Can an irrational number have a finite number of a certain digit?

Question

This question came up because I was wondering the following: If the digits of PI are placed in ascending order, what is the <insert-large-finite-number-here>th digit?

I believe that the answer is 0, but I am not sure. The reason why I'm not sure is dependent on the answer to whether an irrational number can have a finite number of a certain digit. If this is the case, then it is conceivable that there could be a finite number of zeros. Taking an example, let us say that we are going to encode every digit of PI through the function $$ f(x) = \left\{ \begin{array}{c} 10+x,\quad when \quad x≠0\\ 99,\quad when \quad x=0 \end{array} \right. $$ That is, for example if I am encoding the digits of $20$, then I'd replace the $2$ with $f(2)$ and the $0$ with $f(0)$ I'd get, as my end result, $1299$. This mapping function seems to totally eliminate zeros from the answer. By doing the same thing to PI, I can totally eliminate its zeros. It also seems conceivable that I can still generate a unique mapping if I apply the function to digits only after the first zero.

So: does my argument hold water? Is it possible to turn an irrational number into one with a finite number of a certain digit? Additionally, is it possible to prove whether an arbitrary irrational number has a finite number of a certain digit?

Answer 1

The number $0.010010001000010000010000001\ldots$ is irrational and has no instances of any digit other than $0$ and $1$. More generally, pick any two digits $d$ and $e$, and form the number

$$0.\mathrm{dedeedeeedeeeedeeeeed}\dots\;;$$

it never becomes periodic, so it’s irrational. Of course one can insert any finite string of digits between the decimal point and the expansion of an irrational number and still have an irrational number, e.g., $0.345345010010001000010000010000001\ldots$.

Answer 2

Yes. For an extreme example, let $a$ have decimal expansion $$0.101001000100001000001\dots.$$ The number of $0$'s between consecutive $1$'s is $1$, $2$, $3$, $4$, $5$, and so on. Thus the decimal expansion of $a$ cannot be ultimately periodic, and therefore $a$ is irrational.

By the way, it is not known whether the decimal expansion of $\pi$ has infinitely many $0$'s. The same applies to any other digit.

Answer 3

Yes, and without giving you another example let me instead give you the vocabulary. An irrational number is said to be a "normal" number if each digit, 0 through 9 in base-10, and in each base numbering system, occurs in equal quantities. All irrational numbers must contain at least two unique digits in infinite quantities, but not necessarily in equal proportion.