Statement I: $A = \{(x, y) \in R \times R: y - x \text{ is an integer }\}$. Reflexivity: For all $x \in R$, $x - x = 0$ which is an integer. Symmetry: For all $x, y \in R$, if $y - x$ is an integer, then $x - y = -(y - x)$ is also an integer. 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. Therefore, $A$ is an equivalence relation on $R$. Statement II: $B = \{(x,y) \in R \times R: x = \alpha y \text{ for some rational number } \alpha\}$. Reflexivity: For all $x \in R$, $x = 1 \cdot x$. 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, Statement I is true, and Statement II is false.