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Math Help - Analysis proof

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    Analysis proof

    How would I prove there exists a positive real number x such that x^2=2? I need to use the Archimedean property.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tharris View Post
    How would I prove there exists a positive real number x such that x^2=2? I need to use the Archimedean property.
    Suppose there wasn't a root, appeal to the completeness of \mathbb{R} and derive a contradiction using the Archimedean principle.
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    Do I start with: suppose x^2 does not equal 2? Or do I begin with assuming either a lower bound or an upper bound?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tharris View Post
    Do I start with: suppose x^2 does not equal 2? Or do I begin with assuming either a lower bound or an upper bound?
    Assume that x^2\ne 2,\text{ }\forall x\in\mathbb{R} and think about the set \left\{x\in\mathbb{R}:x^2>2\right\}. By assumption it has a infimum, right? But what then?
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    this implies that the infimum is 0 which is a contradiction since x^2 >2?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tharris View Post
    this implies that the infimum is 0 which is a contradiction since x^2 >2?
    Well, the point is that it must have an infimum \gamma. But, \gamma\in\mathbb{R} and so \gamma^2\ne 0 and so argue that \gamma^2>2\implies \gamma^2-2>0 but Archimedean implies what?
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    I guess I don't see the connection. The arcimedian principle states that if x>0, then there exists a positive integer n such that 1/n <x.
    How does this relate to the infimum?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tharris View Post
    I guess I don't see the connection. The arcimedian principle states that if x>0, then there exists a positive integer n such that 1/n <x.
    How does this relate to the infimum?
    Think about it like this.

    The Archimedean principle says (geometrically) that

    0--------|------------------

    in the below we can always find some n\in\mathbb{N} such that

    0----- \frac{1}{n}-----|-------------

    So, now we're claiming that \gamma is the infimum of this set. But, \gamma^2\ne 2 and so if you argue correctly we have this scenario

    0---- \gamma^2-2-----------

    So, we should be able to "fit something" in between zero and \gamma^2-2 (by the A.P.)

    but, what's the problem with that?
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    Does this imply that that "something" must be between 2 and "lambda" squared which is a contradiction since "lamda" squared is > 2?
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  10. #10
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tharris View Post
    Does this imply that that "something" must be between 2 and "lambda" squared which is a contradiction since "lamda" squared is > 2?
    A) \gamma=\text{gamma},\lambda=\text{lambda}

    B) No. The point is that you can fit something squared in between \gamma-2 and zero and with a little algebra that something squared will be strictly greater than two but less than \gamma.


    P.S. I don't know why I've been writing \gamma^2, it should always have been \gamma
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    Sorry, my greek symbols are rusty. You have been a great help. Does this stuff get more understandable? I hope so. So far I have that:
    0 < 1/n < gamma - 2
    If I add 2 to everything I get:
    2 < 1/n + 2 < gamma

    How did I show that 1/n + 2 is a square?
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  12. #12
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tharris View Post
    Sorry, my greek symbols are rusty. You have been a great help. Does this stuff get more understandable? I hope so. So far I have that:
    0 < 1/n < gamma - 2
    If I add 2 to everything I get:
    2 < 1/n + 2 < gamma

    How did I show that 1/n + 2 is a square?
    Of course it's a square \frac{1}{n}+2=\sqrt{\frac{1}{n}+2}^2
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