Addition, negation, and subtraction of Dedekind cuts

Let $A, B \in \mathbb{R}$ be Dedekind cuts of $\mathbb{Q}$ (Dedekind cut, The real numbers $\mathbb{R}$ as Dedekind cuts).

Sum. The sum is the Minkowski sumset in $\mathbb{Q}$: $A + B := \{\, a + b : a \in A,\ b \in B \,\}.$

Additive identity. $0^{*} := \{\, q \in \mathbb{Q} : q < 0 \,\}$, the cut of the rational $0$ under the embedding $q \mapsto q^{*} = \{\, r \in \mathbb{Q} : r < q \,\}$ (The real numbers $\mathbb{R}$ as Dedekind cuts).

Additive inverse. For a cut $A$, $-A := \{\, p \in \mathbb{Q} : \exists\, r \in \mathbb{Q},\ r > 0,\ \text{with } -p - r \notin A \,\}.$ Equivalently, $p \in -A$ iff there is a rational $s \notin A$ with $s < -p$ (set $s = -p - r$; conversely $r = -p - s > 0$). Intuitively $-p$ is bounded away from $A$ from below: some rational strictly beneath $-p$ already fails to lie in $A$.

Subtraction. $A - B := A + (-B)$.

The sum $A + B$ is again a cut, and $(\mathbb{R}, +)$ is an abelian group with identity $0^{*}$: closure, commutativity, associativity, and the identity law are Cut addition: $A+B$ is a cut, commutative and associative, with identity $0^{*}$, and existence of inverses is For a cut $A$, $-A$ is a cut and $A + (-A) = 0^{*}$.

The $r > 0$ slack in the definition of $-A$ is essential and is not cosmetic. Neither $\{-a : a \in A\}$ nor $\{-a : a \notin A\}$ is a cut in general: the first need not be downward closed, and the second can acquire a greatest element. Excising the boundary rational (the "$-p - r$ with $r > 0$" clause) makes $-A$ a genuine cut with no greatest element and forces the exact identity $A + (-A) = 0^{*}$, not merely $A + (-A) \subsetneq 0^{*}$ (For a cut $A$, $-A$ is a cut and $A + (-A) = 0^{*}$).