prove using the completeness axiom in real Nos that the equation has a root

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- Apr 20th 2009, 05:02 PMxalkroot of x^2 -3=0
prove using the completeness axiom in real Nos that the equation has a root

- Apr 21st 2009, 12:27 AMShowcase_22
- Apr 21st 2009, 04:41 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 , .

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 and to prove that we assume 1st and then .

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

?? - Apr 21st 2009, 05:26 AMShowcase_22
Can't the proof that the set is non-empty be done like this:

Therefore and the set is non-empty.

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

Is it necessary to analyse those separate cases? - Apr 21st 2009, 06:35 AMxalk