Find the flaw in the following argument. The claim is false. Do not give a counterexample. Instead, point out the mistake in the reasoning. Claim: If a relation R over set A is symmetric and transitive then R is reflexive. "Proof:" Select any element (a, b) E R. Because R is symmetric it follows that (b, a) E R. Furthermore, since R is transitive, (a, b) E R ^ (b, a) ER= (a, a) E R. It follows that R is reflexive.
Find the flaw in the following argument. The claim is false. Do not give a counterexample. Instead, point out the mistake in the reasoning. Claim: If a relation R over set A is symmetric and transitive then R is reflexive. "Proof:" Select any element (a, b) E R. Because R is symmetric it follows that (b, a) E R. Furthermore, since R is transitive, (a, b) E R ^ (b, a) ER= (a, a) E R. It follows that R is reflexive.
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Question
Struggling on this problem, I can see that it's incorrect and am close but I'm overlooking the flaw in reasoning somewhere.

Transcribed Image Text:**Title:** Analyzing Argument Flaws in Mathematical Proofs
**Content:**
**Objective:** Find the flaw in the following argument. The claim is false. Do not provide a counterexample; instead, identify the mistake in the reasoning.
**Claim:** If a relation \( R \) over set \( A \) is symmetric and transitive, then \( R \) is reflexive.
**"Proof:"** Select any element \((a, b) \in R\). Because \( R \) is symmetric, it follows that \((b, a) \in R\). Furthermore, since \( R \) is transitive, \((a, b) \in R \land (b, a) \in R \Rightarrow (a, a) \in R\). It follows that \( R \) is reflexive.
**Discussion:** Spotting Logical Errors in the Proof
1. **Understanding Symmetry and Transitivity:**
- Symmetric means if \((a, b) \in R\), then \((b, a) \in R\).
- Transitive means if \((a, b) \in R\) and \((b, c) \in R\), then \((a, c) \in R\).
2. **Flaw in Reasoning:**
- The conclusion incorrectly assumes that because \((a, b)\) and \((b, a)\) are present, \((a, a)\) will be too.
- Reflexivity requires every \((a, a)\) to be in \( R\) for all \(a\) in set \(A\), which isn't established here.
The error lies in assuming reflexivity from a few elements being related; reflexivity must hold universally across all elements.
Expert Solution

Step 1
Step 1:-
Introduction:-
First of all, I am talking about the reflexive, symmetric, and transitive relation.
Reflexive is defined as a, b are the elements of the relation R. (a, a) and (b,b) both are in reflexive.
Symmetric defines as a, b are the elements of the relation R.
(a,b) is the element of the relation R and if (b, a) is also the element of the relation R that means R is symmetric
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