Question

Let $R = \{(a,b): a \leq b^2\}$ be a relation on the set of all real numbers. Then $R$ is

• A. symmetric but not transitive
• B. transitive but not symmetric
• C. both symmetric and transitive
• D. neither symmetric nor transitive
• E. having finite range

Hint:

To disprove properties, a single counterexample is sufficient.

Answer 1

The Correct Option is D

Concept:
• Symmetric: $(a,b)\in R \Rightarrow (b,a)\in R$
• Transitive: $(a,b)\in R$ and $(b,c)\in R \Rightarrow (a,c)\in R$

Step 1: Check symmetry.
Given: \[ (a,b)\in R \Rightarrow a \leq b^2 \] For symmetry, we need: \[ (b,a)\in R \Rightarrow b \leq a^2 \] Counterexample:
Take $a=1$, $b=2$: \[ 1 \leq 4 \Rightarrow (1,2)\in R \] But, \[ 2 \leq 1^2 = 1 \text{false} \] Thus, not symmetric.

Step 2: Check transitivity.
Suppose: \[ a \leq b^2, b \leq c^2 \] We need: \[ a \leq c^2 \] Counterexample:
Let $a=4$, $b=2$, $c=1$: \[ 4 \leq 4 \text{true} \] \[ 2 \leq 1 \text{false} \] Try another: \[ a=1, b=1, c=0 \] \[ 1 \leq 1 \text{true} \] \[ 1 \leq 0 \text{false} \] Try valid pair: \[ a=1, b=1, c=1 \] Then works, but not generally. Thus, transitivity fails in general.
Final Answer: \[ \boxed{\text{neither symmetric nor transitive}} \]