For $n, m\in N, n\mid m$ means that $n$ is a factor of $m,$ then relation$\mid$is

Consider the following relations:
$\mathrm{R}:\left\{\right(\mathrm{x}, \mathrm{y})\mid \mathrm{x}, \mathrm{y}$ are real numbers and $x=wy$ for some rational number $w\}$

$S=\{\left(\frac{m}{n},\frac{p}{q}\right)\mid m, n, p$ and $q$ are integers such that $n, q\ne 0$ and $qm=pn\}$
Then

Let $R$ be the set of real numbers.

Statement- $1: A=\left\{\right(x, y)\in R\times R: y-x$ is an integer} is an equivalence relation on $R.$

Statement- $2: B=\left\{\right(x, y)\in R\times R: x=\alpha y$ for some rational number $\alpha )$ is an equivalence relation on $R.$

Consider the following relation $\mathrm{R}$, on the set of real square matrices of order $3$.

$\mathrm{R}=\{\left(\mathrm{A}, \mathrm{B}\right)\mid \mathrm{A}={\mathrm{P}}^{-1}BP$, for some invertible matrix $P\}$.

Statement - $1 : R$, is equivalence relation.

Statement - $2 :$ For any two invertible $3\times 3$ matrices $\mathrm{M}$ and $\mathrm{N}, {\left(\mathrm{MN}\right)}^{-1}={\mathrm{N}}^{-1}{\mathrm{M}}^{-1}.$