Paper Punches

Paper Punches
Paper Punches

A Paper Punch Problem?

Consider a paper punch that can be centered at any point of the plane, and that, when operated, removes from the plane precisely those points whose distance from the center is irrational. How many punches are needed to remove every point?

Original solutions please! :)

Thanks for answering! :)

I think the idea of the question is that every point which is ANY irrational distance from the center is removed (a rather odd looking hole-punch). Then we want to recenter and apply again, etc.

If we stick with standard coordinates, we may as well assume that our first punch is centered at (0,0). Then of course we don’t actually remove (0,0), so let’s take our second center to be, say, (pi, 0) [now here there might be a difference in what point we pick...]. Then we’ve now removed every point of the x-axis: rationals are removed since their distance to pi is irrational; irrationals were removed since the distance to 0 is irrational.

Then the question is which other points of the plane are left out, and how many extra punches we need. [And perhaps we ought to have chosen a different second point, perhaps an algebraic, or maybe off of the x-axis.]

If we assume that (a,b) in R^2 such that sqrt(a^2+b^2) and sqrt((a-pi)^2+b^2) are rational, then we can find that a=pi/2, b=sqrt(r^2-(pi/2)^2) work out for almost any rational r. So we are missing at least some points, along the vertical line x=pi/2.

Mer, late for work…

EDIT: slow day at work :)
I think perhaps it would be best to start with an algebraic point, say A=(sqrt(2), 0). If we punch there and at O=(0,0), assume that there is a point (a,b) such that its distance to O and A is rational, i.e.
sqrt(a^2+b^2) in Q
sqrt(a^2 + b^2 + 2 – 2a*sqrt(2)) in Q
Let r^2=a^2+b^2 (which is rational). Then we need
r^2 + 2 – 2a*sqrt(2)
to be rational (indeed, it must be the square of a rational); this implies that a=s*sqrt(2) for some rational s.

So there may be some points which we have missed, but if they do exist they all are on the family of vertical lines x=s*sqrt(2). Now we can take the point B=(sqrt(3), 0), and if there is a point (a,b) with rational distance from B and O, we similarly obtain that
r^2 + 3 – 2sqrt(3)*a
is rational. If it further has rational distance to A, then we know a=s*sqrt(2), so
r^2 + 3 – 2s*sqrt(6)
is rational, which says that sqrt(6) is rational, which is nonsense.

Thus [unless I've missed something], three punches can suffice. I strongly suspect two cannot do the job; I think that given two centers, there will be points on the perpendicular bisector of the two which were not removed.

EDIT: I’m certain that two cannot do the job. For any two points in the plane, pick some rational number larger than half the distance between them, and construct the isosceles triangle(s) with the segment joining our two points as the base and equal legs of length that chosen rational.

One begins to wonder about higher dimensions, and perhaps even different metrics on the space…

EDIT: holdm: “then sqrt(x^2+y&2) is rational
in this case, sqrt((x-e)^2 + y^2) is irrational”
Consider (x,y)=(e/2, sqrt(25-(e/2)^2)). Then
sqrt(x^2+y^2) = 5 is rational, but
sqrt((x-e)^2 + y^2) = 5 is also rational.

Also, I think Gerry was thinking of just the real line instead of the plane. He raises the interesting question though whether we can center a punch at an already-removed point. If we can’t, we’ll need infinitely many punches. Suppose instead that there is a last punch. Then the last punch is centered at some point X which has not yet been removed. But the last punch won’t remove X, being rational distance away from itself. An interesting question then is to have our punch also remove the point on which it is centered. I’m not sure how to go about choosing our points then…

EDIT (after Vasek’s answer): The higher dimensional question seems to be actually rather uninteresting. The same three punches will suffice for any dimensional space (where any new coordinates are taken to be zero for our punches). The argument is the same as above, and the only time we can get away with fewer than three punches is when we’re dealing with just the real line. (And I suppose you can’t remove everything in R^0, if you take that to be a single point…)

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