Is it possible to have a mixed parentage of 10% 90% Is it possible to have a mixed parentage such that you are, for example, 10% Dutch and 90% English? One generation up your mother would be 100% English and your father 20% Dutch and 80% English etc etc. How do you figure this out back up to the ancestors that were 100% Dutch / 100% English?
 A: First I claim that all the proportions we can reach starting from ancestors who were purely English or purely Dutch are of the form $(\sum 2^{\eta_i}, \sum 2^{\delta_i})$ where both sums contain finitely many (possibly zero) terms. This can be proven by induction: Suppose that parents of some person have proportions of form $(\sum 2^{\eta^1_i}, \sum 2^{\delta^1_i})$ and $(\sum 2^{\eta^2_i}, \sum 2^{\delta^2_i})$. Then the person has proportions of $(\sum 2^{\eta^1_i - 1} + \sum 2^{\eta_i^2 - 1}, \sum 2^{\delta_i^1 - 1} + \sum 2^{\delta_i^2 - 1})$, which are of the wanted form.
Next I'll show that $1/10$ can't be represented in this form. Suppose in contrary that there exist integers $a_i > 0$, $i=1,\dots,m$, such that
$$\frac{1}{10} = \sum_{i=1}^m 2^{-a_i}.$$
Let $a_m$ be the largest of $a_i$. Then we get
$$2^{a_m - 1} = 5 \sum_{i=1}^m 2^{a_m - a_i}.$$
But this is a contradiction because now both sides are integers and the right hand side is divisible by $5$ whereas the left hand side is not.
A: J. J.'s answer and André Nicolas's second comment have shown why you cannot have exactly "10% Dutch, 90% English". This is because anyone born to parents who are a fraction $y$ Dutch and fraction $z$ Dutch respectively is $\displaystyle x = \frac{y+z}{2}$ Dutch, so we can inductively prove (assuming we start in the sufficiently distant past with ancestors who are either 100% Dutch or 0% Dutch) that $x$ must be a rational number whose denominator (in lowest terms) is a power of $2$. And in the fraction $\frac{1}{10}$, the denominator $10$ is not a power of $2$.
However, note that you can get arbitrarily close to "10% Dutch, 90% English", with enough generations. Specifically, with $k$ generations, since you want $\displaystyle \frac{p}{2^k} \approx \frac{1}{10}$, you can pick $p$ to be the closest integer to $\displaystyle \frac{2^k}{10}$. So you can have, after


*

*3 generations: $(1/8, 7/8) = (12.5\%, 87.5\%)$

*5 generations: $(3/32, 29/32) = (9.375\%, 90.625\%)$

*7 generations: $(13/128, 115/128) = (10.15625\%, 89.84375\%)$

*9 generations: $(51/512 , 461/512) = (9.9609375\%, 90.0390625\%)$

*11 generations: $(205/2048 , 1843/2048) = (10.009765625\%, 89.990234375\%)$

*13 generations: $(819/8192 , 7373/8192) = (9.99755859375\%, 90.0024414062\%)$


etc.
(Aside: Perhaps there's a way to do these approximations using continued fractions, but I can't think of anything.)
