Page 2 - mathsvol1ch1to3ans

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(i) A × (B ∩ C) = (A × B) ∩ (A × C).
Solution:
Let (a, b) ∈ A × (B ∩ C) ⇔ a ∈ A, b ∈ (B ∩ C)

Since a ∈ A, b ∈ B ⇔ (a, b) ∈ A × B
and a ∈ A, b ∈ C ⇔ (a, b) ∈ A × C
We have (a, b) ∈ (A × B) ∩ (A × C).

Thus A × (B ∩ C) = (A × B) ∩ (A × C).
(ii) A × (B ∪ C) = (A × B) ∪ (A × C).
Solution:
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Let (a, b) ∈ A × (B ∪ C) ⇔ a ∈ A, b ∈ (B ∪ C)
Since a ∈ A, b ∈ B ⇔ (a, b) ∈ A × B
or a ∈ A, b ∈ C ⇔ (a, b) ∈ A × C

We have (a, b) ∈ (A × B) ∪ (A × C).
Thus A × (B ∪ C) = (A × B) ∪ (A × C).
(iii) (A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A).
Solution:
Let (a, b) ∈ (A × B) ∩ (B × A)
We have (a, b) ∈ (A × B) ⇔ a ∈ A, b ∈ B
and (a, b) ∈ (B × A) ⇔ a ∈ B, b ∈ A

Since a ∈ A, a ∈ B ⇔ a ∈ A ∩ B
and b ∈ A, b ∈ B ⇔ b ∈ B ∩ A

We have (a, b) ∈ (A ∩ B) × (B ∩ A).
Thus (A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A).
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(iv) C − (B − A) = (C ∩ A) ∪ (C ∩ B ).
Solution:
Let a ∈ C − (B − A)
Let a ∈ C, a /∈ (B − A) ⇔ a ∈ C, a /∈ B, a ∈ A
Since a ∈ C, a ∈ A ⇔ a ∈ C ∩ A

and a ∈ C, a /∈ B ⇔ a ∈ C ∩ B 0
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We have a ∈ (C ∩ A) ∪ (C ∩ B ).
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Thus C − (B − A) = (C ∩ A) ∪ (C ∩ B ).