2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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!). $\endgroup$ – Noah Schweber Jan 2 '17 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 '17 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.) $\endgroup$ – Noah Schweber Jan 2 '17 at 17:49
  • $\begingroup$ R is a field with respect to addition and multiplication. $\endgroup$ – Student Jan 2 '17 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.) $\endgroup$ – Noah Schweber Jan 2 '17 at 18:05
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 . . .

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.