Can you be 1/12th Cherokee?

I was watching an old Daily Show clip 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):

1. Nobody can trace their lineage infinitely
2. Incest is OK, including transgenerational, but someone can't be their own parent.

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. $\alpha=\frac1{12}$.

No, nobody can be $1/12th$ Cherokee.

I'll prove a stronger statement:

For any natural $p$ and $q$, one can't be a $p/q$ Cherokee if $p$ is not divisible by $3$, and $q$ is divisible by $3$.

First, the statement that a person can be only "rational number of a Cherokee" can be easily proven - I won't waste space for the proof here.

Now, let's suppose a person was such $p/q$ Cherokee, then, supposing that his/hers parents were $p_1/q_1$ and $p_2/q_2$ Cherokee (fractions are reduced to their minimal form), the following would be valid

$$p/q = (p_1/q_1 + p_2/q_2)/2$$

or

$$2\times p\times q_1\times q_2=q\times (p_1\times q_2+p_2\times q_1)$$

Since $q$ is divisible by $3$, and $p$ is not divisible by $3$, this means that one of $q_1$ and $q_2$ must be divisible by $3$. In other words, one of the parents have the same property as the original person.

And than you can go ad infinitum - but since the oldest ancestor can be either 100% or 0% Cherokee, and that person does not satisfy the property of being "$p/q$ Cherokee if $p$ is not divisible by $3$, and q is divisible by $3$", this is a contradiction.

This means nobody can be $1/12th$ Cherokee.

Answer:

Base 2 numbers! If someone is $a$ of race A ($a\in[0,1]$) and someone else is $b\in[0,1]$ race A, their offspring is $\frac{1}{2}(a+b)=\frac{1}{2}a+\frac{1}{2}b$.

This immediately screams "base 2" because you'll get this recursive half pattern.

So let's write something in base 2, take 101 this is 1/2+1/8 (there's a 0 in the 1/4 column) which is 5/8ths as binary, we halve it, which in base 2 is just shifting everything right once, to get 0101 and then add it with the other parent (after shifting theirs).

For example: 1x0 (where x means "have baby with") is 01+00=01=0.5.

The operation of x is closed with finite binary strings - which I will call "race strings".

Adding two terminating numbers is a terminating number.

That means this is closed. It's like the integers in the real numbers, using adding you cannot escape the integers, from inside them. Same sort of closure.

To be 1/3rd is not a finite race-string, so cannot have come from two finite race strings. QED.

Here's how I got to the answer

It could be a fraction so close to 1/12 it's easier to say 1/12 than say "21/256ths"

Lets take "race A", if someone who is x/y race A and someone who is a/b of race A, their offspring is $0.5\frac{x}{y} + 0.5\frac{a}{b} = \frac{1}{2}(\frac{x}{y}+\frac{a}{b})$

But $\frac{x}{y}$ and $\frac{a}{b}$ must also come from this relation.

Now 1/3 (1/12 = 1/4 * 1/3) is a recurring number expressed in base 2 (in base 10 it goes 0.1s then 0.01s, then 0.001s and so forth, in base 2 it goes 1/2, 1/4, 1/8...)

So say we wanted someone who was 1/2 + 1/8 of race A, that'd be "101" in binary in this form, it terminates, the 3/4 used in the comments, that's "110" it terminates.

Remember the relation above, if someone is "0110" (3/8) and someone else is "1100" (3/4) say, we get the result by shifting one right and adding, in this case

 "00110"   
+"01100" which is "01001" or 9/32, 

So to be 1/12th would mean someone who was a quarter, and someone who is a third, but as you can see no one can be a third (in finite steps) starting from 1 or 0 of race A

To sum up! To be 1/3rd something (an infinite string of 0s and 1s) you can't have come from the "product" of two people who have finite strings representing their race. We've seen that "finite strings" are closed (2 people of finite-race string produce someone of finite race string) and thus can't have been produced by two people of finite strings.