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Exercise 2.1 · Q7

Q.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).
(ii) A×CA \times C is a subset of B×DB \times D.
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The Cartesian product distributes over intersection, so A×(B∩C)=(A×B)∩(A×C)A \times (B \cap C) = (A \times B) \cap (A \times C) holds. Also, since A⊆BA \subseteq B and C⊆DC \subseteq D, every ordered pair in A×CA \times C is also in B×DB \times D, making A×CA \times C a subset of B×DB \times D.

Why This Works — The Core Idea

The Cartesian product X×YX \times Y is the set of all ordered pairs (x,y)(x, y) where x∈Xx \in X and y∈Yy \in Y. When we intersect two sets before taking the product, we are restricting the second coordinate to elements that belong to both sets. On the other hand, taking the product first and then intersecting keeps only those pairs whose second coordinate is common to both original sets. These two processes yield exactly the same collection of pairs — that’s the distributive law.

For the subset part, if every element of AA is also in BB, and every element of CC is also in DD, then any pair (a,c)(a, c) automatically has its first coordinate in BB and its second in DD, so it belongs to B×DB \times D.

Let’s verify both statements concretely.


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

Step 1: Compute B∩CB \cap C

B={1,2,3,4}B = \{1, 2, 3, 4\}, C={5,6}C = \{5, 6\}. These sets have no common element, so

B∩C=∅B \cap C = \varnothing

Step 2: Compute A×(B∩C)A \times (B \cap C)

The Cartesian product with an empty set is empty:

A×∅=∅A \times \varnothing = \varnothing

Step 3: Compute A×BA \times B

A={1,2}A = \{1, 2\}, B={1,2,3,4}B = \{1, 2, 3, 4\}

A×B={(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)}A \times B = \{(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4)\}

Step 4: Compute A×CA \times C

A={1,2}A = \{1, 2\}, C={5,6}C = \{5, 6\}

A×C={(1,5),(1,6),(2,5),(2,6)}A \times C = \{(1,5), (1,6), (2,5), (2,6)\}

Step 5: Intersect A×BA \times B and A×CA \times C

Look for pairs that appear in both lists. The first set has second coordinates 1,2,3,41,2,3,4; the second set has second coordinates 5,65,6. No pair can match because the second coordinates are disjoint.

(A×B)∩(A×C)=∅(A \times B) \cap (A \times C) = \varnothing

Step 6: Compare

Both sides are ∅\varnothing, so the equality holds.

Watch out

A common mistake is to think B∩CB \cap C might be non-empty just because the sets look similar. Always check actual elements — here BB and CC share nothing.


(ii) A×CA \times C is a subset of B×DB \times D

Step 1: Check the sets …

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