Can we define the complex numbers without using the real numbers?

Question

Every definition of $\mathbb{C}$ that I can think of uses $\mathbb{R}$. We can let $\mathbb{C}=\mathbb{R}^2$ equipped with a particular multiplication. We can let $\mathbb{C}$ be the algebraic closure of $\mathbb{R}$.

The best way I could come up with to define $\mathbb{C}$ without $\mathbb{R}$ is to let $\mathbb{C}$ be the "completion" of the algebraic closure of the rational numbers. I'm not really sure what the "completion" would mean in this case. Can anyone finish this idea, or suggest a different approach?

Answer 1

If all you want is to define $\mathbb C$, you could do it along this path:

• Start with $\mathbb Q$.
• Algebraically adjoin a root of $x^2+1$, giving $\mathbb Q[i]$.
• Define the squared distances $d^2(p+qi,r+si)=(p-r)^2+(q-s)^2$.
• The usual definition of metric spaces wants real distances, but squared distances are still enough to define Cauchy sequences anyway.
• Even though $d^2$ itself doesn't satisfy the triangle equality, it ought to be possible to rephrase the usual proofs of the properties of Cauchy sequences to work with $d^2$ instead. In particular we want the sum and product of Cauchy sequences to be Cauchy, the sum of two null sequences to be null, and so forth.
• We can then define $\mathbb C$ as equivalence classes of Cauchy sequences in $\mathbb Q[i]$.

I haven't done the details, but I suppose we can show reasonably painlessly that $\mathbb C$ is a field extension of $\mathbb Q[i]$.

On the other hand it could be a real challenge to prove the fundamental theorem of algebra -- that is, that our $\mathbb C$ is algebraically closed -- without introducing $\mathbb R$ and its topology through a back door somewhere along the way. Withough $\mathbb R$ there would be no intermediate value theorem, for example.