show that $0, 10100100010000100000 · · · \notin \mathbb{Q}$

Question

I'd like to proof $0.10100100010000100000... \notin \mathbb{Q}$

I've tried to go above as a serie I have that $1/10+1/10^3+1/10^6=0.101001$ (the first terms) but I couldn't do, my idea is to have a serie of the form:

$a=\displaystyle\sum_{i=1}^{\infty}\frac{1}{10^{f(j)}}$

then

$a_n=\displaystyle\sum_{i=1}^{n}\frac{1}{10^{f(j)}}$

so

$a_{n+1}-a_n \rightarrow 0$

in the context of real numbers as an equivalence of rational series:

$(a_{n+1}) \sim (a_n)$

$[(a_n)] \in \Re$

I think I would've to prove that there is no injection mapping $\left[(a_{1})\right]+_{\Re}\left[(a_{2})\right]+_{\Re}+...$ to a rational number, where

$[(x_n)]+_{\Re}[(y_n)]=[(x_n+y_n)]$

This is usually done by proof by contradiction, you say that "suppose $z\in\mathbb{Q}$ then $z=p/q$ where $p$ and $q$ are integers, then you show that no such $q$ can exist because we have infinite number of small additions of $1/10^{n}$. — Suzu Hirose, Sep 17 '22 at 22:34
Why not simply use the fact that rational numbers have decimal expansions that terminate or repeat (this is not so hard to prove). It should be then simple to explain why your given decimal number neither terminates nor repeats. If you wanted something 'fancy', an approach using Dirichlet's Approximation Theorem springs to mind: https://en.wikipedia.org/wiki/Dirichlet%27s_approximation_theorem — mathematics2x2life, Sep 17 '22 at 22:37
Does this answer your question? $0.101001000100001000001$... is irrational. - found using an Approach0 search. Note this other question is closed as a duplicate of $2$ other closed related questions, i.e., Is $0.1010010001000010000010000001 \ldots$ transcendental? and Prove that 2.101001000100001... is an irrational number.. — John Omielan, Sep 17 '22 at 22:58
That the decimal expansion does not terminate is obvious. If it would be periodic , it could not have an arbitary number of consecutive zeros followed by a $1$ , since both the string before the period and the period itself have finite length inevitably bounding the number of consecutive zeros (unless we allow the artificial $\bar 0$-period, but then we never have a $1$ anymore) — Peter, Sep 19 '22 at 13:49

show that $0, 10100100010000100000 · · · \notin \mathbb{Q}$

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