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Principal nth Root of a Real Number

  1. Feb 18, 2008 #1

    cepheid

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    According to Wikipedia, if n is odd, then every real number A has a unique real nth root having the same sign as A and known as the principal nth root of A. It is denoted by

    [tex] \sqrt[n]{A} [/tex]

    My question is, how do we know that this is true i.e. that [itex] \sqrt[n]{A} [/itex] exists for all real numbers if n is odd, and for all positive real numbers if n is even?

    Note: I have studied Complex Analysis. I am not interested in the other n-1 complex roots of the number.
     
  2. jcsd
  3. Feb 18, 2008 #2

    cepheid

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    I had a discovery:

    Let y be a negative real number:

    [tex] y = A e^{i(\pi + 2k\pi)}, \ \ \ \ \ k \in \mathbb{Z} [/tex]

    Then

    [tex] y^{1/n} = \sqrt[n]{A} e^{i(\pi/n + 2k\pi/n)}, \ \ \ \ \ k \in \{0,...,n-1\} [/tex]

    We can prove that one of the roots must be real by equating the argument to pi so that we just get a negative real number. Solving for k, we get:

    k = (n-1)/2

    which is an integer in the allowable range of k values IF n is odd.

    Conclusion: [itex] \sqrt[n]{y} [/itex] exists and is equal to -[itex] \sqrt[n]{A} [/itex] where A = |y| (provided n is odd)

    So my question reduces to, "how do we know that every POSITIVE real number has a unique, real nth root?"
     
  4. Feb 18, 2008 #3
    How about using the infimum axiom the superimum axiom and the squeeze theorem.
     
  5. Feb 18, 2008 #4

    morphism

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    x^n is a continuous bijection on the appropriate domain and codomain.
     
  6. Feb 18, 2008 #5

    cepheid

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    I'm not familiar with those axioms. I haven't been taught any number theory formally (if that is indeed what this is), so maybe I won't be able to understand the answer to this question.
     
  7. Feb 18, 2008 #6

    cepheid

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    Haha...ironically, I have a problem asking me to determine whether f(x) = x^3 + 1 is a bijection, given a domain of R and a codomain of R, and THAT problem is what raised this question in my mind in the first place! I have no doubt that if the claim I was asking about is true, then it follows that what you have said is true, and vice versa. I just have no idea how to go about proving either statement.
     
  8. Feb 18, 2008 #7

    Hurkyl

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    This real analysis; in fact, all you need is calc I. You said you took complex analysis -- well, you could use that too, can't you?
     
    Last edited: Feb 18, 2008
  9. Feb 18, 2008 #8
    The axioms are necessary for the existence of the real numbers.
     
  10. Feb 18, 2008 #9

    morphism

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    A sketch of the graph of f(x) indicates that it's increasing. Can you think of how we can prove this formally? This should give us that f(x) is 1-1. To prove that f(x) is onto, we can use the intermediate value theorem.
     
  11. Feb 18, 2008 #10
    Well, at one time we believed all numbers were constructed by ratios. I suppose calc I may have a theorem that could be applied but the result won't be from first principles. Perhaps we can argue the existence of the inverse since x^n=y is one to one and onto.
     
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