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 Total Downloads : 214
 Authors : Dr. M. Jeyalakshmi
 Paper ID : IJERTV1IS7232
 Volume & Issue : Volume 01, Issue 07 (September 2012)
 Published (First Online): 25092012
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Conditions On DR – Group
Dr. M. Jeyalakshmi Assistant Professor in Mathematics
Alagappa Govt. Arts College, Karaikudi 630 003
Abstract: In this paper we introduce that an – group, Brouwerian Algebra and Boolean ring can be realized from a DR – group by some specialization.
Key words: – group, Brouwerian Algebra, Boolean ring, Residuated lattice, DR – group.

Preliminaries Definition 1.1 [4]
A non empty set G is called an group if and only if

(G, +) is a group

(G, ) is a lattice

If x y, then a + x + b a + y + b, for all a, b, x, y in G.
(or) (a + x + b) (a + y + b) = (a + x y + b)
(a + x + b) (a + y + b) = (a + x y + b), for all a, b, x, y in G.
Definition 1.2 [4]
An group G is called a commutative group if a + b = b + a, for all a, b in G.
Definition 1.3 [4], [5]
A lattice L is called a residuated lattice if

(L, .) is an group.

Given a, b in L, there exist the largest x, y such that bx a and yb a.
Definition 1.4 [1], [4]
A non empty set B is called a Brouwerian Algebra if and only if

(B, ) is a lattice

B has a least element

To each a, b in B, there is a least x = a b in B such that b x a

Definition 1.5 [4]
A ring (R, +, . ) is called a Boolean ring if and only if a . a = a, for all a in R.
Definition 1.6 [4]
A system A = (A, +, ) is called dually residuated lattice ordered group (simply DR group) if and only if

(A, +) is an abelian group.

(A, ) is a lattice.

b c a + b a + c, for all a, b, c in A

Given a, b in A, there exist a least element x = a – b in A such that b + x a.
Definition 1.7 [4]
A system A = (A, +, , ) is called a DR group if and only if

(A, +) is an abelian group.

(A, , ,) is a lattice.

a + (b c) = (a + b) (a + c),
a + (b c) = (a + b) (a + c), for all a, b, c in A.

x + (y x) y,
x y (x z) y,
(x + y) y x, for all x, y, z in A.
Remark [4]
Two definitions for DR group are equivalent.
Examples 1.1 [4]
Commutative – group, Brouwerian Algebra and Boolean ring are DR – groups.
Properties of a DR – group
Now we see the properties of a DR – group which is already established in [4] .
1. [(a – b) 0] + b = a b,



a b a c b c and c b c a,
3. (a b) c = (a c) (b c),
4. a – (b c) = (a b) (a c),
5. a – (b c) = (a b) (a c),
6. (b c) a = (b – a) (c – a), 7. a b (a – b) + b = a,
8. a b + a b = a + b, 9. (a – b) 0 + a b= a,
10. a b a b = (a – b) (b – a),
11. a (b – c) (a – b) + c and (a + b) c (a – c) + b, for all a, b, c in A.
Theorem 1.2 [4]
The direct product of the DR – group is also a DR – group.
Theorem 1.3[4]
Any DR group is a distributive lattice.

Realization from a DR – group Theorem : 2.1
If A is a DR group and a + b = a b, for all a, b in A, then A is a Brouwerian Algebra.
Proof :
Given A is a DR group and a + b = a b, for all a, b in A. To prove A is a Brouwerian Algebra.
By given, we have

(A, ) is a lattice

A has a least element 0.

To each a, b in A, there exist a least element x in A such that b x = b + x a.
Hence A is a Brouwerian Algebra.
Theorem : 2.2
If A is a DR group and (A, , ) is a Brouwerian Albegra, then a + b = a b, for all a, b in A.
Proof :
Given A is a DR group and a Brouwerian Algebra. To prove a + b = a b, for all a, b in A.
Let a, b in A be arbitrary.
there exist a least element x = a b in A such that b x a.
We have y (xy) = y x in any Brouwerian Algebra (1)
a (a – a) = a a
a a = a
0 is the least element.
By property 8, we have
a + b = a b + a b
a b + 0 = a b
a + b a b (2)
By property 8 a + b a b = a b a a b
a b = (a b) [( a + b) (a b)]
= (a b) (a + b), by (1)
= a + b, by (2)
Proposition : 2.1
If A is a DR group, then

a b 0

a b = 0 a = b

a b = b a

(a b) (a b) = a b, for all a, b in A.


Proof :
Let a, b in A be arbibrary.

By property 3.10, we have
a b a b = (a – b) (b – a)
= a b
a b = a b a b (1) we have a b a b, for all a, b in A.
a b a b 0 (2)
Using (2) in (1), we get a b 0.

Assume that a b = 0 To prove a = b
Now a b = 0 a b a b = 0, by (1)
a b = a b
a = b. Conversely, assume that a = b To prove that a b = 0
Now, a = b a b = a b
a b a b = 0
a b = 0, by (1)

a b = (a – b) (b – a)
= (b – a) (a – b)
= b a, for all a, b in A.
(iv) (a b) (a b) = [(a b) (a b)] [(a b) (a b)]
= [(a – b) (b – a)] [(b – a) (a – b)], by property 10 [4]
= (a – b) (b – a)
= a b, for all a, b in A.
Theorem 2.3
If the symmetric difference is associative in a DR group A, then (A, *, ) is a Boolean ring and further
a + b = a b = a * b * (a b)
a – b = a * (a b), for all a, b in A.
Proof :
Given that the symmetric difference is associative in a DR group A. To prove (1) (A, *, ) is a Boolean ring.
(2) a + b = a b = a * b * (a b),
a b = a * (a b), for all a, b in A.
For (1) :

For all a, b in A a * b in A: Let a, b in A be arbitrary. Then a * b = (a – b) (b – a)
Now a, b in A a b, b a in A
(a b) (b a) in A
a * b in A.

a * (b * c) = (a * b) * c, for all a, b, c in A:
This follows by assumption.

There exist an element 0 in A such that a * 0 = 0 * a = a, for all a in A:
Let a in A be arbitrary.
Since A is a DR group we have 0 in A.
By associativity of *, we have
a * (a* 0) = (a * a) * 0
= 0 * 0
= 0
Therefore a * 0 = a, by proposition 2.1 (a * 0 ) * a = (0 * a) * a
= 0 * (a * a)
= 0 * 0 = 0
Therefore 0 * a = a, by proposition 2.1

To each a in A, there exist an element a in A such that a * a = 0:
Let a in A be arbitrary.
Then a * a = (a a) (a a) = 0
Thus a * a = 0

a * b = b * a, for all a, b in A:
Let a, b in A be arbitrary.
Then a * b = b * a, for all a, b in A, by proposition 2.1 Therefore (A, *) is an abelian group.

For all a, b in A a b in A:
Let a, b in A be arbitrary.
Then a b in A, since A is a DR group Thus a, b in A a b in A.

a (b c) = (a b) c, for all a, b, c in A :
Let a, b, c in A be arbitrary.
Then a (b c) = (a b) c, for all a, b, c in A, since A is a DR group

a (b * c) = (a b) * (a c):
(b * c) a = (b a) * (c a), for all a, b, c in A :
Let a, b, c in A be arbitrary.
For any x, y we have
(x y) * (x ) = x * y, by proposition 2.1 (1) put x = b a and y = a in (1), we have
[a (b a)] * [a (b a)] = a * (b a)Premultiply a (b a) on both sides, we get
[a (b a)] * ([a (b – a)] * [a (b – a)]) = [a (b – a)] * [a*(b – a)]([a (b – a)] * [a (b – a)]) * [a (b – a)] = [a (b – a)] * [a*(b a)],
by associative law
0 * [a (b – a)] = [a (b – a)] * [a *(b – a)], since a * a = 0
[a (b – a)] = [a (b – a)] * [a *(b – a)], since a * 0 = a (2)
But [a * (b – a)] = [a (b – a)] – [a (b – a)]
= (a b) – [a (b – a)]
= [(a b) – a] [(a b) – (b – a)], by property 4
= (b – a) [(a b) (b – a)] by property 3
= (a b) (b – a)
= a b (ie) a * (b – a) = a b (3)
Using (3) in (2), we get
a (b – a) = [a (b – a)] * (a b)
= (a b) * (a b) = 0, since a * a = 0 Hence a (b – a) = 0 (4)
so that [a (b – c)] [(a b) c]
= (a – [(a b) – c]) ((b – c) – [(a b) – c]) by property 6
= (a – [(a – c) (b – c)]) [(b – c) (a – c)] by property 6 and a b a b = a
= (a – [(a – c) (b – c)]) [b – (c a)]
= (a[(a – c) (b – c)]) [(b – c) (b – a)], by property 5
< a (b – a) (b – c)
= 0 (b – c), by (4)
= 0
so that a (b – c) < (a b) c
But we always have (a b) c < a (b – c),
since (a b) c = (a – c) (b – c) < a (b – c)
Hence a (b – c) = (a b) c (5)
Now, a (b * c) = a [(b – c) (c – b)]
= [a (b – c)] [a (c – b)]
= [(a b) c] [(a c) b], by (5)
= ([(a b) a] [(a b) c]) ([(a c) a] [(a c) b]), since 0 = a (b – a) = (a b) a
0 = a (c – a) = (a c) a
= [(a b) (a c)] [(a c) (a b)]
= (a b) * (a c)
Thus a (b*c) = (a b) * (a c), for all a, b, c in A. Also, (b * c) a = [(b – c) (c – b)] a
= a [(b – c) (c – b)]
= [a (b – c)] [a (c – b)]
= [(a b) c] [(a c) b]
= ([(a b) a] [(a b) c]) ([(a c) – a] [(a c) – b]), since 0 = a (b – a) = (a b) a
0 = a (c – a) = (a c) a
= [(a b) (a c)] [(a c) (a b)]
= [(b a) (c a)] [(c a) (b a)]
= (b a) * (c a)
Thus (b * c) a = (b a) * (c a), for all a, b, c, in A.

a a = a, for all a in A:
Let a in A be arbitrary.
Then a a = a, for all a in A, since A is a DR group.
Thus (A, *, ) is a Boolean Algebra.
For (2):
Let a, b in A be arbitrary.
Then a * (a b) = [a (a b)] [a (a b)]
= a (a b), by absorption and associative laws
= (a – a) (a – b), by property 5
= 0 (a – b) = a b Thus a b = a * (a b), for all a, b in A. Now, a * 0 = a a 0.
so that a + a a, for every a and 0 a 0, for every a. Also, a + a = (a + a) * 0
= (a + a) * (a * a)
= [(a + a) * a] * a
= ([(a + a) a] [(a + a) a]) * a, by property 10
= [(a + a) a] * a, since a * a a.
= ([(a + a) a] a) (a [(a + a) – a])
= a [(a + a) a] Hence (a + a) a = (a – [(a + a) a]) a
= (a – a) [(a + a) a]
= 0 [(a + a) a]
= 0 a 0
a + a a Hence a + a = a
Now, (a + b) (a b) = (a + b) [(a b) + (a b)]
(a [(a b) + (a b)]) + b, by property 11
= [a – (a b)] (a b) + b
= [a – (a b)] + [b – (a b)]
0 + 0 = 0
so that a + b a b (6)
Since a 0 and b 0 and by property 8, we have a + b = a b + a b
(ie) a + b a b (7)
From (6) and (7), we have a + b = a b
Now, a * b * (a b) = a * [b * (b a)]
= a * (b – a), by property 10 and by property 5
= [a (b – a)] [a (b – a)]
= (a b) 0, by (4)
= a b = a + b
Thus a + b = a b = a * b * (a b), for all a, b in A.
References:
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