$A$ relation $R$ is defined from $\{2, 3, 4, 5\}$ to $\{3, 6, 7, 10\}$ by $xRy \Leftrightarrow x$ is relatively prime to $y$. Then domain of $R$ is
$\{2, 3, 5\}$
$\{3, 5\}$
$\{2, 3, 4\}$
$\{2, 3, 4, 5\}$
If $R_{1}$ and $R_{2}$ are equivalence relations in a set $A$, show that $R_{1} \cap R_{2}$ is also an equivalence relation.
The minimum number of elements that must be added to the relation $R =\{( a , b ),( b , c )\}$ on the set $\{a, b, c\}$ so that it becomes symmetric and transitive is:
Let $R$ be an equivalence relation on a finite set $A$ having $n$ elements. Then the number of ordered pairs in $R$ is
Let $A =\{1,2,3,4, \ldots .10\}$ and $B =\{0,1,2,3,4\}$ The number of elements in the relation $R =\{( a , b )$ $\left.\in A \times A : 2( a - b )^2+3( a - b ) \in B \right\}$ is $.........$.
Let $R =\{( P , Q ) \mid P$ and $Q$ are at the same distance from the origin $\}$ be a relation, then the equivalence class of $(1,-1)$ is the set