Existence of Unique Inverse Element for Addition of Cuts

Theorem
Let $\alpha$ be a cut.

Let $0^*$ be the rational cut associated with the (rational) number $0$:
 * $0^* = \set {r \in \Q: r < 0}$

Then there exists a unique cut $\beta$ such that:
 * $\alpha + \beta = 0^*$

where $+$ denotes the operation of addition of cuts.

Proof of Uniqueness
Suppose $\alpha + \beta_1 = \alpha + \beta_2 = 0^*$.

We have:

So if such a $\beta$ exists, it is indeed unique.

Proof of Existence
Let $\beta$ be the set of all rational numbers $p$ such that $-p$ is an upper number of $\alpha$, but not the smallest upper number, were it to exist.

We verify that the conditions are satisfied for $\beta$ to be a cut.

It follows directly from the definition that $\beta \ne \O$ and $\beta \ne \Q$.

Thus condition $(1)$ is satisfied.

Let $p \in \beta$.

Let $q \in \Q$ be a rational number such that $q < p$.

Then:
 * $-p \notin \alpha$

and:
 * $-q > -p$

Thus $-q$ is an upper number of $\alpha$, but not the smallest.

Thus $q \in \beta$.

Thus condition $(2)$ is satisfied.

Let $p \in \beta$.

Then by definition $-p$ is an upper number of $\alpha$, but not the smallest.

Thus there exists a rational number $q$ such that $-q < -p$ and such that $-q \notin \alpha$.

Let:
 * $r = \dfrac {p + q} 2$

By Mediant is Between:
 * $-q < -r < -p$

so $-r$ is also an upper number of $\alpha$, but not the smallest.

So we have found a rational number $r$ such that $r > p$ where $r \in \beta$.

Thus condition $(3)$ is satisfied.

We have demonstrated that $\beta$ is a cut.

We now verify that $\alpha + \beta = 0^*$.

Suppose $p \in \alpha + \beta$.

Then:
 * $p = q + r$

for some $q \in \alpha, r \in \beta$.

Hence:
 * $-r \notin \alpha$
 * $-r < q$
 * $q + r < 0$

and:
 * $p \in 0^*$

Suppose $p \in 0^*$.

Then:
 * $p < 0$

By Existence of Upper and Lower Numbers of Cut whose Difference equal Given Rational, there exist $q \in \alpha, r \notin \alpha$ such that:
 * $r$ is not the smallest upper number of $\alpha$

$r - q = -p$

Because $-r \in \beta$:

Thus by definition of set equality:
 * $\alpha + \beta = 0^*$

and the proof is complete.