# Some Properties of Intuitionistic L-Fuzzy Subnearrings of a Nearring

DOI : 10.17577/IJERTCONV4IS24003

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#### Some Properties of Intuitionistic L-Fuzzy Subnearrings of a Nearring

B. Thenmozhi*

Department of Mathematics, Syed ammal Engineering College Ramanathapuram

S. Karthikeyan

Department of Mathematics Velammal College of Engineering and Technology,

1. Naganathan

Department of Mathematics, Government Arts College Pollachi, Udumalapet, Tamil Nadu, India.

AbstractIn this paper, we study some of the properties of intuitionistic L-fuzzy subnearring of a nearring and prove some results on these. With the expectations that these results may be applied in different fields.

KeywordsL-fuzzy subset, intuitionistic L-fuzzy subset, intuitionistic L-fuzzy subnearring, intuitionistic L-fuzzy relation, Product of intuitionistic L-fuzzy subsets.

1. INTRODUCTION

After the introduction of fuzzy sets by L.A.Zadeh[15], several researchers explored on the generalization of the notion of fuzzy set. The concept of intuitionistic L-fuzzy subset was introduced by K.T.Atanassov[4,5], as a generalization of the notion of fuzzy set. Azriel Rosenfeld[6] defined a fuzzy groups. Asok Kumer Ray[3] defined a product of fuzzy subgroups. We introduce the concept of intuitionistic L-fuzzy subnearring of a nearring and established some results.

2. PRELIMINARIES

Definition 2.1. Let X be a non-empty set and L = (L, ) be a lattice with least element 0 and greatest element 1. A L-fuzzy subset A of X is a function A : X L.

Definition 2.2. Let (L, ) be a complete lattice with an involutive order reversing operation N : L L.

An intuitionistic L-fuzzy subset (ILFS) A in X is defined as an

(i) A(x y) A(x) A(y)

1. A(xy) A(x) A(y)

2. A(xy) A(x) A(y)

3. A(xy) A(x) A(y), for all x and y in R.

Definition 2.4. Let A and B be any two intuitionistic L-fuzzy subnearrings of nearrings R1 and R2 respectively. The product of A and B denoted by AxB is defined as AxB={(x, y),AxB( x, y), AxB( x, y)/for all x in R1 , y in R2}, where AxB(x,y) =A(x) B(y) and AxB(x, y) = A(x) B(y).

Definition 2.5. Let A be an intuitionistic L-fuzzy subset in a set S, the strongest intuitionistic L-fuzzy relation on S, that is an intuitionistic L-fuzzy relation on A is V given by

V( x, y) = A(x) A(y) and V(x, y) = A(x) A(y), for all x and y in S.

Definition 2.6. Let X and X be any two sets. Let f : X X be any function and A be a intuitionistic L-fuzzy subset in X, V be an intuitionistic L-fuzzy subset in f(X) = X, defined by

V(y) = sup A(x) and V(y) = inf A(x), for all

object of the form A={< x, A(x), A(x) > / x in X}, where

x f 1 ( y )

x f 1 ( y )

A : X L and A : X L define the degree of membership and the degree of non-membership of the element xX respectively and for every xX satisfying A(x) N( A(x) ).

Definition 2.3. Let ( R, +, ) be a nearring. A intuitionistic L-fuzzy subset A of R is said to be an intuitionistic L-fuzzy subnearring (ILFSNR) of R if it satisfies the following axioms:

x in X and y in X. A is called a pre image of V under f and is denoted by f-1(V).

3. SOME PROPERTIES OF INTUITIONISTIC L- FUZZY SUBNEARRINGS OF A NEARRING

Theorem 3.1. Intersection of any two intuitionistic L-fuzzy subnearrings of a nearring R is a intuitionistic L-fuzzy subnearring of R.

Proof. Let A and B be any two intuitionistic L-fuzzy subnearrings of a nearring R and x and y in R. Let A = {(x,A(x),A(x))/xR} and B = { (x, B(x), B(x) ) /

xR} and also let C = AB = { ( x, C(x), C(x) ) / xR}, where A(x) B(x) = C(x) and A(x) B(x) = C(x).

Now, C( xy) [A(x) A(y)] [B(x) B(y)] = [A(x)

B(x)] [A(y)B(y)] = C(x) C(y). Therefore, C( x

y) (x) (y), for all x and y in R.

Proof. Let AxB be an intuitionistic L-fuzzy subnearring of R1xR2. By contraposition, suppose that none of the statements

(i) and (ii) holds. Then we can find a in R1 and b in R2 such that A(a) B(e), A(a) B(e) and B(b) A(e), B(b)

C C (e). We have,

(a, b) (e) (e) = (e) (e)

And, C(xy) [A(x) A(y)] [B(x) B(y)] = A

AxB

B A A B

C C

C C

[ (x) (x)] [ (y) (y)] = (x) (y). Therefore,

= AxB(e, e). And, AxB ( a, b) B(e) A(e) = A(e)

A B A B

B(e) = AxB (e, e). Thus AxB is not an intuitionistic L-fuzzy

C(xy) C(x) C(y), for all x and y in R. Also, C(x y)

[A(x)A(y)][B(x)B(y)]=[A(x)B(x)][A(y)B(y)]

subnearring of R1xR2. Hence either B(e) A(x) and B(e)

(x), for all x in R or (e) (y) and (e) (y), for

= C(x) C(y). Therefore, C(xy) C(x) C(y), for all x and y in R. And, C(xy) [A(x) A(y)] [B(x) B(y)] =

A

all y in R2.

1 A B A B

[A(x) B(x)] [A(y) B(y)] = C(x) C(y). Therefore,

C(xy) C(x) C(y), for all x and y in R. Therefore, C is an intuitionistic L-fuzzy subnearring of a nearring R.

Thorem 3.2. Let ( R, +, ) is a nearring. The intersection of a family of intuitionistic L-fuzzy subnearrings of R is an intuitionistic L-fuzzy subnearring of R.

Proof. It is trivial.

Theorem 3.3. If A and B are any two intuitionistic L-fuzzy subnearrings of the nearrings R1 and R2 respectively, then AxB is an intuitionistic L-fuzzy subnearring of R1xR2..

Proof. Let A and B be two intuitionistic L-fuzzy subnearrings of the nearrings R1 and R2 respectively. Let x1 and x2 be in R1 and y1, y2 be in R2. Then (x1, y1) and (x2, y2) in R1xR2. Now, AxB[ (x1, y1) (x2, y2) ] = A( x1x2) B(y1y2)

[A(x1)A(x2)][B(y1)B(y2)]=[A(x1)B(y1)] [A(x2)B(y2)] = AxB(x1, y1) AxB(x2, y2). Therefore,

AxB [(x1, y1) (x2, y2)] AxB ( x1, y1) AxB(x2, y2), for all (x1, y1) and (x2, y2) in R1xR2. Also, AxB[ (x1, y1)(x2, y2)] =

A(x1x2)B(y1y2) [ A(x1)A(x2)] [ B(y1)B(y2)] =

[A(x1) B(y1)][A(x2)B(y2)]= AxB(x1, y1)AxB(x2, y2). Therefore, AxB[(x1, y1)(x2, y2)] AxB(x1, y1)AxB(x2, y2), for all (x1, y1), (x2, y2 ) in R1xR2. And, AxB[(x1, y1)(x2, y2)]

=

A( x1 x2) B( y1 y2) [A(x1) A(x2) ][B(y1) B(y2)

]=[A(x1) B(y1)][ A(x2)B(y2)] = AxB (x1, y1)AxB (x2, y2). Therefore,AxB[(x1, y1)(x2, y2)] AxB( x1, y1) AxB (x2, y2) , for all (x1, y1), (x2, y2) in R1xR2. Also, AxB [(x1, y1)(x2,

y2)] = A(x1x2) B(y1y2) [A(x1) A(x2) ][ B(y1)

B(y2)] = [A(x1)B(y1) ] [ A(x2) B(y2)]= AxB(x1, y1)

AxB (x2, y2). Therefore, AxB [(x1, y1)(x2, y2)] AxB (x1, y1)

AxB(x2, y2), for all (x1, y1), (x2, y2) in R1xR2. Hence AxB is an intuitionistic L-fuzzy subnearring of R1xR2.

Theorem 3.4. Let A and B be intuitionistic L-fuzzy subnearrings of the nearrings R1 and R2 respectively. Suppose that e and e are the identity element of R1 and R2 respectively. If AxB is an intuitionistic L-fuzzy subnearring of R1xR2, then at least one of the following two statements must hold.

1. B(e) A(x) and B(e) A(x), for all x in R1,

2. A(e) B(y) and A(e ) B(y), for all y in R2.

Theorem 3.5. Let A and B be two intuitionistic L-fuzzy subsets of the nearrings R1 and R2 respectively and AxB is an intuitionistic L-fuzzy subnearring of R1xR2. Then the following are true:

if A(x) B(e) and A(x) B( e), then A is an intuitionstic L-fuzzy subnearring of R1.

1. if A(x) B(e) and A(x) B( e), then A is an intuitionistic L-fuzzy subnearring of R1.

2. if B(x) A(e) and B(x) A(e), then B is an intuitionistic L-fuzzy subnearring of R2.

3. either A is an intuitionistic L-fuzzy subnearring of R1 or B is an intuitionistic L-fuzzy subnearring of R2.

Proof. Let AxB be an intuitionistic L-fuzzy subnearring of R1xR2, x, y in R1and e in R2. Then (x, e) and (y, e) are in R1xR2. Now, using the property that A(x) B(e) and

A(x) B(e), for all x in R1, we get, A(xy) = AxB [(xy), (e+e)] AxB (x, e) AxB (y, e) = [A(x) B (e) ]

[ A(y)B(e)]=A(x) A(y) A(x) A(y). Therefore,

A(x y) A(x)A(y), for all x and y in R1. Also, A(xy)

=AxB [(xy), (ee )] AxB(x, e) AxB(y, e) = [A(x)B(e]

[A(y)B(e)] = A(x)A(y). Therefore, A(xy)

A(x) A(y), for all x and y in R1. And, A(xy) = AxB

[(xy), (e+e)]AxB(x, e) AxB(y, e) = [A(x) B(e)] [A(y)B(e) ] = A(x) A(y) A(x)A(y). Therefore,

A(xy) A(x) A(y), for all x and y in R1. Also,

A(xy)=AxB[(xy), (ee )] AxB(x, e) AxB(y, e) =

[A(x) B(e) ][A(y)B(e) ] = A(x) A(y). Therefore,

A(xy) A(x) A(y), for all x and y in R1. Hence A is an intuitionistic L-fuzzy subnearring of R1. Thus (i) is proved. Now, using the property that B(x) A(e) and B(x)A(e), for all x in R2. Let x and y in R2 and e in R1. Then (e, x) and (e, y) are in R1xR2. We get, B(xy) = AxB[(e+e), (xy)]

AxB(e, x)AxB(e, y) = [A(e)B(x)][A(e)B(y)] =

B(x)B(y) B(x)B(y). Therefore, B(xy)

B(x)B(y), for all x and y in R2. Also, B(xy) = AxB[(ee ), (xy)] AxB(e, x)AxB(e, y) = [ A(e)B(x)][ A(e)

B(y)] = B(x)B(y). Therefore, B(xy)B(x)B(y), for all

x and y in R2. And, B(xy) = AxB[(e+e ), (xy)]AxB(e, x)

AxB(e, y)= [A(e) B(x)] [A(e) B(y)] = B(x)

B(y) B(x)B(y). Therefore, B(xy) B(x) B(y), for all x and y in R2. Also, B(xy)= AxB[(ee ), (xy)] AxB(e, x)

AxB(e, y)=[ A(e)B(x) ][ A(e) B(y) ] = B(x)

B(y). Therefore, B(xy) B(x) B(y), for all x and y in R2. Hence B is an intuitionistic L-fuzzy subnearring of a nearring R2. Thus (ii) is proved. (iii) is clear.

Theorem 3.6. Let A be an intuitionistic L-fuzzy subset of a nearring R and V be the strongest intuitionistic L-fuzzy relation of R. Then A is an intuitionistic L-fuzzy subnearring of R if and only if V is an intuitionistic L-fuzzy subnearring of of RxR.

Proof. Suppose that A is an intuitionistic L-fuzzy subnearring of a nearring R. Then for any x = (x1, x2) and y = (y1, y2) are in RxR. We have, V(xy) = V[( x1y1 , x2y2)] [A(x1)A(y1)] [A(x2) A(y2)] = [A(x1)

A(x2)][A(y1)A(y2)] = V(x1, x2) V(y1, y2)=

V(x)V(y). Therefore, V(xy) V(x)V (y), for all x and y in RxR. And, V(xy) = V[( x1y1, x2y2)] [A(x1) A(y1)] [A(x2)A(y2)] = [A(x1)A(x2)] [A(y1)A(y2)]= V(x1, x2) V(y1, y2)= V(x) V(y). Therefore, V(xy) V(x)

V(y), for all x and y in RxR. Also we have,

V(xy)=V[(x1y1, x2y2)] [A(x1)A(y1)] [A(x2)

A(y2)] = [A(x1) A(x2) ][A(y1) A(y2)] = V(x1, x2) V

(y1, y2) = V(x) V(y). Therefore, V(xy) V (x) V (y), for all x and y in RxR. And, V(xy)= V(x1y1, x2y2) [A(x1)

A(y1)][A(x2) A(y2)] = [A(x1) A(x2)] [A(y1)

(y )] = (x , x ) (y , y ) = (x) (y). Therefore,

4. CONCLUSION

We tried to prove some results to use in the field of Intuitionistic L fuzzy subnearring of a nearing

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V(xy) V(x) V(y), for all x and y in RxR. This proves that V is an intuitionistic L-fuzzy subnearring of RxR. Conversely assume that V is an intuitionistic L-fuzzy subnearring of RxR, then for any x=(x1, x2) and y=(y1, y2) are

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y2)]= V (xy) V(x) V(y)= V(x1, x2) V(y1, y2) = [A(x1) A(x2)] [A(y1) A(y2)]. If we put x2= y2= 0, we get, A(x1 y1) A(x1) A(y1), for all x1 and y1 in R. And,

A(x1y1) A(x2y2) = V[(x1, x2)(y1, y2)] = V(xy)

V(x)V(y)= V(x1, x2)V(y1, y2)=[A(x1)A(x2)] [A(y1)A(y2)]. If we put x2=y2=0, we get, A(x1y1)

A(x1)A(y1), for all x1,y1 in R. Also we have,

A(x1y1) A(x2y2) = V[(x1, x2) (y1, y2)]= V(xy)

V (x) V(y) = V(x1, x2) V(y1, y2) = [A(x1)A(x2)]

[A(y1) A(y2)]. If we put x2 = y2 = 0, we get,

A(x1y1)A(x1)A(y1), for all x1 and y1 in R. And,

A(x1y1)A(x2y2) = V[(x1, x2) (y1, y2)] = V (x y)

V(x)V(y)=V(x1,x2)V (y1, y2) = [A(x1)

A(x2)][A(y1) A(y2)]. If we put x2 = y2 = 0, we get,

A(x1y1) A(x1)A(y1), for all x1 and y1 in R. Hence A is a A is an intuitionistic L-fuzzy subnearring of a nearring R.

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