Question

Is it true that for any sets A and B, P ( A ) ∪ P ( B ) = P ( A ∪ B )? Justify your answer. DONT SPAM HERE.

Answer 1

Answer:

P(A)∪P(B)  ≠ P(A∪B)

Step-by-step explanation:

ANSWER

Let A={0,1} and B={1,2}

∴A∪B={0,1,2}

P(A)= {ϕ,{0},{1},{0,1}}

P(B)={ϕ,{1},{2},{1,2}}

P(A∪B)={ϕ,{0},{1},{2},{0,1},{1,2},{0,2},{0,1,2}}

P(A)∪P(B)={ϕ,{0},{1},{0,1},{2},{1,2}}

∴P(A)∪P(B)  ≠ P(A∪B)

Hope you understand..

Answer 2

Answer:

Given statement is false.

Step-by-step explanation

Hey Please Check The Example

Let A = {1, 2} and B = {2, 3}

A ∪ B = {1, 2} ∪ {2, 3} = {1, 2, 3}

P(A) = {Φ, {1}, {2}, {1, 2}}

P(B) = {Φ, {2}, {3}, {2, 3}}

P(A ∪ B) = {Φ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}}

P(A) ∪ P(B) = {Φ, {1}, {2}, {3}, {1, 2}, {2, 3}}

We can observe that P(A) ∪ P(B) ≠ P(A ∪ B).

Hence given statement is false.

Hope it helps You!!!!