How to prove $\sqrt{2}\in \Bbb{R}$ with Dedekind cuts? 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.
 A: Your goal is to show that $x\cdot x = 2$ where $x = A|A^\mathrm{c}$ and $A = \{ r \in \Bbb{Q} \mid r \leq 0 \mbox{ or } r^2 < 2 \}$.
Define $E = \{ r \in \Bbb{Q} \mid r \leq 0 \mbox{ or } r = ac \mbox{ for some } a,c > 0 \mbox{ and } a,c\in A\}$. Then $E|E^\mathrm{c} = A|A^\mathrm{c}\cdot A|A^\mathrm{c}=x\cdot x$ by definition, since $x$ is a positive cut.
Now if $a>0\in A$, then $a^2 < 2$. Thus for $a,c>0\in A$, we have $a^2\cdot c^2 < 2\cdot 2 = 4$. This gives that if $r>0\in E$, then $r=a\cdot c$ for some $a,c>0\in A$, and therefore $r^2 < 4$. We know that $\sqrt{4}=2$, so we can say that $r<2$ for all $r\in E$.
This means the cut $E|E^\mathrm{c}$ is certainly no greater than $2$. And we can assume it is a cut since we must have already proven that multiplication is well-defined. So we need to show it is not less than 2. For every $\varepsilon > 0$ we know there must be some $a>0\in A$ with $2-\varepsilon < a^2 < 2$. And by the definition of multiplication, $a^2 \in E$. This means that $E|E^\mathrm{c} \geq 2 - \varepsilon$. By the $\varepsilon$ principle then, we have that $E|E^\mathrm{c} \geq 2$, and we are done.
I'm sorry, but I had trouble following your proof. Hopefully this will be helpful for finding your mistake.
