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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?

joeb
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  • any of three must occur because of definition of linear order: any two are compatible – Anatoliy R Nov 15 '19 at 02:06
  • How so? The definition of linear order ensures that $b > 0$ or $b < 0$ or $b = 0$ must occur, but it is not obvious from my definition of ordered field that $b < 0$ is the same as $- b > 0$. – joeb Nov 15 '19 at 02:12
  • no it's not, my comment is for you first question: "but why must any occur at all" – Anatoliy R Nov 15 '19 at 02:14
  • @AnatoliyR: I suspect you meant "any two are incompatible." The OP seems to appreciate that at most one of the trichotomy cases holds, and wants to prove at least one case holds (for any $a$). The last edit was shortly before your Comment, so perhaps you were looking at an older revision? – hardmath Nov 15 '19 at 02:18
  • I am trying to prove that a < b => a + c < b + c and I cannot. It does not seem that a, b > 0 => a + b > o is equivalent – Anatoliy R Nov 15 '19 at 02:40
  • The equality $b-a = (b+c) - (a+c)$ comes to mind, and I have a feeling the validity of the implication $a < b \Longrightarrow a+c < b+c$ could be known if the validity of an equivalence like $x < y \Longleftrightarrow y - x > 0$ were known. For instance, if said equivalence were known, you might argue that "$a < b$ implies $(b+c) - (a+c) = b - a >0$, further implying $a+c < b+c$". Of course, I am also led to believe that the desired equivalence very much depends on the validity of $(\ast\ast)$ – joeb Nov 15 '19 at 02:53
  • @hardmath I mean exactly what I meant, but I did not recognize the question properly, so my comment was wrong - see my answer and what I think about this definition for ordered field. – Anatoliy R Nov 15 '19 at 03:18
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    What is your source for that bad definition? Did you find that in a textbook? – bof Nov 15 '19 at 04:32
  • I saw it here https://ncatlab.org/nlab/show/ordered+field – joeb Nov 15 '19 at 12:44

2 Answers2

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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.

Anatoliy R
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  • Really it doesn't matter how the irrationals are ordered at all, as long as you just make them all negative. – Eric Wofsey Nov 15 '19 at 03:31
  • Right, but then you have to prove that such a linear order exists :-) actually, it is, because cardinality is continuum, so you can make it isomorphic to a set of real numbers – Anatoliy R Nov 15 '19 at 03:46
  • also need to specify rule between negative rational and irrational, e.g. any irrational is less than any rational, but either way it works; finally you can replace rational with any other strict sub-field – Anatoliy R Nov 15 '19 at 03:57
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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 $-\frac12\lt0$.

bof
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