### Parity of the Real Numbers: Part I

In some recent posts, here and here we discussed the extension of the concept of parity (Odd v. Even) from the integers to the rational numbers. We found that it is natural to consider three parity classes, determined by the parities of the numerator and denominator of a rational number ${q = m / n}$ (in reduced form):

• q Odd: ${m}$ odd and ${n}$ odd.
• q Even: ${m}$ even and ${n}$ odd.
• q None: ${m}$ odd and ${n}$ even.

or, in symbolic form,

$\displaystyle \mbox{Odd} = \frac{odd}{odd} \,, \qquad \mbox{Even} = \frac{even}{odd} \,, \qquad \mbox{None} = \frac{odd}{even}$

Here, ${None}$ stands for “Neither Odd Nor Even”.

The density of a subset of the rational numbers depends on the ordering. We found that, for several standard ways of ordering ${\mathbb{Q}}$, the three parity classes have equal density ${\frac{1}{3}}$. The question now is: can any or all of this be extended to the real numbers ${\mathbb{R}}$?

The Challenge

For simplicity, we confine attention to the unit interval ${[0,1]}$. Can this be split up in a meaningful way into three subsets ${O}$, ${E}$ and ${N}$, reflecting the usual properties of parity (e.g, ${odd + odd = even}$), each having measure ${\frac{1}{3}}$ and, more crucially, each having uniform density ${\frac{1}{3}}$. Uniform density of a set ${A}$ with density ${\rho(A)=d}$ in ${[0,1]}$ means that, for any subinterval ${[a,b] \subset [0,1]}$ we require ${\rho(A\cap[a,b]) = \frac{1}{3}(b-a)}$.

We seek a (disjoint) partition ${[0,1] = O \uplus E \uplus N}$ and an appropriate ordering of the real numbers in the unit interval, such that

• ${O \uplus E}$ have the usual properties of odd and even numbers.
• ${\lambda(O) = \lambda(E) = \lambda(N) = \frac{1}{3} }$, where ${\lambda}$ is the Lebesgue measure.
• ${\rho(O) = \rho(E) = \rho(N) = \frac{1}{3}}$.
• Each of ${O}$, ${E}$ and ${N}$ has uniform density in ${[0,1]}$.

In case this seems too easy, reference is made to discussions on Math Stack Exchange over a long period considering variations on this problem. In particular, a recent discussion indicates that it is impossible to partition the unit interval into two disjoint sets, each of measure ${\frac{1}{2}}$ in ${[0,1]}$. Some references are given below.

Sources

${\bullet}$ Can the real numbers be equally split into two sets of same measure? Here.

${\bullet}$ For a set of positive measure there is an interval in which its density is high, ${\mu(E\cap I) > \rho \mu( I )}$. Here.

${\bullet}$ Lynch, Peter & Michael Mackey, 2022: Parity and Partition of the Rational Numbers. Preprint on arXiv.

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