If $R$ is an equivalence relation on a set $A$, then ${R^{ - 1}}$ is

Give an example of a relation. Which is Reflexive and transitive but not symmetric.

Let $S$ be set of all real numbers ; then on set $S$ relation $R$ defined as $R = \{\ (a, b) : 1 + ab > 0\ \}$ is

Let $A =\{2,3,4,5, \ldots ., 30\}$ and $^{\prime} \simeq ^{\prime}$ be an equivalence relation on $A \times A ,$ defined by $(a, b) \simeq (c, d),$ if and only if $a d=b c .$ Then the number of ordered pairs which satisfy this equivalence relation with ordered pair $(4,3)$ is equal to :

Let $R = \{(a, a)\}$ be a relation on a set $A$. Then $R$ is

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.