This is a question from Pugh's Real Mathematical Analysis.
My attempt:
Let $x = A|B$ where $\forall a\in A, \forall b\in B$, $a<b$.
Then $-x = A'|B' = \{r\in\mathbb{Q}:$ for some $b\in B$, not the smallest in $B$, $r=-b\}$|rest of $\mathbb{Q}$.
Suppose $0\in A+A'$. Then, there exists $a\in A$ and $a'\in A'$ such that $a+a' = 0$. But,
$a+a'=0 \implies -a=a' \implies -a\in A' \implies$ for some $b\in B$, not the smallest in $B$, $-a=-b$ $\implies a \in B$.
This contradicts our earlier assumption that $\forall a\in A, \forall b\in B$, $a<b$. So $0 \notin A+A'$.
As the sum of cuts is a cut itself, this means $\forall a''\in A+A'$, $a''<0$. Therefore, $A+A'|$rest of $\mathbb{Q}$ is equivalent to $0*$.
I'm unsure if this is correct as I feel like the way I've "proven" it is quite trivial, which shouldn't be the case as the book implies.
Any help would be appreciated!
Proof is not correct. Sure, $0 \notin A + A'$ but that's true if $A+A' = -1$. You must show that the cut actually equals $0$, i.e. that the two sets are equal.
Showing that $A+A' \subseteq0^*$ is easy, every element in $A+A'$ is a negative rational number since $b>a \implies a-b = a+(-b) < 0$, and $0^*$ is all the negative rationals.
You must also show that $0^* \subseteq A+A'$. Since $x \in 0^*$ be a negative rational number, claim $\exists a \in A$ such that $ a-x \in B$ that is not minimal. Suppose A is a cut such that $a - x \in A, \, \forall a \in A$. Then $A = \mathbb{Q}$, an impossibility. Thus there is an element $a \in A$ such that $a- x \in B$. If $a-x$ isn't minimal in $B$ then $a - (a - x) = x \in A + A'$. If $a-x$ is minimal, just take $a' = a - \frac x2 \in A$. Then $a' - (a' - x) = x\in A+A'$.
Therefore, $0^* = A + A'$.