Question

The region R on the set \(A=\{x \in Z:0\leq x\leq 12\}\) , given by \(R=\{(a,b):|a-b|\)  is a multiple of 4 \(\}\)  is :

• A. Reflexive but not Symmetric
• B. Reflexive but not Transitive
• C. Symmetric but not Transitive
• D. Equivalence relation

Answer 1

The Correct Option is D

The given set \( A = \{ x \in Z : 0 \leq x \leq 12 \} \) consists of integers from 0 to 12. The relation \( R = \{ (a, b) : |a-b| \text{ is a multiple of 4} \} \) is defined on this set. We need to determine whether this relation is an equivalence relation by checking if it is reflexive, symmetric, and transitive.

Reflexive: A relation is reflexive if every element is related to itself. For any \( a \in A \), \( |a-a| = 0 \) is a multiple of 4. Thus, \( (a, a) \in R \) for all \( a \). Hence, \( R \) is reflexive.

Symmetric: A relation is symmetric if \( (a, b) \in R \) implies \( (b, a) \in R \). If \( |a-b| \) is a multiple of 4, then clearly \( |b-a| \) (which is also \( |a-b| \)) is a multiple of 4. Thus, if \( (a, b) \in R \), then \( (b, a) \in R \). Hence, \( R \) is symmetric.

Transitive: A relation is transitive if \( (a, b) \in R \) and \( (b, c) \in R \) implies \( (a, c) \in R \). Assume \( |a-b| \) and \( |b-c| \) are multiples of 4. Thus, there exist integers \( n_1, n_2 \) such that \( |a-b| = 4n_1 \) and \( |b-c| = 4n_2 \). Then \( |a-c| \leq |a-b| + |b-c| = 4(n_1 + n_2) \), also a multiple of 4. Therefore, \( (a, c) \in R \). Hence, \( R \) is transitive.

Since \( R \) is reflexive, symmetric, and transitive, it is an equivalence relation.