Question 7 of 33intermediate🔧 ApplyShort Answer4 marks

Let A={1,2}A = \{1, 2\}, B={1,2,3,4}B = \{1, 2, 3, 4\}, C={5,6}C = \{5, 6\} and D={5,6,7,8}D = \{5, 6, 7, 8\}. Verify that

(i)

A×(B∩C)=(A×B)∩(A×C)A \times (B \cap C) = (A \times B) \cap (A \times C).

Answer

LHS: B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = φ. Therefore, A × (B ∩ C) = A × φ = φ.

RHS: A × B = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4)} and A × C = {(1,5), (1,6), (2,5), (2,6)}. Since no ordered pair is common in both, (A × B) ∩ (A × C) = φ.

Hence, A × (B ∩ C) = (A × B) ∩ (A × C) = φ. Verified.

(ii)

A×CA \times C is a subset of B×DB \times D.

Answer

A × C = {(1,5), (1,6), (2,5), (2,6)}.

B × D = {(1,5), (1,6), (1,7), (1,8), (2,5), (2,6), (2,7), (2,8), (3,5), (3,6), (3,7), (3,8), (4,5), (4,6), (4,7), (4,8)}.

Since every element of A × C is present in B × D, we have A × C ⊆ B × D. Verified.

Explanation

This question verifies two important properties of Cartesian products. Part (i) demonstrates the distributive property of Cartesian product over intersection. Since B and C are disjoint sets (B ∩ C = φ), both sides equal the empty set. Part (ii) verifies the subset property: since A ⊆ B and C ⊆ D, the Cartesian product A × C must be a subset of B × D.

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