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Math Help - show the root exists

  1. #1
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    show the root exists

    Greetings, I hope someone can help on this problem

    If n is an odd positive integer and a < 0, prove that there is exactly one negative b such that b^n = a

    Thanks
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  2. #2
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    Quote Originally Posted by Recklessid View Post
    Greetings, I hope someone can help on this problem

    If n is an odd positive integer and a < 0, prove that there is exactly one negative b such that b^n = a

    Thanks
    Nothing to do with number theory.

    You are asking whether \sqrt[n]{a} exists.

    Indeed it does, consider the function f(x) = x^n and apply the indetermediate value theorem. Note, this function is one-to-one because f'(x) = nx^{n-1} >0.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    Nothing to do with number theory.

    You are asking whether \sqrt[n]{a} exists.

    Indeed it does, consider the function f(x) = x^n and apply the indetermediate value theorem. Note, this function is one-to-one because f'(x) = nx^{n-1} >0.
    Actually it is asking that you show that only one such \sqrt[n]{a} exists. I did this in the set theory review in my Topology book, but I can't remember how I did it.

    -Dan
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    Quote Originally Posted by topsquark View Post
    Actually it is asking that you show that only one such \sqrt[n]{a} exists. I did this in the set theory review in my Topology book, but I can't remember how I did it.

    -Dan
    But that is easy to show! Just show this function is strictly monotone. That will imply it is one-to-one.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    But that is easy to show! Just show this function is strictly monotone. That will imply it is one-to-one.
    Actually now that I've had time to think about it the proof was part of a many step problem which started with the order properties of the real line and went into things like "can you always insert a rational number between two other rational numbers" and such. I was dealing with functions, but not their graphs.

    It wasn't that hard to do, I just can't remember the method.

    -Dan
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    The TPH solution is correct and complete!
    But letís say we need to do it without calculus.
    We know that n is odd. Suppose that the are two values b & c such that b^n  = a = c^n
    Note that \left( {b^n  - c^n } \right) = \left( {b - c} \right)\left( {\sum\limits_{k = 0}^{n - 1} {b^{n - k + 1} c^k } } \right).
    So if k is even then n-k+1 is even so that  {b^{n - k + 1} c^k } is a positive term. Moreover, if k is odd then  {b^{n - k + 1} c^k } is also a positive term. Thus the only way \left( {b^n  - c^n } \right) = 0 is that b=c, in other words, b is a unique root.
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