Question

Show that the set of all natural number form a semi group under the composition of addition?​

Answer 1

Answer:

Step-by-step explanation:

Let E={2x:x∈N} be the set of even positive integers.

(i)Closure:Consider (E,+)

Let 2x,2y∈E then 2x+2y=2(x+y)∈E

The closure property is satisfied.

(ii)Associativity:For 2x,2y,2z∈E

(2x+2y)+2z=2x+(2y+2z) since these are natural numbers and addition of natural numbers is associative.

∴(E,+) is a semi-group.

(iii)Consider (E,.)

Closure :For 2x,2y,2z∈E

(2x)(2y)(2z)=2(4xyz)∈E

(The product of even numbers is again an even number)

The closure property is satisfied.

(iv)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.

Hence, E is a semi-group under additions as well as multiplication.

Since E⊂N, then {E,+} and {E,.} are monoids with the identity elements 0 and 1 respectively.

Monoids are semigroups with identity. Hence from the above E is a monoid.