a. Yes b. No a. If alpha is rational, then the sequence $ d_i$ is eventually periodic. In particular, there is a periodic sequence $ e_i$ such that only finitely many $ i$ have $ d_i\ne e_i$. It is clear that for such a sequence, $ \Sum_{i=1}^{\infty}(d_i-e_i)r^i$ is rational (being the sum of finitely many rationals), and hence we just need to show that $ \Sum_{i=1}^{\infty}e_ir^i$ is rational. Note that since $ |r|<1$ and $ e_i$ is bounded, this sum is convergent. Let it converge to $ x$, say. Suppose $ e_i$ has period $ k$. Then $ x=e_1r+e_2r^2+\ldots+e_kr^k+xr^k$. Thus $ x=\frac{(e_1r+e_2r^2+\ldots+e^kr^k)}{1-r^k}$, and so is rational. b. Ignoring the possibility of $ r=0$, let $ r=\frac 23$ and consider the above argument for eventually periodic $ d_i$. If $ x=\Sum_{i=1}^{\infty}e_ir^i$, then $ x=\frac{e_1\frac23+e_2\frac{2^2}{3^2}+\ldots+e_k\frac{2^k}{3^k}}{1-\frac{2^k}{3^k}}= \frac{e_12^13^{k-1}+e_22^23^{k-2}+\ldots+e_k2^k}{3^k-2^k}$. Note that $ x$ has an even numerator, and since $ k\ge 1$ an odd denominator. Note also that if two rational numbers have even numerators and odd denominators, so does there sum. Thus, since $ \Sum_{i=1}^{\infty}d_ir^i=x+\Sum_{i=1}^{\infty}(e_i-d_i)r^i$, this sum is a rational number with even numerator and odd denominator. Now, given any number $ \beta$ in the range $ [0..18]$, let $ \beta_0=\beta$, and let $ d_i$ be the largest integer in the range $ [0..9]$ such that $ \frac{2d_i}3<\beta_{i-1}$, and let $ \beta_i=\frac{3\beta_{i-1}}{2}-d_i$. By a clear induction $ 0\le\beta_i\le 18$ for all $ i$. By another clear induction, $ \beta=\Sum_{i=1}^kd_i\frac{2}{3}^i+\beta_k\frac{2}{3}^k$. Letting $ k$ tend to 0, since $ \beta_k$ is bounded and $ \frac{2}{3}^k\to 0$, it follows that $ \beta=\Sum_{i=1}^{\infty}d_i\frac{2}{3}^i$. We have shown that any number $ \beta$ has an expansion of the form $ \beta=\Sum_{i=1}^{\infty}d_i\frac{2}{3}^i$, but that if $ \beta$ is not a rational number with an even numerator and odd denominator, the the sequence $ d_i$ cannot be periodic, so the associated real number $ \alpha=\Sum_{i=1}^{\infty}d_i10^{-i}$ cannot be rational. In particular, if we take this sequence for $ \beta=1$, we get an irrational number $ \alpha=0.10100000100100101000\ldots$. Luke See my puzzle blog at http://bozzball.blogspot.com/[/url]

Comments: Say a rational $ r$ with $ |r|<1$ is well-behaved if $ \Sum_{i=1}^{\infty}d_ir^i$ being rational with $ d_i\in\{0,1,2,3,4,5,6,7,8,9\}$ implies that $ \Sum_{i=1}^{\infty}d_i10^{-i}$ is rational. We can ask which rational numbers are well-behaved. Theorem I: $ \frac1n$ is well-behaved for $ n\ge 10$ $ \Sum_{i=1}^{\infty}d_ir^i$ is the base $ n$ expansion of a rational number, and is hence eventually periodic. Theorem II: $ \frac mn$ is not well-behaved for $ m<n<10$. For $ i\ge 1$, let us write $ d_{2i-1}=m$ and $ d_{2i}=0$ if $ i$ is prime and $ d_{2i-1}=0$ and $ d_{2i}=n$ if $ i$ is composite. Then $ d_{2i-1}\frac{m}{n}^{2i-1}+d_{2i}\frac{m}{n}^{2i}=\frac{m^{2i}}{n^{2i-1}}$. Thus $ \Sum d_i\frac{m}{n}^i=\Sum \frac{m^{2i}}{n^{2i-1}}=\frac{m^2}{n-m}$ even though $ d_i$ is not periodic. It is clear that 0 is not well-behaved. The argument I provided for $ \frac23$ actually works for any rational number $ r$ with $ \frac1{10}<r<1$. Theorem III: $ r$ is not well-behaved if $ 1>r>\frac1{10}$. If $ r$ is of the form $ \pm\frac1n$, then use Theorem II. Otherwise, $ r$ can be written in a form where there is a prime $ p$ that divides the numerator but not the denominator. Then any expression $ \Sum d_ir^i$ which is eventually periodic gives a rational number in which $ p$ divides the numerator but not the denominator (by the same argument as in the previous post). However, all numbers in the range $ [0,1/(r-1)]$ can be expressed in such a form. Thus 1 can be expressed in such a form, but only in non-periodic ways. But what of rational numbers in the range $ (0,\frac1{10})$ that are not of the form $ \frac1n$? The same argument shows that for all such numbers $ q$ there is a prime $ p$ that divides the numerator of all eventually periodic sums $ \Sum_{i=1}^{\infty}d_iq^i$. Thus if there is any rational number with a numerator not divisible by $ p$ that can be written in this form, $ d_i$ is non-periodic and hence $ q$ is not well-behaved. However, the set of numbers which are expressible in this way is of cantor type, and it seems difficult to show that there definitely is such a rational in the set. I weakly conjecture that there is no such rational in the set, and that $ q$ is in fact well-behaved if $ q<\frac1{10}$. Luke See my puzzle blog at http://bozzball.blogspot.com[/url]