# Doing concrete multiplication of two positive rational Dedekind cuts using definition(s)

According to what I have learned about rational Dedekind cut,
the multiplication of two positive rational cuts $$u$$, $$v$$ is defined as the following:

In Rudin's Principles of Mathematical Analysis: $$uv\triangleq\{q\in\mathbb{Q}|\text{there exist }{r}\in{u}\text{ and }s\in{v}\text{ such that }r>0\text{ and }s>0\text{ and }q.

and,

In Pugh's Real Mathematical Analysis: $$uv\triangleq\{q\in\mathbb{Q}|q<0\text{ or there exist }r\in{u}\text{ and }s\in{v}\text{ such that }r>0\text{ and }s>0\text{ and }q=rs\}$$.

Now consider:
$$a\triangleq\{q\in\mathbb{Q}|q<0\text{ or }q^2<2\}$$ is a rational cut representing the real number $$\sqrt{2}$$,
$$b\triangleq\{q\in\mathbb{Q}|q<2\}$$ is a rational cut representing the real number $$2$$,

Question: How to prove that $$aa=b$$ using the above fancy definition(s) for the multiplication of two positive rational cuts?

• It comes directly from the fact that $x^2$ is monotonously increasing on the non-negative rationals. It's pretty easy to show once you know this. Sep 10, 2019 at 14:08

If $$q \in aa$$ and $$q>0$$ (WLOG, if $$q<0$$, $$q <2$$ and we're done), we know there are $$r,s\in a$$ with $$r,s>0$$ and $$q < rs$$ (Rudin style). By definition of $$a$$ we know that r^2 < 2 and $$s^2 < 2$$ so $$q^2 < r^2s^2 < 4$$. This implies $$q< 2$$ (as $$q \ge 2$$ would imply $$q^2 \ge 4$$ etc., by standard ordered field properties of $$\Bbb Q$$). So $$aa \subseteq b$$ (With $$q=rs$$ the argument is almost identical).