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

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

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.

Attribution
Source : Link , Question Author : FireGarden , Answer Author : Marc van Leeuwen

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