Question

A relation \( R \) on set \( A = \{-4, -3, -2, -1, 0, 1, 2, 3, 4\} \) is defined as: \[ R = \{(x, y) : x + y \text{ is an integer divisible by 2}\}. \]

Show that \( R \) is an equivalence relation. Also, write the equivalence class \([2]\).

Hint:

To verify an equivalence relation, check reflexivity, symmetry, and transitivity. For equivalence classes, solve the defining condition for all elements of the set.

Answer 1

1. Reflexivity: For any \( x \in A \), \( x + x = 2x \), which is divisible by 2. Hence, \( (x, x) \in R \). Therefore, \( R \) is reflexive. 
2. Symmetry: If \( (x, y) \in R \), then \( x + y \) is divisible by 2. This implies \( y + x \) is also divisible by 2 (since addition is commutative). Hence, \( (y, x) \in R \). Therefore, \( R \) is symmetric. 
3. Transitivity: If \( (x, y) \in R \) and \( (y, z) \in R \), then \( x + y \) and \( y + z \) are both divisible by 2. Adding these equations: \[ (x + y) + (y + z) = x + 2y + z. \] Since \( 2y \) is divisible by 2, \( x + z \) is also divisible by 2. Hence, \( (x, z) \in R \). Therefore, \( R \) is transitive. Since \( R \) is reflexive, symmetric, and transitive, \( R \) is an equivalence relation.
4. Equivalence class \([2]\): The equivalence class of \( 2 \), \([2]\), includes all elements \( y \in A \) such that \( (2, y) \in R \). This means \( 2 + y \) is divisible by 2: \[ 2 + y \equiv 0 \pmod{2}. \] Thus, \( y \) must also be even. The even elements in \( A \) are: \[ [2] = \{-4, -2, 0, 2, 4\}. \] 
Final Answer: \( R \) is an equivalence relation. The equivalence class \([2] = \{-4, -2, 0, 2, 4\}\).