# A number is normal base b iff it is simply normal in bases $b^k$

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.

• 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) – coffeemath Dec 4 '18 at 2:45
• @coffeemath done – Sasha Dec 4 '18 at 3:00
• This is not set-theory. Please do not add the tag back. – Andrés E. Caicedo Dec 4 '18 at 5:50