03 - Unique real root for 2

Exists a unique positive real number $\alpha$ such that ${\alpha }^2=2$

Proof - Existence

Let $S=\{s\in \mathbb{R}:s>0,s^2<2\}$
As $1\in S$ then $S$ is non-empty

As for $x>2$ then $x^2>4$ and hence $s\notin S$ therefore $2$ is an upper bound of $S$

By completeness axiom then $\sup S$ exists and let $\alpha =\sup S$
As $1\in S$ so $\alpha \ge 1>0$
Then by trichotomy then

$${\alpha }^2<2\text{ or }{\alpha }^2=2\text{or }\alpha >2$$

1. Suppose for contradiction that ${\alpha }^2<2$ then

$${\alpha }^2=2-ϵ\text{ for some }ϵ>0$$

Remember that $\alpha \le 2$ as $2$ is an upper bound for $S$
For $h\in (0,1)$ then

$$\begin{array}{rl}(a+h{)}^2& ={\alpha }^2+2\alpha h+h^2\\ & =2-ϵ+2\alpha h+h^2\\ & \le 2-ϵ+4h+h\\ & \le 2-ϵ+5h\end{array}$$

Note that on the third line we are using $\alpha <2$ and $h^2<h$ for $h\in (0,1)$
Let $h=\min \left(\frac{ϵ}{10},\frac{1}{2}\right)$
We pick $\frac{1}{2}$ just to ensure that $h$ is between $0$ and $1$ and $\frac{ϵ}{10}$ so that $-ϵ+5h<0$ (safely)

2. …
   

Proof - Uniqueness $\beta$ is also a positive real number such that ${\beta }^2=2$ then

Suppose that

$$0={\alpha }^2-b^2=(\alpha -\beta )(\alpha +\beta )$$

As $\alpha +\beta >0$ then $\alpha -\beta =0$ hence $\alpha =\beta$