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 $[a(\zeta)]_{1}^{\infty} \ne \mathbb{N}$?

• Sure, any irrational whose expansion uses just two digits. – 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.

$$0.101001000100001000001000000100000001\dots$$

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 $0$).

• You might want to link to an explanation of why we know it's irrational: mathworld.wolfram.com/LiouvillesConstant.html – JonathanZ Nov 28 '16 at 7:41
• @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 $n\in\mathbb N$) – 5xum Nov 28 '16 at 7:43