Question

A relation \(R\) is defined on \(Z\), the set of integers, as \[ R = \{(x,y) : |x-y| \text{ is divisible by a prime number } 'p', x,y \in Z\}. \] Check whether \(R\) is an equivalence relation or not.

Hint:

Modulo arithmetic rules: If \( a \equiv b \pmod{p} \) and \( b \equiv c \pmod{p} \), then \( a \equiv c \pmod{p} \).
0 is divisible by every non-zero integer.

Answer 1

Concept:

• Equivalence Relation: A relation is an equivalence relation if it is reflexive, symmetric, and transitive.
• Divisibility: \( a \) is divisible by \( p \) if \( a = kp \) for some integer \( k \).

Step 1: Check for Reflexivity
For any \( x \in Z \), \( |x - x| = 0 \).
Since \( 0 = 0 \cdot p \), where \( 0 \) is an integer, \( 0 \) is divisible by \( p \).
Thus, \( (x, x) \in R \). \( R \) is reflexive.

Step 2: Check for Symmetry
Let \( (x, y) \in R \). Then \( |x - y| \) is divisible by \( p \).
Since \( |x - y| = |y - x| \), it follows that \( |y - x| \) is also divisible by \( p \).
Thus, \( (y, x) \in R \). \( R \) is symmetric.

Step 3: Check for Transitivity
Let \( (x, y) \in R \) and \( (y, z) \in R \).
Then \( |x - y| = k_1 p \) and \( |y - z| = k_2 p \) for integers \( k_1, k_2 \geq 0 \).
This implies \( x - y = \pm k_1 p \) and \( y - z = \pm k_2 p \).
Adding the equations: \( x - z = (x - y) + (y - z) = (\pm k_1 \pm k_2) p \).
Let \( K = \pm k_1 \pm k_2 \), which is an integer. Then \( x - z = Kp \).
Taking magnitude: \( |x - z| = |K| p \), which is divisible by \( p \).
Thus, \( (x, z) \in R \). \( R \) is transitive.
Since \( R \) is reflexive, symmetric, and transitive, it is an equivalence relation.