Prove that $ x= \sum_{k=1}^\infty\frac{\varepsilon_k}{3^k}$

Question

How can I prove that all number $ x\in \mathbb{R}:0 <|x|\leq \frac{1}{2}$ is possible to write as:

$$ x= \sum_{k=1}^\infty\frac{\varepsilon_k}{3^k} $$ $$\varepsilon_k \in \{-1,0, 1\} \, \forall \,k \in \mathbb{N}$$
There is any relationship with p-adic number? wehre $x = \frac{a}{b} = \sum_{k=1}^\infty a_kp^{-k}$

I'm not sure how the problem to treat...
Thank you in advance.

Answer 1

With $\epsilon_k\in\{0,1,2\}$ instead, you'd have teh base-3 expansion of an arbitrary number $\in[0,1]$ on the right (with allowing the 3-ary equivalent of $0.9999\ldots =1$). But we need $\{-1,0,1\}$, so we substract $\sum_k \frac1{3^k}$ and that happens to be $\frac12$. Thus we turn the interval of representable numbers $[0,1]$ into the interval $[-\frac12,\frac12]$.