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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: $uv\triangleq\{q\in\mathbb{Q}|\text{there exist }{r}\in{u}\text{ and }s\in{v}\text{ such that }r>0\text{ and }s>0\text{ and }q<rs\}$.

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?

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  • $\begingroup$ 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. $\endgroup$ – Don Thousand Sep 10 '19 at 14:08
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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).

Now try to do the other inclusion yourself.

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