I went over a few practice problems from my Real analysis class and encountered this statement:
Show that the set $A:=~${$x \in \mathbb{Q}_{>0}~|~x^2 \geq 2$} has no greatest lower bound in $\mathbb{Q}$.
I spent more a lot of time on it, because I couldn't make sense of the provided (overcomplicated) solution. Hence I tried to proove it another way. Here is my proof, I would really appreciate any comments and critics:
Let's assume that $y=\inf(A)$ is the infinum. We have 2 possibilites: $y \in A$ or $y \notin A$
Case 1: $y \in A \Rightarrow y^2 \geq 2$
There is no rational number $q \in \mathbb{Q}$ for which we have $q^2 =2$. Therefore we have that $y^2 > 2$.
Let $p=y - \frac{1}{n}$ with $n \in \mathbb{N}_{>0}$ be a lower bound. $y$ is the greatest lower bound, so we must have that $p < y$ and $p \notin A$. Let $y-\sqrt{2}$ be the distance between $y$ and $\sqrt{2}$.
$$p \in A \Leftrightarrow \frac{1}{n} \leq y - \sqrt{2} \Leftrightarrow n \geq \frac{1}{y-\sqrt{2}}$$
We have that $y-\sqrt{2} > 0$, so there are $n \in \mathbb{N}_{>0}$ for which this is the case. Hence $\exists p \in A$ with $p < y$, which contradicts the fact that $y=\inf(A)$.
Case 2: $y \notin A \Rightarrow y^2 < 2$ $y = \inf(A)$, so there can't be any greater lower bounds. Let's assume that $q = y + \frac{1}{n}$ was a lower bound with $q > y$. Let $\sqrt{2} - y$ be the distance between $\sqrt{2}$ and $y$. $q$ is a lower bound if and only if
$$ \frac{1}{n} < \sqrt{2} - y \Leftrightarrow n > \frac{1}{\sqrt{2}-y} $$
If $y < 0 \Rightarrow 0 < \frac{1}{\sqrt{2}-y} < 1$ and $1 \leq n$ for all $n \in \mathbb{N}_{>0}$.
If $0 \leq y < \sqrt{2} \Rightarrow \frac{1}{\sqrt{2}-y} \geq \frac{1}{\sqrt{2}}$ and $n > \frac{1}{\sqrt{2}}$ for all $n \in \mathbb{N}_{>0}$.
Hence, we always find a lower bound greater than $y$ which contradicts the fact that $y = \inf (A)$.
In conlusion, the set $A$ has no greatest lower bound.
If $x^2>2$ then, applying Newton's iteration, if
$\begin{array}\\ y &=\dfrac{x+\frac{2}{x}}{2}\\ &=\dfrac{x^2+2}{2x}\\ \text{then}\\ y^2 &=\dfrac{(x^2+2)^2}{(2x)^2}\\ &=\dfrac{x^4+4x^2+4}{4x^2}\\ \text{so}\\ y^2-2 &=\dfrac{x^4+4x^2+4}{4x^2}-2\\ &=\dfrac{x^4-4x^2+4}{4x^2}\\ &=\dfrac{(x^2-2)^2}{4x^2}\\ &\gt 0\\ \text{and}\\ x-y &=x-\dfrac{x+\frac{2}{x}}{2}\\ &=\dfrac{x-\frac{2}{x}}{2}\\ &=\dfrac{x^2-2}{2x}\\ &\gt 0\\ \text{so}\\ y &\lt x\\ \end{array} $
Therefore $y < x$ and $y^2 > 2$ so $y$ is rational if $x$ is rational and $x$ is not the smallest rational such that $x^2 > 2$.
The nice thing about this proof is that it works completely in the rationals and does not need the concept of $\sqrt{2}$.