Set of Rational Cuts forms Ordered Field

Theorem
Let $\RR$ denote the set of rational cuts.

Let $\struct {\RR, +, \times, \le}$ denote the ordered structure formed from $\RR$ and:
 * the operation $+$ of addition of cuts
 * the operation $\times$ of multiplication of cuts
 * the ordering $\le$ of cuts.

Then $\struct {\RR, + \times, \le}$ is a totally ordered field.

Proof
We demonstrate that $\struct {\RR, +, \times}$ is a field by showing it is a subfield of the structure $\struct {\CC, +, \times}$, where $\CC$ denotes the set of all cuts.

We do this by establishing that all $4$ criteria of the Subfield Test are satisfied.

We note that $0^* \in \RR$, where $0^*$ is the rational cut associated with the (rational) number $0$:
 * $0^* = \set {r \in \Q: r < 0}$

So $\RR \ne \varnothing$.

Thus criterion $(1)$ is satisfied.