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From the Fun Fact files, here is a Fun Fact at the Advanced level:


Two sets A and B are said to be equidecomposable if you can partition set A into a finite number of subsets and reassemble them (by rigid motions only) to form set B.

Let A be a unit circle, and let B be a unit circle with one point X missing (called a "deleted circle"). Are sets A and B equidecomposable?

Believe it or not, yes! In fact you can do it using just 2 subsets. Can you figure out how?

Presentation Suggestions:
Some students will be tempted to "push together" the ends of the deleted circle, but this is not a rigid motion, and because of the openness of the endpoints, the ends will never "meet" unless they intersect.

The Math Behind the Fact:
Consider set B and let U be the subset consisting of all points that are a positive integer number of radians clockwise from X along the circle. This is a countably infinite set (the irrationality of Pi prevents two such points from coinciding). Let set V be everything else.

If you pick set U up and rotate it counterclockwise by one radian, something very interesting happens. The deleted hole at X gets filled by the point 1 radian away, and the point at the (n-1)-th radian gets filled by the point at the n-th radian. Every point vacated gets filled, and in addition, the empty point at X gets filled too!

Thus, B may be decomposed into sets U and V, which after this reassembling, form set A, a complete circle!

This elementary example forms the beginnings of the idea of how to accomplish the Banach-Tarski paradox.

How to Cite this Page:
Su, Francis E., et al. "Equidecomposability." Math Fun Facts. <>.

Keywords:    set theory
Subjects:    other
Level:    Advanced
Fun Fact suggested by:   Francis Su
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