Let the relation $R_{1}$ be defined on $R$ as $a R_{1}b$ if $1+a b > 0 .$ Then

We observe the following properties: Reflexivity : Let a be an arbitrary element of $R$, Then, $a \in R$ $\Rightarrow 1+a \cdot a=1+a^{2}>0 \quad\left[\because a^{2}>0\right.$ for all $\left.a \in R\right]$ $\Rightarrow \quad(a, a) \in R_{1}$ $\left[\right.$ By def. of $\left.R_{1}\right]$ Thus, $(a, a) \in R_{1}$ for all $a \in R$. So, $R_{1}$ is reflexive on $R$. Symmetry : Let $(a, b) \in R$ Then, $(a, b) \in R_{1}$ $\Rightarrow \quad 1+a b>0$ $[\because a b=b a$ for all $]$ $\Rightarrow \quad 1+b a>0 \Rightarrow(b, a) \in R_{1}$ Thus, $(a, b) \in R_{1} \neq(b, a) \in R,$ for all $a, b \in R$ So, $R_{1}$ is symmetric on $R$. Transitivity : we observe that $\left(1, \frac{1}{2}\right) \in R_{1}$ and $\left(\frac{1}{2},-1\right) \in R_{1}$ but (1,-1)$\notin R_{1}$ because $1+1 \times(-1)=0 \gg 0$ So, $R_{1}$ is not transitive on $R$.