1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rational exponents (was: Math Discussion)

  1. Sep 12, 2015 #1
    1. The problem statement, all variables and given/known data
    (-64)^(3/2)

    2. Relevant equations
    None.

    3. The attempt at a solution
    There is no answer that can be reached and it is supposed not be a real number. I was wondering why that is. How is it that there is no "real" answer to this problem?
     
  2. jcsd
  3. Sep 12, 2015 #2

    Mark44

    Staff: Mentor

    ##(-64)^{3/2} = [(-64)^{1/2}]^3##
    Does that answer your question?
     
  4. Sep 13, 2015 #3
    Alternatively: (−64)3/2 = √(-64)3

    You're right in that there is no real solution. The square root of a negative number is an imaginary number, in which case you must use "i" to express √(-1).

    i2 = -1 and √(-1) = i.
     
    Last edited: Sep 13, 2015
  5. Sep 14, 2015 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    When it says "there is no real solution to the problem" it means there is no real number. Do you understand the difference between "real numbers" and "imaginary numbers"?
     
  6. Sep 19, 2015 #5

    FeDeX_LaTeX

    User Avatar
    Gold Member

    The negative sign makes it so that the stated number isn't real.
     
  7. Sep 19, 2015 #6

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Not exactly.

    Real numbers can be positive or negative. However, taking the square root of a negative number does not give a real number result. Instead, imaginary numbers were invented to overcome this problem. The imaginary unit i is defined such that i2 = -1.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted