1. Limited time only! Sign up for a free 30min personal 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!

Question About Absolute Value

  1. Sep 17, 2015 #1
    If ##\sqrt{x^2} = |x|##, why ##\sqrt{16} ≠ |4|## instead of 4 (please see below image)?

    absolute_value.png
     
  2. jcsd
  3. Sep 17, 2015 #2
    The absolute value of 4 is indeed 4. The absolute value is only important when x is negative.
     
  4. Sep 17, 2015 #3
    I mean if ##\sqrt{16} = 4## why ##\sqrt{x^2}## is not x?
     
  5. Sep 17, 2015 #4
    Because when x is negative, ##\sqrt{x^2} = -x##
     
  6. Sep 18, 2015 #5

    Mark44

    Staff: Mentor

    But is ##\sqrt{(-4)^2} = -4##?
     
  7. Sep 18, 2015 #6
    Yes indeed.

    Because ##\sqrt{(-4)^2} = (-4)^{\frac{2}{2}} = (-4)^1 = -4##
     
  8. Sep 18, 2015 #7

    Mark44

    Staff: Mentor

    Absolutely not! ##\sqrt{(-4)^2} = \sqrt{16} = 4 = |-4|##
     
  9. Sep 18, 2015 #8
    What about this?

    ##\sqrt{(-4)^2} = (-4)^{\frac{2}{2}} = (-4)^1 = -4##

    Isn't from the above shows that ##\sqrt{(-4)^2} = -4##?
     
  10. Sep 18, 2015 #9

    Mark44

    Staff: Mentor

    No, it doesn't.

    The exponent properties you are using apply only to numbers that are nonnegative.

    ##\sqrt{(-4)^2} = [(-4)^2]^{1/2} = 16^{1/2} = + 4##
    The rule that you are misusing says that ##(a^m)^n = a^{mn}##, provided that ##a \ge 0##.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Question About Absolute Value
  1. Absolute value (Replies: 3)

  2. Absolute value (Replies: 1)

Loading...