Suppose I have a binary constant $q = 0.1010000000000000000000000000000000001001..._2$. In base 10 this number is $q $~$ .6250000000077325..$ and is defined as

$$q = \sum_{\rho}^{\infty} \frac{1}{2^{\rho}}$$ Where $\rho$ is taken as the places where the decimal expansion of $\pi$ is the digit $1$. The first few digits of $\pi$ are $3.14159265...$ so the first few digits of $q$ in binary are $.101000..._2$. How can I prove this number is a Liouville number? Or even that it is transcendental?## 1 Answers

I would guess that it is not a Liouville number. The easy way to prove that a number is Liouville is to have lengthening strings of zeros in the expansion (in any base). You construction will have the strings of zeros averaging $9$ in length, so for large numbers of places it may or may not be approximable by a rational. They are measure zero in the reals, so it would be surprising that a number not constructed to be Liouville would be so. On the other hand, the transcendentals are of measure $1$, so you would expect this number to be transcendental. Proving it is irrational is hard already. We know that $\pi$ is irrational, but it could be that (after the first quadrillion places or so) every tenth place is a $1$ and no others are. The intervening places would be enough to make sure it is transcendental, but your number would be rational. If we prove $\pi$ to be a normal number you at least know yours is irrational.

- 362,355
- 27
- 241
- 432