I was watching an old

Daily Showclip and someone self-identified as “one twelfth Cherokee”. It sounded peculiar, as people usually state they’re “1/16th”, or generally $1/2^n, n \in \mathbb{N}$.Obviously you could also be some summation of same to achieve 3/32nds, etc., but will an irreducible fraction with the numerator 1 always need a power-of-two denominator? More generally does it always require a power-of-two denominator?

Assumptions (as per comments):

- Nobody can trace their lineage infinitely
- Incest is OK, including transgenerational, but someone can’t be their
ownparent.

**Answer**

This depends on the model. Instead of arguing that we have only $46$ chromosomes and cross-overs or whatever the mechanism is called are not *that* common, let us assume a continuous model.

That is, a priori, everybody can be $\alpha$ Cherokee for any $\alpha\in[0,1]$ and the rules are as follows

- Everybody has exactly two parents.
- If the parents have Cherokee coefficients $\alpha_m, \alpha_f$, then the child has $\alpha=\frac{\alpha_m+\alpha_f}2$
- In a sufficiently large but finite number of generations ago, people had $\alpha\in\{0,1\}$

It follows by induction, that $\alpha$ can always be expressed as $\alpha=\frac{k}{2^n}$ with $k,n\in\mathbb N_0$ and $0\le k\le 2^n$. Consequently $\alpha=\frac1{12}$ is not possible exactly (though for example $\frac{85}{1024}\approx\frac1{12}$ would be possible).

It doesn’t matter if there is any type of inbreeding taking place anywhere in the tree (or then not-tree) of ancestors.

The only way to obtain $\alpha$ not of this form would involve time-travel and genealogical paradoxes: If you travel back in time and paradoxically become your own grandparent and one of the other three grandparents is $\frac14$-Cherokee and the others are $0$-Cherokee, you end up as a solution to $\alpha=\frac{\frac14+\alpha}4$, i.e. $\alpha=\frac1{12}$.

**Attribution***Source : Link , Question Author : Nick T , Answer Author : Hagen von Eitzen*