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Math Help - root of x^2 -3=0

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    root of x^2 -3=0

    prove using the completeness axiom in real Nos that the equation x^2-3=0 has a root
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    Super Member Showcase_22's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    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??
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    Super Member Showcase_22's Avatar
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    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<b^2<3

    3-\epsilon<b^2<3< 3+\epsilon

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

    |b^2-3|< \epsilon

    Is it necessary to analyse those separate cases?
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    Quote Originally Posted by Showcase_22 View Post
    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<b^2<3

    3-\epsilon<b^2<3< 3+\epsilon

    -\epsilon<b^2-3<\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<b^2<3??

    WHAT is your opinion??
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