# How to state Pythagorean theorem in a neutral synthetic geometry?

In some lists of statements equivalent to the parallel postulate (such as Which statements are equivalent to the parallel postulate?), one can find the Pythagorean theorem. To prove this equivalence one has first to state the pythagorean theorem in neutral geometry (I name 'neutral geometry' a geometry in which parallel lines do exist but with the parallel postulate removed).

If one start with an axiom system like Birkhoff's postulates which assume reals numbers and ruler and protractor from the beginning then there is no problem stating the Pythagorean theorem.

My question is how one can one state the Pythagorean theorem in a neutral synthetic geometry based on axioms such as Hilbert's axioms group I II III or Tarski's axioms $A_1-A_9$ ?

It is possible to define segment length in neutral Tarski's or Hilbert's geometries as an equivalence class using the congruence ($\equiv$) relation. It is also possible to define the congruence of triangles.

However, the geometric definition of multiplication as given by Hilbert assume the parallel postulate. The existence of a square is equivalent to the parallel postulate.

• What is "neutral synethic geometry"? In particular, does it contain a notion of segment length and angle measurement? And if so, why can't you just then simply state the Pythagorean theorem in the ordinary way? – Lee Mosher May 21 '15 at 16:50
• @LeeMosher I edited the question. Neutral geometry is the geometry without the parallel postulate. One can define the segment length as the equivalent class using the congruence of segments. But to state that $a^2 + b^2 = c^2$ you can not use the multiplication because the usual definition of multiplication use the parallel postulate. – Julien Narboux May 22 '15 at 7:29
• How about area measurement, does that exist in your system? If so, then perhaps, without the parallel postulate, one can define and construct the geometric square on a given side, i.e. a regular quadrilateral. And then use the areas of squares to state the Pythagorean theorem. It will be false, of course, if the parallel postulate fails, but you can state it. – Lee Mosher May 22 '15 at 13:47
• Victor Pambuccian pointed me to the following note which give a partial answer to the question: link.springer.com/article/10.1007/s00283-010-9169-0 – Julien Narboux Feb 21 '17 at 18:43
• This link shows that Pythagorean Theorem implies the parallel postulate: cut-the-knot.org/triangle/pythpar/PTimpliesPP.shtml – ray lin Jun 28 '18 at 17:56

## 1 Answer

Within synthetic approach for geometry (such as Hilbert's axioms) one can define the notion of segment measure:

Definition. Denote the set of segments by $$\mathcal{S}$$. We say that a function $$\mu:\mathcal{S}\rightarrow \mathbb{R}$$ is a segment measure whenever:

1. $$\mu(ab)>0$$.
2. $$ab\equiv a_1b_1$$ $$\implies$$ $$\mu(ab)=\mu(a_1b_1)$$.
3. $$B(abc)$$ $$\implies$$ $$\mu(ab)+\mu(bc)=\mu(ac)$$.

$$B(abc)$$ stands for "point $$b$$ lies between $$a$$ and $$c$$", which is a primitive notion.

The definition involves real numbers but they are regarded independently from the theory of geometry.

The following two theorems serve to answer the question about the Pythagorean theorem:

Theorem 1. If $$\mu,\mu_1$$ are segment measures, then there exists $$\lambda>0$$ such that $$\mu=\lambda\mu_1$$

Theorem 2. There exists a segment measure $$\mu$$ (In fact it can be chosen in such way that $$\mu(ab)=x$$ for arbitrary segment $$ab$$ and $$x>0$$).

Asssume we are given a triangle $$\triangle abc$$. Combining these two theorems we see that we can state $$\mu(ab)^2=\mu(bc)^2+\mu(ac)^2$$ and prove that its logical value does not depend on the choice of the measure since assuming above sentence is true for some measure $$\mu$$, then for arbitrary measure $$\mu_1$$ we have $$\mu=\lambda\mu_1$$ for some $$\lambda>0$$ and $$\lambda^2\mu_1(ab)^2=\lambda^2\mu_1(bc)^2+\lambda^2\mu_1(ac)^2$$ $$\mu_1(ab)^2=\mu_1(bc)^2+\mu_1(ac)^2$$

For the proof of theorems 1 and 2 see "Foundations of geometry" by Borsuk and Szmielew.