# Can the b-ary power series of an irrational number evaluate to a rational number?

Assume we have a base $$b$$ and a power series $$f(x)$$ with the coefficient of each power $$x^n$$ being equal to the nth non-zero $$b$$-ary digit times $$b^{-i}$$ (where $$i$$ is the index of the b-ary digit) in the expansion of an irrational number $$m$$.

For instance, in base three: $$\pi = 10.010211012222..._3$$, so:

$$a_0 = 3^{-1}, a_1 = 2 \cdot 3^{-3}, a_2 = 3^{-4}, ...$$

and:

$$f(x) = 3^{-1} + 2 \cdot 3^{-3} \cdot x + 3^{-4} \cdot x^2 \cdots$$

My question is, can $$f(x \in \mathbb{Q}, x \neq 0) \in \mathbb{Q}$$?

Suppose, for example, $$b > 3$$ and $$(b-1) x + 1 = x + 2$$, which is true if $$x = 1/(b-2)$$. Let $$m = \sum_{i=1}^\infty c_i b^{-2i}$$, where each $$c_i$$ is either $$(b-1)b+1$$ or $$b + 2$$. Thus the base-$$b$$ expansion of $$m$$ consists of pairs $$((b-1),1)$$ and $$(1,2)$$. There are uncountably many possible $$m$$ (corresponding to two choices of $$c_i$$ for each $$i \in \mathbb N$$), so some of them will be irrational (in fact, those where the sequence of $$c_i$$ is not eventually periodic). But since $$(b-1) x+1 = x+2$$, the value of $$f(x)$$ is the same for all such $$m$$: in fact it is $$\sum_{i=1}^\infty (x + 2) x^{2i} = \frac{x^3 + 2 x^2}{1-x^2}$$ which is rational.
• Great answer, and very smart technique! However, what I was after was the case that the coefficients of the power series are the digit times the power of the digit in the base-b expansion (e.g. 0.101 in binary --> 0.5 + x*0.125 + ...) -- sorry for the lack of clarity. So I believe the left sum would have an additional term: ${(\frac{1}{x} + 2)}^{2i}$. Sep 25, 2019 at 9:06