# Thread: root of x^2 -3=0

1. ## root of x^2 -3=0

prove using the completeness axiom in real Nos that the equation $x^2-3=0$ has a root

2. Originally Posted by Showcase_22
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 , $S =\{ x: x>0 \wedge x^2<3\}$.

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 $b^2=3$ and to prove that we assume 1st $b^2<3$ and then $b^2>3$.

THE question now is how do we prove that the above two cases are not true and we are forced to accept that:

$b^2 =3$??

3. Can't the proof that the set is non-empty be done like this:

$x=\frac{1}{2} \Rightarrow \ x^2=\left( \frac{1}{2} \right)^2=\frac{1}{4} <3$

Therefore $\frac{1}{2} \in S$ and the set is non-empty.

For the next part, since you know it has a supremum by the axiom of completeness $\exists \ b$ such that:

$3-\epsilon

$3-\epsilon

$-\epsilon

$|b^2-3|< \epsilon$

Is it necessary to analyse those separate cases?

4. Originally Posted by Showcase_22
Can't the proof that the set is non-empty be done like this:

$x=\frac{1}{2} \Rightarrow \ x^2=\left( \frac{1}{2} \right)^2=\frac{1}{4} <3$

Therefore $\frac{1}{2} \in S$ and the set is non-empty.

For the next part, since you know it has a supremum by the axiom of completeness $\exists \ b$ such that:

$3-\epsilon

$3-\epsilon

$-\epsilon

$|b^2-3|< \epsilon$

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 $x^2<3$ and

hence $(b-\epsilon)^2<3$ and the question now is :

Does $(b-\epsilon)^2<3$ imply that $3-\epsilon??