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.

  • $\begingroup$ There are orderings of the complex numbers. for example, we can declare $a+bi \lt c+di$ if $a\lt c$ or $a=c$ and $b\lt d$. This is a total order. But orderings of the complex numbers do not "play nice" with the addition and multiplication. $\endgroup$ Aug 28, 2012 at 15:06
  • $\begingroup$ You are right. I just wanted to use a surface to represent ccomplexes to be available for a comparing without restrictions.(actually the range of real value is restriction but not problem) $\endgroup$ Aug 28, 2012 at 15:15

1 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


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .