Constructing the binary fractions via the Laurent polynomials $\Bbb Z[x,x^{-1}]$.

Question

Are the claims/proof sketch made in the next section valid?

My work

I've become interested in other ways of (ultimately) constructing the real numbers.

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Consider the ring $\mathcal R$ of Laurent polynomials over the integers,

$\tag 1 \mathcal R = \{ \displaystyle{\sum_{k={-n}}^n a_k x^k} \mid a_k \in \Bbb Z \}$

Define a set $\mathcal I$ given by

$\tag 2 \mathcal I = \{ (x-2) \displaystyle{\sum_{k={-n}}^n a_k x^k} \mid \text{ where } a_k \in \Bbb Z \}$

The set $\mathcal I$ forms an ideal. the quotient

$\tag 3 \mathcal B = \frac{\mathcal R}{\mathcal I}$

contains the integers. Moreover, $2^{-1} \in \mathcal B$.

An immediate problem now is that a number in $x \in \mathcal B$ can have different representations of the form

$\tag 4 x = \displaystyle{\sum_{k={-n}}^n a_k 2^k}\quad \text{where } a_k \in \Bbb Z$

Lemma 1: Let $x \in \mathcal B$ satisfy $\text{(4)}$. Then an algorithm can be applied transforming $x$ into the form

$\tag 5 x = \displaystyle{\sum_{k={-m}}^m b_k 2^k}\quad \text{where } |b_k| \lt 2$

Lemma 2: Let $x \in \mathcal B$ satisfy $\text{(5)}$ with $b_m \gt 0$. Then an algorithm can be applied transforming $x$ into the form

$\tag 6 x = \displaystyle{\sum_{k={p}}^q c_k 2^k}\quad \text{where } p \le q \land c_k \in \{0,1\}$

Lemma 3: Let $x \in \mathcal B$ satisfy $\text{(5)}$ with $c_p \ne 0$ and $c_q \ne 0$.
Then all the coefficients $c_p, \dots, c_q$ are uniquely determined.
Proof
If we have two representations then using an argument similar to the one found here can be used to show that $x-2$ can't divide the difference so by ideal theory we have uniqueness. $\quad \blacksquare$

Theorem 4: Every number nonzero number $x \in \mathcal B$ has one and only representation, either in the form

$\tag 7 x = \displaystyle{\sum_{k={p}}^q c_k 2^k}\quad \text{where } p \le q \land c_p = +1 \land c_q = +1 \land c_k \in \{0,+1\}$ xor $\tag 8 x = \displaystyle{\sum_{k={p}}^q c_k 2^k}\quad \text{where } p \le q \land c_p = -1 \land c_q = -1 \land c_k \in \{0,-1\}$

Moreover, $\mathcal B$ is naturally endowed with a dense total ordering.

Note: The proofs are similar no matter what 'base' is chosen. So you can also construct, say, decimal fractions in this manner.

Answer 1

The Wikipedia article Construction of the real numbers gives several constructions. The specific one you suggest seems to be somewhat related to an approach referred to in reference 8 "The Real Numbers as a Wreath Product" by F. Faltin, N. Metropolis, B. Ross and G.-C. Rota. This is discussed in the MathOverflow question 339348 "The real numbers as a wreath product?" with an answer giving some insight into the construction. The main problem there seems to be how to handle required "carries" to ensure uniqueness.

You only deal with a finite number of digits, but this is somewhat related to MSE question 1437992 "Multiplication of doubly-infinite decimal numbers" whose answer gives a link to the original "Wreath product" article.