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
$$\textrm{Vol}(f(P_1))+\ldots \textrm{Vol}(f(P_n)) &gt; \textrm{Vol}(P_1)+\ldots \textrm{Vol}(P_n) \;.$$
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.
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 $\mathrm{Vol}(f(P_i))&gt;\mathrm{Vol}(P_i)$, 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 $\mathrm{Vol}(P_1)+\dots+\mathrm{Vol}(P_n)$.