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I've read other posts about comparing complex numbers, but I didn't find an answer for this: is it possible to define a $\leq$ relation such as: $$ a+bi \leq c+id \iff b = d \quad\text{ and }\quad a \leq c. $$ This means that for any complex number, such as $a = 2+3i$, and $b = 1+4i$ then $a \leq b$ will be false, but so is $b \leq a$. Is that okay, or does having $a \leq b$ being false implies that $b \leq a$ (or $a > b$) should always be true ?

Clayton
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Benoit
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    Nope! You can't order impose an order on $\mathbb{C}$. The Boas text on complex analysis describes this... I think. – Sean Roberson Jul 05 '18 at 16:37
  • You might find the article on Wikipedia interesting, where it talks about posets (Partially Ordered SETS). What everyone keeps insisting on is that you can't have a total order on the complex numbers (I'm guessing you already knew this from the other posts). – Clayton Jul 05 '18 at 16:40
  • We can't define order on the complex numbers like we do the reals, but if you wanted to, you can compare their absolute values, which are just their distances from the origin on the complex plane – WaveX Jul 05 '18 at 16:44
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    You can define anything you want. And, yes, that is an order. But it is not an ordered field (which requires that $0\ge a; 0\ge b \implies a*b \ge 0$.) Making C an ordered field is impossible. – fleablood Jul 05 '18 at 16:48
  • Possible duplicate of Proof of Non-Ordering of Complex Field – amWhy Jul 05 '18 at 16:50
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    It's not a total order. as $a \not \le b$ and $a \ne b$ and $b \not \le a$. That's not possible for a total order. You can define the lexigraphical order as $a+bi \le c + di \iff (a < c) OR (a=c; b \le d)$. But this will not make an ordered FIELD. It finding an order that makes C and ordered FIELD that is impossible. An ordered field requires $x \ge 0;y\ge 0\implies xy \ge 0$. That is impossible on C because that would mean $z^2 \ge 0$ for all $z$. So both $1^2 = 1$ and $i^2 = -1$ must both be $\ge 0$ and that is impossible. – fleablood Jul 05 '18 at 17:14
  • Recall: $\Bbb{C}$ is not totally ordered. – homosapien Oct 02 '23 at 18:33

2 Answers2

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The order that you suggest is indeed an order, but a partial order, meaning that two complex numbers are not necessarily comparable. In fact, it is isomorphic to a product order, if we use the usual order for real parts, and set all imaginary parts incomparable.

An order such that all elements are comparable (that is, at least one of $a \leq b$ and $b \leq a$ is true) is called a linear order. The usual order on $\mathbb R$ is linear, the one you suggest for $\mathbb C$ is not. Whether one needs a linear order or a general partial order depends on the context.

When one says that there is no order in $\mathbb C$, what is meant is that no order can respect the algebraic structure. This is easily seen: $-1$ is less than $0$ in any ordered field, and $x^2$ is always $\geq 0$, so in $\mathbb C$ we must have $i^2 \geq 0$ and $-1<0$, a contradiction.

lisyarus
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  • Isn't it a totatal order? Given $a + bi$ and $c + di$ either i)$a=c;b=d$ ii)$a<c$ or $a=c$ and $b<d$ or iii) $a >c$ or $a=c$ and $b > d$. Those are the three only mutually exclusive options. But what it isn't is an ordered field so that $a\ge 0; b\ge 0$ implies $ab \ge 0$. – fleablood Jul 05 '18 at 16:51
  • @fleablood In the order suggested by OP two complex numbers $a+bi$ and $c+di$ are comparable *only if* $b=d$. Thus, it cannot be total. – lisyarus Jul 05 '18 at 16:54
  • Oh... you are right. I assumed this was the lexagraphical order in that $a + bi \le c + di$ if either $a < c$ or $a=c$ and $b \le d$. That would be a total order. But One in which $i > 0$ and $i^2 < 0$ (which prevents $\mathbb C$ from being an ordered field). – fleablood Jul 05 '18 at 17:03
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Yes, that is an order partial and it would even be a total order if you extended it to allow $a + bi \le c + di$ if either $a < b$ or if $a = b$ and $c \le d$. However it is not an order than allows $\mathbb C$ to be an ordered field.

An ordered field is a field with an order but ALSO that has the axioms that i) $a > b\implies a+c > b+c$ and that ii) $a > 0$ and $b > 0\implies ab > 0$.

This order does not satisify those conditions: $0 + i > 0$ but $(0+i)(0+i) = -1 < 0$.

And it is impossible to find an order than will allow $\mathbb C$ to be an ordered field. In short $a > 0\implies a^2 > 0$ and and $a = 0 \implies a^2 = 0$ and (although this must be proven) $a < 0 \implies -a > 0 \implies a^2 = (-a)^2 > 0$. But in $\mathbb C$ we have $i^2 = -1$ and $1^2 = 1$ so we must have $1 > 0$ and $-1 > 0$ and that is (again it must be proven) is impossible.

fleablood
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