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
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Chapter1: Introduction
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Struggling on this problem, I can see that it's incorrect and am close but I'm overlooking the flaw in reasoning somewhere.

**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.
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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