A relation R is defined on R by a R b if a – b e Z. Prove that R is an equivalence relation and determine the equivalence classes [1/2] and [v2].
A relation R is defined on R by a R b if a – b e Z. Prove that R is an equivalence relation and determine the equivalence classes [1/2] and [v2].
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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![**Title: Understanding Equivalence Relations**
**Introduction**
In mathematics, an equivalence relation is a binary relation that is reflexive, symmetric, and transitive. This concept is fundamental in set theory and has applications across various branches of mathematics.
**Problem Statement**
A relation \( R \) is defined on the set of real numbers \( \mathbb{R} \) by \( a \, R \, b \) if \( a - b \in \mathbb{Z} \), where \( \mathbb{Z} \) represents the set of all integers. The task is to prove that \( R \) is an equivalence relation and determine the equivalence classes \([1/2]\) and \([\sqrt{2}]\).
**Solution**
1. **Reflexivity:**
For any real number \( a \), \( a - a = 0 \), and 0 is an integer. Therefore, \( a \, R \, a \), validating reflexivity.
2. **Symmetry:**
Assume \( a \, R \, b \); then \( a - b \in \mathbb{Z} \). This implies \( b - a = -(a - b) \), which is also an integer. Thus, \( b \, R \, a \), proving symmetry.
3. **Transitivity:**
Suppose \( a \, R \, b \) and \( b \, R \, c \); thus, \( a - b \) and \( b - c \) are both integers. Therefore, \( (a - b) + (b - c) = a - c \) is an integer, so \( a \, R \, c \). This establishes transitivity.
Since the relation \( R \) is reflexive, symmetric, and transitive, it is an equivalence relation.
**Determining Equivalence Classes**
- **Equivalence class \([1/2]\):**
The equivalence class of \( 1/2 \) is all numbers \( x \in \mathbb{R} \) such that \( x - 1/2 \in \mathbb{Z} \). Therefore, \([1/2] = \{ x \mid x = 1/2 + n, \, n \in \mathbb{Z} \}\).
- **Equivalence class \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffa923b6f-81dd-482c-8885-6de6bc295751%2Fc541ef79-0c44-46cb-bbfe-ba14d489e1d9%2F5fqv9u_processed.png&w=3840&q=75)
Transcribed Image Text:**Title: Understanding Equivalence Relations**
**Introduction**
In mathematics, an equivalence relation is a binary relation that is reflexive, symmetric, and transitive. This concept is fundamental in set theory and has applications across various branches of mathematics.
**Problem Statement**
A relation \( R \) is defined on the set of real numbers \( \mathbb{R} \) by \( a \, R \, b \) if \( a - b \in \mathbb{Z} \), where \( \mathbb{Z} \) represents the set of all integers. The task is to prove that \( R \) is an equivalence relation and determine the equivalence classes \([1/2]\) and \([\sqrt{2}]\).
**Solution**
1. **Reflexivity:**
For any real number \( a \), \( a - a = 0 \), and 0 is an integer. Therefore, \( a \, R \, a \), validating reflexivity.
2. **Symmetry:**
Assume \( a \, R \, b \); then \( a - b \in \mathbb{Z} \). This implies \( b - a = -(a - b) \), which is also an integer. Thus, \( b \, R \, a \), proving symmetry.
3. **Transitivity:**
Suppose \( a \, R \, b \) and \( b \, R \, c \); thus, \( a - b \) and \( b - c \) are both integers. Therefore, \( (a - b) + (b - c) = a - c \) is an integer, so \( a \, R \, c \). This establishes transitivity.
Since the relation \( R \) is reflexive, symmetric, and transitive, it is an equivalence relation.
**Determining Equivalence Classes**
- **Equivalence class \([1/2]\):**
The equivalence class of \( 1/2 \) is all numbers \( x \in \mathbb{R} \) such that \( x - 1/2 \in \mathbb{Z} \). Therefore, \([1/2] = \{ x \mid x = 1/2 + n, \, n \in \mathbb{Z} \}\).
- **Equivalence class \
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