Why is (-2)^(-2)^(-2) an imaginary number?

• Aug 13th 2010, 05:21 AM
narko
Why is (-2)^(-2)^(-2) an imaginary number?
I'm kind of curious.. Shouldn't (-2)^(-2)^(-2)= (-2)^0,25..?
• Aug 13th 2010, 06:11 AM
mr fantastic
Quote:

Originally Posted by narko
I'm kind of curious.. Shouldn't (-2)^(-2)^(-2)= (-2)^0,25..?

It is. At least, that's how a calculator interprets it. And (-2)^(1/4) = 0.841 + 0.841i (correct to three decimal places).

But another interprettation might well be [(-2)^(-2)]^(-2) = (1/4)^(-2), which will give a different answer.
• Aug 13th 2010, 06:20 AM
narko
hmm, so negative numbers cannot be exponentiated by fractions (like -10^0,5=imaginary number).. well OK, thanks. We haven't learned yet how to calculate things like 10^0,5 without calculators or programs yet. Thank you, again.

PS. Heh, these are interesting numbers. So basically (-2)^(-2)^(-2) have multiple answers depending on how u interpret it (I always thought you had to solve these issues in "(-2)^(-2)^(-2)= (-2)^0,25" this way, and not this; "0,25^-2" way)..
• Aug 13th 2010, 06:25 AM
mr fantastic
Quote:

Originally Posted by narko
hmm, so negative numbers cannot be exponentiated by fractions (like -10^0,5=imaginary number).. well OK, thanks. We haven't learned yet how to calculate things like 10^0,5 without calculators or programs yet.

Wrong.

If n is odd, then (-ve)^1/n = -ve. eg. (-8)^1/3 = -2. (There are two other answers that are both non-real).

If n is even, then (-ve)^1/n is not real. eg. (-4)^1/2 = -2i. (There are two other answers that are also non-real).
• Aug 13th 2010, 06:46 AM
undefined
Quote:

Originally Posted by mr fantastic
But another interprettation might well be [(-2)^(-2)]^(-2) = (1/4)^(-2), which will give a different answer.

The standard interpretation is (-2)^[(-2)^(-2)].

Reference: Wikipedia - Tetration

Quote:

Originally Posted by mr fantastic
If n is even, then (-ve)^1/n is not real. eg. (-4)^1/2 = -2i. (There are two other answers that are also non-real).

Is this a typo? I thought the only solutions were 2i and -2i.
• Aug 13th 2010, 07:04 AM
narko
hmm, OK. I think I understand this allot better. for example:

10^1/5 solving this would need:

x^5=10
and then u use the 5√ 10 to solve it.

Thinking of it, we HAVE went through this. heh I just forgot it. 10^2/6= 3/1√10 and so on.

And yeah, now that I can see this it is only logical that even exponent fractions are imaginary (square root of -1?) while odd ones are not (3√(-1)=-1*-1*-1=-1).
• Aug 13th 2010, 02:33 PM
Soroban
Hello, narko!

Quote:

$\text{Why is }\,(-2)^{(-2)^{(-2)}}\:\text{ an imaginary number?}$

$\text{Shouldn't }\:(-2)^{(-2)^{(-2)}} \:=\:(-2)^{0.25}\;?$

You are correct . . .

Now how do you take the 4th root of a negative number?