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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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Related: math.stackexchange.com/questions/2550/… –  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
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marked as duplicate by Cameron Buie, Chris Eagle, Beni Bogosel, tomasz, TMM Nov 1 '12 at 13:33

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1 Answer

up vote 8 down vote accepted

Three.

1) Convince yourself two won't be enough.

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

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Would you mind including a proof that the three points are sufficient? –  EuYu Nov 1 '12 at 5:12
    
Yes, please do. Thanks! –  Haskell Curry Nov 1 '12 at 5:19
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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
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Nevermind! I missed the second answer. –  EuYu Nov 1 '12 at 5:43
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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
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