# First-order Euclidean Geometry

I gather that the axiomatisation of Euclidean Geometry, e.g., by Hilbert, can be used with First-order Logic and has good properties. Is it fairly complete in the sense, say, that it proves everything Euclid did? E.g., Eudoxus' theory of the reals and the area of the circle....

• The interesting thing about the question, of course, is that even Euclid's axiomatization was not strong enough to prove everything that Euclid did. – Carl Mummert Dec 5 '13 at 18:19
• @Carl Mummert That's an interesting comment. Would you elaborate further please? Thank you. – mlchristians Jun 29 at 3:13

## 2 Answers

Hilbert's axiomatization is second-order. However, there is a first-order axiomatization by Tarski, which is complete. (And because the theory is recursively axiomatized, it is decidable.)

We cannot develop the full second-order theory of real numbers within the Tarski geometry. For instance, the "plane" $\mathbb{A}^2$, where $\mathbb{A}$ is the field of real algebraic numbers, is a model of Tarski's geometry. The Eudoxean theory of equality of ratios, which comes so strikingly close to the modern definition of real number, cannot be developed within Tarski's geometry, for the set of integers is not definable in that geometry.

However, for example all of Book I consists of theorems of Tarski's geometry.

For a complete list of Tarski's axioms, with additional information and references, please see this.

• I know of Tarski's axioms, but I'm unfamiliar with them. I read somewhere that you can prove all of Hilbert's axiom groups I-IV using Tarski, but that the group V (continuity) is where Hilbert requires second-order logic. Is that an accurate picture? And what would you add? – rschwieb Dec 5 '13 at 18:03
• Yes, it is the continuity axiom which is the problem. It is replaced by geometric versions of every non-negative number has a square root and every odd degree polynomial has a root. This turns out to be doable using an axiom scheme (an infinite list of axioms with a common structure). – André Nicolas Dec 5 '13 at 18:09
• Thanks :) If not for the internet I would have a severe shortage of conversation time about geometry. – rschwieb Dec 5 '13 at 18:11
• The interesting thing to me is the decidability. There is a royal road to geometry: any king with a sufficiently high computing budget can bypass having to master Euclid's proof techniques. – André Nicolas Dec 5 '13 at 18:15
• Tarski planes are isomorphic to planes coordinatizable by real-closed fields. Any real-closed field is elementarily equivalent to the reals. Euclidean field is a much weaker notion. – André Nicolas Dec 5 '13 at 18:20

No, Pythagorean planes, i.e. planes satisfying Hilbert's fourteen axioms up through the Archimedian axiom), form a strictly weaker geometry than Euclidean planes.

Hilbert planes correspond to constructions with straightedge and "segment transporter," which almost but can't quite do the job of a straightedge and a compass. Because of this, I get the impression that you cannot do as much with circles as you can in Euclidean geometry. You'll notice a distinct absence of circles in Hilbert's Foundations of geometry.

Old and New Results... by Greenberg gives a particularly helpful summary of how they are related.

• It seems from Greenberg's article that "Hilbert planes" only have to satisfy some of the axioms that Hilbert proposed, e.g. the "fourteenth" axiom on page 201. According to p. 203, Hilbert's eventual list of 16 axioms has only one model up to isomorphism, so it surely proves everything that Euclid could. – Carl Mummert Dec 5 '13 at 18:27
• Dear @CarlMummert : Thanks, I forgot that Greenberg uses the term "Pythagorean field" for what I mean, and uses "Hilbert field" for a particular Pythagorean field. I'll make an adjustment. While Hilbert mentioned the completeness axiom in a paragraph at the end of his axiom V section, he promptly said it would not be used, and did not make any use of it throughout the Euclidean propositions he proved. I don't think it is appropriate to include that axiom as one of "Hilbert's axioms." – rschwieb Dec 5 '13 at 20:30
• Thanks for that explanation - this is not an area of expertise of mine, so I have to rely sometimes on the face value of things. – Carl Mummert Dec 5 '13 at 20:50
• @CarlMummert : I appreciate your feedback... thank you! :) – rschwieb Dec 5 '13 at 20:56
• and I'm in Marvin's article. I'm not a complete failure! – Will Jagy Dec 5 '13 at 21:00