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 equivalence relations, carefully verify each property (reflexivity, symmetry, transitivity) using the definition of the relation.

Answer 1

To prove that \( R \) is an equivalence relation, we need to show that \( R \) satisfies: 1. Reflexivity, 2. Symmetry, 3. Transitivity. Step 1: Prove reflexivity.
For any \( (a, b) \in \mathbb{N} \times \mathbb{N} \), check if: \[ (a, b) \, R \, (a, b). \] From the definition of \( R \): \[ a - a = b - b \quad \Rightarrow \quad 0 = 0. \] Thus, \( (a, b) \, R \, (a, b) \), and \( R \) is reflexive. Step 2: Prove symmetry.
For \( (a, b), (c, d) \in \mathbb{N} \times \mathbb{N} \), assume: \[ (a, b) \, R \, (c, d). \] This implies: \[ a - c = b - d. \] Rearranging: \[ c - a = d - b. \] Thus: \[ (c, d) \, R \, (a, b). \] Hence, \( R \) is symmetric. Step 3: Prove transitivity.
For \( (a, b), (c, d), (e, f) \in \mathbb{N} \times \mathbb{N} \), assume: \[ (a, b) \, R \, (c, d) \quad \text{and} \quad (c, d) \, R \, (e, f). \] From the definition of \( R \): \[ a - c = b - d \quad \text{and} \quad c - e = d - f. \] Adding these equations: \[ (a - c) + (c - e) = (b - d) + (d - f) \quad \Rightarrow \quad a - e = b - f. \] Thus: \[ (a, b) \, R \, (e, f). \] Hence, \( R \) is transitive. Conclusion:
Since \( R \) satisfies reflexivity, symmetry, and transitivity, \( R \) is an equivalence relation: \[ \boxed{\text{R is an equivalence relation.}} \]