Direct construction of the real numbers using only the integers (c.f. Eudoxus reals) Let the natural numbers $\Bbb N = \{0,1,2,3,\dots\}$ and the integers $\Bbb Z$ be given.
We define a function $\gamma: \Bbb Z \to \Bbb Z$ by
$$   
    \gamma(n) = \left\{\begin{array}{lr}
        \frac{n}{2} , & \text{when }\; n \text{ is even}\\
        \frac{n-1}{2}, & \text{when }\;  n \text{ is odd}\\
        \end{array}\right\} 
$$
A binary relation $R$ on $\Bbb N$ and $\Bbb Z$ is said to be a r-locator if it satisfies the following four properties:
$$\tag 1 \text{The domain of } R \text{ is equal to } \Bbb N$$
$$\tag 2 \text{For every integer } n \ge 0, \text{ if } nRm \text{ then } (n+1)R2m$$
$$\tag 3 \text{For every integer } n \gt 0, \text{ if } nRm \text{ then } (n-1)R\gamma(m)$$
$$\tag 4 \text{For every integer } n \ge 0, \text{ the image } R(n) \text{ is bounded above}$$
We can associate to any r-locator a function $\; \mathtt M(R): \Bbb N \to \Bbb Z$ by writing
$$\tag 5 \mathtt M(R): n \mapsto \text{Max}\big(R(n)\big)$$
A function $\alpha: \Bbb N \to \Bbb Z$ is said to be a binary tick specification if it satisfies the following two properties,
$\tag 6 \text{For every } n \in \Bbb N, \; \big [ \, \alpha(n+1) = 2\alpha(n) \text{ or } \alpha(n+1) = 2\alpha(n) + 1 \,\big ]$
$\tag 7 \text{For every } N \in \Bbb N \text{ there exists a } n  \ge N \text{ such that } \alpha(n+1) = 2\alpha(n)$
Lemma 1: If $R$ is a r-locator then the function $\alpha = \mathtt M(R)$ satisfies $\text{(6)}$.
In general, when a function $\rho$ satisfies only $\text{(6)}$, there is a fix:
Find the smallest $K$ such that for all $k \ge K$, $\rho(k)$ is odd. Then redefine the function by writing $\rho^{'}(k) = \rho(k) + 1$ for $k \ge K$. Also, if $K$ has a predecessor, define $\rho^{'}(K-1) = \rho(K-1) + 1$ and retain the remaining $\rho$ definitions (if any) for $\rho^{'}$. The new function $\rho^{'}$ satisfies both $\text{(6)}$ and $\text{(7)}$.
Example: $\rho = (-1,-1,-1,\dots)$ satisfies $\text{(6)}$ but not $\text{(7)}$. Applying the fix, $\rho^{'} = (0,0,0,\dots)$.
Lemma 2: Let $\alpha$ and $\beta$ be two binary tick specifications. The smallest subset $[R(\alpha,\beta)]$ of $\Bbb N \times \Bbb Z$ containing the graph of $\alpha + \beta$ (pointwise addition) and satisfying $\text{(1)}$ thru $\text{(3)}$ also satisfies $\text{(4)}$.
If necessary we apply the fix to $\mathtt M([R(\alpha,\beta)])$ and define the addition of the two specifications,
$$\tag 8  \alpha + \beta = \mathtt M([R(\alpha,\beta)])$$
giving another binary tick specification.
Let $\Bbb B$ denote the set of all binary tick specifications with this binary operation $+$.
Theorem 3: The structure $(\Bbb B, +)$ is a commutative group. Moreover, it is isomorphic to $(\Bbb R, +)$.
Example: Pointwise addition of
$\quad +\frac{1}{4} = (0,0,1,2,4,\dots)$
$\quad -\frac{1}{4} = (-1,-1,-1,-2,-4,\dots)$
gives
$\quad \quad \;\, = (-1,-1,0,0,0,\dots)$
If this sum generates the r-locator $R$, then $\mathtt M(R)$ returns $(0,0,0,0,0,\dots)$, as expected.
My Work
I've been working through some of the theory details, but felt it would be beneficial to present these rough ideas now rather than attempting to supply complete proofs.

Does this theory hold together?

The motivation for this work came from the desire to find a model for Tarski's axiomatic formulation of the real numbers; see this.

The exposition of the above theory is a direct route to the real numbers that does not require the construction of the rational numbers. Nor the definition of a limit or floor function. However, the following relations hold true:
Every $\alpha$ specifies a real number $a$ as follows,
$\quad a = {\displaystyle \lim _{n \to +\infty} \frac{\alpha(n)}{2^n}}$
The inverse mapping is given by
$\quad \alpha(n) = \lfloor a 2^n \rfloor$

There is a way to construct all the real numbers in $[0, 1]$ using just one number, namely $\frac{2}{3}$. It goes as follows. The binary expansion of $\frac{2}{3}$ is $0.10101010\cdots$. Any real number in $[0, 1]$ can be obtained by performing an infinite permutation on these binary digits. For instance $1 = 0.1111111\cdots$ is obtained by moving all the $1$'s to the left.
