Definition 1   For every $x \in \mathbb{F}$ we define $$ \rho (x) = \left( \{y \in \mathbb{Q} \mid y > x \} , \{y \in \mathbb{Q} \mid y \le x \} \right) $$

Theorem 1   For every $x \in \mathbb{F}$ we find that $\rho (x)$ is a Dedekind cut.

Proof. Because, for all $y \in \mathbb{Q}$ we have either $y > x$ or $y \le x$ the two sets of $\rho (x)$ form a partition of $\mathbb{Q}$ Furthermore, every element of the latter set is less than every element of the former set. By the Archimedean property, there exists an integer $n$ such that $x < n$ hence the former set is not empty. Likewise, there exists and integer $m$ such that $-x < m$ or $x > -m$ so the latter set is also not empty.

Theorem 2   The map $\rho \colon \mathbb{F} \to \mathbb{R}$ is a monomorphism.

Proof. Let $p$ and $q$ be elements of $\mathbb{F}$ set $(A,B) = \rho(p)$ set $(C,D) = \rho (q)$ and set $(E,F) = \rho (p+q)$ Since $a > p$ and $b > q$ implies $a + b > p + q$ for all rational numbers $a$ and $b$ it follows that $a \in A$ and $b \in C$ implies that $a + b \in E$ Likewise, since $a \le p$ and $b \le q$ implies $a + b \le p + q$ for all rational numbers $a$ and $b$ it follows that $a \in B$ and $b \in D$ implies $a + b \in F$ Hence, $\rho (p) + \rho (q) = \rho (p + q)$

Since a rational number is positive if and only if it is greater than $0$ it follows that $\rho (0) = 0$ Together with the fact proven in the last paragraph, this implies that $\rho (-x) = - \rho (x)$ for all $x \in \mathbb{F}$

Suppose that $p$ and $q$ are positive elements of $\mathbb{F}$ As before, set $(A,B) = \rho(p)$ set $(C,D) = \rho (q)$ and set $(E,F) = \rho (p \cdot q)$ Since $a > p$ and $b > q$ implies $a \cdot b > p \cdot q$ for all rational numbers $a$ and $b$ it follows that $a \in A$ and $b \in C$ implies that $a\cdot b \in E$ Likewise, since $a \le p$ and $b \le q$ implies $a \cdot b \le p \cdot q$ for all rational numbers $a$ and $b$ it follows that $a \in B$ and $b \in D$ implies $a \cdot b \in F$ Hence, $\rho (p) \cdot \rho (q) = \rho (p \cdot q)$

By using the fact demonstrated previously that $\rho (-x) = -\rho (x)$ we may extend what was shown above to the statement that $\rho (p \cdot q) = \rho (p) \cdot \rho (q)$ for all $p,q \in \mathbb{F}$ Thus, $\rho$ is a morphism of fields. Since $\mathbb{F}$ is Archmiedean, if $p \neq q$ there must exist a rational number $r$ between $p$ and $q$ hence $\rho (p) \neq \rho(q)$ so $\rho$ is a monomorphism.