distributive and non-commutative binary operation on real numbers I've been searching for a binary operation that is distributive over addition but not commutative on $\mathbb{R}$; to be more exact I want to have $$a(b+c)=ab+ac$$ $$(a+b)c=ac+bc$$ for all $a,b,c \in \mathbb{R}$ and also that $\exists a,b \in \mathbb{R};$ $ab\neq ba$.
I've found that $a.0=0.a=0$ and also $(-a)b=a(-b)=-(ab)$ and similar properties by plugging real numbers in; but that's it. I also tried constructing this operation using addition, multiplication, subtraction and division but I failed since it seems like there has to be symmetry; but with symmetry comes commutativity.
 A: As an additive group, $\mathbb{R}$ is just isomorphic to a direct sum of $2^{\aleph_0}$ copies of $\mathbb{Q}$ (since it is a $\mathbb{Q}$-vector space so you can pick a basis).  In particular, $\mathbb{R}\cong\mathbb{R}^4$ as additive groups.  But there is an easy distributive and noncommutative operation on $\mathbb{R}^4$, namely matrix multiplication of $2\times 2$ matrices.  So, transporting this operation along the additive isomorphism $\mathbb{R}\cong\mathbb{R}^4$ you get such an operation on $\mathbb{R}$.
This is highly nonconstructive, since you can't actually write down a basis for $\mathbb{R}$ over $\mathbb{Q}$ or an additive isomorphism $\mathbb{R}\cong\mathbb{R}^4$.  In fact, it is impossible to prove the existence of such a binary operation on $\mathbb{R}$ without using the axiom of choice.  This follows from the fact that there exists a model of ZF in which every additive homomorphism $\mathbb{R}\to\mathbb{R}$ is given by multiplication by a real number.  So, if you have your binary operation $*$, for each $a\in\mathbb{R}$, the map $b\mapsto a*b$ is an additive homomorphism and thus must have the form $a*b=f(a)b$ for some $f(a)\in\mathbb{R}$.  But now the other distributive law says that $a\mapsto f(a)$ is a homomorphism, so it is multiplication by some constant $c\in \mathbb{R}$.  That is, your multiplication must have the form $a*b=cab$ for a constant $c$, which is commutative.
