So my last post didn't really answer your call for a single first-order sentence.  But you can take the conjunction of sentences (1) and (2).  Neither Q nor C satisfies both of them, but R does.

I kon't know how enlightening this is but R is the only one of the three that is real closed; i.e., satisfying the conjunction of : (1) a^2+b^2=0 implies a=b=0 (formally real); (2) for every a, either it or its negative is a square; and (3)every odd degree polynomial has a root.   (The last condition, of course, is an infinite list of first-order sentences:  Ax_0...A_x_n Ey( x_0+x_1y+x_2y^2+...+x_ny^n=0), n=1,3,5,...)

So my last post didn't really answer your call for a single first-order sentence.  But you can take the conjunction of sentences (1) and (2).  Neither Q nor C satisfies both of them, but R does.

Right Paul. But Q is also formally real, so we can forget (1). An example inspired by axioms (3) is "x^3 = 2 has no solution or three solutions" [optionally: "x^3 = 2 has no solution or has at least two different solutions"] because in R it has exactly one solution. An example inspired by axiom (2) is "there is an element, which is not 0, such that (both x and -x are squares) or (both x and -x are not squares)". We are still facing examples which are disjunctions like [(rational behavior) or (complex behavior)], but as they come from axioms, they are already more natural as my original example. Thank you. It is still interesting if we can find examples which are not this kind of disjunction.

It would be interesting just how much you can restrict the syntax and make the question more precise.  Here's a wild conjecture:  There is a sentence S, the universal closure of a conjunction of atomic and negated atomic formulae in the language of fields, such that both Q and C satisfy S, but R does not.

Paul, purely universal sentences are true in all substructures. Also, purely existential sentences are true in all overstructures. So because Q \subset R \subset C, a purely universal sentence true in C will be true also in R and Q, and a purely existential sentence true in Q will be also true in R and C. So we must look for sentences given by some more complicated prefix (at least AE or EA) followed by a conjunction of negated or positive atomic formulae. (also our examples so far have more complicated prefixes...)

Yes Paul, and all other examples are AE...

I made another step about pure existential and pure universal sentences. Let Field_0 be the theory of fields of characteristic 0. All fields are substructure in a ACF and universal sentences are inherited by substructures. It follows: a universal sentence is provable by Field_0 iff it is true in some ACF iff it is true in C. Also, all fields of characteristic 0 contain Q and purely existential sentences are inherited by extensions. So a pure existential sentence is provable by Fields_0 iff it is true in Q.

Sentences true in Q and C but not in R of the form PREFIX (any complexity) followed by Conjunction of positive and negated atomic formulae. This is a difficult question. I really would like to see such a sentence...

Hi Mihai,  Admittedly I was a bet reckless in my "conjecture" above.  The electrical engineers call a conjunction of atomic and negated-atomic formulas a minterm.  So the not-so-wild conjecture is that there is a sentence S--of the form "quantifier prefix + minterm"--such that S is true for the fields Q and C, but not for the sandwiched-in field R.   

The problem of your example is what is 2? Namely, there is a theorem which says

that the first order theories for Q and R are equal. However, maybe we can enrich

the language of fields with constants!?

Dear Sinisa,

1. The symbol 2 is a notation for the logical term (1+1). The language of  fields is L = {+, -, . , 0, 1} and this term can be always built using the constant "1" and the operation +.

2. The first order theories of Q and R ARE NOT EQUAL. The most easy example is that in R the sentence Ex x^2 = 2 is true and in Q the same statement is false. On the other hand all examples given in this discussion thread are true in Q and false in R. Equal first order theories means that they satisfy the same first order statements, which is not the case.

3. If we enrich the language of R with new constants from R which cannot be interpreted in Q, we cannot speak anymore about this problem.

Thank you for the interest in this problem.

Yes, I was wrong. Check this out

(there is x)(for every y)(xy=y and (there is z)zsquare=x+x)implies(for every u)(there is v)

vsquare=u). This should be OK. Regards.

I see, Sinisa. You want to formalize the sentence """If 2 has a square root, then all elements have square roots.""" without making use of constants. This is surely possible because the constants are definable. I don't know if your form is really the translation. But a translation would be surely the following:

As [   (At st = t) ---> [(Ex x^2 = s + s) ---> (Au Ev v^2 = u)]   ]

In other words I believe that the right translation must be with universal quantifier in front of the prefix. However in this particular situation, due to the unicity of 1, it might really mean the same. Thank you very much.

It seems easy to prove that your formula is equivalent to mine. Regards.

Is there a result in mathematics  which is true in Q and C but not in R ? Give concrete example ? 

A possibility is to state by a sentence in the first order logic that "a

Dear Antonin,

thank you very much for this answer. This answer brings us closer to what we want, but still not completely. Namely, it still contains both conjunctions and disjunctions, and its negation (which is true in Q and C but false in R) will also contain both. But the idea is very nice, makes mathematical sense and is maybe the best example given here so far - because has more mathematical content than the other examples. 

The question to find an example given by any prefix followed by a conjunction of negated and not negated atomic formulae to be true in Q and C and false in R is still open in this discussion thread.

Dear Mihai, I didn't get exactly the question. Then you can discriminate R from Q and C by saying that each element or(exclusive) its opposite has a square root. But it is not so different from your original construction.

The following sentence should do the trick:

A x E y

x=0

OR

(x=y*y) and not (-x = y*y)

Or

(not x = y*y) and (-x =  y*y)

Antonin, you write:

Ax E y [ x = 0 \/ (x = y^2 /\ x =/= -y^2) \/ (x = - y^2 /\ x =/= y^2)]

FIRST REMARK: Read it carrefully. This HAPPENS in C! Every non-zero element is a square and every non-zero element is minus a square! And moreover if something is not 0 and equal y^2, it will be of course different from -y^2, excepting for characteristic 2, which is not the case.

SECOND REMARK: Maybe you wanted to write:

Ax [x=0 \/ (Ey x = y^2 /\ Az x=/= -z^2) \/ (Ey x = -y^2 /\ Az x=/= z^2)]

OK. This separates R from Q and C indeed, but is still a mix-up of conjunctions and disjunctions, as already your formula. We want in prenex normal form a prefix followed by a conjunction of positive and negated atomic formulas to be true in Q and C and false in R - or followed by a disjunction of atomic formulas to be true in R and false in Q and C. OK, the other way around will be also interesting. But all examples so far had both disjunctions and conjunctions. 

Ah, you are right, I wrote too fast!

I am note sur to understand why you want an only conjunction sentence. But you can try this (written in a loosy way):

E a E b E c

A t A u A v A w

E y A z

t + u y + v y^2 + w y^3  = 0

and

a + b z + c z^2 =/= 0

(in words : there is a polynomial of degree 2 that has no root, but all degree 3 polynomial have a root).

I do not know if it feels natural to you!

Antonin

Dear Mihai:

Quoting from your answers:  "We want in prenex normal form a prefix followed by a conjunction of positive and negated atomic formulas to be true in Q and C and false in R"

The answer is yes, such a formula exists. The reason is that any sentence can be written this way (standard triks). In fact, can be written (in any field) as a prefix followed by a conjunction of  polynomials=0 (in many variables).

As an example, your sentence is equivalent (in any field)  to the following:

 Ax  Au  Ev [  ( (x^2-2)v-1 )(v^2-u)=0  ]

Rafel

Dear Antonin,  this sentence is true in R and false C and Q. It contains a conjunction. The negation, true in C and Q and false in R, contains a disjunction. It is almost well. thank you.

Dear Rafel,

I think that your post closes for the moment the discussion thread. Indeed, if you have functions in the language, terms can become very complicated and every polynomial equation is an atomic formula! Very right! Your formula is really equivalent over all fields with my sentence. This shows that we cannot always impose "non-trivial content" using syntactic restrictions. And "we cannot always" is just a sequence of words. In most cases we cannot do this. 

Mihai, Here is an idea based on your sentence in the original post.

Let "P" denote the sentence

P: Ax((Ey y^2=x)-> (Ez z^3=x))

That is P: Every square is also a third power.  P is true of the reals, but not true of the rationals.

Let "Q" denote the sentence

Q: AxEy y^2=x

That is Q: Everything is a square.  Q is true of the complex, false for rationals and reals.

Therefore the sentence P->Q satisfies your question, and avoids the "what does 2 mean" criticism made in earlier comments.

 Of course, in Q the premise is false, BUT conclusion : (Au Ev v^2 = u) also is false in Q. now we have  : false ---> false. on the Truth of  logical connectivity for " ---> " , the sentence is false in Q

No Maryam. Please check again. (false ---> false) is always true. Only (true ----> false) is false.

It's a classical mistake with implication. I always taught my students that implication functions as a promise: if P, get Q. The only way a promise P->Q can be broken occurs when the condition P holds and the promised Q doesn't happen.

Oh my God ! what a terrible mistaken ! You're right

this is a bit strange, coz at least Q is a subfield, and it inherits it's properties form R, How a set that is dense in the greater set, can be richer than ! ( algebraically )

Dear Maryam, the theory of Q is neither richer nor more poor than the theory of R. For any existing structure it is true that, given a sentence psi, exactly one of psi and not psi is true in the structure. So half of all possible sentences are true in Q and the other half are false. The same happens also for R. But it is just not the same half which is true for Q. The half for R is cut differently. 

For example it is sure that in Q one of the following sentences is true and the other one is false:

Statement A:

(E x, y, z    2 z^12 xy + 123 = 17 xy^2 z^3 ) \/ ( A x, y   54 x^7 z + 12345 =/= 54321)

Statement B:

(A x,y,z 2 z^12 xy + 123 =/= 17 xy^2 z^3) /\ (E x,y 54 x^7 z + 12345 = 54321)

Statement B is the negation of statement A. So I know that in Q as also in R one is true and the other is false. Just that right now I don't know which of them is true, and in which structure... :-)  .-) :-)

Yes,  you're right. so my mistaken is that I don't have a right to compare structures R and Q " Model theoretically ", because every L-structure has it's terms and so has  it's sentences, so  to compare them on based of number true or false sentences is wrong (and call rich or poor). but I think R is very richer ! than Q in terms of "Analysis or Algebra". because for example  R is ACF,... or contains an uncountable set with zero measure ( cantor),  and so on. . almost every strange set lives in R, so I thought R is richer than !   I mybe Model theorists compare richness or poverty of a structure in study of  realization of types ,  in number of realized types ?

Dear Maryam,

R is not ACF (algebraically closed) but RCF (real closed). This means an ordered field such that every positive element is a square and every polynomial of odd degree has a root. ACF like the complex numbers are not ordered fields, but just fields. So we speak about different languages.

The other things like Cantor set, etc... cannot be described in terms of "first order logic" because in order to define them, one must quantify not only elements, but subsets. So they cannot be taken as arguments here.

There is however something which can allow us to say that Q is richer than R.

The first order theory of Q is undecidable (Julia Robinson) while the first order theory of R is decidable (Alfred Tarski). This means that when we study Q we need an infinity of non-trivial new ideas in order to prove things, while when studying R there is a mechanical decision procedure to establish the truth of some first order sentence (like those written by me in my last postation).

So in this sence it is really true that Q is richer than R.

I had another answer here were I wrote that Q is richer than R because has an undecidable theory. while R has a decidable theory, so the problems to solve when doing mathematics for Q are much more difficult. We need an infinity of non-trivial ideas for solving problems about Q. My answer disappeared from RG, but OK, it was more or less exactly this.

Another point is that most things which make R difficult do not depend of first order logic. Like the Cantor set, or the Vitali set, etc... All those constructions use quantifications over subsets, so they are not first order. They become first order when doing set theory, but then are treated in a different way.

Well, the question of realized types is itself something like this. It is a question ABOUT models of first order theories but is not itself a first order property, not in most of the case. If we think about Dedekind cuts, nobody realizes so many types as R does. That's sure! But this does not make the 1-order theory richer, just the model.

Oh,  sorry, another mistake, Yes R is not RCF, cause fundamental theorem of algebra doesn't hold in it. I tried to compare them in Analysis, any way. Really, I didn't know " to be decidable" make richer a theory. Now if we axiomatize R in higher order language, is there possibility to say R is richer than, cause there will be many expressible sentences, then more true sentences in R that don't hold in Q ? by the way, you compare them in ordered fields language, isn't it?

what was your main idea about type? you meant  the notion of " type" depends on model, not theory. so it can not be an index to show richness of a theory, yes?

No, the thesis was that "to be decidable" makes the theory poorer, not richer. If there is a machine able to decide the truth, than you, as mathematician, has not so much to do anymore. And this happens for R but not for Q. ACF means algebraicaly closed. So is for example C. R is real closed (RCF). It is not so important if you compare R and Q as ordered fields or just as fields because: (1) they both have just one possible ordering and (2) this ordering is definable in the language of fields in both of them. Moreover, in R: x \geq 0 iff x is a square and in Q: x \geq 0 iff x is sum of four squares. The definition of order true in Q defines also the unique ordering of R, if applied to R. So you can compare them just as fields, it does not make any difference!

Thank you about ordering description, perfect..... of course I think we should define what does it mean exactly : to be richness of a theory,  let define: in same language, we say  theory T_1 is richer than theory  T_2, if it  contains  much expressible and ture sentences, and much means cardinal. so R should be richer in higher order logic?

 I found a definition by chance ! : let $ \ Sigma $ be a set of sentences, $ \ Sigma $ is rich, if it contains a sentence of the form $ \ E x phi (x) $ , then there is a constant symbol c in L such that  $ \ phi (c) $ is in $ \Sigma $ .   

Hello Maryam. I did not look at this site for longer time. You are starting a discussion about richness of theories. Well, my use of the term was extensive. I just said something like "rich in ideas" "rich in difficulties", and so on. You found a definition which seems to mean just "rich in constants". This particular kind of richness is done very easily in every model. Just take a constant (one says "a name") for every element of the model. Then you have a lot of constants, and in particular this kind of richness is fulfilled.

Decidable theories may "feel" as poor ones because some algorithm decides it all. In practice, however, the decision process is usually extremely complex. This has been shown a.o. for Pressburger Arithmetic (addition and equality only). See link.

https://en.wikipedia.org/wiki/Presburger_arithmetic

Thank you Marcel for this remark. I am impressed by one particular paper that one can freely download there (in Wikipedia). It is:

http://www.sciencedirect.com/science/article/pii/0022000078900211/pdf?md5=0415089a2d692fcece18b43b5f63c67d&pid=1-s2.0-0022000078900211-main.pdf

In this paper the author Derek Oppen proves a bound like 2^2^2^(pn) for the decision procedure in Pressburger Arithmetic and explains why (most probably) this cannot be improved. It is a nice reading for everyone!

Hi Dr. Yes you're right, in fact that's  a construction model  by constants. however I meant , I have n't seen the notion of " rich" exactly, before. but I found a definition in the text. and I was interesting for me , when it is" added constants" to model, it is applied this word for it's theory. by the way I saw it in "A Guide to Classical and Modern Model Theory" written by Annalisa Marcja.

No problem Maryam. adding constants is one of the most old and nice methods in Model Theory. It plays an important role also for the upwards Löwenheim-Skolem theorem.

Your examples are not the propositions of the theory of first-order predicates, they are propositions of propositional logic. You take propositions which logical value (true or false) are proved in real and complex analysis, rather than in theory of first-order predicates, and demonstrate the properties of implication.

Comme on, dear Frolov. You cannot say that Rafel's example:

Ax Au Ev [ ( (x^2-2)v-1 )(v^2-u)=0 ]

does not belong to the predicate calculus. The method that I presented in the question is so primitive just because Th(Q), Th(R) and Th(C) are very different, and we have no better idea. But exactly this is the scope of the question: what better idea do you have to produce examples?

Dear Mihai, thank you, you're right, I'm sorry, I hurried.