While constructing the real numbers, we come across the order axioms, used to define the positive real numbers, i.e there exists a set $P$ such that the properties of trichotomy, closure under addition and multiplication hold. The axioms guarantee the existence of such a set. Will such a set be unique as well? I've tried proving this, and it should be trivial, but I can't seem to get it.

  • $\begingroup$ Incidentally, I'll be able to give a more detailed answer if you specify exactly what axioms you're using (there are many different axiomatizations that yield the same theorems, but do so in different ways; for instance, sometimes my characterization of the positive reals is taken as an axiom about order!). $\endgroup$ Jan 2, 2017 at 17:39
  • $\begingroup$ I'm referring to the following axiom: There exists a set of real numbers P such that: 1. Exclusively a is in P or -a is in P or 0 is in P 2. If a,b are in P, a+b is in P 3. If a,b are in P, ab is in P Is this set P unique? That was my question. It certainly exists (axiomatically). But can we guarantee uniqueness? $\endgroup$
    – Student
    Jan 2, 2017 at 17:42
  • $\begingroup$ What other axioms do you have? E.g. you have some axioms about $+$ and $\times$ . . . (And I understand what your question is asking - my answer is that yes, it is unique, and the way to prove this is to show that it equals the set in my answer - although to give a complete proof of this I will need to know exactly what axioms you are using.) $\endgroup$ Jan 2, 2017 at 17:49
  • $\begingroup$ R is a field with respect to addition and multiplication. $\endgroup$
    – Student
    Jan 2, 2017 at 17:50
  • 1
    $\begingroup$ I know we're talking about the uniqueness of the positive reals. My point is that, without any additional axioms, your statement is false - e.g. the field $\mathbb{Q}(\pi)$ satisfies all the properties you've mentioned, but does not have a unique set of positive elements. So you need some additional axiom(s). (You are right that it isn't completeness, though.) $\endgroup$ Jan 2, 2017 at 18:05

2 Answers 2


For general orderable fields, $P$ isn't unique — all you can say is that it must contain all sums of nonzero squares.

To see that $P$ must contain squares, note that if $x \neq 0$, then you have two cases:

  • $x \in P$, and thus $x^2 \in P$
  • $-x \in P$, and thus $(-x)^2 \in P$

Since $x^2 = (-x)^2$, either way we conclude $x^2 \in P$.

I believe there is a theorem that says that if neither $y$ nor $-y$ is a sum of squares, then there is a choice for $P$ containing $y$, and another choice for $P$ containing $-y$.

The reals are a special case, since the set of all nonzero squares already satisfies trichotomy, so $P$ can't contain anything else, thus $P$ is unique.


I assume "satisfies trichotomy" means "every real is positive, negative (= its additive inverse is positive), or zero.

Yes, if you have a reasonable set of axioms, you can show that the set of positive reals is uniquely defined. Specifically, you can prove that the set of positive reals is exactly $$\{x: \mbox{ for some $y\not=0$, $x=y\cdot y$}\}.$$ Note that this can be used to define the order relation: $x_1<x_2$ iff for some $y$, we have $y\cdot y+x_1=x_2$. So in fact, "$<$" does not need to be included as a primitive symbol.

Note: the list of axioms you have given so far are not enough - indeed, the field $\mathbb{Q}(\pi)$ satisfies them, but does not have a unique set of positive elements (since it has an automorphism swapping $\pi$ and $-\pi$). So I think you're missing an axiom or two . . .

How do you prove this?

Well, first suppose $x=y\cdot y$ for some $y$. Then

  • Can you show that $x=(-y)\cdot (-y)$ as well?

  • What does that tell you about $x$ (think about whether $y$ is positive or negative)?

This shows that the set I've defined is a subset of the positive reals. Now we want to show that it contains the positive reals. To do this, you need to argue that for any real $r$, either $r$ or $-r$ has a square root. You can't do that with only the axioms you've listed. I think you have another axiom or two lying around . . .

  • $\begingroup$ If you take $P$ to be the positive rationals and claim that it satisfies trichotomy, then is $\sqrt{2}$ positive, zero, or negative? $\endgroup$
    – user14972
    Jan 2, 2017 at 17:46
  • $\begingroup$ @Hurkyl Ah, I read "satisfies trichotomy" as meaning "is totally ordered," but you're right, your interpretation is probably what the OP means (since every set of reals is totally ordered). $\endgroup$ Jan 2, 2017 at 17:47

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