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 Post subject: 2^?
PostPosted: Tue Apr 02, 2013 8:26 pm 
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What is your favorite power of 2?

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 Post subject: Re: 2^?
PostPosted: Tue Apr 02, 2013 9:37 pm 
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The power of magic.


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 Post subject: Re: 2^?
PostPosted: Tue Apr 02, 2013 9:50 pm 
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.5


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 Post subject: Re: 2^?
PostPosted: Wed Apr 03, 2013 12:21 am 
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-i


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 Post subject: Re: 2^?
PostPosted: Wed Apr 03, 2013 12:44 am 
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I usually think about the bit pattern.

Code:
   00000001   Quite OK
   00000100   Bad
   00001000   OK
   10000000   Nice


Let's just say 128.

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 Post subject: Re: 2^?
PostPosted: Wed Apr 03, 2013 7:10 am 
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iansjack wrote:
-i

Wouldn't that be 1?

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 Post subject: Re: 2^?
PostPosted: Wed Apr 03, 2013 7:21 am 
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m12 wrote:
Wouldn't that be 1?


Only if i was zero and then you don't need the minus sign ;)


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 Post subject: Re: 2^?
PostPosted: Wed Apr 03, 2013 7:31 am 
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I thought you meant i as in √(-1). the square root of √1 is 1, so logically -√(-1) = √1 = 1

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 Post subject: Re: 2^?
PostPosted: Wed Apr 03, 2013 7:44 am 
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You can't just multiply by negative 1 to covert complex-real numbers


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 Post subject: Re: 2^?
PostPosted: Wed Apr 03, 2013 7:50 am 
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2 to the power of -i is most definitely not 1! (And -i is most definitely not 1 either.)


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 Post subject: Re: 2^?
PostPosted: Wed Apr 03, 2013 7:53 am 
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Why not? 0 + i and 0 - i are valid complex numbers.

By the way, how about 2^e


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 Post subject: Re: 2^?
PostPosted: Wed Apr 03, 2013 8:05 am 
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pi-th root of 2 (2^(1/pi))

I'm sure there will be some of you who get the reference


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 Post subject: Re: 2^?
PostPosted: Wed Apr 03, 2013 8:14 am 
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m12 wrote:
I thought you meant i as in √(-1). the square root of √1 is 1, so logically -√(-1) = √1 = 1

This is one of the most frequent problems with understanding complex numbers. i IS NOT sqrt(-1). This is NOT its definition. x = sqrt(-1) is NOT i.

i is defined as follows:

i^2 = -1

No other definition of i is correct, and all of them create the bullshit you wrote before, when you said -i = 1. -i is conceptually the same as (0, -1) in R2, which is *completely different thing* than 1, which is, again conceptually, (1, 0) in R2.


Please educate yourself on such basic things before diving into talking about it.

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 Post subject: Re: 2^?
PostPosted: Wed Apr 03, 2013 8:23 am 
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Hi,

Griwes wrote:
i^2 = -1

No other definition of i is correct


i^6 = -1?

Cheers,
Adam


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 Post subject: Re: 2^?
PostPosted: Wed Apr 03, 2013 8:26 am 
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That ain't a definition. You used the fact that i^2 is -1, which is *the* definition, and you just used that fact to write (i^2)^3 = -1, which is the same as (-1)^3 = -1.

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