# Why does the definition for the multiplication of dedkind cuts explicitly include the negative rationals?

If A and B are both dedekind cuts. Then $$A \times B=\{ab \mid a \in A, b \in B, a \geq 0, b \geq 0 \} \cup \{q \in \mathbb{Q} \mid q <0 \}$$. Can someone explain why this definition doesn't work: $$A \times B=\{ab \mid a \in A, b \in B\}$$.

$$\{ab|a\in A, b\in B\}$$ is not always a Dedekind cut. For example, if $$A=B=\{q\in \mathbb Q: q < -2\}$$, then $$\{ab|a\in A, b\in B\} = \{q\in Q: q > 4\}$$

• Ok that makes sense. On a related note, if I take two sets such that A: q<2 and B: q<8 for example, with my definition does AxB = Q? – Jack Moresy Aug 27 '19 at 9:07
• @JackMoresy Indeed, because for every $q\in Q$, you have $q=(-1)\cdot (-q)$ and since $-1\in A$ and $-q\in B$, you have $(-1)\cdot (-q)\in \{ab| a\in A, b\in B\}$. – 5xum Aug 27 '19 at 9:12