Step 1: Understanding the Concept:
To classify a relation on the set of real numbers \(\mathbb{R}\), we evaluate it against three standard properties:
1. Reflexivity: Does every element relate to itself? \((a, a) \in R \iff a \leq a^2\).
2. Symmetry: Does \((a, b) \in R\) imply \((b, a) \in R\)? \(a \leq b^2 \implies b \leq a^2\).
3. Transitivity: Does \((a, b) \in R\) and \((b, c) \in R\) imply \((a, c) \in R\)? \(a \leq b^2, b \leq c^2 \implies a \leq c^2\).
For a property to hold, it must be true for *every* real number. To prove it does *not* hold, we only need to provide a single counterexample.
Step 2: Key Formula or Approach:
We systematically check each property using specific values (fractions, negative numbers, and integers) to look for counterexamples.
Step 3: Detailed Explanation:
1. Reflexivity: Check if \(a \leq a^2\) for all \(a \in \mathbb{R}\).
Consider \(a = \frac{1}{2}\). Then \(a^2 = \frac{1}{4}\).
Is \(\frac{1}{2} \leq \frac{1}{4}\)? No, this is false.
Therefore, the relation is not reflexive.
2. Symmetry: Check if \(a \leq b^2 \implies b \leq a^2\).
Consider \(a = 1, b = 2\).
Here \(1 \leq 2^2\) is true (\(1 \leq 4\)), so \((1, 2) \in R\).
Now check the converse: Is \(2 \leq 1^2\)? No, \(2>1\).
Therefore, \((2, 1) \notin R\), so the relation is not symmetric.
(Thus, statement B is correct).
3. Transitivity: Check if \(a \leq b^2\) and \(b \leq c^2 \implies a \leq c^2\).
Consider \(a = 10, b = 4, c = 2\).
\(10 \leq 4^2 \implies 10 \leq 16\) (True, so \((10, 4) \in R\)).
\(4 \leq 2^2 \implies 4 \leq 4\) (True, so \((4, 2) \in R\)).
Now check \((10, 2)\): Is \(10 \leq 2^2\)? No, \(10>4\).
Therefore, the relation is not transitive.
(Thus, statement C is correct because it is neither reflexive nor transitive).
Summary: Statements (B) and (C) are correct.
Step 4: Final Answer:
The relation is not symmetric, not reflexive, and not transitive. The combination (B) and (C) is the accurate description. This corresponds to option (C).