prove using the completeness axiom in real Nos that the equation $\displaystyle x^2-3=0$ has a root
O.K showcase ,apart from your proof that uses the IVT AND which is correct ,the rest is not very clear .
So to avoid mixing up threads,let us start here by taking the problem in steps:
First we define the set ,$\displaystyle S =\{ x: x>0 \wedge x^2<3\}$.
THEN we prove that is non empty and bounded from above and using the axiom of completeness we conclude that it has a supremum ,b.
THEN we claim that $\displaystyle b^2=3$ and to prove that we assume 1st $\displaystyle b^2<3$ and then $\displaystyle b^2>3$.
THE question now is how do we prove that the above two cases are not true and we are forced to accept that:
$\displaystyle b^2 =3$??
Can't the proof that the set is non-empty be done like this:
$\displaystyle x=\frac{1}{2} \Rightarrow \ x^2=\left( \frac{1}{2} \right)^2=\frac{1}{4} <3$
Therefore $\displaystyle \frac{1}{2} \in S$ and the set is non-empty.
For the next part, since you know it has a supremum by the axiom of completeness $\displaystyle \exists \ b$ such that:
$\displaystyle 3-\epsilon<b^2<3$
$\displaystyle 3-\epsilon<b^2<3< 3+\epsilon$
$\displaystyle -\epsilon<b^2-3<\epsilon$
$\displaystyle |b^2-3|< \epsilon$
Is it necessary to analyse those separate cases?
YES ,1/2 belongs to S AS you have shown.
But since b=SupS you can say that for any ε>0 there exists an x belonging to S and such that:
b-ε< x and since x belongs to S WE have $\displaystyle x^2<3$ and
hence $\displaystyle (b-\epsilon)^2<3$ and the question now is :
Does $\displaystyle (b-\epsilon)^2<3$ imply that $\displaystyle 3-\epsilon<b^2<3$??
WHAT is your opinion??