Identity 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:
 * $\alpha + 0^* = \alpha$

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

Proof
Let $r \in \alpha + 0^*$.

$\alpha + 0^*$ is the set of all rational numbers of the form $p + q$ such that $p \in \alpha$, $q \in 0^*$, that is, $q < 0$.

It follows that:
 * $p + q < p$

and so:
 * $p + q \in \alpha$

that is:
 * $r \in \alpha$

Let $r \in \alpha$.

Let $s \in \Q$ be a rational number such that $s > r$ and $s \in \alpha$.

This is possible because $r$ is not the greatest element of $\alpha$ by definition of a cut.

Let $q = r - s$.

Then $q < 0$ and so $q \in 0^*$.

Thus we have:
 * $r = s + q$

and so:
 * $r \in \alpha + 0^*$

We have that:
 * $r \in \alpha \implies r \in \alpha + 0^*$

and:
 * $r \in \alpha + 0^* \implies r \in \alpha$

and so by definition of set equality:
 * $\alpha + 0^* = \alpha$