From: Mike Deeth Subject: Re: Hilbert's Geometry Date: Tue, 12 Jan 1999 01:26:31 -0600 Newsgroups: sci.math Keywords: Hilbert's Axioms for geometries Jim Nastos wrote: > Hello, > I remember once reading about Hilberts axiomatization of geometry, in > which his first few propositions were things like "there exists at least > 3 points" and "there exists at least two lines" ... and continued in > that manner. > I'm wondering if anyone could provide any reference (even web) that > would have his geometry, or parts of. I'll also be grateful if someone > could just respond or follow-up with the axioms he used and the first 5 > or so propositions. > Thanks, > Here are the axioms you requested: :-)I. Hilbert's Axioms of Incidence: (1) Any two points are on at least one line. (2) Any two points are on at most one line. (3) Any three points not all on one line are on at least one plane. (4) Any three points not all on one line are on at most one plane. (5) If two points on a line are on a plane, then every point on the line is on the plane. (6) If a point is on each of two planes, then there is another point on each of the two planes. (7) There are at least two points on each line; there are at least three points on each plane; and there are four points not all on one plane (and not all on one line). II. Hilbert's Axioms of Order: (1) If point B is between points A and C, then A, B, C are three collinear points and B is between C and A. (2) If A and C are two points, then there is at least one point B that is between A and C and there is at least one point D such that C is between A and D. (3) Among any three points on a line, exactly one is between the other two. (4) Let A, B, C be three points not on a line. Let l be a line off A, B, C but in the plane containing A, B, C. Let l contain a point that is between A and B. Then l contains either a point between A and C or a point between B and C. III. Hilbert's Axiom of Parallels: If point P is off line l, then there exists exactly one line in the plane containing P and l that does not intersect l. IV. Hilbert's Axioms of Congruence: (1) Given segment AB and a ray with vertex A', there exists one and only one point B' on the ray such that segment AB is congruent to segment A'B'; every segment is congruent to itself. (2) If segment AB is congruent to segment A'B' and to segment A''B'', then A'B' is congruent to A''B''. (3) If point B is between points A and C, point B' is between points A' and C', AB is congruent to A'B', and BC is congruent to B'C', then AC is congruent to A'C'. (4) Given angle {h,k}, ray h', and a halfplane H of the line containing h', then there exists one and only one ray k' on H such that angle {h',k'} is congruent to angle {h,k}; every angle is congruent to itself. (5) If angle {h,k} is congruent to angle {h',k'}and to angle {h'',k''}, then {h',k'} is conguent to {h'',k''}. (6) Given triangle ABC and A'B'C' such that AB is congruent to A'B', angle BAC is congruent to angle B'A'C', and AC is congruent to A'C', then angle ABC is congruent to angle A'B'C'. Remark. Hilbert's fifth and final set of axioms originally contained only Archimedes' axiom. This assures that there are not *too many* points on a line. However, Archimedes' axiom alone is not sufficient to give a categorical axiom system, as Hilbert pointed out. In order to assure that there are enough points on a line to have a categorical axiom system with Cartesian three-space as a model, one more axiom is required. Hilbert's own *completeness axiom*, takes the somewhat awkward form of requiring that it be impossible to properly extend the sets and relations satisfying the other axioms so that all the other axioms still hold. V. Hilbert's Axiom of Continuity: (1) If points A(1) is between points A and B, then there exist points A(2),A(3),...,A(n) such that (i) A(k) is between A(k-1) and A(k+1) for k=1,2,...,n-1 with A(0)=A, (ii) segment A(k)A(k+1) is congruent to segment AA(1) for k=1,2,...n-1, and (iii) point B is between A and A(n). Nathaniel Deeth Age 11 ============================================================================== From: Aakash Mehendale Subject: Re: Axiomatisation of Euclidean geometry Date: Fri, 27 Aug 1999 10:39:20 +0100 Newsgroups: sci.math "Kyle R. Hofmann" wrote: > > On Thu, 26 Aug 1999 11:42:13 +0100, Dimitris wrote: > > > > Hello ! Does anyone know anywhere in the Web where I could find Hilbert's > > axioms of Euclidean geometry ? > > I can list them for you right here. He notes that all lines, points, and > triangles, are assumed to be distinct unless otherwise noted. > [snip] Lots of them, aren't there? Some points: > Axioms of order: > 4. Let A, B, C be three points that do not lie on a line and let a be a > line in the plane ABC which does not meet any of the points A, B, C. If > the line a passes through a points of the segment AB, it also passes > through a point of the segment AB, or through a point of the segment BC. ^ typo for C? > Axiom of parallels: > 1. Let a be any line and A a point not on it. Then there is at most one ^^^^^^^^^^^ > line in the plane, determined by a and A, that passes through A and does > not intersect a. Doesn't that have to be 'exactly one' to match Euclid's fifth? -- Aakash Mehendale email:aakash@popaccount.com ICQ:26396017 Gnus don't kill people Any views and opinions expressed herein are mine alone ============================================================================== From: rhofmann@crl.com (Kyle R. Hofmann) Subject: Re: Axiomatisation of Euclidean geometry Date: 27 Aug 1999 00:55:59 GMT Newsgroups: sci.math On Thu, 26 Aug 1999 11:42:13 +0100, Dimitris wrote: > > Hello ! Does anyone know anywhere in the Web where I could find Hilbert's > axioms of Euclidean geometry ? I can list them for you right here. He notes that all lines, points, and triangles, are assumed to be distinct unless otherwise noted. Axioms of incidence: 1. For every two points A, B there exists a line a that contains each of the points A, B. 2. For every two points A, B there exists no more than one line that contains each of the points A, B. 3. There exist at least two points on a line. There exists at least three points that do not lie on a line. 4. For any three points A, B, C that do not lie on the same line there exists a plane alpha that contains each of the points A, B, C. For every plane there exists a point which it contains. 5. For any three points A, B, C that do not lie on one and the same line there exists no more than one plane that contains each of the three points A, B, C. 6. If two points A, B of a line a lie in a plane alpha then every point of a lies in the plane alpha. 7. If two planes alpha, beta have a point A in common then they have at least one more point B in common. 8. There exist at least four points which do not lie in a plane. Axioms of order: 1. If a point B lies between a point A and a point C then the points A, B, C are three distinct points of a line, and B then also lies between C and A. 2. For two points A and C, there always exists at least one point B on the line AC such that C lies between A and B. 3. Of any three points on a line there exists no more than one that lies between the other two. 4. Let A, B, C be three points that do not lie on a line and let a be a line in the plane ABC which does not meet any of the points A, B, C. If the line a passes through a points of the segment AB, it also passes through a point of the segment AB, or through a point of the segment BC. Axioms of congruence: 1. If A, B are two points on a line a, and A' is a point on the same or on another line a' then it is always possible to find a point B' on a given side of the line a' through A' such that the segment AB is congruent or equal to the segment A'B'. In symbols, AB (symbol for is congruent to) A'B'. 2. If a segment A'B' and a segment A''B'', are congruent to the same segment AB, then the segment A'B' is also congruent to the segment A''B'', or briefly, if two segments are congruent to a third one they are congruent to each other. 3. On the line a let AB and BC be two segments which except for B have no point in common. Furthermore, on the same or on another line a' let A'B' and B'C' be two segments which except for B' also have no point in common. In that case, if AB is congruent to A'B' and BC is congruent to B'C', then AC is congruent to A'C'. 4. Let theta be an angle in the plane alpha given by the rays h and k and a' a line in a plane alpha' and let a definite side of a' in alpha' be given. Let h' be a ray on the line a' that emanates from the point O'. Then there exists in the plane alpha' one and only one ray k' such that the angle that is congruent or equal to the angle theta' given by the rays h' and k' and at the same time all interior points of the angle theta' lie on the given side of a'. Symbolically, theta (is congruent to) theta'. Every angle is congruent to itself, i.e., theta (is congruent to) theta is always true. 5. If for two triangles ABC and A'B'C' the congruences AB (is congruent to) A'B', AC (is congruent to) A'C', (the angle) BAC (is congruent to) (the angle) B'A'C', then the congruence (the angle) ABC (is congruent to) (the angle) A'B'C' is also satisfied. Axiom of parallels: 1. Let a be any line and A a point not on it. Then there is at most one line in the plane, determined by a and A, that passes through A and does not intersect a. Axioms of continuity: 1. If AB and CD are any segments then there exists a number n such that n segments BC constructed contiguously from A, along the ray from A through B, will pass beyond the point B. 2. An extension of a set of points on a line with its order and congruence relations that would preserve the relations existing among the original elements as well as the fundamental properties of line order and congruence that follows from all the previous axioms but the axiom of parallels is impossible. -- Kyle R. Hofmann | "...during the years between 960 and 1000 there was great activity in the production of homilies ... [ The Blickling Homilies ] voice the almost universal belief that the world would end in the year 1000." -- The Concise Cambridge History of English Literature ============================================================================== From: ndm@shore.net (Norman D. Megill) Newsgroups: sci.math Subject: Re: Axiomatisation of Euclidean geometry In article <7q5gpr$ghs$1@mail.pl.unisys.com>, Clive Tooth wrote: >Kyle R. Hofmann wrote... > >>On Thu, 26 Aug 1999 11:42:13 +0100, Dimitris >wrote: >>> >>> Hello ! Does anyone know anywhere in the Web where I could find Hilbert's >>> axioms of Euclidean geometry ? >> >>I can list them for you right here. > > > >Please excuse my ignorance in these matters. >Can these axioms be proved to be consistent? >Can undecidable propositions arise in this geometry? > Tarski's version of these axioms (and I'm not sure whether they are exactly equivalent to Hilbert's) is decidable and therefore consistent. See "What is Elementary Geometry?" by Tarski. @inproceedings{Tarski, author = "Alfred Tarski", title = "What is Elementary Geometry", pages = "16--29", booktitle = "The Axiomatic Method, with Special Reference to Geometry and Physics (Proceedings of an International Symposium held at the University of California, Berkeley, December 26, 1957 --- January 4, 1958)", editor = "Leon Henkin and Patrick Suppes and Alfred Tarski", year = 1959, publisher = "North-Holland Publishing Company", address = "Amsterdam"} Tarski's axioms are included in my "Metamath Solitaire" applet as a first-order theory. http://www.shore.net/~ndm/java/mm.html Click on "Metamath Solitaire"; then in the applet window choose "Select Logic Family" then "Euclidean Geometry" then "Axiom Information". --Norm ============================================================================== From: ah170@FreeNet.Carleton.CA (David Libert) Subject: Re: Axiomatisation of Euclidean geometry Date: 30 Aug 1999 07:31:43 GMT Newsgroups: sci.math Norman D. Megill (ndm@shore.net) writes: > Tarski's version of these axioms (and I'm not sure whether they are > exactly equivalent to Hilbert's) is decidable and therefore consistent. > See "What is Elementary Geometry?" by Tarski. The primitive notions in Hilbert's axiomatization are point, line and plane, and the inclusion relations among these. Tarski's axiomatization is a true first order theory, in which individual variables range over points. In Tarski's, line and plane are not primitive notions. Tarski's axiomatization is over first order logic with =. As well as = it has two relation symbols. One is a ternary betweeness predicate B (I am writing from memory so I may not get letters right), s.t. for all points x,y,z B(x,y,z) <-> x,y,z are distinct and collinear and y is between x and z. Tarski also has a congruence relation used to formalize intuitive congruence notions like congruent angles and similar triangles. I seem to think for some reason he might have called this I. It is a 4-ary relation. I(x,y,w,z) <-> the distance from x to y = the distance from w to z. Tarski gives a first order axiomatization of the theory in this language of the standard Euclidean plane. This is finitely many axioms and a single axiom schematum. This theory is not finitely axiomatizable, there is no clever alternative way to eliminate the schematum with finitely many axioms. Consider an alternative congruence relation, C, a ternary relation. C(x,y,z) <-> the distance from x to y = the distance from y to z. There is one way to define I (4-ary) from C (ternary). It corresponds to the intuitive picture of walking a fixed protractor. I(x,y,w,z) <-> there is a finite sequence of points p0, p1, ..., pn s.t. p0 = x, p1 = y, p(n-1) = w , pn = z and for all 0 <= i < n C(pi, p(i+1), p(i+2)). This definition could be formalized in weak second order logic over Euclidean geometry, where we can quantify over finite sets of individuals and so talk about {p0, p1, ..., pn}. (There is some trickiness because the intuitive definition has a sequence of points instead of a set, but this can be handled.) This intuitive definition of I from C is not readily formalizable in first order Euclidean geometry though. We can make a first order formula representing the above for each fixed n, by using explicit quantifier to name the intermediate points, but longer lists require more quantifiers. However, there is a completely different definition (not "walking a protractor") that is formalizable in first order with variables ranging over points. So I is definable from C. Obviously C is definable from I: C(x,y,z) <-> I(x,y,y,z). So this is enough to rework Tarksi's axiomatization to be over B & C. Tarski's main presentation was B & I though. Tarski's axiomatization is true first order. To compare this to Hilbert, one complication is Tarski's main version was the plane and Hilbert's was 3-space. I don't remember if Tarski discussed extending it to higher dimensions. If not, I can't imagine it would be too difficult once Tarski did all the work to get started. So supposing we have extended Tarski, or taken a plane version of Hilbert, we can try comparing them. The first problem is they are different languages: Tarski is variables ranging over points with relations B & I. Hilbert is a 3 sorted logic with distinct variable types for points, lines and planes. These are inter-interpretable in the obvious way: ie when Hilbert talks about a line Tarski talks about two distinct points etc. In Tarski the powerful axioms are that single schematum. In Hilbert the powerful axioms are the last two: one saying for each short line segment and each long line segment finitely many copies of the short laid end to end span the long: ie an Archimedian property. The other powerful Hilbert axiom is a sort of completeness axiom: saying no model of the all the other axioms than this one can be properly extended to another model of all the axioms than this one. Not that I have checked recently, but I believe when you interpret the languages back and forth, the weak axioms of each theory prove the weak axioms of the other one. So all those Hilbert axioms about basic incidences of lines and points etc. This is as far as it can go. Tarski's axioms are first order and so subject to the completeness theorem and Lowenhiem Skolem. So Tarski's axioms must admit models with infinitesimals violating the Archimedian axiom in Hilbert. Also Tarski's axioms must have countable models (Lowenhiem Skolem), while Hilbert's completeness axiom forces all models to be continuum sized. It does go the other way though. On re-interpretation Hilbert's full theory including the powerful axioms yields Tarski's schematum. And Hilbert's power axioms are needed to get Tarski's schematum. Tarski's schematum is a first order approximation to the completeness property. -- David Libert (ah170@freenet.carleton.ca) 1. I used to be conceited but now I am perfect. 2. "So self-quoting doesn't seem so bad." -- David Libert 3. "So don't be a morron." -- Marek Drobnik bd308 rhetorical salvo IRC sig ============================================================================== From: ah170@FreeNet.Carleton.CA (David Libert) Subject: Re: Axiomatisation of Euclidean geometry Date: 30 Aug 1999 07:56:51 GMT Newsgroups: sci.math Kyle R. Hofmann (rhofmann@crl.com) writes: > Hilbert proves that they are consistent in his _Grundlagen der Geometrie_ > (Foundations of Geometry). That's also where I took them from. I would > suspect that they are also incomplete, but I have no evidence for it. They are complete, since they are categorical: all models are isomorphic to the standrad one. This is impossible for first order theories, but Hilbert's axiomatization is not first order. All but the last two axioms are. Those can be axiomatized in a 3 sorted logic with three types of variables ranging over points, line and planes respectively. Hilbert's second last axiom is an Archimedian property (finitely many copies of a short line segment can span each long line segment). This axiom rules out infinitesimals. A first order subtheory of the theory of the standard structure must have models with infinitesimals by the completeness theorem for first order logic. Hilbert's last axiom is that strange completeness axiom. Normally second order logic involves variables ranging over subsets of the universe. Hilbert's axiom is a lot stranger than this, because as phrased it quantifies over all models of the other axioms. This axiom has the effect of Dedikind completeness. The simple axioms preceding these last two guarantee midpoints of line segments and so guarntee a countable dense set of points. The completeness axiom guarantees Dedikind closure, so a model must include the standard one up to isomorphism. The Archemedian axiom guarantees it doesn't include more than the standard model, ie if two points are in the same bounded Dedikind cut of "binary points" along a line: those produced by finitely many midpoint operations, then the two points determining the same Dedikind cut are infinetisimally close. If there is a point at infinity beyond all the integer points along a line then Archmedian fails using the origin to that point as the long line and a standard line segment as the short. There is no place to stick extra points. -- David Libert (ah170@freenet.carleton.ca) 1. I used to be conceited but now I am perfect. 2. "So self-quoting doesn't seem so bad." -- David Libert 3. "So don't be a morron." -- Marek Drobnik bd308 rhetorical salvo IRC sig