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I am trying to prove that a number is normal base b $\iff$ it is simply normal in all bases $b^k$ for every integer $k \geq 1$.

I'm a little confused on this because if for example we take a number that is normal is base 3 how would that be simply normal in base 9 as it would not have any digits greater than 3.

These are my definitions for normal and simply normal:

We say a number $x$ in decimal expansion form base-$b$ is simply normal base-$b$ if $\lim_{n \to \infty} \frac{N_n^b(x;{w})}{n} = \frac{1}{b},$ $\forall w \in \{0,1,2,...b-1\}$.

A number $x$ in decimal expansion form base $b$ is normal base-$b$ if for any arbitrary finite string (or word) $w$ with letters from the alphabet $\{0,1,2,...,b-1\}$ $ lim_{n \to \infty}\frac{N_n^b(x;w)}{n} = \frac{1}{b^{|w|}}$ where $|w|$ denotes the lengh of the word.

If somebody could help me get started on how to prove this if and only if statement that would be great.

Update: I have figured out the forward direction, I am still confused on the reverse.

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  • $\begingroup$ In base 9 it would have digits 0--8. That is, block it off in subparts of length 2 and convert each to a number from 1 to 8. Example 122122=(12)(21)(22)=(5)(7)(8) base 9. Could you put definitions of normal and simply normal in question? (not as comm4ent or link) $\endgroup$ – coffeemath Dec 4 '18 at 2:45
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    $\begingroup$ @coffeemath done $\endgroup$ – Sasha Dec 4 '18 at 3:00
  • $\begingroup$ This is not set-theory. Please do not add the tag back. $\endgroup$ – Andrés E. Caicedo Dec 4 '18 at 5:50

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