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

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.

• I am not sure I understand your question. Base is just representation. Why would the way in which we write some number on a piece of paper changes its properties? Mar 30, 2019 at 23:41
• @Brian: Good question! But notice that $n_{10}$ and $n_2$ are rather different numbers. It is not that $n_2$ is the binary of $n_{10}$. Rather the $n_{10}$ digits are mapped to even/odd binary. Mar 30, 2019 at 23:43
• I see, thanks for the clarification! I overlooked the distinction between converting between bases and concatenating the base-converted digits of a number. Mar 30, 2019 at 23:47

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.
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: $$n_{10}=0.123456789012123434565678789090121212\dots.$$
• In fact, you could just take $n_{10}$ to be the string of even digits and get the rational number $n_2 = 0$. Mar 30, 2019 at 23:50