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I apologize if this question appears to be a dumb one. However given my preliminary knowledge in real analysis, I am unable to resolve the issue; it’s about the order axiom (of reals).

It can be shown that algebraic properties of reals follows from field axioms that work as a premise for further reasoning. Since order properties of reals cannot (?) be shown from algebraic properties, we assume the existence of a positive set based on which we can define order of reals. That’s the way Bartle-Sherbert text introduced the concepts of order (of reals).

Image: Bartle-Sherbert Introduction to Order

But why do we just assume the existence of a positive set the way we assume an axiom? Can’t we really define a positive set $P$ based on algebraic properties as: $P=\{y^2:y\in\mathbb R, y\ne 0\}$. For then we could define order relation $<$ as: $x<y$ iff $y – x\in P$.

Based on this observation, do we really need extra (order) axioms, whereas we could define order based on algebraic properties?

s a
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    No, we don't need the order axioms to define the order relation of real numbers. The one you suggested is a possibility. Commonly the order is defined, when the real numbers are constructed (for example as equivalence classes of Cauchy sequences). Also, at that point in the development the order axioms can be verified (most notably also the least upper bound property). – Jyrki Lahtonen Aug 04 '17 at 18:41
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    This would require you to know something odder - that every non-zero real number can be written as $y^2$ or $-y^2$ for some non-zero $y$, but not both. And also, you'd need that $x+y\in P$ if $x,y\in P$. – Thomas Andrews Aug 04 '17 at 18:41
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    The key is to look at other fields. What if you did the same, but the field was in the complex numbers? Then, your $P$ would contain all the non-zero complex numbers. (This is because the complex numbers have no useful order.) Or if you were in $\mathbb Q$, the rationals, this definition would miss a lot of positive rational numbers. – Thomas Andrews Aug 04 '17 at 18:46
  • @JyrkiLahtonen: If those axioms can be verified what is the point of calling them axioms? – s a Aug 04 '17 at 18:50
  • @ThomasAndrews: I didn't get it. I would choose a real number $y$ and consider the multiplication $y.y$ and will thus forms a set of all such numbers. What else do I need to know to do this other than algebraic properties? – s a Aug 04 '17 at 18:52
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    Depends on the other axioms, but how do you know that for any $x\in \mathbb R$, that exactly one of the following is true: $x=0,x\in P,-x\in P$? What in your axioms prevents the axioms to give the complex numbers, for example? – Thomas Andrews Aug 04 '17 at 18:56
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    The point in having axioms is economy of thinking. You prove statements that are consequences of axioms. Then if you can verify the axioms for some structure, you know that those consequences are valid in that structure. There are other fields where you can define an order relation using squares. The ordering in such a system may or may not have all the properties of the ordering of reals. For example, the real algebraic numbers have an ordering definable by declaring the squares to be the positive numbers, but that ordering lacks the least upper bound property. – Jyrki Lahtonen Aug 04 '17 at 19:03
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    (cont'd) The reals are kinda exceptional in that they are for most purposes the only number system satisfying all those axioms. Normally that is not the case, and the economy provided by listing which axioms imply what shows. – Jyrki Lahtonen Aug 04 '17 at 19:06
  • How can $\mathbb R$ be defined without defining order on $\mathbb Q$ first? We can assume $\mathbb R$ is just any field, or we could even assume $\mathbb R$ is any field with $\mathbb Q$ as a subfield, but we can't really specify anything more without declaring order on $\mathbb Q$ and defining $\mathbb R$ as the limits. – fleablood Aug 04 '17 at 21:51
  • How would you prove it is transitive. If $y-x = k^2$ and $z-y = m^2$ how does it follow that $z-x = m^2 - k^2 = j^2$ for some $j$? Or that if $y-x = k^2$ that it is impossible for $x - y = v^2$? – fleablood Aug 04 '17 at 22:00

2 Answers2

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You are right that you can define the order on the reals in terms of its algebraic structure. However, you can't prove all of its properties using just the other axioms of real numbers. So no matter what, if you want to describe the real numbers axiomatically, you will need axioms about order in some form or another. For instance, even if you define $P=\{y^2:y\in\mathbb{R},y\neq 0\}$ instead of just postulating that some set $P$ exists, you will still need axioms stating that this set $P$ satisfies properties (i) and (iii) (it turns out property (ii) can be proved from the other axioms for real numbers, but (i) and (iii) cannot).

Another reason to axiomatize the set $P$ is that often we are interested in variants of the real numbers: structures that have some but not all of the properties of real numbers. For instance, the rational numbers $\mathbb{Q}$ have many of the properties of real numbers, but not all of them. In the rational numbers, we have a notion of order, but defining it by $P=\{y^2:y\in\mathbb{Q},y\neq 0\}$ would not work, since not every positive rational number has a rational square root. However, the positive rational numbers are still a subset of the rational numbers satisfying axioms (i), (ii), and (iii). So by axiomatizing the order of the real numbers in this way, we give axioms that are also useful in more general settings.

Eric Wofsey
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You could, and there are many similar definitions equivalent to the usual axiomatic one. It seems from the context, though, that the reference you mentioned is constructing $\mathbb{R}$ from a standard set of axioms, and the particular properties of $\mathbb{R}$ necessary for that equivalence haven't been introduced or proved yet. In one direction, any nonzero $x$ has either $x\in P$ or $-x\in P$, forcing $x^2 = (-x)^2$ to lie in $P$ by axiom (ii). But why does the converse hold? It doesn't hold for $\mathbb{Q}$; it depends on (for example) the completeness of $\mathbb{R}$, which is nontrivial. You may be thinking of a specific definition of model of $\mathbb{R}$, but that's not necessarily how it's defined or constructed in that text. For all I know, the author may be planning to work with hyperreals and is defining $\mathbb{R}$ to eventually lead to that goal.

Anyway, the relevant criterion for a definition is whether it's useful--- not (or at least not necessarily) in the sense of a real-world application, but in understanding what's actually going on and helping to develop more math. Unlike the standard definition, your definition doesn't generalize; number fields (e.g., $\mathbb{Q}$) have elements that are positive but aren't squares, and every complex number is a square. It isn't obvious in the definition that you give that $\mathbb{R}^{>0}$ and $\mathbb{R}^{\geq 0}$ are closed under addition, multiplication, or division. There are notions of positivity in other contexts that we want to have similar properties to this one (e.g., the sum of positive elements is positive) but have no inherent notion of squaring.

Ultimately, math is a game with an undefined goal but very strictly defined rules. You're allowed to define the rules you want to play with, but once you set them, you're stuck with them and their inevitable logical conclusions. The definition you propose isn't inherently bad, but it's probably not the right one for the particular game the reference you mention is playing. It's like printing the instruction manual for a board game in Esperanto: The same information is there, but you've obfuscated the game without any real benefit.

anomaly
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