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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.

E. Joseph
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Student
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  • 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!). – Noah Schweber Jan 02 '17 at 17:39
  • 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?

     – Student Jan 02 '17 at 17:42
  • 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.) – Noah Schweber Jan 02 '17 at 17:49
  • R is a field with respect to addition and multiplication. – Student Jan 02 '17 at 17:50
  • And that's about it. – Student Jan 02 '17 at 17:50
  • You definitely need more axioms - you need some way to show that for every real number, either it or its additive inverse has a square root. Maybe you have a completeness axiom lying around? – Noah Schweber Jan 02 '17 at 17:56
  • Yes, the completeness axiom gives uniqueness to the real numbers. But we are talking about uniqueness of P as a subset of R. Also, won't one need uniqueness of P for the definition of the completeness axiom? As '<' as a strict total order relation is well-defined only if P is unique. – Student Jan 02 '17 at 18:03
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    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.) – Noah Schweber Jan 02 '17 at 18:05
  • Well, I can't think of any other axioms. I only have so much to go on. – Student Jan 02 '17 at 18:29
  • If those are the only axioms you've been given, then the statement you're trying to prove is false. I would double-check your book or notes. – Noah Schweber Jan 02 '17 at 18:33
  • Well, my book/notes don't explicitly mention uniqueness. However, when we go on to define the order relation ">", we need uniqueness of P, as we start by saying a>0 iff a is in P. This comes directly after the axioms I've stated. Could you please help me figure out what is amiss? – Student Jan 02 '17 at 18:37
  • I don't know what to tell you other than what I've already told you. The fact of the matter is, the axioms you've listed aren't enough to prove the thing you want. I can think of three possibilities: either you missed an axiom; or your text doesn't actually need the set of positive reals to be unique (you can define an order relation by fixing some set of positive reals, and maybe that's enough for the text, but I doubt it - you won't be able to prove that every positive real has a square root); or your text made a mistake (sadly this is possible). If you have an instructor I'd email them. – Noah Schweber Jan 02 '17 at 18:41
  • Let us continue this discussion in chat. – Student Jan 02 '17 at 18:42

2 Answers2

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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 . . .

Noah Schweber
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  • If you take $P$ to be the positive rationals and claim that it satisfies trichotomy, then is $\sqrt{2}$ positive, zero, or negative? –  Jan 02 '17 at 17:46
  • @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). – Noah Schweber Jan 02 '17 at 17:47
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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.