# how to proof the uniqueness of the subset of positive numbers

I am considering the following problem: Is the subset of positive real numbers unique, according to the Positivity Axiom (cf: Page 8, Fitzpatrick, Real Analysis, 4th edition):

... There is a set of real numbers, denoted by $\mathcal{P}$, called the set of positive numbers. It has the following two properties:

P1 If $a$ and $b$ are positive, then $ab$ and $a+b$ are also positive.

P2 For a real number a, exactly one of the following three alternatives is true: $$a\text{ is positive, -a is positive, a=0.}$$

In order to prove the uniqueness of the subset of positive numbers, I have tried like this: Assume there are two subsets of positive numbers, denoted by $\mathcal{P}_1,$ and $\mathcal{P}_2$, then I have proved that $1\in\mathcal{P}_1\cap \mathcal{P}_2$, then I do not know what to do. Can anyone help me?

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Its not unique. From your axioms, its easy to determine which rational numbers are positive and which arent, because you can "reach" them from $1$ by applying the algebraic operations of (P1). But pick any transcedent number, say $\pi$, and fix its sign whatever you like. This will fix the sign of all numbers that can in some algebraic way constructed from the rational numbers and $\pi$ (this field is usually called $\mathbb{Q}[\pi]$), but all remaining real numbers are still arbitrarily signed, and this is indeed a "positive set" as described by those axioms.

This drives home the point that transcendent numbers are a strange and arbitrary addition to the number system, as long as we only talk algebra. To make sense of them (to define them in the usual, or maybe all sensible ways), one must make some mention of the topological properties of the real line.

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P1 If a and b are positive, then ab and a+b are also positive.

P2 For a real number a, exactly one of the following three alternatives is true: a is positive, −a is positive, a=0.

I think that the uniqueness you are searching for follows from 'P2' itself. Maybe a formalization is superfluous and I don't know how to write it rigorously, since we cannot even say "the set of $all$ positive numbers" if we are discussing the uniqueness. Anyway I'll try to explain better what I mean:

Suppose that the set of positive numbers is not unique, so there are $\cal{S}_1$ and $\cal{S}_2$, sets of positive numbers. If $\cal{S}_1$ and $\cal{S}_2$ represent different sets, then they must differ at least for an element, say a number $a\ne 0$ satisfying 'P1'. This number would appear as $a$ in $\cal{S}_1$ and as $-a$ in $\cal{S}_2$, or vice versa. But this is not possible because of 'P2': only one between $a$ and $-a$ can be considered positive.

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@amueller: I'm trying to understand your argument. Cannot we prove that one of these "arbitrarly signed numbers", say $b$, lies between two rationals $a$ and $c$? Then, if $a$ and $c$ are positive according to the axiom, also $b$ is positive? Moreover, this wouldn't be a further assumption, but just a consequence of "P1" and "P2", in fact the positive axiom implies a notion of ordering on the real numbers, because we can define $a < b$ when $b − a$ is positive. About the topological property of $\mathbb{R}$, it seems to me that the axiomatic definition of the real line is sufficient here. – the_elder Nov 20 '12 at 12:12
I assumed, in my previous comment, the existence of such arbitrary signed numbers, but actually I didn't really understand from where they come from. – the_elder Nov 20 '12 at 13:28