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tence or two what is wrong with the follo Let~ be a relation on a set S. If ~is reflexive. ~ is symmetric and transitive, and let with a a and b. Since ~ is symmetric, we h- is transitive, we have that a ~ a

tence or two what is wrong with the follo Let~ be a relation on a set S. If ~is reflexive. ~ is symmetric and transitive, and let with a a and b. Since ~ is symmetric, we h- is transitive, we have that a ~ a

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Question 7.** Explain in a sentence or two what is wrong with the following proof. **Proposition.** Let \( \sim \) be a relation on a set \( S \). If \( \sim \) is symmetric and transitive, then \( \sim \) is reflexive. **Proof.** Suppose \( \sim \) is symmetric and transitive, and let \( a \in S \). We will show that \( a \sim a \). Suppose \( b \in S \) with \( a \sim b \). Since \( \sim \) is symmetric, we have that \( b \sim a \). Since \( a \sim b, b \sim a \) and \( \sim \) is transitive, we have that \( a \sim a \). \( \square \) **Explanation of the Error:** The proof incorrectly assumes the existence of an element \( b \) such that \( a \sim b \) without any justification. To prove reflexivity, \( a \sim a \) must be demonstrated for all elements \( a \in S \) unconditionally. The approach relies on an unwarranted assumption, thus invalidating the proof.
Transcribed Image Text:**Question 7.** Explain in a sentence or two what is wrong with the following proof. **Proposition.** Let \( \sim \) be a relation on a set \( S \). If \( \sim \) is symmetric and transitive, then \( \sim \) is reflexive. **Proof.** Suppose \( \sim \) is symmetric and transitive, and let \( a \in S \). We will show that \( a \sim a \). Suppose \( b \in S \) with \( a \sim b \). Since \( \sim \) is symmetric, we have that \( b \sim a \). Since \( a \sim b, b \sim a \) and \( \sim \) is transitive, we have that \( a \sim a \). \( \square \) **Explanation of the Error:** The proof incorrectly assumes the existence of an element \( b \) such that \( a \sim b \) without any justification. To prove reflexivity, \( a \sim a \) must be demonstrated for all elements \( a \in S \) unconditionally. The approach relies on an unwarranted assumption, thus invalidating the proof.
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