Let `R={(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)}` be a relation the set `A= {1, 2, 3, 4}` . The relation R is (a). a function (b). reflexive (c). not symmetric (d). transitive

Options

A. a function
B. reflexive
C. not symmetric
D. transitive

Answer

To solve the problem, we will analyze the relation \( R = \{(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)\} \) with respect to the properties of functions and relations.

### Step 1: Check if R is a function
A relation \( R \) is a function if every element in the domain (set \( A \)) has a unique image in the codomain.

- The elements in set \( A \) are \( \{1, 2, 3, 4\} \).
- From the relation \( R \):
- \( 1 \) maps to \( 3 \)
- \( 2 \) maps to \( 4 \) and \( 3 \) (two images)
- \( 3 \) maps to \( 1 \)
- \( 4 \) maps to \( 2 \)

Since \( 2 \) has two images (\( 4 \) and \( 3 \)), \( R \) is **not a function**.

### Step 2: Check if R is reflexive
A relation \( R \) is reflexive if every element in \( A \) is related to itself, meaning \( (a, a) \) must be in \( R \) for all \( a \in A \).

- We need to check for the pairs \( (1, 1), (2, 2), (3, 3), (4, 4) \).
- None of these pairs are present in \( R \).

Since none of the pairs \( (1, 1), (2, 2), (3, 3), (4, 4) \) are in \( R \), \( R \) is **not reflexive**.

### Step 3: Check if R is symmetric
A relation \( R \) is symmetric if for every \( (a, b) \in R \), the pair \( (b, a) \) is also in \( R \).

- Check the pairs:
- \( (1, 3) \) is in \( R \), and \( (3, 1) \) is also in \( R \) (symmetric).
- \( (4, 2) \) is in \( R \), but \( (2, 4) \) is also in \( R \) (symmetric).
- \( (2, 4) \) is in \( R \), and \( (4, 2) \) is also in \( R \) (symmetric).
- \( (2, 3) \) is in \( R \), but \( (3, 2) \) is not in \( R \) (not symmetric).

Since \( (2, 3) \) does not have \( (3, 2) \) in \( R \), the relation \( R \) is **not symmetric**.

### Step 4: Check if R is transitive
A relation \( R \) is transitive if whenever \( (a, b) \in R \) and \( (b, c) \in R \), then \( (a, c) \) must also be in \( R \).

- Check the pairs:
- From \( (1, 3) \) and \( (3, 1) \), we should have \( (1, 1) \) which is not in \( R \).
- From \( (2, 3) \) and \( (3, 1) \), we should have \( (2, 1) \) which is not in \( R \).
- From \( (2, 4) \) and \( (4, 2) \), we should have \( (2, 2) \) which is not in \( R \).
- From \( (4, 2) \) and \( (2, 3) \), we should have \( (4, 3) \) which is not in \( R \).

Since there are several cases where transitivity fails, the relation \( R \) is **not transitive**.

### Conclusion
- \( R \) is **not a function**.
- \( R \) is **not reflexive**.
- \( R \) is **not symmetric**.
- \( R \) is **not transitive**.