Real numbers mapped onto a sphere

Question

We can compare real values if they were greater, lesser but we cannot do same for complex numbers.

What if we map real values(within some small range) onto a sphere and declare each one of them as complexed, can we use their "angle($phi$) angle($rho$)"/"distance to origin" as components of a complex number? Because the mapped values are real, is this legal(because we can compare them)? I dont know if there exists such canonical thing. Do you know something similar?

Thank you.

Answer 1

Let's look at $i$, since adding it is what causes all the trouble. An order on $\mathbb{C}$ that plays nicely with the rules of addition and multiplication (an ordered field) must have the following always be true:

$$\text{if} \ \ a<b, \ \text{ then } \ a+c<b+c$$ $$\text{if}\ \ a>0\ \ \text{and}\ \ b>0\ \ \text{then}\ \ ab>0$$

Now where does $i$ fit? If it's greater than zero, then the second part doesn't work, since

$$i>0\implies i\cdot i>0\implies -1>0$$

which clearly isn't true. So is $i<0$? If it is, then $-i>0$. Since a positive times a negative is negative, we get that

$$i\cdot-i<0\implies1<0$$

which is also not true. So $\mathbb{C}$ can't be ordered in a way that works with the operations you'd want. If you don't care about that then you can order them tons of different ways! Maybe comparing first by $x$, then $y$? Or by modulus, then argument? You can also find whatever bijection from $\mathbb{C}$ to $\mathbb{R}$ that you like, and order the complex values by the reals they map to (which seems to be your suggestion) and it works perfectly as a total order. But $\mathbb{C}$ will never be an ordered field, because such an ordering can never work nicely with the operations we want it to.