Introduction: This is a homework assignment of mine, first I want to mention that I am aware of that there are many proofs all over the internet (including this site) about the existence of $\sqrt{2}$. However, in my assignment I am somewhat 'guided' through it with some vague steps and I would enjoy to understand them, or at least have an idea about what they are trying to tell me.
Problem: Show the existence of $\sqrt{2}$, which means that: $\exists b \in \mathbb{R}, b \geq 0, b^2=2$ by using the following:
(a) Define $Y=\lbrace a \in \mathbb{R} : a \geq 0 , a^2 \geq 2 \rbrace$
(b) Show that $b:= \inf Y$ exists
(c) Show that $b \geq 0$ and $ b^2=2$
hint: Consider $a=\frac{1}{2}\left(1-\frac{2}{b^2}\right), \ a'=\frac{1}{2}\left(1-\frac{b^2}{2}\right)$ and the values $b(1-a)$ and $\frac{b}{1-a'}$
My approach: The given set $Y= \lbrace a \in \mathbb{R}: a \geq 0, a^2 \geq 2 \rbrace$ clearly is nonempty, for instance $5 \in Y$, thus $Y \neq \emptyset$. For (b) I simply said that i.e. $1$ is a lower bound for $Y$, hence $Y$ is closed from below and there exists a greatest lower bound $b:= \inf Y$.
From all the resources I have read I have an idea how to proceed, I must assume that $b^2 > 2$ and lead this to a contradiction and do the same with $b^2 <2$. If both $b^2 >2$ and $b^2 < 2$ are false then $b^2=2$. In the papers I have read they do it as follows (usually with the Supremum). They subtract a positive number from the supremum such that no member of the Set is larger than it, this would lead to a contradiction (since the Supremum is the least upper bound)
In my case I would have to increase the Infimum by a positive number and show that no member of the set $Y$ is smaller than it.
My approach (cont.) with the given hints: I considered the first case where $b^2>2$ then I came up with the following inequalitites: \begin{align}b^2>2 \implies 1> \frac{2}{b^2} \implies 1>1-\frac{2}{b^2}>0 \\\text{s.t. }\frac{1}{2}> \frac{1}{2}\left(1-\frac{2}{b^2}\right)>0 \end{align} So I could substitute $a=\frac{1}{2}\left(1-\frac{2}{b^2}\right)$ and say $a \in \left]0, \frac{1}{2}\right[$
Unfortunately I am stuck here, it looks to me like $a$ is another lower bound of $Y$ but it doesn't seem to me to be strictly greater than the Infimum of $Y$. I don't see how I can bring this information up to be a contradiction.
I would appreciate some hints/comments about how I could continue from here, and whether or not the given values $b(1-a)$ and $\frac{b}{1-a'}$ can be helpful. I apologize that this problem seems so vague, but I'd much rather continue with the given informations than copying a already existent proof from the internet.
