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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.

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    $\begingroup$ The sets are the same, the addition is the same, but the multiplications work differently. So it depends on whether you see them as sets, groups or rings. $\endgroup$ – Arthur Jul 7 '14 at 3:28
  • $\begingroup$ Actually, the convention is that $\mathbb{C}=(\mathbb{R}^2,+,.)$. That is whenever we use $\mathbb{C}$, merely we do not consider only the set $\mathbb{R}^2$ but together with algebraic operations. $\endgroup$ – Chellapillai Jul 7 '14 at 3:30
  • $\begingroup$ Just as when one writes $\mathbb{R}$ one often means the set $\mathbb{R}$ along with the usual operations, when one writes $\mathbb{C}$ one usually means the set $\mathbb{R}^2$ along with the usual (complex) operations. (There are other ways of defining the complex numbers, of course). So, depending on your degree of informality (or sloppiness), the answer is yes or no :-). Being completely formal is notationally cumbersome (if not impossible). $\endgroup$ – copper.hat Jul 7 '14 at 3:57
  • $\begingroup$ possible duplicate of True or False: $i^2 = -1$, $\mathbb{C} = \mathbb{R}^2$ $\endgroup$ – Moishe Kohan Jul 7 '14 at 4:03
  • $\begingroup$ @studiosus There's some interesting stuff there, but I think the top answer is not quite right. They say "The idea that $\mathbb{C}=\mathbb{R}^2$ is false. In $\mathbb{R}^2$ there is no meaning to $(x_1,x_2)\cdot(y_1,y_2)$. It is true that $\mathbb{C}$ satisfies all of the axioms for a field (associative, commutative, etc.) so it is a field." But this glosses over the fact that we can talk about $\mathbb{C}$ the set or $\mathbb{C}$ the field or $\mathbb{C}$ the group, etc. We can define such a multiplication on $\mathbb{R}^2$ if we desire to. $\endgroup$ – AmadeusDrZaius Jul 7 '14 at 4:08
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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.

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  • $\begingroup$ That's silly. Everyone knows that ${\bf C}=[0,2\pi)\times (0,\infty)\cup \{0\}$. ;-) $\endgroup$ – tomasz Jul 7 '14 at 5:09
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    $\begingroup$ @tomasz: Nonsense. $\Bbb C$ is the unique algebraically closed field of characteristics $0$ which has the cardinality of $\mathcal P(\Bbb N)$. ;-) $\endgroup$ – Asaf Karagila♦ Jul 7 '14 at 5:30
  • $\begingroup$ @AsafKaragila: Isn't the algebraic closure of $\mathbb{Q}_p$ also of cardinality $2^{\aleph_0}$? $\endgroup$ – Najib Idrissi Jul 7 '14 at 14:20
  • $\begingroup$ @Najib: It is. And also the algebraic closure of the metric completion of the algebraic closure of $\Bbb Q_p$. Why? $\endgroup$ – Asaf Karagila♦ Jul 7 '14 at 14:21
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    $\begingroup$ @Najib: As long as it restarts working, it's fine. There are people out there whose brains stopped momentarily but never started working again. $\endgroup$ – Asaf Karagila♦ Jul 7 '14 at 14:32
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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.

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  • $\begingroup$ "We often like to think about $\mathbb{R}$ as a subset of $\mathbb{C}$." For this to be formally true, I guess one should think of $\mathbb{R}$ as the set of all complex numbers with no imaginary part, i.e. $\mathbb{R} = \{(x, y) \in \mathbb{C} \mid y = 0 \}$. But then this is circular, since the ordered pairs are pairs of real numbers. So are you saying instead that we take $i$'s existence as an assumption and define complex numbers in terms of this, or simply that we should "think" of real numbers as a subset, for illustrative purposes? $\endgroup$ – AmadeusDrZaius Jul 7 '14 at 3:36
  • $\begingroup$ No, it's not circular. It depends on what is the primitive notion that you begin with. How do you get $\Bbb R$ anyway? Some people like $\Bbb N$ as a primitive notion, others prefer $\Bbb C$. Also, note the circularity only exists if you begin by defining $\Bbb C$ as ordered pairs of real numbers. $\endgroup$ – Asaf Karagila♦ Jul 7 '14 at 3:38
  • $\begingroup$ If we weren't to define $\mathbb{C}$ as ordered pairs of real numbers, would we just have to accept $i$ as a primitive object? I have trouble seeing how we would fit $i$ into the framework of ZFC. Integers are usually defined using ordered pairs... as are rational numbers. Obviously real numbers have many constructions, Dedekind sets being one that uses only sets and the ordering of rational numbers. What sort of construction could one use for $i$? $\endgroup$ – AmadeusDrZaius Jul 7 '14 at 3:43
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    $\begingroup$ When you define everything from bottom up in $\sf ZFC$ You have that the intersection of any two sets in $\Bbb{N,Z,Q,R,C}$ is empty (not to mention great dependence on the way you choose to define each). When you talk to people about math, we often think about $\Bbb Q$ as a subset of $\Bbb R$, don't we? How come? Well, we define $\Bbb R$ and then we "switch" the copy of $\Bbb Q$ with the canonical embedding of $\Bbb Q$. Similarly, you reach all the way to $\Bbb C$ somehow and then you redefine $\Bbb R$ in a particular fashion as a subset of $\Bbb C$. $\endgroup$ – Asaf Karagila♦ Jul 7 '14 at 3:48
  • $\begingroup$ I see what you're saying now. It's not circular - it's that there are two different sets that we call $\mathbb{R}$ depending on the context and what we want to accomplish. There's the $\mathbb{R}$ that the complex numbers are formally constructed from, and then there's the $\mathbb{R}$ which is a subset of complex numbers. $\endgroup$ – AmadeusDrZaius Jul 7 '14 at 3:50
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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.
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  • $\begingroup$ Good answer; thank you for making the distinction between the different versions of $\mathbb{C}$. $\endgroup$ – AmadeusDrZaius Jul 8 '14 at 11:39
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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.

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    $\begingroup$ As I was discussing with Asaf, defining $i = \sqrt{-1}$ makes intuitive sense, but formally such a definition is incomplete, since it's equivalent to saying $i$ is the only member of the singleton set $\{x \in X \mid x^2 = -1 \}$ ... and we need some set $X$ to use the axiom of separation like that. We could let $X = \mathbb{C}$, but then we're defining $\mathbb{C}$ in terms of itself. Reading your comment about misunderstanding subtleties, I feel bad for being so pedantic about this, but it is a very pedantic question. $\endgroup$ – AmadeusDrZaius Jul 7 '14 at 3:54
  • $\begingroup$ @AmadeusDrZaius Don't feel bad, you were much nicer about it than most. And to pile on the pedantry, would you say $\Im(c) \in \mathbb{R}$ is true or false? $\endgroup$ – Robert Soupe Jul 7 '14 at 4:04
  • $\begingroup$ Is that a capital fraktur J? I don't actually know what that means. EDIT: Is ℑ(c) equivalent to Im(c), the imaginary part of a complex number? If so, I would say that yes, that's in $\mathbb{R}$. EDIT #2: Well, that is, in the $\mathbb{R}$ from which $\mathbb{C}$ is built. But not the other version of $\mathbb{R}$: $\{(x, y) \in \mathbb{C} \mid y = 0\}$. $\endgroup$ – AmadeusDrZaius Jul 7 '14 at 4:11
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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?

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    $\begingroup$ I enjoyed the anecdote about the magician and the mathematician. It's an important one. But I wouldn't say it's a redefinition. Defining $\mathbb{C} = \mathbb{R}^2$ (set-wise) seems to be the most traditional definition since the era of mathematical rigor began. But this does not consider $\mathbb{C}$ as a field, where it is not "equal" to $\mathbb{R}^2$. The question of whether they are equal in the sense that I originally asked is not so relevant, since they are equal or not equal depending on the perspective we wish to take. $\endgroup$ – AmadeusDrZaius Jul 8 '14 at 11:42
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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 $$(0, 1)^2=-1.$$

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    $\begingroup$ Exactly; he'd also have to define all the operators. And then what's the point of defining something that behaves like something we already have but is called something different? $\endgroup$ – Robert Soupe Jul 8 '14 at 0:22
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    $\begingroup$ @RobertSoupe The point is being able to define everything in terms of things we already have. Multiplication is not difficult to define and this is how it is done in most advanced textbooks on the subject. For two complex numbers $z_1 = (x_1, y_1)$ and $z_2 = (x_2, y_2)$, their complex product is $z_1z_2 = (x_1x_2 - y_1y_2, x_1y_2 + x_2y_1)$. Then the proof that $(0, 1)^2 = -1$ follows easily: $(0, 1)(0, 1) = (0*0 - 1*1, 0*1 - 1*0) = (-1, 0) = -1$. The fact that this definition exists and works so well is much more beautiful to me than asserting $i$'s existence and properties as assumed facts. $\endgroup$ – AmadeusDrZaius Jul 8 '14 at 11:36
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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.

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    $\begingroup$ You're confusing equivalence relations with bijections. $\endgroup$ – tomasz Jul 7 '14 at 5:13
  • $\begingroup$ You know what, I probably am. $\endgroup$ – ConMan Jul 7 '14 at 5:28

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