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1. Let A, B abd C be sets. Prove that Au (B-C) = (AUB) - (C-A).

1. Let A, B abd C be sets. Prove that Au (B-C) = (AUB) - (C-A).

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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**Problem Statement:** 1. Let \(A\), \(B\) and \(C\) be sets. Prove that \(A \cup (B - C) = (A \cup B) - (C - A)\). **Explanation:** This problem requires you to prove an equality between two set expressions using set theory. - **Union (\(\cup\))**: Represents all elements that are in either of the sets. - **Difference (\(-\))**: Represents elements that are in the first set but not in the second set. To prove the given equation, you can typically follow these steps: 1. **Express Elements**: Break down the elements belonging to each side of the equation. 2. **Logical Arguments**: Use known properties and operations of sets (such as distributivity, associativity, and De Morgan’s laws) to manipulate and compare both expressions. ### Proof: **Step 1: Analyze \(A \cup (B - C)\)** An element \(x\) is in \(A \cup (B - C)\) if and only if: - \(x \in A\) or - \(x \in (B - C)\) Which means: - \(x \in (B - C)\) implies \(x \in B\) and \(x \notin C\). Therefore: - \(x \in A\) or \((x \in B \land x \notin C)\). **Step 2: Analyze \((A \cup B) - (C - A)\)** An element \(x\) is in \((A \cup B) - (C - A)\) if and only if: - \(x \in (A \cup B)\) and - \(x \notin (C - A)\) Which means: - \(x \in (A \cup B)\) implies \(x \in A\) or \(x \in B\). - \(x \notin (C - A)\) implies \(x \notin C\) or \(x \in A\). Therefore: - \((x \in A \lor x \in B) \land (x \notin C \lor x \in A)\). **Step 3: Compare Both Sides** The expressions: 1. \(A \cup (B - C)\) simplifies to \(
Transcribed Image Text:**Problem Statement:** 1. Let \(A\), \(B\) and \(C\) be sets. Prove that \(A \cup (B - C) = (A \cup B) - (C - A)\). **Explanation:** This problem requires you to prove an equality between two set expressions using set theory. - **Union (\(\cup\))**: Represents all elements that are in either of the sets. - **Difference (\(-\))**: Represents elements that are in the first set but not in the second set. To prove the given equation, you can typically follow these steps: 1. **Express Elements**: Break down the elements belonging to each side of the equation. 2. **Logical Arguments**: Use known properties and operations of sets (such as distributivity, associativity, and De Morgan’s laws) to manipulate and compare both expressions. ### Proof: **Step 1: Analyze \(A \cup (B - C)\)** An element \(x\) is in \(A \cup (B - C)\) if and only if: - \(x \in A\) or - \(x \in (B - C)\) Which means: - \(x \in (B - C)\) implies \(x \in B\) and \(x \notin C\). Therefore: - \(x \in A\) or \((x \in B \land x \notin C)\). **Step 2: Analyze \((A \cup B) - (C - A)\)** An element \(x\) is in \((A \cup B) - (C - A)\) if and only if: - \(x \in (A \cup B)\) and - \(x \notin (C - A)\) Which means: - \(x \in (A \cup B)\) implies \(x \in A\) or \(x \in B\). - \(x \notin (C - A)\) implies \(x \notin C\) or \(x \in A\). Therefore: - \((x \in A \lor x \in B) \land (x \notin C \lor x \in A)\). **Step 3: Compare Both Sides** The expressions: 1. \(A \cup (B - C)\) simplifies to \(
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