Results 1 to 3 of 3

Math Help - Rational Powers

  1. #1
    MHF Contributor harish21's Avatar
    Joined
    Feb 2010
    From
    Dirty South
    Posts
    1,036
    Thanks
    10

    Rational Powers

    Course: Advanced Calculus aka Real Analysis

    Prove that, for a positive number  x and integers m and n,

    (x^{\frac{1}{n}})^m = (x^m)^{\frac{1}{n}}

    How is this proof done?

    I tried this:

    (x^{\frac{1}{n}})^m = (x^{\frac{1}{kn}})^{km}

    for u > 0

    u^k = [(u^{\frac{1}{n}})^n]^k = (u^{\frac{1}{n}})^{nk}

    then i tried to set u = x^m and got completely lost.

    Can anyone tell if I was doing it correctly or show some steps on how to do this ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by harish21 View Post
    Course: Advanced Calculus aka Real Analysis

    Prove that, for a positive number  x and integers m and n,

    (x^{\frac{1}{n}})^m = (x^m)^{\frac{1}{n}}

    How is this proof done?

    I tried this:

    (x^{\frac{1}{n}})^m = (x^{\frac{1}{kn}})^{km}

    for u > 0

    u^k = [(u^{\frac{1}{n}})^n]^k = (u^{\frac{1}{n}})^{nk}

    then i tried to set u = x^m and got completely lost.

    Can anyone tell if I was doing it correctly or show some steps on how to do this ?
    You need to realize that there are different levels at which this function is introduced. Give us exactly how you're class has defined the function.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor harish21's Avatar
    Joined
    Feb 2010
    From
    Dirty South
    Posts
    1,036
    Thanks
    10
    Quote Originally Posted by Drexel28 View Post
    You need to realize that there are different levels at which this function is introduced. Give us exactly how you're class has defined the function.
    This is what the book has stated:

    Definition: For x>0 and rational number r= m/n, where m and n are integers, with n positive, we define

    x^r \equiv (x^m)^{\frac{1}{n}}

    since rational numbers can be expressed in different ways as quotients of integers, we need to establish that if m is an integer and n and k are natural numbers then,
    (x^m)^{\frac{1}{n}} = (x^{km})^{\frac{1}{kn}} .......(1)

    Since, for u>0,

    u^k = [(u^{\frac{1}{n}})^n]^k = (u^{\frac{1}{n}})^{kn},

    setting u = x^m, we have

    x^{km} = (x^m)^k = [(x^m)^{\frac{1}{n}}]^{kn}, from which (1) follows and so rational powers are properly defined.

    Thats all the book has given about rational power and asked us to verfify the question I posted.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prove rational raised to a rational is rational.
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: February 15th 2011, 09:12 PM
  2. Negative and rational powers
    Posted in the Algebra Forum
    Replies: 2
    Last Post: July 24th 2009, 05:52 PM
  3. Simplifying Rational Powers
    Posted in the Algebra Forum
    Replies: 2
    Last Post: July 13th 2009, 03:14 PM
  4. Replies: 6
    Last Post: May 5th 2009, 07:49 AM
  5. Replies: 4
    Last Post: March 27th 2009, 09:19 PM

Search Tags


/mathhelpforum @mathhelpforum