The relation $R$ defined in $N$ as $aRb \Leftrightarrow b$ is divisible by $a$ is
Reflexive but not symmetric
Symmetric but not transitive
Symmetric and transitive
None of these
Show that the relation $R$ in the set $\{1,2,3\}$ given by $R =\{(1,2),(2,1)\}$ is symmetric but neither reflexive nor transitive.
Let $S$ be the set of all real numbers. Then the relation $R = \{(a, b) : 1 + ab > 0\}$ on $S$ is
If $R$ be a relation $<$ from $A = \{1,2, 3, 4\}$ to $B = \{1, 3, 5\}$ i.e., $(a,\,b) \in R \Leftrightarrow a < b,$ then $Ro{R^{ - 1}}$ is
The relation $R$ defined on the set $A = \{1, 2, 3, 4, 5\}$ by $R = \{(x, y)$ : $|{x^2} - {y^2}| < 16\} $ is given by
Let $\mathrm{T}$ be the set of all triangles in a plane with $\mathrm{R}$ a relation in $\mathrm{T}$ given by $\mathrm{R} =\left\{\left( \mathrm{T} _{1}, \mathrm{T} _{2}\right): \mathrm{T} _{1}\right.$ is congruent to $\left. \mathrm{T} _{2}\right\}$ . Show that $\mathrm{R}$ is an equivalence relation.