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: 1
x0
(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.