2^?
Re: 2^?
I usually think about the bit pattern.
Let's just say 128.
Code: Select all
00000001 Quite OK
00000100 Bad
00001000 OK
10000000 Nice
Re: 2^?
Only if i was zero and then you don't need the minus signm12 wrote:Wouldn't that be 1?
- Griwes
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Re: 2^?
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.m12 wrote:I thought you meant i as in √(-1). the square root of √1 is 1, so logically -√(-1) = √1 = 1
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.
Reaver Project :: Repository :: Ohloh project page
<klange> This is a horror story about what happens when you need a hammer and all you have is the skulls of the damned.
<drake1> as long as the lock is read and modified by atomic operations
<klange> This is a horror story about what happens when you need a hammer and all you have is the skulls of the damned.
<drake1> as long as the lock is read and modified by atomic operations
Re: 2^?
Hi,
Cheers,
Adam
i^6 = -1?Griwes wrote:i^2 = -1
No other definition of i is correct
Cheers,
Adam
- Griwes
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Re: 2^?
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.
Reaver Project :: Repository :: Ohloh project page
<klange> This is a horror story about what happens when you need a hammer and all you have is the skulls of the damned.
<drake1> as long as the lock is read and modified by atomic operations
<klange> This is a horror story about what happens when you need a hammer and all you have is the skulls of the damned.
<drake1> as long as the lock is read and modified by atomic operations