Let Q be the set of rational number and R be the relation on Q defined by `R={(x,y):x,y in Q , x^(2)+3y^(2)=4xy}` check whether R is reflexive, symmetric and transitive.

Answer

To determine whether the relation \( R \) defined on the set of rational numbers \( Q \) by \( R = \{(x, y) : x, y \in Q, x^2 + 3y^2 = 4xy\} \) is reflexive, symmetric, and transitive, we will analyze each property step by step.

### Step 1: Check if \( R \) is Reflexive

A relation \( R \) is reflexive if for every element \( a \in Q \), the pair \( (a, a) \) is in \( R \). This means we need to check if:

\[ a^2 + 3a^2 = 4a \cdot a \]

Calculating both sides:

- Left-hand side: \( a^2 + 3a^2 = 4a^2 \)
- Right-hand side: \( 4a \cdot a = 4a^2 \)

Since both sides are equal, we conclude that:

\[ (a, a) \in R \quad \text{for all } a \in Q \]

Thus, \( R \) is reflexive.

### Step 2: Check if \( R \) is Symmetric

A relation \( R \) is symmetric if whenever \( (a, b) \in R \), then \( (b, a) \in R \) as well. We need to find a counterexample to prove that \( R \) is not symmetric.

Let’s take \( a = 3 \) and \( b = 1 \):

1. Check if \( (3, 1) \in R \):
\[ 3^2 + 3 \cdot 1^2 = 9 + 3 = 12 \]
\[ 4 \cdot 3 \cdot 1 = 12 \]
Thus, \( (3, 1) \in R \).

2. Now check if \( (1, 3) \in R \):
\[ 1^2 + 3 \cdot 3^2 = 1 + 27 = 28 \]
\[ 4 \cdot 1 \cdot 3 = 12 \]
Since \( 28 \neq 12 \), we conclude that \( (1, 3) \notin R \).

Thus, \( R \) is not symmetric.

### Step 3: Check if \( R \) is Transitive

A relation \( R \) is transitive if whenever \( (a, b) \in R \) and \( (b, c) \in R \), then \( (a, c) \in R \) must also hold. We will find a counterexample to show that \( R \) is not transitive.

Let’s take \( a = 9 \), \( b = 3 \), and \( c = 1 \):

1. Check if \( (9, 3) \in R \):
\[ 9^2 + 3 \cdot 3^2 = 81 + 27 = 108 \]
\[ 4 \cdot 9 \cdot 3 = 108 \]
Thus, \( (9, 3) \in R \).

2. Check if \( (3, 1) \in R \):
\[ 3^2 + 3 \cdot 1^2 = 9 + 3 = 12 \]
\[ 4 \cdot 3 \cdot 1 = 12 \]
Thus, \( (3, 1) \in R \).

3. Now check if \( (9, 1) \in R \):
\[ 9^2 + 3 \cdot 1^2 = 81 + 3 = 84 \]
\[ 4 \cdot 9 \cdot 1 = 36 \]
Since \( 84 \neq 36 \), we conclude that \( (9, 1) \notin R \).

Thus, \( R \) is not transitive.

### Conclusion

The relation \( R \) is:
- Reflexive: Yes
- Symmetric: No
- Transitive: No