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

I have been studying complex numbers. But I don't understand why we are following almost all the rules of real number system, since this is a new system. Why so?

share|cite|improve this question
See, not all rules are same $\sqrt{ab} \neq \sqrt{a} \times \sqrt{b}$ in complex numbers. – Nick Mar 7 '14 at 14:12
@Nick Yes, that is why I say almost. – Theluqfzhluswcpzflabucheecatne Mar 7 '14 at 17:42
up vote 8 down vote accepted

Basically it's by design.

The complex numbers are deliberately constructed such that all of the rules for real numbers -- except ones that pertain to ordering -- will hold for them.

These rules are, among other things, what allows us to reason from "hey, let's assume there's some number whose square is $-1$" to the particular rules for adding and multiplying complex numbers, as well as the fact that a complex number has exactly two components and so forth.

If we want to (and in some situations we do!) we could well look for extensions of the real numbers that respect fewer of the usual rules than the complex numbers do. Depending on what we need and what we're willing to let go of, we could then end up with matrices, or polynomial rings, or fraction fields, or vectors, or whatever.

So "complex numbers" satisfy the field properties simply because "complex numbers" means the thing we get when we decide we want the field properties to hold, together with having a square root of $-1$.

Addendum: Actually the only thing we need to decide is to keep those rules that speak about addition, subtraction and multiplication. It's an interesting fact when we want to extend the reals with a $\sqrt{-1}$ while also preserving the rules for addition, subtraction and multiplication we get a basically unique solution that just happens to allow a division operation with the same properties as the division of real numbers. I have no intuitive explanation a priori for why this would necessarily happen. But it does.

share|cite|improve this answer
It doesn't necessarily happen because not all complex rings of integers have a division algorithm. I prefer to think of it as a geometric thing - ideals in $\mathbb{Z}[i]$ are related to square lattices, and because the lattices are square no interior point is too far away from the corners. – Stella Biderman Mar 6 '14 at 20:12
@Stella: Oops, I meant to assume we want all of the reals too. Edited. (For futher nitpicking, I'm deliberately not going to care about the possibility of also adding spurious transcendentals at this level of sophistication). – Henning Makholm Mar 6 '14 at 20:14

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.