0
$\begingroup$

Can the complex numbers be totally ordered, and if so, how is this different to $\mathbb{C}$ being an ordered field?

We can totally order $\mathbb{C}$ can we not, for example by saying that:

$$a>c\implies (a+bi>c+di) \quad\forall a,b,c,d,$$

and by complementing that with:

$$(a=c)\land (b>d)\implies (a+bi>c+di) \quad\forall a,b,c,d$$

But $\mathbb{C}$ is not an ordered field so clearly I am missing something somewhere; presumably in what exactly it means to be an ordered field.

  • 4
    To be an ordered field, the order has to be compatible with the field operations. – 2017-02-05
  • 3
    Well, your definition says that $i >0$, but $i^2=-1 <0$. The problem is that this order is not compatible with mutipication. In practice, $( \Bbb{C} , < )$ is an ordered set, $(\Bbb{C}, + ,\cdot )$ is a field, but $(\Bbb{C} , +, \cdot , <)$ is not an odered field. – 2017-02-05
  • 1
    Yes both multiplication and addition need to perserve the order but the multiplication example given by Crostul does not. – 2017-02-05
  • 0
    Thanks @littleO I knew that when I wrote the question, but "compatibility" isn't defined in the material I was reading either. – 2017-02-05
  • 0
    Thanks @Crostul what does it mean for multiplication and addition to preserve the order? Does it mean $a\geq b\implies ac\geq bc\quad \forall a,b,c$? And I guess the same for $+$. – 2017-02-05
  • 0
    @RobertFrost Yes, something like that. See https://en.wikipedia.org/wiki/Ordered_field for more details. – 2017-02-05
  • 0
    @Crostul ok thanks. So coming back to the question, the mistake I make is that it is not a true statement that $\mathbb{C}$ is not an ordered field, since such a statement may only be made in reference to the addition and multiplication functions in operation. Is that correct? – 2017-02-05

1 Answers 1

3

By your ordering, $i > 0$ and $1 > 0$. But if that ordering made $\mathbb{C}$ an ordered field, then

\begin{align*} &i > 0\\[6pt] \implies\; &i^2 > 0\\[6pt] \implies\; &-1 > 0\\[6pt] \implies\; &-1 + 1 > 0 + 1\\[6pt] \implies\; &0 > 1 \end{align*}

contradiction.