# Does every normal number have irrationality measure $2$?

A normal number is a number whose digit expansion in any base is "uniform" in the sense that all finite digit strings occur with the "statistically expected" frequency.

I read a sentence somewhere which could be understood as implying that every normal number had irrationality measure $2$.

Is it known if all normal numbers have irrationality measure (approximation exponent) $2$?

Also, are there some non-normal irrational numbers whose irrationality measures are known?

Also, are there some non-normal numbers with irrationality measure $2$ known?

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Found the post that inspired this question, where it says "The normality of $\pi$ (which would imply the irrationality measure $2$)...". So is that implication known for any normal number, or does it require some special properties that $\pi$ has? –  Jeppe Stig Nielsen Dec 23 '13 at 8:47
Isn't that answered (on the affirmative) here? math.stackexchange.com/a/100915/312 –  leonbloy Feb 21 '14 at 12:09
@leonbloy Not that I understand the answer you refer to, entirely, but it seems that it uses other properties of $\Omega$, possibly its incomputability, and therefore I am not sure if that proof works for arbitrary normal numbers? –  Jeppe Stig Nielsen Feb 21 '14 at 13:37
I don't understand it fully either, but I think the key is that the sequence of digits is incomprenssible (which is a consequence of its incomputability), and the statistical properties of a normal number should imply incomprenssibility (some of this is discussed in Wikipedia). –  leonbloy Feb 21 '14 at 14:07

normal $\; \not\Rightarrow \;$ irrationality measure $2$

There exist numbers that are normal with irrationality measure $>2$. In fact, there exist normal numbers (meaning normal with respect to every base) that have irrationality measure $\infty.$ This is Theorem 2 in Bugeaud [2] (2002). For related results, see [1] and [3].

irrationality measure $2$ $\; \not\Rightarrow \;$ normal

There exist numbers with irrationality measure $2$ that are not normal. In fact, there exist numbers with irrationality measure $2$ that fail to be simply normal in every base. Note that this is stronger than failing to be simply normal in base $10,$ which in turn is stronger than failing to be normal in base $10,$ which in turn is stronger than "normal" in the sense that you are asking about. Shallit [4] (1979) showed that the continued fraction expansion of

$$\sum_{n=0}^\infty 2^{-2^{n}}$$

has bounded partial quotients (see Theorem 3 on p. 213 for meaning of $B(u,\infty)$ and Theorem 9 on p. 216 for the result), and thus has irrationality measure $2.$ [Indeed, for this we only need the $n$th partial quotient to be bounded by a linear function of $n.$ See Robert Israel's answer to the math overflow question Numbers with known irrationality measures.] However, it is clear that this number is not simply normal in any base. Indeed, in this number's expansion in any base, the proportion of $0$'s approaches $1,$ and hence the proportions for each of the other digits approaches $0.$

[1] Yann Bugeaud, Nombres de Liouville et nombres normaux [Liouville numbers and normal numbers], Comptes Rendus Hebdomadaires des Séances de l'Académie des Sciences 335 #2 (2002), 117-120.

[2] Verónica Becher, Pablo Heiber and Theodore A. Slaman, A computable absolutely normal Liouville number, preprint, 30 January 2014, 14 pages.

[3] Satyadev Nandakumar and Santhosh Kumar Vangapelli, Normality and finite-state dimension of Liouville numbers, arXiv:1204.4104v2, 21 January 2014.

[4] Jeffrey Outlaw Shallit, Simple continued fractions for some irrational numbers, Journal of Number Theory 11 #2 (April 1979), 209-217.

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Great answer with references that are only one month old. I was beginning to believe that neither inclusion was true. Actually, can you say if it is known if the famous Champernowne constant (which is normal, at least in one base) has irrationality measure exceeding $2$? Also, is it actually false that (one knows that) the normality of $\pi$ would imply irrationality measure $2$ for $\pi$ (see first comment to my question)? –  Jeppe Stig Nielsen Feb 26 '14 at 19:30
Googling led me to this 5 December 1996 sci.math post by Shallit (author of [4] above), where he says Amou found the irrationality measure of Champernowne's constant in 1991. Using the MR Lookup web page (I don't have access to Mathematical Reviews), I found the title of the paper, which then led me to Amou's 1991 paper, where the abstract says the irrationality measure is shown to be $10.$ I don't know about your $\pi$ question. –  Dave L. Renfro Feb 26 '14 at 21:19
About your $\pi$ question, I doubt that any of the known properties of $\pi$ in conjunction with knowing whether $\pi$ is (or is not) normal would lead to any additional conclusions about the irrationality measure of $\pi.$ In fact, I doubt that any reasonable proof (which might contain new ideas) concerning the normality of $\pi$ would help, where I include "reasonable" to exclude a proof where you cheat and just include a statement in the proof about the irrationality measure for no good reason. –  Dave L. Renfro Feb 26 '14 at 21:45
The stuff about Champernowne's constant is really cool, and I just included it in the linked Wikipedia article in the hope that it will be even easier to google in the future. With your reference. It appears that the remark about $\pi$ which originally made me ask the question, is unsupported. –  Jeppe Stig Nielsen Feb 26 '14 at 22:06
@DaveL.Renfro Would you elaborate on where the implication "normal" $\implies$ "incromprenssible" $\implies$ "rationality measure=2" fails? –  leonbloy Mar 2 '14 at 23:51

The Dirichlet transcedentals, which are of the form $$a=\sum_{n\in\mathbb N} a_n 10^{-n!}, \quad a_n\in\{1,2\},$$ are as many the real numbers.

The irrationality measure of these numbers is infinite, and hence they are not normal!

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@AndréNicolas: You are right. I have corrected it. –  Yiorgos S. Smyrlis Dec 23 '13 at 0:01
I know that, and actually I "stroke out" the line in my original post where I asked about that, a long time before your answer. Your answer does not address the main question I posed, about a relation between normality and $\mu=2$, since your example is non-normal with $\mu=\infty$. (Here, $\mu$ means irrationality measure.) –  Jeppe Stig Nielsen Dec 23 '13 at 0:11