# Can the Banach-Tarski paradox be "realized" by floating-point round-off?

The Banach-Tarski paradox says that a ball in $$\mathbb{R}^3$$ can be partitioned into a finite number of pieces, whose rearrangement has a larger volume than the original. It occurred to me that it might be possible to define the pieces $$P_1,P_2,\ldots,P_n$$ by a finite amount of geometric data, such that the applied rearrangement function $$f(P_i)$$ actually does increase the volume of $$P_i$$, because of floating-point round-off error: $$f$$ is an isometry but when implemented in floating-point hardware it is not an exact isometry. So it might be possible that $$\textrm{Vol}(f(P_1))+\ldots \textrm{Vol}(f(P_n)) > \textrm{Vol}(P_1)+\ldots \textrm{Vol}(P_n) \;.$$ Repeating, this might lead to a type of "realization" of the paradox in terms of floating-point computations.

Whether you think this would be as interesting as I do is in the eye of the beholder.

• It’s worth defining what you mean by two pieces being “the same” in this picture. Do you mean that the one piece can be turned into the other by rounding an isometric and applying it to the first piece? May 25, 2020 at 14:34

This seems to have little to do with Banach-Tarski. In your setting, f is simply not an isometry due to floating-point errors, and in particular there must be a single piece $$i$$ such that $$\mathrm{Vol}(f(P_i))>\mathrm{Vol}(P_i)$$, so no need to cut into several pieces.
Banach-Tarski works because the notion of volume is not well-defined on the pieces. This clashes with your requirement that the pieces are defined by "a finite amount of geometric data", and with your writing of $$\mathrm{Vol}(P_1)+\dots+\mathrm{Vol}(P_n)$$.