TITLE: Writing real numbers as sums of zeros and ones
QUESTION [26 upvotes]: Call a number $\delta\in(0,1)$ "good" if it satisfies the following property:

Every real number $a\in(0,1)$ can be written as an infinite sum of the form:
  $$a = \sum_{i=1}^\infty \delta^i a_i$$
  where $a_i\in\{0,1\}$.

What numbers are good?
I know that $1/2$ is good, since when $\delta=1/2$, the $a_i$ are just the digits in binary representation of $a$.
On the other hand, $1/3$ is not good, since in order to represent a real number in ternary, we also need the digit 2.
It is easy to prove that every $\delta<1/2$ is not good, since the maximum number that can possibly be represented is $\frac{\delta}{1-\delta}$, and it is strictly smaller than 1. 
My conjecture is that every $\delta\in [1/2,1)$ is good. Is this true?

REPLY [2 votes]: Following the hint in Brian Tung's answer, I am trying to write a formal proof. 
Let $\delta\in[1/2,1)$ and $D=\frac{\delta}{1-\delta}\geq 1$.
Claim: Given a number $a\in[0, D)$, for every integer $N\geq 0$ there exists a sequence of numbers $a_1,\dots, a_N$, all in $\{0,1\}$, such that:
$$D\cdot \delta^N > a-\sum_{i=1}^N{\delta^i\cdot a_i}  \geq 0$$
Proof: By induction on $N$. For $N=0$, the sequence is empty and we get:
$$ D > a \geq 0 $$ which is given.  Suppose the claim is true for $N$. Mark:
$$b_N = a - \sum_{i=1}^N{\delta^i\cdot a_i}$$
so by assumption:
$$D\cdot \delta^N >  b_N \geq 0$$
Define $a_{N+1}$ in the following way:

If $D\cdot \delta^{N+1} > b_N$, then set $a_{N+1}=0$. Then $b_{N+1} = b_N$ and it satisfies the requirement: $D\cdot \delta^{N+1} >  b_{N+1} \geq 0$.
Otherwise, set $a_{N+1}=1$. Then $b_{N+1} = b_N - \delta^{N+1}$ and:


$b_{N+1} \geq D\cdot \delta^{N+1} - \delta^{N+1} = (D-1)\delta^{N+1} \geq 0$, since $D\geq 1$.
$b_{N+1} < D\cdot \delta^N - \delta^{N+1} = D\cdot\delta^N(1-\delta) = \delta^{N+1} \leq D\delta^{N+1}$, again since $D\geq 1$.
Hence: $b_{N+1} < D\cdot \delta^{N+1}$ as required.  $\square$


Corollary:
$$  a = \sum_{i=1}^\infty{\delta^i\cdot a_i} $$