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.

Let $R$ be a relation defined on $N$ as a $R$ b is $2 a+3 b$ is a multiple of $5, a, b \in N$. Then $R$ is

Let $R_{1}$ and $R_{2}$ be relations on the set $\{1,2, \ldots, 50\}$ such that $R _{1}=\left\{\left( p , p ^{ n }\right)\right.$ : $p$ is a prime and $n \geq 0$ is an integer $\}$ and $R _{2}=\left\{\left( p , p ^{ n }\right)\right.$ : $p$ is a prime and $n =0$ or $1\}$. Then, the number of elements in $R _{1}- R _{2}$ 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 $N$ be the set of natural numbers and a relation $R$ on $N$ be defined by $R=\left\{(x, y) \in N \times N: x^{3}-3 x^{2} y-x y^{2}+3 y^{3}=0\right\} .$ Then the relation $R$ is: