If we draw all possible lines connecting points with integer coordinates, will the plane be completely filled?
I know the answer is no, because the task of drawing lines is a countable task, while filling a plane with lines is an uncountable task, but I would be happy if someone could point out the coordinates of a point through which none of these lines would pass and prove it.
Assume the point $(\sqrt2,\sqrt3)$ lies on line $l$, which passes through distinct lattice points $(a,b)$ and $(c,d)$.
If $a=c$ then $l$ has equation $x=a$. Plug in $(\sqrt2,\sqrt3)$ and we get $a=\sqrt2$. But $a\in\mathbb{Z}$. $\therefore a\ne c$.
Therefore $l$ has equation $y=\frac{b-d}{a-c}(x-a)+b$.
Plug in $(\sqrt2,\sqrt3)$ and we get $\sqrt3=\frac{b-d}{a-c}(\sqrt2-a)+b$, which is of the form $\sqrt3=\sqrt2 p-q$ where $p,q\in\mathbb{Q}$.
Square both sides: $3=2p^2-2pq\sqrt2+q^2$.
If $p=q=0$ then we get $3=0$, contradiction.
If $p=0$ and $q\ne 0$ then we get $q=\sqrt3$. LHS is rational, RHS is irrational, contradiction.
If $p\ne 0$ and $q=0$ then we get $p=\sqrt{\frac32}$. LHS is rational, RHS is irrational, contradiction.
If $p\ne 0$ and $q\ne 0$ then we get $\sqrt2=\frac{2p^2+q^2-3}{2pq}$. LHS is irrational, RHS is rational, contradiction.
So the point $(\sqrt2,\sqrt3)$ does not lie on any line through distinct lattice points $(a,b)$ and $(c,d)$.