Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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. –  André Nicolas Aug 28 '12 at 15:06
    
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) –  huseyin tugrul buyukisik Aug 28 '12 at 15:15

1 Answer 1

up vote 5 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.