## Welcome!

A spot for all things TestTube
A Random Player
Posts: 523
Joined: Mon Jun 03, 2013 4:54 pm

### Re: Welcome!

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! \$1 = 100¢ = (10¢)^2 = (\$0.10)^2 = \$0.01 = 1¢ 
Always check your units or you will have no money!

robly18
Posts: 413
Joined: Tue Jun 04, 2013 2:03 pm

### Re: Hyperbolic geometry is awesome

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)
Convincing people that 0.9999... = 1 since 2012

A Random Player
Posts: 523
Joined: Mon Jun 03, 2013 4:54 pm

### Re: Hyperbolic geometry is awesome

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)
\$1 = 100¢ = (10¢)^2 = (\$0.10)^2 = \$0.01 = 1¢ 
Always check your units or you will have no money!

robly18
Posts: 413
Joined: Tue Jun 04, 2013 2:03 pm

### Re: Hyperbolic geometry is awesome

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.
I clicked latest post and saw mathemathics, and instantly assumed it was our previous discussion :D

(Edit from Andy: moved to correct thread)
Convincing people that 0.9999... = 1 since 2012

exfret
Posts: 585
Joined: Sun Jul 28, 2013 8:40 pm

### Re: Welcome!

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).
Nobody ever notices my signature. ):

robly18
Posts: 413
Joined: Tue Jun 04, 2013 2:03 pm

### Re: Welcome!

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).
A lazy man's way to disprove his own theories.

EDIT-URL isn't working, so just copypaste it .-.
Last edited by robly18 on Sun Sep 08, 2013 10:19 am, edited 3 times in total.
Convincing people that 0.9999... = 1 since 2012

A Random Player
Posts: 523
Joined: Mon Jun 03, 2013 4:54 pm

### Re: Welcome!

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. \$1 = 100¢ = (10¢)^2 = (\$0.10)^2 = \$0.01 = 1¢ 
Always check your units or you will have no money!

exfret
Posts: 585
Joined: Sun Jul 28, 2013 8:40 pm

### Re: Disproofs

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)
Nobody ever notices my signature. ):