Let R be the set of all real numbers.
Statement 1: $A=\left\{\right(x,y)\in R\times R:y-xisan\text{\hspace{0.33em}integer}\}$ is an equivalence relation on R.
Statement 2: $B=\left\{\right(x,y)\in R\times R:x=\alpha yforsomerationalnumber\alpha \}$ is an equivalence relation on R.
 

$x-y$ is an integer

$x-x=0$ is an integer  $\Rightarrow A$ is Reflexive

$x-y$ is an integer $\Rightarrow y-x$ is an integer $\Rightarrow A$ is symmetric $x-y,y-z$ are integers

As sum of two integers is an integer. 

$\Rightarrow (x-y)+(y-z)=x-z$ is an integer

$\Rightarrow A$ is transitive. Hence statement 1 is true Also $\frac{x}{x}=1$ is a rational number $\Rightarrow B$ is reflexive $\frac{x}{x}=\alpha$ is rational  $\Rightarrow \frac{y}{x}$ need not be rational i.e., $\frac{0}{1}$ is rational $\Rightarrow \frac{1}{0}$ is not rational. Hence B is not symmetric $\Rightarrow B$ is not an equivalence relation.