The cut property for real numbers states the following:
Let $A,B\subset\mathbb{R}$ be nonempty and disjoint sets such that $a<b$ for any $a\in A$, $b\in B$, and $A\cup B=\mathbb{R}$. Then, there exists a number $c\in\mathbb{R}$ such that $x\leq c$ for every $x\in A$, and $x\geq c$ for any $x\in B$.
My task is to find a counterexample of the cut property in $\mathbb{Q}$, that is, to find two sets $A,B\subset\mathbb{Q}$ that do not satisfy the cut property.
Here is my attempt:
I chose $A=\lbrace \frac{n}{m} | n,m\in\mathbb{Z}; n<m\rbrace\cup (-\infty,0)$ and $B=\lbrace r\in\mathbb{Q} | r>1\rbrace$. Not too sure if this combination works.
That doesn't work; for instance, $1$ is neither is $A$ nor in $B$, and even if it were in one of them, it would be a "cut number".
Hint: consider something like $A = \{ x \in \mathbb Q : x^2 < 2 \text{ or } x < 0\}$, $B = \mathbb Q - A$.