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The other day I got to thinking about the decimal expansion of $\sqrt{2}$, and I stumbled upon a somewhat embarrassing problem.

There cannot be only one digit that occurs infinitely often in the decimal expansion of $\sqrt{2}$, because otherwise it would be rational (e.g. $\sqrt{2} = 1.41421356237\ldots 11111111\ldots$ is not possible).

So there must be at least two digits that occur infinitely often, but are there more? Is it possible that e.g. $\sqrt{2} = 1.41421356237\ldots 12112111211112\ldots$?

Mankind
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    I think that every algebraic irrational number contains each one of the digits (on any natural base!!!) infinitely many times. – barak manos Dec 15 '15 at 13:45
  • If there is a "general" first part $g$ of the digit expansion of $\sqrt 2$ which contains somehow all digits (but has only finitely many digits (let its number be called "m") and is thus rational) and then a "special" second part of infinite length which contains, say only the digits $1$ and $2$ then we would have $\sqrt 2 = g + 10^{-m} s $ and $2 = g^2 + 2 gs + s^2$ and I would try then to find configurations in $s$ which disallow consecutive zeros at the end. – Gottfried Helms Dec 16 '15 at 10:58
  • see this similar question:

    https://math.stackexchange.com/q/5004475/870236

     – Math Admiral Nov 30 '24 at 15:46

2 Answers2

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This problem is wide open. It is conjectured that every irrational algebraic number is absolutely normal (i.e. in every base, digits appear asymptoticaly with the same density). However, it is not even known whether there is any algebraic irrational with some three digits appearing infinitely many times in any base! Hence, to the best of our knowledge, every irrational algebraic number could eventually have only zeroes and ones in every base.

Edit way too many years later: the situation is bad but not nearly as bad as my comment above describes. It is definitely not the case that it could happen for all algebraic irrationals (e.g. as Elliot points out in a comment, it obviously can't happen for both $\sqrt{2}$ and $2\sqrt{2}$!). With a little effort you can also show that $\sqrt{2}+q$ for some rational $q$ has at least $3$ distinct digits appearing infinitely often, but my argument is pretty ad hoc and I don't know the extent to which it has been generalized.

However, for any specific algebraic irrational, our understanding is nearly void - I do think it is consistent with our state of knowledge that for instance $\sqrt{2}$ has, in any integer base, only digits 0 and 1 from some point on.

Wojowu
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  • This is amazing. I mean, a problem that seems so simple to say something reasonable about turns out to be quite the opposite. I am very surprised about this. How in the world does mathematics not know this? This has made my day. – Mankind Dec 15 '15 at 14:06
  • I think this is a bit quick-shot. Recently (ok: "recently") Plouffe and the Borweins found that algorithm which produces the digits of $\pi$ (in hex-base) and by such a formula one should be able to say something about frequencies. I don't know actually about the same in dec-base however... – Gottfried Helms Dec 16 '15 at 10:52
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    @GottfriedHelms I assume this is the one: https://en.wikipedia.org/wiki/Bailey%E2%80%93Borwein%E2%80%93Plouffe_formula I haven't seen that before, but maybe you are right that it is possible to figure out when the algorithm returns a certain digit, and if you can do this for all digits, then conclude that the hex-expansion of $\pi$ contains all digits infinitely often. Maybe you can even say something about the frequency of each. I just checked, and it looks like though that $\pi$ is still not known to be normal in any base. – Mankind Dec 16 '15 at 11:15
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    What would be a good reference for the current status of this problem? – Vanessa Sep 30 '18 at 16:57
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    The last sentence can be easily refuted. If $\sqrt{2}$ is eventually all 0's and 1's in decimal, then $2\sqrt{2}$ is eventually all 0's and 2's. – Elliot Glazer Apr 10 '24 at 23:09
  • @ElliotGlazer Thank you! It was pointed out to be a while ago but I forgot what post I made this claim in so thanks for bringing it up. I will amend my post. – Wojowu Apr 11 '24 at 10:41
  • I bet this sort of argument can be pushed to show there is $x_n \in \mathbb{Q}[\sqrt{2}] \setminus \mathbb{Q}$ s.t. for all $d < b \le n,$ there are infinitely many d’s in the base b expansion of $x_n.$ I haven’t been able to find any writeup of such a result, to my surprise. – Elliot Glazer Apr 11 '24 at 23:03
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I will prove something a little weaker: there are nonzero $a, b \in \mathbb{Q}$ such that $a+b\sqrt{2}$ has more than 2 digits which appear infinitely in its decimal expansion.

Suppose $\sqrt{2}$ does not have this property. Let $c<d$ be the two digits which appear infinitely often and $n$ the last placement of a digit which is not $c$ or $d.$ We can easily compute from $c, d, n$ a pair of nonzero rationals $a', b'$ such that $0<x=a'+b'\sqrt{2}<0.01$ and the only decimal digits in $x$ are 0 and 1.

Assume wlog that there are infinitely many pairs of consecutive 0's in $x$ (otherwise replace $x$ with $\frac{1}{9}-x$). Then $a=15a'$ and $b=15b'$ are as desired, because multiplication by 15 replaces the 0 in a 01 with 1, the first 0 in a 00 with 0, and each 1 with a 5 or 6.

Elliot Glazer
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