ITPEC FE Morning October 2021 Question 3

Intersection distributes over union — (A ∩ C) ∪ (B ∩ C) = (A ∪ B) ∩ C, just as ac + bc = (a + b)c in algebra.

An element is on the left side when it is in C and in A or B; that is exactly membership of (A ∪ B) ∩ C. With A = {1, 2}, B = {2, 3}, C = {1, 3}, both sides are {1, 3}; the same sets disprove the other options.

Answer (d)

Why not others:

• •(a) — the left side simplifies to A ∩ C = {1}, the right side B ∩ (A ∪ C) is {2, 3}
• •(b) — mixes the two distributive laws: (A ∪ C) ∩ (B ∪ C) equals (A ∩ B) ∪ C = {1, 2, 3}, not (A ∪ B) ∩ C = {1, 3}
• •(c) — the left side is A ∩ (B ∪ C) = {1, 2} and the right side is B ∩ (A ∪ C) = {2, 3}; the common factor differs

Key rule: Factor out the set common to both terms, as in algebra (though ∪ also distributes over ∩), and test a claimed identity with small concrete sets: one counterexample disproves it.

Basic Theory · AI-written, may contain errors