Further
i) Give a geometric description of the equivalence class to which(0,0) belongs.
ii) Is the number of equivalence classes finite?
My approach is as below:
Equivalence Relation
Reflexive: $\forall (x,y) \in \mathbb{R}^2, (x,y)\sim(x,y)$ since $x-x=0\in \mathbb{Z}$
Symmetric: $(x_1,y_1)\sim(x_2,y_2)\Rightarrow x_1-x_2\in \mathbb{Z}$. Now since $x_2-x_1=-(x_1-x_2)\in \mathbb{Z}$ we have $(x_2,y_2)\sim(x_1,y_1)$
Transitive: ($(x_1,y_1)\sim(x_2,y_2)$ and $(x_2,y_2)\sim(x_3,y_3)) \Rightarrow$ ($x_1-x_2\in \mathbb{Z}$ and $x_2-x_3\in \mathbb{Z}$). Now since sum of any two integers is again an integer we have $(x_1-x_2)+(x_2-x_3)=(x_1-x_3)\in \mathbb{Z}$ And so $(x_1,y_1)\sim(x_3,y_3)$
Geometric Description
We know that the equivalence class to which $(0,0)$ belongs is equal to $[(0,0)]$. Now $0-x \in \mathbb{Z} \Rightarrow x \in \mathbb{Z}$. Hence $[(0,0)]=\{(x,y) \in \mathbb{R}^2 \mid x \in \mathbb{Z}\}$.
Geometrically ~ contains every line through integer values of $x$ and parallel to $y$ axis.
Number of Equivalence classes
This is where I am stuck. How do I determine whether the number of classes is finite or infinite?
Please let me know if I have made any mistake in my approach. Also please help with finite/infinite aspect of the number of classes