Quaternions

Complex numbers were invented to solve x^2 +1 = 0
Quaternions where invented for ???????

Some kind of onions

http://en.wikipedia.org/wiki/Quaternions#History

This quote is interesting then:
"...the thing about a Quaternion 'is' is that we're obliged to encounter it in more than one guise. As a vector quotient. As a way of plotting complex numbers along three axes instead of two. As a list of instructions for turning one vector into another..... And considered subjectively, as an act of becoming longer or shorter, while at the same time turning, among axes whose unit vector is not the familiar and comforting 'one' but the altogether disquieting square root of minus one. If you were a vector, mademoiselle, you would begin in the 'real' world, change your length, enter an 'imaginary' reference system, rotate up to three different ways, and return to 'reality' a new person. Or vector..." Thomas Pynchon

Does anyone have examples of problems that are only to do with the reals that require complex numbers or quaternions to solve?

>>5
You never *need* to use complex numbers to prove things about real numbers, since you can always rewrite proofs to use the ring \mathbb{R}[x]/(x^2+1) instead, which is isomorphic to \mathbb{C} and doesn't actually require imaginary numbers to define and use.

So complex numbers are a fraud?

Care to explain what this notation means?

>>8
Polynomials with real coefficients modulo x^2+1.  So essentially you work with polynomials instead of numbers, but you set x^2+1 to be zero.  So for instance

x^3+x+1 = x*(x^2+1)+1 = x*0+1 = 1

Search for "modular arithmetic" on wiki.  It's not the exact same thing, but the general idea is the same.

>>6 used a lot of words to not say very much

>>8 should not interest himself in mathematics since he lacks basic knowledge.

>>9
Is the isomorphism with the complex numbers easy to show?  I should probably dig up my abstract algebra book again.

>>10
What he said was pretty interesting, faggot.

Yes: f(p(x)) = p(i). Clearly this is a suprayective ring morphism from R[x] to C.

As x^2 + 1 is an element of kerf (and it is an ideal) and is an irreducible polynomial, implies that kerf = <x^2 + 1>.

To conclude, just recall the first isomorphism theorem. You get that R[x]/<x^2 + 1> is isomorphic to C