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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.

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Yes, this is true. It's enough to look at the first digit they do not have in common and the next digit after that to show that they cannot be equal. –  Qiaochu Yuan Jun 25 '12 at 0:12

1 Answer 1

up vote 3 down vote accepted

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 $a_i = b_i$.

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I don't see where you use your ability to rearrange terms. It seems like the first equation ought to be valid even when the series are only conditionally convergent (except that the RHS might converge when the two infinite sums don't, but we assume they do). You're just using linearity of limits. Later on, when you want to show all the $c_i$ are zero, well, how about adding $0.222\dots$ and then just looking at the decimal expansion. –  Ben Millwood Jun 25 '12 at 2:54

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