Question about irrational numbers.

Question

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

Answer 1

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 $a_{10}$?

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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

$$0.101001000100001000001000000100000001\dots$$

shows why.

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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 $0$).