Question

Using laws of algebra of sets, show that (i) (AUB) N (AUB) = A
(ii) AU (B-A)= AUB

please solve and I will mark you as brainliest!!!

Answer 1

Step-by-step explanation:

Commutative Laws:

For any two finite sets A and B;

(i) A U B = B U A

(ii) A ∩ B = B ∩ A

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.

3. Idempotent Laws:

For any finite set A;

(i) A U A = A

(ii) A ∩ A = A

4. 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.

5. De Morgan’s Laws:

For any two finite sets A and B;

(i) A – (B U C) = (A – B) ∩ (A – C)

(ii) A - (B ∩ C) = (A – B) U (A – C)

De Morgan’s Laws can also we written as:

(i) (A U B)’ = A' ∩ B'

(ii) (A ∩ B)’ = A' U B'

More laws of algebra of sets:

6. For any two finite sets A and B;

(i) A – B = A ∩ B'

(ii) B – A = B ∩ A'

(iii) A – B = A ⇔ A ∩ B = ∅

(iv) (A – B) U B = A U B

(v) (A – B) ∩ B = ∅

(vi) A ⊆ B ⇔ B' ⊆ A'

(vii) (A – B) U (B – A) = (A U B) – (A ∩ B)

7. For any three finite sets A, B and C;

(i) A – (B ∩ C) = (A – B) U (A – C)

(ii) A – (B U C) = (A – B) ∩ (A – C)

(iii) A ∩ (B - C) = (A ∩ B) - (A ∩ C)

(iv) A ∩ (B △ C) = (A ∩ B) △ (A ∩ C)

Please mark me as Brainliest if it helps and follow me