question
For each $n$ natural number we denote by $a_n$ the first digit a of the number $n^3$.
Show that the number $A = 0, a_1a_2 ... a_n ...$ is irrational.
my idea
A thing is clear....between $a_1$ and the last digit of $A$ is an infinity of numbers.
This means we have to show that the first digit of $n^3$ doesn't follow a rule.
I thought of showing it with modular arithmetic. I don't know where to start. I also did a table with the first 20 numbers raised to the 3rd power, but I didn't find anything that could help me.
EDIT WITH MY IDEA FOR @5xum
As I said, we have to show that if $10^x≤a^3≤2∗10^x$, $10^y≤b^3≤2∗10^y$ and $x<y$, then b has more solutions than a.
We can write the 2 inequality as $\sqrt[\leftroot{10} \uproot{5} 3]{10^x} \leq a \leq \sqrt[\leftroot{10} \uproot{5} 3]{10^x*2} $ and $\sqrt[\leftroot{10} \uproot{5} 3]{10^y} \leq b \leq \sqrt[\leftroot{10} \uproot{5} 3]{10^y*2}$, which means that we basically have to show that
$\sqrt[\leftroot{10} \uproot{5} 3]{10^x} -\sqrt[\leftroot{10} \uproot{5} 3]{10^x*2} \leq \sqrt[\leftroot{10} \uproot{5} 3]{10^y} -\sqrt[\leftroot{10} \uproot{5} 3]{10^y*2}$
$\sqrt[\leftroot{10} \uproot{5} 3]{10^x} (\sqrt[\leftroot{10} \uproot{5} 3]{2}-1) \leq \sqrt[\leftroot{10} \uproot{5} 3]{10^y}(\sqrt[\leftroot{10} \uproot{5} 3]{2}-1) $
$\sqrt[\leftroot{10} \uproot{5} 3]{10^x} \leq \sqrt[\leftroot{10} \uproot{5} 3]{10^y}$, which is true because $x<y$.
Is my idea right?
Hope one of you can help me! Thank you!
I think 5xum's hint is helpful.
Let $f(n)$ be the number of integers $m$ such that $$10^n\le m^3\lt 2\cdot 10^n\tag1$$
In the following, let us prove that for $n\ge 1$, $f(n+1)\gt f(n)$ holds.
$(1)$ is equivalent to $$10^{\frac n3}\le m\lt 2^{\frac 13}\cdot 10^{\frac n3}\tag2$$ so we can write $$f(n)=\left\lfloor 2^{\frac 13}\cdot 10^{\frac n3}\right\rfloor-\left\lceil 10^{\frac n3}\right\rceil +1$$ where $\lfloor x\rfloor$ represents the greatest integer less than or equal to $x$, and $\lceil x\rceil$ represents the least integer greater than or equal to $x$.
Since $f(1)=0,f(2)=1$ and $f(3)=3$, we have $$f(1)\lt f(2)\lt f(3)$$
For $n\ge 3$, we have $$\begin{align}&f(n+1)-f(n) \\\\&=\left\lfloor 2^{\frac 13}\cdot 10^{\frac {n+1}3}\right\rfloor-\left\lceil 10^{\frac {n+1}3}\right\rceil +1 \\&\quad -\bigg(\left\lfloor 2^{\frac 13}\cdot 10^{\frac n3}\right\rfloor-\left\lceil 10^{\frac n3}\right\rceil +1\bigg) \\\\&=\left\lfloor 2^{\frac 13}\cdot 10^{\frac {n+1}3}\right\rfloor-\left\lceil 10^{\frac {n+1}3}\right\rceil \\&\quad -\left\lfloor 2^{\frac 13}\cdot 10^{\frac n3}\right\rfloor+\left\lceil 10^{\frac n3}\right\rceil\tag3\end{align}$$
Using $$x-1\lt \lfloor x\rfloor\le x$$ $$x\le \lceil x\rceil\lt x+1$$ we have $$\begin{align}(3)&\gt 2^{\frac 13}\cdot 10^{\frac {n+1}3}-1-\bigg(10^{\frac {n+1}3}+1\bigg) \\&\quad -\bigg(2^{\frac 13}\cdot 10^{\frac n3}\bigg)+10^{\frac n3} \\\\&=2^{\frac 13}\cdot 10^{\frac n3} (10^{\frac 13}-1) \\&\quad -10^{\frac n3}(10^{\frac 13}-1)-2\\\\&=10^{\frac n3}(10^{\frac 13}-1)(2^{\frac 13}-1)-2\tag4\end{align}$$
Since $10^{\frac 13}\gt 2.1$ (which is equivalent to $10\gt 9.261$ which is true) and $2^{\frac 13}\gt \frac 54$ (which is equivalent to $2\gt \frac{125}{64}$ which is true), we have
$$\begin{align}(4)&\gt (2.1)^n\times 1.1\times \frac 14-2 \\\\&\gt 2^n\times 1.1\times\frac 14-2 \\\\&\ge 2^3\times 1.1\times \frac 14-2 \\\\&=2.2-2 \\\\&\gt 0\end{align}$$
Therefore, we can say that for $n\ge 1$, $f(n+1)\gt f(n)$ holds.
Another solution :
For every positive integer $k$, we have $$a_n=\begin{cases}1&\text{if $10^k\le n\le 10^k+10^{k-1}$} \\\\ 8&\text{if $n=2\cdot 10^k$}\end{cases}$$
So, we can say that $A$ is irrational. (I think you see why we can say so.)