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In Rudin's Analysis, chapter 1, page 3, definition 1.6 states

$\mathbb Q$ is an ordered set if $r<s$ is defined to mean that $s-r$ is a positive rational number.

But the very definition of 'positive' requires the notion of greater than or less than, so it seems all circular to me.

user10354138
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Shiladitya Mukherjee
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  • if you can distinguish $\Bbb N$ from $\Bbb Z,$ you can distinguish positive from negative ($\Bbb Q$ can be defined as ratios) – J. W. Tanner Jun 24 '19 at 16:28
  • You are right I guess. The problem is Rudin does not build the rationals axiomatically, and for a beginner like me it is hard(impossible) to know what assumptions to take. Should I study Tao first? – Shiladitya Mukherjee Jun 24 '19 at 16:33

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Original Answer No. We have what a "positive rational number" means (it is the quotient of a positive integer by another positive integer), but we don't yet have the ordering $<\subseteq\mathbb{Q}^2$. This definition defines the ordering on $\mathbb{Q}$.

Addendum OK, checking my copy of baby Rudin. Definition 1.6 is

1.6 Definition An ordered set is a set $S$ in which an order is defined.

What you quoted isn't Definition 1.6, but an example in the paragraph after Definition 1.6. Anyway, the circularity is avoided if you do it properly as suggested in my original answer.

user10354138
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  • How do we know what a positive rational number means without first defining the notion of inquality? – Shiladitya Mukherjee Jun 24 '19 at 16:29
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    He isn't defining inequality. The $<$ sign in definition $1.5$ refers to a general order relation. The remark just means that the ordinary relation < on the rationals, which you are presumed to be familiar with, is in fact an order relation. – saulspatz Jun 24 '19 at 16:31
  • @ShiladityaMukherjee We have a definition of "positive rational number" without first defining the notion of inequality. We construct $\mathbb{Z}$ from $\mathbb{N}$ and $\mathbb{Q}$ from $\mathbb{N}$ or $\mathbb{Z}$ as ordered pairs (and ordered pairs of ordered pairs) and we have constructed an ordering on $\mathbb{Z}$ along the way, so we can talk about "represented by a positive integer divided by another positive integer" as definition of positive rational number. – user10354138 Jun 24 '19 at 16:44
  • @saulspatz We are not assumed to be familiar with the notion of inequality among rationals. Read the question. – Shiladitya Mukherjee Jun 24 '19 at 16:59
  • @user Should I do T.Tao Analysis first? – Shiladitya Mukherjee Jun 24 '19 at 17:00
  • @ShiladityaMukherjee 1, Don't be rude. 2. I've not only read the question, I've read the book, and understood it. – saulspatz Jun 24 '19 at 17:08
  • @ShiladityaMukherjee Probably the first four chapters of Tao's book (i.e., up to construction and properties of $\mathbb{Q}$) or something like A.G. Hamilton's Numbers, Sets and Axioms: The apparatus of Mathematics. Then both Tao and baby Rudin start covering the same material in more-or-less the same order, and which one to read depends on what you like. – user10354138 Jun 24 '19 at 17:14
  • @user10354138 thanks a lot. – Shiladitya Mukherjee Jun 25 '19 at 09:27
  • @saulspatz sorry. But we are NOT assumed to be familiar because the example defines the meaning of ">". – Shiladitya Mukherjee Jun 25 '19 at 09:29
  • No, I think you misinterpret the example, as I said. It's true that the same symbol is used for the usual order on the rationals, and the generic order symbol that Rudin is talking about. All he's saying though, is "The usual order on the rationals is an order in the sense of Definition 1.5." He isn't defining the meaning of a positive number here. It would be clearer if he'd used other symbol like $\prec$ in the definition. Then he could have said, "$r\prec s$ if and only if $s-r$ is positive, is an order on $\mathbb{Q}$." – saulspatz Jun 25 '19 at 13:44
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You can define a positive rational number with an algebraic property : it is a rational number which is a non zero square of a real number. That is why he is doing this.

EDIT : sorry, it is not a natural way to do it, since we construct $\mathbb{R}$ after $\mathbb{Q}$. So, the natural way is to say that a rational $p/q = (p,q)$ (rational defined by a equivalence relation on 2-uplets) is positive if $p$ and $q$ are positive, or $p$ and $q$ are negative.

Dlem
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  • You can probably do it in a million other ways, once you first go through the axiomatic approach, for which you have to do Analysis first, for which you have to study Rudin.. – Shiladitya Mukherjee Jun 24 '19 at 16:51
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The positive rational numbers can be defined without reference to order by

  1. $1\in \mathbb Q^+$

  2. $a,b\in \mathbb Q^+ \rightarrow a+b\in\mathbb Q^+\land ab^{-1}\in\mathbb Q^+$

  3. $a\in \mathbb Q^+ \rightarrow -a\notin \mathbb Q^+$

1 and 2 ensure $\mathbb Q^+$ contains all positive rationals: any positive rational can be written as the ratio of two positive integers, which can themselves be written as finite sums of $1$s. 3 then ensures that no negative rationals or $0$ can be elements of $\mathbb Q^+$.

eyeballfrog
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I have an analysis textbook which lays out axioms for the Reals, in which the term 'positive' is an undefined term satisfying the following axioms:

1) For any Real number r, one of the following mutually-exclusive conditions holds: (a) r is positive; (b) r is zero; (c) -r is positive.

2) Any finite sum of positive Real numbers is positive, and any finite product of positive Real numbers is positive.

From these it is not too hard to prove basic properties of positive numbers, e.g: 1 is positive; -1 is NOT positive; N is positive for each natural number N; etc. Then the ordering r < s (for Real r, s) is DEFINED by '(s - r) is positive'. No circularity.

PMar
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