Given that $V = \mathbb{R}^2 - \mathbb{Q}^2$ and ${(x_0,y_0),(x_1,y_1)}\in E$ iff there is no rational point on line connecting $(x_0,y_0)$ and $(x_1,y_1)$ , what is the diameter of G ?
Such that $G = <V,E>$
I know it's bigger than $1$ for instance $(0,-\sqrt{2})$ and $(1,\sqrt{2})$ have no edge between them.
And it's less or equal to $3$ , i think its $2$ but i don't know how to prove it.
On
If the diameter were $3$, there would be a rational point on each of the red line segments in the picture below. But there are uncountably many red line segments, and they are disjoint (except for one non-rational point). This leads to a contradiction, because there are not uncountably many rational points in the plane.
On
The diameter is 2, as conjectured.
Any line described by the function $y=qx+a$ with $q \in \mathbb Q, a \notin \mathbb Q$ has no rational point on it. If $(x_r,y_r)$ was such a point, we would have $a=y_r-qx_r \in \mathbb Q$.
Fix a point $(x_0,y_0) \in V$. For any $q \in \mathbb Q$ there will be exactly one line of the form $y=qx+b_q$ through $(x_0,y_0)$. At most one of them has a rational $b_q$. That's because if there were 2 (with $q_1, q_2$ and rational $b_{q_1}, b_{q_2})$, we'd have a system of 2 linear equations to calculate $(x_0,y_0)$
$$ y_0=q_1x_0+b_{q_1}, y_0=q_2x_0+b_{q_2} $$
which has determinant $q_1 - q_2 \neq 0$ and thus has a unique solution. Since all coefficients are rational, so would be the solution.
To sum up, through every point $(x_0,y_0) \in V$ runs an infinte amount of lines with rational slope, and all but at most one contain no rational point.
Now for the final part of the proof: If you have 2 points $(x_0,y_0), (x_1,y_1) \in V$, select one such line with rational slope and no rational point through $(x_0,y_0)$, and another through $(x_1,y_1)$. Make sure you choose different slopes for those lines. Then they intersect in a point $(x_2,y_2)$, which is not a rational point, so $(x_2,y_2) \in V$.
By construction, the entire line through $(x_0,y_0), (x_2,y_2)$ contains no rational point, so $(x_0,y_0), (x_2,y_2) \in E$. For the same reason, $(x_1,y_1), (x_2,y_2) \in E$.
That means you can get from $(x_0,y_0)$ to $(x_2,y_2)$ in $G$ in at most 2 steps. Since you already proved that the diameter is bigger than 1, that concludes the proof.
For the graph $G = \langle V, E \rangle$ with given vertices $V$ and edges $E$, the diameter is $2$.
Aside from the case
$$(x_0,y_0), (x_1,y_1) \in \mathbb{Q} \times (\mathbb{R}\setminus\mathbb{Q}) \cup (\mathbb{R}\setminus\mathbb{Q}) \times \mathbb{R} $$ it is obvious how to construct a path between $(x_0,y_0), (x_1,y_1)$ with length as most $2$.
By scaling and translation using rational numbers and exchange the role of $x,y$ if needed. It comes down to how to find a path with length $\le 2$ between $(0,p)$ and $(1,q)$ when $p, q \in \mathbb{R} \setminus \mathbb{Q}$.
Recall $\mathbb{R}$ can be viewed as a infinite dimension vector space over $\mathbb{Q}$. If one look at the line segment $(0,p) \to (1,q)$ in $\mathbb{R}^2$, it will intersect $\mathbb{Q}^2$ when and only when the $3$ numbers $\{ 1, p, q \}$ are linear dependent over $\mathbb{Q}$.
When $\{ 1, p, q \}$ are linear independent over $\mathbb{Q}$, the line segment $(0,p) \to (1,q)$ is disjoint from $\mathbb{Q}^2$. In this case, the distance between $(0,p)$ and $(1,q)$ is $1$.
When $\{ 1, p, q \}$ are linear dependent, pick another $r \in \mathbb{R}$ which is linear independent from $1, p, q$ over $\mathbb{Q}$. The line segments $(0,p) \to (\frac12, r)$ and $(\frac12,r) \to (1,q)$ will be disjoint from $\mathbb{Q}^2$. In this case, the distance between $(0,p)$ and $(1,q)$ is $2$.