Comparing complex number (again)

Question

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 ?

Answer 1

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.

Answer 2

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.