# Is a commutative and associative with neutral element operation and inverses on R^2 necessarily componentwise sum?

I'm trying to make a derivation of the standard operations of complex numbers from field axioms and the condition that operations on real numbers work the same way. One part of that work is proving that if an operation between pair that "behaves" like a sum then it is the pairwise sum.

Let ° be a closed binary operation on $$\mathbb{R}^2$$ such that ° is commutative, associative, has a neutral element and every pair has an inverse. Assume that for all pairs (a,0) ,(b,0) $$(a,0)°(b,0)=(a,0)+(b,0)$$ where $$+$$ is the usual componentwise sum for pairs. Proving that + and ° are equal reaches my objectives.

I've managed to prove that the neutral element of ° is (0,0) but I don't know how to continue from this.

(Post edit note: before the edit in the question I forgot to put the condition that ° must have inverses. Without that condition @j-dmaths counterexample works)

• Welcome to Mathematics SE. Take a tour. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. Feb 11 at 22:34

Let $$\pi$$ be a bijection on $$\mathbb R$$ with $$\pi(0)=0$$ and $$\pi(-x) = -\pi(x)$$. Define our operation by $$(a,c)\circ(b,d) = \big(a+c,\pi^{-1}(\pi(b)+\pi(d))\big)$$ So the second coordinate is "disguised" addition, possibly not addition itself.
Example: $$\pi(x) = x^3$$.
• Yes, I thought of a similar idea , letting c*d=($c^5+d^5)^{\frac{1}{5}}$ in the second component. This was a group I had looked at in one of my classes that seemed to satisfy the required properties.
What about if $$(a,c)°(b,d)= (a+b,c*d)$$?