# Uniqueness of the infinite expansion in base $b$

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 $a_k \in \{0,1,\ldots ,b-1\}$. When $x=\frac{1}{b^n}$, for some $n\in \mathbb{N}$, there is two representations. One of them is always $0$ beginning with some index, that is, is finite. How can I prove that the infinite representation is unique?

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If I recall correctly you can prove that the partial sums of the expansion satisfy the following inequality

$$0 \leq x - \sum_{k = 1}^{n} \frac{a_k}{b^k} < \frac{1}{b^n}$$

and from this inequality you can just prove by induction that if you have two expansions for $x$ then they are actually the same by proving that all the "digits" $a_k$ agree.

Edit:

The inequality can be proved as follows:

$$x - \sum_{k = 1}^{n} \frac{a_k}{b^k} = \sum_{k = n+1}^{\infty} \frac{a_k}{b^k} \leq \sum_{k = n + 1}^{\infty} \frac{b - 1}{b^k} = (b - 1) \sum_{k = n+1}^{\infty} \frac{1}{b^k}$$

$$= (b-1) \left ( \frac{1}{b^{n+1}} + \frac{1}{b^{n+2}} + \cdots \right ) = \frac{(b-1)}{b^{n+1}} \left ( 1 + \frac{1}{b} + \frac{1}{b^{2}} + \cdots \right ) = \frac{b-1}{b^{n+1}}\frac{1}{1 - \frac{1}{b}} = \frac{1}{b^n}$$

Then to prove the uniqueness by induction you use the inequality. So suppose that you have two representations for $x$, say

$$x = \sum_{k = 1}^{\infty} \frac{a_k}{b^k} = \sum_{k = 1}^{\infty} \frac{c_k}{b^k}$$

Then you want to prove that $a_k = c_k$ for every $k \geq 1$. For example for $n = 1$ you have the two inequalities

$$0 < x - \frac{a_1}{b} < \frac{1}{b} \quad 0 < x - \frac{c_1}{b} < \frac{1}{b}$$

Then by subtracting them you get

$$\frac{-1}{b} < \frac{a_1}{b} - \frac{c_1}{b} < \frac{1}{b} \implies -1 < a_1 - c_1 < 1 \implies a_1 = c_1$$

since $a_1 - c_1$ is an integer. I'll leave the inductive step to you.

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