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