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

Printable View

- Apr 20th 2009, 04:02 PMxalkroot of x^2 -3=0
prove using the completeness axiom in real Nos that the equation $\displaystyle x^2-3=0$ has a root

- Apr 20th 2009, 11:27 PMShowcase_22
- Apr 21st 2009, 03:41 AMxalk
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:

$\displaystyle b^2 =3$?? - Apr 21st 2009, 04:26 AMShowcase_22
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? - Apr 21st 2009, 05:35 AMxalk
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??