We know from Euler's e^{i x} formula that e^ip = -1. But, we also know that e^i2p = 1, so why can't we take the two away using roots, and get e^ip = 1 or -1? Similarly, why can't we say e^i4p = 1, and use the fourth root to get 4 answers? Why can't we do this an infinite amount of times and get an infinite number of answers? I know we could say we can't do this because of contradiction, but what is the reason for the contradiction?
Brad

This is like saying, suppose x = -1, then x² = 1. If we now take roots we find x = +-1, so -1 = 1 or -1. Then we can consider xⁿ = 1 and there are n roots for x and similarly we could do this an infinite number of times and get an infinite number of answers for -1.

You just have to remember that some operations aren't reversible (ok the wording isn't right there, but I can't think of the proper term).

Regards,
Olof.

When we solve xⁿ = -1 we look for complex numbers x which satisfy the equation.

e^ip is a complex number x which satisfies x²=-1. This does not mean all other complex numbers such numbers x are equal to e^ip.
A similar argument might go...

Let S be the set of all people. My mum is a person therefore she is in S, therefore every person is my mum.

I see. But, then, what significance does e^ip have if there are an infinite other number of ways to evaluate e^ip (all of which are different)) Is it just because that solution is a more "natural" solution that we usually use it with it? Is it just easier to work with?
Brad

There is not an infinite number of ways of evaluating e^ip. e^ip=-1 is well defined (start talking about log on the complex numbers and you have a point).
There are n unique solutions to the equation xⁿ = -1. Do not confuse the two statements.

I'm not sure that's what I'm saying, though. If
e^{i 2 n p} = 1, Then e^{i 2 np/2n} = e^ip = 1^1/2n. Of course, -1 is one solution to this, but it is not the only. It is the only that will appear no matter what n is (unless it's 0). Perhaps I'm not understanding though (and that very could be the case)...

Sorry Brad, but this is *exactly* analogous to the example given by the first Anonymous post, i.e.

1ⁿ = 1, so 1^n/n = 1 = 1^1/n , so there are n different values for 1!

The problem is that you assume (aⁿ )^1/n = a, when it doesn't necessarily.

James.

I see now. Thanks to Olof, James, and Anon. Sorry if I sounded rude earlier-if I did it was solely because of a lack of comprehension on my part. Thanks once again,

Brad