Is $\mathbb{C}$ equal to $\mathbb{R}^2$? Complex numbers are usually formally defined as pairs of real numbers. Although there are operations on $\mathbb{C}$, such as complex multiplication, which are not found in operations usually applied to $\mathbb{R}^2$, the sets themselves seem to be the same. Each consists of pairs of real numbers.
So is it okay to say that $\mathbb{C} = \mathbb{R}^2$? It seems formally correct, but something doesn't feel quite right about it.
 A: You can define the set of complex numbers in different ways. One of those ways defined $\mathbb C$ to be $\mathbb R^2$ and then goes on to define the algebraic structure of the complex numbers. If that is the way you define the complex numbers, then it is certainly correct to write $\mathbb C = \mathbb R^2$ as sets.
A: It depends on what you mean by "equal". 


*

*They are "equal" as sets, in the sense that both can be seen as a Cartesian product of $\mathbb R$ with itself.

*They are equal as vector spaces, where equality is interpreted as a linear isomorphism (both are real vector spaces of dimension 2)

*They are not equal when you consider $\mathbb C$ as a field, because when we think of $\mathbb R^2$ we don't assign it a ring structure. 

A: The answer is both yes, and no.
$\Bbb C$ and $\Bbb R^2$ are both sets with the same cardinality, and they have a very natural bijection between them which preserves a lot of nice properties. So much that we can almost say that these two sets are the same for a lot of purposes.
But these two carry very different structure as a natural structure. $\Bbb C$ is a field and $\Bbb R^2$ is not (because pointwise multiplication does not form a field).  One can even argue that formally $\Bbb C$ is in fact $\Bbb R[x]/(x^2+1)$, and not $\Bbb R^2$, and one would be at least partially correct.
Personally, I'd support the "no" answer more than the "yes" answer. And here's why. We often like to think about $\Bbb R$ as a subset of $\Bbb C$. Namely $x\in\Bbb C$ is a real number if and only if $\overline x=x$. But $\Bbb R$ is not a subset of $\Bbb R^2$, instead there is an obvious embedding $x\mapsto(x,0)$, but still real numbers are generally not ordered pairs of real numbers (you can even notice that this approach takes $\Bbb C$ as sort of a primitive notion, and not quite as $\Bbb R[x]/(x^2+1)$ as others might see it).
Although, as I said, it depends on how you define things, because "formally" things can be done in plenty of different ways.
A: No, it's not, not unless you play some semantics games to make it so. A complex number $c$ can be defined as $c = a + bi$, and indeed $\{a, b\} \in \mathbb{R}$. But what is $i$? That's the imaginary unit, $i = \sqrt{-1}$. And $i \not\in \mathbb{R}$. That's kinda the point of calling it "imaginary."
So you do have a pair of real numbers, but one of those numbers is multiplied by the imaginary unit. For example, $-\frac{1}{2} + \frac{\sqrt{3}}{2}i$ (I would much prefer to write $\frac{\sqrt{-3}}{2}$ but that doesn't quite suit my purpose here). Both $-\frac{1}{2}$ and $\frac{\sqrt{3}}{2}$ are real numbers. But $\frac{\sqrt{3}}{2}i$ is an imaginary number. Now, you could do $\left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right)i$ to obtain $-\frac{\sqrt{3}}{2} -\frac{i}{2}$. What is the number in $\mathbb{R}^2$ that would allow us to perform a similar switcheroo? And how would we be able to consider that equivalent to $i$?
But I misunderstand subtleties as often as I point them out, so if anything's even the slightest bit wrong with what I've said here, we'll hear about it in spades.
A: It really depends on how you define $\mathbb{R}^2$, though you'll probably also have to exercise great care in how you define $\mathbb{C}$ as well. It also depends on what you mean by "okay." If by "okay" you mean "it fits well with widely accepted definitions of $\mathbb{R}$ and $\mathbb{C}$," I'd have to say that no, it's not okay.
Have you heard the one about the magician and the mathematician who were locked in a cage together? As soon as the door closed, the magician started tapping on the wall and listening to the sounds that made. "You're just going to sit there while I do all the work of getting us out?" the magician asked. "I already got us out," the mathematician replied, "but you were too busy knocking on the wall to notice." The mathematician had redefined "outside" to mean inside the cage.
The kernel of truth to that joke is that you can redefine anything.
But for a redefinition to have value, it has to pass two tests. First, it has to be consistent with all the other definitions you provide, and in this case, it looks like you have to provide a definition for $\mathbb{C}$ even though most of the time there would be no need to. Second, the redefinition must accomplish something, even if it's something small.
Every few months, someone decides that there is a fundamental problem with the Fibonacci sequence and that it is up to them to fix it. Sometimes their redefinition fixes the problem without seriously affecting the various well-known identities for Fibonacci numbers. But the problem their redefinition fixes is so small and inconsequential that it does not really justify any efforts to disseminate the new definition.
So the question you have to ask yourself is this: Does definiting $\mathbb{R}^2$ and $\mathbb{C}$ so that $\mathbb{C} =\mathbb{R}^2$ accomplish anything you consider worthwhile?
A: This is a comment (since I cannot write comments yet): 
If $$\Bbb C=\Bbb R^2,$$ how can one write, for example, $i$? 
If you say $$i=(0, 1),$$
then how $$(0, 1)^2=-1.$$
A: If you want to be super formal, you'd say that there exists an equivalence relation between $C$ and $R^2$ defined as $x+yi \sim (x, y)$. It's a very useful and intuitive relationship (in fact one of the reasons it's useful is because it's intuitive), but as you note there are operations that usually only make sense in one or other of the sets, and you'd have to explicitly define that operation in the other set and then show that it's preserved across the relation.
