Question

How to verify when A is set of natural numbers till 5 and B is set of all odd numbers till 5 falls under a commutative ,associative , closure ,distributive law with verification

Answer 1

A= {1,2,3,4,5}

B={1,3,5}

1. Commutative Laws:

For any two finite sets A and B;

(i) A U B = B U A

(ii) A ∩ B = B ∩ A

Here A U B = {1,2,3,4,5} = B U A

A ∩ B = {1,3,5} = B ∩ A

So the set A and B are commutative.

2. Associative Laws:

For any three finite sets A, B and C;

(i) (A U B) U C = A U (B U C)

(ii) (A ∩ B) ∩ C = A ∩ (B ∩ C)

Thus, union and intersection are associative.

LET C = {1,2,3}

Here (A U B) U C={1,2,3,4,5} U {1,2,3)
= {1,2,3,4,5}

A U (B U C) = {1,2,3,4,5} U {1,2,3,5}
= {1,2,3,4,5}

So (A U B) U C = A U (B U C)


(A ∩ B) ∩ C = {1,3,5} ∩ {1,2,3}

= {1,3}


A ∩ (B ∩ C) = {1,2,3,4,5} ∩ {1,3}

= {1,3}

SO (A ∩ B) ∩ C = A ∩ (B ∩ C)

3. Distributive Laws:

For any three finite sets A, B and C;

(i) A U (B ∩ C) = (A U B) ∩ (A U C)

(ii) A ∩ (B U C) = (A ∩ B) U (A ∩ C)

Thus, union and intersection are distributive over intersection and union respectively.


A U (B ∩ C) = {1,2,3,4,5} U {1,3}
= {1,2,3,4,5}

(A U B) ∩ (A U C) = {1,2,3,4,5} ∩ (1,2,3,4,5}
= {1,2,3,4,5}

SO A U (B ∩ C) = (A U B) ∩ (A U C)


A ∩ (B U C) = {1,2,3,4,5} ∩ {1,2,3,5}
= {1,2,3,5}

(A ∩ B) U (A ∩ C) = {1,3,5} U {1,2,3}
={1,2,3,5}

SO A ∩ (B U C) = (A ∩ B) U (A ∩ C)


HENCE VERIFIED