Consider some irrational number. For example we could take $\pi$. As we know $\pi = 3.1415...$. Now let's consider a sequence $[a]_{1}^{\infty}$. We build this sequence using this rule : $a_{i} = \beta_{j} \dots \beta_{k}$ , where $\beta_{j} \dots \beta_{k}$ is consecutive numbers in our irrational number and for $k = [1 \dots i-1]$ $a_{k}$ doesn't contain $\beta_{j} \dots \beta_{k}$.

For $\pi$ our sequence is : $a_{1} = 3$, $a_{2} = 1$, $a_{3} = 4$, $a_{4} = 15$ etc.

So my question is : does there some irrational number $\zeta$ , for which $[a(\zeta)]_{1}^{\infty} \ne \mathbb{N}$?

  • $\begingroup$ Sure, any irrational whose expansion uses just two digits. $\endgroup$ – Ittay Weiss Nov 28 '16 at 7:40

First of all, there is some confusion about how exactly you construct the numbers $a_i$:

  1. What is $j$ in the definition, and how is $k$ quantified? In saying $a_i=\beta_j\dots\beta_k$, you make it look like $k$ is a fixed number (dependent only on $i$), but then you say "for all $k=[1,\dots,i-1]$, which is very confusing.
  2. What happens after all the digits appear? For example, what if the number you have is $0.123456789 + \frac{\pi}{10000000000}$? Then you have $a_1=1,\dots, a_9=9$, but what is $a_{10}$?

But, no matter that, two things are clear from your answer and they are enough to answer it:

  1. You take an irrational number
  2. You construct integers out of the decimal expansion of that number.

and the question is do you thus construct all integers.

The answer is no, and


shows why.

To see that the number I provided is irrational, you can prove that it cannot have repeating decimals, as it contains $n$ consequtive zeros for each $n\in\mathbb N$.

To see that the number cannot generate all integers, you can clearly see it cannot generate $2$ (or any other number with digits other than $1$ and $0$).

  • $\begingroup$ You might want to link to an explanation of why we know it's irrational: mathworld.wolfram.com/LiouvillesConstant.html $\endgroup$ – JonathanZ Nov 28 '16 at 7:41
  • $\begingroup$ @user275313 I don't think it's that hard to see it's irrational, since it clearly has no repeating decimals (it has sequences of zeros of length $n$ for each $n\in\mathbb N$) $\endgroup$ – 5xum Nov 28 '16 at 7:43
  • $\begingroup$ Really? A downvote? Can the downvoter please explain what is so bad and wrong about this answer? $\endgroup$ – 5xum Nov 28 '16 at 7:46
  • $\begingroup$ You're right. I looked too quickly and though it was Liouville's constant (which it isn't), which is famous for being the first proven transcendental number. Thought I'd throw in a link 'cause it's cool. $\endgroup$ – JonathanZ Nov 28 '16 at 7:47
  • 1
    $\begingroup$ @mathbeing Thanks for answering. Overall, I think your downvote is an over-reaction, as downvotes typically mean that you think the answer is wrong or harmful, however if you only think that the answer could be improved, it is more common to first mention this in comments. I did edit my answer now, however. $\endgroup$ – 5xum Nov 28 '16 at 8:10

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