In the article "The Eudoxus Real Numbers", R.D. Arthan proved from the definition of the Eudoxus real numbers in terms of almost linear homomorphisms that the Eudoxus real numbers form an ordered field.
The rational numbers are an ordered subfield of every ordered field, and are initial in the category of ordered fields, and so they can be characterised as the initial ordered subfield of the Eudoxus real numbers.
Edit: one does not need the entire ordered field structure to characterise the rational numbers. Arthan gives the following theorems:
Theorem 9: The multiplication on $\mathbb{E}$ induced by composition of almost homomorphisms makes $\mathbb{E}$ into a commutative ring with unit.
and
Theorem 11: For any non-zero $x$ in $\mathbb{E}$, there is an element $x^{-1}$ such that $x x^{-1} = 1^\mathbb{E}$. Thus the commutative ring $\mathbb{E}$ is a field.
The elements $\mathbb{E}$ are equivalence classes of almost linear homomorphisms. We shall follow Arthan in denoting, for every almost linear homomorphism $f$, $[f]$ as the class of all almost linear homomorphisms equivalent to $f$.
The multiplicative identity element $1^\mathbb{E}$ of $\mathbb{E}$ is of course given by the equivalence class $[\mathrm{id}_\mathbb{Z}]$, where $\mathrm{id}_\mathbb{Z}$ is the identity function on the integers. There is a canonical commutative ring homomorphism $h:\mathbb{Z} \to (\mathbb{Z} \to \mathbb{Z})$ which takes every integer $x$ to the linear function $n \to h(x)(n) = x n$. The equivalence classes $[h(x)]$ of $\mathbb{E}$ are the integers in $\mathbb{E}$.
Instead of Arthan's theorem 11 showing that the Eudoxus real numbers form a field, we could use the weaker theorem
Theorem 11': For any positive $x$ in $\mathbb{Z}$, and thus for any linear function $h(x)(n) = x n$, there is an almost linear homomorphism $g_x$ such that $[h(x)] [g_x] = 1^\mathbb{E}$.
The proof of Theorem 11' has the same structure as Arthan's proof of Theorem 11, only except instead of using all non-zero elements in $\mathbb{E}$, one only uses the positive elements of the image of the commutative ring homomorphism $h:\mathbb{Z} \to \mathbb{E}$. Thus we copy Arthan's proof over here with the minor modifications mentioned above:
First let us assume that $x$ is a positive integer, which means there is a function $n \mapsto h(x)(n) = x n$. By lemma 5, for all natural numbers $m$, $x n > m$ for all sufficiently large n, and so we may define a function $g_x$ from $\mathbb{Z}$ to $\mathbb{Z}$ as follows:
$$g_x(p) := \begin{cases}
\min\{n:\mathbb{N}|x n \geq p\} & \mathrm{if} p \geq 0 \\
-g_x(-p) & \mathrm{otherwise}
\end{cases}$$
I claim $g_x$ is an almost linear homomorphism. By lemma 10, it suffices to check that $d_{g_x}(m, n)$ is bounded for $m, n$ in $\mathbb{N}$. To see this note that for all but finitely many $m, n$ in $\mathbb{N}$, $p = g_x(m)$ and $q = g_x(n)$ are both positive, and then certainly $r = g_x(m + n)$ is also positive. What we have to prove is that $d_{g_x}(m, n) = r − p − q$ is bounded as $m$ and $n$ range over $\mathbb{N}$. By the definition of $g_x$, if $m$ and $n$ are large enough for $p$ and $q$ to be positive, we have:
$$x p \geq m \geq x (p - 1)$$
$$x q \geq n \geq x (q - 1)$$
$$x r \geq m + n \geq x (r - 1)$$
From these inequalities, we can derive:
$$x r - x (p - 1) - x (q - 1) \gt (m + n) - m - n = 0$$
$$x (r - 1) - x p - x q \lt (m + n) - m - n = 0$$
For each of the above two inequalities, the difference between the left-hand side and $x (r − p − q)$ is bounded (independently of $p$, $q$ and $r$) because $n \mapsto x n$ is linear and thus an almost linear homomorphism. It follows that $x (r − p − q)$ is bounded as $m$ and $n$ range over N, but then $r − p − q$ must be be bounded, since we are assuming that $n \mapsto x n$ is positive, so that if $t$ ranges over an unbounded set of integers, so also does $t \mapsto x t$, by lemma 5 applied to $n \mapsto x n$ and $p \mapsto x (−p)$ (using the fact that $x (p) + x (−p)$ is bounded). Thus $d_{g_x}(m, n) = r − p − q$ is bounded as $m$ and $n$ range over $\mathbb{N}$, and so, by lemma 10, $g_x$ is indeed an almost linear homomorphism.
For large enough $m$, we have
$$x g_x(m) \geq m > x (g_x(m) − 1) \geq x g_x(m) − C$$
where $C$ is independent of $m$. Since $1^\mathbb{E} = (m \mapsto m)$, and $g_x$ is an almost linear homomorphism, it follows that $[h(x)][g_x] − 1^\mathbb{E} = [h(x) \circ g_x] − 1^\mathbb{E} = 0^\mathbb{E}$. Thus $[g_x]$ is a multiplicative inverse for the element $[h(x)]$.
Thus, given an integer $x$ and a positive integer $y$ such that $x$ and $y$ are relatively prime, $h(x) \circ g_y$ is an almost linear homomorphism and the element $[h(x) \circ g_y]$ in $\mathbb{E}$ is a rational number.
