = f(x) ◦B f(y).If ϕ is strictly decreasing, then

ϕ(i ∧ j) = ϕ(i) ∨ ϕ(j) and ϕ(i) ∨ ϕ(j) = ϕ(i) ∧ ϕ(j);

30

ϕ is also a bijection, so

f(x ◦A y) =⋃

ϕ(i∨j)≤ϕ(k)≤ϕ(i)∨ϕ(j)

H ′ϕ(k) =

⋃

ϕ(i)∧ϕ(j)≤s≤ϕ(i)∨ϕ(j)

H ′s,

whence again f(x ◦A y) = f(x) ◦B f(y).Therefore, we have obtained that f is a homomorphism and now the theorem

is proved.

3.3 The sequence of join spaces associated with

a hypergroupoid

A third connection between the class of hypergroups and the class of fuzzy setshas been established by Corsini in [17], which opened a new field of researchand its study represents the aim of this thesis. Every hypergroup H determines asequence of fuzzy sets and of join spaces associated to H. M.Stefanescu, P.Corsini,V.Leoreanu and I.Cristea have proved some properties of this sequence and havestudied it in some particular cases: for the i.p.s. hypergroups of order less or equalto 7, 1-hypergroups, finite complete hypergroups or for hypergroups obtained froma rough set.

3.3.1 The construction of the sequence of join spaces andfuzzy sets

In the following, we denote by 〈H, ◦〉 a hypergroupoid.For any (x, y) ∈ H2 and for any u ∈ H, we consider

(ω)

µx,y(u) = 0 iff u /∈ x ◦ y

µx,y(u) =1

|x ◦ y| iff u ∈ x ◦ y

A(u) =∑

(x,y)∈H2

µx,y(u)

Q(u) = {(a, b) ∈ H2 | u ∈ a ◦ b}q(u) = |Q(u)|µ(u) = A(u)/q(u).

Then we obtain a join space 1H as follows (see [9]):

∀(x, y) ∈ H2, x ◦1 y = {z | µ(x) ∧ µ(y) ≤ µ(z) ≤ µ(x) ∨ µ(y)}.

By using the same procedure as in (ω), from 1H we can obtain a membershipfunction µ1 and the associated join space 2H and so on. A sequence of fuzzy setsand join spaces (〈rH, ◦r〉, µr)r≥1 is determined in this way. We denote µ0 = µ,

31

0H = H.Then, for any i, there are r, namely r = ri, and a partition π = {iCj}r

j=1 of iH

such that for any j ≥ 1 and x ∈ iCj,iCj = µ−1

i−1(µi−1(x)) (this means x, y ∈iCj ⇐⇒ µi−1(x) = µi−1(y)).For x ∈ H, we denote λ(x) = ij, when x ∈ iCj.On the set of classes {iCj}r

j=1 we define the following ordering relation :ij < ik if for elements x ∈ iCj and y ∈ iCk, µi−1(x) < µi−1(y)(therefore λ(x) < λ(y)).

We shall use some notations, for all j, s :

kj =| iCj |, sC =⋃

1≤j≤s

iCj,

sC =⋃

s≤j≤r

iCj, sk =|sC|, sk = | sC | and

I(i, j) = {v ∈ I∗m+1 | i ∧ j ≤ v ≤ i ∨ j}.With any ordered chain (iCj1 ,

iCj2 , ...,iCjr) we may associate an ordered r-tuple

(kj1 , kj2 , ..., kjr), where kjl=| iCjl

|, for all l, 1 ≤ l ≤ r.In particular, with any hypergroupoid H we can associate an ordered r-tuple

(k1, k2, ..., kr).

Theorem 3.3.1. (see [17]) For any z ∈ Cs, s = 1, 2, ..., r,

µr(z) =

2

i,j∑i≤s≤ji6=j

kikj∑i≤t≤j

kt

+

k2s

ks

2sksk − k2s

.

Proof. Indeed, for any (x, y) ∈ H2, we have

x ◦r−1µ y =⋃

t∈I(λ(x),λ(y))

Ct.

If x ∈ Cs, then

q(x) = 2sksk − k2

s , A(x) =∑

a◦r−1µb3x

µa,b(x), where µa,b(x) =1

| a ◦r−1µ b | .

Setting, for any z ∈ H and for any (x, y) ∈ H2,

1x,y(z) = 1 ⇐⇒ z ∈ x ◦r−1µ y

1x,y(z) = 0 ⇐⇒ z /∈ x ◦r−1µ y

we obtain

rµ(z) =

∑

(x,y)∈H2

1x,y(z)kλ(x)kλ(y)∑

t∈I(λ(x),λ(y))

kt

2sksk − k2s

.

32

From this, the theorem follows straightforward.

Remark 3.3.2. It is easy to observe that if with a hypergroupoid H1 we associatethe r-tuple (k1, k2, ..., kr) and to an other hypergroupoid H2 we associate the r-tuple (kr, kr−1, ..., k2, k1), then the join spaces 1H1 and 1H2 associated with H1

and, respectively, with H2 are identical.

Theorem 3.3.3. (see [17]) Let 〈H, ◦〉 be a hypergroup, µ : H −→ [0, 1] a fuzzysets on H and 1H the associated join space. Let Reµ be the equivalence defined by

xReµy ⇐⇒ µ(x) = µ(y).

Then Reµ is a regular equivalence and a congruence in 1H.

Proof. For any (a, b) ∈ H2, a ◦eµ b = {x ∈ H | µ(x) ∈ [µ(a), µ(b)]}; if aReµa′ andbReµb′ then a′ ◦eµ b = a ◦eµ b = a ◦eµ b′, so Reµ is regular in 1H.

Set now z′Reµz, where z ∈ u ◦eµ v, with uReµx, vReµy. Then we have µ(z) =µ(z′) ∈ [µ(u), µ(v)] = [µ(x), µ(y)] so, Reµ(z) ⊂ Reµ(x) ◦eµ Reµ(y) whence Reµ is acongruence.

3.3.2 Properties of the membership function µi

It is easy to prove the following remark:

Remark 3.3.4. From the relation (ω), for each i ≥ 0 and u ∈ H and for µi(u),we obtain

A(u) =p1

m1

+p2

m2

+ · · ·+ ps

ms

,

with m1 < m2 < ... < ms and p1, p2, ..., ps ≥ 1,

q(u) = p1 + p2 + ... + ps.

Proposition 3.3.5. For any u ∈ H and i ≥ 0, µi(u) = 1 if and only if for all(x, y) ∈ H2 such that u ∈ x ◦i y, | x ◦i y |= 1.

Proof. Assume µi(u) = 1; then A(u) = q(u).By using the Remark 3.3.4, we get the relation:

p1(1− 1

m1

) + p2(1− 1

m2

) + ... + ps(1− 1

ms

) = 0

which implies that, for any j, 1 ≤ j ≤ s, pj(1− 1

mj

) = 0.

But pj 6= 0, for all 1 ≤ j ≤ s, therefore mj = 1, for all 1 ≤ j ≤ s.As m1 < m2 < ... < ms, it results that s = 1,m1 = 1 and for all (x, y) ∈ H2 withu ∈ x ◦i y, | x ◦i y |= 1.

33

Conversely, suppose that for any (x, y) ∈ H2 with u ∈ x◦i y, | x◦i y |= 1; thenµx,y(u) = 1 and

A(u) =∑

(x,y)∈2H

µx,y(u) = p, where p =| {(x, y) ∈2 H | u ∈ x ◦i y} |= q(u),

thus µi(u) = 1.

This result allows us to prove the following proposition.

Proposition 3.3.6. For any u ∈ H and i ≥ 1, µi(u) 6= 1.

Proof. For each u ∈ H and i ≥ 1 there are only two possibilities:

(i) For any v ∈ H, v 6= u, µi(u) = µi(v).In this case one obtains that i+1H is a total hypergroup (this means that

for any (x, y) ∈ H2, x ◦i+1 y = H) and then µi+1(u) =1

n, where | H |= n.

(ii) There exists v ∈ H, v 6= u such that µi(u) 6= µi(v). Then µi(u) < µi(v) orµi(u) > µi(v) which implies that u, v ∈ u ◦i+1 v, so | u ◦i+1 v |≥ 2 and by theProposition 3.3.5, µi+1(u) 6= 1.

Remark 3.3.7. There are hypergroups H such that

(i) A(x) = A(y) and q(x) 6= q(y), for some x, y ∈ H.

(ii) A(x) 6= A(y) and q(x) = q(y), for some x, y ∈ H.

In the both cases µ(x) 6= µ(y).An example for the case (i) is the hypergroup H1 and the case (ii) is illustratedby the hypergroup H2 given bellow.

H1 x1 x2 x3 x4

x1 x1, x2 x1, x2 H \ {x4} Hx2 x1, x2 H \ {x4} Hx3 x3 x3, x4

x4 x4

where A(x1) = A(x3) =13

3, q(x1) = 12, q(x3) = 11 and by consequence

µ(x1) = 0, 36 6= 0, 39 = µ(x3).

H2 x1 x2 x3 x4 x5

x1 x1 x1, x2 H \ {x5} H \ {x5} Hx2 x2 x2, x3, x4 x2, x3, x4 H \ {x1}x3 x3, x4 x3, x4 x3, x4, x5

x4 x3, x4 x3, x4, x5

x5 x5

34

where q(x1) = q(x5) = 9, A(x1) =17

5, A(x5) =

97

30and by consequence

µ(x1) = 0, 38 6= 0, 36 = µ(x5).

Remark 3.3.8. Assume that the hypergroupoid H has the decomposition

H =r⋃

i=1

Ci, where Ci = µ−1(µ(x)), for any x ∈ H.

If x ∈ Ci and y ∈ Cj, i 6= j, with | Ci |=| Cj |, then q(x) = q(y) if and only if

ik = kj if and only if ik = kj, as we can easily see from the previous notationsand definitions.If i = 1 and j = r, that is x ∈ C1 and y ∈ Cr, then µ(x) < µ(u), for any u ∈ H\C1

and µ(y) > µ(v), for any v ∈ H \ Cr. Therefore 1k = k1 = kr = rk. It followsthat q(x) = q(y).

3.3.3 Fuzzy grade of the hypergroups

Definition 3.3.9. (see [19]) A hypergroupoid H has the fuzzy grade m, m ∈ IN∗,and we write f.g.(H) = m if, for any i, 0 ≤ i < m, the join spaces iH and i+1Hassociated with H are not isomorphic (where 0H = H) and, for any s, s > m, sHis isomorphic with mH.We say that the hypergroupoid H has the strong fuzzy grade m and we writes.f.g.(H) = m if f.g.(H) = m and, for all s, s > m, sH =mH.

First we determine the fuzzy grade of a hypergroupoid H for which |H/Reµ|∈{2, 3}.

Case I. If the hypergroupoid H has only two classes of equivalence, that isH = C1

⋃C2, where |C1|= k1 and |C2|= k2, k1 + k2 = n = |H|, then there are

only two possibilities:

a) k1 = k2 and then the join space 1H is a total hypergroup and therefores.f.g.(H) = 2. Indeed, by the Theorem 3.3.1, µ1(x) = µ1(y), for any x ∈ C1

and for any y ∈ C2.

b) k1 < k2 with µ1(x) =k1 + 3k2

2n2, for any x ∈ C1 and similarly, for any y ∈ C2,

µ1(y) =k2 + 3k1

2n2; this means µ1(x) > µ1(y) and thus 2H is isomorphic with

1H, s.f.g.(H) = 1.

Case II. If the hypergroupoid H has three classes of equivalence, that isH = C1

⋃C2

⋃C3, where |C1|= k1, |C2|= k2 and |C3|= k3 with k1 + k2 + k3 =

= n =|H|, then there are the following possibilities:

a) k1 = k2 = k3 = k; by the Theorem 3.3.1 we have

35

for x ∈ C1, A(x) =8

3k, q(x) = 5k2 and µ1(x) =

8

15k;

for y ∈ C2, A(y) =11

3k, q(y) = 7k2 and µ1(y) =

11

21k;

for z ∈ C3, A(z) =8

3k, q(x) = 5k2 and µ1(z) =

8

15k.

Therefore, µ1(x) = µ1(z) > µ1(y) and with the join space 1H we associatethe pair (k, 2k). By the Case I, we conclude that s.f.g.(H) = 2.

b) k1 = k3 6= k2; using again the Theorem 3.3.1 we obtain µ1(x) = µ1(z) 6=6= µ1(y) as in the case a) and therefore f.g.(H) = 2.

c) k1 < k2 = k3; by the Theorem 3.3.1 we obtain

for x ∈ C1, A(x) = k1 + 2k1k2

k1 + k2

+ 2k1k2

k1 + 2k2

, q(x) = k1(k1 + 4k2)

and µ1(x) =

1 + 2k2

(1

k1 + k2

+1

k1 + 2k2

)

k1 + 4k2

;

for y∈C2, A(y) = 2k2

(1+

k1

k1 + k2

+k1

k1 + 2k2

), q(y) = k2(4k1 + 3k2)

and µ1(y) =

2 + 2k1

(1

k1 + k2

+1

k1 + 2k2

)

4k1 + 3k2

;

for z ∈ C3, A(z) = 2k2

(1 +

k1

k1 + 2k2

), q(z) = k2(2k1 + 3k2)

and µ1(z) =2 + 2k1

1

k1 + 2k2

2k1 + 3k2

.

We verify if µ1(x) > µ1(y); it is equivalent with

1 + 2k2

(1

k1 + k2

+1

k1 + 2k2

)

k1 + 4k2

>

2 + 2k1

(1

k1 + k2

+1

k1 + 2k2

)

4k1 + 3k2

⇐⇒

8k32 − 2k3

1 − 5k21k2 + k1k

22 > 0 ⇐⇒

5k2(k22 − k2

1) + 3k32 − 2k3

1 + k1k22 > 0 which is true for k1 < k2.

Now we verify if µ1(x) > µ1(z); that is

1 + 2k2

(1

k1 + k2

+1

k1 + 2k2

)

k1 + 4k2

>2 + 2k1

1

k1 + 2k2

2k1 + 3k2

⇐⇒

36

8k32 − 2k3

1 − 7k21k2 + k1k

22 > 0 ⇐⇒

(k2 − k1)(2k21 + 9k1k2 + 8k2

2) > 0; it is true for k1 < k2.

Finally, we verify if µ1(y) = µ1(z), that is

2 + 2k1

(1

k1 + k2

+1

k1 + 2k2

)

4k1 + 3k2

=2 + 2k1

1

k1 + 2k2

2k1 + 3k2

⇐⇒

2k21 + k1k2 − 2k2

2 = 0.

We consider this equation as an equation of second grade in k1 ∈ IN∗.

The solutions are−k2 ±

√17k2

2which are not natural numbers, so that

µ1(y) 6= µ1(z).In conclusion, µ1(x) > µ1(y) and µ1(x) > µ1(z), with µ1(y) 6= µ1(z); there-fore to the join space 1H we associate the triple (k2, k2, k1) and, by conse-quence, the join spaces 2H and 1H associated with H are isomorphic andthus s.f.g.(H) = 1.

d) k1 = k2 < k3; again by the same theorem one obtains that

for x ∈ C1, µ1(x) =

2

(1 +

k3

2k1 + k3

)

3k1 + 2k3

;

for y∈C2, µ1(y) =

2

(1 +

k3

k1 + k3

+k3

2k1 + k3

)

3k1 + 4k3

;

for z ∈ C3, µ1(z) =

1 + 2k1

(1

k1 + k3

+1

2k1 + k3

)

4k1 + k3

.

It is simple to verify that µ1(x) 6= µ1(y) and µ1(x) > µ1(z).

Now we see when µ1(y) > µ1(z); this is equivalent with

P = 2k33 − 8k3

1 − k21k3 + 5k1k

23 > 0.

If P > 0, then µ1(z) < µ1(y) and µ1(z) < µ1(x), with µ1(x) 6= µ1(z). As inthe previous case, we have s.f.g.(H) = 1.

If P < 0, then µ1(y) < µ1(z) < µ1(x) and the triple associated with thejoin space 1H is (k1, k3, k1). By the case a), with the join space 2H weassociate the pair (2k1, k3), with 2k1 6= k3 (if 2k1 = k3, then P > 0) andthus s.f.g.(H) = 3. For example, if k1 = k2 = 10 and k3 = 11, then P < 0and s.f.g.(H) = 3.

37

e) k1 6= k2 6= k3; we have two possibilities: k1 < k2 < k3 or k1 < k3 < k2. Usingthe Theorem 3.3.1 we obtain, in the both situations,

for x ∈ C1, µ1(x) =1 +

2k2

k1 + k2

+2k3

k1 + k2 + k3

k1 + 2k2 + 2k3

;

for y∈C2, µ1(y) =k2 + 2

k1k2

k1 + k2

+ 2k1k3

k1 + k2 + k3

+ 2k2k3

k2 + k3

2k1k2 + 2k1k3 + k22 + 2k2k3

;

for z ∈ C3, µ1(z) =1 +

2k1

k1 + k2 + k3

+2k2

k2 + k3

2k1 + 2k2 + k3

.

It is simple to verify that µ1(x) > µ1(y) and µ1(x) > µ1(z), for any x ∈C1, y ∈ C2, z ∈ C3. We can not say that for any triple (k1, k2, k3) there isthe same relation between µ1(y) and µ1(z), because in the first situation,

for (k1, k2, k3) = (10, 11, 12) we have µ1(y) > µ1(z);

for (k1, k2, k3) = (20, 21, 22) we have µ1(y) < µ1(z).

Similarly, in the second situation,

for (k1, k2, k3) = (1, 3, 2) we have µ1(y) < µ1(z);

for (k1, k2, k3) = (10, 20, 19) we have µ1(y) > µ1(z).

So, with the join space 1H one associates the triple (k3, k2, k1) and thens.f.g.(H) = 1, or the triple (k2, k3, k1); in this situation, with the join space2H one associates the triple (k2, k3, k1) and thus s.f.g.(H) = 2 or the triple(k3, k2, k1) and therefore f.g.(H) = 2.

It remains an open problem to see if and when µ1(y) = µ1(z). In thiscase the join spaces 1H and 2H associated with H are not isomorphic andtherefore f.g.(H) = 2.

Theorem 3.3.10. Let H be the hypergroupoid H = {a0, a1, ..., an} with n = 2p+1,p ∈ IN \ {0}, endowed with the fuzzy set µ such that

µ(a0) < µ(a1) < ... < µ(ap) < ... < µ(an).

Then we obtain that

µ1(ap) < µ1(ap−1) < ... < µ1(a1) < µ1(a0)‖ ‖ ‖ ‖

µ1(ap+1) µ1(ap+2) ... µ1(an−1) µ1(an).

38

Proof. The join space 1H associated is

1H a0 a1 a2 · · · ap ap+1 · · · an−1 an

a0 a0 a0, a1 a0, a1, a2 · · · a0 → ap a0 → ap+1 · · · Han H

a1 a1 a1, a2 · · · a1 → ap a1 → ap+1 · · · a1 → an−1 Hao

.... . . · · · · · · · · · · · · · · · · · ·

ap ap ap, ap+1 · · · ap → an−1 ap → an

ap+1 ap+1 · · · ap+1 → an−1 ap+1 → an...

. . . · · · · · ·an−1 an−1 an−1, an

an an

where Hai= H\{ai} and ai → aj = {ai, ai+1, . . . , aj} with i < j.

For i ∈ {0, 1, . . . , n} one calculates µ1(ai) and one obtains:

A(a0) = 1 +2

2+

2

3+ . . . +

2

n + 1, q(a0) = 1 + 2n

A(an) = 1 +2

2+

2

3+ . . . +

2

n + 1, q(an) = 1 + 2n

A(a1) = 1 +4

2+

4

3+ . . . +

4

n+

2

n + 1, q(a1) = 4n− 1

A(an−1) = 1 +4

2+

4

3+ . . . +

4

n+

2

n + 1, q(an−1) = 4n− 1

A(a2) = 1 +4

2+

6

3+

6

4+ . . . +

6

n− 1+

4

n+

2

n + 1, q(a2) = 6n− 7

A(an−2) = 1 +4

2+

6

3+

6

4+ . . . +

6

n− 1+

4

n+

2

n + 1, q(an−2) = 6n− 7

More generally, for 1 ≤ j ≤ p,

A(aj) = 1 +4

2+

6

3+ . . . +

2(j + 1)

j + 1+

2(j + 1)

j + 2+ . . . +

+2(j + 1)

n− j + 1+

2j

n− j + 2+

2(j − 1)

n− j + 3+ . . . +

4

n+

2

n + 1

q(aj) = 2(j + 1)(n− j + 1)− 1.

We observe that µ1(aj) = µ1(an−j) for 0 ≤ j ≤ p.We will prove that, for 0 ≤ j ≤ p, it is satisfied the relation µ1(aj) < µ1(aj−1).

We have:

A(aj−1) = 1+4

2+ . . . +

2j

j+

2j

j+1+ . . . +

2j

n−j+2+

2(j−1)

n−j+3+

2(j−2)

n−j+4+ . . . +

4

n+

2

n+1

q(aj−1) = 2j(n− j + 2)− 1,

39

so

A(aj) = A(aj−1) + 2

(1

j + 1+

1

j + 2+ . . . +

1

n− j + 1

).

Thus, we arrive at the following inequality:

A(aj−1)

2nj − 2j2 + 4j − 1>

A(aj−1) + 2

(1

j + 1+

1

j + 2+ . . . +

1

n− j + 1

)

2nj − 2j2 + 2n + 1⇐⇒

(1) A(aj−1) >2nj − 2j2 + 4j − 1

n− 2j + 1

(1

j + 1+

1

j + 2+ . . . +

1

n− j + 1

).

Rewriting A(aj−1) in the form

A(aj−1) = 2j − 1 + 2j

(1

j + 1+

1

j + 2+ . . . +

1

n− j + 1+

1

n− j + 2

)+

+ 2

(j − 1

n− j + 3+ . . . +

2

n+

1

n + 1

)

it follows that the inequality (1) is equivalent with the following one

2j−1+2j

n−j+2+2j

(1

j+1+ . . . +

1

n−j+1

)+2

(j−1

n−j+3+ . . . +

2

n+

1

n+1

)>

>2nj − 2j2 + 4j − 1

n− 2j + 1

(1

j + 1+

1

j + 2+ . . . +

1

n− j + 1

)⇐⇒

(2)

2nj − 2j2 + 7j − n− 2

n− j + 2+ 2

(j − 1

n− j + 3+ . . . +

2

n+

1

n + 1

)>

>2j2 + 2j − 1

n− 2j + 1

(1

j + 1+

1

j + 2+ . . . +

1

n− j + 1

),

with n− 1 ≥ 2j.

We shall prove the relation by induction in j.For j = 3 we obtain

5n + 1

n− 1+ 2

(2

n+

1

n + 1

)>

23

n− 5

(1

4+

1

5+ . . . +

1

n− 2

)⇐⇒

1

4+

1

5+ . . . +

1

n− 2<

(n− 5)(5n3 + 12n2 − n− 4)

23n(n2 − 1).

It is obvious (by induction in n) that, for n ≥ 6, the relation is true.

40

Now let’s suppose that the relation (2) is true for j and let’s prove it for j +1:

2nj + n− 2j2 + 3j + 3

n− j + 1+ 2

(j

n− j + 2+

j − 1

n− j + 3+ . . . +

1

n + 1

)>

>2j2 + 6j + 3

n− 2j − 1

(1

j + 2+

1

j + 3+ . . . +

1

n− j

).

Using the induction hypothesis, one obtains:

2nj+n−2j2+3j+3

n−j+1+2

(j

n−j+2+

j−1

n−j+3+ . . . +

1

n+1

)>

2nj+n−2j2+3j+3

n−j+1+

+2j

n−j+2+

2j2+2j−1

n−2j+1

(1

j+1+ . . . +

1

n−j+1

)− 2nj−2j2+7j−n−2

n−j+2.

Therefore it remains to verify that

2nj+n−2j2+3j+3

n−j+1−2nj−2j2+5j−n−2

n−j+2+

2j2+2j−1

n−2j+1

(1

j+1+ . . . +

1

n−j+1

)>

>

(2j2 + 6j + 3

n− 2j − 1− 2j2 + 2j − 1

n− 2j + 1

)(1

j + 2+

1

j + 3+ . . . +

1

n− j

).

Calculating, it results that

(2n3j+2n3−4n2j2+6n2j+9n2+2nj3−12nj2+3nj+12n+4j3−8j2−2j+4)

(j+1)(n−j+1)(n−j+2)(4nj+4n−4j2+2)>

>1

n−2j−1

(1

j+2+

1

j+3+ . . . +

1

n−j

), for all j, 3 ≤ j ≤ p.

The first member of the inequality can be made smaller thann

3(j+1)and for this

reason we must prove that

n

3(j + 1)>

1

j + 2+

1

j + 3+ . . . +

1

n− j, ∀j, 3 ≤ j ≤ p.

But,

1

j + 2+

1

j + 3+ . . . +

1

n− j<

1

j + 2+ . . . +

1

3j + 2+

1

3j + 3+ . . . +

1

n− j<

<1

j + 2+

1

j + 3+ . . . +

1

3j + 2+

n− 4j − 2

3j + 3<

n

3j + 3

if and only if1

j + 2+

1

j + 3+ . . . +

1

3j + 2<

4j + 2

3j + 3.

41

It is easy to make sure that the last relation is true (it can be verified by inductionin j), for every j with 3 ≤ j ≤ p.

Finally, we have to prove that

µ1(a2) < µ1(a1) and µ1(a1) < µ1(a0).

We have that

A(a2) = A(a1) + 2

(1

3+

1

4+ . . . +

1

n− 1

)

and denoting

S =1

3+

1

4+ . . . +

1

n− 1

we have thatA(a2) = A(a1) + 2S.

So

µ1(a2) < µ1(a1) ⇐⇒ A(a1) + 2S

6n− 7<

A(a1)

4n− 1⇐⇒

(4n− 1)S < (n− 3)A(a1) ⇐⇒

(n− 3)

(3 +

4

n+

2

n + 1+ 4S

)> (4n− 1)S ⇐⇒

11S < (n− 3) · 3n2 + 9n + 4

n(n + 1).

But

S =1

3+

1

4+ . . . +

1

n− 1<

1

3+

n− 4

4=

3n− 8

12,

then

11S <33n− 88

12< (n− 3)

3n2 + 9n + 4

n(n + 1)

if and only if

(33n− 88)(n2 + n) < (12n− 36)(3n2 + 9n + 4) ⇐⇒3n3 + 55n2 − 188n− 144 > 0.

The relation is true for n ≥ 6.

Next, to prove that µ1(a1) < µ1(a0) we write

A(a1) = A(a0) + 2

(1

2+

1

3+ . . . +

1

n

)= 2A(a0)− n + 3

n + 1.

So we have

A(a0)

2n + 1>

2A(a0)− n + 3

n + 14n− 1

⇐⇒

42

(3) A(a0) <(2n + 1)(n + 3)

n + 1.

But

A(a0) = 1 + 2

(1

2+

1

3+ . . . +

1

n

)< 1 + 2 · n

2= n + 1,

then, to prove (3) we must verify that

n + 1 <(2n + 1)(n + 3)

n + 1⇐⇒ n2 + 5n + 2 > 0,

true for every n ≥ 1.

Remark 3.3.11. If p = 2r + 1, then we can repeat this process and we obtain

µ2(an) µ2(an−1) . . . µ2(an−r+1) µ2(an−r)‖ ‖ ‖ ‖

µ2(a0) > µ2(a1) > . . . > µ2(ar−1) > µ2(ar)‖ ‖ ‖ ‖

µ2(ap) µ2(ap−1) . . . µ2(ar+2) µ2(ar+1)‖ ‖ ‖ ‖

µ2(ap+1) µ2(ap+2) . . . µ2(an−r−2) µ2(an−r−1)

and so on.Generally, if n + 1 = 2s, we can continue this process till when µs(ak) = µs(a`),for every k and `, and in this way we obtain that the join space s+1H associatedis a total hypergroup.

Example 3.3.12. For the hypergroupoid H = {a0, a1, a2, a3, a4, a5, a6, a7} withµ(a0) < µ(a1) < µ(a2) < µ(a3) < µ(a4) < µ(a5) < µ(a6) < µ(a7) the join

space 1H associated is

1H a0 a1 a2 a3 a4 a5 a6 a7

a0 a0 a0, a1a0

a1, a2

a0, a1

a2, a3

a0

a1, a2

a3, a4

a0, a1

a2, a3

a4, a5

Ha7 H

a1 a1 a1, a2a1

a2, a3

a1, a2

a3, a4

a1

a2, a3

a4, a5

a1, a2

a3, a4

a5, a6

Ha0

a2 a2 a2, a3a2

a3, a4

a2, a3

a4, a5

a2

a3, a4

a5, a6

a2, a3

a4, a5

a6, a7

a3 a3 a3, a4a3

a4, a5

a3, a4

a5, a6

a3

a4, a5

a6, a7

a4 a4, a4, a5a4, a5

a6

a4, a5

a6, a7

a5 a5 a5, a6a5, a6

a7

a6 a6 a6, a7

a7 a7

43

Therefore one obtains:

µ1(a0) = µ1(a7) = 0, 296

µ1(a1) = µ1(a6) = 0, 293

µ1(a2) = µ1(a5) = 0, 272

µ1(a3) = µ1(a4) = 0, 267

whence

µ1(a3) < µ1(a2) < µ1(a1) < µ1(a0)‖ ‖ ‖ ‖

µ1(a4) µ1(a5) µ1(a6) µ1(a7)

The join space 2H associated is

2H a0 a1 a2 a3 a4 a5 a6 a7

a0 a0, a7a0, a1

a6, a7

a0, a1

a2, a5

a6, a7

H Ha0, a1

a2, a5

a6, a7

a0, a1

a6, a7a0, a7

a1 a1, a6a1, a2

a5, a6

a1, a2

a3, a4

a5, a6

a1, a2

a3, a4

a5, a6

a1, a2

a5, a6a1, a6

a0, a1

a6, a7

a2 a2, a5a2, a3

a4, a5

a2, a3

a4, a5a2, a5

a1, a2

a5, a6

a0, a1

a2, a5

a6, a7

a3 a3, a4 a3, a4a2, a3

a4, a5

a1, a2

a3, a4

a5, a6

H

a4 a3, a4a2, a3

a4, a5

a1, a2

a3, a4

a5, a6

H

a5 a2, a5a1, a2

a5, a6

a0, a1

a2, a5

a6, a7

a6 a1, a6a0, a1

a6, a7

a7 a0, a7

Again one obtains that

µ2(a0) = µ2(a7) = µ2(a3) = µ2(a4) = 0, 226

µ2(a1) = µ2(a6) = µ2(a2) = µ2(a5) = 0, 219

44

whenceµ2(a1) < µ2(a0)

‖ ‖µ2(a2) µ2(a3)

‖ ‖µ2(a5) µ2(a4)

‖ ‖µ2(a6) µ2(a7)

The join space 3H associated is

3H a0 a1 a2 a3 a4 a5 a6 a7

a0a0, a3

a4, a7H H

a0, a3

a4, a7H H

a0, a3

a4, a7

a1a1, a2

a5, a6

a1, a2

a5, a6H H

a1, a2

a5, a6

a1, a2

a5, a6H

a2a1, a2

a5, a6H H

a1, a2

a5, a6

a1, a2

a5, a6H

a3a0, a3

a4, a7

a0, a3

a4, a7H H

a0, a3

a4, a7

a4a0, a3

a4, a7H H

a0, a3

a4, a7

a5a1, a2

a5, a6

a1, a2

a5, a6H

a6a1, a2

a5, a6H

a7a0, a3

a4, a7

We notice that µ3(a0) = µ3(a1) = . . . = µ3(a7) and therefore the join space 4Hassociated is a total hypergroup.

Theorem 3.3.13. In the same conditions of the Theorem 3.3.10 but for n = 2p,p ∈ IN \ {0}, we obtain that

µ1(ap) < µ1(ap−1) < µ1(ap−2) < ... < µ1(a1) < µ1(a0)‖ ‖ ‖ ‖

µ1(ap+1) µ1(ap+2) ... µ1(an−1) µ1(an)

Corollary 3.3.14. Given a natural number n ∈ IN∗ \ {1} there is a hypergroupH such that s.f.g.(H) = n. Moreover, H is a join space.

45

Proof. We consider the hypergroupoid H = {x1, x2, ..., xp}, p = 2n, with thehyperoperation:

xi ◦ xi = xi, for i ∈ {1, 2, ..., p}xi ◦ xj = xj ◦ xi = {xi, xi+1, ..., xj}, for 1 ≤ i < j ≤ p,

that is, H is the join space 1H associated with the hypergroupoid of the previoustheorem and therefore the join space nH is a total hypergroup and s.f.g.(H)=n.

Theorem 3.3.15. Let us consider the decomposition iH =r⋃

l=1

Cl, where, for all

l, 1 ≤ l ≤ r and for all x ∈ Cl, Cl = µ−1i−1(µi−1(x)) and let (k1, k2, ..., kr) be the

r-tuple associated with the join space iH.If (k1, k2, ..., kr) = (kr, kr−1, ..., k1), then the join spaces i+1H and iH are notisomorphic.

Proof. The condition (k1, k2, ..., kr) = (kr, kr−1, ..., k1) can be written in the form

(1) kl = kr+1−l, for any l, 1 ≤ l ≤[r

2

].

To prove the theorem we shall show that, for any x ∈ Cl and any y ∈ Cr+1−l,

µi(x) = µi(y). It follows, then, that i+1H =r′⋃

l=1

C ′l , where r′ =

[r + 1

2

].

This means that the join space i+1H is not isomorphic with iH.Let us take two arbitrary elements x ∈ Cl and y ∈ Cr+1−l.

First we prove that

(2) q(x) = q(y).

By the condition (1) we have kl = kr+1−l and by the Theorem 3.3.1, we obtainq(x) = 2lk

lk − k2l and q(y) =r+1−l k r+1−lk − k2

r+1−l

where

lk = k1 + k2 + ... + kl = kr + kr−1 + ... + kr+1−l = r+1−lk andlk = kl + kl+1 + ... + kr = kr+1−l + kr−l + ... + k1 = r+1−lk,so lk = r+1−lk and lk = r+1−lk.It is clear that q(x) = q(y).

Now we prove that

(3) A(x) = A(y).

First we remark that, for any j, j′, we havekl+j = kr−(l+j)+1 = kr−l−j+1 = k(r−l+1)−j

and then

(4)

(r−l+1)−j′∑t=j+1

kt =

r−j∑

t=l+j′kt

46

because

(r−l+1)−j′∑t=j+1

kt = kj+1 + kj+2 + ... + k(r−l+1)−j′ =

= kr−j + kr−j−1 + ... + kr−(r−l+1−j′)+1 =

= kr−j + kr−j−1 + ... + kl+j′ =

r−j∑

j+j′kt

We shall determine the formulas of A(x), A(y), according to the Theorem 3.3.1.One calculates:

A(x) = kl +

i,j∑i≤l≤ji6=j

kikj∑i≤t≤j

kt

= kl +l−1∑i=1

r∑

j=l

kikj∑i≤t≤j

kt

=

= kl +l−1∑i=1

r−l∑

j=l

kikj∑i≤t≤j

kt

+l−1∑i=1

r∑

j=r−l+1

kikj∑i≤t≤j

kt

+r−l∑

j=l+1

kikj

j∑

t=l

kt

+r∑

j=r−l+1

kikj

j∑

t=l

kt

=

= kl +l∑

i=1

r∑

j=r−l+1

kikj∑i≤t≤j

kt

+l−1∑i=1

r−l∑

j=l

kikj∑i≤t≤j

kt

+ kl

r−l∑

j=l+1

kj

j∑

t=l

kt

.

Similarly we find

A(y) = kr−l+1 +

i,j∑i≤r−l+1≤ji6=j

kikj∑i≤t≤j

kt

=

= kr−l+1 +r−l∑i=1

r∑

j=r−l+1

kikj∑i≤t≤j

kt

+r∑

j=r−l+2

kr−l+1kj

j∑

t=r−l+1

kt

+r−l∑

j=l+1

kikj

j∑

t=l

kt

=

= kr−l+1 +l∑

i=1

r∑

j=r−l+1

kikj∑i≤t≤j

kt

+r−l∑

i=l+1

r∑

j=r−l+1

kikj∑i≤t≤j

kt

+ kr−l+1

r∑

j=r−l+2

kj

j∑

t=r−l+1

kt

.

Whence, since kl = kr−l+1, we conclude that A(x) = A(y) if and only if

l−1∑i=1

r−l∑

j=l

kikj∑i≤t≤j

kt

+ kl

r−l∑

j=l+1

kj

j∑

t=l

kt

=r−l∑

i=l+1

r∑

j=r−l+1

kikj∑i≤t≤j

kt

+ kl

r∑

j=r−l+2

kj

j∑

t=r−l+1

kt

47

if and only if

k1

kl

l∑t=1

kt

+kl+1

l+1∑t=1

kt

+...+kr−l

r−l∑t=1

kt

+ k2

kl

l∑t=2

kt

+kl+1

l+1∑t=2

kt

+...+kr−l

r−l∑t=2

kt

+...+

+kl−1

kl

l∑

t=l−1

kt

+kl+1

l+1∑

t=l−1

kt

+...+kr−l

r−l∑

t=l−1

kt

+ kl

kl+1

l+1∑

t=l

kt

+kl+2

l+2∑

t=l

kt

+...+kr−l

r−l∑

t=l

kt

=

= kr

kl+1r∑

t=l+1

kt

+kl+2r∑

t=l+2

kt

+...+kr−l+1

r∑

t=r−l+1

kt

+ kr−1

kl+1

r−1∑

t=l+1

kt

+...+kr−l+1

r−1∑

t=r−l+1

kt

+...+

+kr−l+2

kl+1

r−l+2∑

t=l+1

kt

+...+kr−l+1

r−l+2∑

t=r−l+1

kt

+ kr−l+1

kl+1

r−l+1∑

t=l+1

kt

+...+kr−l

r−l+1∑

t=r−l

kt

.

By the relation (4), the two terms of the identity are equal and thereforeA(x) = A(y). Now the proof is complete.

Remark. The condition expressed in the Theorem 3.3.15 is only a sufficientcondition and not a necessary one, as we can see from the following example.

Example 3.3.16. Let us consider the hypergroupoid H = {0, 1, 2, 3, 4, 5} endo-wed with the fuzzy set µ which verifies the relation

µ(0) < µ(2) < µ(3) < µ(5).‖ ‖

µ(1) µ(4)

Clearly, the 4-tuple corresponding to H is (2, 1, 2, 1), which does not verify thecondition of the Theorem 3.3.15. Using the usual formula for calculating themembership function µ1 one obtains:

A(0) = 2 +4

3+

8

5+

4

6=

28

5; q(0) = 20 ; µ1(0) = µ1(1) = 0, 28

A(2) = 1 +8

3+

8

5+

4

6+

2

4=

193

30; q(2) = 23 ; µ1(2) = 0, 279

A(3) = 2 +8

5+

4

6+

8

3+

1

2=

223

30; q(3) = 26 ; µ1(3) = µ1(4) = 0, 285

A(5) = 1 +4

6+

1

2+

4

3=

7

2; q(5) = 11 ; µ1(5) = 0, 318

48

In consequence we have

µ1(2) < µ1(0) < µ1(3) < µ1(5)‖ ‖

µ1(1) µ1(4)

and therefore its associated 4-tuple is (1, 2, 2, 1) and thus the join space 2H is notisomorphic with 1H.

3.3.4 Fuzzy grade of the complete hypergroups

Any complete hypergroup H can be represented as H =⋃g∈G

Ag, where

1) (G, ·) is a group;

2) for any (g1, g2) ∈ G2, g1 6= g2, we have Ag1 ∩ Ag2 = ∅;3) if (a, b) ∈ Ag1 × Ag2 , then a ◦ b = Ag1g2 .

If G is a commutative group, then H is a complete commutative hypergroup, soa join space.

By the above representation, any complete hypergroup of order n is cha-racterized by a m-tuple [k1, k2, ..., km], where m =|G|, 2 ≤ m ≤ n − 1, G ={g1, g2, ..., gm} and, for any i ∈ {1, 2, ..., m}, ki = | Agi

|.

Theorem 3.3.17. Let H be a complete hypergroup. Then for any u ∈ H, we

have µ(u) =1

| Agu |, where u ∈ Agu .

Proof. By hypothesis, ∀u ∈ H, there is a unique gu ∈ G, so that u ∈ Agu . From(ω) we obtain

∀u ∈ H, A(u) =∑

(x,y)∈H2

u∈x◦y

µxy(u) =∑

(x,y)∈H2

u∈x◦y

1

|x ◦ y| =∑

(x,y)∈H2

u∈x◦y

1

|Agxgy |=

=∑

(x,y)∈H2

u∈x◦y

1

|Agu|=

q(u)

|Agu |.

So, µ(u) =1

|Agu |.

We used that u ∈ x ◦ y = Agxgy and gu is unique in G with the propertyu ∈ Agu imply that gxgy = gu.

Definition 3.3.18. Let n be a positive integer, n ≥ 3. A k-decomposition of n,2 ≤ k ≤ n−1, is an ordered system of natural numbers (m1,m2, ..., mk) such thatmi ≥ 1, for any i, 1 ≤ i ≤ k, m1 + m2 + ... + mk = n, m2 ≤ m3 ≤ ... ≤ mk.

49

We denote dk(n) the number of k-decompositions of n.

Theorem 3.3.19. There aren−1∑

k=2

skdk(n) non-isomorphic complete hypergroups

of order n, where sk denotes the number of the non-isomorphic groups of order k.

Proof. We fix an arbitrary group G of order k and let H be a complete hypergroup

of order n. Since H = Ag1 ∪Ag2 ∪ ...∪Agk, we obtain |H |=

k∑i=1

|Agi|. Hence there

are dk(n) possibilities of making a partition of H as in Theorem 1.17.It follows that the class of complete hypergroups of order n has the cardinality

n−1∑

k=2

skdk(n).

Remark. The fuzzy grade of a complete hypergroup does not depend on theconsidered group G, but only on the k-decompositions of n.

Theorem 3.3.20. (see [19]) Let H be a complete hypergroup of order n ≤ 6.

(i) For n = 3, s.f.g.(H) = 1.

(ii) For n = 4, there are 3 hypergroups with s.f.g.(H) = 1 and two hypergroupswith s.f.g.(H) = 2.

(iii) For n = 5, s.f.g.(H) = 1.

(iv) For n = 6, there are 17 hypergroups of s.f.g.(H) = 1 and 4 hypergroups ofs.f.g.(H) = 2.

The proof is detailed in the Appendix A.

Next we determine the fuzzy grades of some complete 1-hypergroups of a givenstructure.

If H is a complete 1-hypergroup, then it is of the type [1, k2, k3, ..., km].

Proposition 3.3.21. If H is a complete 1-hypergroup of the type [1, 1, ..., 1︸ ︷︷ ︸k times

, k],

where n =|H|= 2k, then s.f.g.(H)=2.More general, if the structure of the complete hypergroup H, which is not an 1-hypergroup, is represented by [p, p, ..., p︸ ︷︷ ︸

k times

, kp], where n =|H|= 2kp, then s.f.g.(H)=2.

Proof. We suppose that the structure of H is determined by the (k + 1)-tuple[p, p, ..., p︸ ︷︷ ︸

k times

, kp]. By the Theorem 3.3.17, we obtain that the pair associated with

the hypergroup (H, µ) is (kp, kp), therefore we conclude that the join space 2Hassociated with H is a total hypergroup and s.f.g.(H) = 2.

50

Proposition 3.3.22. Let H be a complete 1-hypergroup of the following type:[1, 1, ..., 1︸ ︷︷ ︸

l times

, k, k, ..., k︸ ︷︷ ︸p times

, l], with 1 < k < l, p ≥ 1, n =|H|= pk + 2l.

(i) If kp = 2l (⇐⇒ n = 4l), then s.f.g.(H) = 3.

(ii) If kp 6= 2l, then f.g.(H) = 2.

Proof. We use again the Theorem 3.3.17.

(i) If kp = 2l, then, with the hypergroupoid (H, µ) is associated the triple(l, 2l, l) and, by the Case II of the section 3.3.3, with the join space 1H isassociated the pair (2l, 2l) and then, by the previous Case I, with the joinspace 2H is associated (4l); thus 3H is a total hypergroup and s.f.g.(H) = 3.

(ii) If kp 6= 2l, then the triple associated with the hypergroup (H, µ) is (l, kp, l);with the join space 1H it is associated the pair (2l, kp) and by consequencef.g.(H) = 2.

Remark 3.3.23. Generalizing the Proposition 3.3.22 for the complete hyper-groups which are not 1-hypergroups, but are of the type [p, p, ..., p︸ ︷︷ ︸

s times

, k, k, ..., k︸ ︷︷ ︸t times

, ps],

with 2 ≤ p < k < ps, n = |H|= 2ps+kt, we obtain that for n = 4ps, s.f.g.(H) = 3and for kt 6= 2ps, f.g.(H) = 2.

Proposition 3.3.24. Let H be a complete 1-hypergroup of the following type:[1, 1, ..., 1︸ ︷︷ ︸

l times

, k, k, ..., k︸ ︷︷ ︸p times

, p, p, ..., p︸ ︷︷ ︸k times

, l], with 1 < k < p < l, n =|H|= 2(l + pk).

(i) If pk = l (⇐⇒ n = 4l), then s.f.g.(H) = 3.

(ii) If pk 6= l, then f.g.(H) = 2.

Proof. We use again the Theorem 3.3.17. With the hypergroupoid (H, µ) isassociated the 4-tuple (l, pk, pk, l) and then, with the join space 1H is associatedthe pair (2l, 2pk).

If l = pk then with the join space 2H is associated (4l); thus 3H is a totalhypergroup and s.f.g.(H) = 3.

If pk 6= l, by the Case I of the section 3.3.3, we have f.g.(H) = 2.

Remark 3.3.25. Generalizing the Proposition 3.3.24 for the complete hyper-groups of the type [k, k, ..., k︸ ︷︷ ︸

l times

, p, p, ..., p︸ ︷︷ ︸s times

, s, s, ..., s︸ ︷︷ ︸p times

, l, l, ..., l︸ ︷︷ ︸k times

], (which are not 1-hyper-

groups), with 2 ≤ k < p < s < l, n = |H| = 2(ps + kl), we obtain that, forkl = ps, s.f.g.(H) = 3 and for kl 6= ps, f.g.(H) = 2.

51

3.3.5 An example of a non-complete 1-hypergroup

We present an example of a non-complete 1-hypergroup and we study the sequenceof join spaces and fuzzy sets associated with it.

We consider the following hypergroup H = Hn = {e}∪A∪B, where | A |= α,| B |= β with α, β ≥ 2 and A ∩ B = ∅, e /∈ A ∪ B. We set A = {a1, ..., aα} andB = {b1, ..., bβ}. The hyperoperation is defined in this way:

∀ a ∈ A, a ◦ a = b1,

∀ (a1, a2) ∈ A2 such that a1 6= a2, a1 ◦ a2 = B,

∀ (a, b) ∈ A×B, a ◦ b = b ◦ a = e,

∀ (b, b′) ∈ B2, b ◦ b′ = A,

∀ a ∈ A, a ◦ e = e ◦ a = A,

∀ b ∈ B, b ◦ e = e ◦ b = B and e ◦ e = e.

Hn is an 1-hypergroup which is not complete.

§1. Let us suppose n = |H6| = 6, where H = H6 = {e} ∪ A ∪ B, α = |A| = 2,β = |B| = 3, A ∩ B = ∅, e /∈ A ∪ B with A = {a1, a2}, B = {b1, b2, b3}. So thestructure in H6 is as follows:

H e a1 a2 b1 b2 b3

e e A A B B B

a1 b1 B e e e

a2 b1 e e e

b1 A A A

b2 A A

b3 A

therefore we have: µ(e)=1; µ(a1)=µ(a2)=0, 5; µ(b1)=0, 467; µ(b2)=µ(b3)=0, 333,whence the join space 1H associated is

1H e a1 a2 b1 b2 b3

e e A ∪ {e} A ∪ {e} H \ {b1, b2} H H

a1 A A A ∪ {b1} H \ {e} H \ {e}a2 A A ∪ {b1} H \ {e} H \ {e}b1 b1 B B

b2 b2, b3 b2, b3

b3 b2, b3

Now we obtain: µ1(e) = 0, 32; µ1(ai) = 0, 286; µ1(b1) = 0, 280 ≈ 0, 2797; µ1(b2) =µ1(b3) = 0, 280 so µ1(b1) < µ1(bj) < µ1(ai) < µ1(e). Therefore the join space 2H

52

associated is the following one

2H e a1 a2 b1 b2 b3

e e A ∪ {e} A ∪ {e} H H \ {b1} H \ {b1}a1 A A H \ {e} H \ {e, b1} H \ {e, b1}a2 A H \ {e} H \ {e, b1} H \ {e, b1}b1 b1 B B

b2 b2, b3 b2, b3

b3 b2, b3

therefore µ2(e) = µ2(b1) = 0, 315; µ2(a1) = µ2(a2) = µ2(b2) = µ2(b3) = 0, 279. Itfollows that the join space 3H associated is:

3H e a1 a2 b1 b2 b3

e e, b1 H H e, b1 H H

a1 H \ {e, b1} H \ {e, b1} H H \ {e, b1} H \ {e, b1}a2 H \ {e, b1} H H \ {e, b1} H \ {e, b1}b1 e, b1 H H

b2 H \ {e, b1} H \ {e, b1}b3 H \ {e, b1}

Now we obtain: µ3(e) = µ3(b1) = 0, 246; µ3(ai) = µ3(b2) = µ3(b3) = 0, 208,therefore ∀ r ≥ 3, we have rH = 3H and s.f.g.(H) = 3.

§2. More generally, set H = Hn = {e} ∪ A ∪ B, |A| = α, |B| = β, α, β ≥ 2,A ∩B = ∅, e /∈ A ∪B, where A = {a1, ..., aα}, B = {b1, ..., bβ}.

First we consider α = β ≥ 2 and therefore the hypergroup H is the followingone:

H e a1 a2 · · · aα b1 b2 · · · bα

e e A A · · · A B B · · · B

a1 b1 B · · · B e e · · · e

a2 b1 · · · B e e · · · e...

. . .aα b1 e e · · · e

b1 A A · · · A

b2 A · · · A...

. . .bα A

whence µ(e) = 1; A(ai) =2α + β + 2(1 + 2 + ... + β − 1)

α=

2α + β2

α; q(ai) =

2α + β2, so µ(ai) =1

α· In the same way we find for i 6= 1, µ(bi) =

1

α· Moreover,

we find A(b1) = 2α + 1, q(b1) = α2 + 2α, whence µ(b1) =2α + 1

α2 + 2α, therefore we

have ∀ i ∈ I(α) = {1, ..., α} and ∀ j ∈ I(α)− {1},µ(e) > µ(b1) > µ(ai) = µ(bj).

53

Set B1 = B \ {b1}, H∗ = H \ {e}.The join space 1H associated is

1H e b1 a1 a2 · · · aα b2 · · · bα

e e e, b1 H H · · · H H · · · H

b1 b1 H∗ H∗ · · · H∗ H∗ · · · H∗

a1 A ∪B1 A ∪B1 · · · A ∪B1 A ∪B1 · · · A ∪B1

a2 A ∪B1 · · · A ∪B1 A ∪B1 · · · A ∪B1...

. . .aα A ∪B1 A ∪B1 · · · A ∪B1

b2 A ∪B1 · · · A ∪B1...

. . .bα A ∪B1

Therefore we have: A(e) =4n− 4

n; q(e) = 2n− 1, whence µ1(e) =

4(n− 1)

n(2n− 1).

A(b1) = 2 +2α + 2(α− 1)

n+

2α + 2(α− 1)

n− 1= 2

3n2 − 6n + 2

n(n− 1); q(b1) = 4n− 5, so

µ1(b1) =2(3n2 − 6n + 2)

n(4n2 − 9n + 5). We have clearly:

µ1(e)>µ1(b1)⇐⇒2n−2

2n−1>

3n2−6n+2

4n2−9n+5⇐⇒ 8n3−18n2+10n−8n2+18n−10>

> 6n3 − 12n2 + 4n− 3n2 + 6n− 2 ⇐⇒ G = 2n3 − 11n2 + 18n− 8 > 0.

We obtain that for n ≥ 5, G = n2(2n − 11) + 2(9n − 4) > 0. On the other side

A(ai) =n3 + n2 − 8n + 4

n(n− 1); q(ai) = (n−2)(n+2), whence µ1(ai) =

n2 + 3n− 2

n(n− 1)(n + 2)·

We have also µ1(b1) > µ1(ai) ⇐⇒ 2(3n2 − 6n + 2)

4n− 5>

n2 + 3n− 2

n + 2⇐⇒ F =

2n3 − 7n2 + 3n− 2 > 0, and since F = n2(2n− 7) + (3n− 2), we have finally, forevery n ≥ 5, ∀ j ∈ I(α) \ {1} and ∀ i ∈ I(α), µ1(b1) > µ1(ai) = µ1(bj). Then

∀ r ≥ 1,rH = 1H and s.f.g.(H) = 1.

§3. Set now α > β ≥ 2, then we have µ(e) = 1, µ(ai) =1

α, µ(bj) =

1

β, for j ∈

I(β)−{1}, so µ(ai) < µ(bj) < µ(e). On the other side µ(b1) =α2 + αβ + 2β − α

β(α2 + 2β)

from which µ(b1) > µ(bj) ⇐⇒ α2 + αβ + 2β − α

β(α2 + 2β)>

1

β, true.

54

Hence µ(ai) < µ(bj) < µ(b1) < µ(e). By consequence, the associated join space is

1H e b1 b2 · · · bβ a1 · · · aα

e e e, b1 H \A · · · H \A H · · · H

b1 b1 B · · · B H∗ · · · H∗

b2 B1 · · · B1 H \ {e, b1} · · · H \ {e, b1}...

. . .bβ B1 H \ {e, b1} · · · H \ {e, b1}a1 A · · · A...

. . .aα A

where B1 = B\{b1}. Then A(e) = 2 +2(β−1)

n−α+

2α

n=

6nβ+2n−2β2−4β − 2

n(β+1),

and q(e) = 2n− 1, it follows µ1(e) =6nβ + 2n− 2β2 − 4β − 2

n(2n− 1)(β + 1)· We have also

A(b1) = 2 +2(β − 1)

n− α+

2α

n+

2(β − 1)

β+

2α

n− 1=

= 25n2β2 − 8nβ2 − 2nβ3 + β3 + 2n2β − 3nβ + 2β2 + β − n2 + n

nβ(n− 1)(β + 1);

q(b1) = 1 + 2β + 4α + 2β − 2 = 4n− 5.

By consequence, µ(e) > µ(b1) if and only if6nβ + 2n− 2β2 − 4β − 2

n(2n− 1)(β + 1)>

> 25n2β2 − 8nβ2 − 2nβ3 + β3 + 2n2β − 3nβ + 2β2 + β − n2 + n

nβ(n− 1)(β + 1)(4n− 5).

Set P = β(n−1)(4n−5)(3nβ +n−β2−2β−1), Q = (2n−1)(5n2β2−8nβ2−2nβ3 + β3 + 2n2β − 3nβ + 2β2 + β − n2 + n). We have

P > Q ⇐⇒ P = n3(12β2 + 4β) + n2(−4β3 − 35β2 − 13β)+

+n(9β3 + 33β2 + 14β)− 5β3 − 10β2 − 5β >

> Q = n3(10β + 4β − 2) + n2(−4β3 − 21β2 − 8β + 3)+

+n(4β2 + 12β2 + 5β − 1)− β3 − 2β2 − β.

This is true if and only if P −Q = n3(2β2 + 2) + n2(−14β2 − 5β − 3) + n(5β3 +21β2 + 9β + 1)− 4β3 − 8β2 − 4β > 0. We have clearly n > 2β + 1.

Let be P −Q = En, so we have

E ′′n = 6n(2β2 + 2)− 28β2 − 10β − 6 > 6(2β + 1)(2β2 + 2)−− 28β2 − 10β − 6 = 24β3 − 16β2 + 14β + 6 > 0,

whence ∀ β ≥ 2. E ′n is a strictly increasing function, that is ∀n : n > 2β + 1

E ′n > E ′

2β+1

55

We obtain for ∀ β : β ≥ 2

E ′n > 3(2β + 1)2(2β2 + 2) + 2(2β + 1)(−14β2 − 5β − 3) + 5β3+

+21β2 + 9β + 1 = 24β4 − 27β3 + 3β2 + 11β + 1 > 0,

therefore En is a strictly increasing function. So ∀n : n > 2β + 1 we have

En > E2β+1.

Finally, we find ∀ β ≥ 2

P −Q >(8β3+12β3+6β + 1)(2β2+2)( 4β2 + 4β + 1)(−14β2 − 5β − 3)+

+ (2β + 1)(5β3 + 21β2 + 9β + 1)− 4β3 − 8β2 − 4β

whence P −Q > 16β5 − 22β4 − 5β3 + 11β2 + 2β > 0, then

µ1(e) > µ1(b1).

On the other side we have

A(bj) =2(β − 1)

n− α+

2α

n+

2(β − 1)

β+

2α

n− 1+ β − 1 +

2α(β − 1)

n− 2=

=

2n(n−1)(n−2)(β−1)(2β+1)+2β(β+1)(n−2)(n−β−1)(2n−1)++n(n−1)β(β+1)(β−1)(3n−2β − 4)

n(n−1)(n−2)β(β+1);

q(bj) = 4(β−1)+4α+(β−1)2+2α(β−1) = 2n−β2+2nβ−2β−5.

Therefore we have µ1(b1) > µ1(bj) if and only if

10n2β2−16nβ2−4nβ3+2β3+4n2β−6nβ+4β2+2β−2n2+2n

nβ(n−1)(β+1)(4n−5)>

>

(2n3−6n2+4n)(2β2−β−1)+(2nβ2+2nβ−4β2−4β)(2n2−3n−2nβ+β+1)++(n2β2+n2β−nβ2−nβ)(3nβ−2β2−2β−3n+4)

n(n−1)(n−2)(β+1)(2n−β2+2nβ−2β−5)β

this is true if and only if, settingP = (2n2 − 9n − nβ2 + 2β2 + 2n2β − 6nβ + 4β + 10)(10n2β2 − 16nβ2 − 4nβ3 +2β3 + 4n2β − 6nβ + 4β2 + 2β − 2n2 + 2n) andQ = (4n− 5)(8n3β2 − n3β − 2n3 − 28n2β2 + 6n2 + 30nβ2 + 6nβ − 4n− 11n2β3 +14nβ3 − 4β3 − 8β2 − 4β − n2β + 3n3β3 − 2n2β4 + 2nβ4)we have P > Q. One finds:P = n4(20β3 +28β2 +4β−4)+n3(−18β4−104β3−156β2−32β +22)+n2(4β5 +64β4 + 198β3 + 302β2 + 78β − 38) + n(−10β5 − 64β4 − 160β3 − 228β2 − 70β +20) + 4β5 + 16β4 + 48β3 + 48β2 + 20β, andQ = n4(12β3 + 32β2 − 4β − 8) + n3(−8β4 − 59β3 − 152β2 + β + 34) + n2(18β4 +

56

111β3+260β2+29β−46)+n(−86β3−10β4−182β2−46β+20)+20β3+40β2+20β.Then one obtains µ1(b1) > µ1(bj) if and only if

P −Q = n4(8β3 − 4β2 + 8β + 4) + n3(−10β4 − 45β3 − 4β2 − 33β − 12)+

+ n2(4β5 + 46β4 + 87β3 + 42β2 + 49β + 8) + n(−10β5 − 54β4−− 74β3 − 46β2 − 24β) + 4β5 + 16β4 + 20β3 + 8β2 > 0.

Set now En = P −Q, ∀ β ≥ 2.

E ′′′n = 24n(8β3 − 4β2 + 8β + 4)− 60β4 − 270β3 − 24β2 − 198β − 72 >

> 24(2β+1)(8β3−4β2+8β+4)−60β4−270β3−24β2−198β−72 =

= 324β4 − 174β3 + 264β2 + 186β + 24 > 0,

whence E ′′n > E ′′

2β+1 = 272β5 − 316β4 + 144β3 + 192β2 + 44β − 8 > 0, thereforeE ′

n is a strictly increasing function. It follows

E ′n > E ′

2β+1 = 152β6 − 222β5 + 24β4 + 137β3 + 50β2 − 9β − 4 > 0,

so En > E2β+1 = 64β7 − 108β6 − 18β5 + 72β4 + 40β3 − 6β2 − 8β > 0, therefore

µ1(b1) > µ1(bj).

On the other side we have

A(ai) =2α

n+

2α

n− 1+

2α(β − 1)

n− 2+ α =

=(n− β − 1)(n3 − n2 − 6n + 2n2β − 2nβ + 4)

n(n− 1)(n− 2)

q(ai) = 4α + α2 + 2α(β − 1) = (n− β − 1)(n + β + 1)

therefore µ1(bj) > µ1(ai) if and only if

8n3β2 − n3β − 2n3 − 28n2β2 + 6n2 + 30nβ2 + 6nβ − 4n− 11n2β3++14nβ3 − 4β3 − 8β2 − 4β − n2β + 3n3β3 − 2n2β4 + 2nβ4

n(n− 1)(n− 2)β(β + 1)(2n− β2 + 2nβ − 2β − 5)>

>n3 − n2 − 6n + 2n2β − 2nβ + 4

n(n− 1)(n− 2)(n + β + 1)

This is true if and only ifP = (n+β +1)(8n3β2−n3β−2n3−28n2β2 +6n2 +30nβ2 +6nβ−4n−11n2β3 +14nβ3 − 4β3 − 8β2 − 4β − n2β + 3n3β3 − 2n2β4 + 2nβ4) >> Q = (2nβ2 + 2nβ − β4 − β3 + 2nβ3 + 2nβ2 − 2β3 − 2β2 − 5β2 − 5β)(n3−n2 −6n + 2n2β − 2nβ+4).One obtainsP = n4(3β3 + 8β2 − β − 2) + n3(β4 − 21β2 − 4β + 4) + n2(−2β5 − 11β4 − 25β3 +

57

β2 + 11β + 2) + n(2β5 + 16β4 + 40β3 + 28β2 − 2β − 4)− 4β4 − 12β3 − 12β2 − 4βandQ = n4(2β3 + 4β2 + 2β) + n3(3β4 + 3β3 − 7β2 − 7β) + n2(−2β5 − 9β4 − 31β3 −31β2 − 7β) + n(2β5 + 12β4 + 40β3 + 68β2 + 38β)− 4β4 − 12β3 − 28β2 − 20β.So P −Q = n4(β3 +4β2− 3β− 2)+n3(−2β4− 3β3− 14β2 +3β +4) +n2(−2β4 +6β3 + 32β2 + 18β + 2) + n(4β4 − 40β2 − 40β − 4) + 16β2 + 16β.

Set En = P −Q, then we have

E ′′′n = 24n(β3 + 4β2 − 3β − 2)− 12β4 − 18β3 − 84β2 + 18β + 24 ≥≥ (48β + 24)(β3 + 4β2 − 3β − 2)− 12β4 − 18β3 − 84β2 + 18β + 24 =

= 48β4 + 216β3 − 48β2 − 168β − 48− 12β4 − 18β3 − 84β2 + 18β + 24 =

= 36β4 + 198β3 − 132β2 − 150β − 24 > 0,

whence E ′′n is a strictly increasing function. It follows

E ′′n > E ′′

2β+1 = 12(4β2 + 4β + 1)(β3 + 4β2 − 3β − 2) + 6(2β + 1)··(−2β4 − 3β3 − 14β2 + 3β + 4)− 4β4 + 12β3 + 64β2 + 36β + 4 =

= 48β5 + 192β4 − 144β3 − 96β2 + 48β4 + 192β3 − 144β2 − 96β+

+12β3 + 48β2 − 36β − 24− 24β5 − 36β4 − 168β3 + 36β2 + 48β−−12β4 − 18β3 − 84β2 + 18β + 24− 4β4 + 12β3 + 64β2 + 36β + 4 =

= 24β5 + 188β4 − 114β3 − 176β2 − 30β + 4 > 0,

and, by consequence, E ′n is a strictly increasing function. So we have

E ′n > E ′

2β+1 = 4(8β3 + 12β2 + 6β + 1)(β3 + 4β2 − 3β − 2)+

+3(4β2 + 4β + 1)(−2β4 − 3β3 − 14β2 + 3β + 4)+

+2(2β + 1)(−2β4 + 6β3 + 32β2 + 18β + 2) + 4β4 − 40β2 − 40β − 4 =

= 8β6 + 108β5 − 66β4 − 109β3 − 14β2 + β + 4 > 0,

therefore En is a strictly increasing function and, by consequence,

En > E2β+1 = (16β4 + 32β3 + 24β2 + 8β + 1)(β3 + 4β2 − 3β − 2)+

+(8β3 + 12β2 + 6β + 1)(−2β4 − 3β3 − 14β2 + 3β + 4)+

+(4β2 + 4β + 1)(−2β4 + 6β3 + 32β2 + 18β + 2)+

+(2β + 1)(4β4 − 40β2 − 40β − 4) + 16β2 + 16β

whence En=P−Q > E2β+1 = 40β6−32β5−34β4+4β3−8β2+2β > 0. Finally,

µ1(bj) > µ1(ai).

We conclude thatµ1(ai) < µ1(bj) < µ1(b1) < µ1(e).

58

It follows ∀ r : r ≥ 1,rH = 1H and s.f.g.(H) = 1.

§4. Let β > α ≥ 2 =⇒ n > 2α + 1. We find in the same way as in §3:

µ(e) = 1, µ(ai) =1

α, for j ∈ {2, 3, ..., β},

µ(bj) =1

β, µ(b1) =

α2 + αβ + 2β − α

β(α2 + 2β)

4.1. Let us consider now the case α = 2, β = n− 3, whence we have:

µ(b1) < µ(ai) iff4β + 2

β(2β + 4)<

1

2iff β2 − 2β − 2 > 0 iff β ≥ 3.

Moreover, for j ∈ I(β),

µ(bj) < µ(b1) iff1

β<

4β + 2

β(2β + 4)iff 2β − 2 > 0, iff ∀ β ≥ 3.

So, we have:µ(bj) < µ(b1) < µ(ai) < µ(e).

The associated join space is the following

1H b2 · · · bβ b1 a1 a2 e

b2 B1 · · · B1 B H∗ H∗ H...

. . ....

......

......

bβ B1 B H∗ H∗ H

b1 b1 A ∪ {b1} A ∪ {b1} H \B1

a1 A A A ∪ {e}a2 A A ∪ {e}e e

We obtain

A(e) = 1 +4

3+

2

4+

2(β−1)

n=

29n−48

6n; q(e)=2n−1; µ1(e)=

29n−48

6n(2n−1)

A(ai)=2+8

3+

2

4+

2(β−1)

n+

4(β−1)

n−1; A(ai)=

67n2−187n+48

6n(n−1); q(ai)=6n−10

whence

µ1(ai) =67n2 − 187n + 48

6n(n− 1)(6n− 10).

Moreover,

A(b1) = 1 +4

3+

1

2+

4(β−1)

n−1+

2(β−1)

n+

2(β−1)

β=

65n3−392n2+615n−144

6n(n−1)(n−3)

q(b1) = 8n− 25

59

whence

µ1(b1) =65n3 − 392n2 + 615n− 144

6n(n− 1)(n− 3)(8n− 25)

Moreover, for j ∈ I(β)− {1}, we have

A(bj) = n− 4 +2(n−4)

n−3+

4(n−4)

n−1+

2(n−4)

n= (n−4)

n3+4n2−19n+6

n(n−1)(n−3)

q(bj) = (n− 4)(n + 4)

whence

µ1(bj) =n3 + 4n2 − 19n + 6

n(n− 1)(n− 3)(n + 4)

For β = 3, that is n = 6, see §1.For β > 3 we have

µ1(e) > µ1(ai)

if and only if29n− 48

6n(2n− 1)>

67n2 − 187n + 48

6n(n− 1)(6n− 10)

which is equivalent with 40n3 − 311n2 + 775n− 432 > 0, true for n ≥ 7. So,

µ1(e) > µ1(ai).

Moreover,

µ1(b1) > µ1(bj) ⇐⇒ 65n3 − 392n2 + 615n− 144

6n(n− 1)(n− 3)(8n− 25)>

n3 + 4n2 − 19n + 6

n(n− 1)(n− 3)(n + 4);

this is equivalent with 17n4 − 174n3 + 559n2 − 822n + 324 > 0. Clearly, theinequivalence is true for any n ≥ 7. So, we obtain

µ1(b1) > µ1(bj).

Finally,

µ1(ai) > µ1(b1) ⇐⇒ 67n2 − 187n + 48

6n(n− 1)(6n− 10)>

65n3 − 392n2 + 615n− 144

6n(n− 1)(n− 3)(8n− 25)

⇐⇒ 146n4 − 1777n3 + 6962n2 − 9363n + 2160 > 0.In conclusion, for α = 2 and β > 3 we have

µ1(e) > µ1(ai) > µ1(b1) > µ1(bj)

therefore

∀ r ≥ 1, rH = 1H, s.f.g.(H) = 1.

60

For α = 2 and β = 3 we have

∀ r ≥ 1, rH = 3H and s.f.g.(H) = 3.

Let us suppose now β > α ≥ 3. One can see that

1 = µ(e) > µ(b1) > µ(bj) and µ(ai) > µ(bj).

Moreover,

µ(b1) > µ(ai) ⇐⇒ α2 + αβ + 2β − α

β(α2 + 2β)≥ 1

α⇐⇒ α2(α− 1) ≥ 2β(β − α).

By consequence, we study two cases:when µ(b1) > µ(ai) and when µ(ai) > µ(b1).

4.2. In the case µ(e) > µ(b1) > µ(ai) > µ(bj) the join space 1H associated is

1H e b1 a1 · · · aα b2 · · · bβ

e e e, b1 H \B1 · · · H \B1 H · · · H

b1 b1 H∗ \B1 · · · H∗ \B1 H∗ · · · H∗

a1 A · · · A H∗ \ {b1} · · · H∗ \ {b1}...

. . ....

......

aα A H∗ \ {b1} · · · H∗ \ {b1}b2 B1 · · · B1...

. . ....

bβ B1

We have

A(e) = 2 +2α

α + 2+

2(n− α− 2)

n=

6nα + 8n− 2α2 − 8α− 8

n(α + 2); q(e) = 2n− 1

whence

µ1(e) =6nα + 8n− 2α2 − 8α− 8

n(2n− 1)(α + 2)·

Moreover

A(b1) = 2 +2α

α + 2+

2α

α + 1+

2(n− α− 2)

n+

2(n− α− 2)

n− 1=

=10n2α2 + 24n2α + 12n2 − 4nα3 − 28nα2 − 50nα− 24n + 2α3 + 10α2 + 16α + 8

n(n− 1)(α + 1)(α + 2)

q(b1) = 4n− 5 so

µ1(b1) =10n2α2 + 24n2α + 12n2 − 4nα3 − 28nα2 − 50nα− 24n + 2α3 + 10α2 + 16α + 8

n(n− 1)(α + 1)(α + 2)(4n− 5)

61

We have also

A(ai) =2α(n− α− 2)

n− 2+

2(n− α− 2)

n− 1+

2(n− α− 2)

n+

2α

α + 2+

2α

α + 1+ α =

=

3n3α3 + 17n3α2 + 24n3α + 8n3 − 2n2α4 − 19n2α3 − 73n2α2 − 98n2α− 36n2++2nα4 + 22nα3 + 84nα2 + 116nα + 48n− 4α3 − 20α2 − 32α− 16

n(n− 1)(n− 2)(α + 1)(α + 2)

q(ai) = α2 + 4α + 4(n− α− 2) + 2α(n− α− 2) = 2nα + 4n− α2 − 4α− 8.

Then

A(bj) =2(β − 1)

n+

2(β − 1)

n− 1+

2α(β − 1)

n− 2+ β − 1 =

= (n− α− 2)n3 + n2 + 2n2α− 2nα− 8n + 4

n(n− 1)(n− 2)

and

q(bj) = 4(β − 1) + 2α(β − 1) + (β − 1)2 = (n− α− 2)(n + α + 2),

whence

µ1(bj) =n3 + n2(1 + 2α) + n(−2α− 8) + 4

n(n + α + 2)(n− 1)(n− 2).

Therefore we have µ1(b1) > µ1(ai) if and only if

10n2α2 + 24n2α + 12n2 − 4nα3 − 28nα2 − 50nα− 24n + 2α3 + 10α2 + 16α + 8

n(n− 1)(α + 1)(α + 2)(4n− 5)>

>

3n3α3 + 17n3α2 + 24n3α + 8n3 − 2n2α4 − 19n2α3 − 73n2α2 − 98n2α− 36n2++2nα4 + 22nα3 + 84nα2 + 116nα + 48n− 4α3 − 20α2 − 32α− 16

n(n− 1)(n− 2)(α + 1)(α + 2)(2nα + 4n− α2 − 4α− 8)

and this is true if and only if

P = (2n2α + 4n2 − nα2 − 8nα− 16n + 2α2 + 8α + 16)(10n2α2 + 24n2α+

+12n2 − 4nα3 − 28nα2 − 50nα− 24n + 2α3 + 10α2 + 16α + 8) >

> Q = (4n− 5)(3n3α3 + 17n3α2 + 24n3α + 8n3 − 2n2α4 − 19n2α3−−73n2α2 − 98n2α− 36n2 + 2nα4 + 22nα3 + 84nα2 + 116nα+

+48n− 4α3 − 20α2 − 32α− 16).

We haveP = n4(20α3 + 88α2 + 120α + 48) + n3(−18α4 − 176α3 − 576α2 − 728α− 288) +

+ n2(4α5 + 84α4 + 494α3 + 1320α2 + 1552α + 608) + n(−10α5− 114α4− 516α3−− 1192α2 − 1312α− 512) + 4α5 + 36α4 + 144α3 + 304α2 + 320α + 128 and

62

Q = n4(12α3+68α2+96α+32)+n3(−8α4−91α3−377α2−512a−18)+n2(18α4+

+183α3 +701α2 +954α+372)+n(−10α4−126α3−500α2−708α−304)+20α3 +

+ 100α2 + 160α + 8, whence

P −Q=n4(8α3+20α2+24α+16)+n3(−10α4−85α3−199α2−216α−104)+

+n2(4α5+66α4+311α3+619α2+598α+236)+n(−10α5−104α4−390α3−692α2−−604α−208)+4α5+36α4+124α3+204α2+160α+48.

Set P −Q = En. Then we find ∀α, α ≥ 3,

E ′′′n = 324α4 + 642α3 + 438α2 + 48α− 240 > 0;

it follows thatE ′′n is a strictly increasing function, soE ′′

n>E ′′2α+1 if and only if

E ′′n > 12(4α2 + 4α + 1)(8α3 + 20α2 + 24α + 16)+

+6(2α + 1)(−10α4 − 85α3 − 199α2 − 216α− 104)+

+8α5 + 132α4 + 622α3 + 1238α2 + 1196α + 472 =

= 272α5 + 396α4 − 68α3 − 388α2 − 292α + 40 > 0.

It follows that E ′n is a strictly increasing function, so ∀n > 2α + 1, E ′

n > E ′2α+1

whence,

E ′n > 4(8α3 + 12α2 + 6α + 1)(8α3 + 20α2 + 24α + 16)+

+3(4α2 + 4α + 1)(−10α4 − 85α3 − 199α2 − 216α− 104)+

+(−10α5 − 104α4 − 390α3 − 692α2 − 604α− 208)+

+2(2α + 1)(4α5 + 66α4 + 311α3 + 619α2 + 598α + 236) =

= 152α6 + 146α5 − 246α4 − 351α3 − 75α2 + 120α + 16 > 0.

We obtain that En is a strictly increasing function and therefore, ∀n > 2α + 1,En > E2α+1. One finds

En = P −Q > (16α4 + 32α3 + 24α2 + 8α + 1)(8α3 + 20α2 + 24α + 16)+

+(8α3 + 12α2 + 6α + 1)(−10α4 − 85α3 − 199α2 − 216α− 104)+

+(4α2 + 4α + 1)(4α5 + 66α4 + 311α3 + 619α2 + 598α + 236)+

+(2α + 1)(−10α5 − 104α4 − 390α3 − 692α2 − 604α− 208)+

+4α5 + 36α4 + 124α3 + 204α2 + 160α + 48 =

= 64α7 + 36α6 − 158α5 − 130α4 + 82α3 + 112α2 − 6α− 12 > 0

and by consequenceµ1(b1) > µ1(ai).

On the other side we have µ1(e) > µ1(b1) if and only if

6nα + 8n− 2α2 − 8α− 8

n(2n− 1)(α + 2)>

>10n2α2+24n2α+12n2−4nα3−28nα2−50nα−24n+2α3+10α2+16α+8

n(n− 1)(α + 1)(α + 2)(4n− 5).

63

This is true if and only if P = (nα+n−α−1)(4n−5)(6nα+8n−2α2−8α−8) >

> Q = (2n− 1)(10n2α2 + 24n2α + 12n2 − 4nα3 − 28nα2 − 50nα− 24n + 2α3 +

+ 10α2 + 16α + 8). We obtain

P = n3(24α2 + 56α + 32) + n2(−8α3 − 94α2 − 190α− 104)+

+ n(18α3 + 120α2 + 214α + 112)− 10α3 − 50α2 − 80α− 40

andQ = n3(20α2 + 48α + 24) + n2(−8α3 − 66α2 − 124α− 60)+

+ n(8α3 + 48α2 + 82α + 40)− 2α3 − 10α2 − 16α− 8

whence

P −Q = n3(4α2 + 8α + 8) + n2(−28α2 − 66α− 44)+

+ n(10α3 + 72α2 + 132α + 72)− 8α3 − 40α2 − 64α− 32.

Set En = P −Q, whence ∀α ≥ 3, we have that

E ′′n = 6n(4α2 + 8α + 8)− 56α2 − 132α− 88 >

> (12α + 6)(4α2 + 8α + 8)− 56α2 − 132α− 88 =

= 48α3 + 96α2 + 96α + 24α2 + 48α + 48− 56α2 − 132α− 88 > 0,

therefore E ′n is a strictly increasing function; from this, one finds ∀α ≥ 3

E ′n > 3(4α2 + 4α + 1)(4α2 + 8α + 8) + 2(2α + 1)(−28α2 − 66α− 44)+

+ 10α3 + 72α2 + 132α + 72 = 48α4 + 42α3 − 44α2 − 56α + 8 > 0,

so En is a strictly increasing function and therefore

En > E2α+1 = (8α3 + 12α2 + 6α + 1)(4α2 + 8α + 8)+

+(4α2+4α+1)(−28α2 − 66α− 44)+(2α + 1)(10α3 + 72α2 + 132α + 72)−−8α3 − 40α2 − 64α− 32 = 32α5 + 20α4 − 46α3 − 24α2 + 26α + 4 > 0.

Then, ∀ i ∈ I(α),

µ1(e) > µ1(b1) > µ1(ai).

Moreover we have µ1(ai) > µ1(bj) if and only if

3n3α3+17n3α2+24n3α+8n3−2n2α4−19n2α3−73n2α2−98n2α−36n2++2nα4+22nα3+84nα2+116nα+48n−4α3−20α2−32α−16

n(n−1)(n−2)(α+1)(α+2)(2nα+4n−α2−4α−8)>

>n3+2n2α+n2−2nα−8n+4

n(n−1)(n− 2)(n+α+2).

64

Set

P = (n + α + 2)(3n3α3 + 17n3α2 + 24n3α + 8n3 − 2n2α4−− 19n2α3 − 73n2α2 − 98n2α− 36n2 + 2nα4 + 22nα3+

+ 84nα2 + 116nα + 48n− 4α3 − 20α2 − 32α− 16) =

= n4(3α3 + 17α2 + 24α + 8) + n3(α4 + 4α3 − 15α2 − 42α− 20)+

+ n2(−2α5 − 21α4 − 89α3 − 160α2 − 116α− 24) + n(2α5 + 26α4+

+ 124α3 + 264α2 + 248α + 80)− 4α4 − 28α3 − 72α2 − 80α− 32

and

Q = (n3+2n2α+n2−2nα−8n+4)(α2+3α+2)(2nα+4n−α2−4α−8) =

= (n3+2n2α+n2−2nα−8n+4)(2nα3+10nα2−α4−7α3−22α2+

+ 16nα−32α+8n−16) = n4(2α3+10α2+16α+8)+

+ n3(3α4+15α3+20α2−8)+n2(−2α5−19α4−87α3−198α2−208α−80)+

+ n(2α5+22α4+108α3+280α2+352α+160)−4α4−28α3−88α2−128α−64

whence

P −Q = n4(α3 + 7α2 + 8α) + n3(−2α4 − 11α3 − 35α2 − 42α− 12)+

+ n2(−2α4 − 2α3 + 38α2 + 92α + 56)+

+ n(4α4 + 16α3 − 16α2 − 104α− 80) + 16α2 + 48α + 32.

Set En = P −Q, so ∀α ≥ 3

E ′′n = 24n(α3 + 7α2 + 8α)− 12α4 − 66α3 − 210α2 − 252α− 72 >

> 48α4 + 336α3 + 384α2 + 24α3 + 168α2 + 192α− 12α4 − 66α3−− 210α2 − 252α− 72 = 36α4 + 294α3 + 342α2 − 60α− 72 > 0,

we obtain

E ′′n > E ′′

2α+1 = 12(4α2 + 4α + 1)(α3 + 7α2 + 8α)+

+ 6(2α + 1)(−2α4 − 11α3 − 35α2 − 42α− 12)−− 4α4 − 4α3 + 76α2 + 184α + 112 =

= 24α5 + 236α4 + 242α3 − 170α2 − 116α + 40 > 0.

It follows

E ′n > E ′

2α+1 = 4(8α3 + 12α2 + 6α + 1)(α3 + 7α2 + 8α)+

+ 3(4α2 + 4α + 1)(−2α4 − 11α3 − 35α2 − 42α− 12)+

+ 2(2α + 1)(−2α4 − 2α3 − 38α2 + 92α + 56)+

+ 4α4 + 16α3 − 16α2 − 104α− 80 =

= 8α6 + 108α5 + 50α4 − 237α3 − 105α2 + 66α− 4 > 0.

65

Therefore En is a strictly increasing function. Then

P −Q > (16α4 + 32α3 + 24α2 + 8α + 1)(α3 + 7α2 + 8α)+

+ (8α3 + 12α2 + 6α + 1)(−2α4 − 11α3 − 35α2 − 42α− 12)+

+ (4α2 + 4α + 1)(−2α4 − 2α3 + 38α2 + 92α + 56)+

+ (2α + 1)(4α4 + 16α3 − 16α2 − 104α− 80) + 16α2 + 48α + 32 =

= 24α6 − 56α5 − 214α4 − 70α3 + 62α2 − 6α− 4 > 0,

so we have, for every β > α ≥ 3, for every i ∈ I(α) and j ∈ I(β) \ {1},

µ1(e) > µ1(b1) > µ1(ai) > µ1(bj)

and therefore, ∀ r ≥ 1, rH = 1H and again we find s.f.g.(H) = 1.

4.3. Set β > α ≥ 3 and

µ(bj) < µ(b1) < µ(ai) < µ(e).

The join space 1H associated is

1H b2 · · · bβ b1 a1 · · · aα e

b2 B1 · · · B1 B H∗ · · · H∗ H...

. . .bβ B1 B H∗ · · · H∗ H

b1 b1 A ∪ {b1} · · · A ∪ {b1} H \B1

a1 A · · · A A ∪ {e}...

. . .aα A A ∪ {e}e e

where B1 = B \ {b1}, H∗ = H \ {e}, α + β + 1 = n. We have

A(e) = 1 +2α

α + 1+

2

n− β + 1+

2(β − 1)

n=

=5α2n− 2α3 − 10α2 + 15nα− 16α + 8n− 8

n(α + 1)(α + 2)

q(e) = 2n− 1, therefore

µ1(e) =5α2n− 2α3 − 10α2 + 15nα− 16α + 8n− 8

n(2n− 1)(α + 1)(α + 2).

Moreover

A(ai) =2α(β − 1)

n− 1+

2(β − 1)

n+

4α

α + 1+

2

α + 2+ α.

66

One obtains

A(ai) =

3n2α3 + 15n2α2 + 22n2α + 6n2 − 2nα4 − 13nα3−−35nα2 − 42nα− 14n + 2α3 + 10α2 + 16α + 8

n(n− 1)(α + 1)(α + 2).

We have also q(ai) = 2α(β−1)+2(β−1)+4α+α2 +2 = 2nα+2n−α2−2α−2,therefore

µ1(ai) =A(ai)

2nα + 2n− α2 − 2α− 2.

Then µ1(e) > µ1(ai) if and only if

5nα2 + 15nα + 8n− 2α3 − 10α2 − 16α− 8

n(2n− 1)(α + 1)(α + 2)>

>

3n2α3 + 15n2α2 + 22n2α + 6n2 − 2nα4 − 13nα3 − 35nα2−−42nα− 14n + 2α3 + 10α2 + 16α + 8

n(n− 1)(α + 1)(α + 2)(2nα + 2n− α2 − 2α− 2).

This is true if and only if

P = (5nα2 + 15nα + 8n− 2α3 − 10α2 − 16α− 8)··(2n2α− 2nα + 2n2 − 2n− nα2 + α2 − 2nα + 2α− 2n + 2) >

> Q = (2n− 1)(3n2α3 + 15n2α2 + 22n2α + 6n2 − 2nα4 − 13nα3−− 35nα2 − 42nα− 14n + 2α3 + 10α2 + 16α + 8).

We obtainP = n3(10α3 +40α2 +46α+16)+n2(−9α4−59α3−140α2−140α−48)+n(2α5 +

+ 23α4 + 89α3 + 160α2 + 142α + 48)− 2α5 − 14α4 − 40α3 − 60α2 − 48α− 16 andQ = n3(6α3 + 30α2 + 44α + 12) + n2(−4α4− 29α3− 85α2− 106α− 34) + n(2α4 +

+ 17α3 + 55α2 + 74α + 30)− 2α3 − 10α2 − 16α− 8.Then

P−Q =n3(4α3 + 10α2 + 2α + 4) + n2(−5α4 − 30α3 − 55α2 − 34α− 14)+

+n(2α5+21α4+72α3+105α2+68α+18)−2α5−14α4−38α3−50α2−32α−8.

Let En be P −Q. We will show that En it is a strictly increasing function.We know that β > α ≥ 2 =⇒ n > 2α + 1 (n ≥ 6)

E ′′n = 6n(4α3 + 10α2 + 2α + 4)− 10α4 − 60α3 − 110α2 − 68α− 28 >

> 6(2α + 1)(4α3 + 10α2 + 2α + 4)− 10α4 − 60α3 − 110α2 − 68α− 28 =

= 38α4 + 84α3 − 26α2 − 8α− 4 > 0.

So, ∀n ≥ 6, E ′′n > 0, whence E ′

n is a strictly increasing function. Since n > 2α+1we have E ′

n > E ′2α+1, therefore

E ′n > 3(4α2 + 4α + 1)(4α3 + 10α2 + 2α + 4)−−2(2α + 1)(5α4 + 30α3 + 55α2 + 34α + 14)+

+2α5 + 21α4 + 72α3 + 105α2 + 68α + 18 =

= 30α5 + 59α4 − 52α3 − 39α2 − 2α + 2 > 0.

67

So En is a strictly increasing function. By consequence, En > E2α+1, that is

P −Q > (8α3 + 12α2 + 6α + 1)(4α3 + 10α2 + 2α + 4)+

+(4α2 + 4α + 1)(−5α4 − 30α3 − 55α2 − 34α− 14)+

+(2α + 1)(2α5 + 21α4 + 72α3 + 105α2 + 68α + 18)−−2α5 − 14α4 − 38α3 − 50α2 − 32α− 8 =

= 16α6 + 30α5 − 34α4 − 22α3 + 14α2 + 8α > 0.

Thenµ1(e) > µ1(ai)

Moreover we have

A(b1) =2(β − 1)

β+

2α(β − 1)

n− 1+

2(β − 1)

n+ 1 +

2α

α + 1+

2

α + 2=

=

(2nα2 + 6nα + 4n− 2α3 − 10α2 − 16α− 8)··(2n2 + n2α− α2n− 2nα− 3n + α + 1)+

+(3α2 + 9α + 4)(n3 − 2n2 − n2α + nα + n)

(α + 1)(α + 2)n(n− 1)(n− α− 1)

q(b1) = 2(β − 1) + 2α(β − 1) + 2(β − 1) + 1 + 2α + 2 =

= 2nα + 4n− 2α2 − 6α− 5.

We have also, for j > 1,

A(bj) = β−1+2(β − 1)

β+

2α(β − 1)

n− 1+

2(β − 1)

n= (β−1)

(2

n+

2α

n− 1+

2

β+ 1

)

q(bj) = (β − 1)2 + 2(β − 1) + 2α(β − 1) + 2(β − 1) = (β − 1)(n + α + 2),

so

µ1(bj) =1

n + α + 2

(2

n+

2α

n− 1+

2

n− α− 1+ 1

)=

=n3 + n2α + 2n2 − 2nα2 − 3nα− 5n + 2α + 2

n(n− 1)(n− α− 1)(n + α + 2)·

Therefore µ1(ai) > µ1(bj) if and only if

3n2α3 + 15n2α2 + 22n2α + 6n2 − 2nα4 − 13nα3 − 35nα2 − 42nα−14n + 2α3 + 10α2 + 16α + 8

n(n− 1)(α + 1)(α + 2)(2nα + 2n− α2 − 2α− 2)>

>n3 + n2α + 2n2 − 2nα2 − 3nα− 5n + 2α + 2

n(n− 1)(n− α− 1)(n + α + 2)

if and only if

P = (n2 + n− α2 − 3α− 2)(3n2α3 + 15n2α2 + 22n2α + 6n2 − 2nα4−−13nα3 − 35nα2 − 42nα− 14n + 2α3 + 10α2 + 16α + 8) >

> Q = (2nα3 + 2nα2 − α4 − 2α3 − 2α2 + 6nα2 + 6nα− 3α3 − 6α2 − 6α+

+4nα + 4n− 2α2 − 4α− 4)(n3 + n2α + 2n2 − 2nα2 − 3nα− 5n + 2α + 2).

68

We obtain

P = n4(3α3 + 15α2 + 22α + 6) + n3(−2α4 − 10α3 − 20α2 − 20α− 8)+

+n2(−3α5 − 26α4 − 84α3 − 127α2 − 88α− 18) + n(2α6 + 19α5 + 79α4+

+175α3 + 220α2 + 142α + 36)− 2α5 − 16α4 − 50α3 − 76α2 − 56α− 16

and

Q = (2nα3 + 8nα2 + 10nα + 4n− α4 − 5α3 − 10α2 − 10α− 4)(n3 + n2α + 2n2−−2nα2 − 3nα− 5n + 2α + 2) = n4(2α3 + 8α2 + 10α + 4) + n3(α4 + 7α3 + 16α2+

+14α+4)+n2(−5α5−29α4−74α3−108α2−86α−28)+n(2α6+13α5+44α4+

+95α3 + 124α2 + 90α + 28)− 2α5 − 12α4 − 30α3 − 40α2 − 28α− 8

whence,

P −Q = n4(α3 + 7α2 + 12α + 2) + n3(−3α4 − 17α3 − 36α2 − 34α− 12)+

+n2(2α5 + 3α4 − 10α3 − 19α2 − 2α + 10) + n(6α5 + 34α4 + 80α3 + 96α2+

+52α + 18)− 4α4 − 20α3 − 36α2 − 28α− 8.

Set P −Q = En. We have

E ′′′n = 24n(α3 + 7α2 + 12α + 2)− 18α4 − 102α3 − 216α2 − 204α− 72 >

> (48α+24)(α3+7α2+12α+2)−18α4−102α3−216α2−204α−72=

= 30α4 + 258α3 + 528α2 + 180α− 24 > 0,

whence E ′′n is a strictly increasing function, so ∀n > 2α + 1 we have

E ′′n > E ′′

2α+1. Since

E ′′n > 12(4α2 + 4α + 1)(α3 + 7α2 + 12α + 2)+

+ 6(2α + 1)(−3α4 − 17α3 − 36α2 − 34α− 12)+

+ 4α5 + 6α4 − 20α3 − 38α2 − 4α + 20 =

= 16α5 + 168α4 + 370α3 + 94α2 − 112α− 28 > 0,

it follows E ′n is a strictly increasing function, so E ′

n > E ′2α+1. We have

E ′n > 4(8α3 + 12α2 + 6α + 1)(α3 + 7α2 + 12α + 2)+

+ 3(4α2 + 4α + 1)(−3α4 − 17α3 − 36α2 − 34α− 12)+

+ (4α + 2)(2α5 + 3α4 − 10α3 − 19α2 − 2α + 10)+

+ 6α5 + 34α4 + 80α3 + 96α2 + 52α + 8 =

= 4α6 + 54α5 + 99α4 − 95α3 − 198α2 − 62α > 0,

therefore En is a strictly increasing, and thus ∀n ≥ 2α + 2 we have En ≥ E2α+2.

69

Then

En ≥ (16α4 + 64α3 + 96α2 + 64α + 16)(α3 + 7α2 + 12α + 2)+

+ (8α3 + 24α2 + 24α + 8)(−3α4 − 17α3 − 36α2 − 34α− 12)+

+ (4α2 + 8α + 4)(2α5 + 3α4 − 10α3 − 19α2 − 2α + 10)+

+ (2α + 2)(6α5 + 34α4 + 80α3 + 96α2 + 52α + 8)−− 4α4 − 20α3 − 36α2 − 28α− 8 =

= 8α6 + 40α5 + 48α4 − 36α3 − 112α2 − 76α− 32 > 0.

By consequence µ1(ai) > µ1(bj).

On the other side, µ1(ai) > µ1(b1) if and only if

3n2α3 + 15n2α2 + 22n2α + 6n2 − 2nα4 − 13nα3 − 35nα2 − 4nα−−14n + 2α3 + 10α2 + 16α + 8

n(n− 1)(α + 1)(α + 2)(2nα + 2n− α2 − 2α− 2)>

>

(2nα2 + 6nα + 4n− 2α3 − 10α2 − 16α− 8)(2n2 + n2α− nα2 − 2nα−−3n + α + 1) + (3α2 + 9α + 4)(n3 − 2n2 − n2α + nα + n)

n(n− 1)(α + 1)(α + 2)(n− α− 1)(2nα + 4n− 2α2 − 6α− 5).

Set P = (2n2α + 4n2 − 2nα2 − 6nα− 5n− 2nα2 − 4nα + 2α3 + 6α2 + 5α−−2nα−4n+2α2+6α+5)(3n2α3+15n2α2+22n2α+6n2−2nα4−13nα3−35nα2−− 42nα− 14n + 2α3 + 10α2 + 16α + 8) andQ = (2nα+2n−α2−2α−2)(2n3α3 +13n3α2 +25n3α+12n3−4n2α4−27n2α3−− 73n2α2− 88n2α− 36n2 + 2nα5 + 14nα4 + 47nα3 + 90nα2 + 87nα + 32n− 2α4−− 12α3 − 26α2 − 24α− 8).We obtain

P = n4(6α4 + 42α3 + 104α2 + 100α + 24) + n3(−16α5 − 130α4 − 417α3−− 647α2 − 466α− 110) + n2(14α6 + 130α5 + 515α4+1101α3 + 1312α2+

+ 802α + 188) + n(−4α7 − 42α6 − 204α5 − 518α4 − 1016α3 − 1063α2−− 604α−142)+4α6+36α5+134α4+264α3+290α2+168α+40

and

Q = n4(4α4 + 30α3 + 76α2 + 74α + 24)+

+ n3(−10α5 − 79α4 − 255α3 − 430α2 − 342α− 96)+

+ n2(8α6 + 67α5 + 257α4 + 562α3 + 732α2 + 506α + 136)+

+ n(−2α7 − 18α6 − 83α5 − 240α4 − 437α3 − 486α2 − 302α− 72)+

+ 2α6 + 16α5 + 54α4 + 124α3 + 108α2 + 64α + 16.

70

So

P −Q = n4(2α4 + 12α3 + 28α2 + 26α)+

+ n3(−6α5 − 51α4 − 162α3 − 217α2 − 124α− 14)+

+ n2(6α6 + 63α5 + 258α4 + 539α3 + 580α2 + 296α + 52)+

+ n(−2α7 − 24α6 − 121α5 − 341α4 − 579α3 − 577α2 − 302α− 70)+

+ 2α6 + 20α5 + 80α4 + 140α3 + 182α2 + 104α + 34.

For n > 2α + 1 and α ≥ 2 we have

P −Q > (16α4 + 32α3 + 24α2 + 8α + 1)(2α4 + 12α3 + 28α2 + 26α)+

+(8α3 + 12α2 + 6α + 1)(−6α5 − 51α4 − 162α3 − 217α2 − 124α− 14)+

+(4α2 + 4α + 1)(6α6 + 63α5 + 258α4 + 539α3 + 580α2 + 296α + 52)+

+(2α + 1)(−2α7 − 24α6 − 121α5 − 341α4 − 579α3 − 577α2 − 302α− 70)+

+2α6 + 20α5 + 80α4 + 140α3 + 182α2 + 104α + 34.

Therefore we find

P −Q > 14α8 + 2α7 − 38α6 + 92α5 + 238α4 + 246α3 + 80α2 − 16α + 2 > 0.

It followsµ1(ai) > µ1(b1)

Therefore we have obtained

µ1(e) > µ1(ai) > µ1(b1), µ1(ai) > µ1(bj)

For j > 1 we have µ1(bj) < µ1(b1) if and only if

n3 + n2α + 2n2 − 2nα2 − 3nα− 5n + 2α + 2

n(n− 1)(n− α− 1)(n + α + 2)<

<

(2nα2 + 6nα + 4n− 2α3 − 10α2 − 16α− 8)··(2n2 + n2α− α2n− 2nα− 3n + α + 1)+

+(3α2 + 9α + 4)(n3 − 2n2 − n2α + nα + n)

n(n− 1)(α + 1)(α + 2)(n− α− 1)(2nα + 4n− 2α2 − 6α− 5)

We set again

P = (2nα+4n−2α2−6α−5)(n3α2+3n3α+2n3+n2α3+5n2α2+8n2α+

+4n2 − 2nα4 − 9nα3 − 18nα2 − 21nα− 10n + 2α3 + 8α2 + 10α + 4)

whence P = n4(2α3 + 10α2 + 16α + 8) + n3(2α3 + 9α2 + 13α + 6) + n2(−6α5 −42α4 − 123α3 − 195α2 − 168α − 60) + n(4α6 + 30α5 + 104α4 + 219α3 + 288α2 +213α + 66)− 4α5 − 28α4 − 78α3 − 108α2 − 74α− 20 and

Q = (n+α+2)(2n3α3+13n3α2+25n3α+12n3−4n2α4−27n2α3−73n2α2−88n2α−36n2+

+2nα5 + 14nα4 + 47nα3 + 90nα2 + 87nα + 32n− 2α4 − 12α3 − 26α2 − 24α− 8),

71

whence Q = n4(2α3 + 13α2 + 25α + 12) + n3(−2α4 − 10α3 − 22α2 − 26α− 12) +

+ n2(−2α5 − 21α4 − 80α3 − 144α2 − 125α− 40) + n(2α6 + 18α5 + 73α4 +

+ 172α3 + 241α2 + 182α + 56)− 2α5 − 16α4 − 50α3 − 76α2 − 56α− 16.Hence

P −Q = n4(−3α2 − 9α− 4) + n3(2α4 + 12α3 + 31α2 + 39α + 18)+

+n2(−4α5 − 21α4 − 43α3 − 51α2 − 43α− 20) + n(2α6 + 12α5 + 31α4+

+47α3 + 47α2 + 31α + 10)− 2α5 − 12α4 − 28α3 − 32α2 − 18α− 4.

Set α = 3. We have

for n = 8, P −Q > 0, whence µ1(bj) > µ1(b1)

for n ≥ 9, P −Q < 0, whence µ1(bj) < µ(b1).

Set α = 4. We have P −Q < 0 if and only if n ≥ 13, so

for n < 13, µ1(bj) > µ1(b1) and for n ≥ 13, µ(bj) < µ1(b1).

Set α = 5. We have P −Q < 0 if and only if n ≥ 25. So

for n < 25, µ1(bj) > µ1(b1) and for n ≥ 25, µ1(bj) < µ1(b1)

The following problem remains open:To find a function n(α) such that P −Q < 0 if and only if n ≥ n(α).

Let us consider now the case α = 3, β = 4, whence n = 8.Set H = A ∪ B ∪ {e}, where A = {a1, a2, a3}, B = {b1, b2, b3, b4} and let us

suppose ∀ i, ∀ j ∈ B′ = B − {b1},

µ(e) > µ(ai) > µ(bj) > µ(b1)

so we have 1H as follows, where H ′ = H \ {b1}.1H b1 b2 b3 b4 a1 a2 a3 e

b1 b1 B B B A,B A,B A,B H

b2 B B′ B′ B′ A,B′ A,B′ A,B′ H ′

b3 B B′ B′ B′ A,B′ A,B′ A,B′ H ′

b4 B B′ B′ B′ A,B′ A,B′ A,B′ H ′

a1 A, B A,B′ A,B′ A,B′ A A A A, e

a2 A, B A,B′ A,B′ A,B′ A A A A, e

a3 A, B A,B′ A,B′ A,B′ A A A A, e

e H H ′ H ′ H ′ A, e A, e A, e e

Therefore A(b1) = 1 +6

4+

6

7+

2

8=

101

28· q(b1) = 15, whence 2µ(b1) = 0, 240.

We have also ∀ j > 1, A(bj) =6

4+

6

7+

2

8+

9

3+

18

6+

6

7=

265

28, q(bj) = 47

72

whence 2µ(bj) = 0, 20136. Moreover, ∀ i, A(ai) =6

7+

2

8+

18

6+

6

7+

9

3+

6

4=

265

28,

q(ai) = 47, whence 2µ(ai) = 2µ(bj) and, finally,

A(e) =2

8+

6

7+

6

4+ 1 =

101

28, q(e) = 15,

whence2µ(e) = 2µ(b1).

One obtains the second join space associated 2H, where B′ ∪ A = C.

2H e b1 b2 b3 b4 a1 a2 a3

e e, b1 e, b1 H H H H H H

b1 e, b1 e, b1 H H H H H H

b2 H H C C C C C C

b3 H H C C C C C C

b4 H H C C C C C C

a1 H H C C C C C C

a2 H H C C C C C C

a3 H H C C C C C C

We have A(e) = A(b1) =4

2+

24

8= 5 and q(e) = q(b1) = 28, whence 3µ(b1) =

3µ(e) = 0, 178. Moreover, ∀ j > 1, ∀ i,

A(bj) = A(ai) =24

8+

36

6= 9, q(bj) = q(ai) = 60,

whence 3µ(bj) = 3µ(ai) = 0, 15. It follows 3H = 2H, hence

∀ r > 2, rH = 2H and s.f.g.(H) = 2.

3.3.6 Fuzzy grade of i.p.s. hypergroups

Corsini proved in [6] that any i.p.s. hypergroup with the cardinal smaller than 9is strongly canonical.

For n ≤ 8, have been determined by Corsini in [6], [7], [8] all the non-isomorphic i.p.s. hypergroups of cardinal n; their fuzzy grades have been studiedby Corsini and Cristea in [18] and [19]. In the following we present the resultsobtained in this direction.

Theorem 3.3.26.

(i) There is a unique i.p.s. hypergroup H of order 3 and s.f.g.(H) = 1.

(ii) There are three non isomorphic i.p.s. hypergroups H1, H2, H3 of order 4 ands.f.g.(H1) = s.f.g.(H2) = 1, s.f.g.(H3) = 2.

73

(iii) There are eight non isomorphic i.p.s. hypergroups of order 5: one of themhas s.f.g.(H) = 2 and the other one have s.f.g.(H) = 1.

(iv) There are nineteen non isomorphic i.p.s. hypergroups of order 6: fourteenof them have s.f.g.(H) = 1, four of them are of s.f.g.(H) = 2 and one iswith s.f.g.(H) = 3.

(v) There are thirty-six non isomorphic i.p.s. hypergroups of order 7: two ofthem have f.g.(H) = 1, twenty-five of them have s.f.g.(H) = 1, eight ofthem are with s.f.g.(H) = 2 and only for of them have s.f.g.(H) = 3

The proof is detailed in the Appendix B.

3.3.7 The sequence of join spaces associated with a roughset

Rough sets theory, proposed by Zdzislaw Pawlak in 1982, has been attractingresearchers and practitioners in various fields of science and technology. Theinterest in rough sets theory and applications is remarkable since the beginningand it is still growing. The ingenious concepts of rough set have been a basefor original developments in both theoretical research, including logics, algebraand topology, and applied research, including knowledge discovery, data mining,decision theory, artificial intelligence. Latter have been found many applicationsof them in medicine, engineering, chemistry, bioinformatics, economy.

Let R be an equivalence relation on a universe H; for any element x ∈ H, wedenote its equivalence class related to R by R(x). For any non-empty subset A ofH, a rough set of H is determined by the pair (R(A), R(A)), where

R(A) =⋃

R(x)⊂A

R(x) and R(A) =⋃

R(x)∩A6=∅R(x).

R.Biswas proved in 1999 (see [2]) that any rough set is a particular type offuzzy set, but conversely not.

Later P.Corsini has associated with a universe H, endowed with a rough set,the following hyperoperation:

∀(x, y) ∈ H2, x ◦ y = R({x, y}) \R({x, y}).Proposition 3.3.27. The partial hyperoperation ”◦” is defined everywhere if andonly if for any x ∈ H, | R(x) | ≥ 3.

Proof. Let us suppose there exists x ∈ H such that R(x) = {x, x′}, with x 6= x′.Then

x ◦ x′ =⋃

R(z)∩{x,x′}6=∅R(z) \

⋃

R(z)⊂{x,x′}R(z) = R(x) \R(x) = ∅.

Let us suppose now there exists x ∈ H such that R(x) = {x}. By the same wayone finds x ◦ x = R(x) \R(x) = ∅. Therefore, for any x ∈ H, |R(x)|≥ 3.

74

We prove now the other implication. For any x ∈ H, set |R(x)| ≥ 3. Then wehave

R({x, y}) =⋃

R(z)⊂{x,y}R(z) = ∅,

whence x ◦ y = R(x) ∪R(y) 6= ∅.

Proposition 3.3.28. The hypergroupoid 〈H, ◦〉 is a join space if and only if, forany x ∈ H, |R(x)| ≥ 3.

Proof. By the previous proposition it is sufficient to prove that if 〈H, ◦〉 is ahypergroupoid, then it is a join space. We remark that the hypothesis x ∈ H,|R(x)| ≥ 3 implies x ◦ y = R(x) ∪ R(y). It follows that 〈H, ◦〉 is a commutativesemihypergroup.

Moreover, since every x is an identity, it follows that the reproducibility lawis satisfies. So 〈H, ◦〉 is a commutative hypergroup.

It remains to prove that the implication a/b ∩ c/d 6= ∅ =⇒ a ◦ d ∩ b ◦ c 6= ∅ issatisfies.

Set x ∈ a/b ∩ c/d, that is a ∈ b ◦ x and c ∈ d ◦ x, whence

a ∈ R(b) ∪R(x), c ∈ R(d) ∪R(x).

We have a ∈ a ◦ d, so, if a ∈ R(b) ⊂ b ◦ c, it results a ∈ a ◦ d ∩ b ◦ c.By the same way c ∈ R(d) implies c ∈ a ◦ d ∩ b ◦ c.We suppose now a /∈ R(b) and c /∈ R(d); it follows a ∈ R(x), whence

x ∈ R(a) ⊂ a◦d and c ∈ R(x), whence x ∈ R(c) ⊂ b◦c. Therefore a◦d∩b◦c 6= ∅.

In the following we consider H a universe of order n endowed with a roughset.

We associate with this join space the fuzzy set µ defined by the relations (ω)from the paragraph 3.3.1. We obtain, for any x, y ∈ H and any u ∈ x ◦ y,

µx,y(u) =

1

| R(x) | , if (x, y) ∈ R

1

| R(x) | + | R(y) | , if (x, y) /∈ R

We denote the factor set of H related to the relation R by

H/R = {R(x1), R(x2), ..., R(xn)} with | R(xi) |= λi.

For any u ∈ H, there exists i ∈ {1, 2, ..., n} such that u ∈ R(xi) and then

µ(u) =

1 + 2

(λ1

λ1 + λi

+λ2

λ2 + λi

+ ... +λi−1

λi−1 + λi

+λi+1

λi+1 + λi

+ ... +λn

λn + λi

)

2n− λi

.

If |R(u)|=|R(v)|, then µ(u) = µ(v).

75

Conversely, if µ(u) = µ(v), then taking u ∈ R(xi) and v ∈ R(xj) with i < jwe obtain

1+2

(λ1

λ1 + λi

+λ2

λ2 + λi

+ ... +λi−1

λi−1 + λi

+λi+1

λi+1 + λi

+ ... +λn

λn + λi

)

2n− λi

=

=

1+2

(λ1

λ1 + λj

+λ2

λ2 + λj

+ ... +λj−1

λj−1 + λj

+λj+1

λj+1 + λj

+ ... +λn

λn + λj

)

2n− λj

⇐⇒

2n− λj + 2(2n− λj)

(λ1

λ1 + λi

+ ... +λi−1

λi−1 + λi

+λi+1

λi+1 + λi

+ ... +λn

λn + λi

)=

=2n− λi + 2(2n− λi)

(λ1

λ1 + λj

+ ... +λj−1

λj−1 + λj

+λj+1

λj+1 + λj

+ ... +λn

λn + λj

).

Calculating, we obtain

(λi − λj) + 4n(λj − λi)

[λ1

(λ1 + λi)(λ1 + λj)+ ... +

λi−1

(λi−1 + λi)(λi−1 + λj)+

+λi+1

(λi+1+λi)(λi+1+λj)+ ... +

λj−1

(λj−1+λi)(λj−1+λj)+

λj+1

(λj+1+λi)(λj+1+λj)+ ...+

+λn

(λn + λi)(λn + λj)+

1

λi + λj

]+ 2(λi − λj)

[λ1(λ1 + λi + λj)

(λ1 + λi)(λ1 + λj)+ ...+

+λi−1(λi−1 + λi + λj)

(λi−1+λi)(λi−1+λj)+

λi+1(λi+1 + λi + λj)

(λi+1+λi)(λi+1+λj)+ ... +

λj−1(λj−1 + λi + λj)

(λj−1+λi)(λj−1+λj)+

+λj+1(λj+1 + λi + λj)

(λj+1 + λi)(λj+1 + λj)+ ... +

λn(λn + λi + λj)

(λn + λi)(λn + λj)+ 1

]= 0.

It is equivalent with

(λi − λj)

[1 +

2λ1(λ1 + λi + λj − 2n)

(λ1 + λi)(λ1 + λj)+ ... +

2λi−1(λi−1 + λi + λj − 2n)

(λi−1 + λi)(λi−1 + λj)+

+2λi+1(λi+1 + λi + λj − 2n)

(λi+1 + λi)(λi+1 + λj)+ ... +

2λj−1(λj−1 + λi + λj − 2n)

(λj−1 + λi)(λj−1 + λj)+

+2λj+1(λj+1+λi+λj−2n)

(λj+1 + λi)(λj+1 + λj)+ ... +

2λn(λn+λi+λj−2n)

(λn + λi)(λn + λj)+

2λi+2λj−4n

λi + λj

]= 0

76

Since all the term of the form2λ1(λ1 + λi + λj − 2n)

(λ1 + λi)(λ1 + λj)is negative and also

1 +2λi+2λj−4n

λi + λj

< 0 we obtain that the previous relation is 0 if and only if

λi = λj.

We conclude that µ(u) = µ(v) ⇐⇒ |R(u)|=|R(v)|.

Moreover, we have µ(u) < µ(v) ⇐⇒ |R(u)|>|R(v)|.

We observe that if R is an equivalence relation on H such that, for any i ∈{1, 2, ..., n}, we have | R(xi) |= k = const, then µ(x) = µ(y), for any x, y ∈ H andtherefore the join space 1H associated is a total hypergroup and s.f.g.(H) = 1.

3.3.8 Properties of the join spaces iH associated with thehypergroupoid H

Besides the fundamental relation β defined on a hypergroup, can be defined otherthree relations, named also fundamental and introduced by Jantosciak in [29].With this relations he introduced the notion of reduced hypergroup. In the follo-wing we find a necessary and sufficient condition for the join spaces iH in orderto be reduced hypergroups.

First, we remind some definitions and results from the paper of Jantosciak.

Definition 3.3.29. (see [29]) Let x, y be arbitrary elements of H.

(i) x and y are called operationally equivalent if x ◦ a = y ◦ a and a ◦ x = a ◦ y,for any a ∈ H;

(ii) x and y are called inseparable if for a, b ∈ H, x ∈ a◦ b if and only if y ∈ a◦ b;

(iii) x and y are called essentially indistinguishable if x and y are operationallyequivalent and inseparable.

Remark. (see [29]) The three relations, operational equivalence, inseparabilityand essential indistinguishability, denoted by ∼o, ∼i and ∼e respectively, areequivalence relations on H.

For any x ∈ H, let xo, xi and xe, respectively denote the equivalence class ofx respect to the relations ∼o,∼i and ∼e.

Proposition 3.3.30. (see [29]) Let be x, y ∈ H. Then

(i) xo yo = x ◦ y;

(ii) (x ◦ y)i = x ◦ y,

(iii) xe = xo ∩ xi and xe ye = (x ◦ y)e = x ◦ y.

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Definition 3.3.31. (see [29]) A hypergroup H is said to be reduced if its e-classesare singletons, that is, for any x ∈ H, xe = {x}.Remark. Any group is a reduced hypergroup.

Remark. A hypergroup H is reduced if, for any x ∈ H, xo = {x} or xi = {x}.Proposition 3.3.32. (see [30]) For any hypergroup (H, ◦), the quotient hyper-groupoid (H/ ∼e, ?) is a reduced hypergroup, where the hyperoperation ? on H/ ∼e

is defined byxe ? ye = {ze | z ∈ x ◦ y}.

Proof. First note that, for all a, b ∈ H, it takes place the relation

a ◦ b = (a ◦ b)e =⋃{ze | z ∈ a ◦ b} =

⋃{ze | z ∈ ae ◦ be} =

⋃ae ? be. (1)

Suppose xe and ye are essential indistinguishable in H/ ∼e. Then xe and ye areoperationally equivalent, thus xe ? ae = ye ? ae for all ae ∈ H/ ∼e. By (1), forevery a ∈ H we obtain

x ◦ a =⋃

xe ? ae =⋃

ye ? ae = y ◦ a.

Similarly, a ◦ x = a ◦ y. Hence x and y are operationally equivalent in H.But xe and ye are also inseparable in H/ ∼e, therefore, for all ae, be ∈ H/ ∼e,

one has xe ∈ ae ? be if and only if ye ∈ ae ? be. By (1), for every a, b ∈ H, z ∈ a ◦ b

if and only if ze ∈ ae ? be. For a, b ∈ H, we have x ∈ a ◦ b if and only if y ∈ a ◦ b.Hence x and y are inseparable in H.

Therefore, x and y are essential indistinguishable in H, that is xe = ye.

Theorem 3.3.33. (see [54]) For all i, i ≥ 1, the quotient hypergroup iH/Reµiis a

reduced hypergroup.

Proof. First, for any x ∈ H, we determine the equivalence class of x in the iHrespect to the equivalence relation ∼e on iH.

By the Definition 3.3.29 we have

xi ={y∈H | y ∈ a ◦i b ⇔ x ∈ a ◦i b, for a, b ∈ H} =

={y∈H | µi−1(y)∈ [µi−1(a), µi−1(b)]⇔ µi−1(x)∈ [µi−1(a), µi−1(b)], a, b∈H}=

={y∈H | µi−1(y) = µi−1(x)}where

[µi−1(a), µi−1(b)] = {c ∈ [0, 1] | µi−1(a) ∧ µi−1(b) ≤ c ≤ µi−1(a) ∨ µi−1(b)}.Then

xo = {y ∈ H | x ◦i a = y ◦i a, for any a ∈ H} =

= {y ∈ H | [µi−1(x), µi−1(a)] = [µi−1(y), µi−1(a)], for any a ∈ H} =

= {y ∈ H | µi−1(y) = µi−1(x)}.

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By consequence, for any x ∈ H, xe = xi ∩ xo = {y∈H | µi−1(y) = µi−1(x)} in iHand therefore iH/ ∼e=

i−1H/Reµi−1. By the Proposition 3.3.32 it follows that the

quotient hypergroup i−1H/Reµi−1is a reduced hypergroup.

We can not say the same thing about the join spaces iH from the sequenceassociated with H, but we obtain the following results.

Theorem 3.3.34. (see [54]) The join space 1H is a reduced hypergroup if andonly if the n-tuple associated with H is (1, 1, ..., 1). Moreover, the join space 1H isthe unique join space of the sequence (〈rH, ◦r〉, µr)r≥1 corresponding to H whichcan be a reduced hypergroup.

Proof. Assume that (iH, ◦i), i > 1, is a reduced hypergroup, that is, for anyx ∈ H, xe = {x} and as we have already seen, xe = {y ∈ H | µi−1(y) = µi−1(x)}.It follows that, for any y ∈ H, y 6= x, µi−1(y) 6= µi−1(x)} and therefore the n-tupleassociated with the join space (i−1H, ◦i−1) is (1, 1, ..., 1).Conversely the proof is clear.

If iH is a reduced hypergroup then for all j < i − 1 the n-tuple associatedwith the join space jH is (1, 1, ..., 1) and by consequence all the join spaces jH,j < i − 1, are isomorphic with H. Therefore, only the join space 1H can be areduced hypergroup.

3.4 Other examples of reduced hypergroups

In this section we present some properties of the three fundamental relations defi-ned by Jantosciak in [29] and we give some examples of the reduced hypergroups.

Proposition 3.4.1.

(i) The operational equivalence is regular, but in general not strongly regular.

(ii) The inseparability is not a regular relation.

(iii) The essential indistinguishability is regular, but in general not strongly re-gular.

Proof. (i) We consider x, y ∈ H such that x ∼o y and u arbitrary in H. Thenx ◦u = y ◦u and there exists b = a ∈ y ◦u such that a ∼o b (since ∼o is reflexive).Therefore ∼o is regular.

Now we give an example of a hypergroup H on which the relation ∼o is notstrongly regular.Let H = {a, b, c} be the following hypergroup

H a b ca a a a, b, cb a a a, b, cc a, b, c a, b, c c

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We observe that a ∼o b and we don’t have a ∼o c.We suppose that ∼o is strongly regular, that is:

a ∼o b =⇒ ∀u ∈ H, ∀x ∈ a ◦ u, ∀y ∈ b ◦ u : x ∼o y.

For u = c, x = a and y = c it would follow a ∼o c, which is a contradiction.(ii) Let H = {a, b, c} be the following hypergroupoid

H a b ca a a a, b, cb a a b, cc a, b, c b, c b, c

We suppose that the relation ∼i is regular, so from b ∼i c it results ∀u ∈ H, ∀x ∈b ◦ u,∃y ∈ c ◦ u such that x ∼i y. For u = b and x = a we obtain that there existsy ∈ {b, c} such that a ∼i b or a ∼i c, which is a contradiction.

(iii) We prove that the relation ∼e is regular, that is,a ∼e b =⇒ ∀u ∈ H, ∀x ∈ a ◦ u,∃y ∈ b ◦ u : x ∼e y.From a ∼e b we have a ∼o b, so ∀u ∈ H, a ◦ u = b ◦ u. Then, there existsy = x ∈ b ◦ u such that x ∼e y (since the relation ∼e is reflexive).

Now we give an example of a hypergroup H on which the relation ∼e is notstrongly regular.Let H = {a, b, c} be the following hypergroup

H a b ca a, b a, b a, b, cb a, b a, b a, b, cc a, b, c a, b, c a, b, c

We easily observe that a ∼e b and we suppose ∼e is strongly regular, then ∀u ∈H, ∀x ∈ a◦u, ∀y ∈ b◦u it results x ∼e y. For u = c, x = a, y = c it follows a ∼e c,which is a contradiction.

Proposition 3.4.2. If H is a hypergroup with a scalar identity, then, for a, b ∈ H,we have:

x ∼o y ⇐⇒ x ∼i y ⇐⇒ x ∼e y ⇐⇒ x = y.

(In this case all the three relations are regular).

Proof. Let e be a scalar identity in H.Let us consider a, b ∈ H such that a ∼o b, that is ∀x ∈ H, a ◦ x = b ◦ x and

x ◦ a = x ◦ b. For x = e we have a = a ◦ e = b ◦ e = b, so a = b.Let us consider now a, b ∈ H such that a ∼i b, that is a ∈ x ◦ y ⇐⇒ b ∈ x ◦ y,

for x, y ∈ H. But a = a ◦ e, so a ∈ a ◦ e and thus b ∈ a ◦ e = a. We obtain a = b.

Proposition 3.4.3. Let (H, ◦) be a hypergroupoid. The relation ∼i is regular onH if and only if a ∼i b =⇒ a ∼o b, for a, b ∈ H.

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Proof. Let us suppose that the relation ∼i is regular on H.We consider a, b ∈ H such that a ∼i b. We have to prove that ∀u ∈ H, a◦u = b◦uand u ◦ a = u ◦ b.For x ∈ a ◦u, by the regularity of ∼i, there exists y ∈ b ◦u such that x ∼i y; sincex ∼i y it follows x ∈ b ◦ u, so a ◦ u ⊂ b ◦ u.Similarly, b ◦ u ⊂ a ◦ u, so a ◦ u = b ◦ u, for any u ∈ H.In the same way we prove u ◦ a = u ◦ b, for any u ∈ H.

Let’s suppose now a, b ∈ H with a ∼i b, so a ∼o b, by hypothesis. We showthat ∀u ∈ H, ∀x ∈ a ◦ u,∃y ∈ b ◦ u such that x ∼i y.We take u ∈ H and x ∈ a ◦ u; it results that there exists y = x ∈ b ◦ u = a ◦ usuch that x ∼i y. So, ∼i is regular on H.

Proposition 3.4.4. Let (H, ◦) be a hypergroup. If, for any x ∈ H, there exist(a, b) ∈ H2 such that a ◦ b = x, then H is a reduced hypergroup.

Proof. We prove that for any x ∈ H, xi = {x} and therefore xe = {x}.Let x be an arbitrary element of H. We have

xi = {y ∈ H | x ∈ a ◦ b ⇐⇒ y ∈ a ◦ b, for a, b ∈ H}.Let’s suppose that there exists x0 ∈ H such that | (x0)i |≥ 2, that is, there exists

y ∈ H \ {x0} with y ∈ (x0)i; it follows that y ∈ a ◦ b = x0, contradiction.Therefore, H is a reduced hypergroup.

Corollary 3.4.5. If H is a hypergroup with scalar identities, then H is a reducedhypergroup.

Example. Any i.p.s. hypergroup is a reduced hypergroup.

Proposition 3.4.6. Let (G, ·) be a commutative group and R an equivalencerelation on G such that, for any x ∈ G, x = {x, x−1}. We define on G/R thehyperoperation

x⊗ y = {xy, xy−1}.Then G/R is a reduced hypergroup.

Proof. Indeed, for any x, y ∈ G we have xy−1 = {xy−1, yx−1} = yx−1, thusx⊗ y = y ⊗ x.

We verify the reproducibility law: for any x ∈ G/R, x⊗G/R = G/R, that is,for any y ∈ G/R, there exists z ∈ G/R such that y ∈ x ⊗ z. For z = x−1y, we

have x⊗ z = {y, xy−1x} 3 y.Now, we verify the associativity law.

For any x, y, z ∈ G/R we have:

(x⊗ y)⊗ z = (xy ⊗ z) ∪ (xy−1 ⊗ z) =

= {(xy)z, (xy)z−1, (xy−1)z, (xy−1)z−1)}x⊗ (y ⊗ z) = (x⊗ yz) ∪ (x⊗ yz−1) =

= {x(yz), x(z−1y−1), x(yz−1), x(zy−1)}

.

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We obtain that (x⊗ y)⊗ z = x⊗ (y ⊗ z), since G is a commutative group.So, G/R is a commutative hypergroup.

For any x ∈ G/R we have x = {x, x−1} = x−1 and e ⊗ x = x, where e is theidentity of the group G. By the Proposition 3.4.4 it results that G/R is a reducedhypergroup.

Example 3.4.7. Let us consider C(n) = {e0, e1, ..., ek(n)}, where k(n) =n

2if

n ∈ 2IN and k(n) =n− 1

2if n ∈ 2IN + 1. For any s, t ≤ k(n) we define

es ◦ et = {ep, er}, where p = min{s + t, n− (s + t)}, r =| s− t | .

It is well known that 〈C(n), ◦〉 is a canonical hypergroup.Moreover, for any t ≤ k(n), we have e0 ◦ et = et, because r =| 0 − t |= t andp = min{t, n − t} = t. Therefore, by the Proposition 3.4.4, H is a reducedhypergroup.

Remark. The condition expressed in the Proposition 3.4.4 is a sufficient condi-tion, but not a necessary one as we can see in the following example.

Example. Let (H, ◦) be the following hypergroup

H x y zx z x, y Hy x, y z Hz H H H

which is a commutative reduced hypergroup and doesn’t verify the condition fromthe Proposition 3.4.4, because there exists x ∈ H such that, for any a, b ∈ H witha ◦ b 3 x, we have | a ◦ b |≥ 2.

Theorem 3.4.8. Any join space (H, ◦) obtained associating to a universe Hendowed with a rough set (R(A), R(A)), the hyperoperation

∀(x, y) ∈ H2, x ◦ y = R({x, y}) \R({x, y})

is not a reduced hypergroup.

Proof. It is known that the above hyperoperation is well defined if and only iffor any x ∈ H, | R(x) |≥ 3, which is equivalent with the fact that (H, ◦) is a joinspace.

To prove that H is not a reduced hypergroup we’ll show that

xRy ⇐⇒ x ∼e y, for x, y ∈ H.

Then R(x) = xe and thus ∀x ∈ H, xe 6= {x}, so H is not reduced.

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First we prove that xRy ⇐⇒ x ∼o y, for x, y ∈ H.Let x, y be arbitrary elements of H such that xRy. For any a ∈ H we havex ◦ a = R(x) ∪R(a) = R(y) ∪R(a) = y ◦ a, so x ∼o y.We consider now x, y ∈ H with x ∼o y, that is, for any a ∈ H, x ◦ a = y ◦ a. Wehave x ◦ x = y ◦ x thus R(x) = R(x) ∪ R(y), so R(y) ⊂ R(x). From x ◦ y = y ◦ yit results R(x) ⊂ R(y). It follows xRy.

It remains to prove that xRy ⇐⇒ x ∼i y, for x, y ∈ H.We suppose that xRy and let us consider u, v ∈ H such that x ∈ u ◦ v = R(u) ∪R(v); it implies x ∈ R(u) or x ∈ R(v), equivalent with xRu or xRv, thus yRu oryRv, so y ∈ R(u) ∪R(v) = u ◦ v.Conversely, we have x ∈ x ◦ x = R(x) and x ∼i y, thus y ∈ x ◦ x = R(x), that is,xRy.

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Chapter 4

Concluding remarks and futurework

Along the previous contributions in the theory of hypergroups and especially toestablish new various connections between hypergroups and fuzzy sets, this thesisconfirms, through the original results obtained in the third chapter, the fact thatthese associations which lead to new algebraic structures open a gate towardsthe large universe of the hyperstructures. Many problems from this domain arestill opened; we present below a few of them which may be a continuation of thisresearch.

The central part of this thesis is formed by the Chapter 3, where we study thesequence of join-spaces and fuzzy sets associated with a hypergroupoid H.

We remember that, with any hypergroupoid H endowed with a fuzzy set, we

can an ordered r-tuple (k1, k2, ..., kr), wherer∑

i=1

ki = n = | H |, as in the subsection

§ 3.3.1.The situation when the r-tuple has the form (k, k, ..., k), k ≥ 1, is discussed

and solved by the Theorem 3.3.10; if the r-tuple is equal with its opposite one,that is (k1, k2, ..., kr) = (kr, kr−1, ..., k1), then we can use the Theorem 3.3.15 todetermine the length of the sequence of join spaces associated with H.

Problem 1. To study the join space 1H associated with the hypergroupoidH, for the r-tuple (1, 2, ..., r).

Problem 2. To verify if, for any k ≥ 1, there exist a hypergroup H such thatthe fuzzy set µ defined by the relations (ω) from the subsection § 3.3.1 determinesexactly the r-tuple (k, k, ..., k).Till now, it is proved the case k = 2, considering on the universe H the hyperpro-duct:

ai ◦ ai = ai, ai ◦ aj = {ai, ai+1, ..., aj}, for i < j

(see the Corollary 3.3.14).One of the fundamental results obtained in this paper is the Theorem 3.3.15,

where we give a sufficient condition, but not a necessary one, such that two con-secutive join spaces from the sequence associated with a hypergroupoid H are notisomorphic.

85

Problem 3. To find a necessary condition such that this join spaces are notisomorphic, that is the sequence associated with H does not end.

To any hypergroupoid corresponds a fuzzy grade, that is the length of thesequence of its associated join spaces. Given a natural number n ∈ IN∗ \ {1}, wecan construct a hypergroup H (which is a join space) such that s.f.g.(H) = n(seethe Corollary 3.3.14).

Problem 4. To find a method which constructs the hypergroup H with thesmaller cardinality such that s.f.g.(H) = n.

Associating to an universe H a rough set (R(A), R(A)), we can construct ajoin space 〈H, ◦〉, where, for any x, y ∈ H, x ◦ y = R({x, y}) \ R({x, y}), if andonly if for any x ∈ H, | R(x) | ≥ 3. Now associating the fuzzy set µ defined bythe relations (ω) of the paragraph 3.3.1, we demonstrated if R is an equivalencerelation such that | R(x) |= k = const, for any x ∈ H, then s.f.g.(H) = 1.

Problem 5. To study the fuzzy grade of the join space 〈H, ◦〉 associated witha rough set (R(A), R(A)), where R is an equivalence relation different by the oneconsidered in the paragraph §3.3.6.

Till now, we have considered only some remarkable classes of hypergroups: thei.p.s. hypergroups, the complete hypergroups or the hypergroups endowed with arough set.

Problem 6. To study the sequence associated with other important types ofhypergroups, for example: the cyclic hypergroups, the hypergroups obtained froma binary relation, or from a hypergraph.

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Appendix A

The sequence of fuzzy sets and ofjoin spaces determined by all thecomplete hypergroups of orderless than or equal to 6

Theorem 3.3.20. (see [17]) Let H be a complete hypergroup of order n ≤ 6.(i) For n = 3, s.f.g.(H) = 1.

(ii) For n = 4, there are 3 hypergroups with s.f.g.(H) = 1 and two hypergroupswith s.f.g.(H) = 2.

(iii) For n = 5, s.f.g.(H) = 1.

(iv) For n = 6, there are 17 hypergroups of s.f.g.(H) = 1 and 4 hypergroups ofs.f.g.(H) = 2.

Proof. (i) Let consider H = {e, a1, a2} a complete hypergroup. In this caseG = ( ZZ 2, +), so we have two structures:

a) H e a1 a2

e e A1 A1

a1 e ea2 e

where A0 = {e}, A1 = {a1, a2}. We note thatH is an 1-hypergroup. By Theorem 3.3.17,µ(e) = 1, µ(a1) = µ(a2) = 0, 5.

The join space 1H associated is

1H e a1 a2

e e H H

a1 A1 A1

a2 A1

hence µ1(e) = 0, 47, µ1(a1) = µ1(a2) = 0, 42.Then rH = 1H, ∀r ≥ 2.

b) H e a1 a2

e A0 A0 A1

a1 A0 A1

a2 A0

where A0 = {e, a1}, A1 = {a2}. We haveµ(e) = µ(a1) = 0, 5 and µ(a2) = 1.

87

It follows that the join space 1H associated is

H e a1 a2

e A0 A0 H

a1 A0 H

a2 a2

with µ1(e) = µ1(a1) = 0, 417,µ1(a2) = 0, 467, so rH = 1H, ∀r ≥ 2.

(ii) Let H = {e, a1, a2, a3} be a complete hypergroup.

a) Setting G = ( ZZ 2, +) we obtain:

a1) H e a1 a2 a3

e e A1 A1 A1

a1 e e ea2 e ea3 e

where A0 = {e}, A1 = {a1, a2, a3}. H is an1-hypergroup. The membership function isµ(e) = 1, µ(ai) = 0, (3), 1 ≤ i ≤ 3.

The join space associated is:

1H e a1 a2 a3

e e H H H

a1 A1 A1 A1

a2 A1 A1

a3 A1

for which µ1(e) = 0, 36, µ1(ai) = 0, 3, 1 ≤i ≤ 3, hence rH = 1H, ∀r ≥ 2.

a2) H e a1 a2 a3

e A0 A0 A1 A1

a1 A0 A1 A1

a2 A0 A0

a3 A0

where A0 = {e, a1}, A1 = {a2, a3}. Weobtain µ(e) = µ(ai), 1 ≤ i ≤ 3 and itfollows that 1H is a total hypergroup ands.f.g.(H) = 1.

a3) H e a1 a2 a3

e A0 A0 A0 A1

a1 A0 A0 A1

a2 A0 A1

a3 A0

with A0 = {e, a1, a2}, A1 = {a3}. Thenµ(e) = µ(a1) = µ(a2) = 0, (3) and µ(a3) = 1.The join space associated is isomorphic withthe one of a1), hence rH = 1H, ∀r ≥ 2.

b) For G = ( ZZ 3, +) we distinguish two hypergroups:

b1) H e a1 a2 a3

e e a1 A2 A2

a1 A2 e ea2 a1 a1

a3 a1

where A0 = {e}, A1 = {a1}, A2 = {a2, a3}.H is an 1-hypergroup.

b2) H e a1 a2 a3

e A0 A0 a2 a3

a1 A0 a2 a3

a2 a3 A0

a3 a2

where A0 = {e, a1}, A1 = {a2}, A2 = {a3}.

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In the last two cases, we obtain that | H/Reµ |= 2 and every class of equivalencehas the same cardinal number, therefore the s.f.g.(H) = 2 and 2H is a totalhypergroup (since the Case I which follows the Definition 3.3.9).

iii) Set H = {e, a1, a2, a3, a4} a complete hypergroup.

a) There are five complete 1-hypergroups of order 5.

a1) H e a1 a2 a3 a4

e e A1 A1 A1 A1

a1 e e e ea2 e e ea3 e ea4 e

where A0 = {e}, A1 = {ai}, 1 ≤ i ≤ 4.One obtains µ(e) = 1, µ(ai) = 0, 25,1 ≤ i ≤ 4. The join space 1H associatedis the following one

H e a1 a2 a3 a4

e e H H H H

a1 A1 A1 A1 A1

a2 A1 A1 A1

a3 A1 A1

a4 A1

hence µ1(e) = 0, 29 and µ1(ai) = 0, 23,1 ≤ i ≤ 4. It follows rH = 1H, ∀r ≥ 2.

a2) H e a1 a2 a3 a4

e e a1 A2 A2 A2

a1 A2 e e ea2 a1 a1 a1

a3 a1 a1

a4 a1

with A0 = {e}, A1 = {a1}, A2 ={a2, a3, a4}. We have µ(e) = µ(a1) = 1and µ(a2) = µ(a3) = µ(a4) = 0, (3).Then

1H e a1 a2 a3 a4

e e, a1 e, a1 H H H

a1 e, a1 H H H

a2 A2 A2 A2

a3 A2 A2

a4 A2

is the join space associated and themembership function is: µ1(e) =µ1(a1) = 0, 28, µ1(ai) = 0, 26, 2 ≤ i ≤4. So rH = 1H, ∀r ≥ 2.

a3) H e a1 a2 a3 a4

e e A1 A1 A2 A2

a1 A2 A2 e ea2 A2 e ea3 A1 A1

a4 A1

where A0 = {e}, A1 = {a1, a2}, A3 ={a3, a4}. We calculate µ(e) = 1, µ(ai) =0, 5, 1 ≤ i ≤ 4. The join space associa-ted is the same as in a1), so rH = 1H,∀r ≥ 1.

a4) H e a1 a2 a3 a4

e e a1 a2 A3 A3

a1 a2 A3 e ea2 e a1 a1

a3 a2 a2

a4 a2

where A0 = {e}, A1 = {a1}, A2 ={a2}, A3 = {a3, a4}. So, µ(e) = µ(a1) =µ(a2) = 1, µ(a3) = µ(a4) = 0, 5; themembership function µ1 is the same asin a2) and we have again s.f.g.(H) = 1.

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a5) For G = (K, ·) the Klein’s group and A0 = {e}, A1 = {a1}, A2 = {a2},A3 = {a3, a4}, we obtain

H e a1 a2 a3 a4

e e a1 a2 A3 A3

a1 e A3 a2 a2

a2 e a1 a1

a3 e e

a4 e

with the join space 1H equal to the oneof a4), hence rH = 1H, ∀r ≥ 2.

b) The following complete hypergroups are not 1-hypergroups :

b1) H e a1 a2 a3 a4

e A0 A0 A1 A1 A1

a1 A0 A1 A1 A1

a2 A0 A0 A0

a3 A0 A0

a4 A0

where A0 = {e, a1}, A1 = {a2, a3, a4}and µ(e) = µ(a1) = 0, 5, µ(a2) =µ(a3) = µ(a4) = 0, (3). The join space1H is the following:

1H e a1 a2 a3 a4

e A0 A0 H H H

a1 A0 H H H

a2 A1 A1 A1

a3 A1 A1

a4 A1

which leads to rH = 1H, ∀r ≥ 1.

b2) H e a1 a2 a3 a4

e A0 A0 A0 A1 A1

a1 A0 A0 A1 A1

a2 A0 A1 A1

a3 A0 A0

a4 A0

where A0 = {e, a1, a2}, A1 = {a3, a4}.µ(e) = µ(a1) = µ(a2) = 0, (3), µ(a3) =µ(a4) = 0, 5.

The join space associated is isomorphic with the one of the case b1) and therefores.f.g.(H) = 1.

b3) H e a1 a2 a3 a4

e A0 A0 A0 A0 a4

a1 A0 A0 A0 a4

a2 A0 A0 a4

a3 A0 a4

a4 A0

with A0 = {e, a1, a2, a3}, A1 = {a4}.We have µ(e) = µ(ai) = 0, 25, 1 ≤ i ≤3, µ(a4) = 1. The join space associatedis

1H e a1 a2 a3 a4

e A0 A0 A0 A0 H

a1 A0 A0 A0 H

a2 A0 A0 H

a3 A0 H

a4 a4

for which µ1(e) = µ1(ai) = 0, 233, 1 ≤i ≤ 3, µ1(a4) = 0, 289. Hence rH = 1H,∀r ≥ 2.

90

b4) H e a1 a2 a3 a4

e A0 A0 a2 A2 A2

a1 A0 a2 A2 A2

a2 A2 A0 A0

a3 a2 a2

a4 a2

where A0 = {e, a1}, A1 = {a2}, A2 ={a3, a4}. We have µ(e) = µ(ai) = 0, 5,i ∈ {1, 3, 4}, µ(a2) = 1. It follows thejoin space

1H e a1 a2 a3 a4

e H2 H2 H H2 H2

a1 H2 H H2 H2

a2 a2 H H

a3 H2 H2

a4 H2

where H2 = H \ {2}, which is iso-morphic with the preceding one. ThenrH = 1H, ∀r ≥ 1.

b5) H e a1 a2 a3 a4

e A0 A0 A0 a3 a4

a1 A0 A0 a3 a4

a2 A0 a3 a4

a3 a4 A0

a4 a3

where A0 = {e, a1, a2}, A1 = {a3},A2 = {a4}. It leads to the join spacefrom the case b1).

b6) H e a1 a2 a3 a4

e A0 A0 a2 a3 a4

a1 A0 a2 a3 a4

a2 a3 a4 A0

a3 A0 a2

a4 a3

with A0 = {e, a1}, A1 = {a2}, A2 ={a3}, A3 = {a4}. The join space is theone of b1).

b7) H e a1 a2 a3 a4

e A0 A0 a2 a3 a4

a1 A0 a2 a3 a4

a2 A0 a4 a3

a3 A0 a2

a4 A0

with A0 = {e, a1}, A1 = {a2}, A2 ={a3}, A3 = {a4} and G = (K, ·) theKlein’s group. Again the join space isthe one from b1).

iv) a) The following complete hypergroups

a1) H e a1 a2 a3 a4 a5

e e a1 a2 A3 A3 A3

a1 a2 A3 e e ea2 e a1 a1 a1

a3 a2 a2 a2

a4 a2 a2

a5 a2

where A0 = {e}, A1 = {a1},A2 = {a2}, A3 = {a3, a4, a5},G = ( ZZ 4, +).

91

a2) H e a1 a2 a3 a4 a5

e A0 A0 A0 a3 a4 a5

a1 A0 A0 a3 a4 a5

a2 A0 a3 a4 a5

a3 a4 a5 A0

a4 A0 a3

a5 a4

where A0 = {e, a1, a2}, A1 ={a3}, A2 = {a4}, A3 = {a5},G = ( ZZ 4, +).

a3) H e a1 a2 a3 a4 a5

e e a1 a2 A3 A3 A3

a1 e A3 a2 a2 a2

a2 e a1 a1 a1

a3 e e ea4 e ea5 e

with G = (K, ·) the Klein’s group,A0 = {e}, A1 = {a1}, A2 = {a2},A3 = {a3, a4, a5}.

a4) H e a1 a2 a3 a4 a5

e A0 A0 A0 a3 a4 a5

a1 a0 A0 a3 a4 a5

a2 A0 a3 a4 a5

a3 A0 a5 a4

a4 A0 a3

a5 A0

with G = (K, ·) the Klein’s group,A0 = {e, a1, a2}, A1 = {a3},A2 = {a4}, A3 = {a5}.

have the property |H/R| = 2 and all the equivalence classes have the same cardi-nality. By the Case I a) of the paragraph 3.3.3, we obtain s.f.g.(H) = 2.

b) The other complete hypergroups of order 6 have s.f.g.(H) = 1. We list them,without writing the join spaces and the membership functions associated.

b1) H e a1 a2 a3 a4 a5

e e A1 A1 A1 A1 A1

a1 e e e e ea2 e e e ea3 e e ea4 e ea5 e

whereA0 = {e},A1 = {a1, a2, a3, a4, a5}.H is an 1-hypergroup.

b2) H e a1 a2 a3 a4 a5

e A0 A0 A1 A1 A1 A1

a1 A0 A1 A1 A1 A1

a2 A0 A0 A0 A0

a3 A0 A0 A0

a4 A0 A0

a5 A0

whereA0 = {e, a1},A1 = {a2, a3, a4, a5}.

92

b3) H e a1 a2 a3 a4 a5

e A0 A0 A0 A1 A1 A1

a1 A0 A0 A1 A1 A1

a2 A0 A1 A1 A1

a3 A0 A0 A0

a4 A0 A0

a5 A0

where A0 = {e, a1, a2}, A1 ={a3, a4, a5}.

b4) H e a1 a2 a3 a4 a5

e A0 A0 A0 A0 A1 A1

a1 A0 A0 A0 A1 A1

a2 A0 A0 A1 A1

a3 A0 A1 A1

a4 A0 A0

a5 A0

where A0 = {e, a1, a2, a3}, A1 ={a4, a5}.

b5) H e a1 a2 a3 a4 a5

e A0 A0 A0 A0 A0 a5

a1 A0 A0 A0 A0 a5

a2 A0 A0 A0 a5

a3 A0 A0 a5

a4 A0 a5

a5 A0

with A0 = {e, a1, a2, a3, a4}, A1 ={a5}.

b6) H e a1 a2 a3 a4 a5

e e a1 A2 A2 A2 A2

a1 A2 e e e ea2 a1 a1 a1 a1

a3 a1 a1 a1

a4 a1 a1

a5 a1

where A0 = {e}, A1 ={a1, a2, a3, a4}, A1 = {a5}. H isan 1-hypergroup.

b7) H e a1 a2 a3 a4 a5

e e A1 A1 A2 A2 A2

a1 A2 A2 e e ea2 A2 e e ea3 A1 A1 A1

a4 A1 A1

a5 A1

where A0 = {e}, A1 = {a1, a2},A2 = {a3, a4, a5}. H is an 1-hypergroup.

b8) H e a1 a2 a3 a4 a5

e A0 A0 a2 A2 A2 A2

a1 A0 a2 A2 A2 A2

a2 A2 A0 A0 A0

a3 a2 a2 a2

a4 a2 a2

a5 a2

where A0 = {e, a1}, A1 = {a2},A2 = {a3, a4, a5}.

93

b9) H e a1 a2 a3 a4 a5

e A0 A0 A1 A1 A2 A2

a1 A0 A1 A1 A2 A2

a2 A2 A2 A0 A0

a3 A2 A0 A0

a4 A1 A1

a5 A1

where A0 = {e, a1}, A1 ={a2, a3}, A2 = {a4, a5}.

b10) H e a1 a2 a3 a4 a5

e A0 A0 A0 a3 A2 A2

a1 A0 A0 a3 A2 A2

a2 A0 a3 A2 A2

a3 A2 A0 A0

a4 a3 a3

a5 a3

where A0 = {e, a1, a2}, A1 ={a3}, A2 = {a4, a5}.

b11) H e a1 a2 a3 a4 a5

e A0 A0 A0 A0 a4 a5

a1 A0 A0 A0 a4 a5

a2 A0 A0 a4 a5

a3 A0 a4 a5

a4 a5 A0

a5 a4

where A0 = {e, a1, a2, a3}, A1 ={a4}, A2 = {a5}.

b12) H e a1 a2 a3 a4 a5

e e a1 A2 A2 A3 A3

a1 A2 A3 A3 e ea2 e e a1 a1

a3 e a1 a1

a4 A2 A2

a5 A2

with A0 = {e}, A1 = {a1}, A2 ={a2, a3}, A3 = {a4, a5}. G =( ZZ 4, +). H is an 1-hypergroup.

b13) H e a1 a2 a3 a4 a5

e A0 A0 a2 a3 A3 A3

a1 A0 a2 a3 A3 A3

a2 a3 A3 A0 A0

a3 A0 a2 a2

a4 a3 a3

a5 a3

where A0 = {e, a1}, A1 = {a2},A2 = {a3}, A3 = {a4, a5}. G =( ZZ 4, +).

b14) H e a1 a2 a3 a4 a5

e e a1 A2 A2 A3 A3

a1 e A3 A3 A2 A2

a2 e e a1 a1

a3 e a1 a1

a4 e ea5 e

where A0 = {e}, A1 = {a1},A2 = {a2, a3}, A3 = {a4, a5}.G = (K, ·) the Klein’s group. His an 1-hypergroup.

94

b15) H e a1 a2 a3 a4 a5

e A0 A0 a2 a3 A3 A3

a1 A0 a2 a3 A3 A3

a2 A0 A3 a3 a3

a3 A0 a2 a2

a4 A0 A0

a5 A0

where A0 = {e, a1}, A1 = {a2},A2 = {a3}, A3 = {a4, a5}. G =(K, ·) the Klein’s group.

b16) H e a1 a2 a3 a4 a5

e e a1 a2 a3 A4 A4

a1 a2 a3 A4 e ea2 A4 e a1 a1

a3 a1 a2 a2

a4 a3 a3

a5 a3

where A0 = {e}, A1 = {a1}, A2 ={a2}, A3 = {a3}, A4 = {a4, a5}.H is an 1-hypergroup.

b17) H e a1 a2 a3 a4 a5

e A0 A0 a2 a3 a4 a5

a1 A0 a2 a3 a4 a5

a2 a3 a4 a5 A0

a3 a5 A0 a2

a4 a2 a3

a5 a4

where A0 = {e, a1}, A1 = {a2},A2 = {a3}, A3 = {a4}, A4 ={a5}.

95

96

Appendix B

The sequence of fuzzy sets and ofjoin spaces determined by all thei.p.s. hypergroups of order lessthan or equal to 7

Theorem 3.3.26. (see [14], [15])

(i) There is only one i.p.s. hypergroup H of order 3 and s.f.g.(H) = 1.

(ii) There are three non isomorphic i.p.s. hypergroups H1, H2, H3 of order 4 ands.f.g.(H1) = s.f.g.(H2), s.f.g.(H3) = 2.

(iii) There are eight non isomorphic i.p.s. hypergroups of order 5: one of themhas s.f.g.(H) = 2 and the other one have s.f.g.(H) = 1.

(iv) There are nineteen non isomorphic i.p.s. hypergroups of order 6: fourteenof them have s.f.g.(H) = 1, four of them are of s.f.g.(H) = 2 and one iswith s.f.g.(H) = 3.

(v) There are thirty-six non isomorphic i.p.s. hypergroups of order 7: two ofthem have f.g.(H) = 1, twenty-five of them have s.f.g.(H) = 1, eight ofthem are with s.f.g.(H) = 2 and only for one of them s.f.g.(H) = 3

Proof. We shall denote, in the following tables, ∀ s, As = H \ {s}.

(i) We have

H 0 1 20 0 1 21 0, 2 12 0

Thenµ(0) = µ(2) = 0, 833, µ(1) = 1,whence the join space associated is

97

1H 0 1 20 0, 2 H 0, 21 1 H2 0, 2

It follows:µ1(0)= µ1(2)=0, 417, µ1(1)=0, 467.Then ∀ r > 1, rH = 1H.

(ii) Let us suppose H of order 4.

a) H 0 1 2 30 0 1 2 31 2 0 32 1 33 0, 1, 2

We have:µ(0) = µ(1) = µ(2) = 0, 833µ(3) = 1.Therefore the join space associa-ted is

1H 0 1 2 30 A3 A3 A3 H

1 A3 A3 H

2 A3 H

3 3

From this, one obtains:µ1(0) = µ1(1) = µ1(2) = 0, 3µ1(3) = 0, 357, whence∀ r > 1, rH = 1H.

b) H 0 1 2 30 0 1 2 31 0, 2 1 32 0 33 0, 1, 2

We have:µ(0) = 0, 708 = µ(2)µ(1) = 0, 867, µ(3) = 1.Therefore the join space associa-ted is

1H 0 1 2 30 0, 2 A3 0, 2 H

1 1 A3 1, 32 0, 2 H

3 3

Hence:µ1(0) = µ2(2) = 0, 361µ1(1) = 0, 394, µ1(3) = 0, 429.One obtains ∀ r > 1, rH = 1H.

c) H 0 1 2 30 0 1 2 31 2 0, 3 12 1 23 0

The membership of H is:µ(0) = µ(3) = 0, 75,µ(1) = µ(2) = 1.We find:

1H 0 1 2 30 0, 3 H H 0, 31 1, 2 1, 2 H

2 1, 2 H

3 0, 3

From this we obtain:µ1(i) = 0, 333, ∀ i.Hence 2H is the totalhypergroup of order 4.

(iii) We consider now the i.p.s. hypergroups of order 5.

98

a1) H 0 1 2 3 40 0 1 2 3 41 0, 2 1 4 32 0 3 43 0, 2 14 0, 2

We have :µ(0) = µ(2) = 0, 7,µ(1) = µ(3) = µ(4) = 1,whence, the join space associa-ted is the following one:

1H 0 1 2 3 40 0, 2 H 0, 2 H H

1 1, 3, 4 H 1, 3, 4 1, 3, 42 0, 2 H H

3 1, 3, 4 1, 3, 44 1, 3, 4

Now:µ1(0) = µ1(2) = 0, 275µ1(1) = µ1(3) = µ(4) = 0, 254.It follows ∀ r > 1, rH = 1H.

a2) H 0 1 2 3 40 0 1 2 3 41 0, 2 1 3 42 0 3 43 0, 1, 2, 4 34 0, 1, 2

We haveµ(0) = 0, 616 = µ(2),µ(1) = 0, 763, µ(4) = 0, 892,µ(3) = 1, whence

1H 0 1 2 3 40 0, 2 0, 1, 2 0, 2 H 0, 1, 2, 41 1 0, 1, 2 1, 3, 4 1, 42 0, 2 H 0, 1, 2, 43 3 3, 44 4

Then µ1(0) = µ1(2) = 0, 320,µ1(1) = 0, 341, µ1(4) = 0, 364,µ1(3) = 0, 385. Therefore ∀ r ≥ 2,we have rH = 1H.

a3) H 0 1 2 3 40 0 1 2 3 41 0, 2 1 3 42 0 3 43 4 0, 1, 24 3

We have:µ(0) = µ(2) = 0, 633,µ(1) = 0, 777, µ(3) = µ(4) = 1.It follows the join space

1H 0 1 2 3 40 0, 2 0, 1, 2 0, 2 H H1 1 0, 1, 2 1, 3, 4 1, 3, 42 0, 2 H H3 3, 4 3, 44 3, 4

whenceµ1(0)= µ1(2)= µ1(3) == µ1(4)= 0, 308, µ1(1) = 0, 309.One obtains the second join space

99

2H 0 1 2 3 40 0, 2, 3, 4 H 0, 2, 3, 4 0, 2, 3, 4 0, 2, 3, 41 1 H H H2 0, 2, 3, 4 0, 2, 3, 4 0, 2, 3, 43 0, 2, 3, 4 0, 2, 3, 44 0, 2, 3, 4

Hence µ2(0) = µ2(2) = µ2(3) = µ2(4) = 0, 217, µ2(1) = 0, 578.Therefore, ∀ r > 2, rH = 2H and s.f.g.(H) = 2.

a4) H 0 1 2 3 40 0 1 2 3 41 0, 2, 3 1 1 42 0, 3 2 43 0 44 A4

We have: µ(0) = 0, 616,µ(1) = 0, 892, µ(2) = 0, 763,µ(3) = 0, 616, µ(4) = 1,whence, the associatedjoin space is the following one

1H 0 1 2 3 40 0, 3 A4 0, 2, 3 0, 3 H1 1 1, 2 A4 1, 42 2 0, 2, 3 1, 2, 43 0, 3 H4 4

One finds:µ1(0) = µ1(3) = 0, 320,µ1(1) = 0, 364, µ1(2) = 0, 341,µ1(4) = 0, 385.Therefore, one obtains∀ r > 1, rH4 = 1H4.

a5) H 0 1 2 3 40 0 1 2 3 41 0, 2, 3 1 1 42 3 0 43 2 44 A4

It results:µ(0) = µ(2) = µ(3) = 0, 716,µ(1) = 0, 892, µ(4) = 1.Therefore we obtain

1H 0 1 2 3 40 0, 2, 3 A4 0, 2, 3 0, 2, 3 H1 1 A4 A4 1, 42 0, 2, 3 0, 2, 3 H3 0, 2, 3 H4 4

From this one finds:µ1(0) = µ1(2) = µ1(3) = 0, 271,µ1(1) = 0, 313, µ1(4) = 0, 355.Then we have:∀ r > 1, rH = 1H.

a6) H 0 1 2 3 40 0 1 2 3 41 A1 1 1 12 3 0, 4 23 2 34 0

Hence:µ(0) = µ(4) = 0, 650,µ(1) = 1, µ(2) = µ(3) = 0, 875.It follows the join space

100

1H 0 1 2 3 40 0, 4 H A1 A1 0, 41 1 1, 2, 3 1, 2, 3 H

2 2, 3 2, 3 A1

3 2, 3 A1

4 0, 4

Then we haveµ1(0) = µ1(4) = 0, 266,µ1(1) = 0, 348,µ1(2) = µ1(3) = 0, 306.Therefore ∀ r > 1, rH = 1H.

a7) H 0 1 2 3 40 0 1 2 3 41 A1 1 1 12 3 0 43 2 44 0, 2, 3

We find:µ(0) = µ(2) = µ(3) = 0, 716,µ(1) = 1, µ(4) = 0, 892.It follows

1H 0 1 2 3 40 0, 2, 3 H 0, 2, 3 0, 2, 3 A1

1 1 H H 1, 42 0, 2, 3 0, 2, 3 A1

3 0, 2, 3 A1

4 4

From this we obtainµ1(0) = µ1(2) = µ1(3) = 0, 271,µ1(1) = 0, 355, µ1(4) = 0, 313.Therefore we have∀ r > 1, rH = 1H.

a8) H 0 1 2 3 40 0 1 2 3 41 A1 1 1 12 4 0 33 4 24 0

Hence:µ(0) = µ(2) = µ(3) == µ(4) = 0, 85, µ(1)=1.We obtain the same join spacewe got from the join space 1Hassociated with H in a3).Therefore, we have s.f.g.(H)=1.

(iv) Now we study the i.p.s. hypergroups of order 6.

b1) H 0 1 2 3 4 50 0 1 2 3 4 51 0 2 3 4 52 3 4 5 0, 13 5 0, 1 24 2 35 4

Calculating the membershipfunction one finds:µ(0) = µ(1) = 0, 6,µ(2) = µ(3) = µ(4) = µ(5) = 1.So the join space associatedwith H is

1H 0 1 2 3 4 50 0, 1 0, 1 H H H H1 0, 1 H H H H2 2, 3, 4, 5 2, 3, 4, 5 2, 3, 4, 5 2, 3, 4, 53 2, 3, 4, 5 2, 3, 4, 5 2, 3, 4, 54 2, 3, 4, 5 2, 3, 4, 55 2, 3, 4, 5

101

From this one obtains: µ1(0) = µ1(1) = 0, 233, µ1(2) = µ1(3) = µ1(4) == µ1(5) = 0, 208. Therefore ∀ r, rH = 1H.

b2) H 0 1 2 3 4 50 0 1 2 3 4 51 0 2 4 3 52 3, 4 5 5 0, 13 1 0 24 1 25 3, 4

We have: µ(0) = µ(1) = µ(3) == µ(4) = 0, 833, µ(2) = µ(5) = 1.Therefore the join space associated is

1H 0 1 2 3 4 50 0, 1, 3, 4 0, 1, 3, 4 H 0, 1, 3, 4 0, 1, 3, 4 H

1 0, 1, 3, 4 H 0, 1, 3, 4 0, 1, 3, 4 H

2 2, 5 H H 2, 53 0, 1, 3, 4 0, 1, 3, 4 H

4 0, 1, 3, 4 H

5 2, 5

whence: µ1(0) = µ1(1) = µ1(3) = 0, 208 = µ1(4), µ1(2) = µ1(5) = 0, 233from which ∀ r, rH = 1H.

b3) H 0 1 2 3 4 50 0 1 2 3 4 51 2 0 3 4 52 1 3 4 53 0, 1, 2 5 44 0, 1, 2 35 0, 1, 2

One finds:µ(0) = µ(1) = µ(2) = 0, 667,µ(3) = µ(4) = µ(5) = 1,whence, we obtain

1H 0 1 2 3 4 50 0, 1, 2 0, 1, 2 0, 1, 2 H H H

1 0, 1, 2 0, 1, 2 H H H

2 0, 1, 2 H H H

3 3, 4, 5 3, 4, 5 3, 4, 54 3, 4, 5 3, 4, 55 3, 4, 5

From this we have: ∀u ∈ H, µ1(u) = 0, 222, that is, 2H is a total hypergroupof order 6 and s.f.g.(H) = 2.

102

b4) H 0 1 2 3 4 50 0 1 2 3 4 51 2 0 3 4 52 1 3 4 53 5 0, 1, 2 44 5 35 0, 1, 2

We have:µ(0) = µ(1) = µ(2) = 0, 667,µ(3) = µ(4) = µ(5) = 1,so the same membershipfunction as for the hypergroup Hfrom b3), whence 2H is a totalhypergroup and s.f.g.(H) = 2.

b5) H 0 1 2 3 4 50 0 1 2 3 4 51 2 3 0 4 52 0 1 4 53 2 4 54 0, 1, 2, 3 55 0, 1, 2, 3, 4

We have:µ(0) = µ(1) = µ(2) == µ(3) = 0, 742,µ(4) = 0, 911, µ(5) = 1,from this we obtainthe join space

1H 0 1 2 3 4 50 0, 1, 2, 3 0, 1, 2, 3 0, 1, 2, 3 0, 1, 2, 3 A5 H1 0, 1, 2, 3 0, 1, 2, 3 0, 1, 2, 3 A5 H2 0, 1, 2, 3 0, 1, 2, 3 A5 H3 0, 1, 2, 3 A5 H4 4 4, 55 5

Now, the membership function is µ1(0) = µ1(1) = µ1(2) = µ1(3) = 0, 333,µ1(4) = 0, 260, µ1(5) = 0, 303, so, we find ∀ r > 1, rH = 1H.

b6) H 0 1 2 3 4 50 0 1 2 3 4 51 0 3 2 4 52 0 1 4 53 0 4 54 0, 1, 2, 3 55 0, 1, 2, 3, 4

One finds:µ(0) = µ(1) = µ(2) == µ(3) = 0, 742,µ(4) = 0, 911, µ(5) = 1,that is the same membershipfunction as for the previous H,so we have ∀ r > 1, 1H = rH.

b7) H 0 1 2 3 4 50 0 1 2 3 4 51 2 3 0 4 52 0 1 4 53 2 4 54 5 0, 1, 2, 35 4

We find:µ(0) = µ(1) = µ(2) = µ(3) = 0, 75,µ(4) = µ(5) = 1.It follows the following join space

103

1H 0 1 2 3 4 50 0, 1, 2, 3 0, 1, 2, 3 0, 1, 2, 3 0, 1, 2, 3 H H1 0, 1, 2, 3 0, 1, 2, 3 0, 1, 2, 3 H H2 0, 1, 2, 3 0, 1, 2, 3 H H3 0, 1, 2, 3 H H4 4, 5 4, 55 4, 5

From 1H one obtains: µ1(0) = µ1(1) = µ1(2) = µ1(3) = 0, 75, µ1(4) =µ1(5) = 1. Therefore ∀ r > 1, rH = 1H.

b8) H 0 1 2 3 4 50 0 1 2 3 4 51 0 3 2 4 52 0 1 4 53 0 4 54 5 0, 1, 2, 35 4

The corresponding µ is as it follows:µ(0)=µ(1)=µ(2)=µ(3)=0, 75,µ(4) = µ(5) = 1, whence we obtainthe same join space as in previous case.So, s.f.g.(H) = 1.

b9) H 0 1 2 3 4 50 0 1 2 3 4 51 2 0 3 4 52 1 3 4 53 4 0, 1, 2 54 3 55 A5

The membership functionassociated with H is:µ(0) = µ(1) = µ(2) = 0, 533,µ(3) = µ(4) = 0, 9.By consequence, we obtain the firstjoin space

1H 0 1 2 3 4 50 0, 1, 2 0, 1, 2 0, 1, 2 A5 A5 H1 0, 1, 2 0, 1, 2 A5 A5 H2 0, 1, 2 A5 A5 H3 3, 4 3, 4 3, 4, 54 3, 4 3, 4, 55 5

From this we obtain:µ1(0) = µ1(1) = µ1(2) = 0, 237,µ1(3) = µ1(4) = 0, 258,µ1(5) = 0, 303,whence we get ∀ r, rH = 1H.

b10) H 0 1 2 3 4 50 0 1 2 3 4 51 2 0 3 4 52 1 3 4 53 0, 1, 2 4 54 5 0, 1, 2, 35 4

We find:µ(0) = µ(1) = µ(2) = 0, 639,µ(3) = 0, 812, µ(4) = µ(5) = 1,whence the associatedjoin space is the following one

104

1H 0 1 2 3 4 50 0, 1, 2 0, 1, 2 0, 1, 2 0, 1, 2, 3 H H1 0, 1, 2 0, 1, 2 0, 1, 2, 3 H H2 0, 1, 2 0, 1, 2, 3 H H3 3 3, 4, 5 3, 4, 54 4, 5 4, 55 4, 5

Then we have: µ1(0) = µ1(1) = = µ1(2) = 0, 241, µ1(3) = 0, 254, µ1(4) =µ1(5) = 0, 267. It follows: ∀ r, rH = 1H.

b11) H 0 1 2 3 4 50 0 1 2 3 4 51 0 2 3 4 52 0, 1 4 3 53 0, 1 2 54 0, 1 55 A5

We have:µ(0) = µ(1) = 0, 617,µ(2) = µ(3) = µ(4) = 0, 886,µ(5) = 1.The corresponding join space is

1H 0 1 2 3 4 50 0, 1 0, 1 A5 A5 A5 H1 0, 1 A5 A5 A5 H2 2, 3, 4 2, 3, 4 2, 3, 4 2, 3, 4, 53 2, 3, 4 2, 3, 4 2, 3, 4, 54 2, 3, 4 2, 3, 4, 55 5

The associated membership function is: µ1(0) = µ1(1) = 0, 253, µ1(2) =µ1(3) = µ1(4) = 0, 255, µ1(5) = 0, 288, whence, we obtain ∀ r, rH = 1H.

b12) H 0 1 2 3 4 50 0 1 2 3 4 51 0 2 3 4 52 4 0, 1 3 53 4 2 54 0, 1 55 A5

One finds that the membershipfunction determined by H coin-cides with the one determined bythe previous hypergroup, thereforewe have ∀ r, rH = 1H.

b13) H 0 1 2 3 4 50 0 1 2 3 4 51 0 2 3 4 52 3 0, 1 4 53 2 4 54 0, 1, 2, 3 55 A5

We have:µ(0) = µ(1) = 0, 575,µ(2) = 0, 778 = µ(3),µ(4) = 0, 911, µ(5) = 1.So, one obtains thefollowing join space

105

1H 0 1 2 3 4 50 0, 1 0, 1 0, 1, 2, 3 0, 1, 2, 3 A5 H

1 0, 1 0, 1, 2, 3 0, 1, 2, 3 A5 H

2 2, 3 2, 3 2, 3, 4 2, 3, 4, 53 2, 3 2, 3, 4 2, 3, 4, 54 4 4, 55 5

The membership function corresponding to 1H is µ1(0) = µ1(1) = 0, 273,µ1(2) = µ1(3) = 0, 279, µ1(4) = 0, 305, µ1(5) = 0, 333, whence, we obtain∀ r > 1, rH = 1H.

b14) H 0 1 2 3 4 50 0 1 2 3 4 51 0 2 3 4 52 3 0, 1 4 53 2 4 54 5 0, 1, 2, 35 4

One finds:µ(0) = µ(1) = 0, 583,µ(2) = µ(3) = 0, 786,µ(4) = µ(5) = 1.The correspondingjoin space is

1H 0 1 2 3 4 50 0, 1 0, 1 0, 1, 2, 3 0, 1, 2, 3 H H1 0, 1 0, 1, 2, 3 0, 1, 2, 3 H H2 2, 3 2, 3 2, 3, 4, 5 2, 3, 4, 53 2, 3 2, 3, 4, 5 2, 3, 4, 54 4, 5 4, 55 4, 5

1H determines the function: µ1(0) = µ1(1) = µ1(4) = µ1(5) = 0, 267,µ1(2) = µ1(3) = 0, 262. From this, one obtains the join space

2H 0 1 2 3 4 50 0, 1, 4, 5 0, 1, 4, 5 H H 0, 1, 4, 5 0, 1, 4, 51 0, 1, 4, 5 H H 0, 1, 4, 5 0, 1, 4, 52 2, 3 2, 3 H H3 2, 3 H H4 0, 1, 4, 5 0, 1, 4, 55 0, 1, 4, 5

hence: µ2(0) = µ2(1) = µ2(4) = µ2(5) = 0, 208, µ2(2) = µ2(3) = 0, 233. Itfollows ∀ r > 2, rH = 2H 6= 1H and s.f.g.(H) = 2.

106

b15) H 0 1 2 3 4 50 0 1 2 3 4 51 0 2 3 4 52 0, 1 3 4 53 0, 1, 2 5 44 0, 1, 2 35 0, 1, 2

We have:µ(0) = µ(1) = 0, 583,µ(2) = 0, 714,µ(3) = µ(4) = µ(5) = 1.The associatedjoin space is the following one

1H 0 1 2 3 4 50 0, 1 0, 1 0, 1, 2 H H H1 0, 1 0, 1, 2 H H H2 2 2, 3, 4, 5 2, 3, 4, 5 2, 3, 4, 53 3, 4, 5 3, 4, 5 3, 4, 54 3, 4, 5 3, 4, 55 3, 4, 5

The corresponding membership function is: µ1(0) = µ1(1) = 0, 267, µ1(2) =0, 254, µ1(3) = µ1(4) = µ1(5) = 0, 24. From this one obtains ∀ r ≥ 2,rH = 1H.

b16) H 0 1 2 3 4 50 0 1 2 3 4 51 0 2 3 4 52 0, 1 3 4 53 5 0, 1, 2 44 5 35 0, 1, 2

One finds the samemembership functionas in the previous case, thereforewe have s.f.g.(H) = 1.

b17) H 0 1 2 3 4 50 0 1 2 3 4 51 0 2 3 4 52 0, 1 3 4 53 0, 1, 2 4 54 0, 1, 2, 3 55 A5

We have:µ(0) = µ(1) = 0, 547,µ(2) = 0, 683, µ(3) = 0, 806,µ(4) = 0, 911, µ(5) = 1.It results the following join space

1H 0 1 2 3 4 50 0, 1 0, 1 0, 1, 2 0, 1, 2, 3 A5 H1 0, 1 0, 1, 2 0, 1, 2, 3 A5 H2 2 2, 3 2, 3, 4 2, 3, 4, 53 3 3, 4 3, 4, 54 4 4, 55 5

We find µ1(0) = µ1(1) = 0, 29, µ1(2) = 0, 302, µ1(3) = 0, 317, µ1(4) = 0, 332,µ1(5) = 0, 348. It follows ∀ r > 1, rH = 1H.

107

b18) H 0 1 2 3 4 50 0 1 2 3 4 51 0 2 3 4 52 0, 1 3 4 53 4 0, 1, 2, 5 34 3 45 0, 1, 2

The membership function isµ(0) = µ(1) = 0, 556,µ(2) = 0, 690,µ(3) = µ(4) = 1, µ(5) = 0, 813.The associated join space is

1H 0 1 2 3 4 50 0, 1 0, 1 0, 1, 2 H H 0, 1, 2, 51 0, 1 0, 1, 2 H H 0, 1, 2, 52 2 2, 3, 4, 5 2, 3, 4, 5 2, 53 3, 4 3, 4 3, 4, 54 3, 4 3, 4, 55 5

From this we have: µ1(0) = µ1(1) = µ1(3) = µ1(4) = 0, 283, µ1(2) = µ1(5) =0, 29. One obtains the following join space

2H 0 1 2 3 4 50 0, 1, 3, 4 0, 1, 3, 4 H 0, 1, 3, 4 0, 1, 3, 4 H

1 0, 1, 3, 4 H 0, 1, 3, 4 0, 1, 3, 4 H

2 2, 5 H H 2, 53 0, 1, 3, 4 0, 1, 3, 4 H

4 0, 1, 3, 4 H

5 2, 5

We have clearly 2H coincides with the second join space associated with thehypergroup from b2). Therefore ∀ r > 2, rH = 2H and s.f.g.(H) = 2.

b19) H 0 1 2 3 4 50 0 1 2 3 4 51 0 2 3 4 52 0, 1 3 4 53 4 0, 1, 2 54 3 55 A5

One finds:µ(0) = µ(1) = 0, 561,µ(2) = 0, 695,µ(3) = µ(4) = 0, 9,µ(5) = 1.One observes

1H 0 1 2 3 4 50 0, 1 0, 1 0, 1, 2 A5 A5 H

1 0, 1 0, 1, 2 A5 A5 H

2 2 2, 3, 4 2, 3, 4 2, 3, 4, 53 3, 4 3, 4 3, 4, 54 3, 4 3, 4, 55 5

Hence we find: µ1(0)=µ1(1)=0, 280, µ1(2)=0, 2797, µ1(3) = µ1(4) = 0, 2858,

108

µ1(5) = 0, 318. Therefore, we have the second join space

2H 0 1 2 3 4 50 0, 1 0, 1 0, 1, 2 0, 1, 3, 4 0, 1, 3, 4 A2

1 0, 1 0, 1, 2 0, 1, 3, 4 0, 1, 3, 4 A2

2 2 A5 A5 H3 3, 4 3, 4 3, 4, 54 3, 4 3, 4, 55 5

From this we have: µ2(0) = µ2(1) = µ2(3) = µ2(4) = 0, 27948, µ2(2) =0, 315 = µ2(5). The function µ2 determines the third join space

3H 0 1 2 3 4 50 0, 1, 3, 4 0, 1, 3, 4 H 0, 1, 3, 4 0, 1, 3, 4 H1 0, 1, 3, 4 H 0, 1, 3, 4 0, 1, 3, 4 H2 2, 5 H H 2, 53 0, 1, 3, 4 0, 1, 3, 4 H4 0, 1, 3, 4 H5 2, 5

Finally, we obtain µ3(0)=µ3(1)=µ3(3)=µ3(4)=0, 208, µ3(2)=µ3(5)=0, 233.Therefore we obtain ∀ r > 3, rH = 3H and s.f.g.(H) = 3.

(v) Finally we study the fuzzy grade of the i.p.s. hypergroups of order 7.

c1) H 0 1 2 3 4 5 60 0 1 2 3 4 5 61 2 3 0 4 5 62 0 1 4 5 63 2 4 5 64 5 K 65 4 66 A6

where K = {0, 1, 2, 3}.We have µ(0) = µ(1) == µ(2) = µ(3) = 0, 666µ(4) = µ(5) = 0, 916, µ(6) = 1.It follows the join space

1H 0 1 2 3 4 5 60 K K K K A6 A6 H1 K K K A6 A6 H2 K K A6 A6 H3 K A6 A6 H4 4, 5 4, 5 4, 5, 65 4, 5 4, 5, 66 6

henceµ1(0)=µ1(1)=µ1(2)=µ1(3)=0, 195µ1(4) = µ1(5) = 0, 223µ1(6) = 0, 267.Then ∀ r > 1,1 H = rH.

109

c2) H 0 1 2 3 4 5 60 0 1 2 3 4 5 61 0 3 2 4 5 62 0 1 4 5 63 0 4 5 64 5 K 65 4 66 A6

In this case the membershipfunction is the same as in c1),therefore we have again ∀ r, rH = 1H.

c3) H 0 1 2 3 4 5 60 4 5 61 4 5 62 IK 4 5 63 4 5 64 6 K 55 6 46 K

where IK ∈ {ZZ 4, ZZ 2 × ZZ 2}. We haveµ(0) = µ(1) = µ(2) = µ(3) = 0, 679.µ(4) = µ(5) = µ(6) = 1.It follows the following join space

1H 0 1 2 3 4 5 60 K K K K H H H1 K K K H H H2 K K H H H3 K H H H4 4, 5, 6 4, 5, 6 4, 5, 65 4, 5, 6 4, 5, 66 4, 5, 6

and we haveµ1(0)=µ1(1)=µ1(2)=µ1(3)=0, 186µ1(4) = µ1(5) = µ1(6) = 0, 195.Then ∀ r, rH = 1H.

c4) H 0 1 2 3 4 5 60 4 5 61 4 5 62 IK 4 5 63 4 5 64 K 6 55 K 46 K

where IK ∈ {ZZ 4, ZZ 2 × ZZ 2}.We have µ and µ1

in the same way as in the case c3);by consequence we have∀ r > 1, 1H = rH.

c5) H 0 1 2 3 4 5 60 4 5 61 4 5 62 IK 4 5 63 4 5 64 K 5 65 6 K ∪ {4}6 5

where IK ∈ {ZZ 4, ZZ 2 × ZZ 2}.We haveµ(0) = µ(1) = µ(2) = µ(3) = 0, 664µ(4) = 0, 865µ(5) = µ(6) = 1.It follows the join space

110

1H 0 1 2 3 4 5 60 K K K K K ∪ {4} H H

1 K K K K ∪ {4} H H

2 K K K ∪ {4} H H

3 K K ∪ {4} H H

4 4 4, 5, 6 4, 5, 65 5, 6 5, 66 5, 6

We have µ1(0)=µ1(1)=µ1(2)=µ1(3)=0, 197, µ1(4)=0, 214, µ1(5)=µ1(6)=0, 234.It follows: rH = 1H, ∀ r > 1.

c6) H 0 1 2 3 4 5 60 4 5 61 4 5 62 IK 4 5 63 4 5 64 K 5 65 K ∪ {4} 66 A6

where IK ∈ {ZZ 4, ZZ 2 × ZZ 2}.We haveµ(0)=µ(1)=µ(2)=µ(3)=0, 66µ(4) = 0, 862, µ(5) = 0, 924,µ(6) = 1.So, we obtain the join space

1H 0 1 2 3 4 5 60 K K K K K ∪ {4} A6 H

1 K K K K ∪ {4} A6 H

2 K K K ∪ {4} A6 H

3 K K ∪ {4} A6 H

4 4 4, 5 4, 5, 65 5 5, 66 6

We have µ1(0) = 0, 202, µ1(4) = 0, 233, µ1(5) = 0, 267 and µ1(6) = 0, 293It follows ∀ r ≥ 2, 1H = rH

c7) H 0 1 2 3 4 5 60 0 1 2 3 4 5 61 2 0 3 4 5 62 1 3 4 5 63 6 0, 1, 2 4 54 5 6 35 3 0, 1, 26 4

We haveµ(0) = 0, 524 = µ(1) = µ(2)µ(3) = µ(4) = µ(5) = µ(6) = 1Therefore the join spaceassociated is the following one

111

1H 0 1 2 3 4 5 60 0, 1, 2 0, 1, 2 0, 1, 2 H H H H

1 0, 1, 2 0, 1, 2 H H H H

2 0, 1, 2 H H H H

3 3, 4, 5, 6 3, 4, 5, 6 3, 4, 5, 6 3, 4, 5, 64 3, 4, 5, 6 3, 4, 5, 6 3, 4, 5, 65 3, 4, 5, 6 3, 4, 5, 66 3, 4, 5, 6

We clearly have:µ1(0) = µ1(1) = µ1(2) = 0, 195, µ1(3) = µ1(4) = µ1(5) = µ1(6) = 0, 186,whence ∀ r ≥ 2, rH = 1H.

c8) H 0 1 2 3 4 5 60 0 1 2 3 4 5 61 2 0 3 4 5 62 1 3 4 5 63 4 0, 1, 2 5 64 3 5 65 6 0, 1, 2, 3, 46 5

We haveµ(0) = µ(1) = µ(2) = 0, 581µ(3) = µ(4) = 0, 822µ(5) = µ(6) = 1It follows that

1H 0 1 2 3 4 5 60 0, 1, 2 0, 1, 2 0, 1, 2 0, 1, 2, 3, 4 0, 1, 2, 3, 4 H H

1 0, 1, 2 0, 1, 2 0, 1, 2, 3, 4 0, 1, 2, 3, 4 H H

2 0, 1, 2 0, 1, 2, 3, 4 0, 1, 2, 3, 4 H H

3 3, 4 3, 4 3, 4, 5, 6 3, 4, 5, 64 3, 4 3, 4, 5, 6 3, 4, 5, 65 5, 6 5, 66 5, 6

Hence: µ1(0) = 0, 216, µ1(3) = 0, 225, µ1(5) = 0, 238.It follows ∀ r ≥ 2, rH = 1H.

c9) H 0 1 2 3 4 5 60 0 1 2 3 4 5 61 2 0 3 4 5 62 1 3 4 5 63 4 0, 1, 2 5 64 3 5 65 0, 1, 2, 3, 4 66 A6

We haveµ(0) = µ(1) = µ(2) = 0, 576µ(3) = µ(4) = 0, 819µ(5) = 0, 924, µ(6) = 1We obtain that the join space1H is the following one

112

1H 0 1 2 3 4 5 60 0, 1, 2 0, 1, 2 0, 1, 2 0, 1, 2, 3, 4 0, 1, 2, 3, 4 A6 H

1 0, 1, 2 0, 1, 2 0, 1, 2, 3, 4 0, 1, 2, 3, 4 A6 H

2 0, 1, 2 0, 1, 2, 3, 4 0, 1, 2, 3, 4 A6 H

3 3, 4 3, 4 3, 4, 5 3, 4, 5, 64 3, 4 3, 4, 5 3, 4, 5, 65 5 5, 66 6

We have µ1(0) = 0, 220, µ1(3) = 0, 239, µ1(5) = 0, 269, µ1(6) = 0, 297.Therefore ∀ r ≥ 1, rH = 1H.

c10) H 0 1 2 3 4 5 60 0 1 2 3 4 5 61 2 0 3 4 5 62 1 3 4 5 63 0, 1, 2 4 5 64 5 0, 1, 2, 3 65 4 66 A6

We haveµ(0) = µ(1) = µ(2) = 0, 571,µ(3) = 0, 741,µ(4) = µ(5) = 0, 917,µ(6) = 1.One obtains

1H 0 1 2 3 4 5 60 0, 1, 2 0, 1, 2 0, 1, 2 0, 1, 2, 3 A6 A6 H

1 0, 1, 2 0, 1, 2 0, 1, 2, 3 A6 A6 H

2 0, 1, 2 0, 1, 2, 3 A6 A6 H

3 3 3, 4, 5 3, 4, 5 3, 4, 5, 64 4, 5 4, 5 4, 5, 65 4, 5 4, 5, 66 6

It follows µ1(0) = µ1(1) = µ1(2) = 0, 223, µ1(3) = 0, 232, µ1(4) = µ1(5) =0, 251, µ1(6) = 0, 284. Therefore ∀ r > 1, rH = 1H.

c11) H 0 1 2 3 4 5 60 0 1 2 3 4 5 61 2 0 3 4 5 62 1 3 4 5 63 0, 1, 2 4 5 64 0, 1, 2, 3 5 65 0, 1, 2, 3, 4 56 A6

We haveµ(0)=µ(1)=µ(2)=0, 564,µ(3)=0, 735, µ(4)=0, 837,µ(5)=0, 936, µ(6)=1.By consequence the joinspace associated is

113

1H 0 1 2 3 4 5 60 0, 1, 2 0, 1, 2 0, 1, 2 0, 1, 2, 3 0, 1, 2, 3, 4 A6 H1 0, 1, 2 0, 1, 2 0, 1, 2, 3 0, 1, 2, 3, 4 A6 H2 0, 1, 2 0, 1, 2, 3 0, 1, 2, 3, 4 A6 H3 3 3, 4 3, 4, 5 3, 4, 5, 64 4 4, 5 4, 5, 65 5 5, 66 6

It follows µ1(0) = µ(1)1 = µ1(2) = 0, 229, µ1(3) = 0, 249, µ1(4) = 0, 272,µ1(5) = 0, 291, µ1(6) = 0, 310, whence we have ∀ r ≥ 1, rH = 1H.

c12) H 0 1 2 3 4 5 60 0 1 2 3 4 5 61 2 0 3 4 5 62 1 3 4 5 63 0, 1, 2 4 5 64 0, 1, 2, 3 6 55 0, 1, 2, 3 46 0, 1, 2, 3

We have µ(0) = µ(1) = µ(2) = 0, 583, µ(3) = 0, 75, µ(4) = µ(5) = µ(6) = 1.Therefore the join space associated is

1H 0 1 2 3 4 5 60 0, 1, 2 0, 1, 2 0, 1, 2 0, 1, 2, 3 H H H

1 0, 1, 2 0, 1, 2 0, 1, 2, 3 H H H

2 0, 1, 2 0, 1, 2, 3 H H H

3 3 3, 4, 5, 6 3, 4, 5, 6 3, 4, 5, 64 4, 5, 6 4, 5, 6 4, 5, 65 4, 5, 6 4, 5, 66 4, 5, 6

We have µ1(0)=µ1(1)=µ1(2)=µ1(4)=µ1(5)=µ1(6)=0, 214, µ1(3)=0, 212.Therefore the second join space associated is

2H 0 1 2 3 4 5 60 A3 A3 A3 H A3 A3 A3

1 A3 A3 H A3 A3 A3

2 A3 H A3 A3 A3

3 3 H H H

4 A3 A3 A3

5 A3 A3

6 A3

It followsµ2(0) = µ2(1) = µ2(2) = µ2(4) == µ2(5) = µ2(6) 6= µ2(3).By consequence ∀ r > 2,rH = 2H and s.f.g.(H) = 2.

114

c13) H 0 1 2 3 4 5 60 0 1 2 3 4 5 61 0 2 3 4 5 62 0, 1 3 4 5 63 0, 1, 2 4 5 64 0, 1, 2, 3 6 55 0, 1, 2, 3 46 0, 1, 2, 3

We have: µ(0)=µ(1)=0, 512, µ(2) = 0, 635 µ(3) = 0, 75, µ(4)=µ(5)=µ(6)=1.The corresponding join space is

1H 0 1 2 3 4 5 60 0, 1 0, 1 0, 1, 2 0, 1, 2, 3 H H H

1 0, 1 0, 1, 2 0, 1, 2, 3 H H H

2 2 2, 3 2, 3, 4, 5, 6 2, 3, 4, 5, 6 2, 3, 4, 5, 63 3 3, 4, 5, 6 3, 4, 5, 6 3, 4, 5, 64 4, 5, 6 4, 5, 6 4, 5, 65 4, 5, 6 4, 5, 66 4, 5, 6

One obtains µ1(0) = µ1(1) = 0, 252, µ1(2) = 0, 250, µ1(3) = 0, 239, µ1(4) =µ1(5) = µ1(6) = 0, 225. So it’s clear that ∀ r > 1, rH = 1H.

c14) One can see that one obtains the same join space as in the previouscase starting from the following i.p.s. hypergroup

H 0 1 2 3 4 5 60 0 1 2 3 4 5 61 0 2 3 4 5 62 0, 1 3 4 5 63 0, 1, 2 4 5 64 0, 1, 2, 3 6 55 4 0, 1, 2, 36 4

c15) H 0 1 2 3 4 5 60 0 1 2 3 4 5 61 0 2 3 4 5 62 0, 1 3 4 5 63 0, 1, 2 4 5 64 5 0, 1, 2, 3 65 4 66 A6

One finds:µ(0) = µ(1) = 0, 5,µ(2) = 0, 625,µ(3) = 0, 741,µ(4) = µ(5) = 0, 917,µ(6) = 1.

By consequence, we obtain the join space

115

1H 0 1 2 3 4 5 60 0, 1 0, 1 0, 1, 2 0, 1, 2, 3 A6 A6 H

1 0, 1 0, 1, 2 0, 1, 2, 3 A6 A6 H

2 2 2, 3 2, 3, 4, 5 2, 3, 4, 5 2, 3, 4, 5, 63 3 3, 4, 5 3, 4, 5 3, 4, 5, 64 4, 5 4, 5 4, 5, 65 4, 5 4, 5, 66 6

Therefore: µ1(0) = µ1(1) = 0, 260, µ1(2) = 0, 263, µ1(3) = 0, 262,µ1(4) = µ1(5) = 0, 265, µ1(6) = 0, 293.

One obtains that the second join space associated to H is isomorphic to thefirst join space associated to H.

By consequence ∀ r > 1, rH ∼ 1H, 2kH = 2H, 1H = 2k+1H and f.g.(H) = 1.

c16) H 0 1 2 3 4 5 60 0 1 2 3 4 5 61 0 2 3 4 5 62 0, 1 3 4 5 63 0, 1, 2 4 5 64 0, 1, 2, 3 5 65 6 0, 1, 2, 3, 46 5

We have: µ(0) = µ(1) = 0, 498, µ(2) = 0, 623, µ(3) = 0, 739, µ(4) = 0, 84,µ(5) = µ(6) = 1, whence the associated join space is

1H 0 1 2 3 4 5 60 0, 1 0, 1 0, 1, 2 0, 1, 2, 3 0, 1, 2, 3, 4 H H

1 0, 1 0, 1, 2 0, 1, 2, 3 0, 1, 2, 3, 4 H H

2 2 2, 3 2, 3, 4 2, 3, 4, 5, 6 2, 3, 4, 5, 63 3 3, 4 3, 4, 5, 6 3, 4, 5, 64 4 4, 5, 6 4, 5, 65 5, 6 5, 66 5, 6

whence: µ1(0) = µ1(1) = µ1(5) = µ1(6) = 0, 262, µ1(2) = µ1(4) = 0, 267,µ1(3) = 0, 271. It follows the second associated join space

2H 0 1 2 3 4 5 60 0, 1, 5, 6 0, 1, 5, 6 A3 H A3 0, 1, 5, 6 0, 1, 5, 61 0, 1, 5, 6 A3 H A3 0, 1, 5, 6 0, 1, 5, 62 2, 4 2, 3, 4 2, 4 A3 A3

3 3 2, 3, 4 H H

4 2, 4 A3 A3

5 0, 1, 5, 6 0, 1, 5, 66 0, 1, 5, 6

116

Then we have: µ2(0)=µ2(1)=µ2(5)=µ2(6)=0, 195, µ2(2)=µ2(4)=0, 223,µ2(3)=0, 267. Therefore ∀ r > 2, rH = 2H and s.f.g.(H) = 2.

c17) H 0 1 2 3 4 5 60 0 1 2 3 4 5 61 0 2 3 4 5 62 0, 1 3 4 5 63 0, 1, 2 4 5 64 0, 1, 2, 3 5 65 0, 1, 2, 3, 4 66 A6

We have: µ(0) = µ(1) = 0, 493, µ(2) = 0, 619, µ(3) = 0, 735, µ(4) = 0, 837,µ(5) = 0, 924, µ(6)=1.

The join space associated with the former µ is the following one

1H 0 1 2 3 4 5 60 0, 1 0, 1 0, 1, 2 0, 1, 2, 3 0, 1, 2, 3, 4 A6 H

1 0, 1 0, 1, 2 0, 1, 2, 3 0, 1, 2, 3, 4 A6 H

2 2 2, 3 2, 3, 4 2, 3, 4, 5 2, 3, 4, 5, 63 3 3, 4 3, 4, 5 3, 4, 5, 64 4 4, 5 4, 5, 65 5 5, 66 6

The corresponding membership function is: µ1(0) = µ1(1) = 0, 265, µ1(2) =0, 274, µ1(3) = 0, 283, µ1(4) = 0, 291, µ1(5) = 0, 303, µ1(6) = 0, 318. Fromthis one obtains: ∀ r > 1, rH = 1H.

c18) H 0 1 2 3 4 5 60 0 1 2 3 4 5 61 0 2 3 4 5 62 0, 1 3 4 5 63 0, 1 4 5 64 0, 1 5 65 0, 1, 2, 3, 4 66 A6

The corresponding µ is: µ(0) = µ(1) = 0, 552, µ(2) = 0, 728, µ(3) = 0, 796,µ(4) = 0, 837, µ(5) = 0, 924, µ(6) = 1. Therefore we have the same joinspace as in the previous case. So, s.f.g.(H) = 1.

117

c19) H 0 1 2 3 4 5 60 0 1 2 3 4 5 61 0 2 3 4 5 62 0, 1 3 4 5 63 4 0, 1, 2 5 64 3 5 65 6 0, 1, 2, 3, 46 5

We obtain µ(0) = µ(1) = 0, 510, µ(2) = 0, 633, µ(3) = µ(4) = 0, 822,µ(5) = µ(6) = 1. Then the associated join space is the hypergroup

1H 0 1 2 3 4 5 60 0, 1 0, 1 0, 1, 2 0, 1, 2, 3, 4 0, 1, 2, 3, 4 H H

1 0, 1 0, 1, 2 0, 1, 2, 3, 4 0, 1, 2, 3, 4 H H

2 2 2, 3, 4 2, 3, 4 2, 3, 4, 5, 6 2, 3, 4, 5, 63 3, 4 3, 4 3, 4, 5, 6 3, 4, 5, 64 3, 4 3, 4, 5, 6 3, 4, 5, 65 5, 6 5, 66 5, 6

One obtains the membership function: µ1(0)=µ1(1)=0, 233, µ1(2)=0, 249,µ1(3)=µ1(4)=0, 247, µ1(5)=µ1(6)=0, 238. Therefore ∀ r > 1, rH = 1H.

c20) H 0 1 2 3 4 5 60 0 1 2 3 4 5 61 0 2 3 4 5 62 0, 1 3 4 5 63 4 0, 1, 2 5 64 3 5 65 0, 1, 2, 3, 4 66 A6

The corresponding membership function is µ(0) = µ(1) = 0, 505, µ(2) =0, 629, µ(3) = µ(4) = 0, 819, µ(5) = 0, 924, µ(6) = 1. The associated joinspace is

1H 0 1 2 3 4 5 60 0, 1 0, 1 0, 1, 2 0, 1, 2, 3, 4 0, 1, 2, 3, 4 A6 H1 0, 1 0, 1, 2 0, 1, 2, 3, 4 0, 1, 2, 3, 4 A6 H2 2 2, 3, 4 2, 3, 4 2, 3, 4, 5 2, 3, 4, 5, 63 3, 4 3, 4 3, 4, 5 3, 4, 5, 64 3, 4 3, 4, 5 3, 4, 5, 65 5 5, 66 6

This determines the following membership function µ1(0) = µ1(1) = 0, 257,

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µ1(2) = 0, 255, µ1(3) = µ1(4) = 0, 261, µ1(5) = 0, 281, µ1(6) = 0, 305 fromwhich one obtains the second join space:

2H 0 1 2 3 4 5 60 0, 1 0, 1 0, 1, 2 0, 1, 3, 4 0, 1, 3, 4 0, 1, 3, 4, 5 A2

1 0, 1 0, 1, 2 0, 1, 3, 4 0, 1, 3, 4 0, 1, 3, 4, 5 A2

2 2 0, 1, 2, 3, 4 0, 1, 2, 3, 4 A6 H3 3, 4 3, 4 3, 4, 5 3, 4, 5, 64 3, 4 3, 4, 5 3, 4, 5, 65 5 5, 66 6

The corresponding membership function is: µ2(0) = µ2(1) = 0, 257, µ2(2) =0, 289, µ2(3) = µ2(4) = 0, 256, µ2(5) = 0, 279, µ2(6) = 0, 304. The thirdassociated join space is

3H 0 1 2 3 4 5 60 0, 1 0, 1 0, 1, 2, 5 0, 1, 3, 4 0, 1, 3, 4 0, 1, 5 0, 1, 2, 5, 61 0, 1 0, 1, 2, 5 0, 1, 3, 4 0, 1, 3, 4 0, 1, 5 0, 1, 2, 5, 62 2 A6 A6 2, 5 2, 63 3, 4 3, 4 0, 1, 3, 4, 5 H4 3, 4 0, 1, 3, 4, 5 H5 5 2, 5, 66 6

Hence, one obtains: µ3(0)=µ3(1)=0, 255, µ3(2)=0, 292, µ3(3)=µ3(4)=0, 252,µ3(5)=0, 270, µ3(6)=0, 311.Therefore, ∀ r > 3, rH = 3H and s.f.g.(H) = 3.

c21) H 0 1 2 3 4 5 60 0 1 2 3 4 5 61 0 2 3 4 5 62 0, 1 3 4 5 63 5 0, 1, 2 4 64 5 3 65 0, 1, 2 66 A6

The associated membership function is:µ(0) = µ(1) = 0, 524, µ(2) = 0, 646, µ(3) = µ(4) = µ(5) = 0, 907, µ(6) = 1.Here is the corresponding join space

119

1H 0 1 2 3 4 5 60 0, 1 0, 1 0, 1, 2 A6 A6 A6 H1 0, 1 0, 1, 2 A6 A6 A6 H2 2 2, 3, 4, 5 2, 3, 4, 5 2, 3, 4, 5 2, 3, 4, 5, 63 3, 4, 5 3, 4, 5 3, 4, 5 3, 4, 5, 64 3, 4, 5 3, 4, 5 3, 4, 5, 65 3, 4, 5 3, 4, 5, 66 6

One finds: µ1(0)=µ1(1)=0, 246, µ1(2)=0, 235, µ1(3)=µ1(4)=µ1(5)=0, 230,µ1(6)=0, 267, whence one obtains:

2H 0 1 2 3 4 5 60 0, 1 0, 1 0, 1, 2 A6 A6 A6 0, 1, 61 0, 1 0, 1, 2 A6 A6 A6 0, 1, 62 2 2, 3, 4, 5 2, 3, 4, 5 2, 3, 4, 5 0, 1, 2, 63 3, 4, 5 3, 4, 5 3, 4, 5 H4 3, 4, 5 3, 4, 5 H5 3, 4, 5 H6 6

The corresponding membership function is as follows: µ2(0)=µ2(1)=0, 251,µ2(2) = 0, 232, µ2(3) = µ2(4) = µ2(5) = 0, 223, µ2(6) = 0, 284. It resultsthat ∀ r > 2, rH = 2H and therefore s.f.g.(H) = 2.

c22) H 0 1 2 3 4 5 60 0 1 2 3 4 5 61 0 2 3 4 5 62 0, 1 3 4 5 63 0, 1, 2 5 4 64 0, 1, 2 3 65 0, 1, 2 66 A6

We obtain the same membership function as for the previous i.p.s. hyper-group and by consequence the same associated join spaces. Therefore wehave: s.f.g.(H) = 2.

c23) H 0 1 2 3 4 5 60 0 1 2 3 4 5 61 0 2 3 4 5 62 0, 1 3 4 5 63 5 0, 1, 2 6 44 6 3 55 4 0, 1, 26 6

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We have: µ(0)=µ(1)=0, 548, µ(2) = 0, 667, µ(3)=µ(4)=µ(5)=µ(6)=1.So, the associated join space has the form

1H 0 1 2 3 4 5 60 0, 1 0, 1 0, 1, 2 H H H H1 0, 1 0, 1, 2 H H H H2 2 2, 3, 4, 5, 6 2, 3, 4, 5, 6 2, 3, 4, 5, 6 2, 3, 4, 5, 63 3, 4, 5, 6 3, 4, 5, 6 3, 4, 5, 6 3, 4, 5, 64 3, 4, 5, 6 3, 4, 5, 6 3, 4, 5, 65 3, 4, 5, 6 3, 4, 5, 66 3, 4, 5, 6

It follows µ1(0)=µ1(1)=0, 234, µ1(2)=0, 214, µ1(3)=µ1(4)=µ1(5)=µ1(6)=0, 197.Therefore, we have ∀ r > 1, rH = 1H.

c24) H 0 1 2 3 4 5 60 0 1 2 3 4 5 61 0 2 3 4 5 62 3 0, 1 4 5 63 2 4 5 64 0, 1, 2, 3 6 55 0, 1, 2, 3 46 0, 1, 2, 3

We have: µ(0) = µ(1) = 0, 536, µ(2) = µ(3) = 0, 719, µ(4)=µ(5)=µ(6)=1.So, the associated join space is

1H 0 1 2 3 4 5 60 0, 1 0, 1 0, 1, 2, 3 0, 1, 2, 3 H H H1 0, 1 0, 1, 2, 3 0, 1, 2, 3 H H H2 2, 3 2, 3 2, 3, 4, 5, 6 2, 3, 4, 5, 6 2, 3, 4, 5, 63 2, 3 2, 3, 4, 5, 6 2, 3, 4, 5, 6 2, 3, 4, 5, 64 4, 5, 6 4, 5, 6 4, 5, 65 4, 5, 6 4, 5, 66 4, 5, 6

We calculate: µ1(0) = µ1(1) = 0, 238, µ1(2) = µ1(3) = 0, 225, µ1(4) =µ1(5) = µ1(6) = 0, 216. It results clearly ∀ r > 1, rH = 1H.

c25) H 0 1 2 3 4 5 60 0 1 2 3 4 5 61 0 2 3 4 5 62 3 0, 1 4 5 63 2 4 5 64 5 6 0, 1, 2, 35 0, 1, 2, 3 46 5

121

Since µ(0) = µ(1) = 0, 536, µ(2) = µ(3) = 0, 719, µ(4) = µ(5) = µ(6) = 1,we conclude ∀ r > 1, rH = 1H as in the previous case.

c26) H 0 1 2 3 4 5 60 0 1 2 3 4 5 61 0 2 3 4 5 62 3 0, 1 4 5 63 2 4 5 64 0, 1, 2, 3 5 65 0, 1, 2, 3, 4 66 A6

The corresponding membership function is:µ(0)=µ(1)=0, 517, µ(2)=µ(3)=0, 702, µ(4)=0, 837, µ(5)=0, 924, µ(6)=1.The function µ determines the join space

1H 0 1 2 3 4 5 60 0, 1 0, 1 0, 1, 2, 3 0, 1, 2, 3 0, 1, 2, 3, 4 A6 H

1 0, 1 0, 1, 2, 3 0, 1, 2, 3 0, 1, 2, 3, 4 A6 H

2 2, 3 2, 3 2, 3, 4 2, 3, 4, 5 2, 3, 4, 5, 63 2, 3 2, 3, 4 2, 3, 4, 5 2, 3, 4, 5, 64 4 4, 5 4, 5, 65 5 5, 66 6

From this, one obtains: µ1(0) = µ1(1) = 0, 252, µ1(2) = µ1(3) = 0, 255,µ(4) = 0, 270, µ(5) = 0, 292, µ(6) = 0, 311. It follows ∀ r > 1, rH = 1H.

c27) H 0 1 2 3 4 5 60 0 1 2 3 4 5 61 0 2 3 4 5 62 3 0, 1 4 5 63 2 4 5 64 0, 1, 2, 3 5 65 6 0, 1, 2, 3, 46 5

The membership function associated with this hypergroup is the following:µ(0)=µ(1) = 0, 521, µ(2)=µ(3) = 0, 706, µ(4)=0, 84, µ(5)=µ(6)=1.One obtains the join space

122

1H 0 1 2 3 4 5 60 0, 1 0, 1 0, 1, 2, 3 0, 1, 2, 3 0, 1, 2, 3, 4 H H

1 0, 1 0, 1, 2, 3 0, 1, 2, 3 0, 1, 2, 3, 4 H H

2 2, 3 2, 3 2, 3, 4 2, 3, 4, 5, 6 2, 3, 4, 5, 63 2, 3 2, 3, 4 2, 3, 4, 5, 6 2, 3, 4, 5, 64 4 4, 5, 6 4, 5, 65 5, 6 5, 66 5, 6

So we have: µ1(0) = µ1(1) = 0, 248, µ1(2) = µ1(3) = 0, 247, µ1(4) = 0, 249,µ1(5) = µ1(6) = 0, 253. From this, one gets the second join space associatedwith H

2H 0 1 2 3 4 5 60 0, 1 0, 1 0, 1, 2, 3 0, 1, 2, 3 0, 1, 4 0, 1, 4, 5, 6 0, 1, 4, 5, 61 0, 1 0, 1, 2, 3 0, 1, 2, 3 0, 1, 4 0, 1, 4, 5, 6 0, 1, 4, 5, 62 2, 3 2, 3 0, 1, 2, 3, 4 H H

3 2, 3 0, 1, 2, 3, 4 H H

4 4 4, 5, 6 4, 5, 65 5, 6 5, 66 5, 6

We clearly have: 2H ' 1H and moreover µ2(0) = µ2(1) = 0, 247,µ2(2) = µ2(3) = 0, 248, µ2(4) = 0, 249, µ2(5) = µ2(6) = 0, 253, that isµ2 = µ, by consequence ∀ k ≥ 1, µ = µ2k and µ1 = µ2k+1. Finally, ∀ k ≥ 1,2kH = 2H ∼ 1H = 2k+1H and therefore f.g.(H) = 1.

c28) H 0 1 2 3 4 5 60 0 1 2 3 4 5 61 0 2 3 4 5 62 0, 1 4 3 5 63 0, 1 2 5 64 0, 1 5 65 6 0, 1, 2, 3, 46 5

We have: µ(0) = µ(1) = 0, 557, µ(2) = µ(3) = µ(4) = 0, 8, µ(5) = µ(6) = 1.Hence one obtains

1H 0 1 2 3 4 5 60 0, 1 0, 1 0, 1, 2, 3, 4 0, 1, 2, 3, 4 0, 1, 2, 3, 4 H H1 0, 1 0, 1, 2, 3, 4 0, 1, 2, 3, 4 0, 1, 2, 3, 4 H H2 2, 3, 4 2, 3, 4 2, 3, 4 2, 3, 4, 5, 6 2, 3, 4, 5, 63 2, 3, 4 2, 3, 4 2, 3, 4, 5, 6 2, 3, 4, 5, 64 2, 3, 4 2, 3, 4, 5, 6 2, 3, 4, 5, 65 5, 6 5, 66 5, 6

123

The corresponding membership function is : µ1(0) = µ1(1) = µ1(5) =µ1(6) = 0, 231, µ1(2) = µ1(3) = µ1(4) = 0, 218. We form the secondassociated join space

2H 0 1 2 3 4 5 60 0, 1, 5, 6 0, 1, 5, 6 H H H 0, 1, 5, 6 0, 1, 5, 61 0, 1, 5, 6 H H H 0, 1, 5, 6 0, 1, 5, 62 2, 3, 4 2, 3, 4 2, 3, 4 H H3 2, 3, 4 2, 3, 4 H H4 2, 3, 4 H H5 0, 1, 5, 6 0, 1, 5, 66 0, 1, 5, 6

From this we find a function µ2 which gives the same join space. Indeed:µ2(0) = µ2(1) = µ2(5) = µ2(6) = 0, 186, µ2(2) = µ2(3) = µ2(4) = 0, 195.Therefore we have ∀ r > 2, rH = 2H and s.f.g.(H) = 2.

c29) H 0 1 2 3 4 5 60 0 1 2 3 4 5 61 0 2 3 4 5 62 4 0, 1 3 5 63 4 2 5 64 0, 1 5 65 6 0, 1, 2, 3, 46 5

One find the same membership function µ associated with the previous i.p.s.hypergroup, and therefore an equal sequence of join spaces and s.f.g.(H)=2.

c30) H 0 1 2 3 4 5 60 0 1 2 3 4 5 61 0 2 3 4 5 62 4 0, 1 5 3 63 5 2 4 64 3 0, 1 65 2 66 A6

We have: µ(0) = µ(1) = 0, 595, µ(2) = µ(3) = µ(4) = µ(5) = 0, 896,µ(6) = 1, whence the associated join space is

124

1H 0 1 2 3 4 5 60 0, 1 0, 1 A6 A6 A6 A6 H1 0, 1 A6 A6 A6 A6 H2 2, 3, 4, 5 2, 3, 4, 5 2, 3, 4, 5 2, 3, 4, 5 2, 3, 4, 5, 63 2, 3, 4, 5 2, 3, 4, 5 2, 3, 4, 5 2, 3, 4, 5, 64 2, 3, 4, 5 2, 3, 4, 5 2, 3, 4, 5, 65 2, 3, 4, 5 2, 3, 4, 5, 66 6

The membership function associated with 1H is µ1(0) = µ1(1) = 0, 218,µ1(2) = µ1(3) = µ1(4) = µ1(5) = 0, 201, µ1(6) = 0, 244, whence the associa-ted join space is as follows:

2H 0 1 2 3 4 5 60 0, 1 0, 1 A6 A6 A6 A6 0, 1, 61 0, 1 A6 A6 A6 A6 0, 1, 62 2, 3, 4, 5 2, 3, 4, 5 2, 3, 4, 5 2, 3, 4, 5 H3 2, 3, 4, 5 2, 3, 4, 5 2, 3, 4, 5 H4 2, 3, 4, 5 2, 3, 4, 5 H5 2, 3, 4, 5 H6 6

From this one obtains: µ2(0) = µ2(1) = 0, 223, µ2(2) = µ2(3) = µ2(4) =µ2(5) = µ2(6) = 0, 267. Therefore we have ∀ r > 2, rH = 2H and s.f.g.(H)=2.

c31) H 0 1 2 3 4 5 60 0 1 2 3 4 5 61 0 2 3 4 5 62 2 5 0, 1 6 33 0, 1 6 2 44 2 3 55 4 0, 16 2

The corresponding membership function is: µ(0) = µ(1) = 0, 643, µ(2) =µ(3) = µ(4) = µ(5) = µ(6) = 1 and the associated join space is

1H 0 1 2 3 4 5 60 0, 1 0, 1 H H H H H1 0, 1 H H H H H2 2, 3, 4, 5, 6 2, 3, 4, 5, 6 2, 3, 4, 5, 6 2, 3, 4, 5, 6 2, 3, 4, 5, 63 2, 3, 4, 5, 6 2, 3, 4, 5, 6 2, 3, 4, 5, 6 2, 3, 4, 5, 64 2, 3, 4, 5, 6 2, 3, 4, 5, 6 2, 3, 4, 5, 65 2, 3, 4, 5, 6 2, 3, 4, 5, 66 2, 3, 4, 5, 6

125

One obtains: µ1(0) = µ1(1) = 0, 202, µ1(2)=µ1(3)=µ1(4)=µ1(5)=µ1(6)=0, 175.By consequence, ∀ r > 1, rH = 1H.

c32) H 0 1 2 3 4 5 60 0 1 2 3 4 5 61 0 2 3 4 5 62 4 0, 1 3 5 63 4 2 5 64 0, 1 5 65 0, 1, 2, 3, 4 66 A6

We have: µ(0)=µ(1) = 0, 552, µ(2)=µ(3)=µ(4)=0, 796, µ(5)=0, 924, µ(6)=1.From µ one obtains the join space

1H 0 1 2 3 4 5 60 0, 1 0, 1 0, 1, 2, 3, 4 0, 1, 2, 3, 4 0, 1, 2, 3, 4 A6 H1 0, 1 0, 1, 2, 3, 4 0, 1, 2, 3, 4 0, 1, 2, 3, 4 A6 H2 2, 3, 4 2, 3, 4 2, 3, 4 2, 3, 4, 5 2, 3, 4, 5, 63 2, 3, 4 2, 3, 4 2, 3, 4, 5 2, 3, 4, 5, 64 2, 3, 4 2, 3, 4, 5 2, 3, 4, 5, 65 5 5, 66 6

The membership function determined by 1H is: µ1(0) = µ1(1) = 0, 235,µ1(2) = µ1(3) = µ1(4) = 0, 228, µ1(5) = 0, 258, µ1(6) = 0, 290. Hence weform the join space:

2H 0 1 2 3 4 5 60 0, 1 0, 1 0, 1, 2, 3, 4 0, 1, 2, 3, 4 0, 1, 2, 3, 4 0, 1, 5 0, 1, 5, 61 0, 1 0, 1, 2, 3, 4 0, 1, 2, 3, 4 0, 1, 2, 3, 4 0, 1, 5 0, 1, 5, 62 2, 3, 4 2, 3, 4 2, 3, 4 A6 H3 2, 3, 4 2, 3, 4 A6 H4 2, 3, 4 A6 H5 5 5, 66 6

and we find: µ2(0) = µ2(1) = 0, 239, µ2(2) = µ2(3) = µ2(4) = 0, 220,µ2(5) = 0, 269, µ2(6) = 0, 297. Therefore we have ∀ r > 2, rH = 2H ands.f.g.(H) = 2.

126

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