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A Random Player
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Re: Welcome!

Post by A Random Player » 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! :P
$1 = 100¢ = (10¢)^2 = ($0.10)^2 = $0.01 = 1¢ [1]
Always check your units or you will have no money!

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robly18
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Re: Hyperbolic geometry is awesome

Post by robly18 » 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)
Convincing people that 0.9999... = 1 since 2012

A Random Player
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Re: Hyperbolic geometry is awesome

Post by A Random Player » 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 :P

(Edit from Andy: moved to correct thread)
$1 = 100¢ = (10¢)^2 = ($0.10)^2 = $0.01 = 1¢ [1]
Always check your units or you will have no money!

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robly18
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Re: Hyperbolic geometry is awesome

Post by robly18 » 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)
Convincing people that 0.9999... = 1 since 2012

exfret
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Re: Welcome!

Post by exfret » 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).
Nobody ever notices my signature. ):

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robly18
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Re: Welcome!

Post by robly18 » 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 .-.
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
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Joined: Mon Jun 03, 2013 4:54 pm

Re: Welcome!

Post by A Random Player » 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. ;)
$1 = 100¢ = (10¢)^2 = ($0.10)^2 = $0.01 = 1¢ [1]
Always check your units or you will have no money!

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

Re: Disproofs

Post by exfret » 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)
Nobody ever notices my signature. ):

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