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I saw a proof submitted in 2018 to ArXiv.org, see here. The author is Pierpaolo Uberti (Department of Economics, University of Unige.) I doubt it is correct, and would like you to check it out, to see if you find some major issues. I spent a lot of time working on this problem myself, and I am nowhere close to a solution.

Here are my comments:

  • The proof does not use anywhere specific properties of $\sqrt{2}$. To the contrary, what I've done so far is deeply connected to very specific, non-trivial properties of $\sqrt{2}$, see here and here.
  • It is short and based on elementary arguments.
  • There is an argument based on permutations of the digits (see proof of his Theorem 2), and while such permutations preserve the proportions of $0$ and $1$ in the binary expansion for the first $n$ digits, it does not remain true as $n\rightarrow\infty$.

Is this proof actually correct (maybe I am missing something), or as I suspect, it is flawed beyond repair?

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    $\begingroup$ Given that there are multiple errors of conflating 'normal' with 'random' in the Introduction itself, I'm immensely leery of any purported proof even before making it to the body of the paper. $\endgroup$ – Steven Stadnicki Jan 15 '20 at 0:12
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    $\begingroup$ @VincentGranville Some authors use “normal in base $2$” to mean just 50/50 zeroes and ones, reserving the term “strongly normal” for the equidistribution of longer binary strings. But you are absolutely right that this paper is unconvincing of anything close to the weaker claim. $\endgroup$ – Erick Wong Jan 15 '20 at 0:20
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    $\begingroup$ Note that there can be a 50/50 distribution of $0$s and $1$s in the binary expansion of a non-normal number. For instance, the Prouhet-Thue-Morse constant has the same amount of $1$s and $0$s by definition, but is not normal because it cannot contain the sub-sequences $111$ and $000$. $\endgroup$ – Klangen Jan 16 '20 at 10:34
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    $\begingroup$ Considering the difficulty of proving that the binary digits in $\sqrt{2}$ are 50/50 distributed (which is far from what the majority of mathematicians would call "normal" in base $2$), odds are clearly against such a proof. In particular , if it is short and based on elementary arguments. $\endgroup$ – Peter Jan 20 '20 at 14:28
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    $\begingroup$ @VincentGranville It is of course your choice, but my advice is that you do not waste your time with such problems. To me it appears that the chance to prove all the three conjectures Riemann Hypothesis, Goldbach's conjecture and the Collatz-conjecture would be much better. $\endgroup$ – Peter Jan 20 '20 at 14:42

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