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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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    Related: http://math.stackexchange.com/questions/2550/irrational-painting-device2012-11-01
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    @Aryabhata: Is this merely related or a duplicate?2012-11-01
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    The problems are equivalent.2012-11-01
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    @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.2012-11-01
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    @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).2012-11-01
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    @Aryabhata: This question asks if there is a minimal number of points, the other question asks what is the minimum number of points. The answers to the linked question solve this problem.2012-11-01
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    @BeniBogosel: Yes, but If you notice, majority of the answers are geared towards part b). This question is mainly about part a and might permit answers not applicable to part b). In any case, I don't feel strongly, so haven't voted to reopen.2012-11-01
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    @Aryabhata: http://math.stackexchange.com/a/80068/7327 A. Walker's answer is perfect for this question.2012-11-01
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    @BeniBogosel: Yes, but my (implicit) point was, any new answer (say using a brand new branch in mathematics yet to be discovered) to this question will have to be added to the other question (because we closed this), but the scope of the other question is mainly geared towards part b).2012-11-01
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    I get it now :)2012-11-01

1 Answers 1

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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?2012-11-01
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    Yes, please do. Thanks!2012-11-01
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    The same answer can be found here, with a couple write-ups: http://answers.yahoo.com/question/index?qid=20090410035513AAgeHjg2012-11-01
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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.2012-11-01
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    Nevermind! I missed the second answer.2012-11-01
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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.2012-11-01