Question

Question #987b8

Answer 1

If the index is even, you get a nonreal complex number.

Explanation:

This is what I call the "even/odd problem."
Simple example using "2" (but any number works).

We see that
`2*2*2 = 8` and `(-2)(-2)(-2) = -8`.
Therefore, `root(3) 8 = 2` and `root(3) (-8) = -2`.
Since this is true for any nonzero real number, x, we see that
`root(3) (-x) = -root(3) (x)`.

However, notice that when we multiply an even number of negative factors, the product is POSITIVE.
`2*2*2*2 = 16` and `(-2)(-2)(-2)(-2) = 16`.

If there are an even number of factors, the product is always positive. Therefore, all real even roots are roots of positive numbers.

What happens with something like `sqrt(-4)`? Should it be 2? No: (2)(2) is not -4. How about -2? No. (-2)(-2) = 4, NOT -4.

As we saw a moment ago, there is no real number whose square is negative. Therefore `sqrt(-4)` is not a real number.
At the lower levels we simply say it is undefined.

Later on, we observe that `sqrt(-4)` would be a zero (root) of the polynomial `f(x) = x^2 + 4`, and we define the basic complex unit, `i` -- also called the "imaginary unit" -- to be `sqrt(-1)`. This definition produces a larger system of numbers called the Complex Numbers. Earlier in Algebra, though, it is enough to know that `root(n) (x)` is not a real number whenever n is even and x < 0.