Just to augment @js21's answer. The question of whether irrational $x_i$ are necessarily transcendent is equivalent to this question. As the answer suggests, even this specific question is wide open, or at least was such in 2012.
As for the other bases $b>2$, the result is similar. Defining $x_i$ as in js21's answer, you get that $$ x_{b-1}=\frac{x+\frac1{b-1}-f(x)}{b} $$ is algebraic though not normal. (Surely, if $x_{b-1}$ is just rational, then $x-(b-1)x_{b-1}$ will be an algebraic irrational though not normal.)