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Figure Eights (Posted on 2005-02-15) Difficulty: 5 of 5
Suppose that I drew an infinite number of disjoint closed curves in the plane (such as circles, squares, etc.). Suppose that I then tell you that there is one curve for each positive real number.

You would not have too much trouble believing my assertions at this point. For example, I could have drawn all circles with center at the origin. They are all disjoint, and for each positive real number x, there is a corresponding circle - namely, the circle of radius x.

But suppose that I also tell you that all the curves I drew were figure eights. Can you believe my assertions now?

(A figure eight is a curve in the plane obtained from the basic "8" shape by any combination of translation, rotation, expansion, or shrinking.)

See The Solution Submitted by David Shin    
Rating: 4.2000 (5 votes)

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still thinking | Comment 8 of 34 |
Another way to do rational numbers would be to place an 8 at the coordinate (x,y) where the fraction x/y is the value of that rational number.  All rational numbers can be expressed as a fraction.  The 8 would be small enough to not impinge on the 8's located at adjacent grid points.

For the irrational numbers, I'm assuming that it's not OK to have a mere point be the representation.  ie that the 8 must have some finite area.   Now the x-axis has a lot of open space on it, since there are no rational numbers with y=0.   So the 8 for pi, for example could just be a small 8 located at x=pi and y=0.  The only problem with this is:  the number if irrational numbers is infinite, so how close together are the closest irrational numbers?  For example "e" and "sqrt(5)" differ by about 0.5 so there is plenty of room for each to have its own 8.  But maybe there are other examples where you can find 2 irrationals closer together than an arbitrarily small number. 
  Posted by Larry on 2005-02-15 21:50:40
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