Rainbow numbers: Can mapping digits to different bases produce different varieties of irrationality?

Question

This is a follow-up to the question, "Irrationality of 0.123456789101112 … and similar numbers."

There I took some decimal number, in one case Champernowne's constant, $$ n_{10} = 0.123456789101112131415161718192021 \ldots \;, $$ and then mapped each base-$10$ digit to a binary digit, $0$ and $1$ for each even and odd digit of $n$: $$ n_{2} = 0.101010101101110111011101110110001 \ldots \;, $$ where $n_{2}$ is to be interpreted as a binary number. My question is, essentially:

Q. Can $n_{10}$ be irrational/transcendental but $n_{2}$ rational? Can you provide examples? Do they work for multiple bases $b$, $n_{b}$?

The ideal would be a number which is trascendental in base-$10$, but when the digits are mapped to base $b$, it becomes rational for some $b_1$, algebraic for some other $b_2$, maybe normal for some other $b_3$, etc.

You see where I'm going with this. I want a rainbow number: a number that exhibits {rational, irrational, algebraic, normal, transcendental} under this digit mapping. It could serve as an educational tool.

Answer 1

The number $$ n_{10} = 0.2772727222227227777.... $$ is transcendental, but $$ n_2 = 0.0110101000001001111... = \sqrt{2}-1 $$ is an algebraic irrational and $$ n_5 = 0.2222222222222222222... = 2/3 $$ is rational.

You should be able to use the Chinese remainder theorem to encode as many numbers into the base $b$ expansion of a number as $b$ has distinct prime factors.

Answer 2

Yes this is possible. For $n_2$ to be rational, the binary expansion must eventually have a repeating pattern of digits. Similarly for $n_{10}$. So simply take $n_2$ to be some repeating pattern, say $n_2=0.101010\dots$ (this is equal to $2/3$). Now by your mapping, $n_{10}$ must consist of alternating odd and even digits in such a way that the pattern never repeats. It's enough to come up with odd and even strings that never repeat, e.g., $$1,3,5,7,9,1,1,3,3,5,5,7,7,9,9,1,1,1,\dots$$ and $$2,4,6,8,0,2,2,4,4,6,6,8,8,0,0,2,2,2,\dots.$$ So then the number $n_{10}$ would be: $$n_{10}=0.123456789012123434565678789090121212\dots.$$