$r=\{(a, b) | a, b \in R \text { and } a-b+\sqrt{3}$ is an irrational number$\}$, then $\mathrm{r}$ is reflexive as, $\mathrm{aRa}=\mathrm{a}-\mathrm{a}+\sqrt{3}=\sqrt{3}$ which is an irrational number $\sqrt{3} \mathrm{R} 1=\sqrt{3}-1+\sqrt{3}=2 \sqrt{3}-1 \mathrm{which}$ is an irrational number but $1R$ $\sqrt{3}=1-\sqrt{3}+\sqrt{3}=1$ which is not irrational number

$\therefore \sqrt{3} \mathrm{R} 1 \Rightarrow 2 \sqrt{3}-1$

$\therefore \mathrm{r}$ is not symmetric Also, $r$ is not transitive

$\because \sqrt{3} \mathrm{Rl}$ and $1 \mathrm{R} 2 \sqrt{3}$ but $\sqrt{3} \mathrm{R} 2 \sqrt{2}$

$\Rightarrow \sqrt{3}-2 \sqrt{3}+\sqrt{3}=0$

Hence, $r$ is not an equivalenece relation