# Extension of a Drl – Group

DOI : 10.17577/IJERTV2IS1270

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#### Extension of a Drl – Group

Dr. M. Jeyalakshmi

Assistant Professor in Mathematics Alagappa Govt. Arts College, Karaikudi 630 003

Abstract: In this paper we introduce the idea of a DR – group is a direct product of a Brouwerian Algebra and a commutative – group.

Key words: commutative – group, Brouwerian Algebra, DR – group.

1. Preliminaries Definition 1.1 [4]

A non empty set G is called a commutative group if and only if

1. (G, +) is an abelian 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 [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 1.3 [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 1.3 [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 1.4 [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.

Extension of a DR – group Theorem : 1.1

Any DR – group A is the direct product of a Brouwerian Algebra B and a commutative – group G if and only if

i) (a + b) (c + c) (a c) + (b c) and

ii) (ma + nb) (a + b) (ma a) + (nb b),

for all a, b, c in A and any pair of positive integers m, n.

Proof :

Assume that

(a + b) (c + c) (a – c) + (b – c) (1)

(ma + nb) (a + b) (ma a) + (nb b) (2) To prove A = B x G

Let a, b, c in A be arbitrary

(a + b) c (a c) + b, by property 11 [4]

[(a + b) c] c [(a c) + b] c

= (a c) + (b c)

(a + b) (c – c) (a c) + (b c) (3)

From (1) and (3), we get

(a + b) (c + c) = (a c) + (b c) (4) Also, (ma + nb) a (ma a) + nb, by property 11 [4]

[(ma + nb) a] b [(ma a) + nb] b

(ma + nb) (a + b) (ma a) + (nb b) (5)

Form (2) and (5), we have

(ma + nb) (a + b) = (ma a) +( nb b) (6) Let B = {a / a + a a = 0}

G = {a / a+ a a = a}

Claim (1): B is a Brouwerian Algebra

1. Closed with respect to and

Let a in B be arbitrary

(a + a ) – a = 0 0

(a + a ) – a 0

[(a + a ) a ] + a 0 + a

a + a a

Now 0 = ( a + a ) a ( a – a ) + a , by property 11 [4]

0 a

0 + a a + a

a a + a

Thus a + a = a (7)

Hence B is closed under + Let a, b in B be arbitrary

Then (a b) + (a b) = (a + a) (b + b), by (4)

a b in B

= a – b, by (7)

(a b) + b in B

a b in B, by property 7 [4]

Also, (a + b) – (a b) = (a + b) – [( a b) + (a b)], since a b in B

= [ a (a b)] + [ b (a b)], by (4)

= [(a a) (a b)] + [(b a) (b b)], by property 4 [4]

= [ 0 (a b)] + [(b a) 0]

= 0 + 0

(a + b) – (a b) = 0

a + b = a b Let a, b in B a + b , a b in B

(a + b) – (a b) in B

a b in B

2. ( B, , ) is a lattice Idempotent law

Let a in B be arbitrary Then a a = a + a

= a

a a = (a + a) (a a)

= a + a a

= a

Thus a a = a; a a = a, for all a in B.

Commutative law:

Let a, b in B be arbitrary Then a b = a+ b

= b + a

= b a

a b = (a + b) (a b), by property 8 [4]

= ( b + a) (b a)

= b a

Thus a b = b a ; a b = b a, for all a, b, in B

Associative Law :

Let a, b, c in B be arbitrary.

Then a (b c) = a + (b + c)

= (a + b) + c

= (a b) c

a (b c) = [a + (b c)] [a (b c)], by property 8 [4]

= [a + (b c)] [a + (b c)]

= 0

(a b) c = [(a b) + c] [(a b) c]

= [(a b) + c] [(a b) + c]

= 0

Thus a (b c) = (a b) c; a (b c) = (a b) c, for all a, b, c in B

Absorption law :

Let a, b in B be arbitrary

Then a (a b ) = a + (a b)

= a + [(a + b) (a b)]

= a + [(a + b) (a + b)]

= a

a (a b) = [a + (a b)] – [a (a b)]

= [a + (a b)] – [(a a) b]

= a + (a b) ( a b)

= a

Thus a (a b ) = a; a (a b ) = a , for all a, b in B Hence ( B, , ) is a lattice.

3. B has a least element :

Let a in B be arbitrary

Then 0 = (a + a ) – a (a a ) + a

0 a, for all a in B Hence B has a least element.

4. To each a, b in B , there exist a least element x = a b in B such that b x a :

Let a, b in B be arbitrary

there exist a least element x = a b in B Now b x = b + x

= b + (a b)

= a a

Thus to each a, b in B, there exist a least element x = a b in B such that b x a Hence B is a Brouwerian Algebra

Claim (2) : G is a commutative – group

1. (G, +) is an abelian group Closure law :

Let a, b in G be arbitrary

Then [(a + b) + (a + b)] – (a + b) = (2a + 2b) (a + b)

= (2a – a) + (2b – b), by (6)

a + b in G

= a + b

Thus a, b in G a + b in G

Clearly, + is both associative and commutative in G, since G is a subset of A.

Existence of Identity:

Let a in G be arbitrary.

Clearly 0 in G, since 0 = 0 + 0 0 Then a + 0 = 0 + a = a , for all a in G.

Existence of Inverse :

Let a in G be arbitrary

Then (-a) + (-a) (-a) = a a + a

= -a

-a in G Now, a + (-a) = (-a) + a = 0

Hence (G, +) is an abelian group

2. (G , ) is a lattice :

Let a, b in G be arbitrary.

a, -b, b in G

a b, b in G

(a b) + b in G

a b in G, by property 7 [4] Also a, b in G a + b , a b in G

(a + b) (a b) in G

a b in G

Idempotent law:

Let a in G be arbitrary

Then a a = ( a a ) + a

= a

a a = ( a + a ) ( a a)

= ( a+ a) a

= a

Thus a a = a ; a a = a, for all a in G

Commutative law:

Let a, b in G be arbitrary

Then a b = (a+ b) (a b)

= [(a + b) – a] [(a + b) – b]

= b a

a b = (a + b) (a b)

= (b + a) (b a)

= b a

Thus a b = b a ; a b = b a, for all a, b in G

Associatve law :

Let a, b, c in G be arbitrary

Then a ( b c) = [ a (b c)] + (b c)

= a

(a b) c = [(a b) c] + c

= a b

= (a b) + b

= a

Therefore, a (b c) = (a b) c, for all a, b, c in G.

Also, a (b c) = [ a + (b c)] [ a (b c)]

= [ a + ( b c) ([a (b c)] + (b c))

= a + ( b c) a

= b c

= (b + c) (b c)

= (b + c) [(b – c) + c ]

= b + c b

= c

(a b) c = [(a b) + c] – [(a b) c]

= [(a b) + c] ([(a b) c] + c)

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

Absorption Law :

Let a, b in G be arbitrary

Then a (a b) = [ a – (a b)] + (a b)

= a

a (a b) = [ a – (a b)] [a (a b)]

= [a + (a b)] [(a a) b]

= a + (a b) (a b) = a

Thus a (a b) = a, a (a b) = a, for all a, b in G Therefore, (G, , ) is a lattice

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

a + (b c) = ( a + b) ( a + c), for all a, b , c in G :

Let a, b, c in G be arbitrary

Then a + (b c) = a + [(b – c) +c]

= a + b

( a + b) ( a + c) = [( a + b) – ( a + c)] + ( a + c)

= a + b

Therefore, a + (b c) = (a + b) (a + c) , for all a, b, c in G Also, a + (b c) = a + [( b + c) – (b c)]

= a + (( b + c) [(b- c) + c])

= a + [( b + c) b]

= a + c

(a + b) (a + c) = [(a + b) + (a + c)] [(a + b) (a + c)], by property 8 [4]

= (a + b) + (a + c) (a + b), by previous result

= ( a + c)

Therefore, a + (b c) = (a + b) (a + c), for all a, b, c in G Hence G is a commutative – group.

Claim (3): A = B x G

For any a in A, y = (a + a) a, x = a – [(a + a) a] implies y in G, x in B Now (y + y) y = [(2a – a) + (2a – a)] (2a a)

= [(2a + 2a) (a + a)] (2a a), by (6)

= ( 4a 2a) (2a a)

4a 2a a , since 2a a a

= a

(y + y) y y

Also, (y + y) y (y y) + y, by property 11 [4]

= y

(y + y) y y Therefore , (y + y) y y

y in G

y = (a + a) a

y a

x 0

x + x 0 + x

x + x x

Now, (a y) + (a y) = (a + a) (y + y), by (4)

= 2a 2y

= 2a 2 ( 2a a)

x + x = 2a ( 4a 2a)

We have

( 4a 2a) + [a (2a a)] = (2a a) + (2a a) + [a (2a a)]

(2a a) + a

= 2a

( 4a 2a) + [a (2a a)] 2a

2a ( 4a 2a) a (2a a) = x

x + x x

x + x = x

x in B

Thus if a in A, then a = x + y , where x in B, y in G Now, let a = x' + y', where x' in B, y' in G Then a + a = (x' + y' ) + (x' + y' )

= (x' + x' ) + (y' + y' )

= x' + 2 y', since x' in B

= (x' + y') + y'

a + a = a + y'

(a + a) a = (a + y' ) a

(a + y' ) a in G

[(a + y' ) a] y' = (a + y' ) (a + y' )

= 0

a + y' = a + y'

a + y' a = y'

(a + a) a = y'

Now, a = x' + y'

a y' = x' x'

a y' x'

Also, x' ( a y') (x' a) + y'

= [x' (x' + y')] + y'

= ( 0 y') + y' = 0

x' (a y' ) 0

x' a y'

Hence x' = a y'

Hence follows that A is the direct product of a Brouwerian Algebra B and a commutative – group G.

Conversely, assume that A = B x G, where B is a Brouwerian Algebra and G is a commutative group.

To prove

(i) (a + b) (c + c) (a c) + (b c),

(ii) (ma + nb) (a + b) (ma a) + (nb b),

for all a, b, c in A and any pair of positive integers m, n. Let a, b, c in A be arbitrary

(i). To each [(a c) + (b c)], (a + b) in B, there exist a least ( c + c ) in B such that (a + b) (c + c) (a c) + (b c),

(ii). To each [(ma a) + (nb b)], (ma + nb) in B, there exist a least element (a + b) in B such that (ma +nb) (a + b) (ma a) + (nb b),

since B is a Brouwerian Algebra

(i) (a + b) (c + c) (a c) + (b c),

(ii) (ma + nb) (a + b) (ma a) + (nb b), for all a, b, c in A and any pair of positive integers m, n.

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).