It's ok, happen with me too I realize a superelegant solution, start doing it, realize ARG, I FORGOT ABOUT THE IRRATIONALS!robly18 wrote:Huh... Well then, I messed up. Thanks for correcting me!
Welcome!

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Re: Welcome!
$1 = 100¢ = (10¢)^2 = ($0.10)^2 = $0.01 = 1¢ [1]
Always check your units or you will have no money!
Always check your units or you will have no money!
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)
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

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Re: Hyperbolic geometry is awesome
http://mathworld.wolfram.com/ExponentLaws.htmlrobly18 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?
Also, wrong threadNote that these rules apply in general only to real quantities, and can give manifestly wrong results if they are blindly applied to complex quantities.
(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!
Always check your units or you will have no money!
Re: Hyperbolic geometry is awesome
Whoops, my bad xPA Random Player wrote:http://mathworld.wolfram.com/ExponentLaws.htmlrobly18 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?Also, wrong thread :PNote 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
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).
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Re: Welcome!
https://www.google.pt/#psj=1&q=%281%29^iexfret 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.
EDITURL 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

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Re: Welcome!
(1)^i = e^(pi) because WolframAlpha says so. Also my calculator.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 = 100¢ = (10¢)^2 = ($0.10)^2 = $0.01 = 1¢ [1]
Always check your units or you will have no money!
Always check your units or you will have no money!
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 counterproofs 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)
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 counterproofs 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. ):