Unique decimal representation using only 0s and 1s (base 10) Say $x \in [0, 1)$ and $\displaystyle x = \sum_{i = 1}^\infty \frac{a_i}{10^i}$ for $a_i \in \{0, 1\}$. Is it true if $\displaystyle x = \sum_{i = 1}^\infty \frac{b_i}{10^i}$ for $b_i \in \{0, 1\}$ that $a_i = b_i$ for all $i$?
It seems to me that yes this is true, since the only times I have seen uniqueness of decimal representation break down is when we introduce repeating 9s. Is there a simple proof of my statement above?
EDIT: I added base 10 to the subject to clarify I don't want to consider $\displaystyle \sum_{i = 1}^\infty \frac{a_i}{2^i}$. I am not sure if that is the correct terminology.
I want to establish $f : \{0, 1\}^\infty \to \mathbb{R}$ by $(a_1, a_2, a_3, \cdots) \mapsto 0.a_1a_2a_3\cdots$ is an injection.
 A: Note that we have $$ 0 = x -x = \sum_{i=1}^{\infty} \dfrac{a_i}{10^i} - \sum_{i=1}^{\infty} \dfrac{b_i}{10^i} = \sum_{i=1}^{\infty} \dfrac{a_i - b_i}{10^i}$$ Note that you are allowed to rearrange terms since $\lvert a_i \rvert + \lvert b_i \rvert \leq 2$ and the sequence $\displaystyle \sum_{i} \dfrac{\lvert a_i \rvert + \lvert b_i \rvert}{10^i}$ is absolutely convergent bounded above by $2x$. Hence, we have $$0 = \sum_{i=1}^{\infty} \dfrac{c_i}{10^i}$$ where $c_i \in \{-1,0,1\}$. Our goal now is to show that all the $c_i$'s are $0$. If not, look at the first non-zero $c_k$. Note that $$\sum_{i=k+1}^{\infty} \dfrac{|c_i|}{10^i} \leq \sum_{i=k+1}^{\infty} \dfrac1{10^i} = \dfrac1{10^{k+1}} \left( \dfrac1{1-1/10}\right) = \dfrac1{9 \cdot 10^{k}} < \dfrac1{10^{k}}.$$  Hence, $$\left \lvert \dfrac{c_k}{10^{k}} \right \rvert = \dfrac1{10^{k}} > \sum_{i=k+1}^{\infty} \dfrac{|c_i|}{10^i}.$$ Hence, $$\left \lvert \sum_{i=1}^{\infty} \dfrac{c_i}{10^i} \right \rvert = \left \lvert \dfrac{c_k}{10^k} - \left(-\sum_{i=k+1}^{\infty} \dfrac{|c_i|}{10^i} \right) \right \vert \geq \left \lvert \dfrac{c_k}{10^k} \right \rvert -  \left \lvert \left(\sum_{i=k+1}^{\infty} \dfrac{|c_i|}{10^i} \right) \right \vert > 0$$
Hence, if $0 = \displaystyle \sum_{i=1}^{\infty} \dfrac{c_i}{10^i} \implies c_i = 0$, which in-turn gives us that $a_i = b_i$.
