Existence of Square Roots

It's time to prove $\sqrt{2}$ is a $R$ !

\sqrt{2} \in R

$\exists α \in R$ such that $α^{2} = 2$ .

Proof
Consider yet again the set $T = {t \in R | t^{2} < 2}$ . Set $α = sup (T)$ . We are going to prove that $α^{2} = 2$ . We'll rule out that $α^{2} ≯ 2$ and $α^{2} ≮ 2$ . The idea is for the former to show that it's opposite violates $α$ being the least upper bound, and for the latter that $α$ isn't an upper bound:

( $α^{2} ≮ 2$ ): Assume the opposite. In search of some element in $T$ that $α$ doesn't bound, notice that if we choose some $α + \frac{1}{n}$ for some $n \in N$ then:

\begin{aligned} {(α + \frac{1}{n})}^{2} & = α^{2} + \frac{2 α}{n} + \frac{1}{n^{2}} \\ < α^{2} + \frac{2 α}{n} + \frac{1}{n} \\ = α^{2} + \frac{2 α + 1}{n} \end{aligned}

But now assuming $α^{2} < 2$ , allows us space, namely the $\frac{2 α + 1}{n}$ part, to fit our new number and stay less than $2$ . Specifically, choose $n_{o} \in N$ large enough such that:

\frac{1}{n_{o}} < \frac{2 - α^{2}}{2 α + 1}

Which implies that $\frac{2 α + 1}{n_{o}} < 2 - α^{2}$ (we constructed this starting from this inequality and working backwards) so consequently:

{(α + \frac{1}{n_{0}})}^{2} < α^{2} + (2 - α^{2}) = 2

Thus $α + \frac{1}{n_{0}} \in T$ contradicting that $α$ is an upper bound for $T$ . Thus $α^{2} ≮ 2$ .

( $α^{2} ≯ 2$ ): Assume $α^{2} > 2$ . Then:

\begin{aligned} {(α - \frac{1}{n})}^{2} & = α^{2} - \frac{2 α}{n} + \frac{1}{n^{2}} \\ > α^{2} - \frac{2 α}{n} \end{aligned}

Continue the argument in More Suprema and Density Practice#1.4.7 (TODO).
☐