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.
Uniqueness of set of Positive real numbers using order axioms
2
$\begingroup$
real-numbers
-
0Incidentally, 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!). – 2017-01-02
-
0I'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? – 2017-01-02
-
0What *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.) – 2017-01-02
-
0R is a field with respect to addition and multiplication. – 2017-01-02
-
0And that's about it. – 2017-01-02
-
0You 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? – 2017-01-02
-
0Yes, 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. – 2017-01-02
-
1**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.) – 2017-01-02
-
0Well, I can't think of any other axioms. I only have so much to go on. – 2017-01-02
-
0If 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. – 2017-01-02
-
0Well, 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? – 2017-01-02
-
0I 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. – 2017-01-02
-
0Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/51085/discussion-between-raghav-talwar-and-noah-schweber). – 2017-01-02
2 Answers
1
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 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 . . .
-
0If you take $P$ to be the positive rationals and claim that it satisfies trichotomy, then is $\sqrt{2}$ positive, zero, or negative? – 2017-01-02
-
0@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). – 2017-01-02
1
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.