Let us define a relation R in R as aRb if `a ge b`. Then, R is

Options

A. an equivalence relation
B. reflexive, transitive but not symmetric
C. symmetric, transitive but not reflexive
D. neither transitive nor reflexive but symmetric

Explanation

To determine the properties of the relation R defined on the set of real numbers \( R \) such that \( aRb \) if \( a \geq b \), we will analyze the relation for reflexivity, symmetry, and transitivity step by step.

### Step 1: Check Reflexivity
A relation is reflexive if every element is related to itself. In mathematical terms, for all \( a \in R \), \( aRa \) must hold true.

- Here, we check if \( a \geq a \).
- Since this is true for all real numbers (as any number is equal to itself), the relation is reflexive.

**Conclusion**: The relation R is reflexive.

### Step 2: Check Symmetry
A relation is symmetric if for all \( a, b \in R \), whenever \( aRb \) holds, then \( bRa \) must also hold.

- We check if \( a \geq b \) implies \( b \geq a \).
- For example, let \( a = 3 \) and \( b = 1 \). Here, \( 3 \geq 1 \) is true, but \( 1 \geq 3 \) is false.
- Since we found a counterexample, the relation is not symmetric.

**Conclusion**: The relation R is not symmetric.

### Step 3: Check Transitivity
A relation is transitive if for all \( a, b, c \in R \), whenever \( aRb \) and \( bRc \) hold, then \( aRc \) must also hold.

- We check if \( a \geq b \) and \( b \geq c \) implies \( a \geq c \).
- For example, let \( a = 2 \), \( b = 1 \), and \( c = 0 \). Here, \( 2 \geq 1 \) and \( 1 \geq 0 \) are both true, and indeed \( 2 \geq 0 \) is also true.
- This holds for any real numbers, so the relation is transitive.

**Conclusion**: The relation R is transitive.

### Final Conclusion
Based on the analysis:
- The relation R is reflexive.
- The relation R is not symmetric.
- The relation R is transitive.

Thus, the relation R is **reflexive and transitive but not symmetric**.

**Answer**: Option B: Reflexive, Transitive but not Symmetric.

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