A relation from $P$ to $Q$ is

For real numbers $x$ and $y$, we write $ xRy \in $ $x - y + \sqrt 2 $ is an irrational number. Then the relation $R$ is

Let $X =\{1,2,3,4,5,6,7,8,9\} .$ Let $R _{1}$ be a relation in $X$ given by $R _{1}=\{(x, y): x-y$ is divisible by $3\}$ and $R _{2}$ be another relation on $X$ given by ${R_2} = \{ (x,y):\{ x,y\}  \subset \{ 1,4,7\} \} $ or $\{x, y\} \subset\{2,5,8\} $ or $\{x, y\} \subset\{3,6,9\}\} .$ Show that $R _{1}= R _{2}$.

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.

Given the relation $R = \{(1, 2), (2, 3)\}$ on the set $A = {1, 2, 3}$, the minimum number of ordered pairs which when added to $R$ make it an equivalence relation is

Let $P ( S )$ denote the power set of $S =\{1,2,3, \ldots, 10\}$. Define the relations $R_1$ and $R_2$ on $P(S)$ as $A R_1 B$ if $\left( A \cap B ^{ c }\right) \cup\left( B \cap A ^{ c }\right)=\varnothing$ and $AR _2 B$ if $A \cup B ^{ c }=$ $B \cup A ^{ c }, \forall A , B \in P ( S )$. Then :