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why imaginary numbers have a a different meaning than $R^2$ if they are semantically equivalent? regardless of the historical perspective, we know that ther is no semantic difference between them

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  • $\begingroup$ What do you mean by "semantically equivalent"? Are you familiar with university algebra, e.g. rings and fields? $\endgroup$ – Math Gems Apr 11 '13 at 4:56
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    $\begingroup$ Yes, I think you are wrong when you use "semantically" this way. In $\mathbb R^2$, which one is $1$, which one is $i$? There is a lot of freedom to choose elements of $\mathbb R^2$ to do that. You might think that the standard $(1,0)=1$ and $(0,1)=i$, but that's just a choice, and an arbitrary one. In particular, $\mathbb R^2$ is missing structure that $\mathbb C$ has - namely, the multiplicative nature. $\endgroup$ – Thomas Andrews Apr 11 '13 at 5:01
  • $\begingroup$ I'm not sure how you can do that unless you give $\mathbb R^2$ a multiplicative structure. Once you do, if you pick exactly the right multiplicative structure, you can do that. But you can do that with the subspace of $\mathbb R^3$ generated by $(1,2,3)$ and $(\pi,e,7)$. $\endgroup$ – Thomas Andrews Apr 11 '13 at 5:06
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    $\begingroup$ I mean that a theorem can be proven in one system if and only if it can be proven on the other. And, Thomas, you are right, but I meant adding to R^2 all the definitions and properties that the complex numbers satisfy $\endgroup$ – noelia Apr 11 '13 at 5:09
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    $\begingroup$ Well, once to add stuff to $\mathbb R^2$, it isn't "just" $\mathbb R^2$ anymore, but something with "more structure." In which case, it is essentially $\mathbb C$. But that wasn't the question you asked. There is other structure we can add to $\mathbb R^2$, or we can send different elements of $\mathbb R^2$ to $1,i$ and get a "different" relationship. Adding the structure implicitly changes the beast you are talking about. It's still $\mathbb R^2$, but it is $\mathbb R^2$ with lots of other features. $\endgroup$ – Thomas Andrews Apr 11 '13 at 5:12
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They are the same as sets and they are isomorphic as (real) vector spaces, however $\mathbb{C}$ is also endowed with a multiplicative structure with respect to which it is a field. Of course we could make $\mathbb{R}^2$ a field as well by definining an appropriate binary multiplication operator but there are many ways to do this.

This lack of multiplicative structure on $\mathbb{R}^2$ by default is what makes holomorphy (differentiability as a map from $\mathbb{C}$ to $\mathbb{C}$) a stronger property than being a differentiable map from $\mathbb{R}^2$ to $\mathbb{C}$ for example.

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