properties that real numbers hold but complex numbers does not

Solution 1:

The relation $<$ on the real numbers is a total order that preserves order under addition and multiplication in the way we're used to, but there is no such order-preserving total order on the complex numbers.

Solution 2:

Good question!

Firstly, I'd like to partially disagree with one of the points made in the comments. Arthur writes:

If $z \in \mathbb{C}$ then $z^2 \geq 0$ isn't just false. It makes no sense to ask about.

Now, I agree with the broader point that "undefined" is different to "false". However, the statement is based on an assumption that the writer of the hypothetical article under question hasn't defined a binary relation $\geq$ on the complex plane. This assumption isn't necessarily justified. Indeed, the writer could have defined $\geq$ on $\mathbb{C}$ in any old weird way, and that would be a valid definition. Furthermore, there's actually a reasonable notion of order for the complex plane, though it seems not to be well-known.

Moving on, what you've got to understand about $\mathbb{C}$ is that algebraically, it's just better than $\mathbb{R}$. There's essentially no reason to use $\mathbb{R}$ instead of $\mathbb{C}$, if all you care about is addition, multiplication, and solving polynomial equations, except perhaps for the added challenge of weird things happening due to a failure of algebraic closedness. Thus, I agree with Matthew Daly and Mark Kamsma point.

The reals form an ordered field, the complex numbers do not.

That is, the real numbers are totally-ordered by a relation that plays well with addition and multiplication. This, together with the completeness of the real line, is key to understanding what $\mathbb{R}$ is all about.

Indeed, using that $\mathbb{R}$ is a complete ordered field, we can prove the following important fact:

Characterization of connected subsets of the real Line. For all non-empty $X \subseteq \mathbb{R}$, the following are equivalent:

1. $X$ is topologically connected
2. For all $a,b \in X$, we have $[a,b] \subseteq X$.

This is untrue for $\mathbb{C}$ with the aforementioned order, and also untrue for $\mathbb{Q}$ with the standard order (because $2$ does not imply $1$ in that case). This, in turn, allows us to prove the all-important intermediate value theorem using the fact that the image of a connected set under a continuous function is connected. The rest of real-analysis largely hinges on this observation. For example, using the least upper bound property, we can prove the existence of the Weierstrass function $f$. And then, using IVT, we can prove e.g. the existence of an $x \in \mathbb{R}$ satisfying $xf(x) = 398173749$. Try doing that using only complex-analytic techniques!

And so, your list of things that are special about $\mathbb{R}$ should include the following:

1. It's a totally ordered field (unlike $\mathbb{C}$)
2. It satisfies the least-upper bound property (unlike $\mathbb{Q})$
3. Connected subsets of $\mathbb{R}$ can be characterized as described above (unlike both $\mathbb{C}$ and $\mathbb{Q}$)
4. The intermediate value theorem holds for $\mathbb{R}$

Solution 3:

Here's an interesting "global" difference, which is unfortunately a bit abstract but hopefully still interesting:

Let's look at just the additive and multiplicative behavior - that is, we're considering $\mathbb{R}$ and $\mathbb{C}$ as fields. An automorphism of a field is a bijection from the field to itself which preserves the structure - e.g. $\alpha(x+y)=\alpha(x)+\alpha(y)$, and so forth.

Every field has at least one automorphism, namely the identity (the trivial automorphism). $\mathbb{C}$ has one obvious nontrivial automorphism, namely conjugation $$a+bi\mapsto a-bi\quad (a,b\in\mathbb{R}),$$ and assuming the axiom of choice it has a lot more (although they're quite wild). By contrast, $\mathbb{R}$ has no nontrivial automorphisms whatsoever! The key observation is that the ordering on $\mathbb{R}$ is definable (which has a precise general meaning, incidentally) just from the field structure: $x\ge y$ iff there is some $z$ such that $y+z^2=x$. From this, together with the fact that each rational must be fixed by each field automorphism (a good exercise) and the density of $\mathbb{Q}$ in $\mathbb{R}$, we rule out any nontrivial automorphisms.

The two points where this breaks down for $\mathbb{C}$ are:

• The relation $\exists z(y+z^2=x)$ does not define an ordering on $\mathbb{C}$ (indeed, $\mathbb{C}$ as a field cannot be definably ordered at all).

• $\mathbb{Q}$ is not in fact dense in $\mathbb{C}$. Even without a definable ordering, if $\mathbb{Q}$ were dense in $\mathbb{C}$ we could at least conclude that there were no continuous automorphisms, but this prevents us from even saying that much (and indeed conjugation is continuous).

Solution 4:

The field of complex numbers is algebraically closed, but the field of real numbers is not.

Solution 5:

Considering the integers (which are included in the reals and complex) an interesting fact is that $5, 13, \cdots$ are prime integers in the real field but they are not prime in the complex field (Gaussian Integers).