3
$\begingroup$

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 $p = x/q < 2$.

$\endgroup$
3
$\begingroup$

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 $p\in\left(\frac x3,2\right)$.

$\endgroup$
2
  • $\begingroup$ Awesome, thank you. It seems to me that I can use this argument for any positive a and b such that a*b = y, and 0 < x < y, is that correct? $\endgroup$ – Franco Victorio Oct 11 '20 at 14:38
  • 1
    $\begingroup$ Sure you can!${}$ $\endgroup$ – José Carlos Santos Oct 11 '20 at 14:50
1
$\begingroup$

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 $2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.