# Rational exponents (was: Math Discussion)

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1. Sep 12, 2015

### goolalklk

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. Sep 12, 2015

### Staff: Mentor

$(-64)^{3/2} = [(-64)^{1/2}]^3$

3. Sep 13, 2015

### Eclair_de_XII

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
4. Sep 14, 2015

### HallsofIvy

Staff Emeritus
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"?

5. Sep 19, 2015

### FeDeX_LaTeX

The negative sign makes it so that the stated number isn't real.

6. Sep 19, 2015

### SteamKing

Staff Emeritus
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.