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I know that, as sets, $\mathbb C$ and $\mathbb R^2$ are exactly the same set, and that the difference is about the structure: $\mathbb C$ is $\mathbb R^2$ with a field structure.

Does this mean that, for example, $i=(0,1)$, or that $3+2i=(3,2)$? Like, are complex numbers exactly the ordered pairs? Or are they different objects, meaning that belonging to an algebraic structure makes them somewhat different?

I can ask this question in a more general formulation: if I have an algebraic structure, are its elements exactly equal to the elements of the underlying set, or does the structure "add something" to them?

Also, is it not possible to define an operation over a set without creating an algebraic structure? If it is possible, then what is the point of algebraic structures? Why put a set and an operation into an ordered tuple (and I have never really understood the formal reasons behind this), instead of just saying "this operation is defined on this set"?

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  • $\begingroup$ The ordered tuple thing is purely linguistic. It is easier to refer to "this structure" than "this set and this and this and this operations". It is always possible to translate between the two without affecting the mathematical content at all. $\endgroup$ Commented Sep 15, 2024 at 15:05
  • $\begingroup$ @Trebor oh, thanks, this answers a question I've had for quite a long time. So, the answer is that the elements of $\mathbb C$ and $\mathbb R^2$ are exactly equal? $\endgroup$ Commented Sep 15, 2024 at 15:07
  • $\begingroup$ No, $\mathbb{C}$ and $\mathbb{R}^2$ are obviously not the same set: for example, the former contains $i$, while the latter does not. But, the natural correspondence $a + b i \mapsto (a, b)$ is a group homomorphism, so in some sense "addition" behaves similarly (independent of the set). $\endgroup$ Commented Sep 15, 2024 at 15:10
  • $\begingroup$ @K.Jiang but, in axiomatic material set theory, $i$ is not an atomic symbol fallen from the sky; it denotes a certain set. You have to explain how $\mathbb{C}$ and $i$ are encoded as sets before you can answer the question, as noted in the answers. $\endgroup$ Commented Sep 15, 2024 at 15:16
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    $\begingroup$ @K.Jiang: There are several ways to construct the complex field. Under one very common construction, $\mathbb C$ and $\mathbb R^2$ are indeed literally equal as sets, and $i=(0,1)$. For an example of how this construction is carried out, see for instance Spivak's Calculus. $\endgroup$ Commented Sep 15, 2024 at 15:21

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I know that, as sets, $\mathbb{C}$ and $\mathbb{R}^2$ are exactly the same set

This isn't necessarily true. There are multiple ways to encode the real numbers, and multiple ways to encode the complex numbers even if you fix an encoding of the reals. While the most common choice is to take the underlying set of $\mathbb{C}$ to be $\mathbb{R}^2$, this choice is not forced and tells you nothing about the structure of $\mathbb{R}^2$ or $\mathbb{C}$. The most you can say is that $\mathbb{C}$ and $\mathbb{R}^2$ are isomorphic as sets.

Does this mean that, for example, $i=(0,1)$, or that $3+2i=(3,2)$?

That is one possible choice of encoding. Another choice would be to swap the components: $i = (1, 0)$ and $3 + 2i = (2, 3)$. Again, this choice is completely immaterial.

These questions are also quite sensitive to the foundations you're working in: one could argue that a decent foundation of mathematics would not let you ask whether an element of $\mathbb{C}$ is an element of $\mathbb{R}^2$. See structural set theory or type theory.

I can ask this question in a more general formulation: if I have an algebraic structure, are its elements exactly equal to the elements of the underlying set, or does the structure "add something" to them?

Usually, one defines an element of an algebraic structure to be an element of the underlying set.

Also, isn't it possible to define an operation over a set without creating an algebraic structure? If it's possible, then what's the point of algebraic structures? Why put a set and an operation into an ordered tuple (and I've never really understood the formal reasons behind this), instead of just saying "this operation is defined on this set"?

That depends on what you mean by "creating an algebraic structure". You can certainly talk about the addition operation on $\mathbb{Z}$, and you can also talk about the additive group of integers, which consists of the set $\mathbb{Z}$ together with the addition operation (and the proofs of the group axioms for this operation).

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  • $\begingroup$ Bit of a nit but re: "the most you can say is that $\mathbb C$ and $\mathbb R^2$ are isomorphic as sets" -- you can also say they are isomorphic as groups, or even vector spaces over $\mathbb R$. The connection is deeper than "they have the same cardinality". $\endgroup$ Commented Sep 15, 2024 at 19:04
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Like, are complex numbers exactly ordered pairs? Or are they different objects, meaning that belonging to an algebraic structure makes them somewhat different?

This is a matter of definition and convention. It certainly "works", and is useful in many context. However, there are other equivalent definitions: you could use polar coordinates with $\mathbb C = \mathbb R \times S_1$, or an algebraist would define $\mathbb C = \mathbb R[x] / (x^2 + 1)$ (see here if unfamiliar).

Generally, mathematicians are willing to switch back and forth between isomorphic representations of an object, sometimes without even mentioning it. In most context, you'll use the representation of $\mathbb C$ that is most convenient in the moment.

I can ask this question in a more general formulation: if I have an algebraic structure, are its elements exactly equal to the elements of the underlying set, or does the strucure "add something" to them?

Formally an algebraic structure is a tuple: in the case of a field, it's $(X, +, \times)$ where $X$ is a set, and $+$ and $\times$ are functions $X \times X \to X$. The set $X$ does not "know" anything about the functions.

However, in commonly used structures like $\mathbb C$ or $\mathbb Q$, we abuse notation and write $X$ to refer to the whole structure. In practice this is usually unambiguous given the wider context.

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We can make the Natural Numbers $\Bbb N$ using Peano Axioms or using Sets.
We can then make the Integers $\Bbb Z$ using those Natural Numbers. Of course , Rational Numbers $\Bbb Q$ can involve 2 Integers. That Set $\Bbb Q$ is not closed when we take limits , hence we introduce Real Numbers $\Bbb R$ which is closed even with limits.

That is one "Potential Way" to come up with $\Bbb R$ , though there are many more ways.

In all those ways , what we mean by $\Bbb R$ is what entities full-fill the Axioms.
In other words : What are Real Numbers ? We do not know , we do not care , we just use those Axioms to talk about that. It is just a "representation" of some abstract entities.

Now , we can make more things with that Eg $(x,y)$ is a tuple where we use a pair of real numbers , we can call it $\Bbb R^2$.

At the same time , we can see that $\Bbb R$ is not closed when we involve roots $\sqrt{x}$ , $\sqrt[3]{x}$ : Hence we can come up with $\Bbb C$ with that new closure , where we still have idea what / where / how those numbers exists.
These new numbers are useful , hence we make various representations : $x+iy$ , ordered tuple $(x,y)$ , Points on Plane , Polar $(r,\theta)$ , Matrix with two rows $(x,y)$ $(-y,x)$ , each suitable in some scenario.

All those are Complex Numbers $\Bbb C$ , though not all representations are $\Bbb R^2$

Hence it is not true that $\Bbb C$ and $\Bbb R^2$ are exactly the same set.
It is true that $\Bbb C$ can be represented by $\Bbb R^2$.
It is also true that $\Bbb C$ can be represented without using $\Bbb R^2$.

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