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

Printable View

- Aug 13th 2010, 04:21 AMnarkoWhy is (-2)^(-2)^(-2) an imaginary number?
I'm kind of curious.. Shouldn't (-2)^(-2)^(-2)= (-2)^0,25..?

- Aug 13th 2010, 05:11 AMmr fantastic
- Aug 13th 2010, 05:20 AMnarko
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, 05:25 AMmr fantastic
- Aug 13th 2010, 05:46 AMundefined
The standard interpretation is (-2)^[(-2)^(-2)].

Reference: Wikipedia - Tetration

Is this a typo? I thought the only solutions were 2i and -2i. - Aug 13th 2010, 06:04 AMnarko
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, 01:33 PMSoroban
Hello, narko!

Quote:

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

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

You are correct . . .

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