Talk:Tarski's axiomatization of the reals

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)[reply]

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)[reply]

Axiom 5

Axiom 5, as stated, does not imply that R is a divisible group. TianDe 21:37, 30 March 2007 (UTC)[reply]

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)[reply]

Axiom 4

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)[reply]

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)[reply]

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)[reply]

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)[reply]

Yeah, that's good. Thanks. Endersdouble 21:50, 1 June 2007 (UTC)[reply]

Format

The axioms are currently formatted using bold fonts like this:

'''Axiom 1'''. "<" is an [[Asymmetric relation|asymmetric]] relation.

'''Axiom 2'''. If ''x'' < ''z'', there exists a ''y'' such that ''x'' < ''y'' and ''y'' < ''z''. In other words, "<" is [[dense order|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|asymmetric]] relation.
;Axiom 2.: If ''x'' < ''z'', there exists a ''y'' such that ''x'' < ''y'' and ''y'' < ''z''. In other words, "<" is [[dense order|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 <dl><dt>…</dt><dd>…</dd></dl> 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)[reply]

It sounds like a good idea. -- EJ 13:00, 1 June 2007 (UTC)[reply]

Field structure

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)[reply]

The almost-homomorphism stuff is actually related, if somewhat tangentally. The trick is that the bootstrapping of multiplication in Tarski's system uses the Eudoxus definition of magnitude, which is also the inspiration for the A-H construction. Therefore it is correct, if somewhat misleading, to say that the A-H reals are a "more elegant" version of Tarski's axioms -- Scott 17:59, 7 July 2007 (UTC)[reply]

I couldn't find any reference to HOW to define multiplication in Tarski's text. Something to do with Eudoxus sounds plausible but it clearly needs a citation to a specific paper that spells it out. Such a paper must exist even though not mentioned in the book. Perhaps Scott can provide? BTW the book does hint at how to define < in an exercise, but not even a hint about multiplication. -- Arthur — Preceding unsigned comment added by 58.175.209.215 (talk) 13:59, 1 February 2012 (UTC)[reply]

PS I suspect from remarks about po-monoids, Eudoxus and Hölder in Birkhoff on Lattice Theory that the proof will be found in Tarski on "Categorical Algebras". I don't have access or time but somebody else might be able to find it in library or with access here:

http://catalog.hathitrust.org/Record/000421833

-- Arthur 58.175.209.215 (talk) 12:41, 3 February 2012 (UTC)[reply]

How Tarski (1994) is possible

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)[reply]

Restored missing first axiom

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)[reply]

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)[reply]

Tarski himself notes that the axiom is redundant two pages later.—Emil J. 11:47, 17 December 2010 (UTC)[reply]

Other citations

Does anyone know any other work by Tarski or others on this system? Specifically, I'm looking for how Tarski defined multiplication in this system. I've got a definition sketched out myself in some private work, but the proofs are getting particularly complicated. Scott (talk) 19:30, 7 July 2014 (UTC)[reply]

I'm also interested in how Tarski defined multiplication in his system. How did you define it? --Andrés — Preceding undated comment added 13:45, 11 August 2014 (UTC)[reply]

I've got

x\cdot y=z

iff

((0<x\leftrightarrow 0<y\leftrightarrow 0<z)\,\wedge \,\forall n\in \mathbb {N} \,\forall m\in \mathbb {N} \,((\sum _{k=0}^{n}|x|<m\leftrightarrow \sum _{k=0}^{n}|z|<\sum _{k=0}^{m}|y|)\wedge (\sum _{k=0}^{n}|x|=m\leftrightarrow \sum _{k=0}^{n}|z|=\sum _{k=0}^{m}|y|)\wedge (\sum _{k=0}^{n}|x|>m\leftrightarrow \sum _{k=0}^{n}|z|>\sum _{k=0}^{m}|y|)))

. The first bit guarantees that the sign of

z

is correct, while the second part sets magnitude. For that element, see Eudoxus' definition. Note that this also only holds for non-zero

x

and

y

. I've special-cased those to make

z=0

. Scott (talk) 15:06, 20 October 2014 (UTC)[reply]

Why is Eudoxus' definition necessary? I mean, I was trying to use a more intuitive approach, first defining inductively

k\cdot A

for

k\in \mathbb {N}

and for

A\in \mathbb {R}

. Then, as Tarski pointed out, we can prove using Axiom 3 that

\mathbb {R}

is divisible, so we can define

{\frac {m}{n}}\cdot A

for

m,n\in \mathbb {N}

. Thereby

C\cdot A

can be seen as the limit of the previosly defined operation. - Andrés 11:29, 4 November 2014 (UTC)[reply]

The problem there is that I'm not sure how to prove that all reals are the limit of a sequence of rationals without using multiplication. Any ideas on a proof of that? Scott (talk) 20:13, 15 November 2014 (UTC)[reply]

Yes, I have some ideas:

1. Multiplication for positive rational numbers can be defined as I indicated above , so we can extend it and to prove that Q is an order field with that. To pass from Q with all the known properties, plus dedekind completeness, to R, is the problem discussed in the construction of the reals. We can use the definitions of any of those constructions.

2. Asking Stefanie Ucsnay.

Andres 20:08, 19 November 20114 (UTC)

Proof

Since $\mathbb {R}$ is a divisible group, $\mathbb {R}$ is a $\mathbb {Q}$ -vector space, and left and right scalar multiplication by elements of $\mathbb {Q}$ is possible in $\mathbb {R}$ . Since $1<1+1$ , $\mathbb {R}$ is not a trivial vector space, and $\mathbb {Q}$ is a $\mathbb {Q}$ -vector subspace of $\mathbb {R}$ .

Since $\mathbb {Q}$ is a field, one could construct the square function $f(x)=x^{2}$ as a partial function restricted on the domain and range to $\mathbb {Q}$ . Since $\mathbb {R}$ is Dedekind complete, the intermediate value theorem is true and so given any non-negative element in $\mathbb {R}$ , the function $f(x)=x^{2}-r$ has a zero, and $f(x)=x^{2}$ is defined to have a zero at $x=0$ . This means that the square function is defined on the entire domain of $\mathbb {R}$ . We define the multiplication operation over the real numbers $x\cdot y$ to be $x\cdot y={\frac {1}{2}}(x+y)^{2}-{\frac {1}{2}}x^{2}-{\frac {1}{2}}y^{2}$ , with canonical embeddings of the left and right scalar multiplication into $x\cdot y$

We could do the same thing with every odd power $f(x)=x^{2n+1}$ for natural number $n$ . Since the intermediate value theorem is true and $f(x)=x^{2n+1}$ is a monotone function for every natural number $n$ , the function $f(x)=x^{2n+1}-r$ has a zero at every real number $r$ , and thus defined on the entire domain of $\mathbb {R}$ , and even powers are defined as $f(x)=x^{2n}={(x^{n})}^{2}$ .

Since the real numbers are Dedekind complete, they are Cauchy complete and sequentially Hausdorff, and one could construct unique limits of Cauchy sequences. In particular, geometric series are Cauchy sequences and the limit of the geometric series $\sum _{n=0}^{\infty }x^{n}$ and $\sum _{n=0}^{\infty }(-1)^{n}x^{n}$ converges in the open interval $(-1,1)$ , and thus let us define functions $f:(-1,1)\to \mathbb {R}$ and $g:(-1,1)\to \mathbb {R}$ as

$f(x):=\sum _{n=0}^{\infty }x^{n}$

$g(x):=\sum _{n=0}^{\infty }(-1)^{n}\cdot x^{n}$

Let us define functions

$f_{2}(x):=\lim _{a\to 0^{-}}(-a)\cdot \sum _{n=0}^{\infty }a^{n}\cdot (x+f(a+1))^{n}$

and

$g_{2}(x):=\lim _{a\to 0^{+}}a\cdot \sum _{n=0}^{\infty }(-a)^{n}\cdot (x+g(a-1))^{n}$

The reciprocal function is piecewise defined as ${\frac {1}{x}}={\begin{cases}f_{2}(x)&{\text{if }}x<0\\g_{2}(x)&{\text{if }}x>0\\\end{cases}}$

All that's left to do is to prove that multiplication is a monoid with unit 1 and distributes over addition (since it is already commutative as defined with an absorbing element 0) and the reciprocal function is the multiplicative inverse function on the domain $\mathbb {R} /\{0\}$ .

Axiom 3

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)[reply]

Something wrong in one of the proofs

The article gives this proof:

Proof

Since every Cauchy net in $\mathbb {R}$ converges to a unique element of $\mathbb {R}$ , for every directed set $A$ and Cauchy net $(a_{i})_{i\in A}$ in the rational numbers, there exists a Cauchy net of linear functions $(f_{i})_{i\in A}$ defined as $f_{i}(x)=a_{i}x$ . The limit of the Cauchy net $\lim _{i\in A}(f_{i})_{i}$ exists and is a unique function $g(x)=\lim _{i\in A}(a_{i})_{i}x$ . Since every real number is the limit of a Cauchy net of rational numbers, there is an $\mathbb {R}$ -action $\alpha :\mathbb {R} \to (\mathbb {R} \to \mathbb {R} )$ which takes a real number $r$ to the linear function $x\mapsto rx$ , with $\alpha (1)=x\mapsto x$ being the idenitity function. The uncurrying of $\alpha$ leads to a bilinear function $(-)(-):\mathbb {R} \times \mathbb {R} \to \mathbb {R}$ called multiplication of the real numbers

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)[reply]