# Conditions On DR – Group

DOI : 10.17577/IJERTV1IS7232

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#### 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.

1. Preliminaries Definition 1.1 [4]

A non empty set G is called an group if and only if

1. (G, +) is a group

2. (G, ) is a lattice

3. 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

1. (L, .) is an group.

2. 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

1. (B, ) is a lattice

2. B has a least element

3. 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

1. (A, +) is an abelian group.

2. (A, ) is a lattice.

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

4. 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

1. (A, +) is an abelian group.

2. (A, , ,) is a lattice.

3. a + (b c) = (a + b) (a + c),

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

4. 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,

2. 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.

1. 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

1. (A, ) is a lattice

2. A has a least element 0.

3. 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 (x-y) = 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

1. a b 0

2. a b = 0 a = b

3. a b = b a

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

Proof :

Let a, b in A be arbibrary.

1. 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.

2. 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)

3. 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) :

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.

2. a * (b * c) = (a * b) * c, for all a, b, c in A:

This follows by assumption.

3. 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

4. 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

5. 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.

6. 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.

7. 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

8. 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)

Pre-multiply 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.

9. 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:

[1]. E. A. Nordhaus and Leo Lapidus, Brouwerian Geometry, Canad. J. Math.,6 (1954).

[2]. G. Birkhoff, Lattice Theory, American Mathematical Society, Colloguium Publications, Volume 25, Providence R. I., 3rd Edition, 3rd Printing, (1979).

[3]. K.L.N. Swamy, Dually Residuated Lattice Ordered Semigroup, Math. Ann. 159, 105 114 (1965).

[4]. M. Jeyalakshmi and R. Natarajan, DR group, Acta Cienca Indica, Vol. XXIX. M, No.4, 823 830 (2003).

[5]. M. Ward and R. P. Dilworth, Residuated Lattice, Trans, Am. Math., Soc. 45, (1939).