A quadratic extension of R is by definition commutative.

you just don't know what this is: a quadratic extension of a field.

The question is not clear, but I will try to answer anyway. Let V denote a 2-dimensional real vector space and let * denote a product, i.e. a bilinear function of V times V into V.  Then the product can be decomposed into a symmetric part *s and an antisymmetric par *a so that x*y=x *s y + x *a y . Let e1 and e2 denote two basis vectors. Then the first coordinate of x *a y is proportional to the determinant of x and y and the second coordinate is proportional to the determinant of x and y.  Therefore there exists a vector v such that  x *a y is proportional to det(x,y) times v. If we choose v as the first basis vector and  choose the second basis vector appropriately we may even get that x *a y equals det(x,y) times v. In this sense the non-commutative part of a 2d-product is essentially the same as the determinant.

@ Abanto: (0.b) x (0,d) = (0,0), for all b,d: not a field!

It is an algebra over the field of real numbers. 

https://en.wikipedia.org/wiki/Algebra_over_a_field

for a field you have: ab = 0, then a=0 or b=0. 

@Peter Harremoës: it is not an algebra! (nonlinear multiplication!)

@ Anton Schober Normally a product is required to be bi-linear and all the examples in this thread are bi-linear. What is the "nonlinear multiplication" that you refer to?

[µ(a,b)]x(c,d) = (µa,µb)x(c,d) = (µ2ab+cd, µ(bc+ad)) unequal µ[(a,b)x(c,d)]

you say: and all the examples in this thread are bi-linear.

nothing here is bi-linear!

Sorry, you are right it is not bi-linear so according to standard definitions it is not a product. I am too used to products so I overlooked that the composition was ab+cd rather than ac+bd. Anyway, since it is not bi-linear it does not define an algebra and the discussion whether it is a field or not becomes irrelevant. 

It is not entirely clear what you mean by number in this case, but I will interprete it to be an algebra over the real numbers. If it is to be an extension of the real numbers the answer is NO, as Anton said in the very first post. (Who was the fool downvoting that post, and why?) Because such an extension must have the form (a + b g), with a and b real numbers. And g must necessarily commute with itself.

However, there is a two-dimensional non-commuting algebra over the reals, without a unit, consisting of elements of the form

(a g1 + b g2)

with a and b real, and

g1 * g1 = g2 * g2 = 0, and g1 * g2 = -g2 * g1 = g2.

You may recognize this as the Lie algebra for the group of upper triagonal 2x2 real matrices with determinant 1. 

 @ Kare: a realization of your “non commuting algebra" in terms of 2x2 matrices is:

g(1) = (a(ik)), a(11) = 1, the other= 0; g(2) = (b(ik)), b(12) = 1, the other = 0.

This is the Lie algebra of the affine group of a line (dilation and translation)

The above mentioned Lie group of upper triagonal 2x2 real matrices with determinant 1 should be 1-dimensional: a(11) = a(22) = 1; a(12) = a, a(21) = 0, the Lie algbra is the above g(2).

Anton> upper triagonal

You can have a(11) = 1/a(22) = a, a(12) = b, a(21)=0. With generators σz/2 and σ+. Your example is also nice.

However the examples given are all, in fact, quaternions. The square of σz/2 is proportional to the 2 x 2 identity matrix. The generalization sought for, in fact, is SL(2,C), if one imposes that the determinant be equal to unity, or GL(2,C) if not.

Stam@ The Lie algebra product, with respect to which every element is nilpotent, is mathematically different from the matrix product.

Note that there is no matrix representation of the algebra, because the product is not associative:

g1 * (g1 * g2) =  g2 , but ( g1 * g1 ) * g2 = 0, 

while matrix multiplication is always associative.

But is it an *algebra* (or a *group*) at all? I.e. closed under addition and/or  multiplication, or the composition rule, in any case? σz;, σ+ and σ- define a 2-dimensional representation of SU(2). This is a non-commutative algebra over the complex numbers and it does have finite-dimensional representations-and it does represent the quaternions. In other words leaving out one of the raising/lowering operators isn't correct, because the structure isn't an algebra any more. Indeed, the statement that it is without unit is a giveaway that this isn't an algebra and that one of its generators does square to a multiple of the identity indicates that something funny's going on.

What's gone wrong in this discussion, as usual, is what are the postulates, what's meant by an extension of the real numbers that's non-commutative and isn't isomorphic to the quaternions. Well group theory abounds with such examples: any non-abelian (the term that means that the group operation is non-commutative) group, that doesn't contain SU(2) (the group of quaternions) as a subgroup and has a two-dimensional representation, is an example. 

An algebra (over the real numbers) is a structure which allow addition of elements, and multiplication of elements by real numbers, equipped with a bilinear "multiplication" operator *. It is a commutative group under addition. One talks about algebras with  a unit, i.e. an element E  such that

E * x = x * E = x for all elements x in the algebra,

but algebras without such an element are also investigated. One also distinguishes between associative algebras, i.e where the multiplication rule satisfy

x * (y * z) = (x * y) * z for all elements x, y, z in the algebra,

and non-associative algebras where this property is not satisfied.

Lie algebras, where the "multiplication" * is usually interpreted as the commutator between the two (operator) operands, are the prime examples of non-associative algebras without a unit. However, this interpretation of the purely algebraic structure is irrelevant for the question of this thread. As is the group(s) generated by such Lie algebras, which I unfortunately mentioned as an interpretation of the algebra, but which only seems to have caused confusion. 

The simplest real non-commuting algebra with a unit and associative multiplication rule is three-dimensional: One example is the algebra of upper triagonal real 2x2 matrices. This is also --- in some sense --- an extension of the real numbers.  In is of course not a number field, which by definition has to be commuting.

Note that SU(2) is not an algebra, because it is not closed under addition.

Kare, 1/2 sigma(z) doesn't generate a(11) = 1/a(22) = a. (this works - in the reals - only with a = 1(!). Anyway, your group is not a Lie group, because it is not defined "at the origin" a = 0. (not differentiable at a = 0)

next: exp(i/2 µ sigma(z) is SU(2)), not a real group. same with sigma +.

Anton> sigma(z) doesn't generate a(11) = 1/a(22) = a.

What is your definition of  σz? According to mine, and I think most others, 

exp(t σz) = diag( exp(t), exp(-t) ) = diag( a, 1/a) with a = exp(t).

I should have qualified the group further by specifying a>0. And the "origin" (the unit element) is, as always,  t=0 or a=1.

You should not be distracted by the missing i in the exponent; this is just a very special convention used by physicists when they want to generate unitary representations from hermitian generators. But Lie groups are much more than that, even to physicists. Of course, if you insist, you may replace σz  -> -iσz, σ+ -> -iσ+ as generators, and keep an i in the exponent. But that would make most mathematicians stare at you in bewilderment.

It is common to talk about various real forms of the same Lie algebra, which upon exponentiation generates very different Lie groups; I was perhaps not precise enough about which real form I meant.

Kara> I don't "insist". sigma(z) and b(12) = 1 generates this group. The group acts on R2, but it is not a "classical group". b(12) generates the translations in x direction and the sigma(z) group has hyperbolas as orbits. very funny, it is a subgroup of the affine group of the plane.

The number systems you propose are not distributive: that is, if z1, z2 and z3 are three generalised complex numbers of the kind you have in mind, then

z1*(z2 + z3) = / z1*z2 + z1*z3

where +, of course, should be defined as usual as sums coordinatewise. That means it is deeply different from what is ordinarily called a number system.

Also, the complex numbers are a field. For a field, we require that for every field element z=/0, the equation

z*x=1

should have a solution.

That is true in the complex: let z=a + bi. Then

x = a/(a^2+b^2)-i b/(a^2+b^2)

satisfies this equation. That condition implies that, if

x*y = 0

then either x = or y = 0, but the two conditions are not equivalent.

if case I is: xy = z, then case III: (c.c.x)y = w. Similar II & IV

``Numbers'' have a set of properties that are constant, and some which we allow to vary. The constant ones are:

for addition: associativity, commutativity, zero elements, existence of additive inverse

for multiplication: associativity

for the pair of operations: distributivity.

Additional properties which may or may not hold for multiplication are

commutativity, existence of a unit, existence of multiplicative inverse for non-zero elements.

The problem with many of your ``numbers' is that they violate the first class of properties.

Your class II I find somehow fascinating: of course, it reduces to

(a + b) * (c + d) = ac + ad + bc + bd

so that, in fact, class II is not an extension of the reals at all!

Just a remark on class II. By introducing e1= (1/2, 1/2) and e2 = (1/2, -1/2) I find

e1 * e1 = e1, e2 * e2 = e2, and   e1 * e2 = e2 * e1 = 0.

So, with respect to this basis it is just two parallel occurrences of the real numbers, a structure which is already very usefully implemented in f.i. the Numpy Array  in Python (only extended to an arbitrary number of parallels). There is some modifications of the axioms of number fields with respect to which "numbers" are invertible, but computer designers have already had to change the arithmetic rules of finite precision numbers somewhat.

As for the classes III and IV: If one invokes a requirement of associative multiplication, it is straightforward to deduce that all numbers must be 0. Which I think formally satisfies the axioms of a number system, but is otherwise not so interesting.

hi, Kare, eventually you have noticed, that I treated ALL cases I - IV in my post above!

All> I treated ALL cases I - IV in my post above!

You said so, but I was not able to extract the meaning in the message you communicated.  

x,y,z,w are complex numbers. (c.c.x) means x overbar, which is conjugate complex. No idea how to create an overbar here (and on all pc's)