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I need to find a few examples about the differences between real numbers and complex numbers like:

1) if $x \in \mathbb R $ then $x^2 \geq0$ is true

if $z \in \mathbb C $ then $z^2 \geq0$ is false

2) let $a \in \mathbb R/\{0, 1\} $ if $a^x =a^y$ then $x=y$ is true

let $a\in \mathbb z/\{0, 1\} \in \mathbb C $ if $a^x =a^y$ then $x=y$ is false

But these examples are not cool enough and feel very trivial. Can you suggest some other properties like these?

Thanks.

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    $\begingroup$ If $z \in \mathbb C $ then $z^2 \geq0$ isn't just false. It makes no sense to ask about. $\endgroup$
    – Arthur
    Sep 2, 2019 at 9:05
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    $\begingroup$ true, but we can say that the statement is false right $\endgroup$ Sep 2, 2019 at 9:08
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    $\begingroup$ No. It is nonsensical. It's neither true nor false. $\endgroup$
    – Arthur
    Sep 2, 2019 at 9:30
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    $\begingroup$ This does get to a valid answer on the question though: the reals form an ordered field, the complex numbers do not. $\endgroup$ Sep 2, 2019 at 9:52
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    $\begingroup$ Agreeing with @Arthur, one might also say the statement is "undefined" (the relation not being defined for complex values). $\endgroup$ Sep 3, 2019 at 0:59

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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.

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    $\begingroup$ It is worth noting that the lexicographical order on $\mathbb{R}^2$ is compatible with its structure as a real vector space; it just isn't compatible with the multiplicative structure of $\mathbb{C}$. See Ordered vector space - Wikipedia. $\endgroup$ Sep 2, 2019 at 16:13
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    $\begingroup$ @Calum Thank you. I edited my answer to clarify the lack of a "well-behaved" total order on $\mathbb C$. You (or anyone else) should feel free to edit it further if you find it still too casual. $\endgroup$
    – user694818
    Sep 2, 2019 at 17:34
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    $\begingroup$ This makes me wonder why we do call complex numbers „numbers“ at all, if we can‘t order them by size. $\endgroup$
    – asmaier
    Sep 3, 2019 at 7:19
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    $\begingroup$ That's a great thing to wonder about, @asmaier. They sure seem numeric when you're finding the roots of quadratic equations. But they get so pathological the further you get into them that I'd prefer if we switched to calling it the complex field like the astronomers kicking Pluto out of the club of planets. $\endgroup$
    – user694818
    Sep 3, 2019 at 10:43
  • $\begingroup$ "Pathological"?! The complex numbers are as pathological as calling them pathological... Not only are they algebraically closed, any differentiable function on an open subset $D$ of the complex plane is not just infinitely differentiable but even has a power series around every point in $D$. In contrast, differentiable real functions are the ones that really can be pathological; for instance they may have zero radius of convergence, or may not have an integrable derivative. $\endgroup$
    – user21820
    Sep 11, 2019 at 8:55
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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 $\mathbb{R}$
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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).

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  • $\begingroup$ Would the same proof show that $\bar{\mathbb{Q}} \cap \mathbb{R}$ has only the trivial automorphism? $\endgroup$
    – Adi Ostrov
    Sep 4, 2019 at 6:05
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The field of complex numbers is algebraically closed, but the field of real numbers is not.

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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).

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Rolle's Theorem does not hold for complexe valued functions.

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Another point is that the complex numbers form a two-dimensional vector space over the field of real numbers, $${\Bbb C}=\{a+ib\mid a,b\in{\Bbb R}\}$$ with basis $\{1,i\}$.

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$\mathbb{C} - \{0\}$ is connected and has a non-trivial fundamental group, while $\mathbb{R} - \{0\}$ is not connected and its components are contractible.

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Polynomials with real coefficients have a property that is not shared by complex polynomials. Here we deal with several variable polynomials.

Let $f_1,f_2,\ldots, f_m$ be $m (>1)$ polynomials in $n(>1)$ complex variables, and so we can talk of the set of all common zero locus of $f_i$'s: that is the collection of points in $n$-dimensional complex space where all the $f_i$'s vanish.

If we consider similar locus for real polynomials in $\mathbf{R^n}$, this can always be realized ALWAYS as the zero locus of a SINGLE polynomial, namely $\sum_if_i^2$.

In the complex case there are complex loci which can't be reduced to single polynomial equation.

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The real numbers satisfy the Archimedian property, complex numbers don't.

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well another difference is that complex numbers are actually on a specific quadrant in the complex plane meanwhile real numbers are just chillin on the real axis

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  • $\begingroup$ everyone is posting here to upvote farm :( $\endgroup$
    – asdf334
    Sep 11, 2019 at 1:43

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