2. Preliminaries
A
d-algebra [
4] is a non-empty set
X with a constant 0 and a binary operation
satisfying the following axioms: (I)
, (II)
, (III)
and
imply
, for all
. For more detailed information we refer to [
5,
6,
7].
A
-algebra [
8] is a
d-algebra
X satisfying the following additional axioms: (IV)
, (V)
for all
.
A
-algebra
is said to be
bounded if there exists an element
such that
for any
. A
-algebra
is said to be
commutative if
for all
, i.e.,
. A
-algebra
is said to be
positive implicative if if
for all
. For more detailed information we refer to [
8].
We denote -algebras or another groupoids (algebras) by usually when we need to emphasize the constant 0, but sometimes we denote it by in short.
Theorem 1. [
8]
For a bounded commutative -algebra , we have- (i)
for all ,
- (ii)
for all ,
- (iii)
for all .
Given a
-algebra
, we define a relation “≤” on
X by
In
-algebras, the relation ≤ is a partial order, and so
forms a partially ordered set with the least element 0, but it need not be a partial order in
d-algebras.
Theorem 2. [
8]
Let be a -algebra. Then- (i)
for all ,
- (ii)
for all .
3. Pre-Commutative Algebras
Given a groupoid (or an algebra) , it is said to be
- (i)
abelian if ;
- (ii)
left-pre-commutative if for some map ;
- (iii)
right-pre-commutative if for some map ;
- (iv)
pre-commutative if for some map ,
for all .
Example 1. (i).
Define a binary operation “*" on a set by the following table.Then it is easy to see that for any . If we let , then for any . Hence is a left-pre-commutative groupoid.(ii). If we define in (i), then is a right-pre-commutative groupoid, since for any .
(iii).
Define a binary operation “*" on a set by the following table.Then it is easy to see that for all . If we define a map by and , then for all . Hence is a pre-commutative groupoid. Remark 1. In Example 1-(i), is not a right-pre-commutative groupoid. In fact, if there exists a map such that for any , then . From the table of Example 1-(i), we obtain . It follows that , a contradiction. This shows that Example 1-(i) is a left-pre-commutative groupoid which is not a right-pre-commutative groupoid. Similarly, Example 1-(ii) is a right-pre-commutative groupoid which is not a left-pre-commutative groupoid.
Remark 2. In Example 1-(i), is not a pre-commutative groupoid. In fact, if there exists a map such that for any , then . From the table of Example 1-(i), we have either or . If , then , a contradiction. Hence we have . Now, since , from the table of Example 1-(i), we have either or . If we take , then from the table, a contradiction. Hence we have . Since , from the table, we obtain either or . If , then , a contradiction. Hence we have . In this fashion, we obtain . It follows that for all , a contradiction. This shows that Example 1-(i) is not a pre-commutative groupoid.
Example 2. Let be a set. If we define a map by and , then the algebra with the following table is left-pre-commutative, but not abelian.where are distinct elements of X. For example, if we let , then the algebra with the following table is left-pre-commutative, but not abelian. Example 3. Given a set with a map φ defined in Example 2, the algebra with the following table is right-pre-commutative, but not abelian, since and .where are distinct elements of X. For example, the algebra with the following table is right-pre-commutative, but not abelian. Proposition 1. If is abelian, then it is (left-, right-) pre-commutative.
Proof. Take . □
Theorem 3. If the groupoid is a group, then (left-, right-) pre-commutativity implies abelian property.
Proof. Let be a group which is left-pre-commutative. Then there exists a map such that for any . It follows that . By cancelation law we obtain . Also, and hence for all , i.e., for all . Hence for all , i.e., is abelian.
If is a group which is right-pre-commutative, then for some . It follows that and hence for all . Hence for all , i.e., is abelian.
If
is a group which is pre-commutative, then
for some
. Since
is a group, we have
. It follows that
. Since
, we obtain
Let
. Then
and
Thus
Since
x is arbitrary,
is arbitrary. If we let
, then
, i.e.,
, i.e.,
is abelian. □
Let K be a field. A groupoid is said to be linear if for any , where are (fixed) elements of K.
Proposition 2. If a linear groupoid is left-pre-commutative, then it is abelian.
Proof. Since
is left-pre-commutative, there exists a map
such that
for any
, i.e.,
If we let
in (1) consequently, then
and
. If
, then
is trivially abelian. If
, then
. By (1) we have
, proving
. Thus,
. Similarly, if
, then
, proving
is abelian. □
Proposition 3. If a linear groupoid is right-pre- commutative, then it is abelian.
Proof. The proof is similar to Proposition 2, and we omit it. □
Note that, in a linear groupoid , the “pre-commutativity” does not imply the “abelian”. In fact, if we define for any , where in K, then it is pre-commutative, but not abelian.
Proposition 4. Let K be a field. Define a binary operation “*” on K byfor all . If is pre-commutative, then . Proof. It is easy to show that
is a
d-algebra. Assume that
is pre-commutative. Then there exists a map
such that
for any
. It follows that
If we let
in (2), then
for all
. Then either
or
for all
. Assume that
for some
. Then
, i.e.,
. Let
. It follows that
for any
. This means that
and
. Since
is pre-commutative,
has at most 4 elements, i.e.,
. Thus
is a possible field. If we let
and
, then
. □
4. Smarandache -Disjoint and Pre-Commutativity
Allen et al. [
9] introduced the notion of a Smarandache disjointness in algebras. We restate the notion of a Smarandache disjointness for its clarification. Simply, two algebras
and
are said to be
Smarandache disjoint [
9] if we add some axioms of an algebra
to an algebra
, then the algebra
becomes a trivial algebra, i.e.,
, or if we add some axioms of an algebra
to an algebra
, then the algebra
becomes a trivial algebra, i.e.,
. Note that if we add an axiom
of an algebra
to another algebra
, then we replace the binary operation “∘” in
by the binary operation “*”. A groupoid
is said to be a
left-zero-semigroup [
10] if
for any
.
Proposition 5. Left-zero-semigroups and (left-, right-) pre-commutative algebras are Smarandache disjoint.
Proof. Assume a left-zero-semigroup is left-pre- commutative. Then for any for some map , which implies for all . This shows that , a contradiction.
Assume a left-zero-semigroup is right-pre-commutative. Then for any for some map . This shows that , a contradiction.
Assume a left-zero-semigroup is pre-commutative. Then for any for some map , which implies for all . This shows that , a contradiction. □
Proposition 6. Left-pre-commutative algebras and d-algebras are Smarandache disjoint.
Proof. Let be a left-pre-commutative d-algebra. Then for any for some map . If we let , then . Since is d-algebra, by applying (I), (II), we obtain for any . Hence for any , which implies by (III), proving the proposition. □
Theorem 4. Non-bounded d-algebras and right-pre-commutative algebras are Smarandache disjoint.
Proof. Let
be a non-bounded right-pre-commutative
d-algebras. Then there exists a map
such that
for any
. If we let
in (3), then
for any
. We claim that
. In fact, if
, then
for any
, i.e.,
X is bounded, a contradiction. Hence
for any
, which shows that
for any
, i.e.,
. □
5. -Algebras and a Pre-Commutativity
Theorem 5. If a -algebra is pre-commutative, then it is bounded.
Proof. If a -algebra is pre-commutative, then for any for some map . It means that the mapping is order-reversing, i.e., implies . Now, since is a -algebra, we have for all . It follows that for all . This means that is the greatest element of the poset . We claim that is the greatest element of the poset . Let . Since for all and is pre-commutative, we obtain for any . It follows from Theorem 2-(ii) and (II) that for any . This shows that is the greatest element of , proving that is bounded by . □
The converse of Theorem 5 need not be true in general. See the following example.
Example 4. Consider a -algebra with the following table:Then is a bounded (positive implicative) -algebra (see [8], p. 243). By routine calculations we see that there is no map satisfying the condition of a pre-commutativity, i.e., is a bounded -algebra having no pre-commutative property. The converse of Theorem 5 is also true if we add the condition of “abelian”, i.e., for all .
Proposition 7. Every bounded commutative -algebra is pre-commutative.
Proof. By Theorem 1-(iii), we have for all , which shows that is pre-commutative. □
Proposition 8. If is a pre-commutative -algebra, then
- (i)
for any ,
- (ii)
.
for some map .
Proof. Since X is pre-commutative, by Theorem 5, X is bounded and is the greatest element of X, say . Since for all , we obtain , for all , proving (i). Since X is pre-commutative, we obtain by (i). □
Theorem 6. Let be a bounded commutative -algebra. If a map satisfies the condition for all , then for all .
Proof. By Proposition 7, if is a bounded commutative -algebra, then it is pre-commutative, i.e., for all . By Proposition 8-(i), we have . Since is commutative and bounded, by Theorem 2-(i), we have for all . This proves the theorem. □
6. Conclusions
We introduced the notions of generalized commutative laws in algebras, and investigated their roles in algebras, and found their interrelationships by using Smarandache disjointness. The notion of pre-commutative law applied to -algebras, and obtained that if a -algebra is pre-commutative, then it is bounded. Moreover, we proved that, in a bounded commutative -algebra , if a map satisfies the condition for all , then for all .
Allen et al. [
3] have developed the generalization of an associative law, and we discussed the generalization of a commutative law by using (deformed) functions. This idea may apply to several axioms appeared in general algebraic structures, and will generalize several algebraic structures.