Problem statement:
Prove that $\sqrt{2} \in \mathbb{R}$ by showing that $x\cdot x=2$ where $x=A|B$ is the Dedekind Cut in $\mathbb{Q}$ where $A=\{r\in\mathbb{Q}:r\leq0\text{ or } r^2 < 2\}$.
Attempt:
$x\cdot x = E|F = \{r\in\mathbb{Q}:r\leq0 \text{ or } \exists a,a' \in A \text{ such that } a\cdot a' = r\}|\text{ rest of }\mathbb{Q}$
Establish that $E\subset2^*$
Two cases to consider:
- $a<a'\implies a\cdot a'<a'\cdot a'<2$
- $a'\leq a\implies a\cdot a'\leq a\cdot a<2$
In both cases, $a\cdot a' < 2$ and so $E\subset 2^*$.
Establish that $2^*\subset E$
This is the same argument from Principles of Mathematical Analysis, Dedekind Cuts, Multiplicative Inverse) but modified to fit my narrative. I've also added some more (hopefully correct) exposition to justify the statements made.
Let's consider $p>0 \in 2^*$ since $p\leq 0 \in E$. Since $2-p>0$, $\exists n,m\in\mathbb{N}$ such that $m\geq n$ and $$ p < 2-\frac{2}{m+1} = \frac{2m}{m+1}$$ by the Archimedean property.
Let's consider positive $a\in A$ such that $0<a^2<2$ and $q$ such that $0<q<\frac{a}{n}$ for some $n\in\mathbb{N}$. Applying the Archimedean property twice, we can get $$\begin{align*} (mq)^2&<2\\ \big((m+1)q\big)^2&>2 \end{align*}$$ so that there is a "last" $m$ such that $mq\in A$ and $(m+1)q\not \in A$.
Then, $$\bigg(\frac{p}{mq}\bigg)^2<\bigg(\frac{2m}{m+1}\cdot\frac{1}{mq}\bigg)^2=\frac{4}{(m+1)^2q^2} $$
Now since $(m+1)q \not\in A$, $\big((m+1)q\big)^2 > 2$ and so $$\frac{4}{(m+1)^2q^2} < \frac{4}{2} = 2\implies\frac{p}{mq} \in A$$
Then let $a=\frac{p}{mq}, a'=mq$ so that $$a\cdot a' = \frac{p}{mq}\cdot mq = p \in E$$
So $2^*\subset E$.
Show that $x\cdot x = 2 \implies \sqrt{2}\in\mathbb{R}$
Since $x=A|B$ is a Dedekind cut, we know that $x=\text{l.u.b }A$.
By the definition of Dedekind cut multiplication, we can safely say that for any two cuts $\alpha=W|X,\beta=Y|Z$, $$\text{l.u.b }W \cdot \text{l.u.b }Y = \text{l.u.b }WY$$
Hence $(\text{l.u.b }A)^2 = \text{l.u.b } E \implies \text{l.u.b }A = \sqrt2$. So $\sqrt2 \in \mathbb{R}$.
Questions:
- Is this proof correct?
- Is there a better way to do this?