11
$\begingroup$

Possible Duplicate:
Irrational painting device

I have a special hole puncher that does the following: When applied to any point $ x \in \mathbb{R}^{2} $, it removes all points in $ \mathbb{R}^{2} $ whose distance from $ x $ is irrational (by this, it is clear that $ x $ is not removed). Is there a minimum number of times that I can apply the hole puncher (to various points in $ \mathbb{R}^{2} $, of course) so as to remove every point in $ \mathbb{R}^{2} $?

  • 0
    I get it now :)2012-11-01

1 Answers 1

8

Three.

1) Convince yourself two won't be enough.

2) Consider $(0, 0), (1,0)$ and $(\pi, 0)$.

  • 1
    It's trivial in the sense that $\pi(2x - \pi) = y$ for $x,y \in \mathbb{Q}$ would imply that $\pi$ is algebraic.2012-11-01