TABLE OF CONTENTS iii
PREFACE vii
FULL PAPERS 1
Are Unschooled Indigenous People Schooled in Mathematics? . . . 2
Development of Performance Task Activity in Teaching Systems of Linear Equation . . . 11
Weighted Improved Hardy-Sobolev Inequality on a Ball Domain . . . 19
Representations of the Positively and Negatively Subscripted Terms of the AB-Generalized Fibonacci Sequence . . . 29
Domination in the Kr - gluing of Complete Graphs and Join of Graphs . . . 38
Global O↵ensive Alliances in Some Special Classes of Graphs . . . 43
The Use of Computation Journals in Reducing Low Achieving Students Errors in Algebraic Rational Expressions . . . 49
Students Level of Understanding of the Basic Concept of Functions and Relations: A Conceptual Analysis . . . 54
Roman Domination Number of the Join and Corona of Graphs . . . 66
The Relationship of BF-algebras and Groups . . . 70
Students Misconceptions in College Algebra . . . 76
PLENARY TALKS 81 Plenary Talk 1 . . . 82 Dr. Fe Annabel N. Yebron
Department of Mathematics, Central Mindanao University, Musuan, Bukidnon
“Are Unschooled Indigenous People Schooled in Mathematics?”
The Relationship of BF-algebras and Groups1 Joemar C. Endama and Jocelyn P. Vilelab
Department of Mathematics, College of Science and Mathematics, MSU-Iligan Institute of Technology, Tibanga, Iligan City, Philippines
e-mail: aayungon [email protected]; b[email protected]
Abstract: In this paper, some relationships of groups and BF-algebras are presented. As a result, the concept of commutativity of BF-algebras is introduced and characterized.
1 Introduction
The concept of B-algebras was introduced by J. Neggers and H.S. Kim [3]. They defined a B-algebra as an algebra (A; ⇤, 0) of type (2, 0) (i.e., a nonempty setA with a binary operation
⇤and a constant 0) satisfying the following axioms: (B1) x⇤x = 0, (B2)x⇤ 0 =x, and (B) (x
⇤y)⇤z =x⇤ (z ⇤(0⇤ y)). In [1], Y.B. Jun, E.H. Roh, and H.S. Kim introduced BH-algebras.
An algebra (A; ⇤, 0) of type (2, 0) is a BH-algebra if it obeys (B1), (B2), and (BH) x ⇤ y = 0 and y ⇤ x = 0 imply x = y. In [2], C.B. Kim and H.S. Kim introduced BG-algebras. A BG-algebra is an algebra (A; ⇤, 0) of type (2, 0) satisfying (B1), (B2), and (BG) x = (x ⇤ y) ⇤ (0⇤y). In 2007, the concept of BF-algebras was introduced by A. Walendziak [4] together with BF1-algebras and BF2-algebras. He defined a BF-algebra as an algebra (A; ⇤, 0) of type (2, 0) satisfying the following axioms: (B1), (B2), and (BF) 0 ⇤ (x ⇤ y) = y ⇤ x. A BF1-algebra is a BF-algebra satisfying (BG) and a BF2-algebra is a BF-algebra satisying (BH). BH-algebras, BG- algebras, and BF-algebras are all generalizations of B-algebras. In this paper, we present some relationships of groups and BF-algebras. Moreover, we introduce and characterize commutative BF-algebras.
2 Some Relationships of Groups and BF-algebras
This section provides conditions to transform all groups into BF-algebras and conditions to transform some BF-algebras into groups.
First, we consider the following example.
Example 2.1 The structure (Z; +, 0) is a group, where Zis the set of integers and the binary operation + is the usual addition. The identity element is 0 and x is the inverse of x 2 Z. Observe that the structure (Z; ⇤, 0) with the binary operation⇤defined byx⇤y =x+yfor all x, y 2 Z is not a BF-algebra since (B1) is not satisfied by (Z; ⇤, 0), that is, x ⇤ x = x + x = 2x 6= 0 if 0 6= x 2 Z. However, (Z; ⇤, 0) with the binary operation ⇤ defined by x ⇤ y =x + ( y) for all x, y 2 Z is a BF-algebra.
This example is generalized in the following theorem.
Theorem 2.2 If (A; •,e)is a group with binary operation •and identity e,then (A; ⇤,e= 0)
with an operation ⇤ defined by x ⇤ y = x • y 1 for all
x, y 2 A is a BF-algebra.
1This research is funded by the Department of Science and Technology-Philippine Council for Advanced Science and Technology Research and Development (DOST-PCASTRD).
Proof: Suppose that (A;•,e) is a group with binary operation•and identitye. Letx,y2A,e=
0, and ⇤ be an operation defined by x ⇤ y = x • y 1.
Then closure property of⇤follows from that of•. Thus,⇤is a binary operation onA. For anyx,y
2 A, x ⇤ x = x • x 1 = e = 0, x ⇤ 0 = x • e 1 = x,
and 0 ⇤ (x ⇤ y) = e • (x • y 1) 1 = (x • y 1) 1 = y • x 1 = y ⇤ x. Therefore, (A; ⇤, e
= 0) satisfies axioms (B1), (B2), and (BF) and so (A; ⇤, e = 0) is a BF-algebra. ⇤ Corollary 2.3 The BF-algebra (A; ⇤, e = 0) in Theorem 2.2satisfies the following:
i. x ⇤ (y ⇤ z) = (x ⇤ (0⇤ z)) ⇤y for all x, y, z 2A, ii. axioms (B), (BG), and (BH).
By Corollary 2.3, the following remark easily follows.
Remark 2.4 Any group can be transformed into a BG-algebra, BH-algebra, BF1-algebra, and BF2-algebra.
Having seen that a group can be made into a BF-algebra, one asks if the converse also holds.
First, we consider the following example.
Example 2.5 Let A = {0,a,b} be a set with a binary operation ⇤ defined by the following table:
⇤ 0 a b
0 0 b a
a a 0 b b b a 0
By routine calculations, (A; ⇤, 0) is a BF-algebra and for all x, y, z 2 A, x⇤(y⇤z) = (x⇤(0⇤z))⇤y. Observe that the structure (A;•, 0) with a binary operation•de- fined byx•y=x⇤yfor allx,y2Ais not a group sinceAhas no identity element. In fact, 0•a
= 0 ⇤ a = b but a • 0 = a ⇤ 0 = a.
This shows that neither 0 nor a can be an identity. Also, b is not an identity since 0 • b
= 0 ⇤b =a but b • 0 =b ⇤ 0 = b. However, (A; •, 0) with a binary operation• defined byx • y = x ⇤ (0 ⇤ y) for all x, y 2 A is a group with 0 as the identity element in A and the inverse of x 2 A is 0 ⇤ x.
This example is generalized in the following theorem.
Theorem 2.6 If (A; ⇤, 0) is a BF-algebra such that for all x, y, z 2 A, x ⇤ (y ⇤ z) = (x ⇤ (0 ⇤ z)) ⇤ y, then the structure (A; •, 0) with operation • defined by x
• y = x ⇤ (0 ⇤ y) for all x,y 2 A is a group called the group induced by the BF-algebra (A; ⇤, 0)and “•”is the group operation induced by ⇤.
Proof: Suppose that (A; ⇤, 0) is a BF-algebra such that for all x, y, z 2 A, x ⇤ (y ⇤ z) = (x ⇤ (0 ⇤ z)) ⇤ y. Let • be an operation on A defined by x • y = x ⇤ (0 ⇤ y). Then closure property of • follows from that of ⇤. Thus, • is a bi- nary operation onA. If x, y 2A, then
x • (y • z) = x • (y ⇤ (0 ⇤ z)) = x ⇤ (0 ⇤ (y ⇤ (0 ⇤ z))) = x ⇤ ((0 ⇤ z) ⇤ y)
and (x • y) • z = (x • y) ⇤ (0 ⇤ z) = (x ⇤ (0 ⇤ y)) ⇤ (0 ⇤ z) = x ⇤ ((0 ⇤ z) ⇤ y).
Hence,•is associative inA. Ifx2A, thenx•0 =x⇤(0⇤0) =x⇤0 =xand 0•x= 0⇤(0⇤x) =
x. Thus, 0 is the identity element in A. Moreover,
x • (0 ⇤ x) = x ⇤ (0 ⇤ (0 ⇤ x)) = x ⇤ x = 0 and (0 ⇤ x) • x = (0 ⇤ x) ⇤ (0 ⇤ x) = 0.
Therefore, 0 ⇤ x is the inverse of x and so (A; •, 0) is a group. ⇤ Remark 2.7 If (A; •, 0) is the group induced by the BF-algebra (A; ⇤, 0), then x ⇤ (y ⇤ z) = (x ⇤(0 ⇤ z))⇤ y for all x,y,z 2 A.
If the condition x ⇤ (y ⇤ z) = (x ⇤ (0 ⇤ z)) does not hold in the BF-algebra (A; ⇤, 0), then the structure (A; •, 0) is not a group as shown in the following example.
Example 2.8 Let (A; ⇤, 0) be the BF-algebra given in Example 2.4 [4], that is, A = [0, 1)
where ⇤ is defined by x ⇤ y = |x y| for all x, y 2 A.
Then there existx,y,z2Asuch thatx⇤(y⇤z)6= (x⇤(0⇤z))⇤y. In particular, 2⇤(4⇤3) = 1
6
= 3 = (2 ⇤ (0 ⇤ 3)) ⇤ 4. Observe that the structure
(A; •, 0) with the binary operation defined by x • y = x ⇤ y is not a group since • is not associative, in fact, (2•3)•4 = 3 but 2 •(3•4) = 1. Also, (A;•, 0) with the binary operation
• defined by x •y = x ⇤ (0⇤ y) is not a group since • is not associative.
In [3], the condition x ⇤ (y ⇤ z) = (x ⇤ (0 ⇤ z)) ⇤ y is satisfied by a B-algebra and adding Corollary 2.3(ii), we obtain the following remark.
Remark 2.9 Any B-algebra can be transformed into a group and vice versa.
3 Characterization of a Commutative BF-algebra
In group theory, a group G is said to be commutative if x • y = y • x for all x, y 2 G, otherwise it is called noncommutative. In the discussions presented in the previous section, a BF-algebra can be transformed into a group by definingx •y =x⇤ (0⇤y).
Thus, we obtain the following definition.
Definition 3.1 A BF-algebraA = (A; ⇤, 0) is said to becommutative if the group induced by A is commutative, that is,x ⇤(0⇤ y) =y ⇤ (0⇤x) for any x,y 2A. If A is not commutative, then we say that A is noncommutative.
Example 3.2 LetA = {0, 1, 2}be a set with the following table:
⇤ 0 a b
0 0 b a
a a 0 b b b a 0
Then by routine calculations, (A; ⇤, 0) is a commutative BF-algebra.
Example 3.3 LetA = (A; ⇤, 0), whereA ={0,a, b, c,d, e}and ⇤ is defined by the following table:
⇤ 0 a b c d e
0 0 b a c d e
a a 0 b d e c b b a 0 e c d c c d e 0 b a d d e c a 0 b e e c d b a 0
Then by routine calculations, A is a BF-algebra. A is noncommutative sinceb ⇤(0 ⇤ c) = e 6= d =c ⇤ (0⇤ b).
In [4], if I is a normal ideal of a BF-algebra A = (A; ⇤, 0), then A/I = (A/I; ⇤0, 0/I) where A/I = {x/I : x 2 A} and ⇤0 is defined by x/I ⇤0 y/I = (x ⇤ y)/I. For x 2 A, x/I is the congruence class containing x, that is, x/I = {y 2 A : x ⇠I y}, where x ⇠I y if and only if x ⇤ y 2 I for any x, y 2 A. The algebra A/I is called the quotient BF-algebra of A modulo I.
From now on, A denotes the BF-algebra (A; ⇤, 0).
Theorem 3.4 Let I be a normal ideal of A. If A is commutative,then A/I is commutative.
Proof: LetA be commutative BF-algebra and let I be a normal ideal of A. If x/I, y/I 2 A/I, then
x/I ⇤0 (0/I⇤0y/I) = x/I ⇤0((0⇤y)/I)
= (x⇤(0⇤y))/I
= (y⇤(0⇤x))/I
= y/I⇤0((0⇤x)/I)
= y/I⇤0(0/I⇤0x/I).
Therefore,A/I is commutative. ⇤
The converse of Theorem 3.4 need not be true as shown in the following example.
Example 3.5 Let A be the BF-algebra in Example 3.3. By routine calculations, I = {0, a, b} is a normal ideal of A and A/I = {0/I, c/I}, where 0/I = {0, a, b} and c/I = {c, d, e}. Moreover, A/I is commutative.
In [4], if A = (A; ⇤, 0) is a BF-algebra, then (P1) x = 0 ⇤ (0 ⇤ x) and (P2) 0 ⇤ x = 0 ⇤ y implies x= y hold for any x, y 2 A.
Theorem 3.6 If A has only two elements, then A is commutative.
Proof: Let 0 and xbe two distinct elements in A. If 0⇤ x= 0, then by (P1) and (B1), x= 0 ⇤ (0⇤ x) = 0 ⇤0 = 0, a contradiction. Thus, 0 ⇤ x= x. Therefore, by (B1), (B2), and (P1),x ⇤ (0⇤ 0) = x ⇤ 0 = x = 0 ⇤ (0⇤ x) and soA is commutative. ⇤ Theorem 3.7 If 0⇤ x = xfor all x 2 A, then A is commutative.
Proof: Let 0⇤ x =x for all x 2A. If x,y 2 A, then by assumption and (BF), x ⇤ (0⇤ y) = x
⇤ y = 0 ⇤ (x⇤ y) = y ⇤ x= y ⇤ (0⇤ x). Therefore, A is commutative. ⇤ As a consequence of the proof of Theorem 3.7, the following remarks immediately follow.
Remark 3.8 1) If x ⇤ y =y ⇤x for all x,y 2A,then A is commutative.
2) Let x, y 2A. Then 0 ⇤ x= x if and only if x ⇤y = y ⇤ x.
The converse of Theorem 3.7 and Remark 3.8(1) need not be true. To see this, if A be the BF-algebra in Example 3.2, then 0 ⇤ a = b 6= a =a ⇤ 0.
The following theorem characterizes a commutative BF-algebra.
Theorem 3.9 The following statements are equivalent: i. A is commutative;
ii. The map A ! A given by x7! 0 ⇤ xis an automorphism;
iii. 0 ⇤(x ⇤ (0⇤ y)) = (0 ⇤ x)⇤ (0 ⇤(0 ⇤ y))for all x, y 2 A;
iv. x ⇤ y = (0 ⇤y)⇤ (0 ⇤x) for all x, y 2 A.
Proof:
(i , ii) Suppose thatAis commutative and the map': A!Ais defined by'(x) = 0⇤xfor allx,
y 2 A. Let x, y 2 A. If x = y, then
'(x) = 0 ⇤ x = 0 ⇤ y = '(y) and so ' is well-defined. If '(x) = '(y), then 0 ⇤ x = '(x) = '(y) = 0 ⇤ y. By (P2), x = y and so ' is one-to-one. Moreover, by (BF), (P1), and commutativity of A, we obtain
'(x⇤y) = 0⇤(x⇤y)
= y⇤x
= y⇤(0⇤(0⇤x))
= (0⇤x)⇤(0⇤y)
= '(x)⇤'(y).
This shows that ' is a homomorphism. To show that ' is onto, let x 2 A. Then 0 ⇤ x 2 A and x = 0 ⇤ (0⇤ x) = '(0 ⇤ x). Therefore,' is onto and so ' is an isomorphism from A ontoA, that is, ' is an automorphism. Conversely, let ' : A! A defined by'(x) = 0
⇤ x be an automorphism. Ifx, y 2 A, then by (P1) and (BF), we obtain '(x⇤(0⇤y)) = '(x)⇤'(0⇤y)
= (0⇤x)⇤(0⇤(0⇤y))
= (0⇤x)⇤y
= 0⇤(y⇤(0⇤x))
= '(y⇤(0⇤x)).
Since ' is one-to-one,x ⇤ (0 ⇤y) = y ⇤ (0⇤ x) and soA is commutative.
(i , iii) Let A be commutative. If x, y 2 A, then by commutativity of A, (BF), and (P1), we obtain
0⇤(x⇤(0⇤y)) = 0⇤(y⇤(0⇤x))
= (0⇤x)⇤y
= (0⇤x)⇤(0⇤(0⇤y)).
Conversely, let 0 ⇤ (x ⇤ (0 ⇤ y)) = (0 ⇤ x) ⇤ (0 ⇤ (0 ⇤ y)) for all x, y 2 A.
Ifx, y 2 A, then by (P1), assumption, and (BF), we obtain x⇤(0⇤y) = 0⇤(0⇤(x⇤(0⇤y)))
= 0⇤((0⇤x)⇤(0⇤(0⇤y)))
= (0⇤(0⇤y))⇤(0⇤x)
= y⇤(0⇤x).
Therefore,A is commutative.
(i, iv) Let A be commutative. Ifx, y 2A, then by (P1) and commutativity, x⇤ y= x⇤ (0⇤(0
⇤ y)) = (0 ⇤ y) ⇤ (0 ⇤ x). Conversely, let x ⇤ y = (0 ⇤ y) ⇤ (0 ⇤ x) for all x, y 2 A. If x, y 2 A, then by (P1), x ⇤ (0⇤ y) = (0 ⇤ (0 ⇤ y))⇤ (0 ⇤ x) = y ⇤ (0 ⇤ x). Therefore, A is
commutative. ⇤
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