Prove equivalence of
R= { (x,y) } € N^2 | x/y=N

To prove the equivalence of the relation \( R = \{ (x, y) \in \mathbb{N}^2 \mid \frac{x}{y} \in \mathbb{N} \} \), we need to show that this defines a valid relation on pairs \((x, y)\) where both \(x\) and \(y\) belong to the set of natural numbers, \( \mathbb{N} \), and that the ratio \( \frac{x}{y} \) results in a natural number. 

### Step-by-Step Proof:

#### 1. **Restating the Problem:**
The relation \(R\) is defined as:
\[
R = \{(x, y) \in \mathbb{N}^2 \mid \frac{x}{y} \in \mathbb{N} \}
\]
This means that for any pair \((x, y)\) in \( \mathbb{N}^2 \), the ratio \( \frac{x}{y} \) must be a natural number. Equivalently, this means that \( y \) divides \( x \) without leaving a remainder.

#### 2. **Condition for the Relation:**
The condition that \( \frac{x}{y} \in \mathbb{N} \) can be rewritten as:
\[
\exists k \in \mathbb{N} \text{ such that } x = k \cdot y
\]
This means that \( x \) is an integer multiple of \( y \), or in other words, \( y \) divides \( x \) (i.e., \( y \mid x \)).

#### 3. **Equivalence:**
The relation \( R \) can thus be equivalently defined as:
\[
R = \{ (x, y) \in \mathbb{N}^2 \mid y \mid x \}
\]
This means that \( R \) consists of all pairs \((x, y)\) such that \( y \) divides \( x \), or equivalently, the ratio \( \frac{x}{y} \) is a natural number.

#### 4. **Reflexivity:**
For reflexivity, we need to check whether \( (x, x) \in R \) for all \( x \in \mathbb{N} \).
\[
\frac{x}{x} = 1 \quad \text{which is a natural number}
\]
Thus, \( (x, x) \in R \) for all \( x \in \mathbb{N} \), so the relation is reflexive.

#### 5. **Transitivity:**
For transitivity, suppose \( (x, y) \in R \) and \( (y, z) \in R \), meaning:
\[
\frac{x}{y} \in \mathbb{N} \quad \text{and} \quad \frac{y}{z} \in \mathbb{N}
\]
This implies that \( x = k_1 y \) and \( y = k_2 z \) for some \( k_1, k_2 \in \mathbb{N} \). Substituting, we get:
\[
x = k_1 (k_2 z) = (k_1 k_2) z
\]
Since \( k_1 k_2 \in \mathbb{N} \), it follows that \( \frac{x}{z} \in \mathbb{N} \), so \( (x, z) \in R \). Thus, the relation is transitive.

#### 6. **Symmetry:**
The relation is not symmetric in general. If \( \frac{x}{y} \in \mathbb{N} \), this does not necessarily mean that \( \frac{y}{x} \in \mathbb{N} \). For example, if \( x = 4 \) and \( y = 2 \), then \( \frac{4}{2} = 2 \in \mathbb{N} \), but \( \frac{2}{4} = 0.5 \notin \mathbb{N} \).

Thus, the relation is not symmetric.

### Conclusion:
The relation \( R = \{ (x, y) \in \mathbb{N}^2 \mid \frac{x}{y} \in \mathbb{N} \} \) defines a relation where \( y \) divides \( x \), and it is reflexive and transitive, but not symmetric.

---
Would you like more details on the transitivity or the reflexivity aspect of the relation?

Here are 5 related questions to further your understanding:
1. How can we modify the relation to make it symmetric?
2. Can we define a similar relation in the set of integers \( \mathbb{Z} \)?
3. What is the inverse of the relation \( R \)?
4. How would this relation look in real numbers \( \mathbb{R} \)?
5. What other properties of relations, like antisymmetry, can we explore for \( R \)?

**Tip:** Transitive relations are the foundation of equivalence classes and partitions in set theory, offering powerful tools for structuring and analyzing sets.

Prove equivalence of R = { (x,y) } ∈ N^2 | x/y ∈ N

Explore the detailed proof of equivalence for the relation R = {(x, y) ∈ N^2 | x/y ∈ N}. This problem covers important concepts in number theory, including reflexivity, transitivity, and divisibility in natural numbers. Suitable for undergraduate students studying relations in mathematics.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation