The decimal expansion of any irrational number $x>0$ is non-repeating. This is well known. So, we have a way of obtaining irrational numbers, such has$$x=0.101\,101\,110\,111\,101\,111\,101\ldots$$(after the decimal point, we have one $1$, one $0$, two $1$'s, one $0$, three $1$'s, one $0$, and so on).

My (admittedly vague) question is this: how to obtain a known irrational number whose decimal expansion (or, for that matter, whose expansion on some base) is easily shown to be non-repeating? Of course, when I write that that the decimal expansion “is easily shown to be non-repeating” what I mean is that it is easy to describe its decimal expansion (and to see that it is non-repeating); otherwise, one could just say that, since the number is irrational, its decimal expansion must be non-repeating. And by “known number” I mean something like, say, $\sqrt[3]2$ or $\pi^e$.

  • 1
    $\begingroup$ "any number $x\in(0,+\infty)$ is non-repeating"? Quite a few have repeating decimal expansions. Have you missed "irrational"? It is very easy to construct artificial non-repeating examples e.g. .101001000100001... See en.wikipedia.org/wiki/Liouville_number for more on this subject. For known numbers, irrationality and hence non-repeatingness is easy to show for some and hard for others. $\endgroup$
    – badjohn
    Jun 26, 2017 at 16:14
  • $\begingroup$ @badjohn I had already added the word “irrational” before this comment appeared. It is easy to construct artificial non-repeating examples? I know! Didn't I do just that in my question? $\endgroup$ Jun 26, 2017 at 16:16
  • $\begingroup$ Sorry, I just noticed that your example is similar to mine but the Liouville numbers are still interesting as examples of transcendental numbers. It is easier to prove that they are transcendental than naturally occurring examples such as $\pi$ and $e$. $\endgroup$
    – badjohn
    Jun 26, 2017 at 16:17
  • $\begingroup$ Sorry, I finished my comment after your correction but I had started it before. Hence my confusion. $\endgroup$
    – badjohn
    Jun 26, 2017 at 16:18
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    $\begingroup$ Well, it is easy to show that $\sqrt 2$ is irrational and hence non-repeating. But, you seem to want a more direct argument: a number that has arisen in a natural context and can be shown to have a non-repeating decimal expansion without invoking its irrationality. I think that is going to be very tough as it relies on our fairly arbitrary way of representing numbers. Something based on continued fractions might be easier. $\endgroup$
    – badjohn
    Jun 26, 2017 at 16:22

1 Answer 1


The example that you posted $$x=0.10110111011110...$$ Can be expressed as an infinite series. As you can see, the position of the $n$th zero in this decimal representation is given by $$\frac{(n+1)(n+2)}{2}-1$$ and so the value of your number is equal to $$x=0.\overline1 -\sum_{n=1}^\infty 10^{1-\frac{(n+1)(n+2)}{2}}$$ Since $0.\overline1=\frac{1}{9}$, then we have $$x=\frac{1}{9}-\sum_{n=1}^\infty 10^{1-\frac{(n+1)(n+2)}{2}}$$ Which can be simplified to $$x=\frac{1}{9}-\sum_{n=1}^\infty (\sqrt{10})^{-n^2-3n}$$ As far as I know, this is as much as it can be simplified without bringing in other non-elementary functions. However, you might be able to use this function to simplify it further.

  • $\begingroup$ Doesn't this share the same "not known" status as my Liouville numbers? $\endgroup$
    – badjohn
    Jun 26, 2017 at 17:11
  • $\begingroup$ Yeah, but it might be able to be expressed in terms of another (non-elementary) function. $\endgroup$ Jun 26, 2017 at 17:12

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