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We've had a class today about decomposing a real number in base $b$ and saw the following theorem :

Let $b\in\mathbb{N},b\geq2$.

For all $x\in\mathbb{R}_+$, there exists a unique sequence $a\in\mathbb{R}^\mathbb{N}$ such as:

  • $\forall k\in\mathbb{N}^*,a_k\in[|0,b-1|]$
  • $\forall n\in\mathbb{N},\sum\limits_{k=0}^n\frac{a_k}{b^k}\leq x\leq\sum\limits_{k=0}^n\frac{a_k}{b^k}+\frac{1}{b^n}$
  • $a$ isn't stationary to $b-1$ (ie there is no $n_0$ such that $a_{k\geq n_0}=b-1$)

$(a_k)_\mathbb{N}$ is called the development of $x$ in the base $b$

$x=\overline{a_0,a_1a_2\cdots a_n}$ is the b-adic decomposition of $x$.

The first two points can be proved quite easily by building the sequence from scratch

$a_0=\lfloor x\rfloor$ in base $b$ and $a_{n+1}=\lfloor b^{n+1}\left(x-\sum\limits_{k=0}^n\frac{a_k}{b^k}\right)\rfloor$.

However, how can the 3rd point be proven easily ?

We went through a really really long proof in class, and i feel like there is an easier way.

Thank you.

Hippalectryon
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  • Feel free to correct any English mistake I might have made. – Hippalectryon Jun 25 '14 at 14:26
  • I think this is the $1=0.9999...$ equality (and its correspondent in other bases) being referenced. – Dustan Levenstein Jun 25 '14 at 15:05
  • @DustanLevenstein Exactly, n°3 is supposed to prevent this kind of decomposition - I still have to prove it holds – Hippalectryon Jun 25 '14 at 15:06
  • Unfortunately by constructing $a_i$ to be a floor, I think you're ensuring that you'll get the $b-1$-stationary form rather than the other one, so what you need to show is that $0.\overline{b-1}_b = 1$, and that generalizes to other numbers with repeating $b-1$'s. – Dustan Levenstein Jun 25 '14 at 15:09

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