Question

A relation \( R \) is defined on \( \mathbb{N} \times \mathbb{N} \) (where \( \mathbb{N} \) is the set of natural numbers) as: \[ (a, b) \, R \, (c, d) \iff a - c = b - d. \] Show that \( R \) is an equivalence relation.

Hint:

To prove a relation is an equivalence relation, carefully verify each property: reflexivity, symmetry, and transitivity, based on the given definition of the relation.

Answer 1

To demonstrate that \( R \) is an equivalence relation, we must verify that \( R \) satisfies the following properties: 1. Reflexivity, 2. Symmetry, 3. Transitivity.
Step 1: Reflexivity. 
Consider any \( (a, b) \in \mathbb{N} \times \mathbb{N} \). To check reflexivity, we need to verify: \[ (a, b) \, R \, (a, b). \] From the definition of \( R \), we have: \[ a - a = b - b \quad \Rightarrow \quad 0 = 0. \] Thus, \( (a, b) \, R \, (a, b) \), proving that \( R \) is reflexive. 
Step 2: Symmetry. 
Consider \( (a, b), (c, d) \in \mathbb{N} \times \mathbb{N} \). Assume: \[ (a, b) \, R \, (c, d). \] This implies: \[ a - c = b - d. \] Rearranging terms: \[ c - a = d - b. \] Hence: \[ (c, d) \, R \, (a, b). \] Therefore, \( R \) is symmetric. 
Step 3: Transitivity. 
Consider \( (a, b), (c, d), (e, f) \in \mathbb{N} \times \mathbb{N} \). Assume: \[ (a, b) \, R \, (c, d) \quad \text{and} \quad (c, d) \, R \, (e, f). \] From the definition of \( R \), we know: \[ a - c = b - d \quad \text{and} \quad c - e = d - f. \] Adding these two equations: \[ (a - c) + (c - e) = (b - d) + (d - f) \quad \Rightarrow \quad a - e = b - f. \] Thus: \[ (a, b) \, R \, (e, f). \] Therefore, \( R \) is transitive. 
Conclusion: 
Since \( R \) satisfies reflexivity, symmetry, and transitivity, we conclude that \( R \) is an equivalence relation: \[ \boxed{\text{R is an equivalence relation.}} \]