Let `R` be a relation on the set of integers given by `aRb<=> a= 2^k.b` for some integer `k.` Then, `R` is

Options

A. an equivalence relation
B. reflexive but not symmetric
C. reflexive and transitive but not symmetric
D. reflexive and symmetric but not transitive.

Explanation

To determine the nature of the relation \( R \) defined on the set of integers by \( aRb \) if and only if \( a = 2^k \cdot b \) for some integer \( k \), we will check if the relation is reflexive, symmetric, and transitive.

### Step 1: Check Reflexivity
A relation \( R \) is reflexive if for every element \( a \) in the set, \( aRa \) holds true.

- For any integer \( a \), we can express \( a \) as:
\[ a = 2^0 \cdot a \]
Here, \( k = 0 \) is an integer. Thus, \( aRa \) holds for all integers \( a \).

**Conclusion**: The relation \( R \) is reflexive.

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

- Assume \( aRb \) is true, which means:
\[ a = 2^k \cdot b \]
for some integer \( k \). We can rearrange this to express \( b \) in terms of \( a \):
\[ b = 2^{-k} \cdot a \]
Since \( -k \) is also an integer, we can say:
\[ bRa \]
holds true.

**Conclusion**: The relation \( R \) is symmetric.

### Step 3: Check Transitivity
A relation \( R \) is transitive if whenever \( aRb \) and \( bRc \) hold true, then \( aRc \) must also hold true.

- Assume \( aRb \) and \( bRc \):
\[ a = 2^k \cdot b \quad \text{(1)} \]
\[ b = 2^m \cdot c \quad \text{(2)} \]
for some integers \( k \) and \( m \). Substituting equation (2) into equation (1):
\[ a = 2^k \cdot (2^m \cdot c) = 2^{k+m} \cdot c \]
Since \( k + m \) is an integer, we have:
\[ aRc \]

**Conclusion**: The relation \( R \) is transitive.

### Final Conclusion
Since the relation \( R \) is reflexive, symmetric, and transitive, it satisfies all the properties of an equivalence relation.

Thus, the final answer is that \( R \) is an equivalence relation.

---