An equivalence relation on a set must satisfy three properties: reflexivity (every element is related to itself), symmetry (if an element is related to a second, the second is related to the first), and transitivity (if a first element is related to a second, and the second is related to a third, then the first is related to the third).

Statement I : $A={(x, y) \in R \times R: }$ y-x is an integer .

• Reflexivity : For all $x$ in $R$, $x - x = 0$ which is an integer. So, every element is related to itself.


• Symmetry : For all $x, y$ in $R$, if $y - x$ is an integer, then $x - y = - (y - x)$ is also an integer. So, if $x$ is related to $y$, then $y$ is related to $x$.


• Transitivity : For all $x, y, z$ in $R$, if $y - x$ and $z - y$ are integers, then $(z - y) + (y - x) = z - x$ is also an integer. So, if $x$ is related to $y$ and $y$ is related to $z$, then $x$ is related to $z$.

Therefore, $A$ is an equivalence relation on $R$.

Statement II : $ B={(x, y) \in R \times R: x=\alpha y}$ for some rational number $\alpha$.

• Reflexivity : For all $x$ in $R$, $x = 1x$. Since 1 is a rational number, every element is related to itself.


• Symmetry : For all $x, y$ in $R$, if $x = \alpha y$ for some rational $\alpha$, then $y = \frac{1}{\alpha}x$. However, if $\alpha = 0$, then $\frac{1}{\alpha}$ is undefined, and therefore, $B$ doesn't satisfy symmetry.


• Transitivity : If $x = \alpha y$ and $y = \beta z$ for some rational numbers $\alpha$ and $\beta$, then $x = (\alpha \beta)z$. Since the product of rational numbers is rational, if $x$ is related to $y$ and $y$ is related to $z$, then $x$ is related to $z$.

Therefore, $B$ is not an equivalence relation on $R$ since it does not satisfy the symmetry property.

In conclusion, the correct answer is

Option B : Statement I is true, Statement II is false.