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Every positive real number has a unique positive n'th root

  1. Sep 12, 2016 #1

    Math_QED

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    1. The problem statement, all variables and given/known data

    Show, using the axiom of completeness of ##\mathbb{R}##, that every positive real number has a unique n'th root that is a positive real number.

    Or in symbols:

    ##n \in \mathbb{N_0}, a \in \mathbb{R^{+}} \Rightarrow \exists! x \in \mathbb{R^{+}}: x^n = a##

    2. Relevant equations

    Axiom of completeness:

    Suppose ##I_n = [x_n,y_n]## is a sequence of closed intervals and they are nested such that:
    ##I_1 \supset I_2 \supset \dots \ I_n \dots##.

    Then, there exists at least one number x, for which:

    ##x \in \bigcap\limits_{n=0}^{\infty} I_n##

    3. The attempt at a solution

    My textbook says that I can prove it as an application of the axiom of completeness. But it doesn't ask to proof this as an exercise, so I'm not sure whether I have covered enough to proof this problem.

    So, I thought of a proof by contradiction. So, suppose that there are ##2## different numbers ##y,z \in \mathbb{R}##, for which: ##y^n = a## and ##z^n = a##. Then I wanted to take them as bounds in ##[x_n, y_n]##, and somehow show that this interval only contains 1 element, but it seems that that's not the correct approach. Any hints in the right direction will be appreciated.
     
    Last edited: Sep 12, 2016
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  3. Sep 12, 2016 #2

    PeroK

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    The axiom of completeness is required for existence of an nth root. I would use the existence of ##sup## and ##inf## for any set.

    For uniqueness, you could show that ##x < y \ \Rightarrow x^2 < y^2## and hence ##x^n < y^n##.
     
  4. Sep 12, 2016 #3

    Math_QED

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    Sup and inf are defined later in the book so I was right that I did not cover enough yet. I will come back later to this thread. Thanks!
     
  5. Sep 12, 2016 #4

    Ray Vickson

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    You don't need to know about sup and inf to do it; using your completeness property is enough.

    You are trying to show the existence of a root of the equation ##y^n = x##, so you can just apply the so-called bisecting search method to do it. See, eg., https://en.wikipedia.org/wiki/Bisection_method
    or http://www.sosmath.com/calculus/limcon/limcon07/limcon07.html . All you need to get started is the existence of a number ##b > 0## such that ##b^n > x##.
     
  6. Sep 12, 2016 #5

    Math_QED

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    I'm sorry but the proof for that method seems to require some analysis (continuous function etc) and at this point I'm not allowed to use that.
     
  7. Sep 12, 2016 #6

    PeroK

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    The idea is that there are, for example, rationals to any number of decimal places, ##k##, such that:

    ##r_1^n \le x \le r_2^n##

    And ##r_2 - r_1 = 10^{-k}##

    You start with integers and add decimal places etc.
     
  8. Sep 12, 2016 #7

    micromass

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    The absolute easiest proof is by using the intermediate value theorem (which is equivalent to completeness) on the function ##f(x) = x^n - a##. But ok, you want to prove it from scratch without using any further analysis. Perhaps you could tell us what textbook you are using, because I don't know what properties you get to accept as true and which not.
     
  9. Sep 12, 2016 #8

    Ray Vickson

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    Are you worried that you cannot use the so-called "intermediate-value theorem"? Well, you do not need it.

    You have ##a_1 \leq a_2 \leq a_3 \leq \cdots \leq b_3 \leq b_2 \leq b_1##, with ##b_k - a_k = b_0 2^{-k}##. Completeness gives
    $$ \bigcap_{j=1}^{\infty} [a_k, b_k] = y \in \mathbb{R}.$$
    What can you say about the value of ##y^n##, using only properties like ##y_1 < y_2 \Rightarrow y_1^n < y_2^n##, etc.?
     
  10. Sep 12, 2016 #9

    Math_QED

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    I'm not really using a textbook but the syllabus from my brother. It's called Analysis 1 and it's in Dutch. I'll send you a link in PM.
     
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