# Math Help - Indices for Negatives

1. ## Indices for Negatives

Why is it that $\sqrt{-1}\sqrt{-1}=-1$ instead of $\sqrt{(-1)(-1)}=1$? What other index laws are different when involving one or more negative numbers?

Also, according to my book, $(-1)^{\frac{2}{6}} \neq (-1)^{\frac{1}{3}}$, but according to my calculator, this is perfectly fine. According to it, $(-1)^{\frac{2}{6}} \neq ((-1)^2)^6$ and $(-1)^{\frac{2}{6}} \neq ((-1)^{\frac{1}{6}})^2$. This is also the case when drawing graphs on my calculator.

2. Originally Posted by DivideBy0
Why is it that $\sqrt{-1}\sqrt{-1}=-1$ instead of $\sqrt{(-1)(-1)}=1$?
[snip]
"sqrt" is defined for real numbers as the positive number whose square is ..... but can't be defined that way for non-real numbers ..... The phrase the positive number no longer makes sense because the complex numbers cannot be ordered so as to make them an ordered field.

By an ordered field I mean:

a < b => a + c < b + c

a < b and c > 0 => ac < bc.

So you can't group complex numbers into positive and negative numbers. Square root, along with many other functions, has to become multi-valued .....

A complication with that is that you can't just define i by $i = \sqrt{-1}$ because -1 now has two square roots and there's no way to distinguish between them.

A more formal definition is to define the complex numbers as the set of all pairs of real numbers (a, b) and deine addition and multiplication by (a, b) + (c, d) = (a + c, b + d) and (a, b) * (c, d) = (ac - bd, ad + bc) respectively. It can then be proved that all the rules of arithmetic hold, that pairs of the form (a, 0) obey the same rules as real numbers and that (0, 1) * (0, 1) = (-1, 0).

If the symbols 1 and i are used to mean (1, 0) and (0, 1) respectively, then (a, b) can be written as (a, 0) * (1, 0) + (b, 0) * (0, 1) = a * 1 + b * i = a + ib.

3. Originally Posted by DivideBy0
Why is it that $\sqrt{-1}\sqrt{-1}=-1$ instead of $\sqrt{(-1)(-1)}=1$? What other index laws are different when involving one or more negative numbers?
Because the mistake is here:
$\sqrt{ab} = \sqrt{a}\sqrt{b}$.

If $a,b\geq 0$ then this is okay. But if $a,b<0$ then it CAN be wrong, like here.

This is Mine 88th Post!!!