Set of Cuts forms Ordered Field

Theorem
Let $\CC$ denote the set of cuts.

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

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

Proof
First we show that $\struct {\CC, + \times}$ is a field by demomstrating that it fulfils the field axioms:

It has been established from Set of Cuts under Addition forms Abelian Group that $\struct {\CC, +}$ forms an abelian group.

Thus $\text A 0$ through to $\text A 4$ are fulfilled.

It remains to verify that $\struct {\CC, + \times}$ fulfils the remaining field axioms.

$\text M 0$: Closure
From Product of Cuts is Cut:


 * $\forall \alpha, \beta \in \CC: \alpha \beta \in \CC$

Thus $\struct {\CC, \times}$ is closed.

$\text M 1$: Associativity
From Multiplication of Cuts is Associative:


 * $\paren {\alpha \beta} \gamma = \alpha \paren {\beta \gamma}$

Thus the operation $\times$ of multiplication of cuts is associative on $\struct {\CC, \times}$.

$\text M 2$: Commutativity
From Multiplication of Cuts is Commutative:


 * $\alpha \beta = \beta \alpha$

Thus the operation $\times$ of multiplication of cuts is commutative on $\struct {\CC, \times}$.

$\text M 3$: Identity
Consider the rational cut $1^*$ associated with the (rational) number $1$]]:
 * $1^* = \set {r \in \Q: r < 1}$

From Cut Associated with 1 is Identity for Multiplication of Cuts:
 * $\alpha \times 1^* = \alpha$

for all $\alpha \in \CC$.

From Multiplication of Cuts is Commutative it follows that:
 * $1^* \times \alpha = \alpha$

That is, $1^*$ is the identity element of $\struct {\CC, \times}$.

$\text M 4$: Inverses
We have that $1^*$ is the identity element of $\struct {\CC, \times}$.

Let $\alpha \ne 0^*$.

From Existence of Unique Inverse Element for Multiplication of Cuts, there exists a unique cut $\dfrac 1 \alpha$ such that:
 * $\alpha \times \dfrac 1 \alpha = 1^*$

From Multiplication of Cuts is Commutative it follows that:
 * $\dfrac 1 \alpha \times \alpha = 1^*$

Thus every element $\alpha$ of $\struct {\CC, \times}$ such that $\alpha \ne 0^*$ has an inverse $\dfrac 1 \alpha$.

$\text D$: Distributivity
From Multiplication of Cuts Distributes over Addition:


 * $\alpha \paren {\beta + \gamma} = \alpha \beta + \alpha \gamma$

for all $\alpha$, $\beta$ and $\gamma$ in $\CC$.

All the field axioms are thus seen to be fulfilled, and so $\struct {\CC, +, \times}$ is a field.

Finally it is shown that $<$ is a total ordering on $\struct {\CC, +, \times}$:

This is demonstrated in Ordering on Cuts is Total.

From Properties of Totally Ordered Field, this is all that is required to be shown.