Proof $r_n := x - \sum_{k=1}^n a_k10^{-k}$

Question

How can one prove that for every $x \in [0,1)$ there are numbers $a_k \in \{0,1,..,9\}, k \in \mathbb{N}$ such that for $r_n := x - \sum_{k=1}^n a_k10^{-k}$ it holds that $10^nr_n \in [0,1)$ for all $n \in \mathbb{N}$?

I think I have to choose $a_1 \leq 10x < a_1 +1 $ and prove via induction $a_k$ so that $a_k \leq r_{k-1}10^k < a_k+1$. The problem I have is that I don't know how it's done with the $r_{k-1}10^k <a_k+1$

Can someone show me how it's done?

Answer 1

Base Case:

$r_1 = x - \frac{a_1}{10}$.

We want to choose $a_1$ such that

$\tag 1 0 \le 10 r_1 \lt 1$ or $\tag 2 0 \le 10 (x - \frac{a_1}{10}) \lt 1$ or $\tag 3 \frac{a1}{10} \le x \lt \frac{a_1+1}{10}$

The inequalities in $\text{(3)}$, as $a_1$ varies, correspond to a segment partitioning of $[0,1)$ and so there is one and only one $a_1$ satisfying the base case $n=1$.

Step Case:

Assume $r_n = x - \sum_{k=1}^n a_k10^{-k} \land 10^nr_n \in [0,1)$ is true.

This can be expressed as follows

$\tag 4 \sum_{k=1}^n a_k10^{-k} \le x \lt \sum_{k=1}^n a_k10^{-k} + 10^{-n}$

We want to choose $a_{n+1}$ such that

$\tag 5 0 \le 10^{n+1} r_{n+1} \lt 1$ or $\tag 6 \sum_{k=1}^n a_k10^{-k} + \frac{a_{n+1}}{10^{n+1}} \le x \lt \sum_{k=1}^n a_k10^{-k} + \frac{a_{n+1}+1}{10^{n+1}} $

The inequalities in $\text{(6)}$, as $a_{n+1}$ varies, correspond to a segment partitioning of the interval

$\tag 7 \big[ \sum_{k=1}^n a_k10^{-k} , \sum_{k=1}^n a_k10^{-k} + 10^{-n} \big)$

and so there is one and only one $a_{n+1}$ satisfying the step case for $n$.

Answer 2

Note that \begin{align*} [0,1/10)\cup[2/10,3/10)\cup\cdots\cup[9/10,1)=[0,1). \end{align*} There exists some $a_{1}\in\{0,1...,9\}$ such that \begin{align*} \dfrac{a_{1}}{10}\leq x<\dfrac{a_{1}+1}{10}, \end{align*} so \begin{align*} 0\leq x-\dfrac{a_{1}}{10}<\dfrac{1}{10}. \end{align*} Moreover, \begin{align*} [0,1/10^{2})\cup[2/10^{2},3/10^{2})\cup\cdots\cup[9/10^{2},10/10^{2})=[0,1/10). \end{align*} There exists some $a_{2}\in\{0,1...,9\}$ such that \begin{align*} \dfrac{a_{2}}{10^{2}}\leq x-\dfrac{a_{1}}{10}<\dfrac{a_{2}+1}{10^{2}}, \end{align*} so \begin{align*} 0\leq x-\dfrac{a_{1}}{10}-\dfrac{a_{2}}{10^{2}}<\dfrac{1}{10^{2}}. \end{align*} Proceed inductively to get the result.