Let $A = \{1, 2, 3\}, B = \{1, 3, 5\}$. If relation $R$ from $A$ to $B$ is given by $R =\{(1, 3), (2, 5), (3, 3)\}$. Then ${R^{ - 1}}$ is
$\{(3, 3), (3, 1), (5, 2)\}$
$\{(1, 3), (2, 5), (3, 3)\}$
$\{(1, 3), (5, 2)\}$
None of these
Give an example of a relation. Which is Symmetric but neither reflexive nor transitive.
Let $A = \{1, 2, 3, 4\}$ and $R$ be a relation in $A$ given by $R = \{(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 1), (3, 1), (1, 3)\}$. Then $R$ is
The relation $R$ defined in $N$ as $aRb \Leftrightarrow b$ is divisible by $a$ is
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.
Show that each of the relation $R$ in the set $A =\{x \in Z : 0 \leq x \leq 12\},$ given by $R =\{(a, b):|a-b| $ is a multiple of $4\}$