Let $R$ be the set of real numbers.

Statement-1: $A=\left\{(x,y)\in R\times R:y-x\mathrm{is}\mathrm{an}\mathrm{integer}\right\}$ is an equivalence relation on R.
Statement-2:   $B=\left\{(x,y)\in R\times R:x=\alpha y\mathrm{for}\mathrm{some}\mathrm{rational}\mathrm{number}\alpha \right\}$ is an equivalence relation on R.

Statement-1: $A=\left\{(x,y\in R\times R:y-xisan\mathrm{int}eger)\right\}$

(a) Reflexive $xRy:(x-x)$ is an integer, Which is true Hence it is reflexive.

(b) Symmetric $xRy:(x-y)$ is an integer. $\Rightarrow -(y-x)$ is also an integer. $\therefore (y-x)$  is also an integer.  $\Rightarrow yRx$ Hence, it is symmetric.

(c)Transitive  $xRyandyRz$

$\Rightarrow (x-y)and(y-z)$ integer and, $\Rightarrow (x-y)+(y-z)$ an integer. $\Rightarrow (x-z)$an integer.

$\Rightarrow xRz$
$\therefore$ it is transitive Hence, it is equivalence relation.

Statement-2 : $B=\{(x,y)\in R\times R:x=\alpha yforsomerationalnumber\alpha \}$

$if\alpha =1,thenxRy:x=y$ ( To check equivalence)

(a) Reflexive $xRy:x=x\left(True\right), \therefore$ Reflexive

(b) Symmetric $xRy:x=y\Rightarrow y=x\Rightarrow yRx,\therefore$ Symmetric

(c) Transitive $xRyandyRz\Rightarrow x=y$ And $y=z\Rightarrow x=z\Rightarrow xRz,\therefore$ Transitive

Hence, it is equivalence relation.
$\therefore$ Both are true but statement-2 is not correct explanation of statement-1