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:

*P*$(AB)$
- $DE cong AB$ $wedge$
*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 $DEcong AB$ and *Bet*$(C,D,E)$. If *M*$(CD)$ for every *P* inductive we have *P*$(CD)$. Furthermore, since $DEcong AB$, 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 $DEcong AB$ , *Bet*$(C,D,E)$ and *M*$(CD)$ then *M*$(CE)$

3 If *M*$(CE)$ and *Bet*$(C,D,E)$ and $DEcong AB$ then $CE ncong AB$

4 If *M*$(CD)$ and *M*$(PQ)$ and *Bet*$(C,D,E)$ and *Bet*$(P,Q,R)$ and $DEcong QR cong AB$ 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*$(AB)$
- $DEcong AB$
*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?