TITLE: Prove that $\sqrt{5}$ exists
QUESTION [6 upvotes]: Prove that $\sqrt{5}$ exists; in other words prove that there exists a positive number $x\in \mathbb R$ satisfying $x^2=5$
Here's what I've done:
I let $A= \{x>0:x^2\leq 5\}$
We know that $A$ is not empty because clearly $2$ is in it: $2^2<5$ and we also know that $A$ is bounded by $3$ because
$$x>3\implies x^2>9$$ so by completeness, $\sup A$ exists. I let $α = \sup A$
So now I'm trying to prove that $α^2=5$ (that there exists a positive number $α\in \mathbb R$ satisfying $α^2=5$) and to do this I'm going to try to show that $α^2<5$ and $α^2>5$ are impossible.
Case 1: $α^2<5$ to do this I'm assuming a proof by contradiction will work.
So, suppose $α^2<5$ then ...
Here's where I'm a bit lost, I don't know how to proceed with a proof by contradiction here. If anyone can give me hints or explain what I should do next that would be really appreciated.

REPLY [2 votes]: I don't think that using the intermediate value theorem we are going to obtain a valid proof. The theorem heavily relies on the completeness of the real line, here we are proving such completeness. So I fear we are producing a circular argument. To support my thesis, if you look in the wikipedia page of the theorem you'll see a counter example if the domain doesn't contain a point ($\sqrt{2} $ in that case)
As you said, there are $2$ cases:
 IF $\alpha^2 <5$ then consider $k=\dfrac{5(\alpha+1)}{\alpha + 5}$. you have that $\alpha < k$ but $k^2 <5$. 
IF $5 < \alpha^2$ then consider again $k$. in this case, $k < \alpha$, but $5 < k^2$ and again an absurd