Express a real number as a product Hi guys if I have a number $x \in [1,2)$ is it possible to express such number as:
$$x = \prod_{j=0}^{+\infty} (1 + \alpha_j 2^{-j})$$
where each $\alpha_j \in \left\{-1,0,1\right\}$?
If yes, how could it be proven? Also assuming such sequence exists, is the sequence $\left\{ \alpha_j \right\}_{j\in \mathbb{N}}$ unique? I tried by starting with the representation
$$x = 1 + \sum_{j=1}^{+\infty} x_j 2^{-j}, x_j \in \left\{0,1\right\}$$
but I didn't end up with anything...

Given $x_0\in[1,2]$, select $α_1=0$ for $x_0<\frac32$ and $α_1=1$ else so that $x_1=(1+α_12^{-1})x_0\in[1,1+2^{-1}]$. Continuing by splitting the interval in half one can guarantee that $x_k\in [1, 1+2^{-k}]$.
Indeed, if $x_k\in [1, 1+2^{-k-1}]$, then do nothing, i.e., set $α_{k+1}=0$. Else if $x_k\in[1+2^{-k-1}, 1+2^{-k}]$, then with $α_{k+1}=1$
$$
1\le x_{k+1}=\frac{x_k}{1+2^{-k-1}}\le \frac{1+2^{-k}}{1+2^{-k-1}}=1+\frac{2^{-k-1}}{1+2^{-k-1}}<1+2^{-k-1}
$$
so that in both cases $x_{k+1}\in [1, 1+2^{-k-1}]$.
Since that implies
$$
\frac1{1+2^{-k}}x_0\le \prod_{j=1}^k(1+α_j2^{-j})\le x_0
$$
one gets the convergence of the infinite product to the original $x_0$.
