# On BRK-Algebras

Authors: Ravi Kumar Bandaru
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## Abstract

The notion of BRK-algebra is introduced which is a generalization of BCK/BCI/BCH/Q/QS/BM-algebras. The concepts ofG-part, p-radical, medial of a BRK-algebra are introduced and studied their properties. We proved that the variety of all medial BRK-algebras is congruence permutable and showed that every associative BRK-algebra is a group.


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## Body

## 1. Introduction

In 1996, Imai and Iséki [1] introduced two classes of abstract algebras: BCK-algebras and BCI-algebras. These algebras have been extensively studied since their introduction. In 1983, Hu and Li [2] introduced the notion of a BCH-algebra which is a generalization of the notion of BCK and BCI-algebras and studied a few properties of these algebras. In 2001, Neggers et al. [3] introduced a new notion, called a Q-algebra and generalized some theorems discussed in BCI/BCK-algebras. In 2002, Neggers and Kim [4] introduced a new notion, called a B-algebra, and obtained several results. In 2007, Walendziak [5] introduced a new notion, called a BF-algebra, which is a generalization of B-algebra. In [6], C. B. Kim and H. S. Kim introduced BG-algebra as a generalization of B-algebra. We introduce a new notion, called a BRK-algebra, which is a generalization of BCK/BCI/BCH/Q/QS/BM-algebras. The concept of G-part, p-radical, and medial of a BRK-algebra are introduced and studied their properties.


## 2. Preliminaries

First, we recall certain definitions from [2–5, 7, 8] that are required in the paper.Definition 2.1.
A BCI-algebra is an algebra(X,*,0) of type (2, 0) satisfying the followingconditions:(B1)
(x*y)*(x*z)≤(z*y),(B2)
x*(x*y)≤y,(B3)
x≤x,(B4)
x≤yandy≤ximplyx=y,(B5)
x≤0impliesx=0,wherex≤yisdefinedbyx*y=0,forallx,y,z∈X.If (B5) is replaced by (B6): 0≤x, then the algebra is called a BCK-algebra. It is known that every BCK-algebra is a BCI-algebra but not conversely.Definition 2.2.
A BCH-algebra is an algebra (X,*,0) of type (2,0) satisfying (B3), (B4), and (B7): (x*y)*z=(x*z)*y.It is shown that every BCI-algebra is a BCH-algebra but not conversely.Definition 2.3.
A Q-algebra is an algebra (X,*,0) of type (2,0) satisfying (B3), (B7),  and  (B8): x*0=x.A Q-algebra is said to be a QS-algebra if it satisfies the additional relation:(B9)
(x*y)*(x*z)=z*y,for anyx,y,z∈X. It is shown that every BCH-algebra is a Q-algebra but not conversely.Definition 2.4.
A B-algebra is an algebra(X,*,0) of type (2,0) satisfying (B3), (B8),  and  (B10): (x*y)*z=x*(z*(0*y)).A B-algebra is said to be 0-commutative ifa*(0*b)=b*(0*a) for any a,b∈X. In [3], it is shown that Q-algebras and B-algebras are different notions.Definition 2.5.
A BF-algebra is an algebra(X,*,0) of type (2,0) satisfying (B3), (B8), and (B11): 0*(x*y)=(y*x).It is shown that every B-algebra is BF-algebra but not conversely.Definition 2.6.
A BM-algebra is an algebra(X,*,0) of type (2,0) satisfying (B8) and  (B12): (x*y)*(x*z)=z*y.Definition 2.7.
A BH-algebra is an algebra(X,*,0) of type (2,0) satisfying (B3), (B4), and (B8).Definition 2.8.
A BG-algebra is an algebra(X,*,0) of type (2,0) satisfying (B3), (B8), and (BG): (x*y)*(0*y)=x.


## 3. BRK-Algebras

In this section, we define the notion of BRK-algebra and observe that the axioms in the definition are independent.Definition 3.1.
A BRK-algebra is a nonempty setA with a constant 0 and a binary operation * satisfying axioms:(B8)
x*0=x,(B13)
(x*y)*x=0*yforanyx,y∈A.For brevity, we also callA a BRK-algebra. In A, we can define a binary relation “≤” by x≤y if and only if x*y=0.Example 3.2.
LetA:=ℝ-{-n},0≠n∈ℤ+ where ℝ is the set of all real numbers and ℤ+ is the set of all positive integers. If we define a binary operation * on A by
(3.1)x*y=n(x-y)n+y,
then (A,*,0) is an BRK-algebra.Example 3.3.
LetA={0,1,2} in which * is defined by
(3.2)*012002211002200.
Then (A,*,0) is a BRK-algebra.We know that every BCK-algebra is a BCI-algebra and every BCI-algebra is a BCH-algebra and every BCH-algebra is a Q-algebra. We can observe that every Q-algebra is a BRK-algebra but converse needs not be true.Example 3.4.
LetA={0,1,2,3} in which * is defined by
(3.3)*012300101110102210133230.
Then (A,*,0) is a BRK-algebra, which is not a BCK/BCI/BCH/Q-algebra.We know that every QS-algebra is a BM-algebra and we can observe that every BM-algebra is a BRK-algebra but converses need not be true.Example 3.5.
LetA={0,1,2,3} in which * is defined by
(3.4)*012300220110022200233110.
Then (A,*,0) is a BRK-algebra, which is not a QS/BM-algebra.It is easy to see that B/BG/BF/BH-algebra and BRK-algebras are different notions. For example, Example3.3 is a BRK-algebra which is not a BH-algebra and Example 3.4 is an BRK-algebra which is not B/BG/BF-algebra. Consider the following example. Let A={0,1,2,3,4,5} be a set with the following table:(3.5)*012345002134511024532210534334502144531025534210.
Then (A,*,0) is a B/BF/BG/BH-algebra which is not an BRK-algebra.We observe that the two axioms (B8) and (B13) are independent. Let A={0,1,2} be a set with the following left table:(3.6)*012001211122212*012001011012010.
Then the axiom (B8) holds but not (B13), since (1*2)*1=2*1=1≠2=0*2. Similarly, the set A={0,1,2} with the above right table satisfies the axiom (B13) but not (B8), since 2*0=0≠2.Proposition 3.6.
If(A,*,0) is a  BRK-algebra, then, for any  x,y∈A, the following conditions hold: (1)
x*x=0,(2)
x*y=0⇒0*x=0*y.Proof.
Let(A,*,0) be a BRK-algebra and x,y∈A. Then(1)
x*x=(x*0)*x=0*0=0 (by B8 and B13),(2)
x*y=0⇒(x*y)*x=0*x⇒0*y=0*x.Proposition 3.7.
Every BRK-algebraA satisfies the following property:
(3.7)0*(x*y)=(0*x)*(0*y),
for any x,y∈A.Proof.
Letx,y∈A. Then
(3.8)0*(x*y)=((0*y)*(x*y))*(0*y)(byB13)=[((x*y)*x)*(x*y)]*(0*y)(byB13)=(0*x)*(0*y).Theorem 3.8.
Every  BRK-algebraA satisfying  x*(x*y)=x*y for all  x,y∈A is a trivial algebra.Proof.
Puttingx=y in the equation x*(x*y)=x*y, we obtain x*0=0⇒x=0. Hence, A is a trivial algebra.Theorem 3.9.
Every  BRK-algebraA satisfying  (x*y)*(x*z)=z*y for all  x,y,z∈A is a  BCI-algebra.Proof.
Let(A,*,0) be a BRK-algebra and (x*y)*(x*z)=z*y for all x,y,z∈A. Then(1)
(x*y)*(x*z)*(z*y)=(z*y)*(z*y)=0,(2)
(x*(x*y))*y=((x*0)*(x*y))*y=(y*0)*y=y*y=0,(3)
x*x=0,(4)
Letx*y=0=y*x. Then x=x*0=x*(x*y)=(x*0)*(x*y)=y*0=y,(5)
x*0=0⇒x=0.Theorem 3.10.
Every 0-commutative B-algebra is a BRK-algebra.Proof.
LetA be a 0-commutative B-algebra. Then x*(x*y)=y for all x,y∈A. Hence, (x*y)*x=x*(x*(0*y))=0*y.The following theorem can be proved easily.Theorem 3.11.
Let(A,*,0) be a BRK-algebra. Then, for any x,y∈A, the following conditions hold.(1)
If(x*y)*(0*(0*y))=(x*y)*y, then 0*(0*(0*y))=0*y.(2)
If(x*y)*(0*y)=(x*y)*y, then 0*(0*y)=0*y.(3)
Ifx*(y*x)=x*(0*(x*y)), then 0*(y*x)=0*(0*(x*y)).


## 4.G-Part of BRK-Algebras

In this section, we defineG-part, p-radical and medial of a BRK-algebra. We give a necessary and sufficient condition for a BRK-algebra to become a medial BRK-algebra and investigate the properties of G-part in BRK-algebras.Definition 4.1.
A nonempty subsetI of a BRK-algebra A is called a subalgebra of A if x*y∈I whenever x,y∈I.Definition 4.2.
A nonempty subsetI of a BRK-algebra A is called an ideal of A if for any x,y∈A:(i)
0∈I,(ii)
x*y∈I and y∈I imply x∈I.Obviously,{0} and A are ideals of A. We call {0} and A the zero ideal and the trivial ideal of A, respectively. An ideal I is said to be proper if I≠A.Definition 4.3.
An idealI of a BRK-algebra A is called a closed ideal of Aif0*x∈I for all x∈I.Example 4.4.
LetA={0,1,2} in which * is defined by
(4.1)*012002211002200.
Then (A,*,0) is a BRK-algebra and the set I={0,2} is a subalgebra, an ideal, and a closed ideal of A.Definition 4.5.
LetA be a BRK-algebra.For any subset SofA, we define
(4.2)G(S)={x∈S∣0*x=x}.
In particular, if S=A, then we say that G(A) is the G-part of a BRK-algebra.For any BRK-algebraA, the set:(4.3)B(A)={x∈A∣0*x=0}
is called a p-radical of A. Clearly, B(A) is a subalgebra and an ideal of A.A BRK-algebraA is said to be p-semisimple if B(A)={0}.The following property is obvious:(4.4)G(A)∩B(A)={0}.Lemma 4.6.
If(A,*,0) is a  BRK-algebra and  a*b=a*c for  a,b,c∈A, then  0*b=0*c.Proof.
Let(A,*,0) be a BRK-algebra and a,b,c∈A. Then by (B13), a*b=a*c⇒(a*b)*a=(a*c)*a⇒0*b=0*c.Theorem 4.7.
Let(A,*,0) be a  BRK-algebra. Then a left cancellation law holds in  G(A).Proof.
Leta,b,c∈G(A) with a*b=a*c. Then, by Lemma 4.6,  0*b=0*c. Since b,c∈G(A), we obtain b=c.Proposition 4.8.
Let(A,*,0) be a  BRK-algebra. If  x∈G(A), then  0*x∈G(A).Proof.
Letx∈G(A). Then 0*x=x and hence 0*(0*x)=0*x. Therefore, 0*x∈G(A).Converse of the above proposition needs not be true. From Example4.4, we can see that 0*1=2∈{0,2}=G(A) but 1∉G(A).Theorem 4.9.
Ifx,y∈G(A), then  x*y∈G(A).Proof.
Letx,y∈G(A). Then 0*x=x and 0*y=y. Hence, 0*(x*y)=(0*x)*(0*y)=x*y. Therefore, x*y∈G(A).Proposition 4.10.
If(A,*,0) is a  BRK-algebra and  x,y∈A, then
(4.5)y∈B(A)⟺(x*y)*x=0.Proof.
Let(A,*,0) be a BRK-algebra and x,y∈A. Then, by (B13),  y∈B(A)⇔0*y=0⇔(x*y)*x=0.Theorem 4.11.
IfS is a subalgebra of a  BRK- algebra (A,*,0), then  G(A)∩S=G(S).Proof.
Clearly,G(A)∩S⊆G(S). If x∈G(S), then 0*x=x and x∈S⊆A. Hence, x∈G(A). Therefore, x∈G(A)∩S. Thus, G(A)∩S=G(S).Theorem 4.12.
Let(A,*,0) be a  BRK- algebra. If  G(A)=A, then  A is  p-semisimple.Proof.
Assume thatG(A)=A. Then {0}=G(A)∩B(A)=A∩B(A)=B(A). Hence, A is p-semisimple.Theorem 4.13.
Every closed ideal of a  BRK- algebra is a subalgebra.Proof.
LetI be a closed ideal of a BRK-algebra (A,*,0) and x,y∈I. Then 0*y∈I. By (B13), (x*y)*x=0*y∈I. Since I is an ideal and x∈I, we have x*y∈I. So I is a subalgebra of A.Note that the converse of the above theorem is not true. In Example3.4, the set {0,1,2} is a subalgebra but not a closed ideal.Theorem 4.14.
LetI be a subset of a BRK-algebra A. Then I is a closed ideal of A if and only if it satisfies (i) 0∈I (ii) x*z∈I,y*z∈I and z∈I imply x*y∈I, for all x,y,z∈A.Proof.
LetI be a closed ideal of A. Clearly, 0∈I. Assume that x*z,y*z,z∈I. Since I is an ideal, we have x,y∈I which implies that x*y∈I because I is a closed ideal and hence a subalgebra of A. Conversely, assume that I satisfies (i) and (ii). Let x*y,y∈I. Since 0*0,y*0,0∈I, by (ii) we have 0*y∈I. From (ii), again it follows that x=x*0∈I so that I is an ideal of A. Now suppose that x∈I. Since 0*0,x*0,0∈I, we obtain 0*x∈I by (ii). This completes the proof.Definition 4.15.
A BRK-algebra(A,*,0) is said to be positive implicative if
(4.6)((x*y)*y)*(0*y)=x*y
for all x,y∈A.The BRK-algebra in Example3.3 is positive implicative.Definition 4.16.
Let(A,*,0) be a BRK-algebra.For a fixed a∈A. The map Ra:A→A given by Ra(y)=y*a for all y∈A is called right translation of A. Similarly the map La:A→A given by La(y)=a*y for all y∈A is called a left translation of A.Definition 4.17.
Let(A,*,0) be a BRK-algebra. For a fixed a∈A. The map Ta:A→A given by Ta(y)=(y*a)*(0*a) for all y∈A is called a weak right translation of A. Similarly, the map Ma:A→A given by Ma(y)=(a*y)*(0*y) for all y∈A is called a weak left translation of A.Theorem 4.18.
A  BRK-algebra(A,*,0) is positive implicative if and only if  Rz=Tz∘Rz for all  z∈A.Proof.
LetA be a BRK-algebra and Rz=Tz∘Rz for z∈A. Then
(4.7)y*z=Rz(y)=(Tz∘Rz)(y)=Tz(Rz(y))=Tz(y*z)=((y*z)*z)*(0*z),∀y,z∈A.
Hence, A is positive implicative BRK-algebra. Conversely, assume that A is positive implicative BRK-algebra. Let x,y∈A. Then
(4.8)Rx(y)=y*x=((y*x)*x)*(0*x)=(Rx(y)*x)*(0*x)=Tx(Rx(y))=(Tx∘Rx)(y).
Hence, Rx=Tx∘Rx.Definition 4.19.
A BRK-algebra(A,*,0) satisfying
(4.9)(x*y)*(z*u)=(x*z)*(y*u)
for any x,y,z and u∈A, is called a medial BRK-algebra.Example 4.20.
LetA:=ℝ-{-n},0≠n∈ℤ+ where ℝ is the set of all real numbers and ℤ+ is the set of all positive integers. If we define a binary operation * on A by
(4.10)x*y=n(x-y)n+y,
then (A,*,0) is a medial BRK-algebra.Theorem 4.21.
IfA is a medial BRK-algebra, then, for any x,y,z∈A, the following hold:(i)
x*(y*z)=(x*y)*(0*z),(ii)
(x*y)*z=(x*z)*y.Proof.
LetA be a medial BRK-algebra and x,y,z∈A. Then(i)
(x*y)*(0*z)=(x*0)*(y*z)=x*(y*z),(ii)
(x*y)*z=(x*y)*(z*0)=(x*z)*(y*0)=(x*z)*y.By the above theorem, the following corollary follows.Corollary 4.22.
Every medial BRK-algebra is a Q-algebra.Theorem 4.23.
LetA be a medial BRK-algebra. Then the right cancellation law holds in G(A).Proof.
Leta,b,x∈G(A) with a*x=b*x. Then, for any y∈G(A),x*y=(0*x)*y=(0*y)*x=y*x. Therefore,
(4.11)a=0*a=(x*a)*x=(a*x)*x=(b*x)*x=(x*b)*x=0*b=b.Now, we give a necessary and sufficient condition for a BRK-algebra to become a medial BRK-algebra.Theorem 4.24.
A BRK-algebraA is medial if and only if it satisfies:(i)
x*y=0*(y*x)forallx,y∈A,(ii)
(x*y)*z=(x*z)*yforallx,y,z∈A.Proof.
Suppose(A,*,0) is medial and x,y,z∈A. Then(i)
0*(y*x)=(x*x)*(y*x)=(x*y)*(x*x)=(x*y)*0=x*y,(ii)
(x*y)*z=(x*y)*(z*0)=(x*z)*(y*0)=(x*z)*y.Conversely, assume that the conditions hold. Then
(4.12)(x*y)*(z*u)=0*((z*u)*(x*y))(by(i))=0*((z*(x*y))*u)(by(ii))=(0*(z*(x*y)))*(0*u)(byProposition3.7)=((x*y)*z)*(0*u)(by(i))=((x*z)*y)*(0*u)(by(ii))=((x*z)*(0*u))*y(by(ii))=(0*((0*u)*(x*z)))*y(by(i))=(0*((z*x)*u))*y(by(ii)&(i))=(u*(z*x))*y(by(i))=(u*y)*(z*x)(by(ii))=0*((z*x)*(u*y))(by(i))=(x*z)*(y*u)(byProposition3.7and(i))
Therefore, A is medial.Corollary 4.25.
A BRK-algebraA is medial if and only if it is a medial QS-algebra.The following theorem can be proved easily.Theorem 4.26.
An algebra(A,*,0) of type  (2,0) is a medial  BRK-algebra if and only if it satisfies: (i)
x*(y*z)=z*(y*x),(ii)
x*0=x,(iii)
x*x=0.Corollary 4.27.
IfA is a medial BRK-algebra, then x*(x*y)=y for all x,y∈A.Corollary 4.28.
The class of all of medial BRK-algebras forms a variety, writtenν(MR).Proposition 4.29.
A varietyν is congruence-permutable if and only if there is a term p(x,y,z) such that
(4.13)ν⊨p(x,x,y)≈y,ν⊨p(x,y,y)≈x.Corollary 4.30.
The varietyν(MR) is congruence permutable.Proof.
Letp(x,y,z)=x*(y*z). Then by Corollary 4.25 and (B8), we have p(x,x,y)=y and p(x,y,y)=x, and so the variety ν(MR) is congruence permutable.The following example shows that a BRK-algebra may not satisfy the associative law.Example 4.31.
LetA={0,1,2} be a set with the following table:
(4.14)*012002211002200.
Then (A,*,0) is a BRK-algebra, but associativity does not hold since (1*2)*1=0*1=2≠1=1*0=1*(2*1).Theorem 4.32.
IfA is an associative BRK-algebra, then, for any x∈B(A),  x=0.Proof.
Letx∈B(A). Then 0=0*x=(x*x)*x=x*(x*x)=x*0=x.Theorem 4.33.
IfA is an associative  BRK-algebra, then  G(A)=A.Proof.
LetA be an associative BRK-algebra. Clearly, G(A)⊆A. Let x∈A. Then 0*x=(x*x)*x=x*(x*x)=x*0=x. Hence, x∈G(A). Therefore, G(A)=A.Now, we prove that every associative BRK-algebra is a group.Theorem 4.34.
Every BRK-algebra(A,*,0) satisfying the associative law is a group under the operation “*”.Proof.
Puttingx=y=z in the associative law (x*y)*z=x*(y*z) and using (B3) and (B8), we obtain 0*x=x*0=x. This means that 0 is the zero element of A. By (B3), every element x of A has as its inverse the element x itself. Therefore, (A,*) is a group.


## 5. Conclusion and Future Research

In this paper, we have introduced the concept of BRK-algebra and studied their properties. In addition, we have definedG-part, p-radical, and medial of BRK-algebra and proved that the variety of medial algebras is congruence permutable. Finally, we proved that every associative BRK-algebra is a group.In our future work, we introduce the concept of fuzzy BRK-algebra, interval-valued fuzzy BRK-algebra, intuitionistic fuzzy structure of BRK-algebra, intuitionistic fuzzy ideals of BRK-algebra, and intuitionistic (T,S)-normed fuzzy subalgebras of BRK-algebras, intuitionisticL-fuzzy ideals of BRK-algebra.I hope this work would serve as a foundation for further studies on the structure of BRK-algebras.



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