Question

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

Hint:

To prove that a relation is an equivalence relation, check that it satisfies reflexivity, symmetry, and transitivity.

Answer 1

To prove that the relation \( R \) is an equivalence relation, we need to show that it satisfies three properties: 1. Reflexivity
2. Symmetry
3. Transitivity
1. Reflexivity:
A relation \( R \) is reflexive if \( (a, a) \in R \) for every \( a \in \mathbb{Z} \).
Since \( a - a = 0 \), and 0 is a multiple of 5, we have \( (a, a) \in R \).
Thus, \( R \) is reflexive. 2. Symmetry:
A relation \( R \) is symmetric if whenever \( (a, b) \in R \), we also have \( (b, a) \in R \).
If \( (a - b) \) is a multiple of 5, then \( (b - a) = -(a - b) \), which is also a multiple of 5.
Hence, if \( (a, b) \in R \), then \( (b, a) \in R \), and thus \( R \) is symmetric. 3. Transitivity:
A relation \( R \) is transitive if whenever \( (a, b) \in R \) and \( (b, c) \in R \), we also have \( (a, c) \in R \).
If \( (a - b) \) is a multiple of 5, and \( (b - c) \) is a multiple of 5, then: \[ (a - c) = (a - b) + (b - c), \] which is the sum of two multiples of 5, and thus is also a multiple of 5.
Hence, if \( (a, b) \in R \) and \( (b, c) \in R \), we have \( (a, c) \in R \), and thus \( R \) is transitive. Conclusion:
Since the relation \( R \) is reflexive, symmetric, and transitive, we conclude that \( R \) is an equivalence relation.