1. R is the relation defined on Z as follows: Vm, n € Z, mRn 5|(m² — n²). (a). Prove that R is an equivalence relation. (b). List all the distinct equivalence classes.
1. R is the relation defined on Z as follows: Vm, n € Z, mRn 5|(m² — n²). (a). Prove that R is an equivalence relation. (b). List all the distinct equivalence classes.
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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Discrete Math Please solve a and b
![### Equivalence Relations Exercise
**Problem Statement:**
1. **R is the relation** defined on \(\mathbb{Z}\) (the set of all integers) as follows:
\[\forall m, n \in \mathbb{Z}, \, mRn \iff 5 | (m^2 - n^2).\]
(a) Prove that \(R\) is an *equivalence relation*.
(b) List all the distinct *equivalence classes*.
---
**Explanation:**
1. **Defining the Relation:**
- The relation \(R\) on the set of all integers \(\mathbb{Z}\) relates two integers \(m\) and \(n\) if and only if the difference of their squares, \(m^2 - n^2\), is divisible by 5.
2. **Proving Equivalence Relation:**
- To prove that \(R\) is an equivalence relation, we need to show that \(R\) is reflexive, symmetric, and transitive.
- **Reflexive:** For all \(m \in \mathbb{Z}\), \(mRm\). This means that \(m^2 - m^2 = 0\) is divisible by 5, which is true since 0 is divisible by any integer.
- **Symmetric:** For all \(m, n \in \mathbb{Z}\), if \(mRn\), then \(nRm\). If \(5 | (m^2 - n^2)\), then \(5 | (n^2 - m^2)\) because \(m^2 - n^2 = -(n^2 - m^2)\). Divisibility is preserved under negation.
- **Transitive:** For all \(m, n, p \in \mathbb{Z}\), if \(mRn\) and \(nRp\), then \(mRp\). If \(5 | (m^2 - n^2)\) and \(5 | (n^2 - p^2)\), then:
\[
5 | [(m^2 - n^2) + (n^2 - p^2)] \implies 5 | (m^2 - p^2).
\]
3. **Listing Equivalence Classes:**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd7e504e0-f273-40ab-8020-eddc4709afa9%2F404ea0f2-c625-49a6-9bb6-842b193b1569%2F28lu1xjn_processed.png&w=3840&q=75)
Transcribed Image Text:### Equivalence Relations Exercise
**Problem Statement:**
1. **R is the relation** defined on \(\mathbb{Z}\) (the set of all integers) as follows:
\[\forall m, n \in \mathbb{Z}, \, mRn \iff 5 | (m^2 - n^2).\]
(a) Prove that \(R\) is an *equivalence relation*.
(b) List all the distinct *equivalence classes*.
---
**Explanation:**
1. **Defining the Relation:**
- The relation \(R\) on the set of all integers \(\mathbb{Z}\) relates two integers \(m\) and \(n\) if and only if the difference of their squares, \(m^2 - n^2\), is divisible by 5.
2. **Proving Equivalence Relation:**
- To prove that \(R\) is an equivalence relation, we need to show that \(R\) is reflexive, symmetric, and transitive.
- **Reflexive:** For all \(m \in \mathbb{Z}\), \(mRm\). This means that \(m^2 - m^2 = 0\) is divisible by 5, which is true since 0 is divisible by any integer.
- **Symmetric:** For all \(m, n \in \mathbb{Z}\), if \(mRn\), then \(nRm\). If \(5 | (m^2 - n^2)\), then \(5 | (n^2 - m^2)\) because \(m^2 - n^2 = -(n^2 - m^2)\). Divisibility is preserved under negation.
- **Transitive:** For all \(m, n, p \in \mathbb{Z}\), if \(mRn\) and \(nRp\), then \(mRp\). If \(5 | (m^2 - n^2)\) and \(5 | (n^2 - p^2)\), then:
\[
5 | [(m^2 - n^2) + (n^2 - p^2)] \implies 5 | (m^2 - p^2).
\]
3. **Listing Equivalence Classes:**
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