Let Q be the set of all rational numbers and R be the relation defined as R={(x,y):1+xy>0,x,y∈Q} Then relation R is

Since $1 + x^{2} > 0$ for all $x \in Q,$ so $R$ is reflexive.

$\begin{array}{l} I f (x, y) \in R \Rightarrow 1 + x y > 0 \\ \Rightarrow 1 + y x > 0 \Rightarrow (y, x) \in R \end{array}$

Thus $R$ is symmetric.

Take $x = 2, y = - \frac{1}{4}, z = - 1$

$1 + x y = 1 - \frac{1}{2} = \frac{1}{2} > 0 so (x, y) \in R$

$1 + y z = 1 + \frac{1}{4} = \frac{5}{4} > 0 so (y, z) \in R$

$1 + x z = 1 - 2 = - 1 \Rightarrow (x, z) \notin R .$

Hence $R$ is not transitive.

Let $Q$ be the set of all rational numbers and $R$ be the relation defined as $R = {(x, y) : 1 + x y > 0, x, y \in Q}$ Then relation $R$ is