I'm trying to prove this original statement :
Let suppose that $f$ is a function that admits a Taylor expansion such that $$ \forall x\in \mathbb{R}, \ > f\left(x\right)=\sum_{n=0}^{+\infty}a_nx^n $$ with $a_n \in \left\{0,1\right\}$ and that $$f\left(\frac{2}{3}\right)=\frac{3}{2}$$
Prove that $\displaystyle f\left(\frac{1}{2}\right)$ is irrational.
My idea was to prove it by supposing the existence of two integers $p$ and $q$ with $p \wedge q=1$ such that $$ f\left(\frac{1}{2}\right)=\frac{p}{q} $$ and to show that there is a contradiction. With the hypothesis i only know that $$ \sum_{n=0}^{+\infty}a_n\left(\frac{2}{3}\right)^n=\frac{3}{2} $$ Maybe I could use that $$ \sum_{n=0}^{+\infty}a_n\left(\frac{2}{3}\right)^n\sum_{n=0}^{+\infty}\frac{a_n}{3^n}=\frac{3}{2}f\left(\frac{1}{2}\right) $$ and to use Cauchy's product but it seems not the right way. Any help ?