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A normal number is a number where no number is favored to appear in the digits. Does this definition imply that all whole numbers appear in its digits? Because the definition involves notions from probability, I was wondering if it might happen that a certain number would not appear in the digits of a normal number without contradicting the definition.

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1 Answer 1

Yes it does. Working in base B, a string with A digits should have the natural density $ \frac {1}{ B^A} $ by definition of a normal number. If the string doesn't appear at all, then it has natural density 0.

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Looking at the first $N$ numbers, let the number of times that the string appears be $N(S)$. Then, the definition of normal means that $\lim_{N \rightarrow \infty} \frac {N(S)}{S} = \frac {1}{B^A}$, which is the natural density. In fact, you can conclude that the string appears infinitely often, since otherwise the density will still be 0. –  Calvin Lin Dec 29 '12 at 4:49
So is it possible to produce a non-normal number where all natural numbers appear anyway or are these notions equivalent? –  user54358 Dec 29 '12 at 5:01
I believe so, but may be wrong. Take a normal number, and then insert $ 2^{n-1}$ 0's in the $2^n$ place for $n\geq 2$. For example, take 0.123456789101112... and transform it into 0.12 0 34 00 5678 0000 91011121 00000000 ... Then for given any string $S$ of length $A$ which isn't 0, $lim_{N \rightarrow \infty} \frac {N(S)} {S} = \frac {1}{2 \times 10^A}$ (which requires a short argument), while the string 0 occurs with natural density $ \frac {1}{2} $. –  Calvin Lin Dec 29 '12 at 5:21
Like if you have the number 0.10203040506070809010011012... the concatenation of natural numbers with 0 inserted in between. What's the density of 0 there? Is 0 already denser or is this still normal? –  user54358 Dec 29 '12 at 5:22
Hey Calvin, if you did that construction wouldn't you end up cutting some of the numbers? Like 13 for example got cut by the zeroes you inserted. –  user54358 Dec 29 '12 at 5:27

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