1. Check Statement I

We have $A=\{(x,y)\in \mathbb{R}\times \mathbb{R}: y-x \text{ is an integer}\}.$

To be an equivalence relation, it must be reflexive, symmetric, and transitive.

(i) Reflexive

For any $x\in \mathbb{R}$, $x-x=0,$ and $0$ is an integer. Hence $(x,x)\in A$.

So, $A$ is reflexive.

(ii) Symmetric

Suppose $(x,y)\in A$. Then $y-x\in \mathbb{Z}.$ Now, $x-y=-(y-x),$ and the negative of an integer is also an integer. Hence $(y,x)\in A$.

So, $A$ is symmetric.

(iii) Transitive

Suppose $(x,y)\in A$ and $(y,z)\in A$. Then $y-x\in \mathbb{Z}, \qquad z-y\in \mathbb{Z}.$ Adding, $(z-y)+(y-x)=z-x\in \mathbb{Z}.$ Hence $(x,z)\in A$.

So, $A$ is transitive.

Therefore, Statement I is true.

2. Check Statement II

We have $B=\{(x,y)\in \mathbb{R}\times \mathbb{R}: x=\alpha y \text{ for some rational number } \alpha\}.$

Again test reflexive, symmetric, transitive.

(i) Reflexive

For any $x\in \mathbb{R}$, $x=1\cdot x,$ and $1\in \mathbb{Q}$. Hence $(x,x)\in B$.

So, $B$ is reflexive.

(ii) Symmetric?

Suppose $(x,y)\in B$. Then $x=\alpha y$ for some $\alpha\in \mathbb{Q}$.

If $x\neq 0$, then $y=\frac{1}{\alpha}x,$ and if $\alpha\neq 0$, then $1/\alpha\in \mathbb{Q}$, so symmetry would hold.

But there is a problem when $\alpha=0$.

Take $x=0, \quad y=\sqrt{2}.$ Then $x=0\cdot y,$ and $0\in \mathbb{Q}$, so $(0,\sqrt{2})\in B$.

For symmetry, we would need $(\sqrt{2},0)\in B$, i.e. $\sqrt{2}=\beta\cdot 0$ for some $\beta\in \mathbb{Q}$. But $\beta\cdot 0=0$ for every $\beta$, so this is impossible.

Hence $(\sqrt{2},0)\notin B$.

Therefore, $B$ is not symmetric.

So, Statement II is false.

3. Conclusion

• Statement I is true.
• Statement II is false.

Hence the correct option is $\boxed{\text{B}}.$

4. Comparison with stored correct answer

Stored correct answer: B

My derived answer: B

They agree.