Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The complex numbers are typically defined as the set of all ordered pairs $(x,y),$ where $x,y \in \mathbb{R},$ along with the usual operations of addition and multiplication (which I won't write out here.)

The standard argument then goes that, by defining $i = (0,1),$ the rule of multiplication on the complex numbers gives $i^2 =(0,1)(0,1) = (-1,0);$ therefore $i^2 = -1.$ But this seems to me to just be $\it wrong.$ Although there clearly exists an isomorphism that maps $i^2$ to $-1,$ it is in fact the case that $i^2$ is an ordered pair; not a real number.

Am I missing something here? Are mathematicians just being lazy/careless when saying that $i^2 = -1?$ This is my first question.

My second question has to do with the definition of $\mathbb C:$

I have always made a distinction between the $\it set$ $\mathbb R$ and the $\it field \ \mathbb R.$ I make a similar distinction with $\mathbb R^2.$ With this in mind, I am slightly confused about $\mathbb C.$ Is $\mathbb C$ only a field, and not a set? Does such a thing as "the set of all complex numbers" even exist? For the underlying set in the field $\mathbb C$ is clearly $\mathbb R^2,$ when $\mathbb C$ is defined as above.

Thank you for your responses.

$\bf Edit:$ Please read my question. Nobody who responded seems to be paying attention to anything I wrote beyond the title of my post.

share|cite|improve this question
$i$ is defined as the $\sqrt{-1}$ – user148697 May 13 '14 at 0:41
Wrong. Did you read what I wrote? Or any text on the subject? – Optional May 13 '14 at 0:53
While $(-1, 0)$ isn't literally $-1$, there is an injection that sends $\mathbb{R}$ to $\mathbb{C}$ given by $a \mapsto (a,0)$. Often in mathematics, we don't care about the representation of an object, just how it behaves. The subset of $\mathbb{C}$ with second component zero behaves exactly like the real numbers, so we say it is "isomorphic" to $\mathbb{R}$. It's a bit weird to get used to at first, but it makes sense once you deal with other groups/rings/fields. – Henry Swanson May 13 '14 at 1:03
I noted this in my post. Hence my question. – Optional May 13 '14 at 1:10
There is something fundamentally odd / strange about the question you are asking and the way you are responding to some of the answers below. What is a "real number" to you? – Braindead May 13 '14 at 4:41

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.

I feel that this excerpt from Conway's "Functions of One Complex Variable" answers your other question better than I ever could.

We will write $a$ for the complex number $(a,0)$. This mapping $a \to (a,0)$ defines a field isomorphism of $\mathbb{R}$ into $\mathbb{C}$ so we may consider $\mathbb{R}$ a subset of $\mathbb{C}$ (Page 1).

share|cite|improve this answer
This is the correct answer. Rudin, Greene and Krantz and many others do it this way as well. – Cameron Williams May 13 '14 at 1:01
Emphasis on isomorphism. The set $(a,0)$ in the complex numbers behaves exactly like the real numbers, so it is not mis-leading/lazy to write $i^2=-1$. And of course, that's how $i$ was originally defined. – NotNotLogical May 13 '14 at 1:01
Brad, perhaps I should have mentioned that Conway was one of my sources for this post. This doesn't answer my question though. I am well aware that $\mathbb C$ and $\mathbb R^2$ are different fields, as noted in my post. – Optional May 13 '14 at 1:12
@Optional I think this answers the part of your question that makes sense. If you understand this answer, then I can't understand what your remaining question is. – Alex Becker May 13 '14 at 2:03
He is saying that without the operations, there is no point talking about the set $\mathbb{C}$, because it's the same as $\mathbb{R}^2$. Take out the operations and think about what you can talk about numbers in $\mathbb{C}$, the point is that in this case you could just talk about numbers in $\mathbb{R}^2$ and would be no difference. – Integral May 13 '14 at 2:40

There is no such thing as the set of complex (or real) numbers. One can say that there exists, unique up to isomorphism, field of complex (or real numbers). These fields have many models. Besides the one you know, one can model complex numbers for instance by certain real 2-by-2 matrices. I can write more details if this is unclear.

Edit: Incidentally, there is no such thing as a complex number $i$ until you fix a model of complex numbers. What is well defined, however, is the set of roots of the polynomial $z^2+1$.

Personally, my favorite definition of C is as the algebraic closure of R. Once you have the fundamental theorem of algebra, you can then identify C with the set of pairs of real numbers (as a model).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.