Certainly! Let's go through the proof using \(x\) and \(y\) instead of \(a\) and \(b\).

### Problem Restatement:

**Consider the set \(\mathbb{Z}\) of integers and an integer \(m > 1\). We say that \(x\) is congruent to \(y\) modulo \(m\), written \(x \equiv y \ (\text{mod} \ m)\), if \(m\) divides \(x - y\). Show that this defines an equivalence relation on \(\mathbb{Z}\).**

### Solution:

To show that the relation \(x \equiv y \ (\text{mod} \ m)\) is an equivalence relation, we need to verify the three properties of equivalence relations: reflexivity, symmetry, and transitivity.

1. **Reflexivity**:
   - For any integer \(x \in \mathbb{Z}\), we need to show that \(x \equiv x \ (\text{mod} \ m)\).
   - \(x \equiv x \ (\text{mod} \ m)\) means that \(m\) divides \(x - x\).
   - Since \(x - x = 0\), and \(m\) divides 0, this property holds true for any \(x\).
   - Therefore, the relation is reflexive.

2. **Symmetry**:
   - We need to show that if \(x \equiv y \ (\text{mod} \ m)\), then \(y \equiv x \ (\text{mod} \ m)\).
   - If \(x \equiv y \ (\text{mod} \ m)\), it means that \(m\) divides \(x - y\), i.e., \(x - y = km\) for some integer \(k\).
   - This can be rewritten as \(y - x = -km\), and since \(-k\) is also an integer, \(m\) divides \(y - x\).
   - Therefore, \(y \equiv x \ (\text{mod} \ m)\), so the relation is symmetric.

3. **Transitivity**:
   - We need to show that if \(x \equiv y \ (\text{mod} \ m)\) and \(y \equiv z \ (\text{mod} \ m)\), then \(x \equiv z \ (\text{mod} \ m)\).
   - If \(x \equiv y \ (\text{mod} \ m)\), then \(m\) divides \(x - y\), i.e., \(x - y = km\) for some integer \(k\).
   - If \(y \equiv z \ (\text{mod} \ m)\), then \(m\) also divides \(y - z\), i.e., \(y - z = lm\) for some integer \(l\).
   - Adding these two equations gives \( (x - y) + (y - z) = km + lm = (k + l)m \), which implies \(x - z = (k + l)m\).
   - Since \(k + l\) is an integer, \(m\) divides \(x - z\), and thus \(x \equiv z \ (\text{mod} \ m)\).
   - Therefore, the relation is transitive.

Since the relation \(x \equiv y \ (\text{mod} \ m)\) satisfies reflexivity, symmetry, and transitivity, it is an equivalence relation on the set \(\mathbb{Z}\).

### Summary:
The relation \(x \equiv y \ (\text{mod} \ m)\) is an equivalence relation on the set of integers \(\mathbb{Z}\) because it satisfies the three necessary properties: reflexivity, symmetry, and transitivity.

Would you like more details or have any questions?

### Related Questions:
1. What are some real-life applications of equivalence relations?
2. How does modular arithmetic play a role in computer science?
3. Can you provide an example of an equivalence class under this relation?
4. How does the concept of congruence modulo \(m\) relate to cyclic groups?
5. How is the Chinese Remainder Theorem related to modular arithmetic?

### Tip:
When working with equivalence relations, always verify the three properties (reflexivity, symmetry, transitivity) as they are key to ensuring the relation behaves as expected in various mathematical contexts.

Explore how modular arithmetic defines equivalence relations on integers. This detailed explanation covers reflexivity, symmetry, and transitivity using examples and proofs. Suitable for advanced high school students and anyone interested in mathematical relations.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation