Today I'm afraid, I found a circular reasoning in Rudin's Principles of Mathematical Analysis( I found no errata that mentions this).
Before actually going through the actual question, I have compiled two documents one on fields(hereafter called document 1) and an incomplete document on construction of real numbers from rational numbers(hereafter called document 2), the second document contains the definitions and proofs that satisfy the criteria that are done before proving that $\mathbb{R}$ is a field (All theorems except the last one is okay).
In the document 2, the last theorem, I have tried to prove is that $\mathbb{R}$ satisfies all the Field axioms for addition.(Referring to step 4 of Apendix in chapter 1 of rudin's book) Rudin's book could only prove the Axiom-5 using the theorem 1.20(in Rudin's book) popularly known as the Archemedian property(You can see page-4 of my document 2 for a proof how it is done). Now I shall show that the proof of theorem 1.20 requires the fact that $\mathbb{R}$ satisfies at least the axioms of addition.
Theorem 1.20
(a) If $x\in \mathbb{R}$, $y\in \mathbb{R}$, and $x>0$, then there is a positive integer $n$ such that $nx>y$.
The proof follows from least-upper-bound property of real numbers(this can be proven with no problem, see my document) to claim that the set $A=\{nx:n\in\mathbb{N}\}$, if doesn't satisfy the theorem has a supremum and hence take $\alpha-x<\alpha$ and using definition of supremum to arrive at a contradiction. Now existence of $-x$ follows from A5 of field, and the property that If $y<z$ then $x+y<x+z$ where $x,y,z$ belong to a field(a part of definition of ordered field) is required for proving that $\alpha-x<\alpha$, to show that $\mathbb{R}$ satisfies this property we require the proposition 1.14-a (in rudin) or proposition 1-a (in document 1) whose proof again requires axioms A4, A5, A3, A2 and A1, existence of A5 is at stake right now.
So to conclude things proving that $\mathbb{R}$ satisfies axiom A5 and the archemedian property are circular proofs.
This is the question:
- Is there any way you can prove A5 without using archemedian property? ( I doubt this)
- If the above mentioned one is not possible, is there any way to prove archemedian property without help of field axiom A5? ( I seriously doubt this)
- Or am I missing something?
If you have not understood my explanation then continue reading.
I'm sorry, I admit I am bad at explaining things, so here is a problem I am encountering. See my document 2, page 4 theorem 2. Proving $A5$ requires a theorem called archemedian property, can you add the proof of Archemedian property, before the beginning of the theorem so that I can complete the proof of theorem 2.
Note:
To talk about the seriousness of the matter, to me right now the whole properties of $\mathbb{R}$ is at stake, especially the universally accepted fact that $\mathbb{R}$ is a field, an ordered one.
Best Answer
The fact that for positive $p,q\in \mathbb{Q}$ and $q<p$ we have an $n\in \mathbb{N}$ which satisfies $nq<p<(n+1)q$ can be proven easily by using the property that $\mathbb{N}$ is unbounded.
The above lemma, which is a corollary of Archimedean property for rational numbers can be proven from the properties of Natural numbers and without referencing to Real numbers.
This lemma plays the role in completing the proof that $\mathbb{R}$ satisfies axiom A5. This saves that proof from being circular.
The presentation of construction of Real numbers from rationals as an appendix in Rudin's book, rather than continuously presenting it throughout the length and breadth of the chapter, may create this confusion.