What is lost when we move from reals to complex numbers?

Question

As I know when you move to "bigger" number systems (such as from complex to quaternions) you lose some properties (e.g. moving from complex to quaternions requires loss of commutativity), but does it hold when you move for example from naturals to integers or from reals to complex and what properties do you lose?

Answer 1

The most important ones as I see it:

• Naturals to integers: lose well-orderedness, gain "abelian group" (and, indeed, "ring").
• Integers to rationals: lose discreteness, gain "field".
• Rationals to reals: lose countability, gain "Cauchy-complete".
• Reals to complexes: lose a compatible total order, gain the Fundamental Theorem of Algebra.

Answer 2

The most important property you loose when moving from real to complex numbers is definitely the notion of an order, i.e. $\mathbb{R}$ is an ordered field whereas $\mathbb{C}$ is not. This follows from the following proposition (Abstract Algebra by P.A. Grillet):

A field $F$ can be ordered if and only if $-1$ is not a sum of squares of elements of $F$.

Moreover, we have to be careful in defining certain functions due to the fact that now even standard functions turn out to be multi-valued rather than single-valued. However, this is paid back immediately by the nice differentiability properties of complex differentiable functions.

Answer 3

Losing order is the most important but we also lose that if $b > 1$ then $b^z$ is injective so $\ln z$ (or $\log_b z$) is no longer a function but an equivalence class.

Answer 4

Comparing naturals to integers, there is a smallest natural (0 or 1) which often makes solving problems easier.

Comparing reals to complex, you can always compare reals but there is no complete ordering of the complex numbers.

Answer 5

The biggest thing you lose when you move from the reals to the complex numbers is the ordering. You can, of course, find some ordering on the complex numbers, but the ordering will have nothing to do with the algebraic structure.

On the real numbers, if $a < b$, and $c$ is positive, then $ac < bc$. And for any numbers at all we have $a < b$ if and only if $a + c < b + c$. You can't build an ordering on the complex numbers that obeys the same properties (You can prove this. The easiest way is to assume i is positive, and find a contradiction, then assume i is negative and find a contradiction).

Answer 6

Many people have said the crucial thing is order. I'd defend another fact we lose: that a number is self-conjugate. It's not obvious this matters, since we don't bother defining a "conjugate" of a real. But there's a reason I bring it up. The Cayley-Dickson construction can be thought of as a dimension-doubling operation on algebras with involution, where if we start with $\mathbb{R}$ we must equip to it the identity map as our "conjugation". Suppose we double from $A$ to $A^\prime$; then there's a nice equivalence between certain properties of $A$ and others of $A^\prime$. As noted in propositions 1-4 here:

• The involution on $A^\prime$ is distinct from the identity;
• Iff this is true also of $A$, $A^\prime$ doesn't commute;
• Iff that is true also of $A$, $A^\prime$ doesn't associate;
• Iff that is true also of $A$, $A^\prime$ doesn't alternate.

In other words, the fact that $\mathbb{C}$ isn't self-conjugate explains all the famous later losses of nice properties, such as $\mathbb{H}$ not commuting.

Answer 7

$\mathbb C$ is more rigid. Holomorphic functions are analytics. This means complex manifolds are polynomial-like and more akin to algebraic varieties than to real manifolds. By contrast $\mathbb R$ manifolds can be glued together using functions like $\exp(-1/x^2)$ that can $C^\infty$-smoothly transition from one function to another.