Question:

Prove that a relation \( R = \{(a, b) : (a-b) \) is a multiple of \( 5 \} \) is an equivalence relation in the set of integers \( \mathbb{Z} \).

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To prove equivalence relations, check reflexivity, symmetry, and transitivity systematically.
Updated On: Mar 3, 2025
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Solution and Explanation

Reflexivity:  To prove \( R \) is an equivalence relation, we verify reflexivity, symmetry, and transitivity: 
 

Symmetry: For any \( a \in \mathbb{Z} \), \( a - a = 0 \), which is a multiple of \( 5 \). Thus, \( (a, a) \in R \). 
If \( (a, b) \in R \), then \( a - b = 5k \) for some \( k \in \mathbb{Z} \). 
 

Transitivity: This implies \( b - a = -5k \), which is also a multiple of \( 5 \). Thus, \( (b, a) \in R \).   If \( (a, b) \in R \) and \( (b, c) \in R \), then \( a - b = 5k \) and \( b - c = 5m \) for \( k, m \in \mathbb{Z} \).
Adding these, \( a - c = 5(k + m) \), which is a multiple of \( 5 \). Thus, \( (a, c) \in R \). Therefore, \( R \) is an equivalence relation.

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