Sqrt(-1)

[Joe[x86]]

Is there one? What is it?

'tis Ergot's question, but he's lazy, so eh?

[Explicit]

i is an imaginary unit, a complex number whose square equals -1.

[rabbit]

i, of course, then leads us into entirely new fields of mathematics altogether...

[Joe[x86]]

i is an imaginary unit, a complex number whose square equals -1.

Then i is basically a logic bomb, because there is no way to get x*x to equal -1. =/

[rabbit]

Yes there is: i * i
Didn't you pay attention to what we just said?

[Joe[x86]]

Well, yeah, i * i, but that just hurts my head. Yes, I fully understood what you said, it just hurts my head thinking about it =p

[Explicit]

Well, yeah, i * i, but that just hurts my head. Yes, I fully understood what you said, it just hurts my head thinking about it =p

It shouldn't hurt your head seeing as to how it's pretty straight forward.

[rabbit]

Well, yeah, i * i, but that just hurts my head. Yes, I fully understood what you said, it just hurts my head thinking about it =p

You don't stick your tongue out when your head hurts.  You squint your eyes and clench your jaw.  You are obviously just trying to toy with us.

[shout]

i = sqrt -1
i² = -1

Now lets say you had something like f(x) = sqrt(x). The only things you can put into f(x) are positive numers and 0. This is called useful mathematics.

[Yoni]

With the introduction of the imaginary number i, it can be proven that every polynomial of degree n can be factored as a product of n polynomials of degree 1. It's called the Fundamental Theorem of Algebra.

http://mathworld.wolfram.com/FundamentalTheoremofAlgebra.html
http://mathworld.wolfram.com/PolynomialFactorization.html

This is a very important result. It means that, by extending numbers onto 2 axes, algebra is "complete" (purely algebraic equations, such as x^2 + 1 = 0, are solvable).

[rabbit]

i = sqrt -1
i² = -1

Now lets say you had something like f(x) = sqrt(x). The only things you can put into f(x) are positive numers and 0. This is called useful mathematics.

Assuming that x is being limited to real numbers, this is true.  However, if complex numbers are allowed, then your statement goes to shit.  Other than that, way to tell Joe off :0

[Rule]

i = sqrt -1
i² = -1

Now lets say you had something like f(x) = sqrt(x). The only things you can put into f(x) are positive numers and 0. This is called useful mathematics

Obviously you have no idea how useful imaginary numbers can be, especially in applied mathematics and Real Analysis.

Also, here's something to think about:

(-1)^(1/2) * (-1)^(1/2) = (-1*-1)^1/2 = 1^(1/2) = 1


As far as "imaginary" numbers not existing, and being hard to think about, they "exist" just as much as negative numbers do.  You just haven't thought about them that much.

[nslay]

Now prove or disprove that R^n can be related to R^(n-1) in the following fashion:
There exists x in R^(n-1) and 1-1 f(x) s.t. f(x)=r is not a member of R^(n-1) (and is not infinite!).
Furthermore f^(-1)(r)=x

Example:
Find a 1-1 f(x) that could generate an imaginary number relative to Z (call it j).

[rabbit]

Hm.

e^(pi)i = -1
e^(pi)i = i^2
0! = 1 = -(i^2)
e^(pi)i = -(0!)
e^(pi)Sqrt(-(0!)) = -(0!)
ln(e^(pi)Sqrt(-(0!))) = ln(-(0!))
(pi)Sqrt(-(0!)) = -ln(0!)
(pi)Sqrt(-(0!)) = -(0)
(pi)i = 0

Somehow I find that wrong... o well.

[dxoigmn]

Hm.

e^(pi)i = -1
e^(pi)i = i^2
0! = 1 = -(i^2)
e^(pi)i = -(0!)
e^(pi)Sqrt(-(0!)) = -(0!)
ln(e^(pi)Sqrt(-(0!))) = ln(-(0!))
(pi)Sqrt(-(0!)) = -ln(0!)
(pi)Sqrt(-(0!)) = -(0)
(pi)i = 0

Somehow I find that wrong... o well.

Um.. ln(-(0!)) != -ln(0!).  ln(-(0!)) = ln(|-(0!)|) + i*arg(-(0!)) = ln(-(0!)) + i*arg(-(0!)) = i*pi