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**1. The problem statement, all variables and given**

Prove that ##\sqrt{2}\in\Bbb{R}## by showing ##x\cdot x=2## where ##x=A\vert B## is the cut in ##\Bbb{Q}## such that ##A=\{r\in\Bbb{Q}\quad \vert \quad r\leq 0 \quad\lor\quad r^2\lt 2\}##.

I believe that I have to show ##A^2=L## however, it seems that ##L\nsubseteq A^2##. What am I doing wrong?

## Homework Equations

Let ##r^*## denote the cut of ##r\in \Bbb{R}##.

## The Attempt at a Solution

By definition of multiplying cuts, if ##x=A\vert B##, then ##x\cdot x## is ##A^2\vert F## such that ##A^2=\{r\in\Bbb{Q}\quad \vert \quad r\leq 0\quad \lor \quad \exists a,a'\in A(r=aa' \land a,a'\gt 0)\}##. Note that ##2^*=L \vert U=\{q\in\Bbb{Q}\vert q\lt 2\}\vert\{q\in\Bbb{Q}\vert 2\leq q\}##. If ##x\in A^2##, then either ##x\leq 0## or ##x\geq 0##. Suppose ##x\geq 0##, then ##\exists ]alpha,\alpha'\in A## s.t. ##x=\alpha\alpha'## and ##\alpha,\alpha'\gt 0##. Keep in mind that ##\alpha,\alpha'\in A## means that ##\alpha^2,(\alpha')^2<2## so ##\alpha^2(\alpha')^2<4 \Longrightarrow \alpha\alpha'\lt 2##. Thus, ##x^2 \lt 2## which implies ##x\in L##. On the other hand, if ##x\leq 0##, then clearly, ##x\in L##. In all cases, ##A^2\subseteq L##.