Can you help me with this?
Let $x=A \mid B$ and $x'=A' \mid B'$ be cuts in $\mathbb Q$. It is defined
$$x+x'=(A+A') \mid \text{ rest of } \mathbb Q$$
Show that although $B+B'$ is disjoint from $A+A'$, it may happen in degenerate cases that $\mathbb Q$ is not the union of $A+A'$ and $B+B'$.
EDIT: As the comment below asked, I'll include the definition of a cut in $\mathbb Q$.
A cut in $\mathbb Q$ is a pair of subsets $A,B$ of $\mathbb Q$ such that
(a) $A\cup B=\mathbb Q$, $A\neq \emptyset$, $B\neq \emptyset$, $A\cap B=\emptyset$
(b) if $a\in A$ and $b\in B$ then $a<b$.
(c) $A$ contains no largest element.
Best Answer
You can simply take the cut corresponding to $\pm\sqrt2$, i.e.
$\newcommand{\Q}{\mathbb{Q}}A=\{x\in\Q; x<\sqrt{2}\}$ and $B=\{x\in\Q; x>\sqrt{2}\}$;
$A'=\{x\in\Q; x<-\sqrt{2}\}$ and $B'=\{x\in\Q; x>-\sqrt{2}\}$.
Clearly $A+A'=\{x\in\Q; x<0\}$ and $B+B'=\{x\in\Q; x>0\}$.
If you prefer to define the cut without any reference to real numbers you can get rid of $\sqrt{2}$ easily. E.g. $A=\{x\in\Q; x\le 0 \lor x^2<2\}$.