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 (n1)th radian
gets filled by the point at the nth 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 BanachTarski paradox.
How to Cite this Page:
Su, Francis E., et al. "Equidecomposability."
Math Fun Facts.
<http://www.math.hmc.edu/funfacts>.
