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I am trying to understand what normal numbers are. Just for simplicity I want to talk about base 10. I understand that a number is normal in base 10 if there a probability of $\frac{1}{10 } $ such that the numbers 0-9 pop up and a probability of $\frac{1}{100}$ that the numbers $0-99$ pop up in the decimal expansion and so on...

However I am wondering, in this case when we talk about normal numbers we are only talking about irrational numbers since numbers like $2$ or $3$ don't have a decimal expansion so there is no sense in talking about them as normal numbers?

Or for example a number like $\frac{1}{3} = 0.3333..$ is not normal base 10 since it's decimal expansion only contains the number 3.

I am just wondering if a normal number has to do anything with the normal form of a number.

Andrés E. Caicedo
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Sasha
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    Numbers like 2 and 3 certainly do have decimal expansions. For example, $$2 = 2.000\overline{0}. $$ – Xander Henderson Dec 01 '18 at 20:42
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    A normal number (even one which is only normal in base ten) must necessarily be irrational, yes. Most real numbers are normal, yet the only numbers we know to be normal are the ones we have explicitly constructed to be normal. – Arthur Dec 01 '18 at 20:43
  • @Arthur Not so. The n-digit sequences must be equiprobable. See http://mathworld.wolfram.com/NormalNumber.html – saulspatz Dec 01 '18 at 20:45
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    Normal doesn't just refer to single digits. Yes, each digit should occur with probability $\frac 1{10}$ but each pair of digits should occur with probability $\frac 1{100}$ and so on. See, e.g., this. But yes, normal implies irrational. – lulu Dec 01 '18 at 20:46
  • @saulspatz You're right. I didn't know that. – Arthur Dec 01 '18 at 20:48
  • @lulu He already mentioned the double digits, and the "and so on ..." – Arthur Dec 01 '18 at 20:48
  • Ok so when I want to explain the definition of a normal number I need to consider every number and it's decimal expansion which is n.0000 for $n \in \mathbb{N}$. Then from there we have to look at how probable it is that certain digits pop up. – Sasha Dec 01 '18 at 20:50
  • @lulu however when we are talking about simply normal we just refer to single digits? – Sasha Dec 01 '18 at 20:51
  • @Arthur: This answer discusses what you're thinking of. And for those interested in lots of links, my answer to Normal Numbers as members of a larger set? (apparently made when I had a bit of time on my hands) will keep you busy for a while. – Dave L. Renfro Dec 01 '18 at 20:52
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    @Sasha Yes. But that does not imply irrational. The rational number $.\overline {0123456789}$ is simply normal in that sense. Proper normality is a much more interesting property. – lulu Dec 01 '18 at 20:57
  • @lulu so when we talk about the normality of numbers even though no natural or rational number is normal almost all real numbers are normal since the set of natural and rational numbers have measure $0$? – Sasha Dec 01 '18 at 21:04
  • @lulu is there a proof or theorem that says that if a number is normal it must be irrational – Sasha Dec 01 '18 at 21:05
  • It's obvious. Suppose you had a normal rational $\alpha$ . Let $P$ denote the length of the period in the decimal expansion of $\alpha$. Then the number $1\underbrace{000\cdots 000}_{\text {2P times}}1$ can not occur in the decimal expansion of $\alpha$. – lulu Dec 01 '18 at 21:08
  • @lulu so this is because the decimal expansion of rational numbers are periodic? – Sasha Dec 01 '18 at 21:11
  • Yes, exactly. $\quad$ – lulu Dec 01 '18 at 21:12
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    Actually my proof should be modified slightly to take into account the fact that the period may not start right away. Easy to handle...suppose there are $Q$ terms before the period of length $P$ starts and replace the $2P$ in my expression by $2(P+Q)$ or such. – lulu Dec 01 '18 at 21:15

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As you noted a normal number ( in some basis) is necessarily irrational. We know (it was proved by E. Borel) that ''almost all'' (in the sense of measure theory) real numbers are absolutely normal normal, that is normal in all basis, but the proof is not constructive, and it is not clear if there is some computable normal number. What we can say is that we dont know if nubers as $\pi, e, \sqrt{2}$ are normal also in base $10$ .

We know that the Champernowne number is normal in base $10$ but it is not normal in all basis.

You can also see here for the problem of construction of a normal number.

Emilio Novati
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  • It actually known that it's not normal in all bases, or just unknown whether it is? I've always thought it was the latter, but I don't have any support for that. If the former, do you know of a reference? – saulspatz Dec 01 '18 at 21:03
  • I did not verify the proof, but here (https://www.google.it/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=2ahUKEwi4r9-Xxf_eAhUFyxoKHXQJCSAQFjAAegQIABAC&url=http%3A%2F%2Fsummit.sfu.ca%2Fsystem%2Ffiles%2Firitems1%2F10167%2Fetd1934.pdf&usg=AOvVaw1t3UAeMdA6fF__ThMY0FfH ) the autors say that Champernowne’s number fails to be strongly normal. – Emilio Novati Dec 01 '18 at 21:20
  • Thank you. I'll enjoy reading this, I'm sure. – saulspatz Dec 01 '18 at 21:37