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

Question

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 $p = x/q < 2$.

Answer 1

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 $\frac x3<p$. Note that $\frac x3<2$ (since $x<6$). So, take $p\in\left(\frac x3,2\right)$.

Answer 2

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 $2$.