If $b>1$ is an integer, is well know that the numbers $x\in (0,1]$, can be written as $$x = \sum_{k=1}^{\infty} \frac{a_k}{b^k}$$ for some integers $a_k \in \{0,1,\ldots ,b-1\} $.
My problem is how to prove that? How to show that exist such integers $a_k$?
Thanks!
What you are trying to do is to write $$ x=0,a_1a_2a_3\dots\text{ in base }b. $$ Then $$\begin{align} b\cdot x&=\phantom{a_2a_3}a_1,a_2a_3\dots\\ b^2\cdot x&=\phantom{a_3}a_1a_2,a_3\dots\\ b^3\cdot x&=a_1a_2a_3,\dots \end{align}$$ and so on. Define $$ a_n=\mod([b^n\cdot x],b) $$ where $[\,\cdot\,]$ is the integer part function. I leave it to you to prove that in fact $$ x=\sum_{n=1}^\infty\frac{a_n}{b^n}. $$