# A hole puncher that hates irrational distances [duplicate]

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 $\mathbb{R}^{2}$?

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## marked as duplicate by Cameron Buie, Chris Eagle, Beni Bogosel, tomasz, TMMNov 1 '12 at 13:33

– Aryabhata Nov 1 '12 at 4:47
@Aryabhata: Is this merely related or a duplicate? – Rahul Nov 1 '12 at 8:53
The problems are equivalent. – Beni Bogosel Nov 1 '12 at 13:03
@Beni / Rahul: I don't think it is a duplicate. The other question is mainly about proving that $(0,0), (1,0), (\sqrt{2}, 0)$ is sufficient. This question allows other proofs. For instance, I believe there might be topological proofs. – Aryabhata Nov 1 '12 at 15:32
@RahulNarain / Beni: I don't think it is (see my earlier comment for why). I do agree that it is not clear from the other question what the intent was (my fault!), and so I am not casting a reopen vote to this (yet). – Aryabhata Nov 1 '12 at 15:36

2) Consider $(0, 0), (1,0)$ and $(\pi, 0)$.
That answer doesn't seem to provide a proof for the three points you specified. It does provide a proof for $(\sqrt{2},\ 0)$. The problem with $\pi$ basically comes down to the statement that $\pi(2x - \pi)$ is irrational for all rational $x$. Certainly believable but I don't think trivial. – EuYu Nov 1 '12 at 5:42
It's trivial in the sense that $\pi(2x - \pi) = y$ for $x,y \in \mathbb{Q}$ would imply that $\pi$ is algebraic. – Benjamin Dickman Nov 1 '12 at 5:47