Question

Which of the following is not correct for relation \( R \) on the set of real numbers?

• A. \( (x, y) \in R \iff |x - y| \leq 1 \) is reflexive and symmetric.
• B. \( (x, y) \in R \iff |x - |y|| \leq 1 \) is reflexive but not symmetric.
• C. \( (x, y) \in R \iff 0<|x - y| \leq 1 \) is neither transitive nor symmetric.
• D. \( (x, y) \in R \iff 0<|x - y| \leq 1 \) is symmetric and transitive.

Hint:

Relations involving absolute differences like \( |x - y| \leq k \) are almost never transitive on real numbers, as small differences can accumulate to exceed the threshold \( k \).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A relation \( R \) is reflexive if \( xRx \), symmetric if \( xRy \implies yRx \), and transitive if \( xRy \) and \( yRz \implies xRz \).
Step 2: Detailed Explanation:
Let's analyze the relation given in (C) and (D): \( R = \{ (x, y) : 0<|x - y| \leq 1 \} \).
1. Symmetry: If \( 0<|x - y| \leq 1 \), then since \( |x - y| = |y - x| \), we have \( 0<|y - x| \leq 1 \). Thus, it is symmetric.
2. Transitivity: Let \( x = 0.8, y = 0, z = -0.8 \).
- \( |x - y| = |0.8 - 0| = 0.8 \), which satisfies \( 0<0.8 \leq 1 \). So \( xRy \).
- \( |y - z| = |0 - (-0.8)| = 0.8 \), which satisfies \( 0<0.8 \leq 1 \). So \( yRz \).
- However, \( |x - z| = |0.8 - (-0.8)| = 1.6 \). This does not satisfy \( \leq 1 \). So \( (x, z) \notin R \).
Therefore, the relation is not transitive.
3. Checking Option (D): Since it incorrectly claims the relation is transitive, option (D) is the statement that is "not correct".
Step 3: Final Answer:
Statement (D) is incorrect.