Question

Relation \( R = \{(x, y) : x<y^2 \text{ where } x, y \in \mathbb{R}\} \) is:

• A. Reflexive but not symmetric
• B. Symmetric and transitive but not Reflexive
• C. Reflexive and Symmetric
• D. Neither reflexive nor symmetric nor transitive

Hint:

For relations involving powers like \( x^2 \) or \( x^3 \), always test with proper fractions (between 0 and 1) to disprove reflexivity and negative numbers to disprove transitivity.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A relation \( R \) on a set of real numbers \( \mathbb{R} \) is checked for three properties:
1. Reflexive: \( (a, a) \in R \) for every \( a \in \mathbb{R} \).
2. Symmetric: If \( (a, b) \in R \), then \( (b, a) \in R \).
3. Transitive: If \( (a, b) \in R \) and \( (b, c) \in R \), then \( (a, c) \in R \).
Step 2: Detailed Explanation:
Check for Reflexivity:
For \( R \) to be reflexive, \( x<x^2 \) must hold for all \( x \in \mathbb{R} \).
Let \( x = \frac{1}{2} \). Here, \( x^2 = \frac{1}{4} \).
Since \( \frac{1}{2}<\frac{1}{4} \) is false (\( 0.5<0.25 \) is false), the pair \( (\frac{1}{2}, \frac{1}{2}) \notin R \).
Hence, \( R \) is not reflexive.
Check for Symmetry:
For \( R \) to be symmetric, if \( x<y^2 \), then \( y<x^2 \) must be true.
Let \( x = 1 \) and \( y = 2 \). Since \( 1<2^2 \) (\( 1<4 \)), \( (1, 2) \in R \).
However, for \( (2, 1) \) to be in \( R \), \( 2<1^2 \) (\( 2<1 \)) must be true, which is false.
Hence, \( R \) is not symmetric.
Check for Transitivity:
For \( R \) to be transitive, if \( x<y^2 \) and \( y<z^2 \), then \( x<z^2 \) must be true.
Let \( x = 3 \), \( y = 2 \), and \( z = 1.5 \).
\( 3<2^2 \implies 3<4 \) (True).
\( 2<(1.5)^2 \implies 2<2.25 \) (True).
But \( 3<(1.5)^2 \implies 3<2.25 \) (False).
Hence, \( R \) is not transitive.
Step 3: Final Answer:
Since the relation fails all three properties, it is neither reflexive nor symmetric nor transitive.