Consider the non-empty set consisting of children in a family and a relation R defined as `aRb`, if `a` is brother of `b`. Then, R is

Options

A. symmetric but not transitive
B. transitive but not symmetric
C. neither symmetric nor transitive
D. both symmetric and transitive

Explanation

To analyze the relation \( R \) defined as \( aRb \) if \( a \) is the brother of \( b \), we will check whether this relation is symmetric, reflexive, or transitive.

### Step 1: Check for Symmetry
A relation \( R \) is symmetric if whenever \( aRb \) holds, then \( bRa \) also holds.

- Given \( aRb \), we know \( a \) is the brother of \( b \).
- If \( a \) is a brother to \( b \), \( b \) can either be a boy or a girl.
- If \( b \) is a girl, then \( b \) is the sister of \( a \), and hence \( bRa \) does not hold because \( b \) is not a brother of \( a \).
- Therefore, the relation is **not symmetric**.

### Step 2: Check for Reflexivity
A relation \( R \) is reflexive if for every element \( a \), \( aRa \) holds.

- For any child \( a \), \( a \) cannot be his own brother.
- Thus, \( aRa \) does not hold for any child \( a \).
- Therefore, the relation is **not reflexive**.

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

- Suppose \( aRb \) (meaning \( a \) is the brother of \( b \)) and \( bRc \) (meaning \( b \) is the brother of \( c \)).
- This implies that \( a \), \( b \), and \( c \) are siblings (children of the same parents).
- Since \( a \) is a brother to \( b \) and \( b \) is a brother to \( c \), it follows that \( a \) must also be a brother to \( c \).
- Therefore, \( aRc \) holds, and the relation is **transitive**.

### Conclusion
The relation \( R \) defined as \( aRb \) if \( a \) is the brother of \( b \) is **not symmetric**, **not reflexive**, but it is **transitive**.

### Final Answer
The relation \( R \) is only transitive.

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