Just to augment @js21's answer. The question of whether irrational $x_i$ are necessarily transcendent is equivalent to [this question][1]. 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.)


  [1]: https://mathoverflow.net/questions/114758/