Let `R={(x,y):x,yepsilonA,x+y=5}` where `A={1,2,3,4,5}` then `R` is

Options

A. Reflexive
B. symmetric
C. Transitive
D. Equivalence

Explanation

To determine the nature of the relation \( R = \{(x, y) : x, y \in A, x + y = 5\} \) where \( A = \{1, 2, 3, 4, 5\} \), we will check if the relation is reflexive, symmetric, and transitive.

### Step 1: Identify the pairs in the relation \( R \)

First, we need to find all pairs \( (x, y) \) such that \( x + y = 5 \) and both \( x \) and \( y \) belong to the set \( A \).

- For \( x = 1 \): \( y = 5 - 1 = 4 \) → pair is \( (1, 4) \)
- For \( x = 2 \): \( y = 5 - 2 = 3 \) → pair is \( (2, 3) \)
- For \( x = 3 \): \( y = 5 - 3 = 2 \) → pair is \( (3, 2) \)
- For \( x = 4 \): \( y = 5 - 4 = 1 \) → pair is \( (4, 1) \)
- For \( x = 5 \): \( y = 5 - 5 = 0 \) → not in \( A \)

Thus, the relation \( R \) consists of the pairs:
\[ R = \{(1, 4), (2, 3), (3, 2), (4, 1)\} \]

### Step 2: Check for Reflexivity

A relation is reflexive if every element \( x \) in \( A \) relates to itself, i.e., \( (x, x) \in R \) for all \( x \in A \).

- Check \( (1, 1) \): \( 1 + 1 = 2 \neq 5 \)
- Check \( (2, 2) \): \( 2 + 2 = 4 \neq 5 \)
- Check \( (3, 3) \): \( 3 + 3 = 6 \neq 5 \)
- Check \( (4, 4) \): \( 4 + 4 = 8 \neq 5 \)
- Check \( (5, 5) \): \( 5 + 5 = 10 \neq 5 \)

Since none of the pairs \( (x, x) \) are in \( R \), the relation is **not reflexive**.

### Step 3: Check for Symmetry

A relation is symmetric if for every \( (x, y) \in R \), the pair \( (y, x) \) is also in \( R \).

- Check \( (1, 4) \): \( (4, 1) \in R \)
- Check \( (2, 3) \): \( (3, 2) \in R \)
- Check \( (3, 2) \): \( (2, 3) \in R \)
- Check \( (4, 1) \): \( (1, 4) \in R \)

Since for every pair \( (x, y) \in R \), the pair \( (y, x) \) is also in \( R \), the relation is **symmetric**.

### Step 4: Check for Transitivity

A relation is transitive if whenever \( (x, y) \in R \) and \( (y, z) \in R \), then \( (x, z) \) must also be in \( R \).

- Check \( (2, 3) \) and \( (3, 2) \): Here \( x = 2, y = 3, z = 2 \) → \( (2, 2) \notin R \)
- Check \( (3, 2) \) and \( (2, 3) \): Here \( x = 3, y = 2, z = 3 \) → \( (3, 3) \notin R \)

Since there are cases where \( (x, y) \in R \) and \( (y, z) \in R \) do not lead to \( (x, z) \in R \), the relation is **not transitive**.

### Conclusion

The relation \( R \) is symmetric but not reflexive and not transitive. Therefore, the final answer is that \( R \) is a **symmetric relation**.

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