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If $(S, +,.)$ is a distributive near-semiring, then prove that the subsemigroup $(S^2, +)$ is additive subcommutative.

[ Hints: A non empty set $S$ with two binary operations $'+'$ and $'.'$ is called near-semirirng if the following conditions are satisfied:

$(i) a+(b+c)=(a+b)+c$

$(ii) a.(b.c)=(a.b).c$

$(iii) a.(b+c)=a.b+a.c$ $\forall$ a, b, c $\in S$ (left distributive law)

Distributive near-semiring means in addition with all above three conditions, $(a+b)c=ac+bc$ $\forall$ $a,b,c \in S $ (right distributive law) also satisfied.

A near-semiring $(S,+,.)$ is said to be additive subcommutative if $a+b+c+d=a+c+b+d$ $\forall a,b,c,d \in S$

$ S^2$ defined as $S^2=\{\sum_{i=1}^na_ib_i \mid a_i, b_i \in S, n=1,2,...\}$]

Petchimuthu Subramanian
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1 Answers1

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Attempt

Let $a,b,c,d \in S$ then

$(a+b)(c+d)=a(c+d)+b(c+d)=ac+ad+bc+bd$.....(1) (use left and right distributive laws)

$(a+b)(c+d)=(a+b)c+(a+b)d=ac+bc+ad+bd$.....(2) (use left and right distributive laws)

From (1) and (2), we have $ac+ad+bc+bd=ac+bc+ad+bd$ $\forall a, b, c, d \in S$

Is it enough to prove the question? I am not sure!. But exactly we need to prove that $ab+cd+ef+gh=ab+ef+cd+gh$ $\forall a,b,c, d, e, f, g, h \in S$. Am I right? Any body help me.

Petchimuthu Subramanian
  • 11