Problem statement: Prove that $\sqrt{2}\in\Bbb{R}$ by showing $x\cdot x=2$ where $x=A\vert B$ is the cut in $\Bbb{Q}$ with $A=\{r\in\Bbb{Q}\quad : \quad r\leq 0\quad \lor \quad r^2\lt 2\}$. Denote the cut of $r\in\Bbb{R}$ by $r^*=X\vert Y$ where $X$ is the left-cut of $r$.
(Attempt:) By definition of multiplying cuts, if $x=A\vert B$, then $x\cdot x$ is $A^2\vert F$ such that $A^2=\{r\in\Bbb{Q}\quad \vert \quad r\leq 0\quad \lor \quad \exists a,a'\in A(r=aa' \land a,a'\gt 0)\}$ because in either case $x\cdot x\gt 0$. 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 $\exists \alpha,\alpha'\in A$ such that $x=\alpha\alpha'$ 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, $A^2\subseteq L$. Conversely we w.t.s. $L\subseteq A^2$. By a known result, if $\forall r\in\Bbb{R}, r^*=X\vert Y$ then $r$ is the l.u.b. for $X$ where $X=\{q\in\Bbb{Q}: q\lt r\}$. Since $\sqrt{2}$ is the l.u.b. of $A$, then $4$ is the l.u.b. of $A^2$. Thus, $L\subseteq A^2$.
I believe I'm doing something wrong because $3/2\in L$, but I don't think $3/2\in A^2$. Please help me do this correctly.