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I state that I am not an expert in formal systems, but I did something as a "self-taught" because I am passionate about it. My question is the following: since geometry hilbert axioms are equivalent to the real numbers axioms from the primes one must deduce the natural numbers, the induction theorem, etc. How you do it? If the following is not wrong we must define segment multiples in the same way that we define natural numbers from real numbers axioms. Premise: as did hilbert in his axioms, since the ancient Greeks used the language of logic (and not that of sets) we use the second order logic language. In this way the first step to introduce the multiples of AB is to define the analogue of the notion of inductive set, i.e. the inductive properties (with respect to the segment AB):

Definition 1 A segments property P is inductive with respect to segment $AB$ when:

  1. P$(AB)$
  2. $DE \cong AB$ $\wedge$ Bet$(C,D,E)\to$ (P$(CD)\to$ P$(CE)$)

where Bet is the betwennes relation.

Exemple 1 The relation $...\geq AB$ is inductive with respect to $AB$.

Proof. in fact $AB \geq AB$ so that the 1 of definition 1 is verified. In addition let C, D, E such that $DE \cong AB$ $\wedge$ Bet$(C,D,E)$. If $CD \geq AB$ then $CE \geq AB$. Thus also 2 of definition 1 is verified. QED

we are now ready for:

Definition 2 ($AB$ multiple) A segment $RS$ is a multiple of $AB$ if for every inductive property P we have P$(RS)$.

The definition given above is the analogue of the definition of natural numbers as the intersection of every inductive sets of R.

Rem. 1 $AB$ is a multiple of itself. In effect for 1 of definition 1 for every P inductive we have P$(AB)$. For definition 2 we have the assert. QED.

Rem.2 The property M(to be a $AB$ multiple) is inductive.

Proof. For Rem.1 we have M$(AB)$. Let $DE\cong AB$ and Bet$(C,D,E)$. If M$(CD)$ for every P inductive we have P$(CD)$. Furthermore, since $DE\cong AB$, one must have also P$(CE)$ for 2 of definition 1, in other words M$(CE)$ QED.

Conseguences of of above facts are:

1 $AB$ is multiple of $AB$

2 If $DE\cong AB$ , Bet$(C,D,E)$ and M$(CD)$ then M$(CE)$

3 If M$(CE)$ and Bet$(C,D,E)$ and $DE\cong AB$ then $CE \ncong AB$

4 If M$(CD)$ and M$(PQ)$ and Bet$(C,D,E)$ and Bet$(P,Q,R)$ and $DE\cong QR \cong AB$ then $CD \cong PQ$

5 (Induction Theorem) Let P a property. If there exist a segment $PQ$ such that M$(PQ)$ and P$(PQ)$ and if:

  • P$(AB)$
  • $DE\cong AB$
  • Bet$(C,D,E)$
  • P$(CD)\to$ P$(CE)$

then P$(HK)$ for every $HK$ such that M$(HK)$.

The problem is that in the geometric proofs, for example Euclid Segement division when the two segment are supposed commensurables , the division of a polygon in n triangles etc.the recursion theorem is not the above (in terms of segment multiples) but it is in terms of integers. Furthermore in the geometric definitions (polygon for example) the induction definition is not the above (in terms of segments multiples) but is in terms of integers. it is therefore necessary "to translate the integers in terms of multiples of a segment". how to do this?

user273366
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    On the one hand, you need the natural numbers already as a given "tool" in Hilbert's Axioms, specifically, the Axiom of Archimedes. On the other hand, the language of the axiomatized theory does not even allow us to (easily) express "This polygon has as many diagonals as this line segment fits repeatedly into that line segment". – Hagen von Eitzen Jul 17 '21 at 13:05

1 Answers1

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Thank you Hagen von Eitzen for your answer. We can express easily the archimede's axiom without referencing natural numbers: "$\forall PQ$ there is $UV$ such that M$(PR)$ (where M is the relation to be $UV$ multiple) and that Bet$(P,Q,R)$".

Hint: when we write $AB = 3 UV$ in geometry it does not mean multiplication between number 3 and the segment $UV$. It's a symbol that means $AB = AH + HK + KB$ where $AH\cong HK\cong KB\cong UV$. It's a recursive definition that we can get using my question's arguments above: i.e.

Def

  1. $1 UV$ =$UV$
  2. $(n+1) UV$ = n$UV$ + $UV$

I want to treat segment lenght not with the segment algebra introdused by Hilbert with Pascal's Theorem, but only in terms of multiples of segments (in the case of segments commensurables). I would like to treat also incommensurability in terms of these arguments (like the eudoxus proportion theory)(evidently the axioms of continuity are also needed )

Hilbert axioms and real numbers axioms are equivalent and therefore we must not add nothing else in geometry for proof the equivalent statements in geometry and with real numbers. Only "the way we express ourselves must change".

user273366
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