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 \( x \in A \), \( x + x = 2x \). Since \( 2x \) is always divisible by 2, \( (x, x) \in R \). Thus, \( R \) is reflexive.
2. Symmetry: If \( (x, y) \in R \), then \( x + y \) is divisible by 2. By the commutative property of addition, \( y + x \) is also divisible by 2. Therefore, \( (y, x) \in R \), proving \( R \) is symmetric.
3. Transitivity: Given \( (x, y) \in R \) and \( (y, z) \in R \), it follows that \( x + y \) and \( y + z \) are both divisible by 2. Summing these divisibility conditions yields \( (x + y) + (y + z) = x + 2y + z \). As \( 2y \) is divisible by 2, for the sum to be divisible by 2, \( x + z \) must also be divisible by 2. Hence, \( (x, z) \in R \), confirming \( R \) is transitive. Because \( R \) satisfies reflexivity, symmetry, and transitivity, it is an equivalence relation.
4. Equivalence class \([2]\): The equivalence class of \( 2 \), denoted as \([2]\), comprises all elements \( y \in A \) for which \( (2, y) \in R \). This condition implies that \( 2 + y \) is divisible by 2, or \( 2 + y \equiv 0 \pmod{2} \). Consequently, \( y \) must be an even number. The even elements within set \( A \) are: \[ [2] = \{-4, -2, 0, 2, 4\}. \]
Final Answer: \( R \) is an equivalence relation. The equivalence class \([2] = \{-4, -2, 0, 2, 4\}\).