Question

Show that the set N of all natural numbers 1, 2, 3, 4, 5… does not form a group with

respect to multiplication or addition but it forms a semi group with respect to the addition

as well as multiplication.​

Answer 1

SOLUTION

TO PROVE

The set N of all natural numbers 1, 2, 3, 4, 5… does not form a group with respect to multiplication or addition but it forms a semi group with respect to the addition as well as multiplication.

EVALUATION

Here the given set is set N of all natural numbers

N = { 1, 2, 3, 4, 5….}

CHECKING UNDER ADDITION

CLOSURE PROPERTY

Clearly for a , b ∈ N we have a + b ∈ N

ASSOCIATIVE PROPERTY

a + ( b + c ) = ( a + b ) + c for every a , b, c ∈ N

Hence N is a semi group under addition

EXISTENCE OF IDENTITY

0 is the identity element such that

0 + a = a + 0 = a for every a ∈ N

EXISTENCE OF INVERSE ELEMENT

Let a ∈ N then - a ∈ N such that

a + ( - a ) = 0 = ( - a ) + a

So - a is the inverse element of a

Hence ( N , + ) is a group

CHECKING UNDER MULTIPLICATION

CLOSURE PROPERTY

Clearly for a , b ∈ N we have a. b ∈ N

So N is closed under multiplication

ASSOCIATIVE PROPERTY

a . ( b . c ) = ( a . b ) . c for every a , b, c ∈ N

Hence N is a semi group under multiplication

EXISTENCE OF IDENTITY

1 is the identity element such that

1 . a = a . 1 = a for every a ∈ N

EXISTENCE OF INVERSE ELEMENT

Every element of N does not have the inverse element in N

For example 3 ∈ N does not have inverse element

Hence ( N , . ) is not a group

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Answer 2

The answer is

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