# Exponent problem

1. Jul 30, 2006

### theumann

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. Jul 30, 2006

x^2 = 16 could allow x = -4

3. Jul 30, 2006

### mathman

Further complication is that y^3=64 has two complex roots as well as the real root 4.

4. Jul 31, 2006

### HallsofIvy

Silly problem anyway!

5. Aug 4, 2006

### Robokapp

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...

6. Aug 4, 2006

### d_leet

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.

7. Aug 4, 2006

### Data

write $y = re^{i\theta}$ with $r>0, \theta \in [0, 2\pi)$, then your equation is

$r^3e^{3i\theta} = 64e^{0i}.$

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

$\theta = 0, \theta= \frac{2\pi}{3}, \theta = \frac{4\pi}{3}$,

so your 3 complex solutions are 4, and $y = 4(\cos{4\pi /3} + i\sin{4\pi /3}) = -2 + 2\sqrt{3}i$ 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).

8. Aug 9, 2006

### Robokapp

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
9. Aug 9, 2006

### HallsofIvy

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.

10. Aug 9, 2006

### mathman

Error: should be (y-4)(y^2+4y+16)

11. Aug 9, 2006

### Robokapp

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