I've convinved myself that negabinary representations of $\Bbb R$ are exhaustive and unique. Can it be proven?
Of course the (relatively minor) problem occurs in conventional base $2$ that $0\overline1=1\overline0$ and therefore there are two representations for many numbers. But that appears not to be replicated in negabinary numbers.
Conventionally the negabinary numbers give unique representations of any given integer $n$ in base $-2$, i.e.:
$$n=\sum_{k=0}^\infty (-2)^ki_k:i\in\{0,1\}$$
Then e.g.
$1000_{-2}=(-2)^3+0+0+0=-8$, and
$10010_{-2}=(-2)^4+0+0+(-2)^1+0=14$.
One way of looking at this, is that every negabinary is the sum of a positive number drawn from the set:
$$\sum_{k=0}^\infty (2)^{2k}i_k:i\in\{0,1\}$$
and a negative number drawn from the set
$$-\sum_{k=0}^\infty (2)^{2k+1}i_k:i\in\{0,1\}$$
We can quickly say e.g. $0.1_{-2}=-\frac12$ and $0.11_{-2}=-\frac14$ but I'm unclear on how infinitely repeating fractions might be treated and whether every number has a (unique) representation.
The largest positive fraction we can write is $1/4+1/16+...=1/3$
and the smallest negative fraction we can write is $-1/2-1/8\ldots=-2/3$
The ability to express numbers in the range $1/3,-2/3$ fits perfectly with the effect of a bit-shift which multiplies by $-2$ so I believe every number is represented.
But what about on the question of whether representations are unique?
e.g. $01_{-2}=1/4\neq3/32=0.00\overline1_{-2}$
Therefore I'm sure a) every real number is represented, and b) no number can have multiple representations as in the binary case. Can this be proven?
I don't think you have uniqueness. For example:
$.\overline{01}=\frac14+\frac{1}{16}+...=\frac13$; and
$1.\overline{10}=1-\frac12-\frac18-...=1-\frac23=\frac13$