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Math Help - Real number system

  1. #1
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    Real number system

    How can I show that if A > 0 and n is a positive, integer there is a positive number c such that c^n = A.

    Here's my start for proving that c is a positive number

    Let S be the set of all positive numbers x such that x^n < A. This set is not empty(how do I show this?). It is a bounded set and has a least upper bound(difficulty on proving ) which is c.

    Then I have to prove that c^n = A.

    If e > 0, there is some x in S s.t. c-e < x. hence (c-e)^n < A. Now let e -> 0 then c^n <= A. If c^n < A then I have prove this is not possible

    I got this far... help !!
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  2. #2
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    [QUOTE=mathsohard;608199]How can I show that if A > 0 and n is a positive, integer there is a positive number c such that c^n = A.

    Here's my start for proving that c is a positive number

    Let S be the set of all positive numbers x such that x^n < A. This set is not empty(how do I show this?).[\quote]
    If A is a positive number then so is [tex]A^{1/n} so there exist postive numbers less than A^{1/n}. 0< x< A^{1/n}, then 0<x^n< A.

    It is a bounded set and has a least upper bound(difficulty on proving )
    A basic property of the real numbers: if a non-empty set of real numbers has an upper bound, it has a least upper bound.
    which is c.

    Then I have to prove that c^n = A.
    Suppose that 0< c^n< A. What can you say about c?
    Suppose that 0< A< c^n. What can you say about c?

    If e > 0, there is some x in S s.t. c-e < x. hence (c-e)^n < A. Now let e -> 0 then c^n <= A. If c^n < A then I have prove this is not possible

    I got this far... help !!
    Good! You might look at c- e/n, say, and compare to (c- e)^n. Remember that you only have to show "<" or ">", not "=" so you have more flexibility.
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  3. #3
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    form a cut.
    if x^n < A, x in L
    if y^n > A, y in R

    c is lub of L

    That's the outline. The details are less interesting.
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  4. #4
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    OK, details:

    if x=A/2, x^n < A
    if y=2A, y^n > A
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