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.

  • $\begingroup$ try it other way around. specialising complex numbers to realsand not generalising reals to complex. $\endgroup$ – Arjang Aug 10 '13 at 13:05
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    $\begingroup$ :-) You have asked some really nice questions here. I am sure someone qualified will give an awesome answer soon enough. Regardless, you should take this chance and read about the formal definition of complex numbers and surreal numbers. Have fun! :-) $\endgroup$ – TenaliRaman Aug 10 '13 at 13:17
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    $\begingroup$ Take a look at "Do complex numbers really exist?" and especially this answer which gives some historical context. When the idea of formal expressions of the form $a+ib$ was first introduced, it was regarded with much suspicion. Suspicion that really took the firm foundations offered by set theory to go away. $\endgroup$ – kahen Aug 14 '13 at 14:45

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.

  • $\begingroup$ While the above is rigorous enough, it is not 'intuitive' or 'straightforward' (so to speak), especially if one looks at the function $g((a,b),(c,d))$. Any reasoning as to why the function$g$ is the one given above would be helpful, because the nonspecialist will always ask- why this function and not something else. I know that the original question is only concerned with the 'non-contradiction' thing. In addition, I think some reasoning would be $nice$, that's all. $\endgroup$ – nb1 Nov 10 '14 at 12:15
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    $\begingroup$ @NikhilBellarykar: It is pretty straightforward if you have already played around with complex numbers in the form $a+bi$ without having a rigorous background for doing so, just under the assumption $i^2=-1$. Then the intention is for the pair $(a,b)$ to represent $a+bi$, and the multiplication rule follows from expanding $(a+bi)(c+di)$ and collecting like terms. $\endgroup$ – Henning Makholm Nov 10 '14 at 12:33
  • $\begingroup$ It is straightforward when one is armed with the assumption $i^2=1$, I agree. But what happens in the case when one is not? I mean, can one make this seem rigorous $and intuitive$ without introducing $i$ at all? $\endgroup$ – nb1 Nov 10 '14 at 12:55
  • $\begingroup$ @NikhilBellarykar: Without introducing $i$ at all, I'm not sure how one would even develop (or formulate) a desire to make complex numbers rigorous. $\endgroup$ – Henning Makholm Nov 10 '14 at 13:02
  • $\begingroup$ hmm maybe, yeah. $\endgroup$ – nb1 Nov 10 '14 at 13:07

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)

  • $\begingroup$ Very interesting. I have never seen this before.+1 $\endgroup$ – Pratyush Sarkar Aug 10 '13 at 14:24
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    $\begingroup$ Note that having no roots in $\mathbb R$ is not a sufficient condition for being irreducible -- for example $x^4+4$ factors in $\mathbb R[x]$. $\endgroup$ – Henning Makholm Aug 11 '13 at 10:46
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    $\begingroup$ It does if the polynom is of degree 2 or 3. (This should be added to the answer.) $\endgroup$ – Artefact2 Aug 11 '13 at 14:42

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.


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 $(0,1)^2=(-1,0)$).


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.


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.


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.

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    $\begingroup$ Gödel's result is only for sufficiently expressive systems. The (first-order) theory of the reals has, indeed, been proved consistent. That theory is far simpler than the theory of the integers... $\endgroup$ – GEdgar Aug 10 '13 at 13:23
  • $\begingroup$ Yes, I didn't go into these details because I assumed ZFC was the framework of all this. $\endgroup$ – Denis Aug 10 '13 at 13:24
  • $\begingroup$ But as Henning's answer points out, the fact that we can construct the complex numbers using ZFC means that if we find a contradiction in the theory of complex numbers, then ZFC was already inconsistent. $\endgroup$ – Nate Eldredge Aug 11 '13 at 17:45
  • $\begingroup$ @GEdgar But the real numbers contain the integers. As far as my understanding goes, the real numbers are only proved to be consistent under the assumption that set theory is consistent (and this is impossible to prove.) $\endgroup$ – Brusko651 Aug 12 '13 at 20:21
  • $\begingroup$ Consistency of the first-order theory of the reals (a result of Tarski) is MUCH EASIER than consistency of set theory (which is not known). $\endgroup$ – GEdgar Aug 12 '13 at 20:59

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.

  • $\begingroup$ Only the sentence "It is consistent... because it preserves addition and multiplication of the real numbers" addresses the OP's question, and it's very unclear in what sense this statement is true. $\endgroup$ – Jesse Madnick Dec 3 '13 at 9:11

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