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### Re: Welcome!

Posted: Fri Sep 06, 2013 6:02 pm
robly18 wrote:Huh... Well then, I messed up. Thanks for correcting me!
It's ok, happen with me too I realize a super-elegant solution, start doing it, realize ARG, I FORGOT ABOUT THE IRRATIONALS! ### Re: Hyperbolic geometry is awesome

Posted: Sat Sep 07, 2013 12:28 pm
Wait a second. I'm a bit confused by this point. I thought one of the main rules of exponents was that a^(bc) = (a^b)^c
If that was true, shouldn't (-1)^(2i) equal ((-1)^2)^i = 1^i = 1, and also making ((-1)^i)^2 equal 1, making (-1)^i equal one of the square roots of 1?

(Edit from Andy: moved to correct thread)

### Re: Hyperbolic geometry is awesome

Posted: Sat Sep 07, 2013 1:10 pm
robly18 wrote:Wait a second. I'm a bit confused by this point. I thought one of the main rules of exponents was that a^(bc) = (a^b)^c
If that was true, shouldn't (-1)^(2i) equal ((-1)^2)^i = 1^i = 1, and also making ((-1)^i)^2 equal 1, making (-1)^i equal one of the square roots of 1?
http://mathworld.wolfram.com/ExponentLaws.html
Note that these rules apply in general only to real quantities, and can give manifestly wrong results if they are blindly applied to complex quantities.
Also, wrong thread (Edit from Andy: moved to correct thread)

### Re: Hyperbolic geometry is awesome

Posted: Sat Sep 07, 2013 1:19 pm
A Random Player wrote:
robly18 wrote:Wait a second. I'm a bit confused by this point. I thought one of the main rules of exponents was that a^(bc) = (a^b)^c
If that was true, shouldn't (-1)^(2i) equal ((-1)^2)^i = 1^i = 1, and also making ((-1)^i)^2 equal 1, making (-1)^i equal one of the square roots of 1?
http://mathworld.wolfram.com/ExponentLaws.html
Note that these rules apply in general only to real quantities, and can give manifestly wrong results if they are blindly applied to complex quantities.
Also, wrong thread :P
Whoops, my bad xP
I clicked latest post and saw mathemathics, and instantly assumed it was our previous discussion :D

(Edit from Andy: moved to correct thread)

### Re: Welcome!

Posted: Sun Sep 08, 2013 9:38 am
So far, I haven't seen any contradictions to the (-1)^i=-1, but I haven't seen any proofs either. Does anyone know the answer? (Keep in mind, (-1)^i may have multiple possible answers).

### Re: Welcome!

Posted: Sun Sep 08, 2013 9:45 am
exfret wrote:So far, I haven't seen any contradictions to the (-1)^i=-1, but I haven't seen any proofs either. Does anyone know the answer? (Keep in mind, (-1)^i may have multiple possible answers).
https://www.google.pt/#psj=1&q=%28-1%29^i
A lazy man's way to disprove his own theories.

EDIT-URL isn't working, so just copypaste it .-.

### Re: Welcome!

Posted: Sun Sep 08, 2013 9:46 am
exfret wrote:So far, I haven't seen any contradictions to the (-1)^i=-1, but I haven't seen any proofs either. Does anyone know the answer? (Keep in mind, (-1)^i may have multiple possible answers).
(-1)^i = e^(-pi) because WolframAlpha says so. Also my calculator. ### Re: Disproofs

Posted: Sun Sep 08, 2013 10:03 am
Wow. You all must have posted right after I did.

First of all, to robly: The URL isn't working because you placed the [/url] before the ^i, which makes it search for (-1) and not (-1)^i. Also, like I said, there could be multiple answers to this. Giving one answer doesn't necessarily disprove the other answers.

Now to A Random Player/ARP: Yes, (-1)^i = e^(-pi). You can also find this through Euler's identity. In fact, I was thinking about posting that in the post before my previous post. Basically, because e^(i*tau/2)=-1, and e^(i*tau/2)^i=(-1)^i, which equals e^(i*i*tau/2)=(-1)^i, meaning e^(-tau/2)=(-1)^i. Even so, like I said, something can have multiple answers when you're working in the strange realm of powers and complex numbers, especially when you're working in both realms. (-1)^i may even be indeterminate. Any full counter-proofs yet?

Also, has anyone found the answer to the puzzle so far? sqrt(-1) think there might be a pattern with what I was doing. (viewtopic.php?f=1&t=106#p700)