Prove that $\sqrt{5}$ exists 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.
 A: 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
A: If we use your strategy, we need to show that we cannot have $\alpha^2\lt 5$, and we cannot have $\alpha^2\gt 5$. We do this with minimal machinery. That makes the argument much longer than if there is some already developed theory to appeal to. 
Suppose that $\alpha^2\lt 5$. Let $\alpha^2+\delta=5$ for some positive $\delta$. Note that $\delta \le 1$ and $\alpha\ge 2$. Then 
$$\left(\alpha +\frac{\delta}{4\alpha}\right)^2=\alpha^2+\frac{\delta}{2}+\frac{\delta^2}{16\alpha^2}=\alpha^2+\delta\left(\frac{1}{2}+\frac{1}{64}\right)\lt 5.$$
It follows that $\alpha +\frac{\delta}{4\alpha}\in A$, contradicting the fact that $\alpha$ is an upper bound for $A$. 
Suppose now that $\alpha^2\gt 5$.  Let $\alpha^2=5+\delta$ for some positive $\delta$. We leave it to you to find a positive $x$ smaller than $\alpha$ such that $x^2\gt 5$, contradicting the fact that $\alpha$ is the least upper bound of $A$. Hint: Something like what we did will work, except that we must subtract some small number from $\alpha$, instead of adding a small number, as we did earlier. 
A: Hint
Show that $f(x) := x^2-5$ is continuous on $\mathbb R$ and use the intermediate value thm. on
$(2, -1) \in {\rm graph}(f)$ and $(3,4) \in {\rm graph}(f)$
