Why is there never a proof that extending the reals to the complex numbers will not cause contradictions?

Question

The number $i$ is essentially defined for a property we would like to have - to then lead to taking square roots of negative reals, to solve any polynomial, etc. But there is never a proof this cannot give rise to contradictions, and this bothers me.

For instance, one would probably like to define division by zero, since undefined values are simply annoying. We can introduce the number "$\infty$" if we so choose, but by doing so, we can argue contradictory statements with it, such as $1=2$ and so on that you've doubtlessly seen before.

So since the definition of an extra number to have certain properties that previously did not exist may cause contradictions, why aren't we so rigourous with the definition of $i$?

edit: I claim we aren't, simply because no matter how long and hard I look, I never find anything close to what I'm looking for. Just a definition that we assume is compatible with everything we already know.

Answer 1

There are several ways to introduce the complex numbers rigorously, but simply postulating the properties of $i$ isn't one of them. (At least not unless accompanied by some general theory of when such postulations are harmless).

The most elementary way to do it is to look at the set $\mathbb R^2$ of pairs of real numbers and then study the two functions $f,g:\mathbb R^2\times \mathbb R^2\to\mathbb R^2$:

$$ f((a,b),(c,d)) = (a+c, b+d) \qquad g((a,b),(c,d))=(ac-bd,ad+bc) $$

It is then straightforward to check that

• $(\mathbb R^2,f,g)$ satisfies the axioms for a field, with $(0,0)$ being the "zero" of the field and $(1,0)$ being the "one" of the field.
• the subset of pairs with second component being $0$ is a subfield that's isomorphic to $\mathbb R$,
• the square of $(0,1)$ is $(-1,0)$, which we've just identified with the real number $-1$, so let's call $(0,1)$ $i$, and
• every element of $\mathbb R^2$ can be written as $a+bi$ with real $a$ and $b$ in exactly one way, namely $(a,b)=(a,0)+(b,0)(0,1)$.

With this construction in mind, if we ever find a contradiction involving complex number arithmetic, this contradiction can be translated to a contradiction involving plain old (pairs of) real numbers. Since we believe that the real numbers are contradiction-free, so are the complex numbers.

Answer 2

One way to define the complex numbers is to take the ring of real polynomials $\mathbb{R}[x]$, and quotient by the ideal generated by the polynomial $(x^2+1)$. This polynomial is irreducible$^{\ast}$, and hence this quotient is a field. Call this new field $\mathbb{C}$, and denote the image $\overline{x}$ of $x$ in the quotient by the symbol $i$.

While this does not shed light on the geometric/analytic properties of complex numbers, it is one construction which shows that such an object exists - which, I believe, is what you want?

$^\ast$ Edit: A polynomial is irreducible if it has no non-trivial factors in $\mathbb{R}[x]$. For a polynomial of degree 2 or 3, this is equivalent to saying that it has no roots in $\mathbb{R}$. (Thanks to Henning Makholm and Artefact for this correction)

Answer 3

In addition to what Prahlad said, I think it is important to point that we never prove the absence of contradiction in mathematics, and indeed Gödel shows that it is impossible to prove that a theory is noncontradictory within this theory.

Answer 4

There is a rigourous definition:

Define on the set ${\mathbb R}\times {\mathbb R}$ the addition $$ (a,b)+(c,d)=(a+b,c+d) $$ and the multiplication $$ (a,b)(c,d)=(ac-bd,ad+bc). $$ Then ${\mathbb R}\times {\mathbb R}$ turns into a field which is isomorphic to ${\mathbb C}$ (in it $(0,1)^2=(-1,0)$).

Answer 5

One way to show some system is consistent, is to exhibit a model for it. The common model for the complex numbers (as pairs of real numbers) does that. Strictly speaking, this is not an absolute proof of consistency, merely a relative proof. That is: given any contradiction about the complex numbers, we may translate it into a contradiction about the real numbers. After that, we would have to go to the theorem of Tarski which shows that the (first-order) theory of the real numbers is consistent.

Answer 6

As far as I remember my history of mathematics, using $i$ began indeed as a trick to solve some equations and later Complex numbers rigorously defined it by defining rules for algebric operations working on pairs of real numbers. As it is built from ground up from pre-existing concepts this should not be expected to lead to any contradiction any more than any other theorem.

I believe you are mistaken about the other part of your question. Defining infinite as an actual number (like in Cantor's Ordinal numbers) does not lead to any contradiction either. The "proofs" leading to $1 = 2$ are usually based on other errors. For instance introducing $\infty$ does not makes division inversible: $1/0 = \infty$ does not imply $\infty.0 = 1$. Other abusive uses of operators are possible with infinite like use of ellipsis $...$ instead of recurrence of continuous fractions. But please exibit any "proof" leading to $1=2$ or any other contradictory statement relying on use of $\infty$ as a number and we should be able to show you where is the actual problem.

Answer 7

Well, actually there are some proofs with contradictions. It's easy to see here where's the "pirate step" that should not happen. I dare you to look for it.

$$i = i$$ that is, $$\sqrt{-1} = \sqrt{-1}$$ $$\sqrt{\frac{-1}{1}} = \sqrt{\frac{1}{-1}}$$ $$\frac{\sqrt{-1}}{\sqrt 1} = \frac{\sqrt 1}{\sqrt{-1}}$$ $$\sqrt{-1} \sqrt{-1} = \sqrt{1}\sqrt{1}$$ $$-1 = 1$$

Of course, some step here is not actually possible to take with complex numbers. Does that mean i is not well defined? Actually, no. But there are some computations we could do that are not possible (or are ill defined) with complex numbers. They are not, of course, as fundamental and they still enable us to build very interesting theory (unlike the divide-by zero example, which is incompatible with the nice properties of rings).

See also: Bernoulli's sophism.

Answer 8

It became obvious early on that the real numbers were not sufficient to proceed with building mathematics. The equation x² + 1 = 0 tells you that right away. Okay, so we need some more numbers. We can define anything we want, but it might not be particularly useful. By defining the complex numbers in the rigorous ways described above we get an enormous amount of usefulness out of a simple definition.

It is consistent as an extension of the real numbers, because it preserves addition and multiplication of the real numbers. The study of functions of a complex variable have led to immense insights into all sorts of things; and straightened out some apparent anomalies on the real line.

Although the complex numbers are closed and complete, I could hypothesize that we may need yet another new extension of the reals. (Yes, there are already many extensions). The reason for this thought is that certain problems, which can be easily stated, are very difficult to prove. As one example, problems about the distribution of the prime numbers have proved quite intractable.

Now sometimes a hard problem is just hard, but we have a history of simplifying hard problems and making them easier by changing our point of view and/or adding more tools to work with. I keep thinking that a different set of numbers, maybe not based on the buildup from integers thru rationals thru reals thru complex, might place a different light on many of these difficult problems, and allow of straightforward solutions.

This is all theorizing, of course.

Would my imagined new number system be "consistent"? If it is to be of any use, it must be consistent in some sense.