Decimal expansion of $x\in [0,1]$

Question

This is an exercise from Royden Real Analysis:

Let $p$ be a natural number greater than 1, and $x$ a real number, $0 \leq x \leq 1$. Show that there is a sequence $\{a_n\}$ of integers with $0 \leq a_n < p$ for each $n$ such that $$ x = \sum_{n=1}^\infty \dfrac{a_n}{p^n}$$ and that this sequence is unique except when $x$ is of the form $q/p^n$, in which case there are exactly two such sequences. Show that, conversely, if $\{a_n\}$ is any sequence of integers with $0\leq a_n<p$ the series above converges to a real number $x$ with $0\leq x \leq 1$.

MY ATTEMPT:

Any number $x\in [0,1]$ can be written out in its decimal form as $x=0.d_1d_2d_3\ldots$ where $d_i$ are integers between $0$ and $9$. So we have $$x = \sum_{n=1}^\infty \dfrac{d_n}{10^n} = \sum_{n=1}^\infty \dfrac{\tfrac{p^n}{10^n}d_n}{p^n} = \sum_{n=1}^\infty \dfrac{a_n}{p^n}$$ However its not readily apparent to me that this construction guarentees $a_n<p$. Furthermore I am a bit stuck on showing this is unique unless $x=q/p^n$.

ATTEMPT 2: (Forward) Let $a_n = (xp^n\mod{p})$. Since for any $y\in\mathbb{R}$, $(y\mod p) \in \left\{0,\ldots,p-1\right\}$ we have $0\leq a_n < p$. Let $x_N$ be the partial sum $$x_N = \sum_{n=1}^N \dfrac{a_n}{p^n}$$ We want to show that $\lim_{n\to\infty}x-x_N=0$. (need to show that) $$x-x_N = r_N < \dfrac{1}{p^N}$$ and so $$\lim_{N\to\infty} x-x_N = \lim_{N\to\infty} \dfrac{1}{p^N}=0$$

(Backward)

Just wanted to point out that you are assuming that a decimal expansion exists; you should not assume this (this is part of the exercise). — parsiad, Sep 03 '15 at 17:12
A solution can be found here: https://math.stackexchange.com/questions/2417982/expansions-of-a-real-number-x — Danilo Gregorin Afonso, Sep 23 '17 at 19:17

score 2 · Accepted Answer · edited Apr 13 '17 at 12:21

Hints: Think about how decimal expansions are obtained. Start with $x\in[0,1]$, multiply it by 10. The integer part of the result yields the first decimal digit $d_1$. Take the fractional part of the result and apply the above algorithm to it (multiplying by 10, taking the integer part of the result) to get the second decimal digit $d_2$, and so on. You show that this expansion $0.d_1d_2d_3\ldots$ converges to $x$ by arguing that the difference between $x$ and the partially obtained expansion goes to zero.

For your problem you are looking for the base $p$ expansion, so you should be multiplying $x$ by $p$ (instead of 10) to obtain $a_1$, and so on.

To study uniqueness, you can adapt the argument in this article: Characterization of non-unique decimal expansions

The "conversely" part is easy because you are proving that an infinite series converges to something in $[0,1]$. Compare the tail of the infinite series to a geometric series to show that the tail goes to zero.

Decimal expansion of $x\in [0,1]$

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