JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Sets Relations and Functions

For every real $a$, $3a-3a+\sqrt{7}=\sqrt{7}$ is irrational, so $(a,a)\in R$ and the relation is reflexive. Taking $a=\frac{\sqrt{7}}{3},b=0$ gives $3a-3b+\sqrt{7}=2\sqrt{7}$ (irrational, so $(a,b)\in R$), while $3b-3a+\sqrt{7}=0$ is rational, so $(b,a)\notin R$ and the relation is not symmetric. Taking $a=\frac{\sqrt{7}}{3},b=1,c=\frac{2\sqrt{7}}{3}$ puts both $(a,b)$ and $(b,c)$ in $R$, yet $3a-3c+\sqrt{7}=0$ is rational, so $(a,c)\notin R$ and the relation is not transitive. Thus the relation is reflexive but neither symmetric nor transitive.