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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


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


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    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


    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
    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^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


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    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


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    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
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