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!

Exponent problem

  1. Jul 30, 2006 #1
    This problem was found in my gre test practice guide and i am having trouble understanding why the answer i chose was wrong.

    x^2 =16
    y^3= 64

    Which is bigger x or y?
    I chose that they were both equal to 4 however the answer stated that the relationship cannot be determined from the information given.

    Help me understand this.
     
  2. jcsd
  3. Jul 30, 2006 #2
    x^2 = 16 could allow x = -4
     
  4. Jul 30, 2006 #3

    mathman

    User Avatar
    Science Advisor
    Gold Member

    Further complication is that y^3=64 has two complex roots as well as the real root 4.
     
  5. Jul 31, 2006 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Silly problem anyway!
     
  6. Aug 4, 2006 #5
    Are you sure? y is just y...so when you expand it (stating the obvious here, don't laugh) you get

    y*y*y=64

    Can 3 complex identical numbers give a 64? I thought they always come in groups of 2 to cancel eachother out...i mean an odd power to the i in each factor would result in...and odd power in the final result...and 64 is having in best case scenario a i to a power that is multiple of 4.

    (i^4 * 4) ^3 =1 * 64 for example.

    So I don't think y^3=64 can get you a complex i. Can it? now I'm curious. My math says no...
     
  7. Aug 4, 2006 #6
    First a complex number is not just some real multiple of i, ex. a*i for real a. A complex number is a number of the form
    a+bi
    where both a and b are real and i2 = -1.

    Secondly if we have y3=64 then we have
    y3 - 64 = 0
    And the fundamental theorem of algebra says that every polynomial equation of degree n has n complex zeroes counting multiplicites, so since we have a polynomial equation of degree 3 we are guarenteed by the fundamental theorem of algebra that this equation had 3 solutions, however in this case only one of them is real, i.e. 4, while the other two are complex.

    If you have a polynomial equation with only real coefficients then and only then will all complex roots come in conjugate pairs, a+bi and a-bi.
     
  8. Aug 4, 2006 #7
    write [itex]y = re^{i\theta}[/itex] with [itex]r>0, \theta \in [0, 2\pi)[/itex], then your equation is

    [itex]r^3e^{3i\theta} = 64e^{0i}.[/itex]

    Since representation of complex numbers in the way I just mentioned for y is unique, you need [itex]3\theta[/itex] to be a multiple of [itex]2\pi[/itex] (and so 0 modulo 2[itex]\pi[/itex]) and you also need [itex]0<r \in \mathbb{R}[/itex] and [itex]r^3 = 64[/itex], ie. r=4. The condition on [itex]\theta[/itex] gives 3 possibilities,

    [itex]\theta = 0, \theta= \frac{2\pi}{3}, \theta = \frac{4\pi}{3}[/itex],

    so your 3 complex solutions are 4, and [itex]y = 4(\cos{4\pi /3} + i\sin{4\pi /3}) = -2 + 2\sqrt{3}i[/itex] and its complex conjugate (keep in mind that if one complex number is the root of a polynomial with real coefficients then its conjugate must also be).
     
  9. Aug 9, 2006 #8
    D-Leet I understand what you mean.

    y^3-64=0 turns into
    (y-4)(y^2+4y+16)=0

    y^2+4y+4=0 => (-4+/- sq(1-64))/2 => -2+31.5i and -2-31.5i

    however, -2+31.5i, -2-31.5i and 4 are not the same answer...Shouldn't they be?

    Edit: What I really am asking is...can you tell if 4 is greater or less than -2+31.5i? How do you compare real with complex numbers?
     
    Last edited: Aug 9, 2006
  10. Aug 9, 2006 #9

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You can't. There is no way to define "< " to make the complex numbers an ordered field.

    Ordered field: A field (usual properties for addition, multiplication) with a transitive relation "<" (transitive: if a< b and b< c, then a< c) satisfying
    a) If a< b then a+ c< b+ c
    b) If a< b and 0< c then ac< bc
    c) For any a, b, one and only one of these must hold
    i) a< b
    ii) b< a
    iii) a= b

    Suppose we have some definition of "<" on the complex numbers
    Obviously 0 is not equal to i (0 is the additive identity and i is not).

    Is 0< i? If so then by (b), 0*i< i*i or 0< -1. That's possible since this doesn't have to be our usual idea of "<" on the real numbers. But then, again by (b), 0*i< -1*i so 0< -i also. But then by (a), 0+ i< -i+ i or i< 0. We can't have both 0< i and i< 0 by (c).

    The only possibility left is i< 0. In that case, by (a) i+ (-i)> 0+ (-i) so 0< -i. By (b) then, 0*(-i)< (-i)(-i) or 0< -1 again. By (b) again, 0*(-i)< (-1)(-i) so 0< i. That also is impossible by (c).

    No, of course not. Just as a quadratic equation may have two different solutions, a cubic equation may have 3 different solutions. In fact, counting "multiplicity" any nth degree equation has exactly n solutions over the complex numbers. More correctly, any nth degree polynomial, with complex coefficients, can be factored into n linear factors, some of which may be the same.
     
  11. Aug 9, 2006 #10

    mathman

    User Avatar
    Science Advisor
    Gold Member

    Error: should be (y-4)(y^2+4y+16)
     
  12. Aug 9, 2006 #11
    You are correct. I'll edit my post immediatelly. Sorry about that.

    Edit: I worked it out, I understand it now. Thank you for the help. I was thinking about imaginary numbers, I wasn't thinknig about complex numbers...and ofcourse, the bi part won't get me where I want if I don't have an a. Thank you.
     
    Last edited: Aug 9, 2006
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Exponent problem
  1. Exponent problem (Replies: 4)

  2. Exponents in Functions (Replies: 15)

  3. Nested Exponents (Replies: 2)

  4. Exponent confusion. (Replies: 2)

Loading...