Given any two normal numbers, can we always find a find a truncation point at which the digits can be rearranged from one to the other?

Question

I just watched a video showing that you can re-arrange (or permute as it is called in the video) the first nine decimal digits of the reciprocal of pi, to give the first nine decimal digits of the reciprocal of phi (the golden ratio).

That is to say: You can rearrange the digits 0.318309886 ($1/\pi$) to form 0.618033988 ($1/\phi$)

This seems pretty much a coincidence. It won't work for the first 8 decimal digits, but presumably might work again for the first [some big number] of digits due to random chance.

This got me thinking - is this inevitable, or is it possible to find two normal numbers where truncating the base-n representation of the digits will never lead to being able to re-arrange the digits of one to form the other?

Some definitions, for a pair of numbers, A and B, let us say that the two are base-n rearrangeable, if for some number of digits, the truncated representation of A in base-n can have the digits re-arranged to form the truncated representation of B.

For most pairs of numbers they are trivially base-n rearrangeable or not rearrangeable. 0.12 and 0.21 are base-10 rearrangeable. 0.1919... and 0.9191… are base-10 rearrangeable. 0.2 and 0.1 are not, nor are 0.333… and 0.666…

The interesting part comes for normal numbers. Is every normal number rearrangeable in every base for some value of x?

For the binary representation, I think this must be possible; my non-rigourous mental argument is as follows:

Let $a(x)$ be the number of 1s in the first x bits of a normal number A, and $b(x)$ be the number of 1s in the first $x$ bits of a normal number B. At large numbers, $a(x)/n$ and $b(x)/n$ tend to 0.5 (by the definition of normal), but sometimes $a(x) > b(x)$, and sometimes $b(x) > a(x)$. Therefore at some point $a(x) = b(n)$ between those two points.

Answer 1

For any base $b$, you can have two base-$b$ normal numbers that are not rearrangeable to each other at any truncation.

Let $x$ be base-$b$ normal with digital representation $0.d_1d_2d_3...$ where $0\le d_i\le b-1$ for all $i$.

Then construct $y$ by taking $e_1$ to be the least nonnegative residue of $d_1+1 \pmod b$ (i.e., increase $d_1$ by $1$ except if $d_1=b-1$, in which case $e_1=0$). Then let $y=0.e_1d_2d_3d_4...$. That is, after the first digit, match the rest of the digits of $y$ to the corresponding digits of $x$.

Certainly $y$ is normal since $x$ is. Also at any truncation, $y$ will have one more digit $e_1$ than $x$, and $x$ will have one more digit $d_1$ than $y$. So $x$ and $y$ are not base-$b$ rearrangeable to each other.

Base-$2$ example:

$\begin{array}{ccc} x&=&0.10011101001...\\ y&=&0.00011101001...\end{array}$

$x$ is always one $1$ ahead of $y$.

Answer 2

It will work in base $b=2$ and $3$, but not higher bases.

In base $b$, the question is whether the number of $0$s is the same, the number of $1$s the same, and so on.
Pick a value of $n$. Let $x_0$ be the number of zeros in the first $n$ digits of $X$, and so on.
So does the vector $(x_0,x_1,...,x_{b-1})$ equal $(y_0,y_1,...,y_{b-1})$?
Let the vector $\vec z=\vec x-\vec y$. This is a random vector that, as $n$ increases, does a random walk on the $b-1$-dimensional subspace $\sum_iz_i=0$. You can rearrange the first $n$ digits in $X$ to those of $Y$ if $\vec z=\vec0$.

It is well-known that random walks return to the origin many times with probability $1$ in one or two dimensions, which is base $2$ and base $3$. But the chance of returning to the origin at all in three or more dimensions is less than $1$.