If you’ve seen the Banach-Tarski paradox, you know that it is possible to cut a solid 3-dimensional ball into 5 pieces and reassemble the pieces using only rigid motions to form two solid balls each the same size as the original. The construction depends on the Axiom of Choice. But it is possible to construct paradoxical decompositions that do not involve Choice.

For instance, did you know there is a subset of the plane, call it S, such that it can be partitioned into two sets A and B such that if you translate A by 1 unit, it becomes the entire S, and if you rotate B by 1 radian, it is also the entire set S? In other words, the set S is equidecomposable to two copies of itself!

**The Math Behind the Fact:**

It seems strange that such a set S exists, but we can explicitly name the set S. Consider the complex number x=e^{i}. It is known to be transcendental, i.e., not the root of any non-zero polynomial with integer coefficients. So let S be the set of all polynomials in x with whole number (non-negative) coefficients. Note that S is a set of points in the complex plane, and because of the transcendental nature of x=e^{i} each polynomial in S corresponds to a unique point in the plane. For instance, one element of S is

3x^{3} + 4x^{2} + 7.

Let A be the subset of S obtained from the polynomials of S that have no constant term. Let B be the subset of S obtained from polynomials that have a non-zero constant term. Thus the union of the points in A and B is S.

Moreover, that if you take A and you rotate all the points in A by 1 radian clockwise (this is the same as multiplying all the points in A by x^{-1}=e^{-i}), you will obtain the entire set S! Why? It’s because the polynomials of S with no constant term are multiples of x, so multiplying by the inverse of x yields all polynomials in S.

Similarly, if you take B and translate all the points of B to the left by 1 unit (this is the same as subtracting 1 from every point in B), you will obtain the entire set S, because the polynomials of S with non-zero constant term must have a constant term that is a positive integer, and subtracting 1 produces all polynomials in S, including ones with constant term 0.

This set is countable, though it is unbounded. It is unknown whether there is any *bounded*subset of the plane that is paradoxical, though it is known that there is no bounded subset that is paradoxical using only two pieces.

**How to Cite this Page:**

Su, Francis E., et al. “Sierpinski-Mazurkiewicz Paradox.” *Math Fun Facts*. <https://www.math.hmc.edu/funfacts>.

**References:**

Stan Wagon, The Banach-Tarski Paradox.

Or see a shorter expository article by F. Su.

**Fun Fact suggested by: **

Francis Su