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This problem is very well documented here at the MSE, e.g. Expansions of a Real Number $x$

Let $p$ be a natural number greater than $1$, and let $x$ be a real number with $0\leq x\leq 1$. Show that there is a sequence of integers $\{a_{n}\}$ with $0\leq a_{n}<p$ for each $n$ such that $$x=\sum_{n=1}^{\infty}\frac{a_{n}}{p^{n}}$$ and that the sequence is unique except when $x$ is of the form $\frac{q}{p^{n}}$, $0<p<q$, in which case there are exactly two such sequences.

But I am struggling to understand something even more basic that those answers, because I don't really get why if $x = \frac{q}{p^n}, $ then there are two expansions for $x$. I don't see why this takes into account the two decimal expansions of 1 in base 10, for example (Characterization of non-unique decimal expansions). So I am lost at seeing why taking $x = \frac{q}{p^n}$ generalizes that behaviour of the 2 possible expansions.

Matthew Leingang
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Trux
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    $1 = \frac{1}{10^0}$, $20.17 = \frac{2017}{10^2}$, etc. – Daniel Schepler Sep 27 '17 at 17:43
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    Hint: $\sum_{n> N} \frac{a_n}{p^n} \le \sum_{n> N} \frac{p-1}{p^n} = \frac{1}{p^N}$. If $a_n$ and $b_n$ are two different expansions, take for $N$ the smallest index such that $a_N<b_N$. – Gribouillis Sep 27 '17 at 17:46

2 Answers2

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I usually understand these things best by starting with a concrete example. Suppose we take $p=2$, so we're talking about binary expansions. Then any number of the form $\frac{q}{2^n}$ is simply one with a terminating expansion. For example, $\frac38=0.011_2$

In such a case, the final "$1$" can be replaced by "$01111\ldots$", because we can write $\frac18=\frac1{16}+\frac1{32}+\frac1{64}+\cdots$. Thus we have:

$$\frac38=0.011_2=0.010111111\ldots_2$$

Can you take that example, and generalize it to other prime bases?

G Tony Jacobs
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    I see... So in base three, $\frac{3}{8}$ doesn't have a $\frac{q}{3^{n_0}}$ representation, because 8 is not a power of 3. So it has only one representation as an infinite sum, which is unique: $\sum_{n=1}^{\infty} \frac{1}{3^{2n-1}}$, whereas a number like $\frac{4}{9}$ is $(0.11)_3$ but is also $(0.1022222...)_3$. An this is what happens for every basis basically. Thanks! – Trux Sep 28 '17 at 00:40
  • Sounds like you got it! – G Tony Jacobs Sep 28 '17 at 00:41
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A favorite example of this phenomenon, which you mentioned in your question, is the equation $$x = 1 = .99999\ldots $$ Although this does not fit your question, where it is required that $x \in (0,1)$, let me alter the example a tiny bit: $$x=.1 = .099999\ldots $$

On the left hand side of this equation one has $x=\frac{q}{p^n}$ where $q=1$, $p=10$, and $n=1$.

And on the right hand side one has the alternate expression $$x = \sum_{i=1}^\infty \frac{a_n}{10^n} $$ where $a_1=0$ and $a_2=a_3=\cdots=9$.

Lee Mosher
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