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The complex numbers as a whole cannot be ordered but could you order the complex numbers of the form $a i$ where $a \in \mathbb{R}$?

pie
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user189059
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3 Answers3

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The complex numbers can be ordered. It cannot be made into an ordered field with the usual addition and multiplication. And that's different.

We can, for example, define $(a+bi)\preceq(c+di)$ if and only if $a<c$ or $a=c$ and $b\leq d$. This defines a linear ordering of the complex numbers.

Another way would be to note that $\Bbb C$ and $\Bbb R$ have the same cardinality, and therefore there is a bijection $f\colon\Bbb C\to\Bbb R$. Now we can define $z_1\leq_f z_2\iff f(z_1)\leq f(z_2)$. And it is not hard to see that this is a linear order as well.

Now the set $\{ai\mid a\in\Bbb R\}$ is linearly ordered as a subset of a linearly ordered set.

But all these orders are incompatible with the field operations. Namely, $\Bbb C$ cannot be made into an ordered field, where $a<b$ implies $a+c<b+c$ for all $c$. The reason is that in an ordered field, we can prove both these statements:

  1. $-1<0$.
  2. $x^2\geq 0$ for every $x$.

And in $\Bbb C$, and in fact in any algebraically closed field, $i^2=-1$.

Asaf Karagila
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You could define an order, though it may not be useful: $a+bi<c+di$ if $a<c$; and $a+bi<a+ci$ if $b<c$. I think that is called the 'long line' http://en.wikipedia.org/wiki/Long_line_%28topology%29 because you rearrange the complex plane into lines of constant real part. EDIT: see comment, (thanks) it isn't the long line.

Empy2
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    No, this is not the 'long line'. The long line has an uncountable well-ordered subset; the lexicographical ordering of $\mathbb R\times\mathbb R$ does not. – bof Nov 21 '14 at 12:13
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Yes. The ordering corresponds to that of the reals.

Define an ordering $<_i$ as follows:

$$ai <_i bi \iff a < b$$

where "$<$" is just as in $\mathbb{R}$.

Since $<$ is an ordering of $\mathbb{R}$, $<_i$ is an ordering of the purely imaginary numbers $\{z \in \mathbb{C} \,\mid\, \exists\, a \in \mathbb{R} \,\,\text{ with } \,\,z=ai\}$

EDIT3: Thanks to Asaf, I think I understand now. $\mathbb{C}$ as a set is totally ordered, but if you consider it as a field it is not.

Zubin Mukerjee
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  • That last line is misleading. – Asaf Karagila Nov 21 '14 at 12:10
  • Yeah I think I'm wrong. Is there a total ordering of $\mathbb{C}$? – Zubin Mukerjee Nov 21 '14 at 12:18
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    Of course there are. Plenty of them. There is no order which makes $\Bbb C$ (with its usual addition and multiplication) into an ordered field, though. But that doesn't mean that there is no order on $\Bbb C$ as a set. – Asaf Karagila Nov 21 '14 at 12:30
  • When people say $\mathbb{C}$ what do they usually mean? The set with the operations, or just the set? (not a rhetorical question, genuinely asking) – Zubin Mukerjee Nov 21 '14 at 14:05
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    I, think it depends on the context, and how accurate the person wants to be. Here, the way the question is asked seems to me as if it means $\Bbb C$ as a set. Because the imaginary axis of the plane is certainly not a subfield of $\Bbb C$. – Asaf Karagila Nov 21 '14 at 14:28
  • Okay, so my $<_i$ thing is a total ordering of the purely imaginary numbers, but there are also total orderings of the set $\mathbb{C}$. But none of it works if we want $\mathbb{C}$ to be a field? – Zubin Mukerjee Nov 21 '14 at 14:30
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    Yes. That is correct. – Asaf Karagila Nov 21 '14 at 14:31
  • One last correction, sets are generally not ordered. They are just sets. But it can be ordered. – Asaf Karagila Nov 21 '14 at 14:39