Question #987b8

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.