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Math Help - prove that :

  1. #1
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    prove that :

     prove  \ that  \  \exists x \in R \ such \ that \ x^2=2
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Do you know the definition of \mathbb{R} as the completion of \mathbb{Q} under taking limits of Cauchy sequences?

    If, so, just show that there exists a Cauchy sequence of rationals whose limit has a square equal to 2.
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  3. #3
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    Quote Originally Posted by Bruno J. View Post
    Do you know the definition of \mathbb{R} as the completion of \mathbb{Q} under taking limits of Cauchy sequences?

    If, so, just show that there exists a Cauchy sequence of rationals whose limit has a square equal to 2.
    i really don't know !!!!
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  4. #4
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    Let A=\{a\in\mathbb R:a>0,\,a^2<2\}. A\ne\O as e.g. 1\in A. Also A is bounded above. \therefore\ x=\sup A exists. Note that 1\leqslant x since 1\in A.


    Suppose x^2>2. Then \frac{x^2-2}{2x}>0 and so \exists\,n\in\mathbb N such that 0<\frac1n<\frac{x^2-2}{2x}. Then

    \left(x-\frac1n\right)^2\ =\ x^2-\frac{2x}n+\frac1{n^2}\ >\ x^2-\frac{2x}n\ >\ x^2-(x^2-2)\ =\ 2

    so a\in A\ \Rightarrow\ a^2<2<\left(x-\frac1n\right)^2\ \Rightarrow\ a<x-\frac1n contradicting the leastness of x.


    Now suppose x^2<2. Choose a natural number n such that 0<\frac1n\leqslant\frac{2-x^2}{4x} and \frac1n<2x. Then

    \left(x+\frac1n\right)^2\ =\ x^2+\frac{2x}n+\frac1{n^2}\ <\ x^2+\frac{2x}n+\frac{2x}n\ \leqslant\ x^2+2-x^2\ =\ 2

    so x+\frac1n\in A contradicting the fact that x is an upper bound for A.


    Since both x^2<2 and x^2>2 lead to a contradiction, we conclude that x^2=2.
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