Does every line segment in $\mathbb{R}^n$ contains a point having only rational coordinates? Let $S$ be the subset of $\mathbb{R}^n$ consisting of all points which have only rational coordinates. I know that $S$ is a dense subset of $\mathbb{R}^n$. Is it true that every line segment in $\mathbb{R}^n$ intersects S?
 A: Here is a complete classification of lines $\ell$ in $\mathbb{R}^2$ with respect to containment of rational points.
0) First we dispose of vertical lines $x = a$.  Evidently these have infinitely many rational points if $a \in \mathbb{Q}$ and no rational points otherwise.  
Now suppose $\ell$ is given by the equation $\ell: y = mx+b$ with $m,b \in \mathbb{R}$.
1) If $m,b \in \mathbb{Q}$ then the rational points on $\ell$ are dense in $\ell$.
2) If $m \in \mathbb{Q}$ and $b \in \mathbb{R} \setminus \mathbb{Q}$, then there are no rational points on $\ell$.
3) If $m \in \mathbb{R} \setminus \mathbb{Q}$ and $b \in \mathbb{Q}$ then there is exactly one rational point on $\ell$: $(0,b)$.
4) If $m,b \in \mathbb{R} \setminus \mathbb{Q}$ and the set $\{1,m,b\}$ is linearly dependent over $\mathbb{Q}$ then there is exactly one rational point on $\ell$.  Example: if $m =\sqrt{2}$, $b = 3 + 4\sqrt{2}$, then the unique rational point is $(-4,3)$.
5) If $\{1,m,b\}$ are linearly independent over $\mathbb{Q}$ then there are no rational points on $\ell$.  Example: $m = \sqrt{2}$, $b = \sqrt{3}$.   
Remark: From the perspective of projective geometry, one should view a vertical line to be a line with rational slope: the point $\infty = [1:0]$ is a $\mathbb{Q}$-rational point on the projective line.  With this convention, the line $x = a$ falls into case 1) if $a \in \mathbb{Q}$ and case 2) if $a \in \mathbb{R} \setminus \mathbb{Q}$.  
A: No, of course not. Consider for example the line $y=\pi$ in $\mathbb{R}^2$.
