A definition of ordered field

Question

Definition: An ordered field is field $F$ equipped with a linear order $<$ such that

• $1 > 0$

• For any $a,b \in F$, $$ a,b > 0 \quad \Longrightarrow \quad a+b,\;ab > 0.$$

I like this definition - it's quite clean - but I'm not convinced it is equivalent to other definitions of the notion ordered field. To elaborate, suppose $F$ is a field, linearly ordered by $<$ as per the definition above. The first result that stands out is $(\ast)$ ... $$a > 0 \Longrightarrow -a < 0.$$ This can be proven by contradiction. For say $a > 0$ but $-a \geqslant 0$. Since $a$ is not itself zero (irreflexivity), its additive inverse is not zero either, and $-a > 0$. Since the linear order with which we are working is such that the sum of elements greater than zero is again greater than zero, we obtain the contradiction $$0 = a + (-a) > 0.$$

Using $(\ast)$, it's not difficult to show that for any $b \in F$, the following relations are mutually exclusive ... $$b > 0, \quad -b > 0, \quad b = 0.$$ For instance, if say $-b > 0$, then applying $(\ast)$ to $a := -b > 0$ we determine $b= -a < 0$, whence by the properties of linear order it is not possible that $b>0$ or $b=0$.

So at most one of the aforementioned three relations can occur (great!), but why must any occur at all. I ultimately need to prove $(\ast\ast)$ ... $$a < 0 \quad \Longrightarrow \quad -a > 0,$$ but I am not sure where to start such a proof, as the definition for ordered field I am working with doesn't say anything about elements less than zero. It is now when I think the definition I have is incomplete. Can $(\ast\ast)$ be deduced?

Answer 1

Let $\mathbb N=\{0,1,2,\dots\}$, the natural numbers including zero. Define a linear order $\lt$ on $\mathbb R$ so that $a\lt b$ has the usual meaning if $a,b\in\mathbb N$ or if $a,b\in\mathbb R\setminus\mathbb N$, but $a\lt b$ whenever $a\in\mathbb R\setminus\mathbb N$ and $b\in\mathbb N$. Your faulty definition of "ordered field" is satisfied, since $a\gt0$ just means that $a$ is a (real) positive integer; but $\frac12\lt0$ and $-\frac12\lt0$.

Answer 2

Consider a field of Real numbers with such an order:

1. All rational number have standard real order.

2. All irrational numbers are less than 0 and for real irrational a, b, when -b < -a < 0 < a < b in standard real order we define -b < b < -a < a < 0

3. Standard order definition between standard negative rational numbers and standard negative irrational numbers.

This field seems to be an ordered field (in your definition), however your (**) is incorrect!

If you find a mistake - tell me, I cannot see it.