Talk:Tarski's axiomatization of the reals

message: Quoting from the article:
The Axioms of:
* Order imply that R is a dense set and a total order;
::The axioms of order imply no such thing. The empty relation satisfies all of them. Or, for a less trivial example, the product order on R×R satisfies the order axioms, but is not total.
*Addition imply that R is an Abelian group under addition;
::The axioms of addition imply no such thing. Any associative right quasigroup, with < defined as the empty relation, satisfies the axioms. It does not have to be commutative, it does not even have to be a group. -- EJ 20:33, 6 August 2006 (UTC)
replies: text: Checking with the original revealed that the problem is in fact deeper, as the real Tarski's axiom 4 is not just associativity, it has commutativity built-in, so to speak. Fixing that. -- EJ 17:59, 9 August 2006 (UTC)

message: Axiom 5, as stated, does not imply that R is a divisible group. TianDe 21:37, 30 March 2007 (UTC)
replies: text: Indeed, one needs more or less all the axioms together to show divisibility. Namely, the axioms imply that R is a dense Dedekind complete ordered Abelian group, and any such is easily seen to be divisible. -- EJ 10:48, 2 April 2007 (UTC)
thread_title:  Axiom 5 

message: Axiom four looks weird.  Is it correct?  It looks like it could either be a typoed statement of associativity--my first guess--but perhaps something more complex which I'm not immediately getting.  Can someone who knows better/has access to the papers on this confirm the correctness of that form?  Thanks. --Endersdouble 03:03, 24 May 2007 (UTC)
replies: text: The axiom is correct. It is not weird, it is clever. It sort of combines associativity and commutativity into one axiom (though by itself it does not imply either, you need at least axiom 5 as well). This kind of concise axiom optimization is in fact the whole point of Tarski's axiomatization. -- EJ 08:32, 24 May 2007 (UTC)
replies: text: That was one thought of mine--looked like some sort of combination.  Didn't know enough about the formalism to be sure, however.  Thanks for assuaging me.  (Should something be added to the page to explain this?) -- Endersdouble 16:57, 24 May 2007 (UTC)
replies: text: I have put a hint to that effect in the lead section. Do you think it's better now? -- EJ 12:12, 1 June 2007 (UTC)
replies: text: Yeah, that's good.  Thanks. Endersdouble 21:50, 1 June 2007 (UTC)
thread_title:  Axiom 4 

message: The axioms are currently formatted using bold fonts like this:
Axiom 1. "<" is an asymmetric relation.
Axiom 2. If x < z, there exists a y such that x < y and y < z. In other words, "<" is dense in R.
Axiom 1. "<" is an asymmetric relation.
Axiom 2. If x < z, there exists a y such that x < y and y < z. In other words, "<" is dense in R.
I would prefer to format them as definition lists, like this:
;Axiom 1.: "<" is an asymmetric relation.
;Axiom 2.: If x < z, there exists a y such that x < y and y < z. In other words, "<" is dense in R.
;Axiom 1.: "<" is an asymmetric relation.
;Axiom 2.: If x < z, there exists a y such that x < y and y < z. In other words, "<" is dense in R.
This Wikitext gets translated into definition lists …… which I think is more of a semantic markup and presumably would be easier to parse for screen readers and the like. — Tobias Bergemann 12:18, 1 June 2007 (UTC)
replies: text: It sounds like a good idea. -- EJ 13:00, 1 June 2007 (UTC)
thread_title:  Format 

message: I have finally taken a look on the three papers mentioned in the last paragraph of our article, and it has confirmed my deep suspicion that this "almost homomorphism" business is completely off-topic. What they do is to give an alternative construction of the field of real numbers from scratch (or rather, from integer sequences), whereas what Tarski needs to do is to show that any structure satisfying his axioms can be uniquely expanded by a multiplication operation which makes it an ordered field, which is a quite different problem. (Well, in fact, one could construct "Tarski multiplication" in a rather indirect way by taking a separately contructed field of real numbers, and showing that all pointed densely completely ordered Abelian groups are isomorphic. But then it is irrelevant how the other reals were actually constructed.) IMO the almost homomorphism stuff does not belong here but to the construction of real numbers article, and, surprise surprise, it is already there. (Another issue is that the construction appears to be due to Stephen Schanuel, not the three people mentioned.)
I thus intend to remove the second paragraph in the "How these axioms imply a field" section of the article, but I ask here first, in case somebody knows better. -- EJ 14:00, 6 June 2007 (UTC)
thread_title:  Field structure 

message: The sole reference for this entry is Tarski's text, published in 1994. Someone has asked how that could be possible, given that Tarski died in 1983. Answer: 1994 is the publication year  of the Dover reprint of a book Tarski published in German in 1936 and in English in 1941 and 1946.132.181.160.42 04:56, 14 September 2007 (UTC)
thread_title:  How Tarski (1994) is possible 

message: This page originally listed only 8 axioms: Tarkski's Axiom 1' (on p. 214 of his book), which says that the order is total, had been omitted for some reason.  As far as I can tell, this axiom really is necessary (as one expect, given that Tarski, who was not one to tolerate redundancy, included it): the set of ordered pairs of real numbers, defining (a,b) < (c,d) iff a<c, (a,b)+(c,d) = (a+c,b+d), 1 = (1,0) are a model for the other axioms.  —Preceding unsigned comment added by 129.67.119.242 (talk) 22:27, 16 December 2010 (UTC)
replies: text: Your model does not satisfy the Dedekind completeness axiom: Take X = {(a,b) : a ≤ 0} and Y = {(a,b) : a > 0}.—Emil J. 11:37, 17 December 2010 (UTC)

text: Tarski himself notes that the axiom is redundant two pages later.—Emil J. 11:47, 17 December 2010 (UTC)
thread_title:  Restored missing first axiom 

message: The claim "The three axioms imply that R is a linear continuum." is very suspicious. If it tries to say that Axioms 1, 2 and 3 alone (using only the usual set-existence principles on top of them) imply that R is a linear continuum, I believe it is incorrect. In the Stefanie Uscnay reference "A Note on Tarski's Note" all axioms 1-8 except 3 were used in proving that (R,+) is an Abelian group before getting to prove that _at least one of the three claims a<b, b<a and a=b holds_. (Proving that _at most one_ of these three claims holds using only axioms 1 and 2 is hunky dory.) If I am wrong and there in fact is a proof of the trichotomy of R w.r.t. < using only 1-3, I would be glad to be corrected. Lapasotka (talk) 15:32, 12 October 2022 (UTC)
thread_title:  Axiom 3 

message: The article gives this proof:
This fails for at least two reasons: 
* we haven't yet established that every real number is the limit of a Cauchy net of rationals;
* a complete metric on the set of \mathbb{Q}-linear functions from \mathbb{R} to \mathbb{R} has not been given, so talking about limits of Cauchy nets of such functions is premature. 
AxelBoldt (talk) 02:33, 13 December 2023 (UTC)
thread_title:  Something wrong in one of the proofs