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Answer:
Solution:
⇒ A ⊂ A ∩ B
⇒ A = A ∩ B
Let us assume that A ∩ B = A
Let X ∈ A
⇒ X ∈ A ∩ B
⇒ X ∈ B and X ∈ A
⇒ A ⊂ B
∴ (i) ⬌ (iv)
∴ (i) ⬌ (ii) ⬌ (iii) ⬌ (iv)
Hence, proved
5. Show that if A ⊂ B, then C – B ⊂ C – A.
Solution:
To show,
C – B ⊂ C – A
According to the question,
Let us assume that x is any element such that X ∈ C – B
∴ x ∈ C and x ∉ B
Since, A ⊂ B, we have,
∴ x ∈ C and x ∉ A
So, x ∈ C – A
∴ C – B ⊂ C – A
Hence, Proved.
6. Assume that P (A) = P (B). Show that A = B
Solution:
To show,
A = B
According to the question,
P (A) = P (B)
Let x be any element of set A,
x ∈ A
Since, P (A) is the power set of set A, it has all the subsets of set A.
A ∈ P (A) = P (B)
Let C be an element of set B
For any C ∈ P (B),
We have, x ∈ C
C ⊂ B
∴ x ∈ B
∴ A ⊂ B
Similarly, we have:
B ⊂ A
SO, we get,
If A ⊂ B and B ⊂ A
∴ A = B
7. Is it true that for any sets A and B, P (A) ∪ P (B) = P (A ∪ B)? Justify your answer.
Solution:
It is not true that for any sets A and B, P (A) ∪ P (B) = P (A ∪ B)
Justification:
Let us assume,
A = {0, 1}
And, B = {1, 2}
∴ A ∪ B = {0, 1, 2}
According to the question,
We have,
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}}
Also,
P (A) ∪ P (B) = {ϕ, {0}, {1}, {2}, {0, 1}, {1, 2}}
∴ P (A) ∪ P (B ≠ P (A ∪ B)
Hence, the given statement is false
8. Show that for any sets A and B,
A = (A ∩ B) ∪ (A – B) and A ∪ (B – A) = (A ∪ B)
Solution:
To Prove,
A = (A ∩ B) ∪ (A – B)
Proof: Let x ∈ A
To show,
X ∈ (A ∩ B) ∪ (A – B)
In Case I,
X ∈ (A ∩ B)
⇒ X ∈ (A ∩ B) ⊂ (A ∪ B) ∪ (A – B)
In Case II,
X ∉A ∩ B
⇒ X ∉ B or X ∉ A
⇒ X ∉ B (X ∉ A)
⇒ X ∉ A – B ⊂ (A ∪ B) ∪ (A – B)
∴A ⊂ (A ∩ B) ∪ (A – B) (i)
It can be concluded that, A ∩ B ⊂ A and (A – B) ⊂ A
Thus, (A ∩ B) ∪ (A – B) ⊂ A (ii)
Equating (i) and (ii),
A = (A ∩ B) ∪ (A – B)
We also have to show,
A ∪ (B – A) ⊂ A ∪ B
Let us assume,
X ∈ A ∪ (B – A)
X ∈ A or X ∈ (B – A)
⇒ X ∈ A or (X ∈ B and X ∉A)
⇒ (X ∈ A or X ∈ B) and (X ∈ A and X ∉A)
⇒ X ∈ (B ∪A)
∴ A ∪ (B – A) ⊂ (A ∪ B) (iii)
According to the question,
To prove:
(A ∪ B) ⊂ A ∪ (B – A)
Let y ∈ A∪B
Y ∈ A or y ∈ B
(y ∈ A or y ∈ B) and (X ∈ A and X ∉A)
⇒ y ∈ A or (y ∈ B and y ∉A)
⇒ y ∈ A ∪ (B – A)
Thus, A ∪ B ⊂ A ∪ (B – A) (iv)
∴