Ask Questions, Get Answers


Show that the set of all positive even integers forms a semi-group under the usual addition and multiplication. Is it a monoid under each of the above operations?

1 Answer

  • A non-empty set S with an operation $*$ i.e., $(S, *)$ is said to be a semi-group if it satisfies the following axioms.
  • (1) Closure axiom : $a,b\in S\Rightarrow a*b\in S$
  • (2) Associative axiom : $(a * b) * c = a * (b * c), \forall a, b, c \in S$.
Step 1:
Let $E=\{2x,x\in N\}$ be the set of ever positive integers.
Closure :Consider $(E,+)$
Let $2x,2y\in E$,then $2x+2y=2(x+y)\in E$
The closure property is satisfied.
Step 2:
Associativity : For $2x,2y,2z\in E$
$(2x+2y)+2z=2x+(2y+2z)$ since these are natural numbers and addition of natural numbers is associative.
$\therefore (E,+)$ is a semi-group.
Step 3:
Consider $(E,.)$
Closure :For $2x,2y,2z\in E$
$(2x)(2y)(2z)=2(4xyz)\in E$
(The product of even numbers is again an even number)
The closure property is satisfied.
Step 4:
Associativity :
The associative property is inherited from the set of natural numbers(as with the associative property of addition over $E$)
Therefore $(E,.)$ is a semi-group.It follows that $E$ is a semi-group under additions as well as multiplication.
answered Sep 13, 2013 by sreemathi.v

Related questions