Discrete Mathematics with Applications 4th Edition

Published by Cengage Learning
ISBN 10: 0-49539-132-8
ISBN 13: 978-0-49539-132-6

Chapter 6 - Set Theory - Exercise Set 6.2 - Page 366: 26

Answer

Proof by contradiction: Let us suppose (A − C) ∩ (B − C) ∩ (A − B) ≠ ∅. That is, suppose there exist sets A, B and C such that (A − C) ∩ (B − C) ∩ (A − B) ≠ ∅. Then there is an element x in (A − C) ∩ (B − C) ∩ (A − B) such that x ≠ ∅. By definition of intersection, we should have, x ∈ (A − C) and x ∈ (B − C) and x ∈ (A − B). Applying the definition of intersection again, we have that Since x ∈ (A - C), x ∈ A and x ∉ C -(i) Since x ∈ (B - C), x ∈ B and x ∉ C -(ii) Since x ∈ (A - B), x ∈ A and x ∉ B -(iii) From (ii) and (iii) we see that x ∈ B and x ∉ B which is a contradiction. So our assumption is false, Therefore (A − C) ∩ (B − C) ∩ (A − B) = ∅.

Work Step by Step

Steps: 1. Start with an assumption saying the statement is false. that is (A − C) ∩ (B − C) ∩ (A − B) ≠ ∅ 2. Breakdown the statements into sub-statements using definition of set intersection. x ∈ (A - C) implies x ∈ A and x ∉ C and so on. 3. Show that the assumption is invalid with the sub-statements.
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