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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)

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2 Answers 2

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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: $\pi(x) = x^3$.

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  • $\begingroup$ 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. $\endgroup$
    – J.D
    Feb 12 at 21:17
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What about if $(a,c)°(b,d)= (a+b,c*d)$?

Then, certainly , it satisfies the fact that ° is commutative(easy to see) and associative (addition and multiplication are associative), and there is an identity element, namely (0,1). Therefore, ° and + are not necessarily equal.

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    $\begingroup$ Yeah, the this counterexample works fine. I actually forgot to specify that the operation needed to have inverses. $\endgroup$ Feb 12 at 21:03

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