# Prove that $2\times3 = 6$ using Dededkind cuts

I'm reading Classic Set Theory by Goldrei, and in Exercise 2.10, after defining real multiplication using Dedekind cuts, I'm asked to prove:

Show that $$2 +_{\mathbb{R}} 3 = 5$$ and $$2 \cdot_{\mathbb{R}}3 = 6$$.

The sum is easy, but I can't do it for the multiplication. I've nailed it down to show that

$$0 < x < 6 \rightarrow \exists p,q \in \mathbb{Q},\ 0 < p < 2,\ 0 < q < 3 \text{ s.t. } p \cdot_{\mathbb{Q}} q = x$$

(Sorry if I'm being slippery with the notation)

This seems like something very elemental to prove. Intuitively, if you take any number between $$x$$ and $$6$$, and then divide it by 2, then that's your $$q$$, and $$p = x / q$$.

One attempt I made was to start with $$q = {{x + 6}\over{2}}$$. Proving that $$q < 3$$ is easy, but I'm not being able to prove that $$p = x/q < 2$$.

Take $$p<2$$ and $$q=\frac xp$$; you want to show that you can do such a choice in such a way that $$q<3$$, which is the same thing as asserting that $$\frac xp<3$$, or $$\frac x3. Note that $$\frac x3<2$$ (since $$x<6$$). So, take $$p\in\left(\frac x3,2\right)$$.
• Sure you can!${}$ – José Carlos Santos Oct 11 '20 at 14:50
You can finish your attempt by observing that since $$x<6$$, $$x=\frac{x}{2}+\frac{x}{2}<\frac{x}{2}+\frac{6}{2}=q$$. So, $$x/q<1$$.
Note that this approach only works because $$2$$ happens to be greater than $$1$$. For an approach that generalizes better, note that you want to choose a $$q$$ such that $$x/q<2$$, or equivalently $$q>x/2$$. So, you can just choose any $$q$$ between $$x/2$$ and $$3$$ (which exists since $$x<6$$), and then $$p=x/q$$ will be between $$0$$ and $$2$$.