Download Information for Students in MATH 348 2003 09

McGILL UNIVERSITY
FACULTY OF SCIENCE
DEPARTMENT OF
MATHEMATICS AND STATISTICS
MATH 348 2003 09
TOPICS IN GEOMETRY
Information for Students
Fall Term 2003 09
Pages 1 through 7 of these notes may be considered
the Course Outline for this course this semester.
W. G. Brown
November 27, 2003
Information for Students in Math 348 2003 09
i
Contents
1 General Information
1
1.1 Instructor and Times . . . .
1
1.2 Calendar Description . . . .
1
1.3 Class Test . . . . . . . . . .
1
1.4 Homework . . . . . . . . . .
2
1.5 Term Mark . . . . . . . . .
2
1.6 Calculators . . . . . . . . .
2
1.7 Final Grade . . . . . . . . .
2
1.8 Text-Book . . . . . . . . . .
2
1.9 Other Published Materials .
4
1.9.1 References to other
textbooks, and to papers in the mathematical literature . .
4
1.9.2 Website . . . . . . .
4
1.9.3 Printed Notes . . . .
5
1.10 Examination information . .
5
1.11 Plagiarism . . . . . . . . . .
5
1.12 Non-prerequisites . . . . . .
5
1.12.1 “Synthetic” geometry
6
1.12.2 Background from Logic 6
1.12.3 Background from Set
Theory . . . . . . . .
7
2 Tentative Timetable
3 Notes to accompany a brief discussion of Euclid’s Elements
3.1 Objectives . . . . . . . . . .
3.2 Definitions in Book I of Euclid’s Elements [9, Volume 1,
pp. 153-154] . . . . . . . . .
3.3 Postulates of Euclid’s Elements [4, §1.2, pp. 4-5] . . .
3.4 “Common Notions” (Axioms) of Euclid’s Elements
[9, Volume 1, pp. 155] . . .
3.5 Some conventions . . . . . .
3.5.1
3.5.2
3.6
3.7
3.8
8
10
10
10
12
13
13
3.9
Line segments vs. lines
Line segments and
their lengths . . . . .
3.5.3 Angles . . . . . . . .
3.5.4 Unit of angular measure . . . . . . . . .
3.5.5 Triangles . . . . . . .
3.5.6 The real numbers . .
Proposition I.1: . . . . . . .
3.6.1 . . . . . . . . . . . .
3.6.2 Enunciation of the
proposition in general terms . . . . . .
3.6.3 Detailed restatement
of the enunciation: .
3.6.4 Construction and Proof
3.6.5 Q. E. F. and Q. E. D.
Proposition I.2. . . . . . . .
3.7.1 . . . . . . . . . . . .
3.7.2 Enunciation
of
Proposition I.2: . . .
3.7.3 Detailed restatement
of the enunciation: .
3.7.4 Proof . . . . . . . . .
Proposition I.3 . . . . . . .
3.8.1 . . . . . . . . . . . .
3.8.2 Enunciation
of
Proposition I.3 . . .
3.8.3 Detailed restatement
of the enunciation . .
3.8.4 Construction
and
sketch of proof . . .
Proposition I.4 . . . . . . .
3.9.1 A
problematic
proposition . . . . .
3.9.2 Enunciation
of
Proposition I.4 . . .
3.9.3 Restatement of the
enunciation in more
modern terms . . . .
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Information for Students in Math 348 2003 09
3.9.4
3.10
3.11
3.12
3.13
Further comments on
Proposition I.4 . . .
A detailed look at Proposition I.5, sometimes called
“Pons Asinorum” . . . . . .
3.10.1 . . . . . . . . . . . .
3.10.2 Enunciation
of
Proposition I.5 in
general terms . . . .
3.10.3 Detailed restatement
of the enunciation . .
3.10.4 A proof of the Pons
Asinorum proposition by Pappus of
Alexandria . . . . . .
3.10.5 What about equilateral triangles? . . . .
Proposition I.6 . . . . . . .
3.11.1 The “converse” to
the first part of
Proposition I.5 . . .
3.11.2 Enunciation
of
Proposition I.6 . . .
3.11.3 Detailed restatement
of the enunciation . .
3.11.4 Proof . . . . . . . . .
3.11.5 Difficulties with this
proof . . . . . . . . .
Proposition I.7 . . . . . . .
3.12.1 Enunciation in general terms (slightly
modernized
from
Euclid’s language) .
3.12.2 Enunciation in detailed terms . . . . .
3.12.3 Proof . . . . . . . . .
3.12.4 Difficulties
with
Proposition I.7 . . .
Proposition I.8 . . . . . . .
3.13.1 Enunciation . . . . .
ii
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3.14
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3.15
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3.16
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3.17
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3.18
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3.13.2 Detailed restatement
of the enunciation . .
3.13.3 Idea of Euclid’s Proof
3.13.4 Philo’s proof . . . . .
3.13.5 A congruence theorem
Proposition I.9 . . . . . . .
3.14.1 Enunciation in general terms . . . . . .
3.14.2 Detailed restatement
of the enunciation . .
3.14.3 Construction . . . .
3.14.4 Proof . . . . . . . . .
Proposition I.10 . . . . . . .
3.15.1 Enunciation in general terms . . . . . .
3.15.2 Detailed restatement
of the enunciation . .
3.15.3 Construction . . . .
3.15.4 Proof . . . . . . . . .
Proposition I.11 . . . . . . .
3.16.1 Enunciation in general terms . . . . . .
3.16.2 Detailed restatement
of the enunciation . .
3.16.3 Construction, based
on Euclid’s but with
some ambiguity removed . . . . . . . .
3.16.4 Proof . . . . . . . . .
Proposition I.12 . . . . . . .
3.17.1 Enunciation in general terms . . . . . .
3.17.2 Detailed restatement
of the enunciation . .
3.17.3 Construction . . . .
3.17.4 Proof . . . . . . . . .
Proposition I.13 . . . . . . .
3.18.1 Enunciation in general terms . . . . . .
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Information for Students in Math 348 2003 09
3.19
3.20
3.21
3.22
3.23
3.24
3.18.2 Detailed restatement
of the enunciation . .
3.18.3 Proof . . . . . . . . .
Proposition I.14 . . . . . . .
3.19.1 Enunciation in general terms . . . . . .
3.19.2 Detailed restatement
of the enunciation . .
3.19.3 Proof . . . . . . . . .
Proposition I.15 . . . . . . .
3.20.1 Enunciation in general terms . . . . . .
3.20.2 Detailed restatement
of the enunciation . .
3.20.3 Proof . . . . . . . . .
Proposition I.16 . . . . . . .
3.21.1 Enunciation in general terms . . . . . .
3.21.2 Detailed restatement
of the enunciation . .
3.21.3 Proof . . . . . . . . .
Proposition I.17 . . . . . . .
3.22.1 Enunciation in general terms . . . . . .
3.22.2 Detailed restatement
of the enunciation . .
3.22.3 Proof . . . . . . . . .
Proposition I.18 . . . . . . .
3.23.1 Enunciation in general terms . . . . . .
3.23.2 Detailed restatement
of the enunciation . .
3.23.3 Proof . . . . . . . . .
Proposition I.19 . . . . . . .
3.24.1 Enunciation in general terms . . . . . .
3.24.2 Detailed restatement
of the enunciation . .
3.24.3 Proof . . . . . . . . .
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iii
3.25 Proposition I.20 – The “Triangle Inequality” . . . . . .
3.25.1 Enunciation in general terms . . . . . .
3.25.2 Detailed restatement
of the enunciation . .
3.25.3 Proof . . . . . . . . .
3.25.4 Proclus’s commentary on I.20 . . . . .
3.26 Proposition I.21 . . . . . . .
3.26.1 Enunciation in general terms . . . . . .
3.26.2 Detailed restatement
of the enunciation . .
3.26.3 Proof . . . . . . . . .
3.27 Proposition I.22 . . . . . . .
3.27.1 Enunciation in general terms . . . . . .
3.27.2 Detailed restatement
of the enunciation . .
3.27.3 Construction . . . .
3.27.4 Proof . . . . . . . . .
3.28 Proposition I.23 . . . . . . .
3.28.1 Enunciation in general terms . . . . . .
3.28.2 Detailed restatement
of the enunciation . .
3.28.3 Construction . . . .
3.29 Proposition I.24 . . . . . . .
3.29.1 Enunciation in general terms . . . . . .
3.29.2 Detailed restatement
of the enunciation . .
3.29.3 Proof . . . . . . . . .
3.30 Proposition I.25 . . . . . . .
3.30.1 Enunciation in general terms . . . . . .
3.30.2 . . . . . . . . . . . .
3.31 Proposition I.26 . . . . . . .
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Information for Students in Math 348 2003 09
3.31.1 Enunciation in general terms . . . . . .
3.31.2 Detailed restatement
of the enunciation . .
3.31.3 Proof of Case 1 . . .
3.31.4 Proof of Case 2 . . .
3.32 Proposition I.27 . . . . . . .
3.32.1 Enunciation in general terms . . . . . .
3.32.2 Detailed restatement
of the enunciation . .
3.32.3 Proof . . . . . . . . .
3.33 Proposition I.28 . . . . . . .
3.33.1 Enunciation in general terms . . . . . .
3.33.2 Detailed restatement
of the enunciation . .
3.33.3 Proof of Case 1 . . .
3.33.4 Proof of Case 2 . . .
3.34 Some other propositions of
Euclid cited in the textbook
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iv
5.3.1
5.3.2
5.3.3
Introductory Examples
Basic Definitions . .
Further investigation
of symmetries of the
square . . . . . . . .
5.3.4 Isometries of the Euclidean plane . . . .
5.3.5 The product of two
reflections in the Euclidean plane . . . .
5.3.6 Half-turns . . . . . .
5.3.7 The composition of
two half-turns. . . .
5.3.8 The composition of
two translations. . .
5.3.9 Composition of three
reflections in the plane
5.3.10 A
simplification:
commuting reflections
50 6 Second Problem Assignment
2
53 7 Symmetries of R (continued)
7.1 Isometries (continued) . . .
5 Symmetries in the Real Plane 54
7.1.1 Miscellaneous results
5.1 Sets and Functions . . . . . 54
on half-turns and
5.1.1 Basic definitions . . . 54
translations . . . . .
5.1.2 Textbook
conven7.2 Groups of isometries . . . .
tion:
functions
7.2.1 The periods of isomecompose “on the
tries of the plane . .
right”. . . . . . . . . 56
7.2.2 Abstract
Groups;
5.1.3 Associativity of funcIsometry Groups . .
tion composition . . 56
7.2.3 The Cayley Table
5.1.4 Inverses . . . . . . . 57
or Composition Ta5.2 Vectors in R2 . . . . . . . . 57
ble of a group . . . .
5.2.1 Example of the use of
7.2.4 The Direct Product
“geometric vectors”:
of Groups . . . . . .
7.2.5 Types of symmetry
Concurrence of the
groups for configuramedians of a triangle 59
5.3 Isometries . . . . . . . . . . 62
tions in R2 . . . . . .
4 First Problem Assignment
62
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Information for Students in Math 348 2003 09
v
8 Solutions to Problems on As14 Brief discussion of (symbolic)
signment 1
88
logic
135
14.1 Statements . . . . . . . . . . 135
9 Third Problem Assignment
92
14.2 Truth values of sentences . . 136
14.3 Tautologies and contradictions137
10 The symmetries of friezes (strip
14.4 Logical Implication . . . . . 137
patterns)
95
14.5 “Rules of Logic” . . . . . . . 138
10.1 Comments on the table of
14.6 Rules of inference . . . . . . 139
symmetry groups of friezes . 95
14.7 Reductio ad absurdum =
10.1.1 What would have to
Proof by Contradiction . . . 139
be proved? . . . . . . 95
14.8 The converse and contrapos10.1.2 How can two “difitive of an implication . . . 140
ferent” friezes both
14.9 The Predicate Calculus . . . 141
have group C∞ ? . . . 96
14.10The axioms, postulates, and
10.1.3 How can three “diftheorems of a logical system 141
ferent” friezes all
have group D∞ ? . . 96 15 Solutions, Third Problem As10.1.4 How can we differsignment
143
entiate the cases of
. . . DDDDD
...
16 Solutions to Problems on the
and
. . . HHHHH
Class Tests
146
. . . from the oth16.1 Version 1 . . . . . . . . . . . 146
ers, and between
16.2 Version 2 . . . . . . . . . . . 152
themselves? . . . . . 97
16.3 Version 3 . . . . . . . . . . . 156
16.4 Version 4 . . . . . . . . . . . 159
11 Solutions, Second Problem Assignment
98 17 Fourth Problem Assignment 164
12 Two results on plane lattices 104 18 Ordered Geometry
166
12.1 Sylvester’s
Problem
of
18.1 The structure of our geometry 166
Collinear Points . . . . . . . 104
18.1.1 Excluding the “triv12.2 The Crystallographic Reial” geometries with
striction . . . . . . . . . . . 105
0 or 1 point . . . . . 167
18.1.2 Axioms concerning
13 Class Tests ‡
107
points on a line . . . 168
13.1 Version 1 . . . . . . . . . . . 107
13.2 Version 2 . . . . . . . . . . . 114 19 Ordered
Geometry
(con13.3 Version 3 . . . . . . . . . . . 121
cluded)‡
174
13.4 Version 4 . . . . . . . . . . . 128
Information for Students in Math 348 2003 09
vi
19.0.3 We require that not
22 Solutions, Fourth Problem Asall the points lie on
signment
186
the same line. . . . . 174
A Solved Assignments from Pre19.0.4 “Pasch’s”
Axiom;
vious Years
1001
Points are “dense”
A.1
2002/2003
.
.
.
.
.
.
.
.
.
.
1001
in any line . . . . . . 175
A.1.1 First
2002/2003
20 Projective Geometries
176
Problem Assignment 1001
20.1 The Real Projective Plane . 176
A.1.2 Second
2002/2003
20.2 “Homogeneous”
coordiProblem Assignment 1004
nates; models of the Real
A.1.3 Third
2002/2003
Projective Plane . . . . . . . 177
Problem Assignment 1007
20.3 The Euclidean plane as a
A.1.4 Fourth
2002/2003
subgeometry of the Real
Problem Assignment 1009
Projective Plane . . . . . . . 179
B Tests and Examinations from
20.4 The points and lines which
Previous Years
1015
are adjoined to the EuB.1 Tests from Fall, 2002 . . . . 1015
clidean Plane to create the
B.1.1 Version 1 . . . . . . 1015
Real Projective Plane . . . . 180
B.1.2 Version 2 . . . . . . 1016
20.5 Some “incidence” properties
B.1.3 Version 3 . . . . . . 1018
of the real projective plane . 180
B.1.4 Version 4 . . . . . . 1020
20.6 Projective planes over other
B.2 Solutions to Problems on the
fields than the reals . . . . . 181
2002 Class Tests . . . . . . . 1022
20.6.1 What is a “field”? . . 181
B.2.1
Problems on all four
20.6.2 Fields of “residues
versions of the test . 1022
modulo a prime” . . 182
B.2.2 The problems that
20.6.3 The projective plane
were each on only
over Fp . . . . . . . . 182
one test . . . . . . . 1024
20.7 The “Fano” Geometry . . . 183
B.3 Final Examination, Decem21 Projective Geometries (continber, 2002 . . . . . . . . . . . 1030
ued)
183
B.4 Supplemental/Deferred Ex21.1 Axiomatic Definition of Proamination, May, 2003 . . . . 1033
jective Planes . . . . . . . . 183
21.2 What can we infer from the 3
given axioms only? The Duality Principle. . . . . . . . 184
21.3 Some simple properties of finite projective planes . . . . 185
Information for Students in Math 348 2003 09
1
1
General Information
Distribution Date: Wednesday, September 3rd, 2003
(all information is subject to change)
1.1
Instructor and Times
OFFICE:
OFFICE HOURS
(subject to change):
OFFICE PHONE:
E-MAIL:
CLASSROOM:
CLASS HOURS:
CRN:
1.2
INSTRUCTOR
Professor W. G. Brown
BURN 1224
W 14:35→15:25;
F 10:00→11:00
or by appointment
398–3836
[email protected]
BURN 1B24
MWF 12:35–13:35
660
Calendar Description
(3 credits) (Fall and Summer) (Prerequisite: Previous course in Mathematics) Selected
topics - the particular selection may vary from year to year. Topics include: isometries
in the plane, symmetry groups of frieze and ornamental patterns, equidecomposability,
non-Euclidean geometry, and problems in discrete geometry.
1.3
Class Test
A test will be administered during the regular class hour on Wednesday, November 5th,
2003.1 No provision is planned for a “make-up” test for a student absent during the test.
Any change in this date will be announced in the lectures.2
In your instructor’s eyes the main purpose of the test is as a “dry run” for the final
examination.3
1
Date changed as of 8 October, 2003.
Note that the date of the test is after the deadline for withdrawal from the course.
3
Notwithstanding the minimal contribution of the test grade to the student’s final grade (cf. §1.5
below), the test is to be considered an “examination” in the sense of the Handbook of Student Rights
and Responsibilities, to be found at the following URL:
2
http://ww2.mcgill.ca/students-handbook/index.html
UPDATED TO November 27, 2003
Information for Students in Math 348 2003 09
1.4
2
Homework
There will be approximately 4 or 5 assignments. While students are not discouraged
from discussing assignment problems with their colleagues, written solutions that are
handed in should be each student’s own work.4
Submitted homework should be stapled with a cover page that contains your NAME,
STUDENT NUMBER, the COURSE NUMBER, and the ASSIGNMENT NUMBER.
Other pages should always include your student number. You can minimize the possibility that your assignment is lost or fragmented.
1.5
Term Mark
Graded out of 30, the TERM MARK will be the sum of the HOMEWORK GRADE
(out of 10) and the CLASS TEST GRADE (out of 20).
1.6
Calculators
The use of calculators, computers, notes, or other aids will not be permitted at the test
or examination.
1.7
Final Grade
The final grade will be a letter grade, computed from the maximum of
• the Examination Mark (out of 100); and
• the sum of the Term Mark (out of 30) and 0.7 times the examination mark (out of
100).
1.8
Text-Book
The primary textbook for the course will be: [4] H. S. M. Coxeter, Introduction to Geometry, Second Edition. John Wiley and Sons (1969). ISBN 0-471-18283-4 hardbound; or
the paperback edition, published 1989, ISBN 0-471-50458-0. (The paperbound edition
is the one to be stocked by the Bookstore.)
4
From the Handbook on Student Rights and Responsibilities:
“No student shall, with intent to deceive, represent the work of another person as his or
her own in any...assignment submitted in a course or program of study or represent as
his or her own an entire essay or work of another, whether the material so represented
constitutes a part or the entirety of the work submitted. cf. ¶15(a)5
UPDATED TO November 27, 2003
Information for Students in Math 348 2003 09
3
Most of the material needed for the course is supported by material to be found in
this textbook, mainly from the following sections of the following chapters; this list is
tentative, and subject to revision:
• Part I
1. Triangles: §6 §1.2, 1.3, 1.4 This part of the course will be supported substantially by notes on the Web. (Added September 19th, 2003: Propositions I.1
through I.1 through I.12 of Euclid’s Elements were discussed in the lectures;
the First Assignment includes problems on other propositions in the range
I.1–I.28; students are expected to read Propositions I.13–I.28 privately.
2. Regular Polygons: §§2.3, 2.4, 2.5, 2.6, 2.7, 2.8
3. Isometry in the Euclidean plane: §§3.1, 3.2, 3.3, 3.4, 3.5, 3.7
4. Two-dimensional crystallography: §§4.1, 4.2, 4.3; (omit §4.4); §4.5; (omit
§4.6); §4.7.
5. Similarity in the Euclidean plane: omit
6. Circles and spheres: omit
7. Isometry and similarity in Euclidean Space: omit
• Part II
8. Coordinates: omit
9. Complex numbers: omit
10. The five Platonic solids: §§10.1, 10.2, 10.3, 10.5
11. The golden section and phyllotaxis: omit
• Part III
12. Ordered geometry: §§12.1, 12.2, 12.3, 12.4(part), 12.5, 12.6
13. Affine geometry: §§13.1, 13.2, to be completed
14. Projective geometry: §§14.1, 14.2, to be completed
15. Absolute geometry: §§15.1, 15.2, 15.3, 15.4, 15.5 to be completed
16. Hyperbolic geometry: omit
• Part IV Omit all chapters, except possibly
6
In this section of these notes, § refers to sections of the textbook; elsewhere in the notes, unless
otherwise indicated, § refers to sections of these notes.
Information for Students in Math 348 2003 09
4
21. Topology of surfaces: part
Not all topics in these chapters will be studied in depth. Some topics will be studied
through a careful development of properties as the axiomatic systems are gradually
expanded. Other topics will be studied only descriptively, since there will not be time
to cover all chapters in depth.
1.9
1.9.1
Other Published Materials
References to other textbooks, and to papers in the mathematical
literature
The following book, which has been used as a reference for some topics during recent
years, will be kept on reserve in the Schulich Library: [8] D. W. Farmer, Groups and Symmetry. A Guide to Discovering Mathematics. Mathematical World, Volume 5. American
Mathematical Society, Providence, R. I. (1995), ISBN 0-8218-0450-2.
1.9.2
Website
These notes, and other materials distributed or posted to students in this course, will be
accessible at the following URL:
http://www.math.mcgill.ca/brown/math348a.html
(1)
The notes will be in “pdf” (.pdf) form, and can be read using the Adobe Acrobat reader,
which many users have on their computers. It is expected that most computers in campus
labs should have the necessary software to read the posted materials.
Where revisions are made to distributed printed materials — for example these information sheets — it is expected that the last version will be posted on the Web.
The notes will also be available via a link from the WebCT URL:
http://www.mcgill.ca/webct/
but not all features of WebCT will be implemented. It is, however, intended to mount
grades on the WebCT site: students should verify that their grades have been properly
recorded.7
7
Always keep graded materials — homework and test — to substantiate any claim that a grade
has not been properly recorded. Of course, these materials are also useful in preparing for the final
examination.
Information for Students in Math 348 2003 09
1.9.3
5
Printed Notes
Typeset notes will be made available from time to time to supplement material in the
text-book or lectures; these notes will be available through WebCT or the URL above
(1); possibly some of the notes will be distributed in hard copy. Normally, when there
are revisions and corrections to printed notes, these will not be distributed in
hard copy, but will be posted on the Web. Unless you are explicitly told otherwise,
you should assume that all printed notes — whether distributed in hard copy or mounted
on the Web or both — are as much part of the required materials in the course as if they
had been written on the chalkboard during a lecture.8
1.10
Examination information
1. “Will there be a supplemental examination in this course.” Yes.
2. “Will students with marks of D, F, or J have the option of doing additional work
to upgrade their mark?” No.
3. “Will the final examination be machine scored?” No.
1.11
Plagiarism
“McGill University values academic integrity. Therefore all students must understand the
meaning and consequences of cheating, plagiarism and other academic offences under the
Code of Student Conduct and Disciplinary Procedures. (See http://www.mcgill.ca/integrity
for more information).
“L’université McGill attache une haute importance à l’honnêteté académique. Il incombe
par conséquent à tous les étudiants de comprendre ce que l’on entend par tricherie, plagiat
et autres infractions académiques, ainsi que les conséquences que peuvent avoir de telles
actions, selon le Code de conduite de l’étudiant et des procédures disciplinaires. (Pour de
plus amples renseignements, veuillez consulter le site http://www.mcgill.ca/integrity).”9
1.12
Non-prerequisites
Some parts of this course will be purely descriptive; but the intention is that some parts
of the course will contain real mathematics, i.e., theorems and proofs. Since the course
description requires only “a previous course in mathematics”, students can expect to be
8
Sometimes the notes may be nothing more than an edited version of material that was written on
the chalkboard, or a preliminary version of materials to be used in a subsequent lecture.
9
Inserted at the request of the Provost of the University.
Information for Students in Math 348 2003 09
6
provided with the background needed to understand the topics under discussion. Some
parts of this background material could be review for students who have studied MATH
318, MATH 240, MATH 235, and certain other courses, but the only material that will
be formally assumed is the pre-calculus contents of a course like MATH 112 or CÉGEP
201-101.
1.12.1
“Synthetic” geometry
[5] “Synthetic geometry is ultimately based on certain primitive concepts and axioms,
appropriate to the particular kind of geometry under consideration (e.g., projective or
affine, real, complex, or finite). Each problem belongs to one kind (or to a few kinds), and
I would call a solution synthetic 10 if it remains in that kind, analytic if it goes outside.
The use of coordinates is one way of going outside; the use of trigonometry is another.”
In part of the course we will try to develop several types of geometry synthetically, and
will use analytic methods only for the provision of examples and models. Elsewhere we
may shift to analytic methods in order to be able to cover more territory in limited time.
Even though we will usually accompany our geometric proofs with a “sketch”, the proof
must be able to stand without any reference to the drawing; the sketch only serves to
help us develop, understand, and remember the proof. For example, in several theorems
we will be concerned with proving that certain lines are concurrent — that they all pass
through a common point. Our proofs will never be based on a drawing in which the
lines appear to have the property, although we may make such a drawing in the course
of trying to understand the proof.
1.12.2
Background from Logic
The textbook is not written in the notation of symbolic logic, and we don’t plan to write
most of the proofs we study symbolically either; however, we need to have some of the
notation of symbolic logic available to streamline proofs when it is convenient; and also
we need to be comfortable with the notion of a proof based on inference from axioms
and hypotheses. The ancient Greeks did not have symbolic logic available when they
developed Euclidean geometry; nor did they have modern notation or modern writing
paraphernalia, nor the theory of sets. However, by using these tools judiciously we will
be able to reduce the study of geometry from a Herculean task requiring the genius of a
Euclid or a Pythagoras to an interesting challenge.
10
italics added
Information for Students in Math 348 2003 09
1.12.3
7
Background from Set Theory
The language and notation of set theory are standard in most mathematical writing
today. We will review the simplest notions as we require them, so that we have them
available when we need them. Students are not expected to have been exposed to these
ideas before, but those for whom they are new should devote the time necessary to
become comfortable with the concepts and notations.
Information for Students in Math 348 2003 09
2
8
Tentative Timetable
Distribution Date: (0th version) Wednesday, September 3rd, 2003
(All information is subject to change.)
MONDAY
1
8
15
22
29
6
13
20
27
LABOUR DAY
3
WEDNESDAY
SEPTEMBER
Introduction
FRIDAY
5
R
§1.1°;§1.2,
Euclid
axioms, etc.
§1.3, Euclid I.1
10 Euclid I.1–I.4
12 Euclid I.5
Course changes must be completed by midnight, September 15
Euclid I.5–I.6
17 Euclid I.6–I.9
19 Euclid I.10–I.28
Deadline for withdrawal with fee refund = September 22
Notes, §§5.1, 5.2
24 Notes §§5.1, 5.2
26 Notes §§5.2, 5.3
Notes §5.3
OCTOBER
P
1 Notes §§5.3.2, 5.3.3 3 Notes §5.3.4 1
NO LECTURE IN 8 §§3.1–3.5, §3.7
10 §§3.1–3.5, §3.7
MATH 348
Deadline for withdrawal (with W) from course = Oct. 13
THANKSGIVING
15 §§3.1–3.5, §3.7
17 Notes §7.2
DAY (Canada)
P
Notes §7.2 2
22 Notes §10.1
24 Notes §10.1
Notes §10.1; text 29 §4.5, §4.7
31 Logic
§4.5, §4.7
P
Notation: n
= Assignment #n due
R
° = Read Only
X = reserved for eXpansion or review
Section numbers refer to the text-book unless Notes are specified.
The next page will not be distributed until the syllabus has been revised.
Information for Students in Math 348 2003 09
MONDAY
3
Logic + Axiomatic
Systems
5
10
17
24
12
19
26
1
3
WEDNESDAY
NOVEMBER
CLASS TEST
9
FRIDAY
7
14
21
28
DECEMBER
Notation:
P
n
R
°
X
= Assignment #n due at midnight on Monday this week
= Read Only
= reserved for eXpansion or review
Section numbers refer to the text-book.
UPDATED TO November 27, 2003
Information for Students in Math 348 2003 09
3
10
Notes to accompany a brief discussion of Euclid’s
Elements
3.1
Objectives
We shall explore some of the simpler geometric results in Euclid11 ’s Elements. We make
no serious attempt to modernize the treatment, beyond the introduction of some modern
terminology; however, we will indicate from time to time features of this remarkable work
that do not meet current standards of scholarship. These comments should not cloud
the profound debt that all modern mathematics owes to Euclid and his predecessors.
(We will omit the last part of Chapter 1 of the textbook, [4, pp. 11-25].)
3.2
Definitions in Book I of Euclid’s Elements [9, Volume 1,
pp. 153-154]
The following “definitions” would need some polishing if we were to use them today.
However, the comments provided after some of them should not detract from an appreciation of the power and coherence of the theory that Euclid and his predecessors
constructed.
In some instances in the sequel we will, when presenting results from Euclid, update
the terminology in whole or in part to current usage.
1. A point is that which has no part.
2. A line is breadthless length.
3. The extremities of a line are points.12
13
4. A straight line is a line which lies evenly with the points on itself.
5. A surface is that which has length and breadth only.
6. The extremities of a surface are lines.
7. A plane surface is a surface which lies evenly with the straight lines on itself.
11
Heath [9] places Euclid “intermediate between the first pupils of Plato (d. 347/346 B.C.) and
Archimedes (287-212 B.C.)”; educated in Athens from pupils of Plato; not to be confused with the
philosopher, Euclid of Megara, Euclid “taught and founded a school at Alexandria”.
12
Modern terminology would reserve the word “line” for one that has no extremities, and does not
admit further extension. Where there is one extremity, modern terminology would use the word “ray”;
where there are two extremities, the word “line segment” would be used.
13
It will be clear immediately that the intention is that the extremities of a line segment must be
different.
Information for Students in Math 348 2003 09
11
8. A plane angle is the inclination to one another of two lines in a plane which meet
one another and do not lie in a straight line.
9. And when the lines containing the angle are straight, the angle is called rectilineal.
10. When a straight line set up on a straight line makes the adjacent angles equal to
one another, each of the equal angles is right, and the straight line standing on
the other is called a perpendicular to that on which it stands.
11. An obtuse angle is an angle greater than a right angle.
12. An acute angle is an angle less than a right angle.
13. A boundary is that which is an extremity of anything.
14. A figure is that which is contained by any boundary or boundaries.
15. A circle is a plane figure contained by one line such that all the straight lines
falling upon it from one point among those lying within the figure are equal to one
another;
16. And the point is called the centre of the circle.
17. A diameter of the circle is any straight line drawn through the centre and terminated in both directions by the circumference of the circle, and such a straight line
also bisects the circle.
18. A semicircle is the figure contained by a diameter and the circumference cut off
by it. And the centre of the semicircle is the same as that of the circle.
19. Rectilinear figures are those which are contained by straight lines, trilateral
figures are those which are contained by three, quadrilateral those contained by
four, and multilateral those contained by more than four straight lines.
20. Of trilateral figures, an equilateral triangle is that which has its three sides
equal, an isosceles triangle is that which has two of its sides alone14 equal, and
a scalene triangle that which has its three sides unequal.
21. Further, of trilateral figures, a right-angled triangle is that which has a right
angle, an obtuse-angled triangle that which has an obtuse angle, and an acuteangled triangle that which has its three angles acute.
14
Modern terminology would not require that an isosceles triangle not be equilateral. In the sequel
we shall normally use the modern definition.
Information for Students in Math 348 2003 09
12
22. Of quadrilateral figures, a square is that which is both equilateral and right-angled;
an oblong that which is right-angled but not equilateral; a rhombus that which is
equilateral but not right-angled15 ; and a rhomboid16 that which has its opposite
sides and angles equal to one another but is neither equilateral nor right-angled.
And let quadrilaterals other than these be called trapezia.17
23. Parallel straight lines are straight lines which, being in the same plane and being produced indefinitely in both directions, do not meet one another in either
direction.18
3.3
Postulates of Euclid’s Elements [4, §1.2, pp. 4-5]
(We shall not be concerned with Euclid’s distinction between “Postulates” and “Axioms”.)
1.21 A straight line may be drawn from any point to any other point.
1.22 A finite straight line may be extended continuously in a straight line.19
1.23 A circle may be described with any centre and any radius.
1.24 All right angles are equal to one another.
1.25 If a straight line meets two other straight lines so as to make the two interior
angles on one side of it together less than two right angles, the other straight lines,
if extended indefinitely, will meet on that side on which the angles are less than
two right angles.
15
Modern terminology would permit the right-angled case — i.e. the square — as being a special type
of rhombus
16
We would call such a figure a parallelogram, and would include the rectangle and rhombus as
special cases.
17
Modern usage would usually reserve the word trapezium for a figure in which a pair of opposite
sides are parallel, and would include all parallelograms.
18
Modern terminology would change this definition in several ways.
19
Note that this translation is ambiguous, as it uses the term straight line with two somewhat different
meanings. Modern usage would reserve the term straight line for the “doubly” infinite object; a “line”
beginning at a point and extending out indefinitely would be called a ray, and a “finite straight line”
would be called a line segment. Note that in the use of the word “finite” here it is assumed that the
length is non-zero, which would not be the current interpretation of that word.
Information for Students in Math 348 2003 09
3.4
13
“Common Notions” (Axioms) of Euclid’s Elements [9, Volume 1, pp. 155]
(Some manuscripts have variations and extensions of this list.20 )
1. Things which are equal to the same thing are also equal to one another.
2. If equals be added to equals, the wholes are equal.
3. If equals be subtracted from equals, the remainders are equal.
4. Things which coincide with one another are equal to one another.
5. The whole is greater than the part.
3.5
3.5.1
Some conventions
Line segments vs. lines
In preparation for future refinements, let us distinguish between the line segment joining
points A and B (which we can denote by AB), and the length of that line segment,
which we can denote by |AB|. Keep in mind that length will be unsigned , that is, where
we wish to talk about positive and negative distances, we shall use other terminology
and another notation. For the present, AB and BA represent the same line segment.
Where a proposition is not concerned with the length of a line segment, we may be
casual about replacing the segment by the line or a ray in the line. This can be justified
by appealing to axiom 1.21 of §3.3.
3.5.2
Line segments and their lengths
In some cases Euclid has not distinguished verbally between a line (segment) and its
length; in the versions of the propositions and their proofs given below we have tried to
sharpen the distinction.
3.5.3
Angles
When we wish to talk about the angle at a vertex A between a ray21 BA and a ray
BC, we may write ∠BAC or ∠CAB. We can think of the angle — intuitively — as the
collection of rays “between” these two bounding rays; but, by modern standards, the
concept is not adequately defined, and additional postulates or definitions are required.
20
21
For example, If equals be added to unequals, the wholes are unequal.
We haven’t defined this term formally.
Information for Students in Math 348 2003 09
14
Eventually — after we leave the study of Euclid’s Elements — we may wish to
introduce a sign convention. One could argue that we should be using a symbol like
|∠A| to distinguish the name of ∠A from its length. We will not do this.
3.5.4
Unit of angular measure
Nowadays we usually measure angles in degrees, or in radians, where the former measure
may be linked to ancient Babylonian practice, and the later measure is linked to the
length of an arc on a circle of unit radius. Euclid’s angles are expressed in terms of right
angles.
3.5.5
Triangles
We will denote the triangle with vertices A, B, C by 4ABC; no orientation is associated
with the vertices in this notation, so
4ABC = 4BCA = 4CAB = 4ACB = 4CBA = 4BAC .
3.5.6
The real numbers
Euclid’s axioms could form the basis of an axiomatization of a number system: the usual
definitions employed today would include axioms equivalent to these, and more. But,
even with the limited machinery that he created, Euclid was able to prove (in later books
of the Elements) non-trivial results about numbers — although he expressed these ideas
in geometric language.
Information for Students in Math 348 2003 09
15
Distribution Date: Monday, September 8th, 2003
3.6
Proposition I.1:
3.6.1
In this first result of his Elements, Euclid proves the validity of a constructive procedure.
The procedure uses Postulate 1.23 of §3.3, which asserts that A circle may be described
with any centre and any radius. If we wish to interpret this postulate in terms of the use
of a modern compass, we have to understand that Euclid does not claim that you can
first set your compass at a particular radius, and then place the point of the compass
on paper and draw a circle. Rather, you are to first place the point on the paper, then
adjust the compass so that the circle will pass through a particular point, and then
describe the circle; think of a compass that has no memory — the moment the point
is lifted from the paper, the radius reverts to zero. In Proposition I.3 he will show
that one can, on the strength of his propositions, indeed set the compass at any desired
radius. Euclid’s economy in his use of postulates is very much in the spirit of modern
mathematics: mathematicians always try to weaken their assumptions to obtain “best
possible” results, while assuming the least.
Euclid’s construction involves the drawing of two circles: the apex of the triangle he
is constructing will be the intersection of these circles. There is a gap in his axioms: he
did not see the need to ensure that the circles actually meet in a point. To see that there
is a problem here, think of the circles in the Cartesian plane with radius 1 and centres
(±1, 0). Suppose that the only points that we can see on these
√ ¢ are those with
¡ circles
rational coordinates. Then the circles will both miss the point 0, 3 where we would
like them to intersect. The property we need here is called continuity and is related to
the concept of continuity that one meets in calculus. It is where, in the following proof,
he says “...the point C, in which the circles cut one another...” that he has made an
inference not justified by his logical system. To modernize this part of his geometry, we
would need to supply some additional axioms; this can be done in various ways, and is
beyond this course.
You should make a sketch and follow every step on your sketch; of course, the sketch
is not part of the proof, but it helps you understand what Euclid is claiming. The proof
given is based on the translation in [9, pp. 241–242].
3.6.2
Enunciation of the proposition in general terms
On a given finite straight line, to construct an equilateral triangle.
Information for Students in Math 348 2003 09
3.6.3
16
Detailed restatement of the enunciation:
Let AB be the given finite straight line. To construct an equilateral triangle ABC, i.e.
to find a point C such that the line segments AB, BC, CA are of equal length.
3.6.4
Construction and Proof
By Postulate 1.23 of §3.3 we may draw a circle with centre A, passing through B; and a
circle with centre B, passing through A. Let C denote a22 point of intersection of these
circles. By Postulate 1.21 of §3.3 line segments may be drawn from C to each of A, B. By
Definition 15, the lengths of the line segments joining the centre of a circle to any of the
points on the circle are equal: in particular, for the circle with centre A, |AC| = |AB|;
and, for the circle with centre B, |BC| = |BA|. By Axiom 1, |AC| = |AB| = |BC|.
By Definition 20, the triangle with vertices A, B, C is equilateral, and, as required, it
has been constructed on the base AB. Q.E.F. (see §3.6.5 next below)
3.6.5
Q. E. F. and Q. E. D.
The proofs of the first three propositions end with abbreviation of words meaning “what
it was required to do”23 ; the next three end with words meaning “what it was required
to prove”24 , etc. The Latin abbreviations, especially Q.E.D., are still in current use
by mathematicians; however, some mathematicians now use some sort of a box symbol
instead of writing Q.E.D., e.g. ¤.
3.7
Proposition I.2.
3.7.1
This second proposition is, like its predecessor and its successor, a construction. Euclid is
proceeding to prove develop results that will enable him to compare lengths. Euclid did
not have the real number system available to him. (Even in later books of the Elements,
where Euclid proved theorems about the integers — e.g. that there exist an infinity of
primes 25 — his proofs were geometric.)
Remember that Euclid’s axioms do not permit a circle to be drawn with a “compasscarried distance”; however, if one foot of the compass is placed at the proposed centre,
the axioms permit the placing of the other foot at a point at any desired distance, and to
22
There are two such points of intersection unless the given points coincide.
Greek oπ²ρ ²δ²ι πoιησαι; Latin Q.E.F. = quod oportebat fieri .
24
Greek oπ²ρ ²δ²ι δ²ιξαι; Latin Q.E.D. = quod erat demonstrandum.
25
Proposition IX.20, [9, Vol 2, p. 412]
23
Information for Students in Math 348 2003 09
17
then draw the circle through that point. It has been suggested that one should “suppose
the compasses to close of themselves the moment they cease to touch the paper”.
3.7.2
Enunciation of Proposition I.2:
To place at a given point (as an extremity) a (straight) line segment whose length is
equal to that of a given (straight) line segment.
The proof below is paraphrased from the version in [9, p. 244].
3.7.3
Detailed restatement of the enunciation:
Let A be the given point, and BC the given line segment. Thus it is required to place
at the point A, as an extremity, a line segment whose length is |BC|.
3.7.4
Proof
On the line segment AB (whose existence is postulated in Postulate 1.21 of §3.3), construct an equilateral triangle (using Proposition I.1 §3.6) whose third vertex will be
called D. Produce line segments DA and DB respectively beyond A and B “sufficiently
far”26 to points E and F . By Postulate 1.23 of §3.3 there exists a circle with centre B
passing through the point C. Assuming that segment DB was extended sufficiently far,
this circle will meet segment BF . Call the point of intersection G. We again appeal to
Postulate 1.23. This time the circle with centre D and passing through the point G will
meet DE, the extension of DA; call the point of intersection L. Now27
|DA| + |AL| =
=
=
=
=
|DL|
|DG|
|DB| + |BG|
|DB| + |BC|
|DA| + |BC| ;
(2)
(3)
(4)
(5)
(6)
by Axiom 3, this implies that |AL| = |BC|, so we have constructed a line segment with
one end at A, whose length is equal to |BC|.
Exercise 3.1 Explain why each of equations (2) – (6) holds.
Solution:
26
The proof depends on these extended line meeting circles, so the line segments have to be “long
enough”.
27
Can you explain why each of the equalities stated should hold?
Information for Students in Math 348 2003 09
18
(2): since L lies on the extension of DA beyond A
(3): since L lies on the circle with centre D and radius DG
(4): since G lies on the extension of DB beyond B
(5): since G lies on the circle with centre B and radius BC
(6): since ABC is an equilateral triangle
Exercise 3.2 In the proof of the preceding proposition, care has not been taken to
accommodate special cases where certain pairs of points coincide. What would happen
if
1. A = B? (Here 4ABD would degenerate to a single point D = A = B.)
2. A = C?
There are two questions here:
• Is the proof, as stated, correct?
• Could one give another proof, different from the general one stated, which could
accommodate the situation.?
Can you see any other possible degeneracies that would require adjustments in the proof?
3.8
Proposition I.3
3.8.1
This constructive proposition is an application of its immediate predecessor. Students
are again reminded that Euclid’s postulates do not include the drawing of a circle with a
“compass-carried distance”. Once the present proposition has been established, however,
that operation will have been proved legitimate: if you wish to construct a circle with
centre A and radius equal to BC, you may first construct any line segment through A,
for example one of AB or AC 28 (by Postulate 1.21 of §3.3), then extend that line segment
(by Postulate 1.22), then apply the present proposition to find a point on the line at a
distance equal to BC from A, and finally apply Postulate 1.23.
3.8.2
Enunciation of Proposition I.3
Given two unequal line segments, to cut off from the greater a line segment whose length
is equal to the less.
28
Note that one of these might be “degenerate”, if the ends of the line segment coincide.
Information for Students in Math 348 2003 09
3.8.3
19
Detailed restatement of the enunciation
Let AB and CD be the two given line segments, of which |AB| is the greater length; it
is required to cut off from AB a part equal in length to |CD|.
3.8.4
Construction and sketch of proof
Use Proposition I.2 to construct a segment AE such that |AE| = |CD|. Then the circle
with centre A, passing through E, will meet AB in a point whose distance from A is
equal to |AE|, hence equal to |CD| by Axiom 1.29
Therefore, given the two straight lines, AB, CD, of which AB is greater in length, a
line segment AE has been cut off, equal in length to CD.
29
Here again one can quibble that there is an assumption of continuity that was not explicit; an axiom
of continuity would state, in technical language, that there could not be “holes” in either the circle or
the line.
One can raise another type of objection also. The axioms have not been precise about what would
be meant by “greater length” — remember that Euclid did not have a formalized real number system
available.
Information for Students in Math 348 2003 09
20
Distribution Date: Sunday, September 14th, 2003
3.9
3.9.1
Proposition I.4
A problematic proposition
By current standards the proof presented for this proposition is inadequate. Coxeter
observes [4, p. 5] that “Euclid’s ‘principle of superposition,’ used in proving Proposition
I.4, raises the question whether a figure can be moved without changing its internal
structure. This principle is nowadays replaced by further explicit assumptions. If the
principle of superposition is a legitimate tool, could it not have been used in the proof
of Proposition I.3?
Some explanation is needed for the language of the enunciation below. Euclid’s
custom was to designate one side of a triangle as the “base”, and the other two as “the
sides”.30 When he says that “the triangle will be equal to the triangle”, he appears to
be referring to a relation that we would be more likely to call congruence today —
asserting that the two triangles are equal in both the lengths of corresponding sides
and the magnitudes of corresponding of angles; triangles related under the relation of
congruence are said to be congruent.
3.9.2
Enunciation of Proposition I.4
If two triangles have two sides equal to two sides respectively, and have the angles
contained by the equal line segments equal, they will also have the base equal to the
base, the triangle will be equal to the triangle, and the remaining angles will be equal to
the remaining angles respectively, namely those which the equal sides subtend.
3.9.3
Restatement of the enunciation in more modern terms
If two triangles have the lengths of two sides of the one equal to the lengths of two sides
respectively, and have the angles contained by the equal line segments equal, they will
also have the base equal to the base, the triangle will be equal to the triangle, and the
remaining angles will be equal to the remaining angles respectively, namely those which
the equal sides subtend.
3.9.4
Further comments on Proposition I.4
Some of the ambiguities in the enunciation could be resolved by more careful use of the
terms; but the intention of the proposition becomes clear when the points and angles are
30
Heath [9, I, p. 249] suggests that the term base “must have been suggested by the practice of
drawing the particular side horizontally...and the rest of the figure above it.”
Information for Students in Math 348 2003 09
21
named. Consider the first sections of the proof31 :
Let ABC, DEF be two triangles having the two sides AB, AC equal in
length to the two sides DE, DF respectively, namely |AB| to |DE| and
|AC| to |DF |, and the angle BAC equal to the angle EDF . I say that the
base BC is also equal to the base EF , the triangle ABC will be equal to
the triangle DEF , and the remaining angles will be equal to the remaining
angles respectively, namely those which the equal sides subtend, that is, the
angle ABC to the angle DEF , and the angle ACB to the angle DF E.
The removal of ambiguities does not remove objections to the basing of a proof on the
questionable concept of superposition. One solution could be to designate the enunciated
statement above as a new axiom, or to introduce new axioms from which the proposition
would be a logical consequence.
3.10
A detailed look at Proposition I.5, sometimes called “Pons
Asinorum”
3.10.1
Euclid’s proof of this proposition repeatedly uses Proposition I.4 and the constructive
propositions I.1, I.2, I.3. We will discuss his proof in detail. The textbook gives a newer
proof, due to Pappus of Alexandria [4, §1.3], which we will also discuss.
The textbook discusses another approach to the theorem, in which the apex — here
the vertex between the two sides of equal length — is joined to the midpoint of the base.
Such a proof would require substantial reworking of other proofs in Euclid, since it is
not shown until I.10 that the midpoint can be found, and the congruence of the two
triangles formed would require not I.4, but I.8. However, the proposal to use the concept
of reflection to prove theorems in geometry is an important one, and will be seen again
in the textbook.32
The name, Pons Asinorum. While the name was used for a proposition in Euclid
over an extended period of time, it has not always been applied to the same proposition!
(cf. §3.25.4).
31
called, by the ancient Greek logicians, the ²κθ²σις = “ekthesis” = “setting out” and the διoρισµoς =
“diorismos” = “specification”. In the sequel we will call this “Detailed restatement of the enunciation”,
and may paraphrase Euclid’s version.
32
It is possible to reconstruct Euclidean geometry by taking reflection as one of the building blocks.
This is done, for example, in the book [2]. This reference is given only for academic completeness —
whenever possible, a reference to a theorem should be substantiated by a citation; students in MATH
348 are not expected to chase down these references, or even to remember that there was a reference.
Information for Students in Math 348 2003 09
3.10.2
22
Enunciation of Proposition I.5 in general terms
In isosceles33 triangles the angles at the base are equal to one another; and, if the equal
sides be extended below the base, the angles under the base will also be equal to one
another.
3.10.3
Detailed restatement of the enunciation
Let ABC be an isosceles triangle, having |AB| = |AC|; and let sides AB and AC be
produced further (away from A) to form line segments ABD and ACE (by Postulate
1.22 of §3.3). I say that ∠ABC = ∠ACB, and that ∠CBD = ∠BCE.
u
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1. Let F be any34 point in line segment BD (the extension of AB). Construct a circle
33
Euclid did not include equilateral triangles among those he called isosceles. His proof is valid for
equilateral triangles as well.
34
We shall see from the construction that there could be some restriction on the choice of F , since
Information for Students in Math 348 2003 09
23
with centre A, passing through F ; let G be the point where the circle intersects
AE.
2. Join BG, CF .
3. We prove that triangles ACF and ABG are congruent, i.e. that their vertices may
be put into one-to-one correspondence so that the angles at corresponding vertices
are equal and the lengths of the sides joining corresponding pairs of points are
equal. The tool we have available for such a conclusion is Proposition I.4, which
requires that there be vertices in the two triangles with equal angles, and that the
sides surrounding these equal angles are also pairwise equal in length.
First consider 4ABG and 4ACF . We know that
|AB| = |AC| (hypothesis)
|AG| = |AF | (construction)
∠BAG = ∠CAF (same angle)
(7)
(8)
and can conclude by the previous proposition that the triangles are congruent.
That means that, in addition to the pairs of sides and angles mentioned above,
|BG| = |F C|
∠ABG = ∠ACF
∠BGA = ∠CF A
(9)
(10)
(11)
4. Now we apply Proposition I.4 to 4BGC and 4CF B. One pair of sides are equal
in length, by (9). We have a pair of equal angles by
∠BGC = ∠BGA (same angle)
= ∠CF A by (11)
= ∠CF B (same angle).
For the second pair of equal sides, we have
|CG| =
=
=
=
|AG| − |AC|
|AF | − |AC| (by construction)
|AF | − |AB| (by hypothesis)
|BF |
we need to ensure that segment ACG be long enough to intersect the circle we shall be constructing.
Think about other ways to describe the construction so that quibbles like this can be avoided.
Information for Students in Math 348 2003 09
24
As the two pairs of equal sides enclose the equal angles, the conditions of I.4 are
satisfied, and we can conclude that 4BGC ≡ 4CF B.35 From this congruence
we may infer the equality in length of the third pair of line segments, and of two
angles:
|BC| = |BC|
∠GCB = ∠F BC
∠GBC = ∠F CB
(12)
(13)
5. Then (12) is the second statement that was to be proved.
6. Finally
∠ABC =
=
=
=
∠ABG − ∠GBC
∠ACF − ∠GBC
∠ACF − ∠F CB
∠ACB
¤
(by (10))
(by (13))
Exercise 3.3 In Euclid’s proof of I.5 points are chosen in the productions of the two
equal sides. Modify that proof by taking points in the sides themselves (not in the
productions). (Do not attempt to prove equality of the “angles under the base”.)
Solution: See §A.1.1, #1, 1001. This proof is due to Proclus, and is reproduced in [9, p.
254].
3.10.4
A proof of the Pons Asinorum proposition by Pappus of Alexandria
Pappus of Alexandria [4, p. 6], [9, Vo. I, p. 254] is reported to have produced the following
very short proof, which can be viewed as requiring no construction at all.
Let us conceive of this one triangle as being two triangles, and let us apply Proposition
I.4 to these two triangles, but with the vertices taken in different orders in the two
triangles; we will prove that 4ABC ≡ 4ACB, where vertex B in each triangle will
correspond to vertex C in the other. The conditions of I.4 are satisfied:
∠BAC = ∠CAB (same angle)
AB = AC (hypothesis, 4 is isosceles)
AC = AB
35
The symbol ≡ is often used to denote congruence of triangles.
Information for Students in Math 348 2003 09
25
Hence the triangles are congruent, implying three other equations:
|BC| = |CB|
∠ABC = ∠ACB
∠ACB = ∠ABC
of which the second (and third) are the equality that we wished to prove.
3.10.5
What about equilateral triangles?
Since Euclid’s definition (#20, §3.2) of isosceles does not permit an equilateral triangle
to be so designated, can we apply I.5 to such triangles? Technically, we cannot do so yet,
since the enunciation does not claim to apply to equilateral triangles. But, if we check
the proofs line by line, we see that the proofs would be perfectly valid for equilateral
triangles — we have never used the hypothesis that the triangle should not be equilateral.
We can cope with this difficulty in two ways:
• We could revise the enunciation to include the equilateral case.
• We could state a new result:
Corollary 3.1 (to Proposition I.5) The results of Proposition I.5 also apply
when the triangle is equilateral.
The term corollary is used in more than one way to describe a theorem linked to another
theorem. In the present use of the term, the new theorem is such that a proof could be
obtained by minor changes to the proof of the original theorem. In fact, in this case, no
changes at all are required! Sometimes the term is used to describe a consequence of the
original theorem, which is not the case here.
3.11
Proposition I.6
3.11.1
The “converse” to the first part of Proposition I.5
As we shall see in §14.8, the converse of a statement “A implies B” is the statement “B
implies A”.
To use the converse here entails a slight abuse of language. The theorem that we wish to
reverse is not just one implication, but asserts the truth of a certain statement in all isosceles
triangles. Since the statement does not have the form of an implication, how can we reverse it?
In practice mathematicians (ab)use the term “converse” in the way it is being (ab)used here,
without any difficulty: the implication which is being reversed is the statement about any one
triangle.
Information for Students in Math 348 2003 09
3.11.2
26
Enunciation of Proposition I.6
q
A
¡T
¡
q
D
¡b
¡
q
B¡
T
b T
b T
b
b
TqC
If, in a triangle, two angles be equal to one another, the sides which subtend the equal
angles will also be equal to one another.
3.11.3
Detailed restatement of the enunciation
Let ABC be a triangle having ∠ABC = ∠ACB. I say that |AB| = |AC|.
3.11.4
Proof
For, if
|AB| =
6 |AC| ,
(14)
one of them is greater. Let |AB| be greater, and from AB — the greater — let DB be
cut off, of in length |AC| (using Proposition I.3); let D and C be joined. We now apply
Proposition I.4 to 4ABC and 4DCB, with corresponding vertices given in corresponding positions in the labelling given:
|AC| = |DB| by construction
|CB| = |BC| same edge, same undirected length
∠ACB = ∠DBC equal angles hypothesis
and conclude that 4ACB ≡ 4DBC. At this point Euclid appeals to Axiom 5: The
whole is greater than the part. The whole is 4ACB; its part is 4DBC.
We have proved that the whole is congruent to its part. Thus, when we begin by
assuming the truth of (14), we arrive at a conclusion that is nonsense. We have to
conclude that the assumption was unjustified, and (14) is false; i.e. that
|AB| = |AC| ,
Q. E. D.
(15)
Information for Students in Math 348 2003 09
3.11.5
27
Difficulties with this proof
Is this the type of application we expected when we read the axiom? We interpreted
“equality” in Proposition I.4 in terms of lengths of sides and magnitudes of angles. Now
we see that Euclid intended something more. Perhaps he is thinking of the areas of the
triangles. By modern standards some more precise axiom is needed. But remember, the
purpose of the preceding comment is not to suggest that Euclid’s development was, at the time,
anything less than brilliant!
A “cleaner” ending to the proof Euclid was comfortable speaking of 4DBC as
being part of 4ABC, but this is a property that can cause us some reservations today,
because it appears to have some spacial assumptions that we may feel lack the necessary
postulates. We can complete the proof in another way, once we know that 4ACB ≡
4DBC. If we list the three equations that are consequences of this congruence, one of
them is
∠ABC = ∠DCB .
When combined with the hypothesis that ∠ABC = ∠ACB, we may infer that ∠ACB =
∠DCB , which is another statement of a whole equal to its part, but in terms of angles
rather than triangles.
Exercise 3.4 Converses are not always true. We can have 2 converse statements such
that
– both are true
– neither is true
– one is true and one is false.
In each of the following cases
• Write the converse.
• Determine whether or not the statement is true, and whether or not the converse
is true.
• Indicate which of the preceding situations obtains: i.e. both true, neither true, just
one true.
(The “universe” is the real numbers, R.)
1. If x3 = y 3 , then x = y.
2. If x2 = y 2 , then x = y.
3. If x = 5, then x = 11.
Information for Students in Math 348 2003 09
28
3.12
Proposition I.7
3.12.1
Enunciation in general terms (slightly modernized from Euclid’s language)
Given two line segments constructed from the ends of a line segment, and meeting in a
point, there cannot be constructed from the ends of the same line segment and on the
same side of it, two other line segments meeting in a different point, respectively of the
same lengths as the former two line segments, namely each equal to that which emanates
from the same end of the fixed line segment.
3.12.2
Enunciation in detailed terms
Let AB be a fixed line segment, and C and D be points on one side of the segment.
Assume that |AC| = |AD| and |BC| = |BD|. It will be proved that C and D cannot be
distinct, by assuming that C 6= D and inferring from that a contradiction.
3.12.3
Proof
Join C and D. Considering 4ACD, we infer from the equality |AC| = |AD| that
∠DCA = ∠ADC
(by Proposition I.5) .
(16)
We will express Euclid’s proof algebraically:
∠CDB = ∠ADC + ∠ADB
decomposition into two summed angles
= ∠DCA + ∠ADB
by(16)
= (∠DCB + ∠ACB) + ∠ADB
decomposition into two summed angles
< ∠DCB
(17)
(18)
(19)
(20)
(21)
(22)
But 4CDB has sides BC and BD of equal length; by Proposition I.5,
∠CDB = ∠DCB
(23)
which contradicts the preceding inequality. We infer from the contradiction the falsity
of our hypothesis that distinct points C and D exist with the stated properties, i.e. we
infer the uniqueness of C.
q
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Information for Students in Math 348 2003 09
3.12.4
29
Difficulties with Proposition I.7
The proof given above was extracted from Euclid’s by expressing the relations of angles
algebraically. But we have not assumed properties that would justify the algebraic
operations we have performed. The inference of (22) from (21) is based on assumptions
about inequalities that have not been made explicit. In Euclid’s proof he argued that
“∠ACD = ∠ADC, therefore ∠ADC is greater than ∠DCB; therefore ∠CDB is much
greater than ∠DCB”. Was Euclid justified in stating the first inequality? The geometer
Proclus36 , observed that a second case was possible, represented by the following figure.
C¡qJ
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J@
¡ ((
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(Proclus is known to have often observed that Euclid’s constructions admitted several
cases; Euclid, it must be assumed, chose to discuss only one case where the other cases
could be proved in a similar way.)
3.13
Proposition I.8
3.13.1
Enunciation
If two triangles have two sides equal in length to two sides respectively, and also have
the bases equal in length, they will also have the angles equal which are contained by
the equal sides.
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3.13.2
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@
Detailed restatement of the enunciation
Let ABC and DEF be two triangles, such that
|AB| = |DE|
|AC| = |DF |
|BC| = |EF |
36
(24)
(25)
(26)
Proclus Lycius “Diadochus”, born approximately 410 A.D., in Lydia; died 485 in Athens, a successor
of Plutarch the Athenian, who had revived the School of Plato; was the author of numerous scientific,
mathematical, and philosophic works, including a commentary on Euclid’s Στ oιχ²ια (Elements).
Information for Students in Math 348 2003 09
30
I say that
∠BAC = ∠EDF .
3.13.3
(27)
Idea of Euclid’s Proof
For, if 4ABC be applied to 4DEF , and if pointB be placed on point E and side BC
on EF , then point C will coincide with point F , by (26). Then Proposition I.7 ensures
that line segments BA and CA can intersect on the same side of EF as D only at D.
Thus ∠BAC = ∠EDF .
3.13.4
Philo’s proof
Proclus quotes the following proof, which does not require the use of Proposition I.7.
Let 4ABC be applied to 4DEF so that B is placed on E and BC on EF , but so that
A falls on the opposite side of EF from D, taking the position G. Then C will coincide
with F , since |BC| = |EF |. Now F G will either be in a straight line with DF , or will
make an angle with it; and, in the latter case, the angle will either be “interior” to the
figure or “exterior”. There are three cases:
1. Case 1: F G is in a straight line with DF . Since |DE| = |EG|, 4DEG is isosceles;
by Proposition I.5, ∠D = ∠G.
2. Case 2: DF and F G form an angle “interior” to the figure. Let DG be joined. Since
|DE| = |EG|, 4DEG is isosceles; by Proposition I.5, ∠EDG = ∠EGD; since
|DF | = |F G|, 4DF G is isosceles; by Proposition I.5, ∠F DG = ∠F GD. By
addition, the whole ∠EDF is equal to the whole ∠EGF .
3. Case 3. DF and F G form an angle “exterior” to the figure. Let DG be joined.
The proof proceeds as in the last case, except that subtraction takes the place
of addition, and the remaining ∠EDF is equal to the remaining ∠EGF .
Therefore, in all three cases,
∠EDF = ∠EGF
= ∠BAC
Information for Students in Math 348 2003 09
Case 1
Case 2
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31
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Case 3
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The proved equality of angles now permits the application of Proposition I.4 to 4ABC
and 4DEF .
3.13.5
A congruence theorem
If we remove the distinction of one side as the “base”, this theorem proves that two
triangles are congruent if their sides are pairwise equal in length. This, after I.4, is the
second congruence theorem we have proved. A third congruence theorem is Proposition
I. 26, which is concerned with the case of one pair of equal sides, and two pairs of equal
angles.
Information for Students in Math 348 2003 09
32
Distribution Date: Tuesday, September 16th, 2003
3.14
Proposition I.9
3.14.1
Enunciation in general terms
To bisect a given angle.37
3.14.2
Detailed restatement of the enunciation
Let the angle BAC be given. Thus it is required to bisect it.
3.14.3
Construction
Let a38 point D be taken on AB. By Proposition I.3, we may locate a point E on ray AC
(or its extension) such that |AE| = |AD|. Let DE be joined, and construct on DE an
equilateral triangle DEF (by Proposition I.1). We need to ensure that the triangle be
constructed on the side of DE opposite from A, (even though Euclid did not explicitly
state that). We claim that AF bisects ∠BAC.
3.14.4
Proof
We apply congruence proposition I.8 to 4AF D and 4AF E:
|AF | = |AF |
|AD| = |AE| by construction
|DF | = |EF | equilateral triangle, construction
hence 4AF D ≡ 4AF E, from which it follows that
∠DAF = ∠EAF
∠DF A = ∠EF A
∠ADF = ∠AEF
the first of which equalities states that ray AF bisects ∠BAC.
Q.E.F.
37
Euclid speaks of a given rectilinear angle: later in the Elements he will be considering the angles
between intersecting curves other than lines. Modern practice would be to consider such “angles” as the
angles between the tangents at the point of intersection; i.e., to restrict the term angle to intersecting
lines.
38
Euclid speaks of a point taken at random; but this is not necessary: nothing is gained by making
the selection of D appear to be completely general, as the construction still gives a bisecting ray.
Information for Students in Math 348 2003 09
3.15
Proposition I.10
3.15.1
Enunciation in general terms
33
To bisect a given line segment.
3.15.2
Detailed restatement of the enunciation
In any given line segment, AB, we wish to find a point D such that |AD| = |DB|.
3.15.3
Construction
By Proposition I.1 we may construct an equilateral 4ABC on the given line segment
AB. By Proposition I.9 we may bisect ∠ACB by a ray, which meets AB in a point we
shall denote by D. We claim that D is the bisecting point — we would now call it the
midpoint of AB.
3.15.4
Proof
Consider triangles ADC and BDC. Since
|DC| = |DC| common side
|AC| = |BC| construction, equilateral 4
∠ACD = ∠BCD construction, Prop. I.9
4ADC ≡ 4BDC by congruence Proposition I.4. It follows that
|AD| = |BD|
∠CAD = ∠CBD
∠CDA = ∠CDB
the first of which equalities states that D bisects the given line segment.
Q.E.F.
3.16
Proposition I.11
3.16.1
Enunciation in general terms
To draw a straight line at right angles to a given line segment, from a given point on it.
3.16.2
Detailed restatement of the enunciation
Let AB be the given line segment, and C be a point on it. Thus it is required to draw
from the point C a straight line at right angles to AB.
Information for Students in Math 348 2003 09
3.16.3
34
Construction, based on Euclid’s but with some ambiguity removed
C divides AB into two subsegments — AC and CB. Take a point D in AC. Apply
Proposition I.339 to find a point E in CB such that |CE| = |CD|. Now apply Proposition
I.1 to construct an equilateral triangle on base DE, and join the third vertex of this
triangle — call it F — to C. We claim that F C is perpendicular to AB, and evidently
it passes through C, the given point on AB.
3.16.4
Proof
Consider triangles DCF and ECF . Since
|CF | = |CF | common side
|DC| = |CE| construction
|DF | = |EF | constructed equilateral 4
4DCF ≡ 4ECF by congruence Proposition I.8. It follows that
∠F CD = ∠F CE
∠F DE = ∠F ED
∠DF C = ∠EF C
the first of which equalities states that line segment AB makes adjacent angles formed
by the ray CF equal; by Definition 10 of §3.2, these are right angles (and ray CF is
perpendicular to AB).
3.17
Proposition I.12
3.17.1
Enunciation in general terms
To a given straight line, from a given point which is not on it, to draw a perpendicular
straight line.
3.17.2
Detailed restatement of the enunciation
Let AB be the given straight line, and C the given point which is not on it; thus it is
required to draw a perpendicular from C to AB.
39
This is the justification given in [9, Vol 1, p. 169]. However, we do not need the full strength of
the Proposition, and could simply take E to be the intersection of the circle with centre C, and passing
through D.
Information for Students in Math 348 2003 09
3.17.3
35
Construction
Take a point D on the opposite side of AB from C. By Postulate 1.23 of §3.3 one may
construct a circle with centre C, passing through D; denote the intersections of that
circle with line AB by E and G. Let the line segment EG be bisected at H, and the line
segments CG, CH, CE be joined. Then we claim that CH is the desired perpendicular
to AB through C.
3.17.4
Proof
Applying Proposition I.8 to 4CHG and 4CHE, which have a common side, we see
that the three pairs of corresponding angles are equal; in particular ∠CHG = ∠CHE.
As these are adjacent angles formed by line AB, and as they are equal, they are right
angles, by definition. Q.E.F.
3.18
Proposition I.13
3.18.1
Enunciation in general terms
A line segment set up on a line makes angles with that line which sum to two right
angles.
3.18.2
Detailed restatement of the enunciation
For let any line segment AB be set up on the line segment CD, making angles CBA,
ABD. I say that ∠CBA and ∠ABD are either both right angles, or, in any case, sum
to two right angles.
3.18.3
Proof
If the angles are equal, then, by Definition 10 they are each right angles, and their sum is
two right angles. If they are not equal, let BE be drawn from the point B perpendicular
to AB. Suppose, without limiting generality, that BE lies in the angle ABD. Then
∠ABD + ∠ABC =
=
=
=
(∠DBE + ∠EBA) + ∠ABD
∠DBE + (∠EBA + ∠ABC)
∠DBE + ∠EBC
2 right angles
(A modern algebraist would be concerned about the removal of parentheses in the preceding computations; this is the property of associativity, that is given by the identity
x + (y + z) = (x + y) + z
Information for Students in Math 348 2003 09
36
for any real numbers. It is clear from Euclid’s proof here and earlier that he is assuming
this property for the addition of angles. In a modern development this property, or
some other properties implying it, would have to be among the axioms/postulates of the
system.)
3.19
Proposition I.14
3.19.1
Enunciation in general terms
If at any point on a line segment, two line segments not lying on the same side make the
adjacent angles equal to two right angles, the two line segments will be in a straight line
with one another.
3.19.2
Detailed restatement of the enunciation
With any line segment AB let the line segments BC and BD, not lying on the same
side, make the adjacent angles ABC, ABD together sum to two right angles. I say that
DB is in a straight line with BC.
3.19.3
Proof
Suppose that DB and BC do not form a line segment, i.e. that, DB, when produced
beyond B, does not pass through C: let E be a point on the production of DB beyond
B.
By “not lying on the same side” Euclid means that the segments are not on the same
side of the ray obtained from AB by extending it beyond B.
Two cases. There are two possible cases here, depending on which side of the extension
of DB the ray BE lies. Euclid discusses only one of these cases; you will be asked to
consider the other case in an exercise.
Aq
qA
Cq
qB
qE
@
@
@qD
Case 1
Cq
Case 2
©qD
©
q©
q
B
E
Information for Students in Math 348 2003 09
37
1. Case 1. Suppose that ray BE is contained in ∠ABD. Then
2 right angles =
=
=
=
∠ABC + ∠ABD hypothesis
∠ABC + (∠ABE + ∠EBD)
(∠ABC + ∠ABE) + ∠EBD
2 right angles + ∠EBD
by Theorem 13 applied to line segment AB making angles with line CBE. Subtracting 2 right angles from both sides of this equation yields a contradiction (to
Axiom 3. We conclude that the hypothesis that BE does not lie on BD is false;
thus
∠ABC + ∠ABD = 2 right angles
¤
2. Case 2. Suppose that ray BE is contained in ∠DBC.
Exercise 3.5 Complete the proof of Cases 2 of the preceding proposition. Proof:
Case 2. By Theorem 13, applied to line segment AB, making angles with line CBE,
2 right angles =
=
=
=
∠ABC + ∠ABE construction
∠ABC + (∠ABD + ∠EBD)
(∠ABC + ∠ABD) + ∠EBD
2 right angles + ∠EBD by hypothesis
From this contradiction we conclude that it is not possible that the production of CB
lie in ∠CBD.
3.20
Proposition I.15
q
c
A c
c
c
D
q
C
E
cq
c
c
q
c
c
Bcq
Information for Students in Math 348 2003 09
3.20.1
38
Enunciation in general terms
If two lines cut one another, they make the vertical angles equal.
3.20.2
Detailed restatement of the enunciation
Let straight lines AB and CD cut one another at the point E; I say that ∠AEC = ∠DEB
and that ∠CEB = ∠AED.
3.20.3
Proof
We can apply Proposition I.13 to the “supplementary” angles on two straight lines:
∠AED + ∠AEC = 2 right angles
(ray AE on 1 side of line DEC)
= ∠AED + ∠DEB
(ray DE on 1 side of line AEB)
By Axiom 1.21 of §3.4, we obtain, after subtraction of ∠AED from the two sums, that
∠AEC = ∠DEB .
Similarly it may be proved that
∠AED = ∠CEB .
3.21
Proposition I.16
3.21.1
Enunciation in general terms
In any triangle, if one of the sides be produced, the exterior angle is greater than either
of the interior and opposite angles.
3.21.2
Detailed restatement of the enunciation
Let one side, BC of 4ABC be produced beyond C to D. I say that the exterior angle
ACD is greater than either of the interior and opposite angles, CBA, BAC.
Information for Students in Math 348 2003 09
3.21.3
39
Proof
A­qB
"
"­
"
­ B "
­
­
Bq"
­
E
­ ""B
­
­""
B ­
Dq
­
Bq­
B q"
CB
B
Bq G
(This proof is based on the construction of a pair of congruent triangles. This might
appear to be a stronger tool than should be necessary to prove the type of inequality
that has been enunciated. But, can you think of a simpler proof?)
Let AC be bisected (by Proposition I.10) at E, and let BE be joined (by Postulate
1.21) and extended past E “sufficiently far”. On that extension a point F is found so
that |EF | = |BE| (applying Proposition I.3). Then F is joined to C (by Postulate 1.21),
and AC is produced beyond C, to a point G.
We apply congruence Proposition I.4 to 4AEB and 4CEF : Since
|AE| = |CE| E is the midpoint of AC
|BE| = |EF | construction of F
∠AEB = ∠CEF “vertical” angles
we may conclude that the triangles are congruent. This implies the equality of one pair
of sides and two pairs of angles. We require only one of these three facts:
∠ECF = ∠EAB
(28)
Hence
∠ACD =
=
=
>
∠ACF + ∠F CD
∠ECF + ∠F CD (same angle)
∠EAB + ∠F CD (28)
∠EAB
by Axiom 5 (§3.4). The inequality ∠DCA > ∠CBA can be proved by a similar construction, beginning with the bisection of BC.
Information for Students in Math 348 2003 09
3.22
40
Proposition I.17
q
A@
A
A@
A @
BAq @q
C
3.22.1
q
D
Enunciation in general terms
In any triangle, two angles taken together in any manner are less than two right angles.
3.22.2
Detailed restatement of the enunciation
Let ABC be a triangle; I say that the sum of two angles of 4ABC is less than two right
angles.
3.22.3
Proof
Let side BC be produced beyond C to D. Then, by Proposition I.16, interior and
opposite angle ABC is less than exterior angle ACD. Adding to both sides of this
inequality ∠ACB yields
∠ABC + ∠ACB < ∠ACD + ∠ACB
= 2 right angles
by Proposition I.13. (We are using properties of inequalities here like the property
mentioned in footnote 20 on p. 13. These properties could be proved from the other
“common notions” by simple consideration of cases, but that will not be done in this
course.) The other 2 inequalities of the same type could be proved analogously, by
extending other edges.
3.23
Proposition I.18
3.23.1
Enunciation in general terms
In any triangle the greater side subtends the greater angle.
3.23.2
Detailed restatement of the enunciation
In 4ABC let |AC| > |AB|. Then ∠ABC > ∠BCA.
Information for Students in Math 348 2003 09
3.23.3
41
Proof
Since |AC| > |AB|, we may, by Proposition I.3, find a point D in AC such that |AD| =
|AB|. Join B and D. Then
∠ABC =
>
=
>
=
∠ABD + ∠DBC
∠ABD (by Axiom 5 of §3.4)
∠ADB (4ABD is isosceles)
∠DCB
∠BCA
since ∠ADB is an exterior angle of 4BDC, and ∠DCB is an interior and opposite angle
(applying Proposition I.16) ¤
3.24
Proposition I.19
3.24.1
Enunciation in general terms
In any triangle the greater angle is subtended by the greater side.
3.24.2
Detailed restatement of the enunciation
Suppose that, in 4ABC,
∠ABC > ∠BCA .
(29)
|AC| > |AB| .
(30)
I say that
3.24.3
Proof
Suppose that (30) is false; i.e. that either
|AC| = |AB| .
(31)
|AC| < |AB| .
(32)
or
Observe first that (31) implies, by Proposition I.5, equality of angles B and C, a contradiction to (29). And (32) implies, by Proposition I.18, that ∠B < ∠C, again contradicting (29). We have shown that (30) must be true.
Information for Students in Math 348 2003 09
3.25
Proposition I.20 – The “Triangle Inequality”
3.25.1
Enunciation in general terms
42
In any triangle, two sides taken together in any manner are greater than the remaining
side.
3.25.2
Detailed restatement of the enunciation
I say that, in any triangle ABC,
|BA| + |AC| > |BC|
|CB| + |BA| > |CA|
and |AC| + |CB| > |AB|
3.25.3
(33)
(34)
(35)
Proof
Applying Axiom 1.22, we extend side BA sufficiently far past A; on the extension, we
cut off a length |AD| = |AC|. We join C and D. Then,
∠BCD = ∠ACB + ∠ACD
= ∠ACB + ∠ADC
(by Proposition I.5)
> ∠ADC = ∠BDC
Applying Proposition I.19 to 4DBC, we infer that |BD| > |BC|. But, by construction,
|BD| = |AB| + |AD| = |AB| + |AC|. This yields (33). The other two inequalities may
be proved in the same way.
3.25.4
Proclus’s commentary on I.20
Proclus reports [11, p. 251]“The epicureans are wont to ridicule this theorem, saying it
is evident even to an ass, and needs no proof... That the present theorem is known to
an ass they make out from the observation that, if straw is placed at one extremity of
the sides, an ass in quest of provender will make his way along the one side and not by
way of the two others.”
3.26
Proposition I.21
3.26.1
Enunciation in general terms
If, on one of the sides of a triangle, from its extremities, there be constructed two line
segments meeting within the triangle, the sum of the lengths of the lines so constructed
Information for Students in Math 348 2003 09
43
will be less than the sum of the lengths of the remaining two sides of the triangle, but the
line segments will contain a greater angle. (This enunciation has been modified from the
original, which did not distinguish between the sides of the triangle and their lengths.)
3.26.2
Detailed restatement of the enunciation
On BC, one of the sides of 4ABC, from its extremities B and C, let line segments BD
and CD be constructed, meeting at D, a point within 4ABC. I say that |BD| + |DC| <
|BA| + |AC|, but that ∠BDC > ∠BAC.
3.26.3
Proof
B
q
AH
Hq E
¡ ©
D©
qXXHH
¡©
XXHH
¡
©
XXH
©
XH
q
Xq
¡
©
C
For let BD be drawn, and extended to meet side AC in E. Then, by Proposition I.20
applied to 4ABE, |BA| + |AE| > |BE|. Adding |EC| to both sides of the inequality
yields |BA| + (|AE| + |EC|) > |BE| + |EC|, which implies that
|BA| + |AC| > |BE| + |EC|
= (|BD| + |DE|) + |EC|
= |BD| + (|DE| + |EC|)
(36)
When we apply Proposition I.20 to 4DEC, we obtain that |DE| + |EC| > |DC|; adding
|BD| to both sides of this inequality yields
|BD| + (|DE| + |EC|) > |BD| + |DC|
which, when combined with (36), yields
|BA| + |AC| > |BD| + |DC| .
Now, applying Proposition I.16 to 4CDE, we conclude that exterior ∠BDC > ∠CED.
By the same proposition, ∠CED = ∠CEB, which is an exterior angle of 4ABE must
therefore be greater than interior and opposite ∠BAE. We have shown that ∠BDC >
∠BAE = ∠BAC, ¤
Information for Students in Math 348 2003 09
3.27
Proposition I.22
3.27.1
Enunciation in general terms
44
Given three line segments, to construct a triangle whose sides have the lengths of the
given segments. Thus, by Proposition I.20, it is necessary that the sum of the lengths of
any two of the given segments exceed the length of the third.
3.27.2
Detailed restatement of the enunciation
Let three line segments be given40 Let the three given line segments be A, B, C, where
all three of the inequalities hold:
|A| + |B| > |C|
|B| + |C| > |A|
|C| + |A| > |B|
It is required to construct a triangle whose sides have lengths equal to |A|, |B|, |C|.
Without limiting generality, let us assume that the segments have been so labelled that
A is (one of) the longest, and C is (one of) the shortest; thus
|A| ≥ |B| ≥ |C|
(37)
Now (37) and (37) are seen to be simple consequences of (37).
3.27.3
Construction
Let DF be a line segment having the length of A, and extend the segment sufficiently
far through F to a point we will call E. Apply the construction of Proposition I.3 to
find a point G in segment F E, such that |F G| = |B|. Then apply the construction a
second time to find a point H in segment GE, such that |GH| = |C|.
By Postulate 1.23 we can describe the circle with centre F , passing through D, and
the circle with centre G, passing through H. Let K be one41 of the points where the two
circles intersect. Join KF and KG. We claim that 4KF G has the desired properties.
40
Euclid’s notation here differs from the modern, in that he chooses to denote his line segments each
by a single capital letter. We will specialize the relative magnitudes of the lengths of the sides, in order
to fill a gap in Euclid’s proof.
41
As we have seen earlier, one needs a postulate enxuring “continuity” in order to prove that these
curves intersect.
Information for Students in Math 348 2003 09
3.27.4
45
Proof
By construction, |F G| = |B|. Since K lies on a circle with radius |GH|, |GK| = |C|.
And, since K lies on a circle with centre F and radius |F D| = |A|, |KF | = |A|. Thus
4F GK has the desired properties. Q. E. F.
3.28
Proposition I.23
3.28.1
Enunciation in general terms
At a given point on a given line segment, to construct an angle equal to a given angle.
3.28.2
Detailed restatement of the enunciation
Let AB be a given line segment, and let ∠DCE be given. We wish to construct a line
segment AF emanating from A such that ∠F AB = ∠DCE.
3.28.3
Construction
Join DE.
Problem with Euclid’s construction. Euclid asks that “4AF G be constructed in
such a way that |CD| = |AF |, |CE| = |AG|, |DE| = |F G|”, where G is a point on
segment AB, and cites the use of Proposition I.22. But the previous proposition does
not provide for the construction of a triangle where one of its vertices, and one of the
sides at that vertex is prescribed. The weakness, however, is not in the construction of
Proposition I.22, but only in its enunciation.
3.29
Proposition I.24
3.29.1
Enunciation in general terms
If two sides of one triangle are respectively equal in length to two sides of another triangle,
and if angle between the two sides of the first triangle is greater than that of the other,
then the base of the first triangle will be longer than that of the second.
3.29.2
Detailed restatement of the enunciation
Let 4ABC and 4DEF be such that
|AB| = |DE|
|AC| = |DF |
∠BAC > ∠EDF
Information for Students in Math 348 2003 09
46
I say that |BC| > |EF |.
3.29.3
Proof
There are actually three cases for this proposition, depending upon the relative magnitude of CB as compared with the lengths of the other sides of 4ABC. Euclid’s practice
was to give the proof of only one of the more difficult cases; in that sense he possibly
followed modern practice. But, in that he did not explicitly state the need to consider
more than one case, his proof can be considered flawed.
By (38) we can apply Proposition I.23 to obtain a line segment DG such that
∠EDG = ∠BAC > ∠EDF ; and such that |DG| = |AC| = |DF |.
Euclid considers the case where F is inside 4DEG. He joins EF , EG.
q
A¶c
¶
¶
q
C¶
q
Dc
c
D¶qc
c
c
c
c B
cq
F
c E
cq
ÃÃ
Ã
q Ã
Ã
c
¶
c
¶
c E
¶
cq
Ã
Ã
G qP
Ã
PPÃ
q Ã
F
He then applies Proposition I.4 to prove congruence of 4DF E with 4ACB from the
equality of the lengths of two pairs of sides, |AB| = |DE| (by hypothesis), and |AC| =
|DG| by construction, and from equality of the angle determined by these pairs of line
segments, ∠EDG = ∠BAC again by virtue of the construction. Congruence of the
triangles implies that
|BC| = |EG|
(38)
and that
∠GF D = ∠F GD
(by Proposition I.5 , since |F D| = |GD|
= ∠F GE + ∠EGD > ∠F GE .
By Proposition I.19, |EG| > |EF |; by (38), |BC| = |EG| > |EF |. The other cases could
be proved in a similar way.
3.30
Proposition I.25
3.30.1
Enunciation in general terms
If two triangles have the two pairs of sides respectively of equal lengths, but have the
base of one longer than the base of the other, they will also have the one of the angles
contained by the sides mentioned greater than the corresponding angle in the other
triangle.
Information for Students in Math 348 2003 09
47
3.30.2
We will not provide discuss this proposition further in this course.
3.31
Proposition I.26
3.31.1
Enunciation in general terms
If two triangles have the two angles at the base equal to two angles, respectively, and
one side of the one equal in length to the corresponding side of the other; namely, either
(Case 1) the side adjoining the equal angles, or (Case 2) that subtending one of the equal
angles, they will also have the corresponding remaining sides equal in length, and the
remaining angle equal to the remaining angle.42
3.31.2
Detailed restatement of the enunciation
Let 4ABC and 4DEF have
∠ABC = ∠DEF
∠BCA = ∠EF D ;
(39)
(40)
|BC| = |EF |,
(41)
|AB| = |DE|.
(42)
in the first case, let
or, in the second case, let
In either case I say that the remaining sides are, respectively, pairwise equal in length,
and the remaining angles are equal.
3.31.3
Proof of Case 1
q
A¡\
¡
¡
B
42
¡
q
¡
\
q
D¡\
\
¡
\
¡
\q
C
E
¡
q
¡
\
Aq
\
\
G¡q¡
c \
\
¡
\q
F
¡
B
q¡
c \
c
\
c
c
\q
C
This is the third congruence theorem we have met, following Proposition I.4 and Proposition I.8.
These three theorems can be interpreted as stating sets of information that determine a triangle. There
is another theorem that can be proved, not found in Euclid, in which a triangle is determined — up to
two possible cases — by the lengths of two sides, and one angle — but not the angle between the sides,
which case is Proposition I.4. This last theorem is sometimes called the “Ambiguous” case.
Information for Students in Math 348 2003 09
48
The proof is by reductio ad absurdum. We assume that (42) is false; without limiting
generality43 , we may assume that |AB| > |DE|. By Proposition I.3 we may find a point
G in AB such that |BG| = |DE|; we then join GC. We may apply Proposition I.4 to
4GBC and 4DEF , since we know the equality of two pairs of sides and the angles they
contain. The inferred congruence of the triangles implies that
|GC| = |DF |
and ∠GCB = ∠EF D
(43)
(44)
Hence ∠BCA = ∠GCB+∠GCA = ∠EF D+∠GCA < ∠EF D, contradicting hypothesis
(40). From this contradicting we infer the falsity of our assumption (that (42) is false);
hence (42) is true. We may now apply Proposition I.4 to 4ABC and 4DEF )as we
have (41), (42) and (39)) to conclude that the triangles are congruent: the remaining
pairs of sides will be equal in length, and the remaining pair of angles will be equal, ¤
3.31.4
Proof of Case 2
q
C¡\
¡
¡
B
¡
q
¡
\
q
F¡\
\
¡
\
¡
\q
A
E
¡
q
¡
\
\
Cq
\
H¡q¡
c \
\
¡
\q
D
¡
B
q¡
c \
c
\
c
\q
c
A
In this case we are assuming in (42) the equality of the lengths of a pair of sides other
than the side adjacent to the two angles given as pairwise equal. Our strategy will be
similar to the preceding: we will prove that (41) holds, and then apply Proposition I.4 to
the two given triangles. We assume that (41) is false; and, without limiting generality,
assume that |BC| > |EF |. We apply Proposition I.3 to find a point H in BC such
that |BH| = |EF |. Applying Proposition I.4 to 4HBA and 4F ED, we infer that
|AH| = |DF | and
∠AHB = ∠DF E .
(45)
Then, Proposition I.16, exterior ∠AHB of 4AHC must be greater than interior and
opposite ∠HCA, i.e. ∠BCA. But this inequality contradicts the equation that results
from the transitivity of equality44 . from this contradiction we infer the falsity of our
assumption that (41) was false: (41) is true. We now have the equalities needed to
apply Proposition I.4 to the original triangles, and may conclude that the triangles are
congruent, ¤
43
We are not limiting generality, because, were we to take the opposite inequality, we could convert
that problem to the one we have solved simply by interchanging the labels of A, B, C with D, E, F
respectively.
44
i.e. Axiom 3.4.1 in (40) and (45)
Information for Students in Math 348 2003 09
3.32
Proposition I.27
3.32.1
Enunciation in general terms
49
If a straight line falling on two straight lines makes alternate angles equal, the straight
lines will be parallel.
3.32.2
Detailed restatement of the enunciation
Let line EF falling on lines45 AB and CD meet them respectively in points E and F ,
and make
∠AEF = ∠EF D .
(46)
Then AB is parallel to CD.
3.32.3
Proof
Since the definition of parallel lines §3.2.23 requires that the lines may not meet when
produced indefinitely in either direction, failure of lines to be parallel entails the meeting
of the lines when extended in one direction.46 The proof is by reductio ad absurdum.
For convenience we will use the symbol k to denote parallelism, and ∦ to denote its
negation. Suppose that AB meets CD when produced in the directions of B and D, in
a point G. Then we have a contradiction to Proposition I.16, in that exterior ∠AEF of
4GEF is equal, by (46) to ∠EF D, which is interior and opposite ∠EF G. From this
contradiction we infer that the lines do not meet when produced in the given direction;
a similar argument shows that they do not meet when produced in the direction of A
and C either. As lines AB and CD do not meet when produced in either direction,
AB k CD.
3.33
Proposition I.28
3.33.1
Enunciation in general terms
If a straight line falling on two straight lines makes an exterior angle equal to the interior
and opposite angle on the same side, or the interior angles on the same side equal to two
45
When we denote a line by a sequence of two or more of its lines, the understanding is that the
line may be extended indefinitely in either direction (by Postulate 1.22, §3.3); the points we choose
to describe the line may have no special significance in the problem at hand, as is the case in this
proposition.
46
Of course, they cannot meet when extended on both sides, as we are implicitly assuming that two
points determine a unique line. This was not explicitly stated as an axiom, although it was intended
in Axiom 1.21. In modern axiomatic treatments of geometry an important issue of this type would
normally appear as an explicit axiom. We will return to this question later in the course.
Information for Students in Math 348 2003 09
50
right angles, the straight lines will be parallel.
3.33.2
Detailed restatement of the enunciation
Let line EF falling on lines AB, CD meet them respectively in points G and H, and
(Case 1) make exterior ∠EGB equal to interior and opposite ∠GHD; or (Case 2) the
interior angles on the same side, namely BGH and GHD sum to two right angles. I say
that AB k CD.
3.33.3
Proof of Case 1
We assume that
∠EGB = ∠GHD .
(47)
∠EGB = ∠AGH ,
(48)
∠AGH = ∠GHD .
(49)
By Proposition I.15,
implying, by Axiom 3.4.1, that
But these are alternate angles, so their equality ensures, by Proposition I.27, that AB k
CD.
3.33.4
Proof of Case 2
We assume that
∠BGH + ∠GHD = 2 right angles.
(50)
∠AGH + ∠BGH = 2 right angles.
(51)
By Proposition I.13,
Subtracting ∠BGH from both members of both equations yields
∠AGH = ∠GHD ,
(52)
which implies that AB k CD, by Proposition I.27.
3.34
Some other propositions of Euclid cited in the textbook
Following are the other propositions listed by Coxeter in [4, §1.3].
1. Proposition III.3. If a diameter of a circle bisects a chord which does not pass
through the centre, it is perpendicular to it; or, if perpendicular to it, it bisects it.
Information for Students in Math 348 2003 09
51
2. Proposition III.20. In a circle the angle at the centre is double the angle at
the circumference, when the rays forming the angles meet the circumference in the
same two points.
3. Proposition III.21. In a circle, a chord subtends equal angles at any points on
the same one of the two arcs determined by the chord.
4. Proposition III.22. The opposite angles of any quadrangle inscribed in a circle
are together equal to two right angles.
5. Proposition III.32. If a chord of a circle be drawn from the point of contact
of a tangent, the angle made by the chord with the tangent is equal to the angle
subtended by the chord at a point on that part of the circumference which lies on
the far side of the chord.
6. Proposition VI.2. If a straight line be drawn parallel to one side of a triangle,
it will cut the other sides proportionately; and, if two sides of a the triangle be
cut proportionately, the line joining the points of section will be parallel to the
remaining side.
7. Proposition VI.4. If corresponding angles of two triangles are equal, then the
sides are proportional.
8. Proposition III.35. If, in a circle, two straight lines cut each other, the rectangle
contained by the segments of the one is equal to the rectangle contained by the
segments of the other.
9. Proposition III.36. If, from a point outside a circle, a secant and a tangent be
drawn, the rectangle contained by the whole secant and the part outside the circle
will be equal to the square on the tangent.
10. Proposition VI.19. Similar47 triangles are to one another in the squared ratio of
their corresponding sides.
11. Proposition I.47. In a right-angled triangle the square on the hypotenuse48 is
equal to the sum of the squares on the two catheti49 .
The converse of the last theorem stated is the last proposition of Book I:
47
Note that the a definition of the term similar was not included in the definitions given above for
Book I. A definition was included with Book VI: [7, Volume 2, p. 188] Similar rectilinear figures are
such as have their angles severally equal and the sides about the equal angles proportional.
48
Euclid does not use the term hypotenuse, but simply speaks of the side opposite the right angle.
49
The two sides, different from the hypotenuse, which meet in the right angle.
Information for Students in Math 348 2003 09
52
12. Proposition I.48. If, in a triangle, the square on one of the sides be equal to
the squares on the remaining two sides of the triangle, the angle contained by the
remaining two sides of the triangle is right.
We will not discuss these theorems now, but may need to return to some of them later
in the course.
Information for Students in Math 348 2003 09
4
53
First Problem Assignment
Distribution Date: on the Web on Thursday, September 18th, 2003;
in hard copy on Monday, September 22nd, 2003.
Solutions are to be submitted by Friday, October 3rd, 2003
Solve each of the following problems using only the propositions in the notes50 from
Euclid’s Elements. Every step of your proof should be justified. Moreover, in some of
these problems you will be asked not to use propositions beyond a given specific location
in Book 1.
1. [13, 2.3H5, p. 61]51 The median from a vertex of a triangle to the opposite side is
a line segment joining the vertex to the midpoint of the opposite side — i.e. to the
point that bisects the side. We will call an edge of an isosceles triangle the base if
the other two sides of the triangle are equal in length: if the triangle is equilateral,
the term will be ambiguous — any one of the sides can be considered the base.
Prove that the median to the base of an isosceles triangle is perpendicular to the
base and bisects the opposite angle.
(Solve this problem without using other propositions than I.1 — I.10.)
2. [13, 2.3I3, p. 64] An altitude of a triangle from a vertex is the line perpendicular to
the opposite side, and passing through the given vertex. Prove that if one of the
altitudes of a triangle is also a median, then the triangle is isosceles.
(Solve this problem without using other propositions than I.1 - I.12.)
3. [13, 2.3O7, p. 73] A quadrilateral is the 4-sided figure formed by joining 4 vertices
cyclically. Thus the quadrilateral ABCD, with vertices A, B, C, D, has sides AB,
BC, CD, DA. The diagonals of ABCD are the line segments AC and BD.(We
are not concerned here whether or not the quadrilateral is convex , i.e. whether line
segments joining points on the quadrilateral lie “inside”.) Prove that the sum of
the lengths of the diagonals of a quadrilateral is less than its perimeter.
4. [13, 2.3Q1, p. 77] Prove that if one of the altitudes of a triangle is also an angle
bisector, then the triangle is isosceles.
50
You may view these notes on WebCT. Some of the versions of the propositions may still undergo
further editing; for that reason, it is suggested that you not print out the full file at this time.
51
These footnotes acknowledge that the problem was taken from a particular source. Students are not
expected to consult the source; there are not likely to be any hints in the source that would be helpful
in solving the problem.
Information for Students in Math 348 2003 09
54
Distribution Date: Monday, September 22nd, 2003
5
Symmetries in the Real Plane
5.1
Sets and Functions
The following definitions and theorems are needed for our discussion of coming geometric
topics. Students are not expected to become experts in the algebra, but should be able
to understand the use of the terminology in the sections following this section.
Some of the material in this and other sections of these notes is not new to many
students. If you are familiar with these concepts, you need not spend any time here. But
be sure that you glance through to be sure that you aren’t missing anything, including
conventions that may be different from your usual practices.52
5.1.1
Basic definitions
We shall not attempt to formalize the basic ideas of set theory. We assume students
have an intuitive understanding of the following concepts and notations
• set, element, member, a ∈ A, subset, superset, B ⊆ A53 , A ⊇ B, A ⊂ B, A ⊃ B
• “set-builder” notation {x|P (x)}, {x : P (x)}
• cardinality |A| of a finite set A. If we need to differentiate between cardinalities of
infinite sets, we will not assume that students are familiar with this topic.
The following definition generalizes the idea of Cartesian coordinates for points in the
real plane.
Definition 5.1
1. The Cartesian product A × B of sets A and B is the set of ordered
54
pairs (a, b) where a and b respectively range over all elements of A and of B.
That is
A × B = {(a, b)|a ∈ A and b ∈ B} ,
or A × B = {(a, b) : a ∈ A and b ∈ B} .
2. The set A × A of ordered pairs of elements of a set A may be denoted by A2 .
52
For example, to be consistent with the textbook, we will write function names to the right of the
variable name, rather than the left, writing xf or xf rather than f (x).
53
Note that we will write B ⊂ A only when B is a proper subset of A, i.e., when there exists at least
one element of A that is not a member of B.
54
It is possible to define what is meant by an ordered pair using only set-theoretic concepts, but we
will not discuss that in this course.
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55
3. For any positive integer n, the set of ordered n-tuples of elements of a set A may
be denoted by An .55
Think of A as replacing the x-axis, and B replacing the y-axis. We generalize the idea
of a real valued function of a real variable:
Definition 5.2 Let A and B be sets. A function f : A → B is a subset of A × B in
which
1. For every point a ∈ A there exists a point (a, b) in the subset.
2. For every point a ∈ A there exists no more than one point (a, b) in the subset.
Where there is no ambiguity, we may refer to the function f : A → B simply as f .
We can think of the function as associating with each point a in A a unique member of
B, called its image; or of mapping A into B. In calculus we are accustomed to denote
this point by f (a); but we will usually use a different notation in this course (cf. §5.1.2).
Definition 5.3 For a function f : A → B
1. The set A is called the domain.
2. The set B is called the codomain.
3. The set {b|(a, b) ∈ f } is called the image 56 of f ; if (a, b) ∈ f , the point b may be
called the image of a, and we may say that f maps a onto b.
4. If the image coincides with the codomain, f is said to be surjective or onto; in that
case we may speak of a function mapping A onto B, rather than simply into.
5. If distinct points are always mapped onto distinct points, the function is said to
be injective or one-to-one.
6. If f is both injective and surjective, it is said to be bijective, or a one to one
correspondence.
55
In practice we shall usually not distinguish between the various ways in obtaining ordered n-tuples.
For example, the sets A2 × A and A × A2 are both different from A3 : their elements have the forms
((a1 , a2 ), a3 ), (a1 , (a2 , a3 )), and (a1 , a2 , a3 ) respectively. Technically these sets are not the same, and
are both different from A3 . If we think of the elements of the sets as being words in the “alphabet”
augmented by the symbols (, ), and the comma, then “corresponding” objects have different numbers
of parentheses, or the parentheses appear in different places.
56
Another term still in use is range. We are avoiding that word because some authors have used it to
mean what we call the codomain.
Information for Students in Math 348 2003 09
56
7. If A = B, a bijective function f : A → A may be called a permutation of A.
8. For any function f : A → A, a point a ∈ A with the property that af = a is said
to be invariant under f ; or, more colloquially, a is a fixed point of the function. If
C ⊆ A, and cf = c for all c ∈ C, then f is said to fix C pointwise. (If we knew
only that xf ∈ C for all x ∈ C, we could say simply that f fixes C.)
5.1.2
Textbook convention: functions compose “on the right”.
Note that the text-book follows the convention that the combination of a mapping f
followed by a mapping g will be denoted by f g. This differs from the notation that
students see in many of their courses, especially in calculus. Students who have studied
calculus at McGill have usually seen the definition
(f ◦ g)(x) = f (g(x))
(53)
and have often suppressed the symbol ◦ and simply named the function f g. But this
notation is inconvenient for algebraists and geometers. To understand the convention,
think of the name of the function as being written after the point on which it acts. That
is, we write x 7→ xf , which could be confusing, because parentheses are not usually used
around the argument x of the function; more often one writes
x 7→ xf ,
writing the name of the function in the upper right hand corner, like an exponent. Then
the composition of two functions can be defined by
¡ ¢g
xf g = xf .
The suppression of parentheses appears to become more difficult if one wants to talk
about functions of several variables. In the usual calculus notation, if we have a function f
from R2 to R, we may denote its image by f (x, y). If we want to follow the “algebraists’ ”
convention, it would appear at first that we should be writing x, y f , which would certainly
be ambiguous: is f acting on the point, or just on its second coordinate? The problem
arises because we should be thinking of the domain of the function as being points
(x, y) ∈ R2 ; that is, the parentheses come from the name of the point, not from the
function notation; so the image is (x, y)f ; but the algebraists’ convention remains to
suppress parentheses when a function has a single variable.
5.1.3
Associativity of function composition
It can be shown easily that, if f, g, h are functions such that the compositions f g and
gh are defined, then
(x)((f g)h) = (x)(f (gh))
Information for Students in Math 348 2003 09
57
and, in practice, we usually suppress parentheses where there is no ambiguity, here
writing simply (x)f gh or xf gh .
5.1.4
Inverses
Definition 5.4
1. A function f : A → B is an injection or is one-to-one if two
distinct points are never carried on to the same point. That is, if, for all points
x, y in the domain, xf = yf ⇒ x = y (where ⇒, when connecting two sentences,
means that the first sentence implies the second, that is, that the second sentence
will be true whenever the first is true).
2. For any set A, the function defined by a 7→ a for all a ∈ A is called the identity
function on A, and is denoted by symbols like ι or ιA or 1A or idA or IA ; the
subscript A may be suppressed where there is no danger of confusion.
3. For a function f : A → B, a function g : B → A is said to be an inverse if both
compositions f g and gf are identity functions.
Theorem 5.1
1. A function cannot have an inverse unless it is injective.
2. A function cannot have an inverse unless it is surjective.
3. A bijection possesses an inverse.
4. If a function has an inverse, that inverse is unique.
A function with an inverse is said to be invertible. Since there can be at most one inverse
for a function f , we may, without ambiguity, use the symbol f −1 for it, when it exists.
Theorem 5.2 If f and g are invertible functions for which the composition f g is defined,
then (f g)−1 = g −1 f −1 .
5.2
Vectors in R2
We may discuss the concept of vectors in the plane at several levels of abstraction. In
[4, §12.6] the author defines a vector as simply a translation which he has defined earlier
in [4, §13.2, p. 195] in the context of “Affine Geometry”; we expect to be studying
“Ordered Geometry”, from which “Affine Geometry” is obtained by the adjoining of
new axioms [4, Axioms 13.11, 13.12, p. 192], but will probably not have time to study
“Affine Geometry” formally. Some students may have met vectors in a course in linear
algebra, defined purely as ordered pairs of real numbers, with familiar operations on the
Cartesian product R × R. In this first encounter in this course, we shall consider the
equivalent concept, of “geometric vectors”. We will minimize the level of formality —
but we may return to secure the underpinnings of the definitions we give.
Information for Students in Math 348 2003 09
58
Geometric vectors in R2 . For a given directed line segment in the plane we can
consider the set of all line segments that can be obtained from it by translating the
segment “in a parallel way” (that is, to segments contained in lines that are parallel to
the line containing the given segment), maintaining its length, its slope, and its “sense”.57
If the segment is directed from point A to point B, we can denote the collection of all
~
these parallel segments of equal length and sense by AB.
If one of the directed line
~
~ the set
segments in AB is directed from C to D, we could also represent the set as CD;
is called a vector , and the directed line segments from A to B and also from C to D
are each a representative of that vector. With operations we are about to define, the set
of vectors is called a vector space. We define two operations: addition of vectors, and
multiplication of vectors by a real number58 .
~ and EF
~ are two given vectors, we define AB
~ + EF
~ as
Addition of vectors. If AB
~ , which has B as its “initial
follows: Select a representative segment from EF
vertex”. That is, you are to find a directed line segment BC that originates at B
and has length |EF |, is parallel to the line segment EF , and has the same “sense”
as the line segment directed from E to F . We define
~ + EF
~ = AC
~ .
AB
(This definition is sometimes called the “triangle rule”; there is an alternative
definition that could also have been used, based on the use of a parallelogram
rather than a triangle.)
Multiplication of a vector by a scalar. Let r be a real number; we will define what
~
~ is the vector of all directed line segments of
we mean by r(AB).
If r = 0, r(AB)
zero length — i.e. where the two ends of the line segment coincide; we may denote
~ is obtained from AB
~ by scaling all directed line
this vector by ~0. If r > 0, r(AB)
segments in the vector by a factor r, without changing the slope or the sense. If
r < 0, change the direction of the directed line segment, then adjust the length by
the factor |r|.
Sometimes we give a vector a name which does not contain the ends of a representative
directed line segment; then we usually still modify the symbol, either using an arrow
~ We may use bold-face type elsewhere
above, as V~ , or boldface type, as V~ or V or V.
when discussing vectors — except for the scalars, which are always denoted by type of
normal thickness. We may use other conventions that are consistent with our symbols
for the algebra of real numbers; for example, we may use a minus sign, as when we write
57
He mean here that the quadrilateral obtained by using the two given segments, and joining the two
“initial” vertices and the two “terminal” vertices is a parallelogram.
58
called a scalar
Information for Students in Math 348 2003 09
59
~−B
~ instead of A+(−1)
~
~ Frequent type changes are impractical in notes like these,
A
B.
so we will often use regular type here for convenience.
We can prove that the operations defined have the properties we would “expect” from
the kind of notation we are using. For example, addition is commutative:
~ =W
~ + V~
V~ + W
~ and associative:
for all vectors V~ and W
~ )+U
~ = V~ + (W
~ +U
~),
(V~ + W
~,U
~.
for all vectors V~ , W
5.2.1
Example of the use of “geometric vectors”: Concurrence of the medians of a triangle
The medians of a triangle are the line segments joining each of the vertices to the
midpoint of the opposite side. The theorem that interests us is
Theorem 5.3 There exists a point in a triangle ABC which lies on all three medians;
equivalently, the medians of a triangle ABC are concurrent.
As is often the case in Mathematics, the theorem is easier to prove if we seek to prove
more. Instead, we shall prove the following
Theorem 5.4 The three medians of a triangle intersect in a point
each vertex to the opposite side.
2
3
of the way from
Students may read a “synthetic” proof of this result in [4, §1.4]. The proof uses Euclid
VI.2 (§3.34.6) and VI.4 (§3.34.7); but, since it also refers to parallelism, it is not surprising
that it requires results that have not been cited yet. In addition to Euclid I.1 — I.28,
discussed earlier in these notes, it may require use of
• Proposition I.29. [9, I, p. 311] A straight line falling on parallel straight lines
make the alternate angles equal to one another, the exterior angle equal to the
interior and opposite angle, and the interior angles on the same side equal to two
right angles.
• Proposition I.30. [9, I, p. 314] Straight lines parallel to the same straight line
are also parallel to one another.
• Proposition I.31. [9, I, p. 315] Through a given point to draw a straight line
parallel to a given straight line .
Information for Students in Math 348 2003 09
60
• Proposition I.32. [9, I, p. 316] In any triangle, if one of the sides be produced,
the exterior angle is equal to the two interior and opposite angles, and the three
interior angles of the triangle are equal to two right angles.
• Proposition I.33. [9, I, p. 307] The straight lines joining equal and parallel
straight lines (at the extremities of which are) in the same directions (respectively)
are themselves also equal and parallel.
Euclid does not include a definition of parallelogram in his formal list of definitions
given in §3.2, but appears to intend a quadrilateral in which alternate sides are parallel
line segments.
The textbook proof, attributed to N. A. Court, requires the property that the diagonals of a parallelogram bisect each other; this could be proved, for example, using some
of the propositions listed above, but we shall not discuss a proof in the lectures.
Students are not expected to be familiar with this “synthetic” proof !
A vector proof of the concurrence of the medians We follow the notation of [4,
Figure 1.4a, p. 10]: the vertices of the triangle are A, B, C, and the midpoints of sides
BC, CA, AB are respectively A0 , B 0 , C 0 . Let O be any point in the plane: we will use
it as the “origin”, and refer the position of all points to it. The position vector of the
point two-thirds of the way from A along the median from A has, relative to the origin
O, the position vector
~ + 2 AA
~ 0 = OA
~ +
OA
3
~ +
= OA
2 ~
~ 0)
(AB + BA
3
1 ~
2 ~
(AB + BC)
3
2
1 ~
1 ~
~ + 1 (OA
~ + AB
~ + BC)
~
=
OA + (OA + AB)
3
3
3
1 ~
~ + OC)
~
=
(OA + OB
3
If we had applied an analogous argument to obtain the position vector of B referred to
~ + OC
~ + OA);
~
O, we would have obtained the position vector 31 (OB
this can be seen to be
equal to the corresponding point on the median from A, since the two sums differ only
in the order of the summands, and we know that vector addition is commutative. Thus
we see that the same point lies on all three medians — i.e. the three medians intersect in
a point; and that, moreover, the point is located two-thirds of the way from the vertex
to the opposite side.
Exercise 5.1 Modify the proofs discussed in §§5.2.1, 5.2.1 to produce a synthetic proof
and then a proof using geometric vectors of the following theorem:
Information for Students in Math 348 2003 09
Definition 5.5 1. Let A, B, C, D be distinct points in R3 . The four triangles whose vertices are 3 of these four vertices are called the faces of
the tetrahedron spanned by A, B, C, D. The tetrahedron is a figure in
R3 which is the convex hull of the points in the triangles — i.e. it is
the union of all line segments obtained by joining by a line segment two
points (not necessarily distinct, nor necessarily in distinct faces). We
usually require that the vertices be in general position, i.e. that
(a) the vertices are distinct;
(b) no three of the vertices are collinear i.e. contained in the same
straight line;
(c) the four vertices are not coplanar , i.e. contained in the same plane.
2. The medians of a tetrahedron are the line segments joining any vertex
to the centroid of the opposite face.
Theorem 5.5 The medians of a tetrahedron are concurrent.
61
Information for Students in Math 348 2003 09
5.3
62
Isometries
Distribution Date: Thursday, September 25th, 2003
5.3.1
Introductory Examples
Example 5.6 Symmetries of the square.
Consider a square in the plane, with
vertices A, B, C, D, and edges AB, BC, CD, DA. We wish to inquire into symmetries
of the square — functions that map the square onto itself and preserve the “essential”
properties. There are several natural sets of properties to examine: we could consider
distances between points on the square, and ask that these remain unchanged under
the mapping; we could consider the angles at the corners; another suggestion would be
to consider only the “adjacencies” between the vertices. All three of these approaches,
or combinations of them, lead to the same set of functions. We will denote them by a
2-rowed array, where the top row simply lists the vertices, in the same convenient order,
and the bottom row shows where each vertex is mapped.
There are the “reflections”, which map each point into the point obtained by projecting along a line perpendicular to a fixed line (the “mirror”), mapping to a point an
equal distance on the other side of the line. Four such reflections are possible, two having
the fixed line perpendicular to two of the sides of the square, and two having the mirror
passing through two of the vertices:
¶
µ
¶
µ
A B C D
A B C D
R2 =
R1 =
D C B A
B A D C
µ
¶
µ
¶
A B C D
A B C D
R3 =
R4 =
A D C B
C B A D
There are also the rotations around the centre of the square, turning through multiples
of 90◦ (= π2 radians); we include the identity mapping here, as a rotation through 0◦ .
These rotations can be expressed as the result of a sequence of reflections. For example,
µ
¶µ
¶ µ
¶
A B C D
A B C D
A B C D
R1 R4 =
=
C B A D
B C D A
B A D C
The other rotations are powers of R1 R4 :
µ
¶µ
A B C D
2
(R1 R4 ) =
B C D A
µ
¶µ
A B C D
3
(R1 R4 ) =
C D A B
µ
¶µ
A B C D
4
(R1 R4 ) =
D A B C
UPDATED TO November 27, 2003
A B C D
B C D A
A B C D
B C D A
A B C D
B C D A
¶
µ
=
¶
µ
=
¶
µ
=
A B C D
C D A B
A B C D
D A B C
A B C D
A B C D
¶
¶
¶
;
Information for Students in Math 348 2003 09
where the last function is simply the
pressed as products in other ways. In
R1 and R4 :
µ
A B C D
R4 R1 R4 =
C B A D
µ
A B C D
=
D C B A
µ
A B C D
R1 R4 R1 =
B A D C
µ
A B C D
=
A D C B
63
identity mapping. These rotations could be exfact, we can even express R2 and R3 in terms of
¶µ
¶
A B C D
B A D C
¶µ
A B C D
C B A D
¶
= R2
¶µ
¶µ
¶
A B C D
A B C D
C B A D
B A D C
¶
= R3 .
We say that the group of symmetries of the square is “generated” by the reflections R1
and R4 . The essential property of the generating reflections is that the angle between
the mirrors be as small as possible, here π4 radians (45◦ ).
Example 5.7 Symmetries of friezes. Consider the infinite strings of symbols
. . . VVV . . .
. . . VDVDVD . . .
. . . HHH . . .
. . . 113113113113 . . .
. . . 31415926589793238462 . . .
. . . IfmusicbethefoodofloveplayonGivemeexcessofitthatsurfeiting . . .
(54)
(55)
(56)
(57)
(58)
(59)
We can ask whether any of these infinite sequences have symmetries, and, if so, how
those symmetries are related. For example, (54) has the following symmetries:
• The infinite string may be shifted to the left or right an integer number of letters.
• The string may be reflected in a vertical mirror lying between two successive V’s.
• The string may be reflected in a vertical mirror passing through the point of any
one of the V’s.
We will be interested in showing how the two reflections together give rise to the shifts
(which we will call translations).
Example (55) has some of the symmetries of the preceding example, but not all
of them; it does not have reflective symmetry in any vertical mirror; but it still has
Information for Students in Math 348 2003 09
64
translational symmetry, even though we cannot arrive at the translations by composing
reflections.
Example (56) has more symmetries than (54), since there is also reflective symmetry
in a horizontal mirror through the bars of the letters H. And, since we have symmetry
in non-parallel mirrors, we also have rotational symmetry, as seen earlier for the square.
Example (57) has translational symmetry and reflective symmetry in a horizontal
mirror (if we assume the 3’s have such symmetry).
Example (58) is not clear. What does the author intend to appear to the left of the
first 3 shown? If it is a string of 0’s, then there will be no symmetry. Could it have been
intended to think of a periodic sequence of digits related to the expansion of π? That
would make sense if π were “periodic”. But it is known that only rational numbers have
periodic decimal expansions, and that π is not rational. So we can’t make sense of this
example.
As for example (59), this would appear to be the beginning of a Shakespeare play,
with spaces and punctuation suppressed. Where does it end? If the intention is that
the play repeats again and again, also to the left, then we have a sequence that has
translational symmetry, and only translational symmetry. There will be infinitely many
powers, both positive and negative, of the “generating” translation, which will be a
translation through one copy of the play.
Example 5.8 The Platonic solids. Read about these five 3-dimensional polytopes in
[4, Chapter 10]. Each of them has a high degree of symmetry, although, in each case,
there are only a finite number of symmetries.
5.3.2
Basic Definitions
An isometry is a function which preserves all distances between pairs of points. We can
consider isometries in any situations where there is a distance function available with the
“usual” properties. Nowadays mathematicians usually associate with the term distance,
or a more technical term, metric, the following properties:
Definition 5.6 Let X be a set. A function ρ : X × X → R is called a metric on X if
it has all of the following properties for all x, y, z in X:
1. (“Non-negativity”) (x, y)ρ ≥ 0;
2. (x, y)ρ = 0 if and only if59 x = y;
3. (“Symmetry”) (x, y)ρ = (y, x)ρ;
4. (“The Triangle Inequality”) (x, z)ρ ≤ (x, y)ρ + (y, z)ρ.
59
usually abbreviated to iff ; French mathematicians may abbreviate si et seulement si to ssi .
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65
Exercise 5.2
1. Let A be any set, and n be a positive integer. On the set An we
propose to define the distance between points (a1 , a2 , ..., an ) and (b1 , b2 , . . . , bn )
to be the number of places in which the two ordered n-tuples differ. That is,
|{i|ai 6= bi }| .60 Prove that the proposed function is a distance function. (It is
called the Hamming distance.)
2. On the set of points with integer coordinates in the real plane R2 , we propose
to define the distance between (x1 , y1 ) and (x2 , y2 ) to be |x1 − x2 | + |y1 − y2 |.61
Show that the proposed function is a distance function. (It is sometimes called the
Manhattan metric, because it resembles the distance between points in mid-town
and uptown Manhattan, New York, measured along a rectangular grid of streets
and avenues.)
3. On any set of points A, define a function f : A × A → R as follows: if a 6= b,
(a, b)ρ = 1; and, for any a ∈ A, (a, a)ρ = 0. Check the definition carefully to see
that ρ is a metric. Mathematicians call this the discrete metric.
Definition 5.7 Suppose that ρ is a metric on X. Then a function f : X → X is an
isometry if, for any two points x, y ∈ X,
(x, y)ρ = (xf, yf )ρ .
(60)
Theorem 5.9 An isometry is an injection.
Proof: Suppose that an isometry f mapped distinct points a, b on to the same point.
Then, by Definition 5.6.2 (“if”), (a, b)ρ 6= 0, but (af, bf )ρ = 0, which contradicts
(60). ¤
60
61
Remember: For any set A we are using the symbol |A| to denote its cardinality.
Here the vertical lines denote absolute value; or, equivalently, distance on the real line.
Information for Students in Math 348 2003 09
5.3.3
66
Further investigation of symmetries of the square
Distribution Date: Wednesday, October 1st, 2003
Continuing the discussion begun in Example 5.6 of these notes, page 62, suppose we set
up a square matrix with the number of rows and columns to be determined, and label
the first four rows and columns by R1 , R2 , R3 , R4 . Does the 4 × 4 submatrix constitute a
“composition table” for these symmetries — i.e., if we compose two of these symmetries,
do we obtain another of them? We have already seen that we do not. Can we expand this
table so as to obtain a set of more than four symmetries that is closed under function
composition? We can! If we define
µ
¶µ
¶ µ
¶
A B C D
A B C D
A B C D
S = R1 R4 =
=
,
B A D C
C B A D
B C D A
we find that all the products of two of R1 , R2 , R3 , R4 are equal either to S or to a power
of S, i.e. to a composition of S with itself a number of times. If we extend our 4 × 4
matrix to an 8 × 8 matrix, with additional columns S, S 2 , S 3 , andµS 4 , where the last
¶
A B C D
4
mentioned symmetry can be seen to be the identity function S =
,
A B C D
the set of 8 transformations is “closed under function composition”. This set has the
following properties:
• Composition is associative, i.e. (T1 T2 )T3 = (T1 T2 )T3 for any three transformations
(not necessarily all different. This is true because our transformations are functions,
and function composition can be seen to always have this property.
• It contains the identity function.
• It contains, with each element, its inverse: the inverse of each of R1 , R2 , R3 , R4
is itself, since the composition of 2 copies of each is the identity; the inverse of
S, S 2 , S 3 , S 4 can be seen to be S 3 , S 2 , S, and the identity.
A set that is endowed with a “binary law of composition” with the preceding properties
is said to constitute a group; we shall return to this concept.
5.3.4
Isometries of the Euclidean plane
Read the discussion in the textbook [4, §3.1, pp. 39-40] on direct and opposite isometries
of the plane. We will not attempt to define these concepts rigorously.
We quote the following theorem from the textbook:
Information for Students in Math 348 2003 09
67
Theorem 5.10 [4, 2.31, p. 29] If an isometry of the plane has more than one invariant
point, the isometry is either the identity function on the plane (which is direct) or is a
reflection in a line (which is opposite).
Proof: Suppose that f : R2 → R2 is an isometry, and that A and B are distinct invariant
points. If C1 and C2 are circles centred at A and B, the points on each circle must be
mapped in their entirety onto the set of points constituting that circle — however, a
point need not be mapped on to the same point on the same circle. A points lying at the
intersection of the two circles will have to be mapped on to the same or another points on
both circles. Two circles meet in either 0 or 1 or 2 points. Points on the line determined
by A and B are characterized by the fact that they are the unique intersection of circles
centred at A and B: this implies that all such points are invariant. Suppose that a
point C not on such a line ` of invariant points can be shown to be invariant; then the
line joining such a point with any point on ` will also be (pointwise) invariant. This
shows that all points in the plane except possibly the points on one line through C 62 are
invariant, and it can then be argued that all points on that line are also invariant, so
the entire plane is invariant — i.e. the isometry is the identity. If there is no invariant
point off the line joining A and B, the argument on [4, 2.31, pp. 29-30] applies: f is a
reflection, and the line AB is the “mirror”, an invariant line. ¤
(Note that the theorem, as stated, does not hold in 3 dimensions. For example,
consider the rotations of 3-space around the z-axis; they have infinitely many fixed
points.)
We state without proof:
Theorem 5.11 [4, 3.13, 3.14, p. 41; etc.]
1. Every isometry of the plane is the product of at most three reflections.
2. The identity mapping may be thought of as the product of 0 reflections.
3. If an isometry of the plane possesses an invariant point, then the isometry is either
• a single reflection (in which case the isometry is opposite);
• the product of two reflections in non-parallel mirrors, which is a rotation (in
which case the isometry is direct).
• the product of two reflections in parallel mirrors. (If the two mirrors are the
same line, the product is the identity.)
62
the line through C parallel to `
Information for Students in Math 348 2003 09
68
Distribution Date: Sunday, October 5th, 2003
5.3.5
The product of two reflections in the Euclidean plane
The product of two reflections in non-parallel lines Analogously to what was
seen in Example 5.6, it can be shown that if Ri is a reflection in line `i (i = 1, 2), then
the composition R1 R2 will be a rotation around the point of intersection — call it `1 · `2
— of `1 and `2 , through an angle which is double the angle between the “mirrors”.
The intersecting lines `1 and `2 divide the plane into 4 sectors, which can be labelled
by their bounding rays, listed in the positive order — which we usually define to be the
counterclockwise order. Two of the sectors begin with a ray of `1 and end with a ray of
`2 ; and two have the reverse order; the first two can be used to specify an angle “from
`1 to `2 ” — by Euclid’s Proposition I.15 these angles are equal. The product R1 R2 ,
(remember, R2 follows R1 in our notation), turns the plane around `1 · `2 through an
angle twice the angle from `1 to `2 in the positive direction.
In particular, the composition of reflections whose mirrors are perpendicular will be
a rotation through an angle of 2 × 90◦ = 180◦ =two right angles; we call such a mapping
a half-turn. Here the order of measuring the angle doesn’t matter!
The product of two reflections in parallel mirrors Suppose that R1 and R2 are
reflections in parallel lines `1 and `2 respectively. To analyze the situation, let’s suppose
that a coordinate system is imposed on the plane so that the mirrors are vertical lines,
`1 having equation x = a1 and `2 having equation x = a2 .63 Then R1 , R2 is given by
½
(ai + (ai − x), y) (x ≤ ai )
(i = 1, 2),
(x, y) 7→
(ai − (x − ai ), y) (x > ai )
i.e.
(x, y) 7→ (2ai − x, y) (i = 1, 2) .
The composition R1 R2 is therefore given by
(x, y) 7→ (2a1 − x, y) 7→ (2a2 − (2a1 − x), y) = (2(a2 − a1 ) + x, y) ,
which is a translation through twice the distance between the mirrors. More precisely,
the translation is in the direction from `1 to `2 , and perpendicular to those mirrors. We
will use the symbol >a,b to denote the function64 given by
(x, y) 7→ (x + a, y + b) ,
63
(61)
These concepts could be developed totally synthetically, based on concepts in Euclidian geometry.
However, as many students lack familiarity with the methods of Euclidean geometry, we will suppress
most proofs, but will occasionally base an argument on an analytic description using an orthogonal
Cartesian coordinate system, as this will be accessible to more students.
64
This is not a standard notation.
Information for Students in Math 348 2003 09
69
so
(x, y)>a,b = (x + a, y + b).65
(62)
R1 R2 = >2(a2 −a1 ),0 .
(63)
and
5.3.6
Half-turns
A half-turn is a rotation through π radians (=180◦ ) about a point. When the rotation is
in a plane that has been endowed with a rectangular cartesian coordinate system, and
the centre of rotation is the origin O, the rotation may be achieved by applying the
mapping
(x, y) 7→ (−x, −y) .
We may need a symbol for this function; let’s denote it by ªO,π : the first subscript will
represent the centre of rotation, and the second will represent the angle through which
the rotation turns — here 180◦ .66 Thus
(x, y) ª(0,0),π = (−x, −y) .67
But remember that this description of the function is given in terms of the coordinates
of a point; if we use a different coordinate system to describe the location of points, then
this description may not be valid.
Let’s find a formula for the half-turn about a general point C(a, b). Imagine a new
cartesian coordinate system, parallel to the old, and with the positive directions in the
same sense, with its origin at C; we will denote the cartesian coordinates referred to the
new system using “wedge” brackets, like hx0 , y 0 i; then the new and old coordinates of a
point P (x, y)hx0 , y 0 i are related by the equations
x = x0 + a
y = y0 + b
(64)
(65)
A half-turn about C, whose coordinates are (a, b) and h0, 0i respectively, would be given
by the mapping
hx0 , y 0 i 7→ h−x0 , −y 0 i .
(66)
if described in terms of the new coordinate system. We can determine a formula describing the action of this function when referred to the original coordinate system:
(x, y) = hx0 , y 0 i
= hx − a, y − bi by (64), (65)
7→ h−(x − a), −(y − b)i by (66)
65
Remember, the symbol for the function is being written to the right of its argument (x, y).
This is not a standard notation!
67
Remember, we chose to write our functions on the right!
66
Information for Students in Math 348 2003 09
70
whose coordinates in the original coordinate system will be (−(x − a) + a, −(y − b) + b),
i.e. (−x + 2a, −y + 2b). In terms of the (x, y) coordinate system, this function is given
by
(x − a, y − b) 7→ (−x − a, −y − b) .
(67)
Thus
(x, y) ª(a,b),π = (−x + 2a, −y + 2b)
5.3.7
(68)
The composition of two half-turns.
Suppose we follow the rotation ª(a1 ,b1 ),π by the rotation ª(a2 ,b2 ),π . The composition
function ª(a1 ,b1 ),π ª(a2 ,b2 ),π will be given by:
(x, y) ª(a1 ,b1 ),π ª(a2 ,b2 ),π
= (−x + 2a1 , −y + 2b1 ) ª(a2 ,b2 ),π
= (−(−x + 2a1 ) + 2a2 , −(−x + 2b1 ) + 2b2 )
= (x + 2(−a1 + a2 ), y + 2(−b1 + b2 ))
(69)
Note that this composition is affected only by the difference between the coordinates
of the two centres of rotation; if we shifted both centres to the right a distance u, and
upwards a distance v, there would be no change in the composition. The composition
is, evidently, the same translation >2(a2 −a1 ),0 that we expressed earlier as a product of
reflections in parallel mirrors x = a1 and x = a2 .
Is this a surprise? In fact, we can prove this without the use of coordinates. The
rotation ª(ai ,bi ),π , being a half-turn, is the product of reflections in two perpendicular
mirrors that intersect in (ai , bi ) (i = 1, 2). All that is required is that each pair of
mirrors be perpendicular; thus, a judicious choice of pairs of mirrors can achieve some
cancellation. Let `2 denote the line through (a1 , b1 ) and (a2 , b2 ), and let R2 denote the
reflection in that line; let `1 be the line through (a1 , b1 ) perpendicular to `2 , and let `3
be the line through (a2 , b2 ) perpendicular to `2 . Then
ª(a1 ,b1 ),π ª(a2 ,b2 ),π =
=
=
=
(R1 R2 )(R2 R3 )
R1 (R2 R2 )R3
R1 IR3
R1 R3 ,
the product of reflections in parallel mirrors `1 and `3 . But such a product has already
been seen to be a translation in the direction from (a1 , b1 ) to (a2 , b3 ), through twice the
distance between those points.
Information for Students in Math 348 2003 09
5.3.8
71
The composition of two translations.
We can determine the result of following translation >a1 ,b1 by translation >a2 ,b2 by substituting in (62), or otherwise (cf. [4, §3.2, p. 41]). We observe that
(x, y)>a2 ,b2 >a1 ,b1 = (x, y)>a1 ,b1 >a2 ,b2
so we may write
>a2 ,b2 >a1 ,b1 = >a1 ,b1 >a2 ,b2
(70)
and say that the operation of composition in the set of translations is commutative.
We also observe that the compositions in (70) are both equal to another translation, >a1 +a2 ,b1 +b2 : we say that the set of translations is closed under the operation of
composition.
Among all the translations, there is just one for which there is a point that is
not moved. We see from equation (62) that, if, for some real numbers x, y, (x, y) =
(x, y)>a,b = (x + a, y + b), then a = b = 0. Thus the only translation that has any fixed
points is the translation >0,0 , and it is the identity function on the plane — mapping
every point in the plane onto itself.
The identity translation has the property that, when it is composed with any other
translation — in either order — it yields no change; that is
>0,0 >a,b = >a,b = >a,b >0,0 .
We say that the translation >0,0 is an identity for the operation of composition on the
set of all translations.
For every translation >a,b there exists another translation that “inverts” that translation; that is, it reverses the effect of the translation; this is the translation >−a,−b ; we
have
>−a,−b >a,b = >0,0 = >a,b >−a,−b .
We may define >−1
a,b = >−a,−b .
We thus have, in the set of translations of the plane, a set with an associated mapping
from ordered pairs to one member of the set which have the following properties:
• It is associative (since all compositions of functions are associative).
• There is an identity element.
• Every element has an inverse.
• Composition is commutative.
We call such a set with associate law of composition an abelian group or a commutative
group.
Information for Students in Math 348 2003 09
5.3.9
72
Composition of three reflections in the plane
Composition of three reflections with parallel mirrors If R1 , R2 , R3 are reflections respectively in parallel mirrors `1 , `2 , `3 , then the product R1 R2 is a translation
(cf. §5.3.5) equal to a product R10 R20 in mirrors `01 and `02 obtained by shifting the original
mirrors in a perpendicular direction, maintaining the distance between the mirrors and
the order of application of the reflections. In particular, we can arrange that `02 = `3 ;
but then R20 = R3 , and the product R3 R3 is the identity. Thus we have
R1 R2 R3 = R10 R20 R3 = R10 R3 R3 = R10 I = R10
so the product is simply a single reflection.
Composition of three reflections whose mirrors are not all parallel Suppose
R1 , R2 , R3 are reflections respectively in mirrors `1 , `2 , `3 , and that `1 is not parallel to
`2 . Then R1 R2 is a rotation, which can be expressed (cf. §5.3.5) as a product of another
pair of mirrors intersecting in the same point as `1 and `2 : we have only to maintain
the angle between the mirrors and the order of applying the reflections. Let’s arrange
that R1 R2 = R10 R20 , where R10 and R20 respectively have mirrors `01 and `02 , and that `02 is
perpendicular to `3 . Now consider the half-turn R20 R3 , and replace the two component
reflections by reflections R200 and R30 in mirrors `002 and `03 , where `002 is perpendicular to `01 .
(3)
Finally, replace the reflections forming the half-turn R10 R200 by reflections R100 and R2 ,
(3)
(3)
respectively in mirrors `001 and `2 , so that `2 is parallel to `03 . We have
(3)
R1 R2 R3 = R10 R20 R3 = R10 R200 R30 = R100 R2 R30
(71)
which is a reflection followed by a translation (cf. §5.3.5) in the direction of its mirror.
Analogously, if `2 is not parallel to `3 , we can prove R1 R2 R3 equal to a similar product
where the translation precedes the reflection. We shall observe a simplification of this
exposition below, in (72).
And what of the two cases we have excluded? We are assuming that not all three
mirrors are parallel, so it must happen that the middle mirror is not simultaneously
parallel to both of the other mirrors. Then the decomposition we called R10 R200 R30 consists
of a middle reflection whose mirror is perpendicular to both of the other two, and we
know that we can always arrive at this decomposition using either the first procedure or
its analogous reversal. From that decomposition we can attain products of a translation
and a reflection in either order, in the obvious way.
This case includes the preceding case of three parallel mirrors, in that the translation
could be the identity mapping. We call an isometry of this type a glide reflection. Except
for the case of the single reflection68 , a glide reflection has no invariant points.
68
Some authors will restrict the term glide reflection to isometries without fixed points.
UPDATED TO November 27, 2003
Information for Students in Math 348 2003 09
73
And what if `1 was parallel to `2 ? We could have applied an analogous argument
beginning with the product R2 R3 — unless R2 was parallel to R3 . But that would take
us to the case treated above, in which the product is equal to a reflection.
5.3.10
A simplification: commuting reflections
Definition 5.8 Two functions F and G are said to commute if they compose in both
orders, and if the compositions are the same function; i.e. if F G and GF are both defined,
and if xF G = xGF for all x in their common domain.
Theorem 5.12 Reflections R1 and R2 in the plane commute if and only
1. R1 = R2 , in which case the composition R1 R2 = R2 R1 is the identity I; or
2. the mirrors of R1 and R2 are perpendicular, in which case the composition R1 R2 =
R2 R1 is the half turn about the point of intersection of the mirrors.
Proof:
Necessity (If ): Suppose first that R1 = R2 . Every reflection is its own inverse; for a
function F which is its own inverse, F F = F F −1 = I.
Next, if R1 and R2 are reflections in perpendicular mirrors, their composition is a
rotation through twice the angle between the mirrors, i.e. through π radians. But
a rotation through this angle is its own inverse!
Sufficiency (Only if ): Suppose that R1 R2 = R2 R1 , but that R1 6= R2 . If the mirrors
of R1 and R2 are parallel, the mirrors cannot coincide, as we assumed the reflections
are different. Then R1 R2 and R2 R1 represent translations in opposite directions,
and they cannot be the same. Thus the mirrors cannot be parallel, must intersect
in a point, and the compositions are rotations about that point, through twice the
angle between the two mirrors, but in opposite (angular) directions; for a rotation
to be its own inverse, the angle of rotation must be π radians. ¤
In our discussion earlier about the composition of three reflections whose mirrors are not
(3)
all parallel, equation (71) can be simplified, since we know that R2 = R10 , and R100 = R200 :
R1 R2 R3 = R10 R20 R3 = R10 R200 R30 = R200 R10 R30
(72)
Information for Students in Math 348 2003 09
6
74
Second Problem Assignment
Distribution Date: Mounted on the Web on Sunday, October 5th, 2003; distributed in
hard copy on Wednesday, October 8th, 2003.
Solutions are to be submitted by Monday, October 20th, 2003
1. Consider a regular hexagon in the Euclidean plane, where the plane has been
endowed with a rectangular Cartesian coordinate system, so that the hexagon is
inscribed in a circle of radius 1 centred at the origin, and its vertices are located at
points making angles of π3 , 2π
, π, 4π
, 5π
, 2π. Now the vertices making the following
3
3
3
π
5π
angles with the centre are joined to the centre by a line segment:
¡ 3 , ππ, 3 .π ¢Label
the vertices cyclically counterclockwise, starting with the vertex cos 3 , sin 3 : A1 ,
A2 , . . ., A6 , so that vertices A1 , A3 , and A5 are joined to the centre O, and the
other three are not.
(a) Make a very rough sketch of the geometric structure we are considering, showing its 6 vertices, the centre, the 6 line segments forming its perimeter, and
the 3 line segments joining boundary vertices to the centre.
(b) Find all isometries of the plane which map the structure on to itself. Denote
the isometry
µ F by a 2-line matrix which denotes its
¶ action on the 6 boundary
A1
A2
A3
A4
A5
A6
vertices:
, where each Ai F is one of
A1 F A2 F A3 F A4 F A5 F A6 F
the vertices of the structure. For each of the isometries, indicate whether it is
a reflection, a rotation, a translation, or a glide reflection. If it is a reflection,
describe the mirror of the reflection; if it is a rotation, describe the centre of
the rotation, and the angle through which the rotation moves, being careful to
take the counterclockwise direction as “positive”; if it is a translation, describe
the direction and the distance; if it is a glide reflection, describe the direction
in which it “glides”, and the line which is mapped onto itself by the isometry.
Include
the “identity” isometry
µ
¶ in your classification, and denote it by either
A1 A2 A3 A4 A5 A6
or I.
A1 A2 A3 A4 A5 A6
(c) Express each of the isometries as a product of a non-negative number of
reflections, always using as few reflections as possible. (There may be more
than one way to do this, but any solution will be acceptable.)
(d) Set up and complete a rectangular “Cayley Table” for your isometries, so that
the product of any two of them, in each order, can be read off from the table.
(e) Determine, for each isometry, its inverse; you may refer to your table in the
preceding part.
Information for Students in Math 348 2003 09
75
(f) Show that there are in your set of isometries two reflections — call them F
and G — such that all the isometries can be expressed as a product of possibly
multiple copies of F and G; that is, that the group of all of your isometries is
generated by F and G.
(g) Show that there are, in your set of isometries one reflection and one rotation
— call them B and C, such that all the isometries can be expressed as a
product of possibly multiple copies of B and C; that is, that the group of all
of your isometries is generated by B and C.
2. Let R1 , R2 and R3 be reflections, respectively in the line y = 1, the line x = y, and
the y-axis. Apply the procedure discussed in §5.3.9 of the notes to express R1 R2 R3
as the product of a reflection R in a mirror ` and a translation in the direction
of the mirror. Show all your work, and describe all the reflections that enter into
your computations carefully, by specifying their mirror in each case. Determine
the distance through which the translation shifts the plane.
The procedure discussed in §5.3.9 produces a product of reflection and translation
in two orders — either the reflection is applied first or last. You are expected to
carry out your calculations also in two orders, and to produce two factorizations,
one with the reflection preceding the translation, and one where it follows the
translation.
Information for Students in Math 348 2003 09
7
76
Symmetries of R2 (continued)
Distribution Date: Friday, October 10th, 2003
7.1
7.1.1
Isometries (continued)
Miscellaneous results on half-turns and translations
Theorem 7.1 If an isometry F has the property that F = R1 R2 , where R1 and R2 are
reflections, then the inverse of F is the composition R2 R1 .
Proof: (R1 R2 )(R2 R1 ) = R1 (R2 R2 )R1 = R1 IR1 = R1 R1 = I. If we precede the compositions above by (R1 R2 )−1 , we obtain
(R1 R2 )−1 (R1 R2 )(R2 R1 ) = (R1 R2 )−1 I
⇒ I(R2 R1 ) = (R1 R2 )−1
⇒ R2 R1 = (R1 R2 )−1
as claimed.
Theorem 7.2 The composition of a half-turn and a translation is a half-turn.
Proof: Without limiting generality, we may take the half-turn to be the composition
R1 R2 of reflections in perpendicular mirrors `1 and `2 ; and the translation as a product
of reflections in mirrors parallel to `2 . Indeed, we may even require that the first of those
reflections be R2 , so we may assume that the translation is R2 R3 , where R3 has mirror
`3 , which is parallel to `2 . Then (R1 R2 )(R2 R3 ) = R1 R3 , which, being the composition
of reflections in perpendicular mirrors `1 and `3 , is a half-turn. ¤
Corollary 7.3 (to Theorem 7.2) Any translation may be expressed as the composition of two half-turns; moreover, either the first or the second of these half-turns may be
prescribed. More precisely, if H0 is a half-turn, then there exist, for any translation T ,
half-turns H1 and H2 such that T = H0 H1 = H2 H0 .
Proof: Let H0 be a prescribed half-turn, and T any translation. By Theorem 7.2 the
composition H0 T is a half-turn; denote it by H1 . Then H0 T = H1 ⇒ H0 H0 T = H0 H1 ⇒
IT = H0 H1 ⇒ T = H0 H1 . An analogous argument shows that there exists a factorization T = H2 H0 , where H2 will depend on T but H0 may be prescribed. ¤
Information for Students in Math 348 2003 09
7.2
7.2.1
77
Groups of isometries
The periods of isometries of the plane
We can now improve on our earlier theorem:
Corollary 7.4 (to Theorem 5.11) Every isometry of the plane is the product of at
most three reflections.
0. The product of 0 reflections is the identity function69 .
1. A single reflection is an isometry70 .
2.(a) The product of two reflections in parallel lines is a translation. This includes the
case where the lines coincide — the identity is also a translation.
2.(b) The product of two reflections in intersecting lines is a rotation through twice the
angle between the mirrors (lines of fixed points) of the reflections; in particular, the
product of two reflections in perpendicular lines is a half-turn. It is convenient to
include the case where the mirrors coincide also: the identity is a rotation through
a zero angle.
3.(a) The product of three reflections in lines that are all parallel is a single reflection.
3.(b) The product of three reflections in lines that are not all parallel is a glide reflection.
We include here the case where the glide reflection is simply a reflection, with no
“glide”.
We can strengthen Theorem 5.9 for isometries of the plane:
Corollary 7.5 (to Theorem 5.11) Every isometry of the plane is a bijection, whose
inverse is also an isometry.
In order the investigate the totality of all symmetries of a particular configuration or
figure in the plane, we introduce the following concept:
Definition 7.1 If, for an isometry f , there exists a positive integer n such that f n = I
— i.e., if the set S = {n|n is a positive integer, f n = I} is not empty, we define
min{n|f n = I, n ∈ N, n > 0}
to be the period or order of f . If S = ∅, we say that f is of infinite period or of infinite
order .
69
which we can denote by ιR2 or IR2 or simply ι or I.
which can be interpreted as being both a translation through a distance of zero, and a rotation
through a zero angle
70
Information for Students in Math 348 2003 09
78
Note that, under this definition, I has period 1. We are using the symbol N for the
natural numbers.71
Rotations through angles that are rational multiples of a whole revolution, i.e. through
angles of the form 2πm
where m and n are integers, n 6= 0, have the property that
n
sufficiently many applications72 yield a rotation through an integer number of revolutions,
which is I. Such revolutions have finite period; all other revolutions have infinite period.
Glide reflections in which the “glide” — the translation — is not “trivial”, i.e. glide
reflections which are not simply a reflection, also have infinite period, since the square
of a glide reflection is a translation.
And, what finite periods are possible? For any positive integer n we can find an
isometry that has n as period:
1. The only isometry with period 1 is the identity.
2. Period 2 is realized as the period of an isometry in two ways: any reflection, and
any half-turn.
n. For any integer n > 2, any rotation through an angle of 2π
radians has period
n
2πm
n; more generally, any rotation through an angle of n radians, where m is any
non-zero integer having no common integer factor with n, except ±1 and ±n.73
Note that, while every isometry of the plane is a product of reflections, whose period
is 2, the period of the product need not be 2. This is because composition is not, in
general, commutative: you cannot move the factors around in order to get two copies of
a reflection to become the identity. For example, if R1 is the reflection in the x-axis, and
R2 is the reflection in the line y = x, R1 R2 is a rotation through an angle of 2 × π4 = π2 ,
whose period is 4; when we consider the product (R1 R2 )(R1 R2 ) we do have the right to
suppress the parentheses; but, in the product R1 R2 R1 R2 , we can’t change the order of
the two middle factors to R1 R1 R2 R2 , which is, in fact, the identity.
7.2.2
Abstract Groups; Isometry Groups
We saw earlier that one particular collection of isometries (the translations of the plane)
had the properties of a group. Let’s now define the concept formally. Where we speak
of a “law of composition”, think of the composition of two functions.
71
As some authors define these to be the non-negative integers, while others (to which camp the
author belongs) define them to be the positive integers, we have qualified the definition with n > 0,
which is redundant with our preferred definition.
72
here n is one such integer
73
We say that m is (relatively) prime to n.
Information for Students in Math 348 2003 09
79
Definition 7.2 Let G be a set, and suppose that there exists a function (called an
operation 74 or law of composition) from G × G to G, given by (g1 , g2 ) 7→ g1 ∗ g2 with the
following properties:
1. For any g1 , g2 , g3 ∈ G,
(g1 ∗ g2 ) ∗ g3 = g1 ∗ (g2 ∗ g3 )
(We call this the property of associativity of ∗.)
2. There exists an element g0 ∈ G such that, for all g ∈ G,
g0 ∗ g = g = g ∗ g0 .
(g0 is called an identity element.)
We will temporarily call the structure consisting of the set G and the operation ∗ a
“semigroup”, until we can prove some properties that must hold. Then we will return
to the definition and add additional structure.
Theorem 7.6 In a “semigroup” an identity function is unique.
Proof: Suppose that, for all g1 ∈ G,
g0 ∗ g
g ∗ g0
g00 ∗ g
g ∗ g00
=
=
=
=
g
g
g
g
(73)
(74)
(75)
(76)
that is, that both g0 and g00 are identities. Note that the “quantification” that permits
g to be any element of G applies independently in the four equations; that is, we are
assuming that
g0 ∗ g
h ∗ g0
g00 ∗ k
` ∗ g00
=
=
=
=
g
h
k
`
(77)
(78)
(79)
(80)
for any g, h, k, ` ∈ S. Then, taking g = g00 , ` = g0 , we obtain
g0 ∗ g00 = g00
g0 ∗ g00 = g0
which together imply that g0 = g00 . ¤
We are now ready to return to our definition to add one more condition:
74
more precisely, a binary operation, because it operates on two elements
(81)
(82)
Information for Students in Math 348 2003 09
80
Definition 7.3 Let G be a set, and suppose that there exists a function (called an
operation or law of composition) from G × G to G, given by (g1 , g2 ) 7→ g1 ∗ g2 with the
following properties:
1. For any g1 , g2 , g3 ∈ G,
(g1 ∗ g2 ) ∗ g3 = g1 ∗ (g2 ∗ g3 )
(We call this the property of associativity of ∗.)
2. There exists an element g0 ∈ G such that, for all g ∈ G,
g0 ∗ g = g = g ∗ g0
(g0 is called an identity element.
3. For every element g ∈ G there exists an element h ∈ G such that g ∗ h = g0 = h ∗ g.
We will call the structure consisting of the set S and the operation ∗ a group. Moreover,
if the following property holds, the group is said to be abelian or Abelian or commutative:
4. For all g, h ∈ G, g ∗ h = h ∗ g.
In practice we often speak of the group G; where there is any doubt about the law of
composition, one should mention it, as in the group (G, ∗).
The number of elements in a group is described as its order .75
For any “configuration” in the plane we can consider all the isometries that map the
configuration onto itself. The set of these isometries will always be a group, since it
will contain the identity mapping, and, for any isometry, it will also contain the inverse
function, which is an isometry. As for the associative law, this can be shown to hold for
any functions that can compose, and isometries are a special class of functions! We can
call such a group the group of symmetries of the configuration.
7.2.3
The Cayley Table or Composition Table of a group
When a group is finite, i.e. when it consists of a finite number of elements, we can
tabulate the law of composition in a square array whose rows and columns are labelled
by the elements of the group, usually both labelled in the same order. For example,
the following is an incomplete Cayley Table for the group of symmetries of the square,
studied in Example 5.6, on page 62 of these notes. The entries in this table could be
expressed entirely in terms of R1 and R4 , which we showed to be generators of the group.
75
This should not be confused with the order of an element of a group, which can be shown to be the
order of the smallest subgroup — a group contained in the given group — containing the given element.
We will also be using the word order in other, totally different, senses. For example, we call the
elements of S × S, the cartesian product of a set S with itself, ordered pairs.
Information for Students in Math 348 2003 09
I
R1 R4
(R1 R4 )2
(R1 R4 )3
R1
R2
R3
R4
I
R 1 R4
I
R 1 R4
R1 R4 (R1 R4 )2
(R1 R4 )2 (R1 R4 )3
(R1 R4 )3
I
R1
R4
R2
R3
R4
81
(R1 R4 )2
(R1 R4 )3
R1 R2
R3
R4
2
3
(R1 R4 )
(R1 R4 )
R1 R2
R3
R4
(R1 R4 )3
I
R 3 R4
R2
R1
I
R 1 R4
R 1 R4
(R1 R2 )2
R4 R1 R4 R4 R1 R4 R1 R4 I R 1 R2 R1 R3 R1 R4
Table 1: (Incomplete) Cayley Table of the Symmetry Group of the Square
Isomorphism; Group presentations
Definition 7.4 Let G1 and G2 be groups, respectively having operations ∗1 and ∗2 . A
bijection f : G1 → G2 is called an isomorphism if, for any s, t ∈ G1 ,
(s ∗1 t)f = sf ∗2 tf .
The groups are said to be isomorphic if there exists an isomorphism from one to the
other.
This definition is a formalization of a situation described earlier: G1 is isomorphic to G2
iff it is possible to relabel the elements of G1 so that the relabelled elements combine
exactly as their counterparts in G2 . Remember that, as mentioned earlier with reference
to the symmetries of the letters N and T, when the groups are made up of isometries,
isomorphism ignores the geometric significance of the elements of the groups.
Example 7.7 Here is an example of two symmetry groups which are isomorphic as
“abstract” groups, yet consist of symmetries of different geometric types: Consider the
symmetries of the letter T and of the letter Z. The former has only reflectional symmetry
(and the identity): its group has 2 elements. The latter has only rotational symmetry
through a half-turn, in addition to the identity, and its group also has 2 elements. It is
easy to see that there is only one possible Cayley table for a group with 2 elements if,
say, we place the identity in the first row and column. Thus there is only one abstract
group. But one of these groups contains an isometry that reverses sense, and the other
does not.
It is possible to exhibit situations where these same two groups appear as “subgroups”
of a single group. For example, consider the group of symmetries of the square: both of
the symmetries of period 2 that we see for the letters T and N are symmetries of the
square.
Information for Students in Math 348 2003 09
82
Definition 7.5 For a group G, a set of its elements, S, with the property that every
element of G is expressible as a composition of its elements and/or their inverses is said
to generate G, and its elements are called generators.
Example 7.8
1. Sometimes a set of generators freely generates a group. For example, consider the configuration consisting of an infinite string of identical copies of
the letter F, equally spaced, with centres at integer points on the x-axis. The only
symmetries possessed by this configuration are translations through an integer distance; and every possible integer distance is realized, both positive and negative. If
we define f to be the translation of R2 to the right one unit, then every power of f
(i.e. composition of f with itself an integer number of times) — including negative
powers, which will be powers of the inverse function, which translates to the left
— is a different isometry of the configuration; and these are the only symmetries
of this configuration. The group of symmetries, which is . . . , f −2 , f −1 , I, f, f 2 , . . .
may be “presented” as hf i; the group is sometimes denoted by C∞ , and could be
called an infinite cyclic group with one generator .
Note that we cannot describe this group in terms of a pair of reflections in parallel
planes. While f is, indeed, the composition of such a pair, the two reflections
will not be symmetries of the configuration, and are not contained in its group of
symmetries.
2. If we now replace the letters F by letters E, the powers of f are still symmetries;
but there are others — the reflection R in the horizontal line through the middle
of the letters, and also the composition of that reflection with the powers of f —
i.e. the glide reflections formed by composing R with powers of f . We know that
R2 = I; there are other relations equations that relate R and f . For example,
our earlier analysis (cf. §5.3.9) shows that Rf = f R, or f = Rf R. Any other
“relations” between the 2 generators can be shown to be derivable from the two
relations given; the group can be presented as hf, R; R2 = I, Rf R = f i. The group
is said to have the structure of C∞ × D1 . Here D1 denotes the group hR; R2 = Ii
which consists of the symmetries of, for example, a single letter E, and is called a
dihedral group. (If we had written only hf, R; R2 = Ii, the group would consist of
infinitely many “words” of the form ...f no Rf n1 Rf n2 ..., and elements Rf and f R
would be considered as different elements.)
The conditions written to the right of the semicolon in a presentation are called relations
of the abstractly presented group. We shall describe the concept of group presentations
only through its use in the preceding and several other examples.
Example 7.9
1. Suppose we wish to consider the symmetries of the array of letters
F placed at the integer lattice points in the plane, that is, at the points with integer
Information for Students in Math 348 2003 09
83
coordinates relative to a fixed orthogonal (perpendicular) pair of coordinate axes.
We are using the letter F in order that the only symmetries are translations of
the plane moving one lattice point to another. If we denote translation along the
x-axis by T1 , and along the y-axis by T2 , the symmetries are all products of the
form T1r T2s , where r and s are integers — positive, negative, or 0. The symmetry
group is not the group hT1 , T2 i: that would be a group where every “string” of
T1 ’s and T2 ’s or their inverses would be a distinct member. But we know76 that
T1 T2 = T2 T1 . A correct presentation would be hT 1 , T 2 ; T1 T2 = T2 T1 i.
2. Suppose we place, at each lattice point in the rectangular lattice of the previous
example, a point (or a perfectly round letter O). In that case the group of symmetries will include all the symmetries of the square with vertices (0, 0), (0, 1), (1, 0),
(1, 1), which we know to be generated by the following reflections:
R1 : (x, y) 7→ (y, x)
whose mirror is the line y = x, and
R2 : (x, y) 7→ (1 − x, y)
whose mirror is the line x = 12 . There is also reflective symmetry in the lines
x = integer, and y = integer, in particular, in the coordinate axes. Let’s define
R3 to be the reflection in x = 0, and R4 the reflection in y = 0. Then T1 = R3 R2
and R1 R2 R1 is the reflection in the line y = 21 ,77 so R2 = R4 (R1 R2 R1 ). But we
don’t need all 4 of these reflections to generate the group, since both R4 R1 and
R1 R3 are the rotation about the origin through an angle of 2 × π4 = π2 . Taking
R3 = R1−1 R4 R1 = R1 R4 R1 , we see that the group is generated by R1 , R2 , R4 ,
76
77
commutativity of addition of vectors in R2
How do we know that R1 R2 R1 is the reflection in the line y = 12 ? First observe that
µ
¶
µ
¶
¡
¢
1 1
1 1
,
R1 R2 R1 =
,
R2 R1 since the point 12 , 12 is on the mirror of R1
2 2
2 2
µ
¶
¡
¢
1 1
=
,
R1 since the point 12 , 12 is on the mirror of R2
2 2
¶
µ
1 1
,
=
2 2
So R1 R2 R1 is a sense-reversing isometry with a fixed point, and the only such isometries are reflections.
If we now examine the action of the composition on the origin, we find that it fixed by R1 , then mapped
on to (1, 0), then reflected on to (0, 1): the mirror of the reflection ¡that ¢is the composition must bisect
the line segment {(0, y) : 0 ≤ y ≤ 1} and pass through the point 12 , 12 , so it is the line y = 12 . (We
could have argued this in other, shorter ways.)
Information for Students in Math 348 2003 09
84
whose mirrors form the sides of the triangle with vertices at (0, 0), ( 12 , 0), ( 12 , 12 ).
The following relations must hold:
R12 = R22 = R42 = I
(R1 R2 )4 = (R1 R4 )4 = I
(R2 R4 )2 = I
While you should be able to see why each of these relations holds between the
generators78 , you are not expected to see why these relations are sufficient to characterize the group of isometries of the integer-point lattice.
7.2.4
The Direct Product of Groups
Suppose that (G1 , ∗1 ) and (G2 , ∗2 ) are groups. We can construct from them a group
whose underlying set is the Cartesian product of sets G1 and G2 (cf. these notes, §5.1).
If we define a binary operation ∗ on G1 × G2 by
((g1 , g2 )(h1 , h2 )) 7→ (g1 ∗1 h1 , g2 ∗2 h2 ) .
Exercise 7.1
1. Prove that the operation ∗ defined above is associative.
2. Prove that, if the identities in G1 and G2 are respectively I1 and I2 , then
(g, h) ∗ (I1 , I2 ) = (g, h) = (I1 , I2 ) ∗ (g, h) ,
i.e. that (I1 , I2 ) is an identity. for the operation ∗.
3. Prove that the inverse of an element (g, h) ∈ G × H is (g −1 , h−1 ).
The group we have defined is often called the direct product of (G1 , ∗1 ) and (G2 , ∗2 ), and
may be denoted by (G1 × G2 , ∗), or simply as G1 × G2 .
We have introduced this concept because students will see it used in [4, §3.7]; we will
not be making extensive use of it at this time.
78
The first relations derive from the fact that each generator is of order 2; the second from the fact
that the two products of reflections each represent rotations through π2 radians; and the last because
the rotation R2 R4 is a half-turn.
Information for Students in Math 348 2003 09
7.2.5
85
Types of symmetry groups for configurations in R2
We wish to distinguish some types of groups that can occur by reference to the kinds
of isometries that they contain. We may also make observations about the “abstract
structure” of the groups — in the sense that two groups can be considered to have the
same structure if the members of one can be relabelled so that the two groups have
precisely the same laws of composition. For example, the letter N and the letter T both
have symmetry groups containing 2 isometries, one of which is the identity. But the
second isometry of N is a rotation (which preserves sense) while the second isometry of
T is a reflection (which reverses sense).
All groups of symmetries will be subgroups of the group of all isometries of R2 ; their
members will be exclusively translations, reflections, and glide reflections; and, as seen
earlier, these can all be expressed as compositions of reflections. But note: even if an
isometry be expressible as a product of reflections, there is no assurance that those
reflections themselves are isometries of the configuration in question.
Example 7.10 Consider the configuration consisting of an equilateral triangle whose
sides have been oriented cyclically. This configuration has only 3 symmetries: vertices
have to be mapped on to vertices; once it is known where a vertex is to be mapped, its
successor in the cyclic orientation must be mapped on to the successor of the image, in
order to map arrowheads on to arrowheads; thus the symmetries are simply the three
rotations through multiples of 2π
radians about the centre of the triangle. While each of
3
these rotations is a product of reflections in a mirror passing through one vertex and perpendicular to the opposite side, those rotations are not symmetries of this configuration,
as they do not preserve the direction of the arrows.
(Students who are troubled by the definition of this configuration in terms of directed
arrows can think, instead, of an equilateral triangle in each of whose sides a point is placed
two-thirds of the way along the side in some cyclic rotation.)
One way to classify the various groups of isometries in the plane is to look at the
types of translations, reflections, and glide reflections that are present in the group. For
some configurations the numbers of these isometries will be infinite, so care is needed
in gaining a “handle” with which to describe the groups. The complete analysis of this
problem is beyond the present course. We will satisfy our desire for a “complete” analysis
by confining our attention to one specific type of configuration, and, even there, will not
present a full proof that we have actually found all isometry groups. Our description
will be stated in terms of two specific abstract types of groups, denoted by Cn , and C∞ ,
and Dn and D∞ , which we have met above, and which will be further described in the
examples below.
Example 7.11 Suppose we wish to study isometry groups containing no translations
or glide reflections, and, in particular, no reflections. Groups of this type contain only
Information for Students in Math 348 2003 09
86
rotations. If the group contains two rotations with different centres, then, by the methods
of §5.3.9, it may be shown to contain a translation. Hence it consists solely of rotations
about the same fixed point.
There are still several possibilities here.
1. If there is a smallest positive (counterclockwise) angle through which these rotations turn, we can prove that the group consists of all (positive and negative)
powers of that rotation; that rotation is a generator , and the group that it generates is said to be cyclic. If some power of that generator is the identity, i.e. if
the element has finite period n, the group has order n, and is often denoted by
Cn ; if there is no element of finite period, the group could be denoted by C∞ , and
is described as being infinite cyclic. One presentation of Cn is hx; xn = Ii; one
presentation of C∞ is simply hx; i or hxi — a “free” group on one generator, having
no relations.
2. If there is no smallest positive angle, the group will be infinite; even here, there is
more than one possible group of symmetries. For simplicity we will not consider
any configurations of this type in this course: we will confine our attention in this
course to groups that are discrete.79 (For example, the symmetries of the circle
x2 + y 2 = 1 would not form a discrete group.)80
Example 7.12 Let’s consider groups generated by two distinct reflections R1 , R2 , in
mirrors `1 and `2 respectively.
Case 1 (Intersecting Mirrors): If `1 and `2 are non-parallel, the product of the generators is a rotation (through twice the angle between the mirrors). If that rotation
is a rational multiple of a full revolution, then some power of it is I; suppose n is
the smallest power with this property. It is beyond this course to show that this
group can be presented as
hR1 , R2 ; R22 = R22 = (R1 R2 )n = Ii .
The group is denoted by Dn and is called a dihedral group; it has exactly 2n
elements — n reflections, and n rotations.
If R1 R2 is not a rational multiple of a complete revolution, then it can be shown
that the group generated by R)1 and R2 is not discrete; nothing more will be said
about these cases in this course.
79
An isometry group G is discrete if the images of any point P under the isometries in a group will
not have a point of accumulation or limit point: this is a point Q with the property that, if one selects
any (positive) distance ², there will be an isometry f such that f (P ) is within a distance of ² from Q.
80
Note that the word is discrete, not discreet.
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87
In the case n = 1
R1 R2 = I .
(83)
We knew that R12 = R22 = I since R1 and R2 are rotations; equivalently, we knew
that
R1 = R1−1 , R2 = R2−1 .
From (83) we can conclude that R1 = R2 = R1−1 = R2−1 ; the group is cyclic of
order 2, so it is isomorphic to C2 .
Case 2 (Parallel Mirrors): In this case R1 R2 is a translation; call it T . It is beyond
this course to show that this group can be presented as
hT, R2 ; R22 = (T R2 )2 = Ii .
This group can be denoted by D∞ .
(84)
Information for Students in Math 348 2003 09
8
88
Solutions to Problems on Assignment 1
Distribution Date: Posted on the Web on Wednesday, October 15th, 2003
Solutions were to be submitted by Friday, October 3rd, 2003
(first version, subject to correction)
Solve each of the following problems using only the propositions in the notes from Euclid’s
Elements. Every step of your proof should be justified. Moreover, in some of these
problems you will be asked not to use propositions beyond a given specific location in
Book 1.
1. [13, 2.3H5, p. 61]81 The median from a vertex of a triangle to the opposite side is
a line segment joining the vertex to the midpoint of the opposite side — i.e. to the
point that bisects the side. We will call an edge of an isosceles triangle the base if
the other two sides of the triangle are equal in length: if the triangle is equilateral,
the term will be ambiguous — any one of the sides can be considered the base.
Prove that the median to the base of an isosceles triangle is perpendicular to the
base and bisects the opposite angle.
(Solve this problem without using other propositions than I.1 — I.10.)
Solution: Let 4ABC be given, and let D be the midpoint of side BC. The median
from A is the line segment AD.
A£qB
B
q£
£
£
£
£
£
£
£
£
£
£ B
B
B
B
qD
B
B
B
B
B
BC
Bq
In triangles ABD and ACD
|AB| = |AC| (hypothesis, 4 is isosceles)
|BD| = |CD| (hypothesis, D is midpoint of BC)
|AD| = |AD| (common edge of 4’s)
81
These footnotes acknowledge that the problem was taken from a particular source. Students are not
expected to consult the source; there are not likely to be any hints in the source that would be helpful
in solving the problem.
Information for Students in Math 348 2003 09
89
Hence, by Proposition I.8, 4ABD ≡ 4ACD. From this congruence follow three
equations, of which two are
∠ADB = ∠ADC
∠BAD = ∠CAD
(85)
(86)
Now DA is a line segment set up on line BC making the adjacent angles ∠ADB,
∠ADC equal. By Definition 3.2.10 AD is perpendicular to BC. Equation (86)
shows that DA bisects ∠BAC.
2. [13, 2.3I3, p. 64] An altitude of a triangle from a vertex is the line perpendicular to
the opposite side, and passing through the given vertex. Prove that if one of the
altitudes of a triangle is also a median, then the triangle is isosceles.
(Solve this problem without using other propositions than I.1 - I.12.)
Solution:
A£qB
B
q£
£
£
£
£
£
£
£
£
£
£ B
B
B
B
B
B
B
B
D
q
B
B
Bq
C
Let 4ABC be given, and let D be a point in side BC such that AD is perpendicular
to BC. By definition of perpendicular
∠ADB = ∠ADC .
(87)
In triangles ABD and ADC we have (87) and
|AD| = |AD| common side of triangles
|BD| = |CD| D is midpoint of BC
Hence 4ABD ≡ 4ADC, so |AB| = |AC|, i.e. 4ABC is isosceles.
3. [13, 2.3O7, p. 73] A quadrilateral is the 4-sided figure formed by joining 4 vertices
cyclically. Thus the quadrilateral ABCD, with vertices A, B, C, D, has sides AB,
BC, CD, DA. The diagonals of ABCD are the line segments AC and BD.(We
Information for Students in Math 348 2003 09
90
are not concerned here whether or not the quadrilateral is convex , i.e. whether line
segments joining points on the quadrilateral lie “inside”.) Prove that the sum of
the lengths of the diagonals of a quadrilateral is less than its perimeter.
Solution: Draw the diagonals AC and BD, thus forming 4 triangles: ABC, ADC,
ABD, CBD. Applying Proposition I.20 to each of these triangles yields the following inequalities:
|AB| + |BC| > |AC|
|AD| + |CD| > |AC|
|AB| + |AD| > |BD|
|BC| + |CD| > |BD|
Summing those inequalities yields, after reduction,
2(|AB| + |BC| + |CD| + |AD|) > 2(|AC| + |BD|)
from which it follows that
|AB| + |BC| + |CD| + |AD| > |AC| + |BD|
i.e., that the perimeter exceeds the sum of the lengths of the two diagonals, ¤.
4. [13, 2.3Q1, p. 77] Prove that if one of the altitudes of a triangle is also an angle
bisector, then the triangle is isosceles.
Solution:
A£qB
B
q£
£
£
£
£
£
£
£
£
£
£ B
B
B
B
qD
B
B
B
B
B
B
Bq
C
Let 4ABC be given, and let D a point of side BC such that AD is perpendicular
to BC. By definition of perpendicular
∠ADB = ∠ADC .
UPDATED TO November 27, 2003
(88)
Information for Students in Math 348 2003 09
91
We apply Proposition I.26 to triangles ABD and ADC, where we have (88) and
|AD| = |AD| common side of triangles
∠BAD = ∠CAD| AD bisects ∠BAC
Hence 4ABD ≡ 4ADC, so |AB| = |AC|, and 4ABC is isosceles.
Information for Students in Math 348 2003 09
9
92
Third Problem Assignment
Distribution Date: Mounted on the Web on Sunday, October 19th, 2003
distributed in hard copy on Wednesday, October 22nd, 2003
Solutions are to be submitted on Monday, November 3rd, 2003
Preceding the list of friezes numbered 1, 2, . . . , a table has been reproduced.
• Associate with each of the listed friezes the frieze in the table which has the same
isometry group, or argue that the isometry group is different from all those listed.
• For each type of isometry possessed by the frieze shown, give a precisely described
example of each isometry of that type. More precisely,
– If the pattern has reflective symmetry in a vertical mirror, describe all types of
vertical mirrors that are possible. For example, for the pattern ...VVVVV...
one would observe that there are two possible types of mirrors:
1. between two successive letters V, as ...VV|VVV...
2. along the axis of symmetry of any one letter V, as ...VVVVV...
|
– If the pattern has rotational symmetry under a half-turn, describe the location(s) of the centre(s) of the half-turn(s).
– If the pattern has symmetry under translations, describe a fundamental interval; e.g. for the pattern ...UVUVUVUVUV... one such interval would
be UV.
– If the pattern has symmetry under a glide reflection, describe the axis of the
translation and a shortest interval that is translated; e.g. in LΓLΓL one such
interval would be LΓ.
• It may happen that the text processor used to produce these notes will show
inconsistent spacing between the symbols used. In these examples we have never
varied the spaces between symbols: treat them all as being of equal length.
Also the intention is that symbols whose “simplest” form has symmetries may
appear to be slightly asymmetric because of variations in the typography used.
Treat all symbols as having the maximum symmetries of the “simplest” forms.82
82
For example, the letter V should be treated as having symmetry in a vertical reflective axis even
though the type being used is thicker on the left side than on the right.
Information for Students in Math 348 2003 09
I
II
III
IV
V
VI
VII
Typical pattern Ref(H) Ref(V) Trans HT GL Abstract Group
...LLLLL...
0
0
∞
0
0
C∞
...LΓLΓL...
0
0
∞
0
∞
C∞
...VVVVV...
0
∞
∞
0
0
D∞
...NNNNN...
0
0
∞
∞
0
D∞
...VΛVΛV...
0
∞
∞
∞ ∞
D∞
...DDDDD...
1
0
∞
0
∞
C∞ × D1
...HHHHH...
1
∞
∞
∞ ∞
D∞ × D1
Symbol Meaning
Ref(H) Reflection in Horizontal mirror
Ref(V) Reflections in Vertical mirrors
Legend:
Trans Translations
HT
Half-Turns
GL
Glide Reflections (with nontrivial translation)
1. . . . LLΓLLΓLLΓ. . .
2. . . . ÀÂÁĂÀÂÁĂÀÂÁĂ. . .
3. . . . HXHXHXHXHX. . .
4. . . . $SZS$$SZS$$SZS$. . .
5. . . . AĀAAĀAAĀA. . .
¯
¯
¯
6. . . . ÀÂÁÀÂÁÀÂÁÀÂÁ. . .
7. . . . ¡¿!?¡¿!?¡¿!?. . .
n X
m X
n X
m X
n X
m
X
8. . . .
...
m
9. . . .
n
m
n
m
n
= X
× X
= X
× X
= X
×
X
×
=
×
=
×
...
=
10. . . . §‡z‡§‡z‡§‡z‡. . .
11. . . . ±♦ ∓ ♦ ± ♦ ∓ ♦ ± ♦ ∓ ♦ ± ♦∓. . .
12. . . . ∈3 >⊥ ∈3 >⊥ ∈3 >⊥ ∈3 >⊥. . .
13. . . . ΓΓΓΓΓΓ=LLLLLL. . . (only one = sign)
93
Information for Students in Math 348 2003 09
14. . . .
= X
× X
= X
× X
= X
×
X
=
×
=
×
=
...
×
15. . . . HHHHHHOHHHHHH. . . (only one O)
16. . . . LLΓΓLLΓΓLLΓΓ. . .
17. . . . 6666660999999. . . (only one 0)
18. . . . HHHHHH=OOOOOO. . . (only one = sign)
94
Information for Students in Math 348 2003 09
10
95
The symmetries of friezes (strip patterns)
Distribution Date: Mounted on the web on Monday, October 20th, 2003
Students should read the definitions and discussion in [4, §3.7, pp. 47-49]. We reproduce
the table of possible frieze groups from this source: Remember that the list contains only
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
Typical pattern
...LLLLL...
...LΓLΓL...
...VVVVV...
...NNNNN...
...VΛVΛV...
...DDDDD...
...HHHHH...
Generators
1 translation
1 glide reflection
2 reflections
2 half-turns
1 reflection + 1 half-turn
1 translation + 1 reflection
3 reflections
Abstract Group
C∞
C∞
D∞
D∞
D∞
C ∞ × D1
D∞ × D1
Table 2: Isometry groups of friezes
the infinite groups of strip patterns.
Exercise 10.1 What finite groups could be realized as the isometry group of a strip
pattern?
10.1
Comments on the table of symmetry groups of friezes
10.1.1
What would have to be proved?
We will not work through the full derivation of Table 10 in this course. If we wished to
do so, what would that entail? We would need
• to show that different strip patterns in the list have isometry groups that are
different;
• to show that there are no other isometry groups for strip patterns which do not appear somewhere in the list. (Remember that the friezes that have finite symmetry
groups are not covered by this analysis — see Exercise 10.1 above.)
We would also wish to prove that the group of each pattern shown is, indeed, generated
by a set of isometries as described in the 3rd column of the table.
Information for Students in Math 348 2003 09
10.1.2
96
How can two “different” friezes both have group C∞ ?
The groups for friezes . . . LLLLL. . . and . . . LΓLΓL. . . are isomorphic as abstract
groups. That means that there is a bijection between the two groups that preserves the
group structure. If we denote translation to the right through 1 letter by T , and the
glide reflection to the right through 1 letter with a mirror bisecting the letters L and Γ
horizontally by G, then one isomorphism is given by T n 7→ Gn ; and another — the only
other — is given by T n 7→ G−n , as n ranges over all integers.
But the sets of isometries are different geometrically: the group for the first frieze
consists exclusively of sense-preserving isometries; the group for the second has as many
sense-reversing isometries as sense-preserving. The “abstract” structures of both of these
groups can be described as C∞ — a free group on 1 generator , or a cyclic group on 1
generator since the groups can be presented as hx; i or simply as hxi.
10.1.3
How can three “different” friezes all have group D∞ ?
The group of . . . VVVVV. . . can be shown to be generated by reflections in two parallel
mirrors; for example by one reflection R1 whose mirror passes through the bottom point
of one V and another, R2 , whose mirror passes between that letters V and the next on
the right. We could also have described this group as, for example, being generated by
R2 and the translation R1 R2 , since we know we could recover R1 if our only generators
were R2 and T from the equation R1 = T R2−1 = T R2 . This is precisely the presentation
of D∞ given above in (84): D∞ is isomorphic to any group generated by two generators
of period 2, with no other relations.
The group of . . . NNNNN. . . has to be different from the preceding, since it contains no sense-reversing isometries: it is generated by half turns centred, for example,
at a point P1 as in [4, Figure 2.4b, p. 30], and at a point P2 half-way up, between two
adjacent N’s. Other sets of generators are possible. For example, the product ªP1 ,π ªP2 ,π
is expressible as a product of 4 reflections: R1 with vertical mirror through P1 , R2 with
horizontal mirror through P1 ; and then the same reflection R2 again, and R3 with vertical mirror between two successive N’s. The product is R1 T , where T is a horizontal
translation — i.e., the product is a glide reflection: let’s call it G. The set {ªP2 ,π , G}
certainly generates the group, since we can recover ªP1 ,π using the relation
ªP1 ,π = G ªP2 ,π
We could show that generators G and ªP2 ,π are related only by
ª2P2 ,π = (G ªP2 ,π )2 = I
to show that the abstract structure of this group is also D∞ ; students in this course are
not expected to carry this out.
Information for Students in Math 348 2003 09
97
The third pattern given with the same abstract isometry group is . . . VΛVΛV. . . .
The group has to differ from those of . . . VVVVV. . . and . . . NNNNN. . . since
it contains glide reflections, which reverse sense; also, it contains half-turns, unlike the
symmetry group of . . . VVVVV . . . . We will not give details of a proof that the group
is again D∞ .
10.1.4
How can we differentiate the cases of . . . DDDDD . . . and . . . HHHHH
. . . from the others, and between themselves?
These two cases are the only ones in which there is reflective symmetry in a horizontal
mirror; this differentiates them from the other 5. The first case has no further reflective
symmetries; but the second has reflective symmetry in vertical mirrors as well — that
differentiates between the two cases.
We tabulate in Table 10.1.4 the multiplicities of members of the various geometrical
types in the isometry groups of the 7 friezes shown in Table 10.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
Typical pattern Ref(H) Ref(V) Trans HT GL Abstract Group
...LLLLL...
0
0
∞
0
0
C∞
...LΓLΓL...
0
0
∞
0
∞
C∞
...VVVVV...
0
∞
∞
0
0
D∞
...NNNNN...
0
0
∞
∞
0
D∞
...VΛVΛV...
0
∞
∞
∞ ∞
D∞
...DDDDD...
1
0
∞
0
∞
C∞ × D1
...HHHHH...
1
∞
∞
∞ ∞
D∞ × D1
Symbol
Meaning
Ref(H) Reflection in Horizontal mirror
Ref(V) Reflections in Vertical mirrors
Legend:
Trans Translations
HT
Half-Turns
GL
Glide Reflections (with nontrivial translation)
Table 3: Membership of Infinite Isometry Groups of Friezes
UPDATED TO November 27, 2003
Information for Students in Math 348 2003 09
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98
Solutions, Second Problem Assignment
Distribution Date: Monday, October 27th, 2003
Solutions were to be submitted by Monday, October 20th, 2003
1. Consider a regular hexagon in the Euclidean plane, where the plane has been
endowed with a rectangular Cartesian coordinate system, so that the hexagon is
inscribed in a circle of radius 1 centred at the origin, and its vertices are located at
points making angles of π3 , 2π
, π, 4π
, 5π
, 2π. Now the vertices making the following
3
3
3
π
5π
angles with the centre are joined to the centre by a line segment:
¡ 3 , ππ, 3 .π ¢Label
the vertices cyclically counterclockwise, starting with the vertex cos 3 , sin 3 : A1 ,
A2 , . . ., A6 , so that vertices A1 , A3 , and A5 are joined to the centre O, and the
other three are not.
(a) Make a very rough sketch of the geometric structure we are considering, showing its 6 vertices, the centre, the 6 line segments forming its perimeter, and
the 3 line segments joining boundary vertices to the centre.
(b) Find all isometries of the plane which map the structure on to itself. Denote
the isometry F by a 2-line matrix which denotes its action on the 6 boundary
vertices:
µ
¶
A1
A2
A3
A4
A5
A6
,
A 1 F A2 F A 3 F A4 F A 5 F A6 F
where each Ai F is one of the vertices of the structure. For each of the isometries, indicate whether it is a reflection, a rotation, a translation, or a glide
reflection. If it is a reflection, describe the mirror of the reflection; if it is
a rotation, describe the centre of the rotation, and the angle through which
the rotation moves, being careful to take the counterclockwise direction as
“positive”; if it is a translation, describe the direction and the distance; if it
is a glide reflection, describe the direction in which it “glides”, and the line
which is mapped onto itself by the isometry. Include
the “identity” isometry
µ
¶
A1 A2 A3 A4 A5 A6
in your classification, and denote it by either
A1 A2 A3 A4 A5 A6
or I.
(c) Express each of the isometries as a product of a non-negative number of
reflections, always using as few reflections as possible. (There may be more
than one way to do this, but any solution will be acceptable.)
(d) Set up and complete a rectangular “Cayley Table” for your isometries, so that
the product of any two of them, in each order, can be read off from the table.
(e) Determine, for each isometry, its inverse; you may refer to your table in the
preceding part.
Information for Students in Math 348 2003 09
99
(f) Show that there are in your set of isometries two reflections — call them F
and G — such that all the isometries can be expressed as a product of possibly
multiple copies of F and G; that is, that the group of all of your isometries is
generated by F and G.
(g) Show that there are, in your set of isometries one reflection and one rotation
— call them B and C, such that all the isometries can be expressed as a
product of possibly multiple copies of B and C; that is, that the group of all
of your isometries is generated by B and C.
Solution:
(a)
(b) The isometries must map the regular hexagon onto itself; and the vertices A1 ,
A3 , A5 must be, as a set, mapped onto themselves. Once the placement
√ of A1
is decided, there are two choices for A3 and A5 , as their distances ( 3) from
A1 must be preserved: there are 3 choices for placing A1 , and then two ways
of extending this mapping to A3 and A5 , following which the positions of the
other 3 vertices are determined. In all there are 6 isometries: 3 reflections,
µ
¶
A1 A2 A3 A4 A5 A6
,
A1 A6 A5 A4 A3 A2
µ
¶
A1 A2 A3 A4 A5 A6
,
A3 A2 A1 A6 A5 A4
µ
¶
A1 A2 A3 A4 A5 A6
,
A5 A4 A3 A2 A1 A6
√
respectively with mirrors the lines through 1 and 4 (i.e.√y = 3x), through 2
and 5 (i.e., the x-axis), and through 3 and 6 (i.e., y = − 3x); and 3 rotations
µ
¶
A1 A2 A3 A4 A5 A6
,
A3 A4 A5 A6 A1 A2
¶
µ
A1 A2 A3 A4 A5 A6
,
A5 A6 A1 A2 A3 A4
¶
µ
A1 A2 A3 A4 A5 A6
,
A1 A2 A3 A4 A5 A6
, 4π
, and 0 radians. There are no glide
respectively through angles of 2π
3
3
reflections (other than the 3 reflections), and no translations (other than the
identity).
Information for Students in Math 348 2003 09
100
(c) To simplify the exposition, we name the reflections
µ
A1 A2 A3 A4 A5
R1 =
A1 A6 A5 A4 A3
µ
A1 A2 A3 A4 A5
R2 =
A3 A2 A1 A6 A5
µ
A1 A2 A3 A4 A5
R3 =
A5 A4 A3 A2 A1
as follows:
¶
A6
A2
¶
A6
A4
¶
A6
A6
Then two of the rotations may be factorized as follows as products of two
reflections, while the identity is the product of 0 reflections. We find it convenient for the coming purpose of the Cayley Table to name the rotations as
follows:
µ
¶
A1 A2 A3 A4 A5 A6
S1 =
= R1 R2 = R2 R3 = R3 R1
A3 A4 A5 A6 A1 A2
µ
¶
A1 A2 A3 A4 A5 A6
S2 =
= R1 R3 = R2 R1 = R3 R2 = S12
A5 A6 A1 A2 A3 A4
µ
¶
A1 A2 A3 A4 A5 A6
I=
= empty product
A1 A2 A3 A4 A5 A6
(d)
I
S1
S2
R1
R2
R3
I
I
S1
S2
R1
R2
R3
S1
S1
S2
I
R2
R3
R1
S2
S2
I
S1
R3
R1
R2
R1
R1
R3
R2
I
S2
S1
R2
R2
R1
R3
S1
I
S2
R3
R3
R2
R1
S2
S1
I
(e) Each of the identity and the reflections is its own inverse. The only isometries
which are not their own inverse are the two non-identity rotations, each of
which is the inverse of the other.
(f) We will show how to generate the group with reflections R1 and R2 . The
rotations can be factorized, for example, as follows:
¶
µ
A1 A2 A3 A4 A5 A6
= R1 R2
A3 A4 A5 A6 A1 A2
¶
µ
A1 A2 A3 A4 A5 A6
= R2 R1
A5 A6 A1 A2 A3 A4
¶
µ
A1 A2 A3 A4 A5 A6
= R10
A1 A2 A3 A4 A5 A6
Information for Students in Math 348 2003 09
101
Since the factorizations of R1 and R2 are obvious, the only missing decomposition is that of R3 . We have seen earlier that R1 R2 = R2 R3 = R3 R1 , and
from this it follows that R2 (R1 R2 ) = R2 (R2 R3 ) = (R1 R1 ) R3 = IR3 = R3 ,
so we have one possible factorization for R3 .
(g) The specific factorization depends on the choice of rotation and reflection; we
give one example, taking R1 as the reflection B, and S1 as the rotation C:
R1 = B
R2 = IR2 = (R1 R1 )R2 = R1 (R1 R2 ) = BC
R3 = (R2 R2 )R3 = R2 (R2 R3 ) = (BC)C = BC 2
or, alternatively,
R3 = R3 (R1 R1 ) = (R3 R1 )R1 = CB
C = C
S2 = C 2
I = I empty product, or
I = BB
2. Let R1 , R2 and R3 be reflections, respectively in the line y = 1, the line x = y, and
the y-axis. Apply the procedure discussed in §5.3.9 of the notes to express R1 R2 R3
as the product of a reflection R in a mirror ` and a translation in the direction
of the mirror. Show all your work, and describe all the reflections that enter into
your computations carefully, by specifying their mirror in each case. Determine
the distance through which the translation shifts the plane.
The procedure discussed in §5.3.9 produces a product of reflection and translation
in two orders — either the reflection is applied first or last. You are expected to
carry out your calculations also in two orders, and to produce two factorizations,
one with the reflection preceding the translation, and one where it follows the
translation.
Solution: We observe that the given reflections have mirrors of which two are
orthogonal; but, as these are not adjacent, we cannot work with this pair, as the
result we wish to apply is that reflections in perpendicular mirrors commute.
(a)
i. First we operate on reflections R1 and R2 , whose mirrors intersect at
(1, 1), replacing them by reflections R10 with mirror x+y = 2, and R20 = R1
with the line y = 1 as its mirror. In this way we have arranged for the
perpendicularity of the mirrors in two adjacent reflections in the product.
ii. Next we operate on reflections R20 and R3 , whose mirrors intersect in the
point (0, 1). We turn both mirrors counterclockwise (i.e., in the positive
Information for Students in Math 348 2003 09
102
π
direction) through an angle
in order to arrange for the new middle
4
reflection to be in a mirror perpendicular to the mirror of the first factor,
R10 ; this rotation yields reflections R200 , with −x + y = 1 as mirror, and R30
with mirror
x + y = 1; the mirrors of R200 and R30 intersect in the point
¶
µ
1 3
,
.
2 2
¡
¢
iii. Finally, we turn mirrors R10 and R200 which intersect in the point 12 , 23
clockwise through an angle of π2 radians, obtaining reflections R100 = R200
(3)
in mirror −x + y = 1, and R2 = R10 in mirror x + y = 2; that is, we
interchange the factors.
iv. The given product therefore factorizes into the reflection in mirror −x +
y = 1, followed by a translation in the direction
¶ of the mirror through a
µ
1
1 3
distance of 2 × √ units, directed from
,
to (0, 1), but twice the
2 2
2
distance between those points.
v. We could have accomplished the transformations even more efficiently if
3π
we had turned through an angle of
; the result would be that the new
4
middle reflection would be in a mirror parallel to the mirror of the first
factor, R10 ; this rotation would yield reflections R30 , with −x + y = 1 as
mirror, and Rµ200 with¶mirror x + y = 1; the mirrors of R30 and R10 intersect
1 3
,
. In this way there are no further transformations
in the point
2 2
needed, as we have expressed the transformation reflection in y = x + 1
preceded by a translation in a direction parallel to that mirror.
(b)
i. We first replace reflections R2 and R3 , whose mirrors intersect in the
origin, by reflections R20 = R3 , whose mirror is the y-axis, x = 0, and R30 ,
whose mirror is y = −x.
ii. Next we operate on reflections R1 and R20 , whose mirrors intersect in
radians, replacing them by reflections R10 ,
(0, 1): we turn through - 7π
4
with mirror x + y = 1, and R200 , whose mirror is −x + y = 1.
iii. Finally, we replace reflections R200 and R30 by turning their mirrors through
an angle of π2 radians — this has µ
the effect
¶ of interchanging the mirrors.
1 1
(3)
The mirrors intersect in the point − ,
. We obtain reflections R2 =
2 2
R30 with mirror y = −x and R300 = R200 with mirror −x + y = 1.
iv. The given product therefore factorizes into a reflection in mirror −x +
y = 1 preceded by a translation in the direction of the mirror through
Information for Students in Math 348 2003 09
103
1
a distance of 2 × √ units, twice the directed distance from (0, 1) to
2
µ
¶
1 1
− ,
.
2 2
v. As seen above, we could have simplified our transformation by turning
about (0, 1) through an additional right angle. We would then have the
mirror of R200 as x + y = 0, parallel to the mirror of R30 ; and the mirror of
R10 as the line y = x + 1, perpendicular to those of both R200 and R30 . This
yields a decomposition in which the translation follows. the reflection.
(c) Once either variant of either approach is followed, and the transformation is
expressed as a product of reflections in which two of the mirrors are parallel
and the third is perpendicular to them, rearranging the order becomes trivial,
since reflections in perpendicular mirrors commute. But, of course, we cannot
reverse the order of the reflections in the parallel mirrors, as this would reverse
the translation.
Information for Students in Math 348 2003 09
12
104
Two results on plane lattices
Distribution Date: Monday, October 27th, 2003
12.1
Sylvester’s Problem of Collinear Points
(These notes are based on [4, §4.7].) Hopefully we will have time to investigate a number
of plane “configurations” in which points are “densely” contained in in a small number
of lines; the textbook gives one example of such a configuration in [4, Figure 4.7a, p.
65], the so-called Pappus configuration. Even in configurations in which many lines pass
through a multiplicity of points, one finds that there are lines connecting 2 points in
the figure that pass through no other points. It was possibly a study of such interesting
configurations that led Sylvester to conjecture (1893):
It is not possible to arrange any finite number of points in R2 so that every
line through 2 of them passes through a third, unless all of the points lie on
the same line.
This conjecture was first proved by Tibor Grünbaum, a Hungarian mathematician whose
later work was under his changed name, Tibor Gallai.[14]; Coxeter gives a version of the
proof which he attributes to L. M. Kelly.
Definition 12.1 Let a set S of points in R2 be prescribed. Relative to this set S, a line
` in R2 is said to be ordinary if it passes through exactly two distinct members of S.
Sylvester’s conjecture may be reformulated as follows:
If n points in the plane are not collinear, then there exists at least one ordinary
line passing through 2 of them.
Solution: Let a set S of points in the plane be prescribed. Since there is nothing to
prove for n = 0, 1, 2, we may assume
µ ¶ |S| > 2. The number of distinct lines joining pairs
n
of these points cannot exceed
. The number of pairs consisting of one of these lines,
2
n(n − 1)(n − 2)
P2 P3 , and a member P1 of S not on P2 P3 cannot exceed
. If this number
2
of pairs is 0, we have nothing to prove — all the points lie on one line; so assume there
exist instances where P1 does not lie on P2 P3 ; the formula itself is of no interest at this
point; what we need is that there are only a finite number of these situations. Among
all such pairs — finite in number — select one for which the distance from P1 to P2 P3 is
minimal. We prove that line P2 P3 is ordinary.
The proof is by contradiction: assume that the claim is false, and that there is a point
P4 ∈ S, distinct from P2 , P3 also on the line. Let Q be the foot of the perpendicular to
Information for Students in Math 348 2003 09
105
P2 P3 from P1 , and consider the two rays into which Q divides the line P2 P3 , where we
consider Q as being a member of both of those rays. Since the union of those rays is to
contain the 3 points P2 , P3 , P4 , it cannot happen that neither of the rays contains more
than 1 of these points; thus we may assume that the points are labelled so that one of
the rays contains P2 and P3 — lying on the same side of Q, and with P2 lying closer to Q
than P3 ; we permit P2 = Q. Let R and T be respectively the feet of the perpendiculars
from Q and P2 to the line P1 P3 . Then 4P3 P2 T is similar to 4P3 QR, so
|P2 T |
|P2 R|
=
<1
|QR|
|QP3 |
and |P2 T | < |QR|. But, in the right-angled 4QP1 R, |QR| < |QP1 |; thus |P2 T | < |P1 Q|.
We have thus shown that the distance from P2 to line P1 P3 is less than the distance
P1 Q that was assumed to be the minimum distance. From this contradiction we infer
that our hypothesis — that P2 P3 contained another point in S — was not true: P2 P3
must be an ordinary line.
12.2
The Crystallographic Restriction
These notes are based on [4, §4.5, pp. 60-61].) Suppose that a configuration of points
in the plane has the property that, among its isometries, there are two translations
in directions that are not parallel. A point of the configuration moves under repeated
applications of these translations and their inverses to an array of infinitely many points,
called a 2-dimensional lattice. If there exist translations of minimum positive distance
in each of the two directions, call them, respectively, X and Y . An essential property
of the geometry of the plane is that XY = Y X. The lattice will have symmetry under
half turns about each of the lattice points. But there may be symmetry under rotations
through smaller angles. The Crystallographic Restriction states that
the only angles that are possible are
2π
, where n = 2, 3, 4, 6.
n
Suppose that a lattice has symmetry of period n, where n is a positive integer, n ≥ 5.
Let P be a point of the lattice, and let Q be another point of the lattice that is as close
to P as any other point of the lattice, i.e. a point for which the distance |P Q| is minimal;
let us scale the lattice so that this minimal distance is taken to be 1. Then there must
radians.
be a lattice point P 0 obtainable from P by a rotation through an angle of 2π
n
π
0
The distance P P is then 2 sin n . This is a decreasing function of n; for n = 6, P P 0 = 1;
thus, for n > 6, the distance P P 0 is less than 1, which was the minimum distance. This
proves that no period n can exceed 6; it remains to prove that n = 5 is impossible.
Information for Students in Math 348 2003 09
106
For n = 5 let Q0 be the rotation of lattice point Q about P 0 . We obtain a trapezoid
= 1 − 0.618 approxwith vertices P, Q, P 0 , Q0 , where the distance |P 0 Q0 | = 1 − 2 cos 2π
5
imately, and hence |P 0 Q0 | < 1; here again we would obtain two lattice points closer
together than the minimum distance; from this contradiction we infer that n 6= 5.
13
Class Tests
13.1
Version 1
CLASS TEST: MATHEMATICS 189–348 2003 09
TOPICS IN GEOMETRY
EXAMINER: Professor W. G. Brown
DATE: Wednesday, 5th November, 2003.
TIME: 12:35 – 13:20
FAMILY NAME:
GIVEN NAMES:
STUDENT NUMBER:
Instructions
1. All your writing — even rough work — must be handed in.
2. Calculators are not permitted.
3. Your neighbour’s version of this test may not be the same as yours.
4. This examination booklet consists of this cover, Pages 108 through 111 containing questions;
and Pages 112, 113, which are blank.
5. Show all your work. All solutions are to be written in the space provided on the page where
the question is printed. When that spaces is exhausted, you may write on the facing page,
on one of the blank pages, or on the back cover of the booklet, but you must indicate any
continuation clearly on the page where the question is printed! (Please inform the invigilator
if you find that your booklet is defective.)
PLEASE DO NOT WRITE INSIDE THIS BOX
1
2(a)
/12
3(d)
2(b)
/3
4(a)
/3
2(c)
/3
2(d)
/3
3(a)
/3
3(b)
/3
3(c)
/3
4(b)
/8
/4
Raw
/48
/20
/3
Information for Students in Math 348 2003 09
108
1. [12 MARKS] Showing all your work, carefully prove the following theorem in Euclidean geometry. Your solution may use any of Euclid’s axioms or postulates, and any of the following
Propositions (but no others) which you should cite by number.
The median to the base of an isosceles triangle is perpendicular to the base
and bisects the opposite angle.
These are the Propositions you may use:
Proposition I.1: On a given finite straight line, to construct an equilateral triangle.
Proposition I.2: To place at a given point (as an extremity) a (straight) line segment
whose length is equal to that of a given (straight) line segment.
Proposition I.3: Given two unequal line segments, to cut off from the greater a line
segment whose length is equal to the less.
Proposition I.4: If two triangles have the lengths of two sides of the one equal to the
lengths of two sides respectively, and have the angles contained by the equal line
segments equal, they will also have the base equal to the base, the triangle will
be equal to the triangle, and the remaining angles will be equal to the remaining
angles respectively, namely those which the equal sides subtend.
Proposition I.5: In isosceles triangles the angles at the base are equal to one another;
and, if the equal sides be extended below the base, the angles under the base will
also be equal to one another.
Proposition I.6 If, in a triangle, two angles be equal to one another, the sides which
subtend the equal angles will also be equal to one another.
Proposition I.7 Given two line segments constructed from the ends of a line segment,
and meeting in a point, there cannot be constructed from the ends of the same line
segment and on the same side of it, two other line segments meeting in a different
point, respectively of the same lengths as the former two line segments, namely
each equal to that which emanates from the same end of the fixed line segment.
Proposition I.8 If two triangles have two sides equal in length to two sides respectively,
and also have the bases equal in length, they will also have the angles equal which
are contained by the equal sides.
Proposition I.9: To bisect a given angle.
Proposition I.10: To bisect a given line segment.
Proposition I.20 In any triangle, two sides taken together in any manner are greater
than the remaining side.
Proposition I.26 If two triangles have the two angles at the base equal to two angles,
respectively, and one side of the one equal in length to the corresponding side of the
other; namely, either (Case 1) the side adjoining the equal angles, or (Case 2) that
subtending one of the equal angles, they will also have the corresponding remaining
sides equal in length, and the remaining angle equal to the remaining angle.
Information for Students in Math 348 2003 09
109
2. Preceding the list of strip patterns numbered (a), (b), (c), (d), a table has been reproduced.
• Associate with each of the listed strip patterns the pattern in the table which has the
same isometry group, or argue that the isometry group is different from all those listed.
• If the pattern has reflective symmetry in a vertical mirror, describe all types of vertical
mirrors that are possible. For example, for the pattern ...VVVVV... one would observe
that there are two possible types of mirrors:
(a) between two successive letters V, as ...VV|VVV...
(b) along the axis of symmetry of any one letter V, as ...VVVVV...
|
• If the pattern has rotational symmetry under a half-turn, describe the location(s) of the
centre(s) of the half-turn(s).
• If the pattern has symmetry under translations, describe an interval of minimum length
that is mapped by the translations so that it covers the strip without overlaps; e.g. for
the pattern ...UVUVUVUVUV... one such interval would be UV.
• If the pattern has symmetry under a glide reflection, describe the axis of the translation
and an interval of minimum length that is mapped by all the symmetries to cover the
strip without overlapping; e.g. in LΓLΓL one such interval would be L.
I
II
III
IV
V
VI
VII
Typical pattern
...LLLLL...
...LΓLΓL...
...VVVVV...
...NNNNN...
...VΛVΛV...
...DDDDD...
...HHHHH...
Legend:
Ref(H)
0
0
0
0
0
1
1
Symbol
Ref(H)
Ref(V)
Trans
HT
GL
Ref(V)
0
0
∞
0
∞
0
∞
Trans
∞
∞
∞
∞
∞
∞
∞
HT
0
0
0
∞
∞
0
∞
GL
0
∞
0
0
∞
∞
∞
Abstract Group
C∞
C∞
D∞
D∞
D∞
C∞ × D1
D∞ × D1
Meaning
Reflection in Horizontal mirror
Reflections in Vertical mirrors
Translations
Half-Turns
Glide Reflections (with nontrivial translation)
(a) [3 MARKS] . . . HAWAHAWAHAWAHAWA. . .
(b) [3 MARKS] . . . MAXIMAXIMAXIMAXIMAXIMAX. . .
(c) [3 MARKS] . . . SOSXSOSXSOSXSOSXSOSXS. . .
(d) [3 MARKS]. . . HODHIDHODHIDHODHIDHODHIDHODH. . .
Information for Students in Math 348 2003 09
110
3. Consider a regular hexagon in the Euclidean plane, where the plane has been endowed with a
rectangular Cartesian coordinate system, so that the hexagon is inscribed in a circle of radius
1 centred at the origin, and its vertices are located at points making angles of 0, π3 , 2π
3 , π,
4π 5π
,
.
Let
the
vertices
be
labelled
cyclically,
counterclockwise
around
the
hexagon
as
1, 2,
3
3
3, 4, 5, 6, with vertex 1 being the point(1, 0); then join each of the following pairs of vertices
by a line segment: 1 and 3, 3 and 5, 5 and 1.
(a) [3 MARKS] Express each of the isometries as a product of a non-negative number of
reflections.
(b) [3 MARKS] Set up and complete a rectangular “Cayley Table” for your isometries, so
that the product of any two of them, in each order, can be read off from the table.
(c) [3 MARKS] Determine, for each isometry, its inverse; you may refer to your table in
the preceding part.
(d) [3 MARKS] Show that there are in your set of isometries two reflections which generate
the group of isometries.
Information for Students in Math 348 2003 09
111
4. Let R1 , R2 and R3 be reflections, respectively in the line x + y = 1, the line x = 1, and the
line y = x − 1.
(a) [8 MARKS] Describe a sequence of procedures in each of which a product of reflections
is replaced by an equivalent product of reflections, in such a way that R1 R2 R3 is
expressed as a product of a reflection R0 followed by a translation in a direction parallel
to the mirror of R0 . Show all your work. Describe all the reflections that enter into
your computations carefully, by specifying their mirror in each case.
(b) [4 MARKS] Determine the distance through which the translation shifts the plane, and
verify that your product transforms the point (0, 2) in the same way as R1 R2 R3 .
Information for Students in Math 348 2003 09
continuation page for problem number
You must refer to this continuation page on the page where the problem is printed!
112
Information for Students in Math 348 2003 09
continuation page for problem number
You must refer to this continuation page on the page where the problem is printed!
113
13.2
Version 2
CLASS TEST: MATHEMATICS 189–348 2003 09
TOPICS IN GEOMETRY
EXAMINER: Professor W. G. Brown
DATE: Wednesday, 5th November, 2003.
TIME: 12:35 – 13:20
FAMILY NAME:
GIVEN NAMES:
STUDENT NUMBER:
Instructions
1. All your writing — even rough work — must be handed in.
2. Calculators are not permitted.
3. Your neighbour’s version of this test may not be the same as yours.
4. This examination booklet consists of this cover, Pages 115 through 118 containing questions;
and Pages 119, 120, which are blank.
5. Show all your work. All solutions are to be written in the space provided on the page where
the question is printed. When that spaces is exhausted, you may write on the facing page,
on one of the blank pages, or on the back cover of the booklet, but you must indicate any
continuation clearly on the page where the question is printed! (Please inform the invigilator
if you find that your booklet is defective.)
PLEASE DO NOT WRITE INSIDE THIS BOX
1
2(a)
/12
3(d)
2(b)
/3
4(a)
/3
2(c)
/3
2(d)
/3
3(a)
/3
3(b)
/3
3(c)
/3
4(b)
/8
/4
Raw
/48
/20
/3
Information for Students in Math 348 2003 09
115
1. [12 MARKS] Showing all your work, carefully prove the following theorem in Euclidean geometry. Your solution may use any of Euclid’s axioms or postulates, and any of the following
Propositions (but no others) which you should cite by number.
If one of the altitudes of a triangle is also a median, then the triangle is
isosceles.
These are the Propositions you may use:
Proposition I.1: On a given finite straight line, to construct an equilateral triangle.
Proposition I.2: To place at a given point (as an extremity) a (straight) line segment
whose length is equal to that of a given (straight) line segment.
Proposition I.3: Given two unequal line segments, to cut off from the greater a line
segment whose length is equal to the less.
Proposition I.4: If two triangles have the lengths of two sides of the one equal to the
lengths of two sides respectively, and have the angles contained by the equal line
segments equal, they will also have the base equal to the base, the triangle will
be equal to the triangle, and the remaining angles will be equal to the remaining
angles respectively, namely those which the equal sides subtend.
Proposition I.5: In isosceles triangles the angles at the base are equal to one another;
and, if the equal sides be extended below the base, the angles under the base will
also be equal to one another.
Proposition I.6 If, in a triangle, two angles be equal to one another, the sides which
subtend the equal angles will also be equal to one another.
Proposition I.7 Given two line segments constructed from the ends of a line segment,
and meeting in a point, there cannot be constructed from the ends of the same line
segment and on the same side of it, two other line segments meeting in a different
point, respectively of the same lengths as the former two line segments, namely
each equal to that which emanates from the same end of the fixed line segment.
Proposition I.8 If two triangles have two sides equal in length to two sides respectively,
and also have the bases equal in length, they will also have the angles equal which
are contained by the equal sides.
Proposition I.9: To bisect a given angle.
Proposition I.10: To bisect a given line segment.
Proposition I.20 In any triangle, two sides taken together in any manner are greater
than the remaining side.
Proposition I.26 If two triangles have the two angles at the base equal to two angles,
respectively, and one side of the one equal in length to the corresponding side of the
other; namely, either (Case 1) the side adjoining the equal angles, or (Case 2) that
subtending one of the equal angles, they will also have the corresponding remaining
sides equal in length, and the remaining angle equal to the remaining angle.
Information for Students in Math 348 2003 09
116
2. Preceding the list of strip patterns numbered (a), (b), (c), (d), a table has been reproduced.
• Associate with each of the listed strip patterns the pattern in the table which has the
same isometry group, or argue that the isometry group is different from all those listed.
• If the pattern has reflective symmetry in a vertical mirror, describe all types of vertical
mirrors that are possible. For example, for the pattern ...VVVVV... one would observe
that there are two possible types of mirrors:
(a) between two successive letters V, as ...VV|VVV...
(b) along the axis of symmetry of any one letter V, as ...VVVVV...
|
• If the pattern has rotational symmetry under a half-turn, describe the location(s) of the
centre(s) of the half-turn(s).
• If the pattern has symmetry under translations, describe an interval of minimum length
that is mapped by the translations so that it covers the strip without overlaps; e.g. for
the pattern ...UVUVUVUVUV... one such interval would be UV.
• If the pattern has symmetry under a glide reflection, describe the axis of the translation
and an interval of minimum length that is mapped by all the symmetries to cover the
strip without overlapping; e.g. in LΓLΓL one such interval would be L.
I
II
III
IV
V
VI
VII
Typical pattern
...LLLLL...
...LΓLΓL...
...VVVVV...
...NNNNN...
...VΛVΛV...
...DDDDD...
...HHHHH...
Legend:
Ref(H)
0
0
0
0
0
1
1
Symbol
Ref(H)
Ref(V)
Trans
HT
GL
Ref(V)
0
0
∞
0
∞
0
∞
Trans
∞
∞
∞
∞
∞
∞
∞
HT
0
0
0
∞
∞
0
∞
GL
0
∞
0
0
∞
∞
∞
Meaning
Reflection in Horizontal mirror
Reflections in Vertical mirrors
Translations
Half-Turns
Glide Reflections (with nontrivial translation)
(a) [3 MARKS] . . . BIDDEBIDDEBIDDEBIDDEBID. . .
(b) [3 MARKS] . . . EΓELEΓELEΓELEΓELEΓ. . .
(c) [3 MARKS] . . . NIBINIBINIBINIBINI. . .
(d) [3 MARKS]. . . INIXINIXINIXINIXINIXINI. . .
Abstract Group
C∞
C∞
D∞
D∞
D∞
C∞ × D1
D∞ × D1
Information for Students in Math 348 2003 09
117
3. Consider a regular hexagon in the Euclidean plane, where the plane has been endowed with a
rectangular Cartesian coordinate system, so that the hexagon is inscribed in a circle of radius
1 centred at the origin, and its vertices are located at points making angles of 0, π3 , 2π
3 , π,
4π 5π
,
.
Let
the
vertices
be
labelled
cyclically,
counterclockwise
around
the
hexagon
as
1, 2,
3
3
3, 4, 5, 6, with vertex 1 being the point(1, 0); then join each of the following pairs of vertices
by a line segment: 1 and 3, 3 and 5, 5 and 1.
(a) [3 MARKS] Express each of the isometries as a product of a non-negative number of
reflections.
(b) [3 MARKS] Set up and complete a rectangular “Cayley Table” for your isometries, so
that the product of any two of them, in each order, can be read off from the table.
(c) [3 MARKS] Determine, for each isometry, its inverse; you may refer to your table in
the preceding part.
(d) [3 MARKS] Show that there are, in your set of isometries one reflection and one rotation
which generate the group of isometries.
Information for Students in Math 348 2003 09
118
4. Let R1 , R2 and R3 be reflections, respectively in the line −x + y = 1, the line x = −1, and
the line y = −x − 1.
(a) [8 MARKS] Describe a sequence of procedures in each of which a product of reflections
is replaced by an equivalent product of reflections, in such a way that R1 R2 R3 is
expressed as a product of a reflection R0 followed by a translation in a direction parallel
to the mirror of R0 . Show all your work. Describe all the reflections that enter into
your computations carefully, by specifying their mirror in each case.
(b) [4 MARKS] Determine the distance through which the translation shifts the plane, and
verify that your product transforms the point (0, 2) in the same way as R1 R2 R3 .
Information for Students in Math 348 2003 09
continuation page for problem number
You must refer to this continuation page on the page where the problem is printed!
119
Information for Students in Math 348 2003 09
continuation page for problem number
You must refer to this continuation page on the page where the problem is printed!
120
13.3
Version 3
CLASS TEST: MATHEMATICS 189–348 2003 09
TOPICS IN GEOMETRY
EXAMINER: Professor W. G. Brown
DATE: Wednesday, 5th November, 2003.
TIME: 12:35 – 13:20
FAMILY NAME:
GIVEN NAMES:
STUDENT NUMBER:
Instructions
1. All your writing — even rough work — must be handed in.
2. Calculators are not permitted.
3. Your neighbour’s version of this test may not be the same as yours.
4. This examination booklet consists of this cover, Pages 122 through 125 containing questions;
and Pages 126, 127, which are blank.
5. Show all your work. All solutions are to be written in the space provided on the page where
the question is printed. When that spaces is exhausted, you may write on the facing page,
on one of the blank pages, or on the back cover of the booklet, but you must indicate any
continuation clearly on the page where the question is printed! (Please inform the invigilator
if you find that your booklet is defective.)
PLEASE DO NOT WRITE INSIDE THIS BOX
1
2(a)
/12
3(d)
2(b)
/3
4(a)
/3
2(c)
/3
2(d)
/3
3(a)
/3
3(b)
/3
3(c)
/3
4(b)
/8
/4
Raw
/48
/20
/3
Information for Students in Math 348 2003 09
122
1. [12 MARKS] Showing all your work, carefully prove the following theorem in Euclidean geometry. Your solution may use any of Euclid’s axioms or postulates, and any of the following
Propositions (but no others) which you should cite by number.
The sum of the lengths of the diagonals of a quadrilateral is less than its
perimeter.
These are the Propositions you may use:
Proposition I.1: On a given finite straight line, to construct an equilateral triangle.
Proposition I.2: To place at a given point (as an extremity) a (straight) line segment
whose length is equal to that of a given (straight) line segment.
Proposition I.3: Given two unequal line segments, to cut off from the greater a line
segment whose length is equal to the less.
Proposition I.4: If two triangles have the lengths of two sides of the one equal to the
lengths of two sides respectively, and have the angles contained by the equal line
segments equal, they will also have the base equal to the base, the triangle will
be equal to the triangle, and the remaining angles will be equal to the remaining
angles respectively, namely those which the equal sides subtend.
Proposition I.5: In isosceles triangles the angles at the base are equal to one another;
and, if the equal sides be extended below the base, the angles under the base will
also be equal to one another.
Proposition I.6 If, in a triangle, two angles be equal to one another, the sides which
subtend the equal angles will also be equal to one another.
Proposition I.7 Given two line segments constructed from the ends of a line segment,
and meeting in a point, there cannot be constructed from the ends of the same line
segment and on the same side of it, two other line segments meeting in a different
point, respectively of the same lengths as the former two line segments, namely
each equal to that which emanates from the same end of the fixed line segment.
Proposition I.8 If two triangles have two sides equal in length to two sides respectively,
and also have the bases equal in length, they will also have the angles equal which
are contained by the equal sides.
Proposition I.9: To bisect a given angle.
Proposition I.10: To bisect a given line segment.
Proposition I.20 In any triangle, two sides taken together in any manner are greater
than the remaining side.
Proposition I.26 If two triangles have the two angles at the base equal to two angles,
respectively, and one side of the one equal in length to the corresponding side of the
other; namely, either (Case 1) the side adjoining the equal angles, or (Case 2) that
subtending one of the equal angles, they will also have the corresponding remaining
sides equal in length, and the remaining angle equal to the remaining angle.
Information for Students in Math 348 2003 09
123
2. Preceding the list of strip patterns numbered (a), (b), (c), (d), a table has been reproduced.
• Associate with each of the listed strip patterns the pattern in the table which has the
same isometry group, or argue that the isometry group is different from all those listed.
• If the pattern has reflective symmetry in a vertical mirror, describe all types of vertical
mirrors that are possible. For example, for the pattern ...VVVVV... one would observe
that there are two possible types of mirrors:
(a) between two successive letters V, as ...VV|VVV...
(b) along the axis of symmetry of any one letter V, as ...VVVVV...
|
• If the pattern has rotational symmetry under a half-turn, describe the location(s) of the
centre(s) of the half-turn(s).
• If the pattern has symmetry under translations, describe an interval of minimum length
that is mapped by the translations so that it covers the strip without overlaps; e.g. for
the pattern ...UVUVUVUVUV... one such interval would be UV.
• If the pattern has symmetry under a glide reflection, describe the axis of the translation
and an interval of minimum length that is mapped by all the symmetries to cover the
strip without overlapping; e.g. in LΓLΓL one such interval would be L.
I
II
III
IV
V
VI
VII
Typical pattern
...LLLLL...
...LΓLΓL...
...VVVVV...
...NNNNN...
...VΛVΛV...
...DDDDD...
...HHHHH...
Legend:
Ref(H)
0
0
0
0
0
1
1
Symbol
Ref(H)
Ref(V)
Trans
HT
GL
Ref(V)
0
0
∞
0
∞
0
∞
Trans
∞
∞
∞
∞
∞
∞
∞
HT
0
0
0
∞
∞
0
∞
GL
0
∞
0
0
∞
∞
∞
Abstract Group
C∞
C∞
D∞
D∞
D∞
C∞ × D1
D∞ × D1
Meaning
Reflection in Horizontal mirror
Reflections in Vertical mirrors
Translations
Half-Turns
Glide Reflections (with nontrivial translation)
(a) [3 MARKS] . . . NONXNONXNONXNONXNONXNO. . .
(b) [3 MARKS] . . . THIHTHIHTHIHTHIHTHIHTHIHT. . .
(c) [3 MARKS] . . . ∪ ∧ ∩ ∨ ∪ ∧ ∩ ∨ ∪ ∧ ∩ ∨ ∪ ∧ ∩ ∨ ∪ ∧ ∩ ∨ ∪∧. . .
(d) [3 MARKS] . . . RADARADARADARADARADARA. . .
Information for Students in Math 348 2003 09
124
3. Consider a regular hexagon in the Euclidean plane, where the plane has been endowed with a
rectangular Cartesian coordinate system, so that the hexagon is inscribed in a circle of radius
1 centred at the origin, and its vertices are located at points making angles of 0, π3 , 2π
3 , π,
4π 5π
,
.
Let
the
vertices
be
labelled
cyclically,
counterclockwise
around
the
hexagon
as
1, 2,
3
3
3, 4, 5, 6, with vertex 1 being the point(1, 0); then join each of the following pairs of vertices
by a line segment: 1 and 3, 3 and 5, 5 and 1.
(a) [3 MARKS] Express each of the isometries as a product of a non-negative number of
reflections.
(b) [3 MARKS] Set up and complete a rectangular “Cayley Table” for your isometries, so
that the product of any two of them, in each order, can be read off from the table.
(c) [3 MARKS] Determine, for each isometry, its inverse; you may refer to your table in
the preceding part.
(d) [3 MARKS] Show that there are in your set of isometries two reflections which generate
the group of isometries.
Information for Students in Math 348 2003 09
125
4. Let R1 , R2 and R3 be reflections, respectively in the line x + y = 1, the line y = 1, and the
line x = y − 1.
(a) [8 MARKS] Describe a sequence of procedures in each of which a product of reflections
is replaced by an equivalent product of reflections, in such a way that R1 R2 R3 is
expressed as a product of a reflection R0 followed by a translation in a direction parallel
to the mirror of R0 . Show all your work. Describe all the reflections that enter into
your computations carefully, by specifying their mirror in each case.
(b) [4 MARKS] Determine the distance through which the translation shifts the plane, and
verify that your product transforms the point (0, 2) in the same way as R1 R2 R3 .
Information for Students in Math 348 2003 09
continuation page for problem number
You must refer to this continuation page on the page where the problem is printed!
126
Information for Students in Math 348 2003 09
continuation page for problem number
You must refer to this continuation page on the page where the problem is printed!
127
13.4
Version 4
CLASS TEST: MATHEMATICS 189–348 2003 09
TOPICS IN GEOMETRY
EXAMINER: Professor W. G. Brown
DATE: Wednesday, 5th November, 2003.
TIME: 12:35 – 13:20
FAMILY NAME:
GIVEN NAMES:
STUDENT NUMBER:
Instructions
1. All your writing — even rough work — must be handed in.
2. Calculators are not permitted.
3. Your neighbour’s version of this test may not be the same as yours.
4. This examination booklet consists of this cover, Pages 129 through 132 containing questions;
and Pages 133, 134, which are blank.
5. Show all your work. All solutions are to be written in the space provided on the page where
the question is printed. When that spaces is exhausted, you may write on the facing page,
on one of the blank pages, or on the back cover of the booklet, but you must indicate any
continuation clearly on the page where the question is printed! (Please inform the invigilator
if you find that your booklet is defective.)
PLEASE DO NOT WRITE INSIDE THIS BOX
1
2(a)
/12
3(d)
2(b)
/3
4(a)
/3
2(c)
/3
2(d)
/3
3(a)
/3
3(b)
/3
3(c)
/3
4(b)
/8
/4
Raw
/48
/20
/3
Information for Students in Math 348 2003 09
129
1. [12 MARKS] Showing all your work, carefully prove the following theorem in Euclidean geometry. Your solution may use any of Euclid’s axioms or postulates, and any of the following
Propositions (but no others) which you should cite by number.
If one of the altitudes of a triangle is also an angle bisector, then the triangle
is isosceles.
These are the Propositions you may use:
Proposition I.1: On a given finite straight line, to construct an equilateral triangle.
Proposition I.2: To place at a given point (as an extremity) a (straight) line segment
whose length is equal to that of a given (straight) line segment.
Proposition I.3: Given two unequal line segments, to cut off from the greater a line
segment whose length is equal to the less.
Proposition I.4: If two triangles have the lengths of two sides of the one equal to the
lengths of two sides respectively, and have the angles contained by the equal line
segments equal, they will also have the base equal to the base, the triangle will
be equal to the triangle, and the remaining angles will be equal to the remaining
angles respectively, namely those which the equal sides subtend.
Proposition I.5: In isosceles triangles the angles at the base are equal to one another;
and, if the equal sides be extended below the base, the angles under the base will
also be equal to one another.
Proposition I.6 If, in a triangle, two angles be equal to one another, the sides which
subtend the equal angles will also be equal to one another.
Proposition I.7 Given two line segments constructed from the ends of a line segment,
and meeting in a point, there cannot be constructed from the ends of the same line
segment and on the same side of it, two other line segments meeting in a different
point, respectively of the same lengths as the former two line segments, namely
each equal to that which emanates from the same end of the fixed line segment.
Proposition I.8 If two triangles have two sides equal in length to two sides respectively,
and also have the bases equal in length, they will also have the angles equal which
are contained by the equal sides.
Proposition I.9: To bisect a given angle.
Proposition I.10: To bisect a given line segment.
Proposition I.20 In any triangle, two sides taken together in any manner are greater
than the remaining side.
Proposition I.26 If two triangles have the two angles at the base equal to two angles,
respectively, and one side of the one equal in length to the corresponding side of the
other; namely, either (Case 1) the side adjoining the equal angles, or (Case 2) that
subtending one of the equal angles, they will also have the corresponding remaining
sides equal in length, and the remaining angle equal to the remaining angle.
Information for Students in Math 348 2003 09
130
2. Preceding the list of strip patterns numbered (a), (b), (c), (d), a table has been reproduced.
• Associate with each of the listed strip patterns the pattern in the table which has the
same isometry group, or argue that the isometry group is different from all those listed.
• If the pattern has reflective symmetry in a vertical mirror, describe all types of vertical
mirrors that are possible. For example, for the pattern ...VVVVV... one would observe
that there are two possible types of mirrors:
(a) between two successive letters V, as ...VV|VVV...
(b) along the axis of symmetry of any one letter V, as ...VVVVV...
|
• If the pattern has rotational symmetry under a half-turn, describe the location(s) of the
centre(s) of the half-turn(s).
• If the pattern has symmetry under translations, describe an interval of minimum length
that is mapped by the translations so that it covers the strip without overlaps; e.g. for
the pattern ...UVUVUVUVUV... one such interval would be UV.
• If the pattern has symmetry under a glide reflection, describe the axis of the translation
and an interval of minimum length that is mapped by all the symmetries to cover the
strip without overlapping; e.g. in LΓLΓL one such interval would be L.
I
II
III
IV
V
VI
VII
Typical pattern
...LLLLL...
...LΓLΓL...
...VVVVV...
...NNNNN...
...VΛVΛV...
...DDDDD...
...HHHHH...
Legend:
Ref(H)
0
0
0
0
0
1
1
Symbol
Ref(H)
Ref(V)
Trans
HT
GL
Ref(V)
0
0
∞
0
∞
0
∞
Trans
∞
∞
∞
∞
∞
∞
∞
HT
0
0
0
∞
∞
0
∞
GL
0
∞
0
0
∞
∞
∞
Abstract Group
C∞
C∞
D∞
D∞
D∞
C∞ × D1
D∞ × D1
Meaning
Reflection in Horizontal mirror
Reflections in Vertical mirrors
Translations
Half-Turns
Glide Reflections (with nontrivial translation)
(a) [3 MARKS] . . . YAMAYAMAYAMAYAMAYAMAY. . .
(b) [3 MARKS] . . . EOXEOCEOXEOCEOXEOCEOXEOC. . .
(c) [3 MARKS] . . . SHSISHSISHSISHSISHSIS. . .
(d) [3 MARKS] . . . ⊕±H⊕∓H⊕±H⊕∓H⊕±H⊕∓H⊕±H⊕ . . .
Information for Students in Math 348 2003 09
131
3. Consider a regular hexagon in the Euclidean plane, where the plane has been endowed with a
rectangular Cartesian coordinate system, so that the hexagon is inscribed in a circle of radius
1 centred at the origin, and its vertices are located at points making angles of 0, π3 , 2π
3 , π,
4π 5π
,
.
Let
the
vertices
be
labelled
cyclically,
counterclockwise
around
the
hexagon
as
1, 2,
3
3
3, 4, 5, 6, with vertex 1 being the point(1, 0); then join each of the following pairs of vertices
by a line segment: 1 and 3, 3 and 5, 5 and 1.
(a) [3 MARKS] Express each of the isometries as a product of a non-negative number of
reflections.
(b) [3 MARKS] Set up and complete a rectangular “Cayley Table” for your isometries, so
that the product of any two of them, in each order, can be read off from the table.
(c) [3 MARKS] Determine, for each isometry, its inverse; you may refer to your table in
the preceding part.
(d) [3 MARKS] Show that there are, in your set of isometries one reflection and one rotation
which generate the group of isometries.
Information for Students in Math 348 2003 09
132
4. Let R1 , R2 and R3 be reflections, respectively in the line x − y = 1, the line y = −1, and
the line x = −y − 1.
(a) [8 MARKS] Describe a sequence of procedures in each of which a product of reflections
is replaced by an equivalent product of reflections, in such a way that R1 R2 R3 is
expressed as a product of a reflection R0 followed by a translation in a direction parallel
to the mirror of R0 . Show all your work. Describe all the reflections that enter into
your computations carefully, by specifying their mirror in each case.
(b) [4 MARKS] Determine the distance through which the translation shifts the plane, and
verify that your product transforms the point (0, 2) in the same way as R1 R2 R3 .
Information for Students in Math 348 2003 09
continuation page for problem number
You must refer to this continuation page on the page where the problem is printed!
133
Information for Students in Math 348 2003 09
continuation page for problem number
You must refer to this continuation page on the page where the problem is printed!
134
Information for Students in Math 348 2003 09
14
135
Brief discussion of (symbolic) logic
Distribution Date: Wednesday, October 29th, 2003
This background may be needed for the discussion which follows on several geometric
systems defined axiomatically. Students not familiar with mathematical logic should
look through these pages and then refer back to them whenever a new concept appears.
The emphasis is on understanding the specific geometries that we will be discussing, not
on general concepts of mathematical logic.
14.1
Statements
We will use Latin letters — usually in the lower case — possibly subscripted, e.g. p, q, r5 ,
to denote “primitive” or “elementary” statements (sentences); the letters T and F will
be reserved to represent the constants “True” and “False”. Statements can be combined
into “compound” statements by using “logical connectives” — 5 types of symbols:
• unary connectives: We will have only one “unary” connective, so named because
it maps one sentence on to one sentence.
¬: From any sentence S — either a primitive sentence or a compound sentence —
we will create a sentence ¬S, read “not S”. A coming definition will give this
new sentence the meaning we expect when we say “not”.
• binary connectives: These connectives create one new sentence from 2 others. It
is convenient to work with 4 distinct connectives, although it is possible to achieve
the same results with fewer.83
∧ (“and”): This connective is called conjunction, and the statements that are
being connected are called conjuncts.
∨ (“or”): This connective is called disjunction, and the statements that are being connected are called disjuncts. The English word “or” used here is the
“‘inclusive’ or”, meaning either the first alternative, or the second alternative,
or both of them.
→ (“implies”):
↔ (“is equivalent to”):
83
In fact, it is possible — in two different ways — to construct the entire logical system with just one
(binary) connective; there is no time in the present course to discuss such questions.
Information for Students in Math 348 2003 09
136
A valid statement will be a “string” of symbols from the alphabet consisting of the
symbols representing primite statements, the logical connectives, and the symbols (, ).
A statement will be a “string” that can be constructed by a finite number of operations
of either of the following types:
• If S is a statement, then so is (¬S).
• If S and U are statements, then so are (S ∧ U ), (S ∨ U ), (S → U ), (S ↔ U ).
In practice we often suppress some of the parentheses, provided we can reconstruct the
formula by a convenient convention. Thus, for example, we may write ¬p, since there
is no doubt about what is intended; however, we need parentheses is a statement like
p ∧ q ∨ r. The valid strings that can be constructed in the manner described are called
well formed formulæ, or wffs.
14.2
Truth values of sentences
Any sentence can be true or false. We can call an assignment of truth values to sentences
an interpretation, and consider it to be a function mapping the set of all statements to
the set {T, F }. T and F are the constants in the logical system we are constructing. We
often wish to consider the truth value of a sentence under several, or even all possible
assignments of truth values. In that case we may tabulate the results in a truth table.
When there are n primitive variables, the number of lines in a complete truth table will
be 2n , and we often list the cases in a systematic way, for example, in “lexicographic”
order84 . The naming of our logical connectives suggest the following definitions for the
truth values associated with compound statements in terms of the truth values of the
statements out of which they are constructed:
S
F
F
T
T
U
F
T
F
T
(¬S) (S ∧ U ) (S ∨ U ) (S → U ) (S ↔ U )
T
F
F
T
T
T
F
T
T
F
F
F
T
F
F
F
T
T
T
T
This table defines the truth values of statements. The third column is the definition
referred to earlier, giving ¬S its expected “negative” meaning. Most parts of this definition should appear “reasonable” to you, except possibly some parts of the definition
for S → U . Note that this proposition is defined to be true always, except when S
84
the order that the words would have in a dictionary
Information for Students in Math 348 2003 09
137
is true and U is false; in particular, it is true when both S and U are false.85 When
we wish to consider statements constructed with multiple connectives, we determine the
truth values recursively, by building the statement in a sequence of applications of the
connectives.
14.3
Tautologies and contradictions
We are particularly interested in statements whose truth value does not depend on the
truth values of the components: when a statement is true for any assignment of values
to the components, it is a tautology; otherwise it is a contradiction. A contradiction can
be said to be unsatisfiable; a statement which is not a contradiction can be said to be
satisfiable. We can also apply these terms to sets of statements: a set is satisfiable if
there is an assignment of truth values under which all members of the set are true, and
unsatisfiable otherwise. In the following table we will determine the truth values of the
propositions (sentences) p ∨ ¬p, i.e. (p ∨ (¬p)), which will be seen to be a tautology,
(((p ∨ q) ∧ (¬p)) ∧ (¬q)), which will be seen to be a contradiction.
p
T
T
F
F
14.4
q
T
F
T
F
S=
¬p ¬q p ∨ q (p ∨ q) ∧ (¬p) S ∧ (¬q)
F F
T
F
F
F T
T
F
F
T F
T
T
F
T T
F
F
F
p ∨ (¬p)
T
T
T
T
Logical Implication
We shall say that a proposition S (logically) implies a proposition U if U is true whenever
S is true. Symbolically, we may abbreviate this situation to S ⇒ U .
Exercise 14.1 Show that S ⇒ U iff S → U is a tautology.
When both S ⇒ U and U ⇒ S, we will say that S is logically equivalent to U , and write
S ⇔ U.
Exercise 14.2 Show that S ⇔ U iff S ↔ U is a tautology.
More generally, we may wish to consider situations where the simultaneous truth of
several statements S1 , S2 , . . ., Sk implies the truth of a statement U . We will write this
85
This part of the definition may appear counter-intuitive to you at first, but is traditional; we might,
of course, have constructed our logical system differently, since there are 24 different ways in which a
binary logical connective can be defined, and we have only exploited 4 of these ways in our system.
Information for Students in Math 348 2003 09
138
as {S1 , S2 , . . . , Sk } ⇒ U , or by
S1
S2
...
Sk
∴U
(89)
Most mathematical theorems can be formulated in terms of implications of this type.
14.5
“Rules of Logic”
Certain logical implications are often called “Rules of Logic”. The following are some of
these. All of these logical equivalences can be proved, for example, by setting up truth
tables to verify that the statements on the two sides of the ⇔ symbol always have the
same truth value, no matter what truth value is associated with the various components.
Name of Rule
Double Negation
Commutativity of ∧
Commutativity of ∨
Associativity of ∧
Associativity of ∨
Distributivity of ∨ over ∧
Distributivity of ∧ over ∨
De Morgan Laws
Exercise 14.3
Logical Equivalence
¬¬S ⇔ S
S∧U ⇔U ∧S
S∨U ⇔U ∨S
((S ∧ U ) ∧ V ) ⇔ (S ∧ (U ∧ V ))
((S ∨ U ) ∨ V ) ⇔ (S ∨ (U ∨ V ))
(S ∨ (U ∧ V ) ⇔ ((S ∨ U ) ∧ (S ∨ V ))
(S ∧ (U ∨ V ) ⇔ ((S ∧ U ) ∨ (S ∧ V ))
(¬(S ∧ U )) ⇔ ((¬S) ∨ (¬U ))
(¬(S ∨ U )) ⇔ ((¬S) ∧ (¬U ))
1. Prove each of the equivalences in the preceding table.
2. Prove the following logical equivalences or logical implications for all statements S
and U :
(a) S ∧ T ⇔ S (T is a “right identity” for the logical connective ∧)
(b) S ∨ F ⇔ S (F is a “right identity” for the logical connective ∨)
(c) S ∧ S ⇔ S (Every proposition is an “idempotent” for the logical connective
∧.)
(d) S ∨ S ⇔ S (Every proposition is an “idempotent” for the logical connective
∨.)
(e) (S → U ) ⇔ ((¬S) ∨ U )
(f) (S → U ) ⇔ ((¬U ) → (¬S))
UPDATED TO November 27, 2003
Information for Students in Math 348 2003 09
14.6
139
Rules of inference
Certain implications of the type of (89) will be called Rules of Inference. Typically these
will be implications that will arise frequently in mathematical proofs. Among these,
some “named” rules of inference that we may wish to refer to are
Name of Rule
Modus Ponens
Modus Tollens
Hypothetical Syllogism
Proof by cases
Addition
Simplification
S→U
S
∴U
S→U
¬U
∴ ¬S
S→U
U →V
∴S→V
S→V
U →V
∴ (S ∨ U ) → V
S
∴S∨U
S∧U
∴S
By virtue of the definition of the truth value of statements constructed with the connective ∧, and of its properties (associativity and commutativity), statement (89) could
also be written as
(S1 ∧ S2 ∧ . . . ∧ Sk ) ⇒ U .
(90)
This could be called the inference rule of Conjunction.
14.7
Reductio ad absurdum = Proof by Contradiction
Instead of proving (89), or, equivalently, (90), we can — if it is convenient — prove,
instead, the logically equivalent implication
S1
S2
...
Sk
¬U
∴F
(91)
Information for Students in Math 348 2003 09
140
i.e.
(S1 ∧ S2 ∧ . . . ∧ Sk ∧ (¬U )) ⇒ F .
(92)
For the last implication is false precisely when all of the “conjuncts” S1 , S2 , . . . , Sk , ¬U
are true, i.e. when all of the Si (i = 1, . . . , k) are true and U is false, which is precisely
when (90) is false. So we can try to prove that, when the hypotheses S1 , . . ., Sk are
all true, the falsity of the desired conclusion U would entail a contradiction, i.e. an
absurdity.
14.8
The converse and contrapositive of an implication
These terms all refer to a given implication of the form S → U .
The converse of S → U is the statement U → S. Where we know that both of the
implications are true, we could combine them into a statement logically equivalent
to their conjunction S ↔ U .
The contrapositive of S → U is the implication (¬U ) → (¬S). We have shown above
(cf. Exercise 14.3.2f) that the contrapositive is logically equivalent to the original
statement.
The inverse of S → U is logically equivalent to the converse.
Exercise 14.4 Let S, U respectively denote the statements 3 < 5, 5 < 3.
1. Determine which of the statements S → U , U → S is true for every r. (Explain
how you know, without examining the specific definitions of S and U , that at least
one of them must be true.)
2. Give each of the following statements, and determine its truth value.
(a) the converse of S
(b) the contrapositive of S
(c) the converse of U
(d) the contrapositive of U
(e) the converse of S → U
(f) the contrapositive of U → S
(g) the converse of F → S
(h) the contrapositive of F → T . (Remember, ¬T has truth value F , ¬F has
truth value T .)
Information for Students in Math 348 2003 09
14.9
141
The Predicate Calculus
Most of the statements we work with in mathematics have some variable elements. If,
for example, we wish to make the statement that
No two distinct lines have more than one point in common,
we are speaking of all the points and all the lines in some “universes”, for example, the
points and lines in the real plane. We represent such statements symbolically using statements like P (x1 , x2 , . . . , xk ), where P represents a predicate, and x1 , x2 , . . . , xk represent
variables. For example, we could define a predicate I to mean “is incident with”, so that
I(P, `) would mean that the point P lies on the line `, or that the line ` passes through
the point P . Then the preceding statement could be formulated in various ways, that
would assert that the four statements I(P1 , `1 ), I(P1 , `2 ), I(P2 , `1 ), I(P2 , `2 ) could not be
true at the same time, unless either P1 = P2 or `1 = `2 . For example, we could say that
(I(P1 , `1 ) ∧ I(P1 , `2 ) ∧ I(P2 , `1 ) ∧ I(P2 , `2 )) ⇒ ((P1 = P2 ) ∨ (`1 = `2 )) .
(93)
This last statement still does not say what we intended, since it speaks of two specific points and two specific lines. To make the general assertion we intend, we have
to quantify the variables. For example, we could prefix to statement (93) the string
(∀P1 )(∀P2 )(∀`1 )(∀`2 ), where the statement ∀x means “for all x”. This notation is sufficient if the “universe” from which x is to be chosen is obvious. Otherwise we might
wish to specify the universe explicitly; for example we might define P and L to be,
respectively, the sets of points and lines in the real plane, and write
(∀P1 ∈ P)(∀P2 ∈ P)(∀`1 ∈ L)(∀`2 ∈ L) .
The statement would then look like ∀P1 ∈ P)(∀P2 ∈ P)(∀`1 ∈ L)(∀`2 ∈ L)((I(P1 , `1 ) ∧
I(P1 , `2 ) ∧ I(P2 , `1 ) ∧ I(P2 , `2 )) ⇒ ((P1 = P2 ) ∨ (`1 = `2 ))) In addition to the quantifier
∀, we will also wish to work with a quantifier ∃, meaning there exists. Other quantifiers
are possible. For example, we could define ∃1 x to mean there exists exactly one.
Negation of Quantifiers: The following logical equivalences hold:
¬((∃x)S(x)) ⇔ ((∀x)(¬S(x))
¬((∀x)S(x)) ⇔ ((∃x)(¬S(x))
14.10
The axioms, postulates, and theorems of a logical system
Our objectives in the various situations that we will be studying will be to prove theorems.
We will infer these theorems from other proved theorems, as well as from statements
Information for Students in Math 348 2003 09
142
whose truth we shall postulate; the latter statements will be called the axioms of the
system. Nowadays most mathematicians use the words axiom and postulate (as a noun)
interchangeably.86
Any statement inferred in this way can be called a theorem. However, mathematicians
have many other words that are also used for theorems. Some of these describe the way in
which the result has been proved87 ; some terms will refer to the way in which a theorem
is going to be used88 ; some will refer to the importance that we give to the result89 ;
some mathematicians are reluctant to use the word theorem for any result that is easy
to prove. In Euclid the word proposition is used for most theorems.
86
However, postulate can also be used as a verb, as we have done in the previous sentence.
e.g., a Corollary is a theorem that is either an immediate consequence of another theorem, or
a theorem whose proof can be obtained through minor changes to the proof of another theorem; a
Corollary is always associated with another theorem.
88
e.g., a Lemma is a theorem that is going to assist in the proof of another, more important theorem.
89
e.g., some theorems have acquired names like The Fundamental Theorem of ....
87
Information for Students in Math 348 2003 09
15
143
Solutions, Third Problem Assignment
Distribution Date: Monday, November 3rd, 2003
corrected Tuesday, November 4th, 2003
corrected Wednesday, November 12th, 2003
(subject to further correction)
Solutions were to be submitted on Monday, November 3rd, 2003
Preceding the list of friezes numbered 1, 2, . . . , a table has been reproduced.
• Associate with each of the listed friezes the frieze in the table which has the same
isometry group, or argue that the isometry group is different from all those listed.
• For each type of isometry possessed by the frieze shown, give a precisely described
example of each isometry of that type. More precisely,
– If the pattern has reflective symmetry in a vertical mirror, describe all types of
vertical mirrors that are possible. For example, for the pattern ...VVVVV...
one would observe that there are two possible types of mirrors:
1. between two successive letters V, as ...VV|VVV...
2. along the axis of symmetry of any one letter V, as ...VVVVV...
|
– If the pattern has rotational symmetry under a half-turn, describe the location(s) of the centre(s) of the half-turn(s).
– If the pattern has symmetry under translations, describe a fundamental interval; e.g. for the pattern ...UVUVUVUVUV... one such interval would
be UV.
– If the pattern has symmetry under a glide reflection, describe the axis of the
translation and a shortest interval that is translated; e.g. in LΓLΓL one such
interval would be LΓ.
• It may happen that the text processor used to produce these notes will show
inconsistent spacing between the symbols used. In these examples we have never
varied the spaces between symbols: treat them all as being of equal length.
Also the intention is that symbols whose “simplest” form has symmetries may
appear to be slightly asymmetric because of variations in the typography used.
Treat all symbols as having the maximum symmetries of the “simplest” forms.90
90
For example, the letter V should be treated as having symmetry in a vertical reflective axis even
though the type being used is thicker on the left side than on the right.
Information for Students in Math 348 2003 09
I
II
III
IV
V
VI
VII
Typical pattern Ref(H) Ref(V) Trans HT GL Abstract Group
...LLLLL...
0
0
∞
0
0
C∞
...LΓLΓL...
0
0
∞
0
∞
C∞
...VVVVV...
0
∞
∞
0
0
D∞
...NNNNN...
0
0
∞
∞
0
D∞
...VΛVΛV...
0
∞
∞
∞ ∞
D∞
...DDDDD...
1
0
∞
0
∞
C∞ × D1
...HHHHH...
1
∞
∞
∞ ∞
D∞ × D1
Symbol Meaning
Ref(H) Reflection in Horizontal mirror
Ref(V) Reflections in Vertical mirrors
Legend:
Trans Translations
HT
Half-Turns
GL
Glide Reflections (with nontrivial translation)
1. . . . LLΓLLΓLLΓ. . .
2. . . . ÀÂÁĂÀÂÁĂÀÂÁĂ. . .
3. . . . HXHXHXHXHX. . .
4. . . . $SZS$$SZS$$SZS$. . .
5. . . . AĀAAĀAAĀA. . .
¯
¯
¯
6. . . . ÀÂÁÀÂÁÀÂÁÀÂÁ. . .
7. . . . ¡¿!?¡¿!?¡¿!?. . .
n X
m X
n X
m X
n X
m
X
8. . . .
...
m
9. . . .
n
m
n
m
n
= X
× X
= X
× X
= X
×
X
×
=
×
=
×
...
=
10. . . . §‡z‡§‡z‡§‡z‡. . .
11. . . . ±♦ ∓ ♦ ± ♦ ∓ ♦ ± ♦ ∓ ♦ ± ♦∓. . .
12. . . . ∈3 >⊥ ∈3 >⊥ ∈3 >⊥ ∈3 >⊥. . .
13. . . . ΓΓΓΓΓΓ=LLLLLL. . . (only one = sign)
144
Information for Students in Math 348 2003 09
14. . . .
= X
× X
= X
× X
= X
×
X
=
×
=
×
=
145
...
×
15. . . . HHHHHHOHHHHHH. . . (only one O)
16. . . . LLΓΓLLΓΓLLΓΓ. . .
17. . . . 6666660999999. . . (only one 0)
18. . . . HHHHHH=OOOOOO. . . (only one = sign)
Solution:
Number
(1)
(2)
(3)
(4)
(5)
(6)
(7)
Axis of
Ref(H)
—
—
yes
—
—
—
—
Axes of
Ref(V)
—
Â, Ă
H, X
—
—
Â, Á|À
—
(8)
—
—
(9)
—
—
(10)
(11)
(12)
(13)
—
—
—
—
—
±, ∓
—
—
(14)
yes
—
Fund.Interval
(Translation)
LLΓ
ÀÂÁĂ
HX
$$SZS
AĀA
¯
ÀÂÁ
¡¿!?
n X
m
X
m n
= X
×
X
×
(15)
(16)
(17)
(18)
yes
—
—
yes
O
—
—
—
UPDATED TO November 27, 2003
Shortest Interval
(Glide-Reflection)
—
—
HX
—
—
—
Isometry
Group
I
III
VII
IV
I
III
I
—
—
I
—
=
X
=
§‡z‡
±♦ ∓ ♦
∈3 >⊥
—
= X
×
X
=
Centres of
HT
—
—
H, X
Z, $|$
—
—
—
×
§, z
±♦∓, ∓♦±
∈ | 3,>|⊥
—
—
±♦ ∓ ♦
—
= X
×
X
=
×
—
LLΓΓ
—
—
II
O
—
0
—
IV
V
IV
I
VI
×
—
LLΓΓ
—
—
D2
II
C2
D1
Information for Students in Math 348 2003 09
16
146
Solutions to Problems on the Class Tests
which were administered on Wednesday, November 5th, 2003
Distribution Date: Wednesday, November 12th, 2003
(subject to correction)
There were 4 versions of the test; these had similar but different problems ##1, 2, 4.
Problem 3 was identical on all versions, except for part (d) thereof, which was the same
on versions ##1, 3 and the same on versions ##2, 4.
Here is how the recorded grade on the test has been computed: The medians on
the four versions were computed, and the grades on those with the lower medians were
scaled upwards so that all 4 versions had the same medians. Then the raw grades out of
48 were ranked. About 7 students had by then obtained a perfect score. The maximum
needed for a perfect score was based on the next grade, and the papers were graded out
of 45, and scaled to a recorded grade out of 20.
16.1
Version 1
1. [12 MARKS] Showing all your work, carefully prove the following theorem in Euclidean geometry. Your solution may use any of Euclid’s axioms or postulates, and
any of the following Propositions (but no others) which you should cite by number.
The median to the base of an isosceles triangle is perpendicular to the
base and bisects the opposite angle.
These are the Propositions you may use:
Proposition I.1: On a given finite straight line, to construct an equilateral triangle.
Proposition I.2: To place at a given point (as an extremity) a (straight) line segment
whose length is equal to that of a given (straight) line segment.
Proposition I.3: Given two unequal line segments, to cut off from the greater a line
segment whose length is equal to the less.
Proposition I.4: If two triangles have the lengths of two sides of the one equal to the
lengths of two sides respectively, and have the angles contained by the equal line
segments equal, they will also have the base equal to the base, the triangle will
be equal to the triangle, and the remaining angles will be equal to the remaining
angles respectively, namely those which the equal sides subtend.
Proposition I.5: In isosceles triangles the angles at the base are equal to one another;
and, if the equal sides be extended below the base, the angles under the base will
also be equal to one another.
Information for Students in Math 348 2003 09
147
Proposition I.6 If, in a triangle, two angles be equal to one another, the sides which
subtend the equal angles will also be equal to one another.
Proposition I.7 Given two line segments constructed from the ends of a line segment,
and meeting in a point, there cannot be constructed from the ends of the same line
segment and on the same side of it, two other line segments meeting in a different
point, respectively of the same lengths as the former two line segments, namely
each equal to that which emanates from the same end of the fixed line segment.
Proposition I.8 If two triangles have two sides equal in length to two sides respectively,
and also have the bases equal in length, they will also have the angles equal which
are contained by the equal sides.
Proposition I.9: To bisect a given angle.
Proposition I.10: To bisect a given line segment.
Proposition I.20 In any triangle, two sides taken together in any manner are greater
than the remaining side.
Proposition I.26 If two triangles have the two angles at the base equal to two angles,
respectively, and one side of the one equal in length to the corresponding side of the
other; namely, either (Case 1) the side adjoining the equal angles, or (Case 2) that
subtending one of the equal angles, they will also have the corresponding remaining
sides equal in length, and the remaining angle equal to the remaining angle.
Solution: (This solution is reproduced from the Solutions to Assignment 1, §8 of
these notes.)
Let 4ABC be given, and let D be the midpoint of side BC. The median from A
is the line segment AD.
A£qB
B
q£
£
£
£
£
£
£
£
£
£
£ B
B
B
B
qD
B
B
B
B
B
BC
Bq
In triangles ABD and ACD
|AB| = |AC| (hypothesis, 4 is isosceles)
|BD| = |CD| (hypothesis, D is midpoint of BC)
|AD| = |AD| (common edge of 4’s)
(94)
(95)
(96)
Information for Students in Math 348 2003 09
148
Hence, by Proposition I.8, 4ABD ≡ 4ACD.
Alternatively, one could first apply Proposition I.5 to hypothesis (94) and conclude
that
∠ABD = ∠ACD ,
(97)
and then apply Proposition I.4 to (97), (94) and (95).
From this congruence follow three equations, of which two are
∠ADB = ∠ADC
∠BAD = ∠CAD
(98)
(99)
Now DA is a line segment set up on line BC making the adjacent angles ∠ADB,
∠ADC equal. By Definition 3.2.10 AD is perpendicular to BC. Equation (99)
shows that DA bisects ∠BAC.
2. Preceding the list of strip patterns numbered (a), (b), (c), (d), a table has been
reproduced.
• Associate with each of the listed strip patterns the pattern in the table which
has the same isometry group, or argue that the isometry group is different
from all those listed.
• If the pattern has reflective symmetry in a vertical mirror, describe all types of
vertical mirrors that are possible. For example, for the pattern ...VVVVV...
one would observe that there are two possible types of mirrors:
(a) between two successive letters V, as ...VV|VVV...
(b) along the axis of symmetry of any one letter V, as ...VVVVV...
|
• If the pattern has rotational symmetry under a half-turn, describe the location(s) of the centre(s) of the half-turn(s).
• If the pattern has symmetry under translations, describe an interval of minimum length that is mapped by the translations so that it covers the strip
without overlaps; e.g. for the pattern ...UVUVUVUVUV... one such interval would be UV.
• If the pattern has symmetry under a glide reflection, describe the axis of
the translation and an interval of minimum length that is mapped by all the
symmetries to cover the strip without overlapping; e.g. in LΓLΓL one such
interval would be L.
Information for Students in Math 348 2003 09
I
II
III
IV
V
VI
VII
Typical pattern
...LLLLL...
...LΓLΓL...
...VVVVV...
...NNNNN...
...VΛVΛV...
...DDDDD...
...HHHHH...
Legend:
Ref(H)
0
0
0
0
0
1
1
Symbol
Ref(H)
Ref(V)
Trans
HT
GL
Ref(V)
0
0
∞
0
∞
0
∞
149
Trans HT GL
∞
0
0
∞
0
∞
∞
0
0
∞
∞
0
∞
∞ ∞
∞
0
∞
∞
∞ ∞
Abstract Group
C∞
C∞
D∞
D∞
D∞
C∞ × D1
D∞ × D1
Meaning
Reflection in Horizontal mirror
Reflections in Vertical mirrors
Translations
Half-Turns
Glide Reflections (with nontrivial translation)
(a) [3 MARKS] . . . HAWAHAWAHAWAHAWA. . .
(b) [3 MARKS] . . . MAXIMAXIMAXIMAXIMAXIMAX. . .
(c) [3 MARKS] . . . SOSXSOSXSOSXSOSXSOSXS . . .
(d) [3 MARKS] . . . HODHIDHODHIDHODHIDHODHIDHODH . . .
Solution:
(a) [3 MARKS] In . . . HAWAHAWAHAWAHAWA. . . there are letters that
do not horizontal reflective symmetry, and letters that do not have symmetry
under a half turn. But the string HAWA does have translational symmetry,
and the interval of minimum length that covers the strip under these translations — a so-called fundamental interval, is any translate of HAWA. The
reasoning given so far shows that the isometry group is one of I, II, III. But I
and II have no reflective symmetry in vertical mirrors, and the given pattern
is symmetric under reflections in vertical mirrors bisecting any H and any W.
The group is III. After this example, we will tabulate the data for the four
cases compactly.
Frieze
. . . HAWAHAWAHAWAHAWA. . .
. . . MAXIMAXIMAXIMAXIMAXIMAX. . .
. . . SOSXSOSXSOSXSOSXSOSXS . . .
. . . HODHIDHODHIDHODHIDHODHIDHODH . . .
Axis of
Ref(H)
no
no
no
yes
Axes of
Ref(V)
H, W
none
none
none
Fund.Interval
(Translation)
HAWA
MAXI
SOSX
HODHID
Centres of
Half-Turns
none
none
O, X
none
Axis
(Glide-Reflection)
none
none
none
HODHID
Isometry
Group
III
I
IV
VI
3. Consider a regular hexagon in the Euclidean plane, where the plane has been
endowed with a rectangular Cartesian coordinate system, so that the hexagon is
inscribed in a circle of radius 1 centred at the origin, and its vertices are located at
, π, 4π
, 5π
. Let the vertices be labelled cyclically,
points making angles of 0, π3 , 2π
3
3
3
counterclockwise around the hexagon as 1, 2, 3, 4, 5, 6, with vertex 1 being the
Information for Students in Math 348 2003 09
150
point(1, 0); then join each of the following pairs of vertices by a line segment: 1
and 3, 3 and 5, 5 and 1.
(a) [3 MARKS] Express each of the isometries as a product of a non-negative
number of reflections.
(b) [3 MARKS] Set up and complete a rectangular “Cayley Table” for your isometries, so that the product of any two of them, in each order, can be read off
from the table.
(c) [3 MARKS] Determine, for each isometry, its inverse; you may refer to your
table in the preceding part.
(d) [3 MARKS] Show that there are in your set of isometries two reflections which
generate the group of isometries.
Solution:
(a) The problem asks you to express the isometries in terms of the reflections.
Thus, in order to provide a correct answer, you need to identify the reflections,
and then to observe that each of them is expressible in terms of itself (or as
some other product of reflections; and, you need to identify the rotations and
express each of them in some way as a product of the reflections, previously
identified.
The symmetries must map the points 1, 3, 5 among themselves, and the points
2, 4, 6 among themselves. While the configuration of line segments is slightly
different from that considered on Assignment 2, the symmetries are the same.
The only symmetries available for the hexagon are those in the group D6 —
6 rotations through multiples of π6 radians, and 6 reflections — three of them
“edge-centred”, and three of them “vertex-centred”. But the “edge-centred”
isometries of the hexagon are not isometries of the present configuration; and
only 3 of the rotations of the hexagon are symmetries here — those through
multiples of 2π
radians. For a compact discussion, I will assign names to the
6
isometries:
The Rotations:
µ
S=
1 2 3 4 5 6
3 4 5 6 1 2
¶
and the others will be powers of S: S 2 , and
µ
¶
1 2 3 4 5 6
3
S =
=I.
1 2 3 4 5 6
Information for Students in Math 348 2003 09
The Reflections:
µ
R1 =
µ
R2 =
µ
R3 =
151
¶
1 2 3 4 5 6
1 6 5 4 3 2
¶
1 2 3 4 5 6
3 2 1 6 5 4
¶
1 2 3 4 5 6
5 4 3 2 1 6
Now the reflections are each written as a product of 1 reflection. The rotations
may be written in infinitely many ways as such a product. For example,
µ
R1 R2 =
1 2 3 4 5 6
1 6 5 4 3 2
¶µ
1 2 3 4 5 6
3 2 1 6 5 4
¶
µ
=
1 2 3 4 5 6
3 4 5 6 1 2
¶
=S
To obtain a decomposition of S 2 we can simply take R1 R2 R1 R2 . But, if we
want the product of a minimum number of reflections, we could observe that,
for example
µ
R2 R1 =
1 2 3 4 5 6
3 2 1 6 5 4
¶µ
1 2 3 4 5 6
1 6 5 4 3 2
¶
µ
=
1 2 3 4 5 6
5 6 1 2 3 4
¶
= S2
and there are other possible decompositions.
(b)
I
S
S2
R1
R2
R3
I
I
S
S2
R1
R2
R3
S
S
S2
I
R2
R3
R1
S2
S2
I
S
R3
R1
R2
R1
R1
R3
R2
I
S2
S
R2
R2
R1
R3
S
I
S2
R3
R3
R2
R1
S2
S
I
(c) Each of the reflections is its own inverse. Of the rotations, the identity is its
own inverse, and either non-identity rotation is the inverse of the other.
(d) We have seen that S = R1 R2 ; hence S2 = R1 R2 R1 R2 , and the identity is
an empty product, or the product R1 R1 , for example. This leaves only one
element, R3 . This can be expressed in infinitely many ways as products of R1
and R2 . It will need to be the product of an odd number of these reflections,
and not of 1 of them; so let’s try 3 of them. And, in any product, we won’t
want to have one reflection followed by itself, as the product of the two of
them would be the identity. So let’s consider R1 R2 R1 and R2 R1 R2 ; we find
that both of these are equal to R3 .
Information for Students in Math 348 2003 09
152
4. Let R1 , R2 and R3 be reflections, respectively in the line x + y = 1, the line x = 1,
and the line y = x − 1.
(a) [8 MARKS] Describe a sequence of procedures in each of which a product
of reflections is replaced by an equivalent product of reflections, in such a
way that R1 R2 R3 is expressed as a product of a reflection R0 followed by a
translation in a direction parallel to the mirror of R0 . Show all your work.
Describe all the reflections that enter into your computations carefully, by
specifying their mirror in each case.
(b) [4 MARKS] Determine the distance through which the translation shifts the
plane, and verify that your product transforms the point (0, 2) in the same
way as R1 R2 R3 .
Solution:
(a) The following is only one way of achieving the desired decomposition; there
are others, some much simpler in this special case.
i. Let us turn the mirrors of R2 and R3 through an angle of − π4 about their
point of intersection, (1, 0), in order that the mirror of the new middle
factor is perpendicular to the mirror of R1 . The product is then replaced
by R1 R20 R30 , where the mirrors of R20 and R30 are, respectively, the line
y = x − 1 and the x-axis.
ii. Now consider the product of R1 and R20 , whose perpendicular mirrors
again intersect in (1, 0). We could turn the mirrors so that the new middle
mirror is parallel to the mirror of R30 , i.e. to the x-axis; this requires a turn
through − π4 radians, and results in reflections R200 and R30 , whose mirrors
are, respectively, the x-axis and the line x = 1.
iii. In the product R10 R200 R30 the last two factors are identical, and cancel one
another, since they are reflections. Another way to view this is that the
product is a translation through distance 0. (Of course, the 0 translation
may be viewed as having all directions!) The product is equal to R10
alone — the reflection in the line x = 1 (followed by a 0 translation in
the direction of that axis, indeed, in any direction).
(b) Under the transformed product, the point (0, 2) moves to (2, 2). Under the
original product, (0, 2) would move first to (−1, 1), then to (3, 1), then to
(2, 2), as before.
16.2
Version 2
1. [12 MARKS] Showing all your work, carefully prove the following theorem in Eu-
Information for Students in Math 348 2003 09
153
clidean geometry. Your solution may use any of Euclid’s axioms or postulates, and
any of the following Propositions (but no others) which you should cite by number.
If one of the altitudes of a triangle is also a median, then the triangle is
isosceles.
These are the Propositions you may use: (see Version 1 for this list)
Solution: (This solution is reproduced from the Solution to Assignment 1, §8 of
these notes.)
A£qB
B
q£
£
£
£
£
£
£
£
£
£
£ B
B
B
B
D
q
B
B
B
B
B
B
Bq
C
Let 4ABC be given, and let D be a point in side BC such that AD is perpendicular
to BC. By definition of perpendicular
∠ADB = ∠ADC .
(100)
In triangles ABD and ADC we have (100) and
|AD| = |AD| common side of triangles
|BD| = |CD| D is midpoint of BC
Hence, by Proposition I.4, 4ABD ≡ 4ADC, so |AB| = |AC|, i.e. 4ABC is
isosceles.
2. Preceding the list of strip patterns numbered (a), (b), (c), (d), a table has been
reproduced.
• Associate with each of the listed strip patterns the pattern in the table which
has the same isometry group, or argue that the isometry group is different
from all those listed.
• If the pattern has reflective symmetry in a vertical mirror, describe all types of
vertical mirrors that are possible. For example, for the pattern ...VVVVV...
one would observe that there are two possible types of mirrors:
Information for Students in Math 348 2003 09
154
(a) between two successive letters V, as ...VV|VVV...
|
(b) along the axis of symmetry of any one letter V, as ...VVVVV...
• If the pattern has rotational symmetry under a half-turn, describe the location(s) of the centre(s) of the half-turn(s).
• If the pattern has symmetry under translations, describe an interval of minimum length that is mapped by the translations so that it covers the strip
without overlaps; e.g. for the pattern ...UVUVUVUVUV... one such interval would be UV.
• If the pattern has symmetry under a glide reflection, describe the axis of
the translation and an interval of minimum length that is mapped by all the
symmetries to cover the strip without overlapping; e.g. in LΓLΓL one such
interval would be L.
I
II
III
IV
V
VI
VII
Typical pattern
...LLLLL...
...LΓLΓL...
...VVVVV...
...NNNNN...
...VΛVΛV...
...DDDDD...
...HHHHH...
Legend:
Ref(H)
0
0
0
0
0
1
1
Symbol
Ref(H)
Ref(V)
Trans
HT
GL
Ref(V)
0
0
∞
0
∞
0
∞
Trans HT GL
∞
0
0
∞
0
∞
∞
0
0
∞
∞
0
∞
∞ ∞
∞
0
∞
∞
∞ ∞
Abstract Group
C∞
C∞
D∞
D∞
D∞
C∞ × D1
D∞ × D1
Meaning
Reflection in Horizontal mirror
Reflections in Vertical mirrors
Translations
Half-Turns
Glide Reflections (with nontrivial translation)
(a) [3 MARKS] . . . BIDDEBIDDEBIDDEBIDDEBID . . .
(b) [3 MARKS] . . . EΓELEΓELEΓELEΓELEΓ. . .
(c) [3 MARKS] . . . NIBINIBINIBINIBINI . . .
(d) [3 MARKS] . . . INIXINIXINIXINIXINIXINI. . .
Solution:
Frieze
. . . BIDDEBIDDEBIDDEBIDDEBID . . .
. . . EΓELEΓELEΓELEΓELEΓ. . .
. . . NIBINIBINIBINIBINI . . .
. . . INIXINIXINIXINIXINIXINI. . .
Axis of
Ref(H)
yes
no
no
no
Axes of
Ref(V)
none
none
none
none
Fund.Interval
(Translation)
BIDDE
EΓEL
NIBI
INIX
Centres of
Half-Turns
none
none
none
N, X
Axis
(Glide-Reflection)
BIDDE
EΓ
no
no
Isometry
Group
VI
II
I
IV
3. Consider a regular hexagon in the Euclidean plane, where the plane has been
endowed with a rectangular Cartesian coordinate system, so that the hexagon is
Information for Students in Math 348 2003 09
155
inscribed in a circle of radius 1 centred at the origin, and its vertices are located at
points making angles of 0, π3 , 2π
, π, 4π
, 5π
. Let the vertices be labelled cyclically,
3
3
3
counterclockwise around the hexagon as 1, 2, 3, 4, 5, 6, with vertex 1 being the
point(1, 0); then join each of the following pairs of vertices by a line segment: 1
and 3, 3 and 5, 5 and 1.
(a) [3 MARKS] Express each of the isometries as a product of a non-negative
number of reflections.
(b) [3 MARKS] Set up and complete a rectangular “Cayley Table” for your isometries, so that the product of any two of them, in each order, can be read off
from the table.
(c) [3 MARKS] Determine, for each isometry, its inverse; you may refer to your
table in the preceding part.
(d) [3 MARKS] Show that there are, in your set of isometries one reflection and
one rotation which generate the group of isometries.
Solution: This problem is the same as the corresponding problem on Version 1,
except for the following, which we solve using the notation given in the solution in
Version 1.
(d) This can be done in many ways. We have shown that S = R1 R2 ; hence
R2 S = R1 R2 R2 = R1 , so R1 is expressible in terms of R2 and S. We can
show, for example, that R3 = R2 S. Of course, S 2 is a product of two of the
given generators; I is the empty product, or, if you prefer, it is the product
R2 R2 or SSS.
4. Let R1 , R2 and R3 be reflections, respectively in the line −x + y = 1, the line
x = −1, and the line y = −x − 1.
(a) [8 MARKS] Describe a sequence of procedures in each of which a product
of reflections is replaced by an equivalent product of reflections, in such a
way that R1 R2 R3 is expressed as a product of a reflection R0 followed by a
translation in a direction parallel to the mirror of R0 . Show all your work.
Describe all the reflections that enter into your computations carefully, by
specifying their mirror in each case.
(b) [4 MARKS] Determine the distance through which the translation shifts the
plane, and verify that your product transforms the point (0, 2) in the same
way as R1 R2 R3 .
Solution:
Information for Students in Math 348 2003 09
156
(a) The following is only one way of achieving the desired decomposition; there
are others, some much simpler in this special case.
i. Let us turn the mirrors of R2 and R3 through an angle of π4 about their
point of intersection, (−1, 0), in order that the mirror of the new middle
factor is perpendicular to the mirror of R1 . The product is then replaced
by R1 R20 R30 , where the mirrors of R20 and R30 are, respectively, the line
y = −x − 1 and the x-axis.
ii. Now consider the product of R1 and R20 , whose perpendicular mirrors
again intersect in (1, 0). We could turn the mirrors so that the new middle
mirror is parallel to the mirror of R30 , i.e. to the x-axis; this requires a turn
through π4 radians, and results in reflections R200 and R10 , whose mirrors
are, respectively, the x-axis and the line x = −1.
iii. In the product R10 R200 R30 the last two factors are identical, and cancel one
another, since they are reflections. Another way to view this is that the
product is a translation through distance 0. (Of course, the 0 translation
may be viewed as having all directions!) The product is equal to R10 alone
— the reflection in the line x = −1 (followed by a 0 translation in the
direction of that axis, indeed, in any direction).
(b) Under the transformed product, the point (0, 2) moves to (−2, 2). Under the
original product, (0, 2) would move first to (1, 1), then to (−3, 1), then to
(−2, 2), as before.
16.3
Version 3
1. [12 MARKS] Showing all your work, carefully prove the following theorem in Euclidean geometry. Your solution may use any of Euclid’s axioms or postulates, and
any of the following Propositions (but no others) which you should cite by number.
The sum of the lengths of the diagonals of a quadrilateral is less than its
perimeter.
These are the Propositions you may use: (see Version 1 for this list)
Solution: (This solution is reproduced from the Solution to Assignment 1, §8 of
these notes.)
Draw the diagonals AC and BD, thus forming 4 triangles: ABC, ADC, ABD,
CBD. Applying Proposition I.20 to each of these triangles yields the following
inequalities:
|AB| + |BC| > |AC|
Information for Students in Math 348 2003 09
157
|AD| + |CD| > |AC|
|AB| + |AD| > |BD|
|BC| + |CD| > |BD|
Summing those inequalities yields, after reduction,
2(|AB| + |BC| + |CD| + |AD|) > 2(|AC| + |BD|)
from which it follows that
|AB| + |BC| + |CD| + |AD| > |AC| + |BD|
i.e., that the perimeter exceeds the sum of the lengths of the two diagonals, ¤.
2. Preceding the list of strip patterns numbered, a table has been reproduced.
• Associate with each of the listed strip patterns the pattern in the table which
has the same isometry group, or argue that the isometry group is different
from all those listed.
• If the pattern has reflective symmetry in a vertical mirror, describe all types of
vertical mirrors that are possible. For example, for the pattern ...VVVVV...
one would observe that there are two possible types of mirrors:
(a) between two successive letters V, as ...VV|VVV...
(b) along the axis of symmetry of any one letter V, as ...VVVVV...
|
• If the pattern has rotational symmetry under a half-turn, describe the location(s) of the centre(s) of the half-turn(s).
• If the pattern has symmetry under translations, describe an interval of minimum length that is mapped by the translations so that it covers the strip
without overlaps; e.g. for the pattern ...UVUVUVUVUV... one such interval would be UV.
• If the pattern has symmetry under a glide reflection, describe the axis of
the translation and an interval of minimum length that is mapped by all the
symmetries to cover the strip without overlapping; e.g. in LΓLΓL one such
interval would be L.
I
II
III
IV
V
VI
VII
Typical pattern
...LLLLL...
...LΓLΓL...
...VVVVV...
...NNNNN...
...VΛVΛV...
...DDDDD...
...HHHHH...
Ref(H)
0
0
0
0
0
1
1
UPDATED TO November 27, 2003
Ref(V)
0
0
∞
0
∞
0
∞
Trans HT GL
∞
0
0
∞
0
∞
∞
0
0
∞
∞
0
∞
∞ ∞
∞
0
∞
∞
∞ ∞
Abstract Group
C∞
C∞
D∞
D∞
D∞
C∞ × D1
D∞ × D1
Information for Students in Math 348 2003 09
Legend:
Symbol
Ref(H)
Ref(V)
Trans
HT
GL
158
Meaning
Reflection in Horizontal mirror
Reflections in Vertical mirrors
Translations
Half-Turns
Glide Reflections (with nontrivial translation)
(a) [3 MARKS] . . . NONXNONXNONXNONXNONXNO . . .
(b) [3 MARKS] . . . THIHTHIHTHIHTHIHTHIHTHIHT . . .
(c) [3 MARKS] . . . ∪ ∧ ∩ ∨ ∪ ∧ ∩ ∨ ∪ ∧ ∩ ∨ ∪ ∧ ∩ ∨ ∪ ∧ ∩ ∨ ∪∧. . .
(d) [3 MARKS] . . . RADARADARADARADARADARA . . .
Solution:
Frieze
. . . NONXNONXNONXNONXNONXNO . . .
. . . THIHTHIHTHIHTHIHTHIHTHIHT . . .
. . . ∪ ∧ ∩ ∨ ∪ ∧ ∩ ∨ ∪ ∧ ∩ ∨ ∪ ∧ ∩ ∨ ∪ ∧ ∩ ∨ ∪∧. . .
. . . RADARADARADARADARADARA . . .
Axis of
Ref(H)
no
no
no
no
Axes of
Ref(V)
none
I, T
no
no
Fund.Interval
(Translation)
NONX
THIH
∪ ∧ ∩∨
RADA
Centres of
Half-Turns
O, X
none
none
none
Axis
(Glide-Reflection)
no
no
∪∧
no
Isometry
Group
IV
III
II
I
3. Consider a regular hexagon in the Euclidean plane, where the plane has been
endowed with a rectangular Cartesian coordinate system, so that the hexagon is
inscribed in a circle of radius 1 centred at the origin, and its vertices are located at
, π, 4π
, 5π
. Let the vertices be labelled cyclically,
points making angles of 0, π3 , 2π
3
3
3
counterclockwise around the hexagon as 1, 2, 3, 4, 5, 6, with vertex 1 being the
point(1, 0); then join each of the following pairs of vertices by a line segment: 1
and 3, 3 and 5, 5 and 1.
(a) [3 MARKS] Express each of the isometries as a product of a non-negative
number of reflections.
(b) [3 MARKS] Set up and complete a rectangular “Cayley Table” for your isometries, so that the product of any two of them, in each order, can be read off
from the table.
(c) [3 MARKS] Determine, for each isometry, its inverse; you may refer to your
table in the preceding part.
(d) [3 MARKS] Show that there are in your set of isometries two reflections which
generate the group of isometries.
Solution: This is the same as the corresponding problem on Version 1.
4. Let R1 , R2 and R3 be reflections, respectively in the line x + y = 1, the line y = 1,
and the line x = y − 1.
Information for Students in Math 348 2003 09
159
(a) [8 MARKS]Describe a sequence of procedures in each of which a product
of reflections is replaced by an equivalent product of reflections, in such a
way that R1 R2 R3 is expressed as a product of a reflection R0 followed by a
translation in a direction parallel to the mirror of R0 . Show all your work.
Describe all the reflections that enter into your computations carefully, by
specifying their mirror in each case.
(b) [4 MARKS] Determine the distance through which the translation shifts the
plane, and verify that your product transforms the point (0, 2) in the same
way as R1 R2 R3 .
Solution:
(a) The following is only one way of achieving the desired decomposition; there
are others, some much simpler in this special case.
i. Let us turn the mirrors of R2 and R3 through an angle of π4 about their
point of intersection, (0, 1), in order that the mirror of the new middle
factor is perpendicular to the mirror of R1 . The product is then replaced
by R1 R20 R30 , where the mirrors of R20 and R30 are, respectively, the line
y = x − 1 and the y-axis.
ii. Now consider the product of R1 and R20 , whose perpendicular mirrors
again intersect in (0, 1). We could turn the mirrors so that the new middle
mirror is parallel to the mirror of R30 , i.e. to the y-axis; this requires a turn
through π4 radians, and results in reflections R200 and R10 , whose mirrors
are, respectively, the y-axis and the line y = 1.
iii. In the product R10 R200 R30 the last two factors are identical, and cancel one
another, since they are reflections. Another way to view this is that the
product is a translation through distance 0. (Of course, the 0 translation
may be viewed as having all directions!) The product is equal to R10
alone — the reflection in the line y = 1 (followed by a 0 translation in
the direction of that axis, indeed, in any direction).
(b) Under the transformed product, the point (0, 2) moves to (0, 0). Under the
original product, (0, 2) would move first to (−1, 1), then to (−1, 1), then to
(0, 0), as before.
16.4
Version 4
1. [12 MARKS] Showing all your work, carefully prove the following theorem in Euclidean geometry. Your solution may use any of Euclid’s axioms or postulates, and
any of the following Propositions (but no others) which you should cite by number.
Information for Students in Math 348 2003 09
160
If one of the altitudes of a triangle is also an angle bisector, then the
triangle is isosceles.
These are the Propositions you may use: (see Version 1 for this list)
Solution: (This solution is reproduced from the Solutions to Assignment 1, §8 of
these notes.)
A£qB
B
q£
£
£
£
£
£
£
£
£
£
£ B
B
B
B
qD
B
B
B
B
B
B
Bq
C
Let 4ABC be given, and let D a point of side BC such that AD is perpendicular
to BC. By definition of perpendicular
∠ADB = ∠ADC .
(101)
We apply Proposition I.26 to triangles ABD and ADC, where we have (101) and
|AD| = |AD| common side of triangles
∠BAD = ∠CAD| AD bisects ∠BAC
Hence 4ABD ≡ 4ADC, so |AB| = |AC|, and 4ABC is isosceles.
2. Preceding the list of strip patterns numbered (a), (b), (c), (d), a table has been
reproduced.
• Associate with each of the listed strip patterns the pattern in the table which
has the same isometry group, or argue that the isometry group is different
from all those listed.
• If the pattern has reflective symmetry in a vertical mirror, describe all types of
vertical mirrors that are possible. For example, for the pattern ...VVVVV...
one would observe that there are two possible types of mirrors:
(a) between two successive letters V, as ...VV|VVV...
(b) along the axis of symmetry of any one letter V, as ...VVVVV...
|
Information for Students in Math 348 2003 09
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• If the pattern has rotational symmetry under a half-turn, describe the location(s) of the centre(s) of the half-turn(s).
• If the pattern has symmetry under translations, describe an interval of minimum length that is mapped by the translations so that it covers the strip
without overlaps; e.g. for the pattern ...UVUVUVUVUV... one such interval would be UV.
• If the pattern has symmetry under a glide reflection, describe the axis of
the translation and an interval of minimum length that is mapped by all the
symmetries to cover the strip without overlapping; e.g. in LΓLΓL one such
interval would be L.
I
II
III
IV
V
VI
VII
Typical pattern
...LLLLL...
...LΓLΓL...
...VVVVV...
...NNNNN...
...VΛVΛV...
...DDDDD...
...HHHHH...
Legend:
Ref(H)
0
0
0
0
0
1
1
Symbol
Ref(H)
Ref(V)
Trans
HT
GL
Ref(V)
0
0
∞
0
∞
0
∞
Trans HT GL
∞
0
0
∞
0
∞
∞
0
0
∞
∞
0
∞
∞ ∞
∞
0
∞
∞
∞ ∞
Abstract Group
C∞
C∞
D∞
D∞
D∞
C∞ × D1
D∞ × D1
Meaning
Reflection in Horizontal mirror
Reflections in Vertical mirrors
Translations
Half-Turns
Glide Reflections (with nontrivial translation)
(a) [3 MARKS] . . . YAMAYAMAYAMAYAMAYAMAY . . .
(b) [3 MARKS] . . . EOXEOCEOXEOCEOXEOCEOXEOC. . .
(c) [3 MARKS] . . . SHSISHSISHSISHSISHSIS . . .
(d) [3 MARKS] . . . ⊕±H⊕∓H⊕±H⊕∓H⊕±H⊕∓H⊕±H⊕ . . .
Solution:
Frieze
. . . YAMAYAMAYAMAYAMAYAMAY . . .
. . . EOXEOCEOXEOCEOXEOCEOXEOC. . .
. . . SHSISHSISHSISHSISHSIS . . .
. . . ⊕±H⊕∓H⊕±H⊕∓H⊕±H⊕∓H⊕±H⊕ . . .
Axis of
Ref(H)
no
yes
no
no
Axes of
Ref(V)
Y, M
no
none
none
Fund.Interval
(Translation)
YAMA
EOXEOC
SHSI
⊕±H⊕∓H
Centres of
Half-Turns
none
none
I, H
none
Axis
(Glide-Reflection)
no
EOXEOC
no
⊕±H
Isometry
Group
III
VI
IV
II
3. Consider a regular hexagon in the Euclidean plane, where the plane has been
endowed with a rectangular Cartesian coordinate system, so that the hexagon is
inscribed in a circle of radius 1 centred at the origin, and its vertices are located at
, π, 4π
, 5π
. Let the vertices be labelled cyclically,
points making angles of 0, π3 , 2π
3
3
3
counterclockwise around the hexagon as 1, 2, 3, 4, 5, 6, with vertex 1 being the
Information for Students in Math 348 2003 09
162
point(1, 0); then join each of the following pairs of vertices by a line segment: 1
and 3, 3 and 5, 5 and 1.
(a) [3 MARKS] Express each of the isometries as a product of a non-negative
number of reflections.
(b) [3 MARKS] Set up and complete a rectangular “Cayley Table” for your isometries, so that the product of any two of them, in each order, can be read off
from the table.
(c) [3 MARKS] Determine, for each isometry, its inverse; you may refer to your
table in the preceding part.
(d) [3 MARKS] Show that there are, in your set of isometries one reflection and
one rotation which generate the group of isometries.
Solution: This is the same as the corresponding problem on Version 2.
4. Let R1 , R2 and R3 be reflections, respectively in the line x−y = 1, the line y = −1,
and the line x = −y − 1.
(a) [8 MARKS] Describe a sequence of procedures in each of which a product
of reflections is replaced by an equivalent product of reflections, in such a
way that R1 R2 R3 is expressed as a product of a reflection R0 followed by a
translation in a direction parallel to the mirror of R0 . Show all your work.
Describe all the reflections that enter into your computations carefully, by
specifying their mirror in each case.
(b) [4 MARKS] Determine the distance through which the translation shifts the
plane, and verify that your product transforms the point (0, 2) in the same
way as R1 R2 R3 .
Solution:
(a) The following is only one way of achieving the desired decomposition; there
are others, some much simpler in this special case.
i. Let us turn the mirrors of R2 and R3 through an angle of − π4 about their
point of intersection, (0, −1), in order that the mirror of the new middle
factor is perpendicular to the mirror of R1 . The product is then replaced
by R1 R20 R30 , where the mirrors of R20 and R30 are, respectively, the line
y = −x − 1 and the y-axis.
ii. Now consider the product of R1 and R20 , whose perpendicular mirrors
again intersect in (0, −1). We could turn the mirrors so that the new
middle mirror is parallel to the mirror of R30 , i.e. to the y-axis; this requires
Information for Students in Math 348 2003 09
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a turn through − π4 radians, and results in reflections R200 and R10 , whose
mirrors are, respectively, the y-axis and the line y = −1.
iii. In the product R10 R200 R30 the last two factors are identical, and cancel one
another, since they are reflections. Another way to view this is that the
product is a translation through distance 0. (Of course, the 0 translation
may be viewed as having all directions!) The product is equal to R10 alone
— the reflection in the line y = −1 (followed by a 0 translation in the
direction of that axis, indeed, in any direction).
(b) Under the transformed product, the point (0, 2) moves to (0, −4). Under the
original product, (0, 2) would move first to (3, −1), then to (0, −4), then to
(0, −4), as before.
Information for Students in Math 348 2003 09
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164
Fourth Problem Assignment
Distribution Date: Mounted on the Web on Sunday, November 9th, 2003
Distributed in hard copy on Wednesday, November 12th, 2003
Solutions are to be handed in on Friday, November 21st, 2003
1. For each of the following statements, indicate whether it is true or false.
• If you indicate that a statement is false, you must prove that it is false — if
possible by describing an example or a theorem in precise detail.
• If you indicate that a statement is true, you are not expected to provide a
proof.
• You should not expect part marks when a statement that you indicate to be
true is false; nor when a statement that you indicate to be false is true.
[THESE PROBLEMS APPEARED ON THE FINAL EXAMINATION OF THIS
COURSE, APRIL, 2003, AND ON THE SUPPLEMENTAL/DEFERRED EXAMINATION, AUGUST, 2003]
(a) [3 MARKS] For every non-identity element g of every group (G, ∗) there exists
a positive integer n such that the composition of n copies of g, g ∗ g ∗ . . . ∗ g ,
|
{z
}
n copies
is equal to the identity element.
(b) [3 MARKS] There exist sets S of points in the Euclidean plane such that there
are no isometries that preserve S.
(c) [3 MARKS] Let A, B, C be three distinct points in the Euclidean plane, not
lying in the same line. Then the medians of 4ABC all pass through the same
point.
(d) [3 MARKS] There cannot exist 13 distinct points in the Euclidean plane with
the property that any 4 of them lie on a straight line.
(e) [3 MARKS] The product of two reflections of the Euclidean plane is always
either a translation or a half-turn.
(f) [3 MARKS] There exists in every group (G, ∗) containing at least 2 elements
a non-identity element g and a positive integer n such that the composition
of n copies of g, g ∗ g ∗ . . . ∗ g , is equal to the identity element.
{z
}
|
n copies
(g) [3 MARKS] For some sets S of points in the Euclidean plane every isometry
that preserves S is either a reflection or a half-turn.
Information for Students in Math 348 2003 09
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(h) [3 MARKS] If T1 , T2 , T3 are three translations in the Euclidean plane, and
if there is a point P such that T1 T2 T3 does not move P , then T1 T2 T3 is the
identity mapping.
(i) [3 MARKS] If there exist 13 points in the Euclidean plane, not all in the
same straight line, which have the property that every line containing 2 of
them contains at least 3 of them, then every isometry of the plane is either a
rotation or a translation.
(j) [3 MARKS] The product of two glide reflections of the Euclidean plane is
always either a translation or a rotation.
Information for Students in Math 348 2003 09
18
166
Ordered Geometry
Distribution Date: Mounted on the Web on Tuesday, November 11th, 2003
This section of the course is intended to give an example of a geometry developed as
a strict axiomatic system. As an illustration, we will also discuss a purely “ordered”,
non-metric solution to Sylvester’s Problem — i.e., a solution that does not employ the
concept of distance, unlike L. M. Kelly’s proof seen earlier [4, §4.7, pp. 65-66].
The significance of “Ordered” Geometry in the development of “Absolute” Geometry
and “Affine” Geometry is discussed in [4, §12.1, pp. 175-177]; it will provide a base on
which one can construct both “affine” and “absolute” geometry. In this course we will
not be studying those topics, beyond what has already been done in connection with
vector geometry. We will, however, consider projective geometry, which can be derived
from affine geometry.
Students are cautioned that [4, p. 176] “It is important to remember that, in this
kind of work, we must define all the concepts used (except the primite concepts), and
prove all the statements (except the axioms), however ‘obvious’ they may seem.”
18.1
The structure of our geometry
When geometries are modelled, we usually have, among the objects we consider, a class of
basic elements that we call points, with the intention that they should behave the way we
intuitively expect points to behave. Sometimes we also postulate the existence of other
classes of objects; for example, we could in this case consider a class of objects called lines,
since the geometry we are trying to construct is intended to include the “non-metric”
properties of Euclidean geometry seen in the first propositions of [9, Volume 1, Book I],
which was described in terms of points, lines, and angles. Here, instead of introducing
this second class of objects, we will build our entire theory with points as our only
primitive kind of object; we will then define lines in terms of these points. The structure
of our geometry will be described by a property of “intermediacy” or “betweenness”.
This will be what we call a ternary relation on P, because it is expressed in terms of
three points91 ; the ternary relation that we denote by Q, said to be “on P”, will be
simply a subset of the set of ordered triples of P. We usually denote P × P × P by P 3 ;
91
This type of construction differs from the kind of relation students have seen hitherto in this course,
and in most other courses. Typically mathematicians try to build structures using binary relations and
operations, where we describe properties in terms of two objects, as ternary and n-ary constructions
are often very difficult to visualize. For example, our definition of a group involved a binary operation;
so also most of our connectives in the sentential calculus. But, in the present construction, it would
probably be less intuitive if we tried to use binary relations and operations — although it would certainly
be possible to do it if we wished.
Information for Students in Math 348 2003 09
167
thus Q ∈ P × P × P. The idea we wish to represent by (A, C, B) ∈ Q is that C lies on
the line segment joining A and B. There are two ideas here:
• The three points are collinear , i.e. they are “incident with” in one line; and
• C lies between A and B on that line.
The rest of the terminology, except that of elementary set theory, remains to be defined.
Thus far words like “line”, “between” should be read as being only suggestive, but they
will be made more precise. The following axioms will impose on Q the properties we
want. The axioms we present are those of Coxeter [4, Chapter 12], based on the axioms
of M. Pasch. The axioms will be introduced progressively, as we need them to prove
theorems. Remember that, even though we may make rough sketches to help visualize
proofs, the sketches have no formal place in our proofs: the correctness of a proof depends
upon the validity of the logical steps that constitute it.
The notation in these notes is slightly different from that in the textbook. What
we write as (A, B, C) ∈ Q, Coxeter writes simply as [A, B, C]; we have adopted the
notation involving ∈, in order to emphasize that (A, B, C) ∈ P is actually a statement
— a sentence — a proposition — because it contains a verb, ∈.
18.1.1
Excluding the “trivial” geometries with 0 or 1 point
Mathematicians often include an axiom of this type in a logical system, where the “trivial” geometries eliminated might violate some of the theorems that can be proved as
consequences of the other axioms. An exception never “proves a rule” in mathematics
— it destroys a rule; where a general result might be destroyed by a trivial exception,
one way of coping is to mention the exception when the rule is stated; another strategy
is to redefine the set of objects to which the rule applies, thereby legislating what would
have been an exception out of existence.92
Axiom 18.1 [4, 12.21] |P| ≥ 2.
Equivalently, we could formulate the axiom as
(∃A)(∃B)((A ∈ P) ∧ (B ∈ P) ∧ (A 6= B)) .
(102)
which avoids the need for the reader to be able to count. If it is certain that the reader
will know that the “universe” of the existential quantifiers is P, we could write simply
(∃A)(∃B)(A 6= B) ;
92
(103)
cf. [16] “EXCEPTION, n. A thing which takes the liberty to differ from other things of its class, as
an honest man, a truthful woman, etc. ‘The exception proves the rule’ is an expression constantly upon
the lips of the ignorant, who parrot it from one another with never a thought of its absurdity. In the
Latin, Exceptio probat regulam (in casibus non exceptis) means that the exception tests the rule, puts
it to the proof, not confirms it...”
Information for Students in Math 348 2003 09
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or we could show the universes when we write the quantifiers, as
(∃A ∈ P)(∃B ∈ P)(A 6= B) .
(104)
To minimize the complications of our statements, we shall understand for the rest of
§18 of these notes that — unless specifically stated otherwise — the universe of the
quantifiers ∀ and ∃ will be P.
18.1.2
Axioms concerning points on a line
The next axiom asserts the “unboundedness” of a line:
Axiom 18.2 [4, 12.22] (∀A)(∀B)((A 6= B) ⇒ (∃C)((A, B, C) ∈ Q))
Note the quantifiers (∀A)(∀B); in practice mathematicians are often casual about quantifiers, and might write only
(A 6= B) ⇒ (∃C)((A, B, C) ∈ Q) ;
(105)
but that would be technically incorrect, as it would assert the existence of C only for
one specific pair of points A and B. If we want the property to hold for all pairs (A, B),
we need to quantify the statement universally. Observe that we have not demanded in
our axiom that there exists a point C as we exit the segment AB in the other direction:
we will be able to prove this property in Exercise 18.1 below, but only after we have an
axiom of reversibility available (Axiom 18.4).
In order to be precise about the concept of “betweenness”, we need to ensure that
the points we are talking about are distinct. We do this with the following axiom, which
is concerned with one of the possible coincidences between points in an ordered triple;
the other possible coincidences will be consequences of this axiom and its successors.
Axiom 18.3 [4, 12.23] ∀A∀B∀C(((A, B, C) ∈ Q) ⇒ (A 6= C))
Given 3 distinct points, A, B, C, there are 3! = 6 possible ordered triples that can be
formed from them. The first of the following two axioms, Axiom 18.4, ensures that a
triple can be “reversed” — justifying our use of the English word “between”; when we
say “between A and C”, we normally mean the same as “between C and A”; that is,
the reader could expect, from our use of the word, that the definition is symmetric; the
axiom asserts just that. The second axiom, Axiom 18.5, will be seen, with the first,
to imply that no more than 2 of those 6 ordered triples that can be formed from three
distinct points will be in Q93 ; note that we say with the first, since we will need both of
these axioms, and their predecessors, to prove that result.
93
Why did we say no more than 2? If 3 points are not in the same line, then none of the six
permutations will yield an ordered triple in Q.
Information for Students in Math 348 2003 09
169
Axiom 18.4 [4, 12.24] ∀A∀B∀C(((A, B, C) ∈ Q) ⇒ ((C, B, A) ∈ Q)) .
/ Q)) .
Axiom 18.5 [4, 12.24] ∀A∀B∀C(((A, B, C) ∈ Q) ⇒ ((B, C, A) ∈
Of course, (B, C, A) ∈
/ Q is an abbreviation for ¬((B, C, A) ∈ Q).
Exercise 18.1 Prove that (∀A)(∀B)((A 6= B) ⇒ (∃C)((C, A, B) ∈ Q) .
Solution:
A 6= B ⇒ B 6= A
⇒ ∃C((B, A, C) ∈ Q)) by Axiom 18.2
⇒ (C, A, B) ∈ Q by Axiom 18.4
Exercise 18.2 Prove that if (A, B, C) ∈ Q, then none of the triples (A, C, B), (B, A, C),
(C, A, B), (B, C, A) is in Q.
Solution: In each of the following 4 cases, (A, B, C) ∈ Q is a hypothesis.
1. (A, C, B) ∈
/ Q: By Axiom 18.4 (A, C, B) ∈ Q ⇒ (B, C, A) ∈ Q, contradicting94
Axiom 18.5; we conclude that (A, C, B) ∈
/ Q. (This is a proof by contradiction 95 .)
2. (B, A, C) ∈
/ Q: We prove this also by contradiction.
(B, A, C) ∈ Q ⇒ (C, A, B) ∈ Q by Axiom 18.4
⇒ ¬((A, B, C) ∈ Q) by Axiom 18.5
which contradicts the hypothesis. We conclude that (B, A, C) ∈
/ Q.
3. (C, A, B) ∈
/ Q: By Axiom 18.5, (C, A, B) ∈ Q ⇒ (A, B, C) ∈
/ Q. From this
contradiction we infer that (C, A, B) ∈
/ Q.
4. (B, C, A) ∈
/ Q: (left to the reader)
We assumed in Axiom 18.3 that the first and third entries of any ordered triple in Q
would be distinct; we are now in a position to prove that all three entries are distinct.
Exercise 18.3 Prove that
94
To say that S contradicts U is defined to mean that S and U are not simultaneously true, equivalently
that S ∧ U is a contradiction — always having truth value F . If S ∧ U ⇒ F , then S ∧ U → F is a
tautology, so S ∧ U cannot be true — i.e. must be false, which entails that at least one of S and U is
false. If U is an axiom, it is being assumed to be true; here S must be false.
95
or reductio ad absurdum.
Information for Students in Math 348 2003 09
170
1. ∀A∀B∀C((A, B, C) ∈ Q ⇒ (A 6= B).
2. ∀A∀B∀C((A, B, C) ∈ Q ⇒ (C 6= B).
Solution:
1. Suppose that (A, A, C) ∈ Q. Then, by Axiom 18.4, (C, A, A) ∈ Q; applying Axiom
18.5 to the point (C, A, A) ∈ Q yields (A, A, C) ∈
/ Q, contradicting our hypothesis.
From that contradiction we infer that (A, A, C) ∈
/ Q, for any points A, C ∈ P.
2. Apply the preceding argument to (C, B, A) ∈ Q.
Definition 18.1 For any distinct points A, B ∈ P, the segment AB is defined to be the
set {C|(A, C, B) ∈ Q}. When C ∈ AB we will say that C is on or in the segment AB.
Exercise 18.4 Let A and B be distinct points. Prove that
1. A ∈
/ AB and B ∈
/ AB.
2. AB = BA.
Solution:
1. A ∈ AB means (A, A, B) ∈ Q; B ∈ AB means (A, B, B) ∈ Q. Neither of these
ordered triples can belong to Q, by Exercise 18.3.
2. C ∈ AB ⇔ (A, C, B) ∈ Q ⇔ (B, C, A) ∈ Q ⇔ C ∈ BA.
We can now introduce more familiar terminology.
Definition 18.2 Let A, B be distinct members of P.
1. AB denotes the set AB ∪ {A, B}, and is called an interval or a closed interval .
2. In contradistinction to the preceding, AB could be called an open inverval .
3. A/B denotes the set {C|(C, A, B) ∈ Q}, and is called the ray from A, away from
B.
4. The line determined by A, B, denoted by L(A, B), is the union A/B ∪ B/A ∪ AB ∪
{A, B}.
5. If C ∈ L(A, B), we may say that the point C lies on the line L(A, B), or that the
line L(A, B) passes through the point C; or that the point C is incident with the
line L(A, B) or the line L(A, B) is incident with the point C.
Information for Students in Math 348 2003 09
171
Exercise 18.5 Let A and B be distinct points.
1. L(A, B) = L(B, A).
2. C ∈ L(A, B) ⇔ (C = A) ∨ (C = B) ∨ ((C, A, B) ∈ Q) ∨ ((A, C, B) ∈ Q) ∨
((A, B, C) ∈ Q)
Solution:
1.
A/B ∪ B/A ∪ AB ∪ {A, B}
= B/A ∪ A/B ∪ AB ∪ {A, B} commutativity of ∪
= B/A ∪ A/B ∪ BA ∪ {A, B} Exercise 18.4.2
= B/A ∪ A/B ∪ BA ∪ {B, A}
(Of course, we are also applying the associativity of the set theory operator ∪
repeatedly, just as earlier we repeatedly applied — tacitly — the associativity of
∨ as a logical connective.
2. This is simply a paraphrasing of the definition of L(A, B) as the union A/B ∪
B/A ∪ AB ∪ {A, B} = A/B ∪ B/A ∪ AB ∪ {A} ∪ {B}.
The next axiom will ensure that there is just one line through any two distinct points.
Axiom 18.6 [4, 12.25] If C, D are distinct points in L(A, B), then A ∈ L(C, D).
The following theorem will show that the axiom has the desired effect.
Theorem 18.1 [4, Theorem 12.251] If C, D ∈ L(A, B) and C 6= D, then L(C, D) =
L(A, B).
Proof: (cf. [4, Theorem 12.251]) Assume that
C, D ∈ L(A, B)
C 6= D
(106)
(107)
1. Suppose that
|{A, B, C, D}| = 2 .
(108)
In using the notation L(A, B) we are implying that A 6= B; by (108),
2 = |{A, B}| = |{A, B, C, D}| ;
(109)
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by (107), (109) can be extended to
2 = |{A, B}| = |{A, B, C, D}| ≥ |{C, D}| = 2 .
(110)
which implies that the inequality is an equality, that |{A, B}| = |{C, D}|, and
hence that {A, B} = {C, D} = 2. Hence, either (A = C) ∧ (B = D), in which case
L(C, D) = L(A, B), or (A = D) ∧ (B = C), in which case L(C, D) = L(B, A),
which is equal to L(A, B), by virtue of Exercise 18.5.1.
2. Suppose that |{A, B, C, D}| = 3; without limiting generality, we assume that D =
B, and that A, B, C are distinct. By Axiom 18.6, A lies on L(B, C); let X be any
other point on L(B, C). By Axiom 18.6, B lies on L(A, X); by the same Lemma,
X lies on L(A, B), since B 6= A. Thus we have shown that L(B, C) ⊆ L(A, B).
Interchanging the roles of A and C in the preceding argument, we can show that
L(A, B) ⊆ L(B, C), so L(A, B) = L(B, C).
3. Suppose that |{A, B, C, D}| = 4. We first apply the argument of part 2 to the
points A, B, C; then the same argument is applied to B, C, D, to conclude that
L(B, C) = L(C, D). It follows that l(A, B) = L(B, C) = L(C, D).
Having proved that two distinct points lie on a unique line, we can now investigate the
intersections of distinct lines. But the reader should keep in mind that we have not
proved that there exist two distinct lines.96
Exercise 18.6 Suppose {E, F } ⊆ L(A, B) ∩ L(C, D), and that L(A, B) 6= L(C, D).
Prove that E = F .
Proof: Suppose that E 6= F . Then, by Theorem 18.1, L(A, B) = L(E, F ), and L(C, D) =
L(E, F ), contradicting the hypothesis that L(A, B) 6= L(C, D). We conclude that E =
F , i.e., that two lines can have no more than one point in common.
Definition 18.3
section.
97
If two lines have a point in common, that point is called their inter-
Our design was to have the ordered triples in Q express the “betweenness” properties
of points on a line. We can now prove that our axioms achieve that:
Theorem 18.2 If A, B, C are distinct points on a line, then exactly one of the following
points is in Q: (A, B, C), (B, C, A), (C, A, B).
96
Indeed, our present system of axioms are satisfied by the one line geometry consisting of the points
on a line — e.g. the real axis in R2 .
97
This definition isn’t really needed, since we defined each line to be a set of points with certain
properties; the set-theoretic intersection of these sets is precisely the point we have called the intersection
of the lines.
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Proof: By Exercise 18.2 not more than one of the points (A, B, C), (B, C, A), (C, A, B)
is in Q.
By Theorem 18.1 the line in question is L(A, B). By definition, the points distinct
from A and B on this line lie in the union A/B ∪ B/A ∪ AB. A point C in subsets B/A,
AB, A/B will respectively have the property that (C, A, B) ∈ Q, (B, C, A) ∈ Q, and
(A, B, C) ∈ Q.
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174
Ordered Geometry (concluded)
Distribution Date: Wednesday, November 19th, 2003
19.0.3
We require that not all the points lie on the same line.
As a consequence of Axiom 18.2 there will always be infinitely many points on a line.
However, the geometry constructed with the axioms given thus far is still rather “uninteresting”. To demonstrate that, consider the following Exercise, which was discussed in
the lecture of November 1st, 2002).
Exercise 19.1 Suppose one takes P to be the integers — positive, negative, and 0, and
defines Q (i.e. P × P × P) to consist of the set
{(A, B, C)|(A < B < C) ∨ (C < B < A)} .
Prove that all the axioms stated thus far are satisfied; that is, that this describes a model
for the axiomatic system we have described. Proof: left to the student
In order to exclude the possibility that our entire geometry reside on one line, we shall
postulate an additional axiom:
Axiom 19.1 [4, Axiom 12.26] ∀A∀B∃C(C ∈
/ L(A, B).
This will permit us to talk about “triangles”.
Theorem 19.1 [4, Theorem 12.261] Either none or all of the following three statements
are true: A ∈ L(B, C), B ∈ L(C, A), C ∈ L(A, B).
Proof: If any one of the statements is true, then, as seen in the proof of Theorem 18.2.
L(A, B) = L(B, C) = L(C, A), and all three of the statements will be true; otherwise
they are all false.
Definition 19.1
1. A subset S ∈ P is said to be collinear if there exists a line
containing all members of S; otherwise the set is said to be non-collinear .
2. A triangle is a non-collinear set of 3 points; we may also say that this is “a set of 3
non-collinear points”. The lines determined by the pairs of these points are called
the sides of the triangle, and the points themselves are the vertices of the triangle.
If A, B, C are the vertices of a triangle, we may speak of the triangle ABC, or of
∆ABC.
We have postulated that our geometry does not consist only of the points on a line.
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19.0.4
175
“Pasch’s” Axiom; Points are “dense” in any line
While axiom 19.1 now forces our geometry to consist of more than points on just one
line, we still do not have all of the properties we would like a plane geometry that is
emulating the Euclidean plane to possess. We have made no assumptions about how
lines need to intersect, except for Axiom 18.6 and its consequences, which ensure that
distinct lines can’t intersect more than once. To see this, consider the following “model”:
Example 19.2 Define P to consist of all ordered pairs of integers (i.e. members of the
set we denote by Z); one can think of the points in R2 having integer coordinates. To
describe this model precisely requires concepts from linear algebra that may not be in
the background of all students in the course; so our description will only be intuitive.
Define Q = P × P × P to consist of all ordered triples
((a1 , b1 ), (a2 , b2 ), (ac , b3 ))
where ai , bi ∈ Z, (i = 1, 2, 3) with the property that there exist integers `, m, n — not all
three of them zero — such that `ai + nbi = n (i = 1, 2, 3). Then all the axioms stated to
date are satisfied.
Here again is a situation that does not conform to our expectations for an axiomatic
geometry that will have the non-metric properties of Euclidean geometry. The next
axiom, that we can associate specifically with Pasch, will enable use to conclude that
the points of a line form a “dense” set, in the sense that there will always be a point
between any two others.
Axiom 19.2 [4, Axiom 12:27] If ABC is a triangle, and (B, C, D) ∈ Q, (C, E, A) ∈ Q,
there exists a point F ∈ L(D, E) such that (A, F, B) ∈ Q.
Theorem 19.3 For any two distinct points A, B, there exists C ∈ P such that (A, C, B) ∈
Q.
Without requiring any further postulates, the structure we have defined is now capable of producing a non-metric proof of Sylvester’s conjecture; the proof will not be
discussed in the course, but may be read in [4, §12.3, pp. 181-182].
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176
Projective Geometries
Distribution Date: Wednesday, November 19th, 2003
20.1
The Real Projective Plane
One technical problem in working in the Euclidean plane is that the possibility that lines
can be parallel sometimes produces variations in the form of theorems, depending upon
whether certain lines are parallel or not.98 It would be so much simpler if one did not have
to treat special cases. Projective geometry permits this by eliminating the possibility of
non-intersecting lines in the plane. But how can this be done? We construct, from the
familiar real plane, an “extended” plane, that has in it “more” points and “more” lines;
when one looks only at the “old” points and lines, one has nothing more or less than
the “old” Euclidean plane; but the “new” objects permit us to transform the facts that
previously were stated in terms of parallelism into terms of intersection of lines in the
extended plane.
The “Real Projective Plane” will consist of two sets — a set that we shall denote
by P, whose members will be called points, and a second set, denoted by L, whose
members will be called lines; the properties of the geometry can be described in terms of
a relation of incidence, each of whose statements involves a single point and a single line:
we could write these statements in some form like Inc(P, `): which would mean “point P
is incident with line `, or P is on `, or ` passes through P . Because the term “incident”
appears to suggest a symmetrical relationship between certain points and certain lines,
it is convenient to be able to say “` is incident with P ” whenever P is incident with `.
We achieve this by a minor change in our definition, defining a predicate I by

(x ∈ P) ∧ (y ∈ L) ∧ (x lies on y)




 i.e. (x ∈ P) ∧ (y ∈ L) ∧ (y passes through x)
or
.
I(x, y) ⇔


(y
∈
P)
∧
(x
∈
L)
∧
(y
lies
on
x)



i.e. (y ∈ P) ∧ (x ∈ L) ∧ (x passes through y)
If we apply this terminology to the Euclidean plane without change we find that there
exist distinct pairs of lines with the property that there exists no point which is incident
with both of them: for example, the lines whose equations in a cartesian coordinate
system are x = 0 and x = 1 are parallel , and there is no point of the plane that is
incident with both of them; indeed, the set of lines
{x = r|r ∈ R}
98
Examples of theorems of this type are the theorems now known as Desargues’ Theorem [4, §14.3] and
Pappus’ Theorem [4, Axiom 14.15, p. 231] which will be discussed in the lectures. It will be explained
why the first appearance of the second of these is in a proposed axiom.
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are such that there exists for no two of them a point lying on both. We call the collection
of all these lines one point at infinity, in this case “the point at infinity corresponding to
the parallel class of lines with infinite slope”. Intuitively we want to think of the lines
as somehow coming together into a single point that is “infinitely far away”. Of course
there is no such point in the Euclidean plane. When we define the meaning of incidence
for this new kind of “point”, we will be extending the meaning of the term to a broader
set of “points” and “lines”; but our definition will not affect the properties that already
are possessed by points and lines in the Euclidean plane. As mentioned, we define each
of the maximal families of parallel classes of lines that have the same slope (including
the family of vertical lines, which have “infinite” slope) as constituting one of our points
at infinity. And we define one new line — the “line at infinity”, which is incident only
with “points at infinity”. Any normal line is defined to be incident with the “point at
infinity” (i.e. the parallel class) of which it is a member.
This new system, with “points at infinity” and a “line at infinity”, has the property
that every two distinct lines meet in exactly one point — i.e. that there exists, for every
two distinct lines, precisely one point that is incident with both of them. When one
of the lines is the “line at infinity”, and the other is a “normal” line, they meet in the
“point at infinity” defined by the slope of the given normal line.
The new system also has the property that there is exactly one line incident with
any two distinct points. Where the points are “normal” points, the line is the one we
had before in the Euclidean plane; where one of the points is “at infinity”, and the
other is not, the line is that having the slope corresponding to the point at infinity, and
passing through the given “old” point; and, when the points are both “at infinity”, the
line through them is the “line at infinity”. Without altering the geometric properties of
“normal” points and lines in the Euclidean plane, the new objects permit us to describe
the properties of the plane without having to address the problems associated with
parallel lines.
20.2
“Homogeneous” coordinates; models of the Real Projective Plane
Let’s begin with the Euclidean plane, endowed with a cartesian coordinate system, where
each point has a unique ordered pair of coordinates (x, y). The new coordinates we
associate with points will be ordered triples (X, Y, Z) of real numbers, which will be
related to the old coordinates by the equations
x=
X
,
Z
y=
Y
.
Z
The new, “homogeneous”, coordinates are not unique; for example, the point (1, 7) has
homogeneous coordinates (α, 7α, α), where α is any non-zero real number; the origin has
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homogeneous coordinates (0, 0, α), where α is any non-zero real number.99 We can also
associate ordered triples of coordinates with lines: the line ax + by + c = 0 can be given
coordinates [a, b, c]; the point (X, Y, Z) can be seen to lie on the line [a, b, c] if and only
if
X
Y
a·
+ b · + c = 0,
Z
Z
i.e. iff
aX + bY + cZ = 0
In the cases of both points and lines the new coordinates are not unique; in both cases it
cannot happen that all three of the coordinates are 0. But, thus far, our points always
have the Z coordinate non-zero; and always have at least one of the first two coordinates
of a line non-zero. We now proceed to generalize these restrictions, while maintaining the
restriction that not all 3 of the coordinates of either a point or a line may be 0. The “point
at infinity” which corresponds to the lines whose equations are ax + by + constant = 0
will be (b, −a, 0), or, more generally, (bα, −aα, 0), where α is any non-zero real number.
The “line at infinity” has coordinates [0, 0, 1]. Thus all points at infinity have their third
coordinate equal to 0, which ensures that they lie on the line [0, 0, 1]; indeed, it ensures
that they are the only points on this line.
These ordered triples of coordinates for points and lines suggest another model for
the Real Projective Plane:
• Associate the ordered triples (X, Y, Z) with lines through the origin in 3-space,
consisting of all points in 3-space whose coordinates are of the form (x, y, z) =
(αX, αY, αZ), where α ranges over the real numbers. In particular, associate
(X, Y, Z) with the two points where the line in question meets the unit sphere
centred at the origin. Thus every point in the Real Projective Plane is associated
with two points in real 3-space, points that are at opposite ends of a diameter of a
unit sphere.
• Now associate the line [a, b, c] with the plane through the origin in 3-space, whose
equation is ax + by + cz = 0, and then with the circle where that plane intersects
the unit sphere centred at the origin.
• When a point (X, Y, Z) lies on a line [a, b, c], the corresponding pair of antipodal
points on the sphere will lie on the great circle of intersection of the plane with the
unit sphere. But remember — the pair of antipodal points correspond to the same
point of the real projective plane.
99
The specific symbols used to denote homogeneous coordinates vary from one author to another;
we have chosen to use CAPITAL letters here to distinguish from the corresponding lower case letters,
which we are using for inhomogeneous coordinates.
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Another way to look at this model is to take only the upper hemisphere of the sphere,
and to require that “identifications” take place between opposite points of the bounding
circle of this hemisphere.
When mathematicians are interested in the “topological” properties of the projective
plane — those properties not affected by stretching or compression — we sometimes
flatten this hemisphere into a disk , where, as before, the pairs of antipodal points on the
diameter are to be identified.
20.3
The Euclidean plane as a subgeometry of the Real Projective Plane
Let’s make precise what we have proposed earlier. The Real Projective Plane will consist
of 2 sets: P is the set of points, which will be represented as ordered triples of real
numbers (X, Y, Z), with the only restriction that not all of the three coordinates can be
zero; we will treat two ordered triples (X1 , Y1 , Z1 ), (X2 , Y2 , Y2 ) as representing the same
point if the coordinates are proportional, i.e. if there exists a non-zero constant k such
that
X2 = kX1
Y2 = kY1
Z2 = kZ1
L is the set of lines, which will be represented as ordered triples of real numbers [a, b, c]
with again the only proviso that not all three coordinates can be zero; here also we
will treat two ordered triples (a1 , b1 , c1 ), (a2 , b2 , c2 ) as representing the same point if the
coordinates are proportional, i.e. if there exists a non-zero constant k such that
a2 = ka1
b2 = kb1
c2 = kc1 .
We will say that
“point (X, Y, Z) is incident with line [a, b, c]”
or “point (X, Y, Z) lies on line [a, b, c]”
or “line [a, b, c] is incident with point (X, Y, Z)”
or “line [a, b, c] passes through point (X, Y, Z) .”
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180
If, in the preceding set of “points” (X, Y, Z), we require that Z 6= 0, then we are free to
define
X
Z
Y
y =
Z
x =
(111)
(112)
and to think of the points as ordered pairs (x, y) of real numbers; if, in the preceding set
of “lines” [a, b, c], we require that a and b be not both zero, we are free to define [a, b, c]
to be the line with equation ax + by + c = 0 (or, equivalently, AX + BY + CZ = 0)100 .
In that case we recover the familiar properties of points and lines in the Euclidean plane,
coordinatized with a rectangular Cartesian coordinate system.
20.4
The points and lines which are adjoined to the Euclidean
Plane to create the Real Projective Plane
The “points” (X, Y, Z) with Z = 0 are not in the Euclidean plane: these are the objects
we have adjoined to the Euclidean plane to eliminate the phenomenon of parallel lines;
we call these points at infinity. The “line” [0, 0, c] (with c 6= 0) is an object added in
order to ensure that any two points are incident with a line — even when the two points
are not both in the “old” part of the plane; we call this the line at infinity. As always
in this discussion, we treat two ordered triples as representing the same point or line if
their coordinates are proportional: this [0, 0, 1], [0, 0, −26], and [0, 0, −π] are the same
line.
20.5
Some “incidence” properties of the real projective plane
The enlarged plane inherits properties that were possessed by the Euclidean plane. We
will label these as “axioms” because we plan to investigate the nature of logical systems
we can construct where we will postulate these properties, and assume as little else as
possible. It may be surprising to see how much we can extract from some apparently
very weak assumptions:
Axiom 20.1 (cf. [4, Axiom 14.11, p. 230]) Any two distinct points are incident with
just one line.
Axiom 20.2 (cf. [4, Axiom 14.12, p. 231]) Any two lines are incident with at least one
point.
100
If we had permitted a = 0 = b, we would have to have the third coordinate c non-zero, and we
would have a line 0x + 0y + c = 0, which would be a contradiction.
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181
We will eventually be investigating the subtle differences between these axioms; for
example, why does the first require that the points be distinct, while the second does
not require that the lines be distinct? Why does one axiom refer to exactly while the
second only to at least...?
The reader should note that these properties hold not only for the points and lines of
the “old” system, but for points and lines which we adjoined to create the Real Projective
Plane as well, and for combinations. Thus, two points at infinity are incident with only
one line — the line at infinity; a point at infinity (X1 , Y1 , 0) and a point not at infinity,
(X2 , Y2 , Z2 ) where Z2 6= 0, are both incident with the line
[Y1 Z2 , −X1 Z2 , X1 Y2 − X2 Y1 ]
Exercise 20.1 Prove the preceding statement — that the two points are, indeed, incident with the line.
That this is the only line with which they are incident follows from some simple linear
algebra that we will not consider in this course.
Two parallel lines from the Euclidean plane, whose equations are, say, ax + by = c1
and ax + by = c2 , where c1 6= c2 , i.e. two lines [a, b, −c1 ] and [a, b, −c2 ] with c1 6= c2 ,
will both be incident with the point (b, −a, −0), which is on the line at infinity. Thus
we have “destroyed” parallelism in extending the Euclidean plane! Finally, the line at
infinity, [0, 0, 1] and another line, say ax + by = c or [a, b, −c], are both incident with the
point [b, −a, 0].
Exercise 20.2 Prove the preceding 2 claims about the intersection of lines in the Real
Projective Plane.
20.6
Projective planes over other fields than the reals
20.6.1
What is a “field”?
Since some students do not have the background to study this topic, we will be satisfied
with saying that a “field” is a system in which we have defined addition, subtraction,
multiplication, division, which contains elements 0 and 1, all with the “familiar” properties. Rather than indulge in a precise definition, we will consider some particular cases
that are easily defined, and which we will simply “declare” to be fields. Among the fields
which are contained in the real numbers are:
R
Q
√
Q[ 2]
the real numbers
the rational numbers =√fractions of integers
a 2+b
numbers of the form √
c 2+d
where c and d are not both 0.
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In each of these cases we can consider ordered triples (X, Y, Z) and [a, b, c] defined as
above, not all 3 coordinates 0, and get an interesting projective geometry. We will
not attempt to show in this course how these geometries differ — but students should
understand that, even though these geometries all have infinitely many points and lines,
there are differences between them that cannot be attributed simply to the way in which
we have defined the geometries.
20.6.2
Fields of “residues modulo a prime”
We can also obtain fields by considering the remainders that we obtain when we divide
integers by a fixed positive “prime” integer, i.e. by an integer p ≥ 2 which cannot be
expressed as a product of two other integers which are also greater than 1. For example,
the smallest primes are 2, 3, 5, 7, 11, 13, 17, 19, 23,.... We will denote the field obtained
using an integer p by Fp , or by GF (p). In these fields we carry out all arithmetic “modulo
p”, in the sense that we always divide by p and take the non-negative remainder. Thus,
for example, in F3 , the elements are the possible remainders 0, 1, 2, and the tables
defining the operations of addition and multiplication are
+
0
1
2
20.6.3
0
0
1
2
1
1
2
0
2
2
0
1
×
0
1
2
0
0
0
0
1
0
1
2
2
0
2
1
The projective plane over Fp
We will define the points of this geometry again to be ordered triples (X, Y, Z), but now
the 3 coordinates must come from the field elements, and cannot all be 0. The lines
will be defined in the same way: ordered triples [a, b, c] of field elements, not all three
of them equal to 0.101 In all of these definitions we require that ordered triples that are
proportional represent the same object. So, in the projective plane over F3 , the lines
[1, 2, 1] and [2, 1, 2] are the same, since multiplication of the three coordinates in the
first set by 2 gives 2 × 1 = 2, 2 × 2 = 4 = 1, and 2 × 1 again = 2. In these kinds of
computations we often use a symbol ≡ instead of =, writing 4 ≡ 1, or, more completely,
4 ≡ 1 (mod 3) ,
which is read as “4 is congruent to 1 modulo 3.”
101
But, in either case, as many as 2 of the coordinates can be 0.
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20.7
183
The “Fano” Geometry
This geometry, named in deference to Gino Fano, who discussed it in 1892, is sometimes
denoted by P G(2, 2), where the first 2 represents the fact that this is a “plane” rather
than a higher-dimensional geometry, and the second 2 represents the field that is used, F2 .
In this field there is only one number that is different from 0, that is 1 — corresponding
to the two possible smallest remainders when an integer is divided by 2. This means
that we need not be concerned about triples of coordinates that are proportional, since
the only possible constant of proportionality is 1. Of the 8 = 2 × 2 × 2 possible ordered
triples of field elements, one consists only of 0’s and is not acceptable: we have 7 points
in this geometry; in the same way we see that there are exactly 7 lines.
Exercise 20.3 List the 7 points of the Fano plane, showing, for each of them the 3
lines passing through them; then list the 7 lines, showing the 7 points passing through
them. For example, the point (1, 0, 1) is contained in the following lines: [1, 0, 1], [0, 1, 0],
[1, 1, 1]. (While there are systematic ways of attacking this problem, one can simply test
each of the 7 lines to see whether the coordinates have the desired “dot product” or
“inner product” property.)
21
21.1
Projective Geometries (continued)
Axiomatic Definition of Projective Planes
We saw earlier102 that the Real Projective Plane had the following properties:
Axiom 20.1: (cf. [4, Axiom 14.11, p. 230]) Any two distinct points are incident with
just one line.
Axiom 20.2: (cf. [4, Axiom 14.12, p. 231]) Any two lines are incident with at least one
point.
The reason for describing the second axiom in a wording that is apparently weaker
than the first, is that the two axioms together can be shown [4, Theorem 14.121, p. 231]
to imply a stronger statement:
Theorem 21.1 Any two lines are incident with exactly one point.
Proof: supplied in the lecture of Friday, November 15th, 2002.
Mathematicians routinely formulate axioms “as weak as possible”.
The planes that we have described above, using homogeneous coordinates over fields,
have other properties in common. Among these is the following, which we shall describe
as an axiom, because we eventually wish to explore what properties of planes derive
102
these notes, page 180
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solely from these properties, and not from the properties, for example, of the fields that
we have used in constructing the examples we have seen so far.103
Axiom 21.1 (cf. [4, Axiom 14.13, p. 231]) There exist four points, of which no three
are collinear.
This axiom is akin to Axiom 18.1 for ordered geometries, which was designed in order
to exclude the possibility of certain “degenerate” geometries, i.e. geometries that satisfy
the other axioms, but are otherwise not interesting, and whose presence would require
frequent exceptional cases in other theorems — a situation that Projective Geometry
was, in part, designed to avoid. So, for the remaining study of Projective Geometry
in this course, let us see what we can extract from these axioms alone, keeping the
geometries constructed using homogeneous coordinates over fields as examples in the
background.
21.2
What can we infer from the 3 given axioms only? The
Duality Principle.
We have already shown that the axioms imply a strengthening of one of them. Note
that Axiom 20.1 and Theorem 21.1 have the property that one can be obtained from the
other by interchanging the words point and line. This property, which could be called
the Duality Principle for projective planes can also be proved to extend to Axiom 21.1,
as we can prove the following theorem:
Theorem 21.2 There exist four lines, of which no three are concurrent104 .
Proof: Suppose that the four points whose existence is postulated by Axiom 18.3 are
P1 , P2 , P3 , P4 . They define, by virtue of Axiom 20.1, 6 lines, which we may denote by
`12 , `13 , ..., `34 . These lines are distinct, since, if two of them coincided, it would contain
more than 2 of the points; each of the lines must contain at least one other point: for
example, lines `12 and `34 must meet — by virtue of Theorem 21.1 — in a point that
must be distinct from P1 , P2 , P3 , P4 . We claim that no three of the lines `12 , `13 , `24 , and
`34 can pass through a common point. While there may be a more elegant proof of this
fact, we can prove it laboriously by considering the 4 sets of 3 lines:
1. {`13 , `24 , `34 }: Lines `13 and `34 meet in P3 ; if the 3 lines had a common point, P3
would lie on `24 . making it collinear with P2 and P4 .
103
The textbook also explores the use of other axioms, including one which excludes the possibility of
the Fano geometry; in the limited time remaining in the term we will not impose these other axioms,
nor investigate why the author finds the need for them.
104
concurrent=passing through the same point.
Information for Students in Math 348 2003 09
185
2. {`12 , `24 , `34 }: Lines `12 and `24 meet in P2 , which cannot be collinear with P3 and
P4 .
3. {`12 , `13 , `34 }: Lines `12 and `13 meet in P1 , which cannot be collinear with P3 and
P4 .
4. {`12 , `13 , `24 }: Lines `12 and `13 meet in P1 , which cannot be collinear with P2 and
P2 .
If we do not postulate any more axioms, we can be assured that the Duality Principle
will remain true.
21.3
Some simple properties of finite projective planes
We first observe that it is not possible for a line to contain only two points.
Theorem 21.3 Every line in a projective plane contains at least three points.
Proof:105 Let ` be a given line, and consider the points Pi (i = 1, 2, 3, 4) whose existence
is postulated in Axiom 21.1. It must happen that at least two of those points do not lie
on `; without limiting generality, let’s label the points so that P1 is one such point not
on `. The lines joining P1 to each of P2 , P3 , P4 must be distinct, and each of them meets
` in a point; those points must be distinct, as there is at most one line through P1 and
any point on `. Hence there exist at least 3 points on `.
By virtue of the Duality Principle, we can immediately conclude
Theorem 21.4 Every point in a projective plane is incident with at least three lines.
We list some interesting properties that can be proved with there is a line with a finite
number n of points:
Theorem 21.5 (cf. [17, Theorem 20.1.1, p. 348]) Let n ≥ 2 be an integer. The following
properties are equivalent in a projective plane:
1. There exists a line containing exactly n + 1 points.
2. There exists a point incident with exactly n + 1 lines.
3. Every line contains exactly n + 1 points.
4. Every point is incident with exactly n + 1 lines.
5. P contains exactly n2 + n + 1 points.
6. L contains exactly n2 + n + 1 lines.
105
added November 20th, 2002
Information for Students in Math 348 2003 09
22
186
Solutions, Fourth Problem Assignment
Distribution Date: Thursday, November 27th, 2003
Solutions were to be handed in on Friday, November 21st, 2003
1. For each of the following statements, indicate whether it is true or false.
• If you indicate that a statement is false, you must prove that it is false — if
possible by describing an example or a theorem in precise detail.
• If you indicate that a statement is true, you are not expected to provide a
proof.
• You should not expect part marks when a statement that you indicate to be
true is false; nor when a statement that you indicate to be false is true.
[THESE PROBLEMS APPEARED ON THE FINAL EXAMINATION OF THIS
COURSE, SPRING, 2003, AND ON THE SUPPLEMENTAL/DEFERRED EXAMINATION, AUGUST, 2003]
(a) [3 MARKS] For every non-identity element g of every group (G, ∗) there exists
a positive integer n such that the composition of n copies of g, g ∗ g ∗ . . . ∗ g ,
{z
}
|
n copies
is equal to the identity element.
(b) [3 MARKS] There exist sets S of points in the Euclidean plane such that there
are no isometries that preserve S.
(c) [3 MARKS] Let A, B, C be three distinct points in the Euclidean plane, not
lying in the same line. Then the medians of 4ABC all pass through the same
point.
(d) [3 MARKS] There cannot exist 13 distinct points in the Euclidean plane with
the property that any 4 of them lie on a straight line.
(e) [3 MARKS] The product of two reflections of the Euclidean plane is always
either a translation or a half-turn.
(f) [3 MARKS] There exists in every group (G, ∗) containing at least 2 elements
a non-identity element g and a positive integer n such that the composition
of n copies of g, g ∗ g ∗ . . . ∗ g , is equal to the identity element.
{z
}
|
n copies
(g) [3 MARKS] For some sets S of points in the Euclidean plane every isometry
that preserves S is either a reflection or a half-turn.
Information for Students in Math 348 2003 09
187
(h) [3 MARKS] If T1 , T2 , T3 are three translations in the Euclidean plane, and
if there is a point P such that T1 T2 T3 does not move P , then T1 T2 T3 is the
identity mapping.
(i) [3 MARKS] If there exist 13 points in the Euclidean plane, not all in the
same straight line, which have the property that every line containing 2 of
them contains at least 3 of them, then every isometry of the plane is either a
rotation or a translation.
(j) [3 MARKS] The product of two glide reflections of the Euclidean plane is
always either a translation or a rotation.
Solution:
(a) This statement is false. There exist groups in which some element has infinite
order. For example, in the additive group of the integers, with identity 0,
operation +, the element 1 has infinite order. A geometric example would be
the group of translations of the strip pattern . . . AAAAAAAAAA. . . : the
translation 1 unit to the right has infinite order.
(b) This statement is false. One isometry that preserves every element in the
plane, hence every set S of points in the plane, is the identity isometry.
(c) This topic was not discussed in the course this year. Students had seen a vector
proof of the theorem, which is true; the medians intersect in the centroid ,
whose coordinates are the averages of the coordinates of the given 3 points.
(d) This negative statement is false: while it appears, at first, to be a special case
of Sylvester’s problem, it does not exclude the possibility that the 13 points
are all on a line. A counterexample to the claim is given by 13 collinear points
in the plane.
(e) This statement is false. While the product of two reflections is a rotation or a
translation, the rotation need not be a half-turn. A counterexample is given
by reflections in mirrors that are not parallel and not perpendicular, e.g. by
reflections in the mirrors x = 0 and x = y, whose products are rotations
through an angle of π2 .
(f) This statement is false; the same counterexample used in the first part can
serve here.
(g) This statement is false. The isometries of a set form a group. A group must
contain an identity. The identity can be viewed as a translation (through a
distance 0) or as a rotation (through a 0 angle), but is not a half-turn; nor is
it a reflection.
Information for Students in Math 348 2003 09
188
(h) This statement is true. The product of three translations is again a translation. Unlike a rotation, a translation has no fixed points, unless the translation
is the identity.
(i) This statement is true. The statement “There exist 13 points in the Euclidean
plane, not all of them in a straight line, which have the property that every
line containing 2 of them contains at least 3 of them,” is false, since it would
contradict Sylvester’s Theorem. The statement “Every isometry of the plane
is either a rotation or a translation,” is also false, since reflections are isometries and are neither rotations nor translations (as they reverse sense, while
the two types named preserve sense). Thus the statement we are to analyze
is logically equivalent to “If FALSE then FALSE.” Such a statement is indeed
TRUE, by our definition of the logical connective →.
(j) This statement is true. The product of two glide reflections preserves sense,
since it is the product of two isometries that reverse sense. We can even be
more specific. If we start with the product of 6 reflections and then apply
the procedures we have studied to products of 2 successive reflections, we can
eventually reduce the product to the product of 2 reflections. If the mirrors
are parallel, we have a translation; if they are not, we have a rotation about
the point of intersection of the mirrors.
Information for Students in Math 348 2003 09
A
1001
Solved Assignments from Previous Years
A.1
2002/2003
A.1.1
First 2002/2003 Problem Assignment
1. (Exercise 3.3 in these notes) In Euclid’s proof of I.5, sketched in §3.10.3, points
are chosen in the productions of the two equal sides. Modify that proof by taking
points in the sides themselves (not in the productions). You are not permitted to
use any propositions in Euclid other than I.1, I.2, I.3, I.4. Do not attempt to prove
equality of the “angles under the base”.
[Hint: First see if you can complete a proof by using adding only the same line
segments as were drawn in Euclid’s proof, i.e. GB and CF , to 4ABC, this time
without any extended sides. If you can’t get through this way, you should then
consider adding the line segment F G.]
Solution:
u A
¢A
¢ A
A
¢
A
¢
A
¢
Au
u
¢
F ¢e
G
%A
¢ e
% A
A
¢
e %
A
¢
e
%
A
¢
% e
A
¢
%
e
A
¢
%
e
A
¢
%
e
A
¢
%
e
A
¢
%
e
¢
A
%
e
¢
A
%
e
¢ %
e A
¢ %
e A
¢ %
e A
¢%
eA
¢%
eA
¢
%
eAu
u
B
C
(a) Beginning with 4ABC, a circle is drawn with centre A and radius less than
Information for Students in Math 348 2003 09
1002
the common length of the sides, |AB| = |AC|; denote the points where this
circle meets AB and AC by F and G respectively.
Now we wish to imitate Euclid’s proof of Pons Asinorum to prove that ∠ABC =
∠ACB. Following precisely the same route as in the earlier proof, we first
prove 4ACF ≡ 4ABG. To apply Proposition I.4. we need the equality of
the lengths of two pairs of sides and of the angles between them. We know
that
|AB| = |AC| (hypothesis)
|AG| = |AF | (construction)
∠BAG = ∠CAF (same angle)
(113)
(114)
and can conclude by I.4 that the triangles are congruent. That means that,
in addition to the pairs of sides and angles mentioned above,
|BG| = |F C|
∠ABG = ∠ACF
∠BGA = ∠CF A
(115)
(116)
(117)
But what can we do with these equalities? We don’t have available Proposition
I.13, which would enable us to conclude that ∠BF C = ∠CGB from (117).
Nor do we have Proposition I.8, which would enable us to prove 4F BC ≡
4GBC and achieve the same result.
(b) Let’s add the line segment F G. It is now not surprising that we should be
able to complete a proof, since we have a miniature of the “Pons Asinorum”
configuration.
(c) Next we shall prove that 4BGF ≡ 4CF G.
∠F BG = ∠ABG (same angle)
= ∠ACF by (116)
= ∠GF C (same angle) .
By (115) we have one pair of equal sides at the equal angles. And
|BF | =
=
=
=
|AB| − |AF |
|AB| − |AG| (by construction)
|AC| − |AG| (by hypothesis)
|CG|
(118)
Information for Students in Math 348 2003 09
1003
gives another pair of equal sides at the equal angles. By Proposition I.4, the
triangles are congruent; consequently we also have
∠F GB = ∠GF C
∠BF G = ∠CGF
(119)
(120)
(d) We will prove 4BF C ≡ 4CGB. We can apply the last two equations to
obtain equality between two angles; in effect, we are “subtracting equals from
equals”, as permitted in Euclid’s Axioms (cf. §3.4.3).
∠BF C = ∠BF G − ∠GF C
= ∠CGF − ∠F GB
= ∠CGB .
(by (120), (119))
One pair of equal sides at these equal angles is |BF | and |CG|, proved equal
in (118). The other sides were proved equal in (115). By Proposition I.4, the
triangles are congruent, so
∠F BC = ∠GCB
¤
(The idea for the preceding proof is contained in a 5th century commentary on
Euclid by Proclus Lycius [11, pp. 193-194].)
2. Use geometric vectors to prove the following
Theorem A.1 Let a quadrangle ABCD be given, and let points E, F, G, H
be, respectively, the mid-points of the sides AB, BC, CD, DA. Then
E, F, G, H — in that cyclic order — are the vertices of a parallelogram.
~ = HG
~ and that F~G = EH.
~
[Hint: You can try to prove that EF
This would
~ =
show that the opposite sides are parallel and of equal length. Show that EF
1 ~
1 ~
1 ~
1 ~
~
~
~
EB + BF = 2 AB + 2 BC = 2 (AB + BC) = 2 AC, etc.]
Solution:
~ = EB
~ + BF
~
EF
definition of addition
1 ~
1 ~
=
AB + BC midpoints
2
2
1 ~
~
=
(AB + BC)
(distributivity of multiplication by scalar)
2
1 ~
AC definition of addition
=
2
~ = 1 AC.
~ It follows that EF
~ = HG.
~ An
In a similar way, we can prove that HG
2
~
analogous argument would show that F~G = EH.
Information for Students in Math 348 2003 09
A.1.2
1004
Second 2002/2003 Problem Assignment
1. (a) (cf. Exercise 5.2.3 in these notes.) Recall Definition 5.6 on page 64 of these
notes, of a distance function or metric on a set. Let N be the set of positive
integers. Define a function f : N × N → R as follows: if m 6= n, (m, n)ρ = 1;
and, for any n ∈ N, (n, n)ρ = 0. Check the various parts of the definition
carefully to prove that ρ is a metric. (Mathematicians call this the discrete
metric.) [Hint: One way to solve this problem, possibly not the most elegant
way, is to divide each of the steps into cases. For example, to prove the triangle
inequality, you may first consider the case when all three points coincide;
then when there are two the same and one different; then when all three
are different. Or it might be possible to merge the last two cases into one.
By dividing up into cases you enable yourself to determine the values of the
various distances that have to be considered.
(b) Define a function f : N → N by n 7→ n + 1 for all n ∈ N. Prove that f is an
isometry relative to the metric ρ.
(c) Determine whether f is surjective.
Solution:
(a)
i. (Non-negativity.) The distance between a point and itself is defined to be
0; this is a requirement of the definition of a distance function. We must
also check that the distance between two distinct points is not zero; but
this is also true, because, in such a case, the distance has been defined to
be 1.
ii. (“Symmetry”.) If x = y, (x, y)ρ = 0 = (y, x)ρ. If x 6= y, (x, y)ρ = 1 =
(y, x)ρ. Thus, in either case, symmetry holds.
iii. (“The Triangle Inequality”.) Let x, y, z be three points. If they are not
all the same point, then (x, y)ρ + (y, z)ρ will be at least 1, possibly even
2. This is not less than (x, y)ρ, which cannot be more than 1.
If x, y, z are all the same point, then (x, z)ρ = 0 = 0+0 = (x, y)ρ+(y, z)ρ.
(b) There are only two possible distances in our metric space: 0 and 1. When
two points coincide, they are mapped to the same point — since an essential
feature of a function is that it maps a point to a uniquely defined point in the
codomain; thus the image points have distance 0, which is the same as the
distance between the two points in the domain. If x 6= y, their distance is 1; f
maps x and y to points which will again be distinct, since x + 1 = y + 1 ⇒ x =
y, which would be a contradiction. Hence 2 points at distance 1 are mapped
onto 2 points at distance 1. Thus f is an isometry of this “discrete metric
space”.
Information for Students in Math 348 2003 09
1005
(c) For every n, nf = n + 1 ≥ 1 + 1 = 2 > 1, so 1 is not the image of any positive
integer, and f is not surjective. (This situation — an isometry that is not
surjective — will not arise in the Euclidean situations we will be considering
from the textbook.)
2. Consider a rectangle in the Euclidean plane, where the plane has been endowed
with a rectangular Cartesian coordinate system, so that the corners of the rectangle
are at the points (±1, 2) and (±1, −2). Unlike the square, which has 8 symmetries,
this rectangle has only 4 symmetries.
(a) Describe each of the 4 isometries of the plane which map the rectangle on to
itself. You are expected to give each of the isometries a name, and to describe
each of the functions precisely, explaining how each of them acts on a point in
the plane with coordinates (x, y). Use names like R1 , . . . for those isometries
that are reflections.
(b) Show how each of the rotations — if any — can be expressed as a product of
the reflections.
(c) Set up a Cayley Table for the symmetries of the rectangle, and fill in all the
entries.
(d) Explain why none of the symmetries of the rectangle is a glide reflection (with
non-trivial translation).
Solution:
(a) Reflection R1 in y-axis: (x, y) 7→ (−x, y)
Reflection R2 in x-axis: (x, y) 7→ (x, −y)
Rotation ª(0,0),π through π radians about the origin: ª(0,0),π = R1 R2 =
R2 R1 : (x, y) 7→ (−x, −y)
Identity function ιR2 : (x, y) 7→ (x, y)
(b) A factorization in terms of reflections R1 and R2 is given above. Note that
there are several way — indeed, infinitely many ways — in which the halfturn can be expressed in terms of the two generating rotations. (You are not
expected to be able to characterize all of them, but you should be able to
exhibit some of them.)
(c)
ι
R1
R2
ª(0,0),π
ι
R1
R2
ª(0,0),π
ι
R1
R2
R1 R2 = R2 R1
ι
R 1 R2 = R2 R1
R2
R1
R2
R1 R2 = R2 R1
ι
R1
R1 R2 = R2 R1
R1
R1
ι
Information for Students in Math 348 2003 09
1006
(d) A glide reflection with non-trivial translation — indeed, any nontrivial translation — has infinite period. But all of the isomorphisms in this group have
period 2 or 1.
(This is not a problem that you could solve easily if we were to change
some of the angles or points. But you should be able to solve it with the
data that are given.)
3.
Let R1 , R2 and R3 be reflections, respectively in the y-axis, the line x + y = 1, and
the x-axis. Apply the procedure discussed in §5.3.9 of the notes to express R1 R2 R3
as products of the form given in equation (71)
(3)
R1 R2 R3 = R10 R20 R3 = R10 R200 R30 = R100 R2 R30 ,
and interpret it as a reflection followed by a translation (cf. §5.3.5) in the direction
of its mirror106 . Then repeat the exercise, this time expressing the product as a
translation followed by a reflection. Show all your work, giving, the equation of
the mirror for each reflection. Where the mirror is not horizontal or vertical, you
need not give a formula for the reflection — it suffices to give the equation for its
mirror.
Solution:
106
(a)
i. First we operate on the two reflections whose mirrors intersect at (0, 1),
replacing them by reflections R10 with mirror y = x + 1, and R2 with the
y-axis, x = 0, as its mirror.
ii. Next we operate on reflections R20 and R3 , which intersect at the origin.
Turning the mirrors counterclockwise through an angle π4 yields rotations
R200 , with mirror x + y = 0, and R30 with mirror y = x.
¢
¡
iii. Finally, we turn mirrors R10 and R200 which intersect at the point 12 , 21
counterclockwise through an angle of π2 radians, obtaining reflections R100
(3)
in mirror x + y = 0, and R2 in mirror y = x + 1.
iv. The given product therefore factorizes into the reflection in mirror x +
y = 0, followed by a translation in the direction of the mirror through a
distance of 1 unit.
(b)
i. We first replace reflections R2 and R3 , whose mirrors intersect in (1, 0),
by reflections R20 , whose mirror is the x-axis, y = 0, and R30 , whose mirror
is y = x − 1.
corrected November 18th, 2002
UPDATED TO November 27, 2003
Information for Students in Math 348 2003 09
1007
ii. Next we operate on reflections R1 and R20 , whose mirrors intersect in
(0, −1): we turn through − π4 radians, replacing them by reflections R10 ,
with mirror y = x, and R200 , whose mirror is x + y = 0.
iii. Finally, we replace reflections R200 and R30 by turning their mirrors counterclockwise through an angle of − π2 radians. The mirrors intersect in the
¡
¢
(3)
point 12 , 21 . We obtain reflections R2 with mirror y = x − 1 and R300
with mirror x + y = 0.
iv. The given product therefore factorizes into a reflection in mirror x+y = 0
preceded by a translation in the direction of the mirror through a distance
of 1 unit.
A.1.3
Third 2002/2003 Problem Assignment
Preceding the list of friezes numbered 1, 2, . . . , Table 10.1.4 of these notes has been
reproduced.
• Associate with each of the listed friezes the frieze in the table which has the same
isometry group, or argue that the isometry group is different from all those listed.
• For each type of isometry possessed by the frieze shown, give a precisely described
example of each isometry of that type. More precisely,
– If the pattern has reflective symmetry in a vertical mirror, describe all types of
vertical mirrors that are possible. For example, for the pattern ...VVVVV...
one would observe that there are two possible types of mirrors:
1. between two successive letters V, as ...VV|VVV...
2. along the axis of symmetry of any one letter V, as ...VVVVV...
|
– If the pattern has rotational symmetry under a half-turn, describe the location(s) of the centre(s) of the half-turn(s).
– If the pattern has symmetry under translations, describe a fundamental interval; e.g. for the pattern ...UVUVUVUVUV... one such interval would
be UV.
– If the pattern has symmetry under a glide reflection, describe the axis of the
translation and a shortest107 interval that is shifted; e.g. in LΓLΓL one such
interval would be LΓ.
107
Corrected 19 Oct 03 from original version; in the original version the word used was “fundamental”,
which could be interpreted as being the smallest interval that, under the isometry in question, would be
mapped to cover the entire frieze. Under such an interpretation the correct interval would be half the
size of the answer under the current wording.
Information for Students in Math 348 2003 09
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
1008
Typical pattern Ref(H) Ref(V) Trans HT GL Abstract Group
...LLLLL...
0
0
∞
0
0
C∞
...LΓLΓL...
0
0
∞
0
∞
C∞
...VVVVV...
0
∞
∞
0
0
D∞
...NNNNN...
0
0
∞
∞
0
D∞
...VΛVΛV...
0
∞
∞
∞ ∞
D∞
...DDDDD...
1
0
∞
0
∞
C∞ × D1
...HHHHH...
1
∞
∞
∞ ∞
D∞ × D1
Symbol Meaning
Ref(H) Reflection in Horizontal mirror
Ref(V) Reflections in Vertical mirrors
Legend:
Trans Translations
HT
Half-Turns
GL
Glide Reflections (with nontrivial translation)
1. . . . AAAAA. . .
2. . . . ABABABABA. . .
3. . . . AFAFAFAF. . .
4. . . . HAHAHAHA. . .
5. . . . HOHOHOHOHO. . .
6. . . . AHHHHAHHHAHHAHAHHAHHHAHHHHA. . .
7. . . . NONONONO. . .
RRRRR
8. . . .
...
RHRHRH
9. . . .
...
QP`PQP`PQP`P
10. . . .
...
11. . . . ± ∓ ± ∓ ± ∓ ±∓. . .
12. . . . ∈3∈3∈3∈3. . .
RRRHRRR
13. . . .
...
14. . . . ¯ ¯ ¯ ⊕ ¯ ¯ ¯. . .
15. . . . McGillMcGillMcGillMcGill. . .
Information for Students in Math 348 2003 09
1009
16. . . . ThinkThinkThinkThinkThimkThinkThinkThink. . .
Solution:
Number
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
(15)
(16)
Axis of
Ref(H)
—
—
—
—
yes
—
—
—
—
—
—
yes
—
yes
—
—
Axes of
Ref(V)
A, A|A
—
—
A, H
O, H
H
—
—
—
—
±, ∓
∈, 3
—
⊕
—
—
Fund.Interval
(Translation)
A
AB
AF
HA
HO
—
NO
R
Centres of
HT
—
—
—
—
H, O
—
N,R O
RH
QP`P
R H
,
—
±, ∓
∈ | 3,H 3 | ∈
±∓
∈3
—
—
McGill
—
⊕
—
—
Fund.Interval
(Glide-Reflection)
—
—
—
—
HO
—
—
—
—
yes
yes
yes
—
—
—
—
Isometry
Group
III
I
I
III
VII
C2
IV
IV
IV
II
V
VII
C2
D2
I
{ι}
Comments: In frieze (16) the symmetry has been destroyed by one letter m; were that
letter an n, the symmetry group would consist of infinitely many translations (I), as in
(15).
A.1.4
Fourth 2002/2003 Problem Assignment
More so than the previous assignments, this assignment was designed as a
learning — rather than a testing — activity. Thus there are some parts of
some questions, #1 in particular, where students did not see comparable
specific problems worked in the lectures. The problems will be graded to
recognize any reasonable attempt at solution, without demanding a particular
level of success. All the material in the assignment should be considered part
of the course, and, therefore, possible examination material.
1. Each of the following sets of coordinates or equations is presented either in inhomogeneous or homogeneous coordinates, where the relationship between the two
, y = YZ .
(for points off the line at infinity) is given by x = X
Z
• Transform from inhomogeneous to homogeneous, or from homogeneous to
inhomogeneous, depending on which system is used in the given data.
Information for Students in Math 348 2003 09
1010
• If the given point is on the line at infinity, state that.
• If a given equation in X, Y, Z could not represent a curve in the plane in
homogeneous coordinates, state that.
• Since homogeneous coordinates are not unique, a conversion from inhomogeneous to homogeneous coordinates can be expected to be non-unique — any
one example will be acceptable.
(a) (x, y) = (0, 6)
(b) (x, y) = (6, 0)
(c) X + Y + Z = 0
(d) x2 + y 2 = 25
(e) Z = 4
(f) X + 2Y + 3Z = 6
(g) Z = 0
(h) XY = Z 2
(i) (X = 12) ∧ (Y = −3)
Solution:
(a)
µ
¶
X Y
(x, y) = (0, 6) ⇔
,
= (0, 6)
Z Z
⇒ (X = 0) ∧ (Y = 6Z)
⇒ One set of homogeneous coordinates is (X, Y, Z) = (0, 6, 1);
more generally, the homogeneous coordinates for this point would be (X, Y, Z) =
(0, 6Z, Z), where Z ranges over R − {0}. (The representation might be “prettier” if we used another symbol for the parameter ; we could state that the
general solution is (X, Y, Z) = (0, 6t, t), where t is any non-zero real number.)
[Students were asked to supply only one solution.]
(b) As in the preceding problem, X = 6Z, Y = 0Z, so the most general ordered
triple of homogeneous coordinates is (X, Y, Z) = (6t, 0, t), where t is any nonzero real number.
(c) When Z 6= 0, X + Y + Z = 0 ⇒ X
+ YZ + 1 = 0, so x + y + 1 = 0, and the
Z
equation is seen to represent the line with intercepts (0, −1), (−1, 0). (The
Information for Students in Math 348 2003 09
1011
equation makes sense even when Z = 0; the intersection of the given line and
the line at infinity is the “point at infinity” with homogeneous coordinates
(X, Y, Z) = (1, −1, 0), or, more generally, (X, Y, Z) = (u, −u, 0), where u is
any non-zero real number.)
¡ ¢2 ¡ Y ¢2
(d) x2 + y 2 = 25 ⇒ X
+ Z = 25 ⇒ X 2 + Y 2 − 25Z 2 = 0.
Z
(e) What points in the real projective plane are described when we restrict Z to
the value 4? Any point not on the line at infinity is representable in this way!
Thus the equation Z = 4 represents all points off the line at infinity, i.e. the
entire Euclidean (or “finite”) plane.
(f) What points in the real projective plane are described when we restrict X, Y, Z
to ordered triples such that X + 2Y + 3Z = 6? There are no points whose
homogeneous coordinates have this property.
The polynomial X + 2Y + 3Z − 6 is not homogeneous, in that it is a sum of
terms whose “total degree” [the sum of the degrees of the various variables]
is not constant. Whenever a polynomial equation in x and y is transformed
by the substitution x = X
, y = YZ , we obtain an equation whose terms are
Z
fractions in which the numerator is a monomial in X and Y whose total degree
is the same as that of a pure power of Z in the denominator. When we clear
these fractions by multiplying through by the largest power of Z appearing
in any denominator, we obtain a sum of monomials in X, Y , Z all of the
same total degree, i.e. what is called a homogeneous polynomial. The given
equation is not of that type, so it cannot have come from the transformation
to homogeneous coordinates of a polynomial equation in x and y.
(g) Z = 0 is the equation whose solutions are precisely the points at infinity; it
represents the collection of all such solutions, i.e. the “line at infinity”.
(h) XY = Z 2 ⇒ xy = 1 if Z 6= 0. The curve represented is a hyperbola. (This
hyperbola also contains points on the line at infinity: they are the points such
that XY = 0, i.e. such that X = 0 or Y = 0 but not both, since we can’t have
all three coordinates equal to 0. When X = 0 we have the point (0, Y, 0) for
any non-zero Y , e.g. (0, 1, 0) — which is the point at infinity on the x-axis;
similarly the curve contains the point at infinity on the y-axis, (1, 0, 0). These
facts are not surprising, since the curve is asymptotic to the x- and y-axes,
and appears to meet these lines “at infinity”.)
(i) From the given information we know that xy = − 14 . Thus the curve is contained
in the line through the origin with slope − 41 . Conversely, any
¢ on that
¡ point
1
line has inhomogeneous coordinates of the form (x,¡ y) = t, −
¢ 4 t , which is
equivalent to homogeneous coordinates (X, Y, Z) = 12, −3, 12t provided t 6=
0. It follows that the given equations describe the line through the origin with
Information for Students in Math 348 2003 09
1012
slope − 14 with the origin removed! (including the point at infinity, (12, −3, 0),
or equivalently, (4, −1, 0)).
2. List the 7 points of the Fano plane, showing, for each of them the 3 lines passing
through them; then list the 7 lines, showing the 3 points on each of them. For example, the point (1, 0, 1) is contained in the following lines: [1, 0, 1], [0, 1, 0], [1, 1, 1].
(While there are systematic ways of attacking this problem, one can simply test
each of the 7 lines to see whether the coordinates have the desired “dot product”
or “inner product” property.)
Solution:
Lines
Coordinates
Equation
[0, 0, 1]
Z=0
[0, 1, 0]
Y =0
[1, 0, 0]
X=0
[0, 1, 1]
Y +Z =0
[1, 0, 1]
X +Z =0
[1, 1, 0]
X +Y =0
[1, 1, 1]
X +Y +Z =0
Coordinates
(0, 0, 1)
(0, 1, 0)
(1, 0, 0)
(0, 1, 1)
(1, 0, 1)
(1, 1, 0)
(1, 1, 1)
points on each line
(0, 1, 0), (1, 0, 0), (1, 1, 0)
(0, 0, 1), (1, 0, 0), (1, 0, 1)
(0, 1, 0), (0, 0, 1), (0, 1, 1)
(0, 1, 1), (1, 1, 1), (1, 0, 0)
(1, 0, 1), (1, 1, 1), (0, 1, 0)
(1, 1, 0), (1, 1, 1), (0, 0, 1)
(1, 1, 0), (0, 1, 1), (1, 0, 1)
Points
lines through each point
[0, 1, 0], [1, 0, 0], [1, 1, 0]
[0, 0, 1], [1, 0, 0], [1, 0, 1]
[0, 1, 0], [0, 0, 1], [0, 1, 1]
[0, 1, 1], [1, 1, 1], [1, 0, 0]
[1, 0, 1], [1, 1, 1], [0, 1, 0]
[1, 1, 0], [1, 1, 1], [0, 0, 1]
[1, 1, 0], [0, 1, 1], [1, 0, 1]
3. Find the coordinates of all points and lines in the geometry P G(2, 3), constructed
using homogeneous coordinates over the field F3 . (Take care not to list a point or
a line more than once. There are 13 points and 13 lines.) In particular, determine
the points on the line [1, 2, 0], and the lines through the point (2, 1, 1).
Solution: We need to consider all ordered triples of field elements — which are
0, 1, 2 — but must recall that proportional triples represent the same point; the
only non-zero constants that can be the constants of proportionality are 1 and
2, and multiplication by 1 yields the same point. We will list all the admissible
ordered triples, followed by the result of multiplying by 2 (and reducing modulo
Information for Students in Math 348 2003 09
1013
3); remember that multiplying by 2 is the same as changing a sign, since −1 is the
same as +2 in this field!
We tabulate the points:
(1,0,0)
(0,1,0)
(0,0,1)
(0,1,1)
(1,0,1)
(1,1,0)
(0,1,2)
(1,0,2)
(1,2,0)
(1,1,1)
(2,1,1)
(1,2,1)
(1,1,2)
alternative coordinates
(2,0,0)
(0,2,0)
(0,0,2)
(0,2,2)
(2,0,2)
(2,2,0)
(0,2,1)
(2,0,1)
(2,1,0)
(2,2,2)
(1,2,2)
(2,1,2)
(2,2,1)
and the lines:
[1,0,0]
[0,1,0]
[0,0,1]
[0,1,1]
[1,0,1]
[1,1,0]
[0,1,2]
[1,0,2]
[1,2,0]
[1,1,1]
[2,1,1]
[1,2,1]
[1,1,2]
alternative coordinates
[2,0,0]
[0,2,0]
[0,0,2]
[0,2,2]
[2,0,2]
[2,2,0]
[0,2,1]
[2,0,1]
[2,1,0]
[2,2,2]
[1,2,2]
[2,1,2]
[2,2,1]
equation
X=0
Y =0
Z=0
Y +Z =0
X +Z =0
X +Y =0
2Y + Z = 0
2X + Z = 0
2X + Y = 0
X +Y +Z =0
2X + Y + Z = 0
X + 2Y + Z = 0
X + Y + 2Z = 0
The line [1, 2, 0], i.e. having equation X + 2Y = 0, contains all points having a set
of coordinates in which X = −2Y , i.e. X = +Y . From our table, we find these to
be (1, 1, 0), (1, 1, 1), (1, 1, 2), and (0, 0, 1).
UPDATED TO November 27, 2003
Information for Students in Math 348 2003 09
1014
The lines through the point (2, 1, 1) are those whose coordinates [`, m, n] satisfy
2` + m + n = 0, i.e. ` = m + n. Again from the table (or otherwise) we find these
to be [1, 1, 0] or X + Y = 0, [2, 1, 1] or 2X + Y + Z = 0, [0, 1, 2] or Y + 2Z = 0,
and [1, 0, 1] or X + Z = 0.108
4. In ordered geometry, suppose one takes P to be the integers — positive, negative,
and 0, and defines Q (i.e. P × P × P) to consist of the set
{(`, m, n)|(` < m < n) ∨ (n < m < `)} .
Prove that Axioms 18.1, 18.2, 18.3, 18.4, 18.5, 18.6 are all satisfied; that is, that
this describes a model for the axiomatic system we have described.
Solution:
Axiom 18.1: We have defined the set P to be Z, the set of all integers, which
contains more than one element.
Axiom 18.2: Given any 2 distinct integers, x and y, let us assume them labelled
so that x < y. Then x < y < y + 1, so (x, y, y + 1) ∈ Q.
Axiom 18.3: If (`, m, n) ∈ Q, then either ` < m < n or n < m < `. In either
case ` 6= n.
Axiom 18.4: Our definition is symmetric: (`, m, n) ∈ Q ⇔ ((` < m < n) ∨ (n <
m < `)) ⇔ ((n < m < `) ∨ (` < m < n)) ⇔ (n, m, `) ∈ Q.
Axiom 18.5: Suppose that both (`, m, n) ∈ Q and (m, n, `) ∈ Q. Since each of
these membership statements corresponds to one or other of two inequality
statements, we have to consider 2 × 2 = 4 cases, where we shall show that —
in every case — a contradiction is implied.
(a) (` < m < n) ∧ (m < n < `) ⇒ ` < m < n < `, a contradiction, since no
number can be less than itself.
(b) (` < m < n) ∧ (` < n < m) ⇒ (m < n < m), a contradiction.
(c) (n < m < `) ∧ (m < n < `) ⇒ (n < m < n), a contradiction.
(d) (n < m < `) ∧ (` < n < m) ⇒ (n < m < ` < n), a contradiction.
Axiom 18.6: We need to determine the nature of lines in this model. Let x, y be
integers such that x < y.
x/y = {n|n < x}
y/x = {n|n > y}
xy = {n|x < n < y}
108
corrected 9 December, 2002
Information for Students in Math 348 2003 09
1015
The union of these three sets with {x, y} is the set Z. This is what is defined
to be L(x, y). Thus, in this model of an ordered geometry, all lines are the
set Z. Thus all points lie in all lines, and this axiom is trivially satisfied.
B
Tests and Examinations from Previous Years
B.1
Tests from Fall, 2002
The four versions of the test are followed by solutions.
B.1.1
Version 1
1. You are given a triangle with vertices A, B, C, in which E is a point on side
AB with the property that |AE| = 2|EB|, and F is a point on side AC with the
property that |AF | = 2|F C|.
(a) [13 MARKS] Use vectors — no other method will be accepted — to prove
that EF is parallel to BC.
(b) [12 MARKS] Determine, by using information from your vector proof of the
|EF |
previous part, the value of the ratio
.
|BC|
2. [10 MARKS] Give a proof of EITHER (a) or (b). In either case, you are permitted
to use only the results that precede the result in Euclid’s Book I. What will be
looked for in your proof is coherence — that your statements fit together in a
logical way — and that you justify each step in the proof you choose to give; you
are not expected to use precisely the same language as in the proof given to you.
(a) Proposition I.2: There exists a construction to place at a point P (as an
extremity) a line segment equal in length to a line segment BC.
(b) Proposition I.5 (part): If, in a triangle ABC, |AB| = |AC|, then the interior angles at B and C are equal. (If you select to prove this proposition, you
may use either Euclid’s proof, or the proof of Proclus shown in the solutions
to an assignment question.)
3. Answer the following questions for each of these strip patterns:
I.
II.
· · · ∀∃∀∃∀∃∀∃∀∃ · · ·
· · · `a`a`a`a`a`a · · ·
Information for Students in Math 348 2003 09
1016
(a) [5 MARKS] If the pattern has reflective symmetry in some vertical mirror,
describe all locations for vertical mirrors that are possible. For example, for
the pattern ...VVVVV... one would observe that there are two possible
locations for mirrors:
i. between two successive letters V, as ...VV|VVV...
ii. along the axis of symmetry of any one letter V, as ...VVVVV...
|
(b) [5 MARKS] If the pattern has rotational symmetry under a half-turn, describe
the locations of the centres of all such half-turn(s).
(c) [5 MARKS] If the pattern has symmetry under translations, describe an interval whose images cover the entire strip, and that is as short as possible;
e.g., for the pattern ...UVUVUVUVUV... one such interval would be UV,
another would be VU; but UVUV would not be acceptable, because it is
not as short as possible.
(d) [5 MARKS] If the pattern has symmetry under a glide reflection, describe the
axis of the translation and a shortest possible interval whose images cover the
entire strip; e.g., in · · · LΓLΓL· · · one such interval would be LΓ.
(e) [5 MARKS] Determine which line of the table reproduced below gives the
numbers of isometries of the given strip pattern, or argue that none of the
rows of the table has that property.
Typical pattern
Ref(H) Ref(V) Trans
...EEEEE...
0
0
∞
...bdbdbdbd...
0
0
∞
...44444...
0
∞
∞
···; ; ; ; ; ···
0
0
∞
· · · 4 5 4 5 45 · · ·
0
∞
∞
· · ·∈∈∈∈∈ · · ·
1
0
∞
· · · mmmmm · · ·
1
∞
∞
Symbol Meaning
Ref(H) Reflection in Horizontal mirror
Ref(V) Reflections in Vertical mirrors
Legend:
Trans Translations
HT
Half-Turns
GL
Glide Reflections (with nontrivial
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
B.1.2
HT
0
0
0
∞
∞
0
∞
GL
0
∞
0
0
∞
∞
∞
Abstract Group
C∞
C∞
D∞
D∞
D∞
C∞ × D1
D∞ × D1
translation)
Version 2
1. The real plane is endowed with the usual orthogonal Cartesian coordinate system
(with perpendicular coordinate axes), and A(0, 2), B(−3, 1), C(0, 0), D(3, 1) are
the vertices of a quadrilateral with sides AB, BC, CD, DA.
Information for Students in Math 348 2003 09
1017
(a) [12 MARKS] List all of the isometries of the quadrilateral. You may use either
of the following forms to specify each isometry:
µ
¶
A B C D
• You may use the form
, where U, V, W, X are the
U V W X
points A, B, C, D in some order.
• Alternatively, you may specify precisely which type of isometry you have
(reflection, rotation, etc.) and describe its action in a precise way so that
the reader can reconstruct the function unambiguously. In this case you
should give the isometry a name, which will be needed in the next part
of the problem.
(b) [13 MARKS] Give a complete a Cayley table for the isometries, where the
function applied second is named in the labels for the rows (horizontal lines)
of the table, and the function applied second is first in the labels for the
columns (vertical lines).
2. [10 MARKS] Give a proof of EITHER (a) or (b). In either case, you are permitted
to use only the results that precede the result in Euclid’s Book I. What will be
looked for in your proof is coherence — that your statements fit together in a
logical way — and that you justify each step in the proof you choose to give; you
are not expected to use precisely the same language as in the proof given to you.
(a) Proposition I.2: There exists a construction to place at a point P (as an
extremity) a line segment equal in length to a line segment BC.
(b) Proposition I.5 (part): If, in a triangle ABC, |AB| = |AC|, then the interior angles at B and C are equal. (If you select to prove this proposition, you
may use either Euclid’s proof, or the proof of Proclus shown in the solutions
to an assignment question.)
3. Answer the following questions for each of these strip patterns:
I.
II.
· · · ∀¬∀¬∀¬∀¬∀¬ · · ·
· · · ∼ °∼ °∼ °∼ °∼ ° · · ·
(a) [5 MARKS] If the pattern has reflective symmetry in some vertical mirror,
describe all locations for vertical mirrors that are possible. For example, for
the pattern ...VVVVV... one would observe that there are two possible
locations for mirrors:
i. between two successive letters V, as ...VV|VVV...
ii. along the axis of symmetry of any one letter V, as ...VVVVV...
|
Information for Students in Math 348 2003 09
1018
(b) [5 MARKS] If the pattern has rotational symmetry under a half-turn, describe
the locations of the centres of all such half-turn(s).
(c) [5 MARKS] If the pattern has symmetry under translations, describe an interval whose images cover the entire strip, and that is as short as possible;
e.g., for the pattern ...UVUVUVUVUV... one such interval would be UV,
another would be VU; but UVUV would not be acceptable, because it is
not as short as possible.
(d) [5 MARKS] If the pattern has symmetry under a glide reflection, describe the
axis of the translation and a shortest possible interval whose images cover the
entire strip; e.g., in · · · LΓLΓL· · · one such interval would be LΓ.
(e) [5 MARKS] Determine which line of the table reproduced below gives the
numbers of isometries of the given strip pattern, or argue that none of the
rows of the table has that property.
Typical pattern
Ref(H) Ref(V) Trans
...EEEEE...
0
0
∞
...bdbdbdbd...
0
0
∞
...44444...
0
∞
∞
···; ; ; ; ; ···
0
0
∞
· · · 4 5 4 5 45 · · ·
0
∞
∞
· · ·∈∈∈∈∈ · · ·
1
0
∞
· · · mmmmm · · ·
1
∞
∞
Symbol Meaning
Ref(H) Reflection in Horizontal mirror
Ref(V) Reflections in Vertical mirrors
Legend:
Trans Translations
HT
Half-Turns
GL
Glide Reflections (with nontrivial
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
B.1.3
HT
0
0
0
∞
∞
0
∞
GL
0
∞
0
0
∞
∞
∞
Abstract Group
C∞
C∞
D∞
D∞
D∞
C∞ × D1
D∞ × D1
translation)
Version 3
1. (a) [5 MARKS] If n is any positive integer such that n > 6, it is not possible to
arrange n (distinct) points in the plane so that a straight line through any
two of them must pass through a third.
(b) [5 MARKS] Let n be any positive integer greater than 6. If n (distinct)
points in the plane are not all on one straight line, then there exists a line
which contains exactly three of them.
(c) [5 MARKS] Let n be any positive integer greater than 6. If n (distinct) points
in the plane are not all on one straight line, then there exists no line which
contains exactly three of them.
Information for Students in Math 348 2003 09
1019
(d) [5 MARKS] If n > 6 (distinct) points in the plane are all on one straight line,
then there exists a straight line which contains exactly three of them.
(e) [5 MARKS] There cannot exist 7 distinct points in the plane with the property
that any three of them lie on a straight line.
2. [10 MARKS] Give a proof of EITHER (a) or (b). In either case, you are permitted
to use only the results that precede the result in Euclid’s Book I. What will be
looked for in your proof is coherence — that your statements fit together in a
logical way — and that you justify each step in the proof you choose to give; you
are not expected to use precisely the same language as in the proof given to you.
(a) Proposition I.2: There exists a construction to place at a point P (as an
extremity) a line segment equal in length to a line segment BC.
(b) Proposition I.5 (part): If, in a triangle ABC, |AB| = |AC|, then the interior angles at B and C are equal. (If you select to prove this proposition, you
may use either Euclid’s proof, or the proof of Proclus shown in the solutions
to an assignment question.)
3. Answer the following questions for each of these strip patterns:
I.
II.
· · · ∀∀∀∀∀∀ · · ·
· · · 6=÷6=÷6=÷6=÷6=÷ · · ·
(a) [5 MARKS] If the pattern has reflective symmetry in some vertical mirror,
describe all locations for vertical mirrors that are possible. For example, for
the pattern ...VVVVV... one would observe that there are two possible
locations for mirrors:
i. between two successive letters V, as ...VV|VVV...
ii. along the axis of symmetry of any one letter V, as ...VVVVV...
|
(b) [5 MARKS] If the pattern has rotational symmetry under a half-turn, describe
the locations of the centres of all such half-turn(s).
(c) [5 MARKS] If the pattern has symmetry under translations, describe an interval whose images cover the entire strip, and that is as short as possible;
e.g., for the pattern ...UVUVUVUVUV... one such interval would be UV,
another would be VU; but UVUV would not be acceptable, because it is
not as short as possible.
(d) [5 MARKS] If the pattern has symmetry under a glide reflection, describe the
axis of the translation and a shortest possible interval whose images cover the
entire strip; e.g., in · · · LΓLΓL· · · one such interval would be LΓ.
Information for Students in Math 348 2003 09
1020
(e) [5 MARKS] Determine which line of the table reproduced below gives the
numbers of isometries of the given strip pattern, or argue that none of the
rows of the table has that property.
Typical pattern
Ref(H) Ref(V) Trans
(i)
...EEEEE...
0
0
∞
(ii)
...bdbdbdbd...
0
0
∞
(iii)
...44444...
0
∞
∞
(iv)
···; ; ; ; ; ···
0
0
∞
(v) · · · 4 5 4 5 45 · · ·
0
∞
∞
(vi)
· · ·∈∈∈∈∈ · · ·
1
0
∞
(vii)
· · · mmmmm · · ·
1
∞
∞
Symbol Meaning
Ref(H) Reflection in Horizontal mirror
Ref(V) Reflections in Vertical mirrors
Legend:
Trans Translations
HT
Half-Turns
GL
Glide Reflections (with nontrivial
B.1.4
HT
0
0
0
∞
∞
0
∞
GL
0
∞
0
0
∞
∞
∞
Abstract Group
C∞
C∞
D∞
D∞
D∞
C∞ × D1
D∞ × D1
translation)
Version 4
1. [10 MARKS] Give a proof of EITHER (a) or (b). In either case, you are permitted
to use only the results that precede the result in Euclid’s Book I. What will be
looked for in your proof is coherence — that your statements fit together in a
logical way — and that you justify each step in the proof you choose to give; you
are not expected to use precisely the same language as in the proof given to you.
(a) Proposition I.2: There exists a construction to place at a point P (as an
extremity) a line segment equal in length to a line segment BC.
(b) Proposition I.5 (part): If, in a triangle ABC, |AB| = |AC|, then the interior angles at B and C are equal. (If you select to prove this proposition, you
may use either Euclid’s proof, or the proof of Proclus shown in the solutions
to an assignment question.)
2. Answer the following questions for each of these strip patterns:
I.
II.
· · · ∀I∀I∀I∀I∀I∀I· · ·
··· =
6 ∝6=∝6=∝6=∝6=∝ · · ·
(a) [5 MARKS] If the pattern has reflective symmetry in some vertical mirror,
describe all locations for vertical mirrors that are possible. For example, for
Information for Students in Math 348 2003 09
1021
the pattern ...VVVVV... one would observe that there are two possible
locations for mirrors:
i. between two successive letters V, as ...VV|VVV...
ii. along the axis of symmetry of any one letter V, as ...VVVVV...
|
(b) [5 MARKS] If the pattern has rotational symmetry under a half-turn, describe
the locations of the centres of all such half-turn(s).
(c) [5 MARKS] If the pattern has symmetry under translations, describe an interval whose images cover the entire strip, and that is as short as possible;
e.g., for the pattern ...UVUVUVUVUV... one such interval would be UV,
another would be VU; but UVUV would not be acceptable, because it is
not as short as possible.
(d) [5 MARKS] If the pattern has symmetry under a glide reflection, describe the
axis of the translation and a shortest possible interval whose images cover the
entire strip; e.g., in · · · LΓLΓL· · · one such interval would be LΓ.
(e) [5 MARKS] Determine which line of the table reproduced below gives the
numbers of isometries of the given strip pattern, or argue that none of the
rows of the table has that property.
Typical pattern
Ref(H) Ref(V) Trans
(i)
...EEEEE...
0
0
∞
(ii)
...bdbdbdbd...
0
0
∞
(iii)
...44444...
0
∞
∞
(iv)
···; ; ; ; ; ···
0
0
∞
(v) · · · 4 5 4 5 45 · · ·
0
∞
∞
(vi)
· · ·∈∈∈∈∈ · · ·
1
0
∞
(vii)
· · · mmmmm · · ·
1
∞
∞
Symbol Meaning
Ref(H) Reflection in Horizontal mirror
Ref(V) Reflections in Vertical mirrors
Legend:
Trans Translations
HT
Half-Turns
GL
Glide Reflections (with nontrivial
HT
0
0
0
∞
∞
0
∞
GL
0
∞
0
0
∞
∞
∞
Abstract Group
C∞
C∞
D∞
D∞
D∞
C∞ × D1
D∞ × D1
translation)
3. For each of the following statements, indicate whether it is true or false.
• If you claim it is false, prove that it is false — if possible by describing an
example or a theorem in precise details.
• If you claim it is true, no proof is required. You should not expect part marks
for a false statement that you indicate to be true, or for a true statement that
you indicate to be false.
Information for Students in Math 348 2003 09
1022
[To say that a function preserves a set S of points is to say that the
function is an isometry of the plane which maps the set S on to itself.]
(a) [5 MARKS] Any rotation which preserves a finite set of points in the plane
must have period 1, 2, 3, 4, or 6.
(b) [5 MARKS] Any set of points in the plane must be preserved by some rotation
whose period is 1, 2, 3, 4, or 6.
(c) [5 MARKS] The only sets of points in the plane that have an isometry of
period 2 are finite.
(d) [5 MARKS] There is no set of points in the plane that has an isometry of
period 7.
(e) [5 MARKS] If the symmetry group of a set of points in the plane contains
translations, then the entire set must be contained in a line.
B.2
Solutions to Problems on the 2002 Class Tests
B.2.1
Problems on all four versions of the test
1. [10 MARKS] Give a proof of ONE of (a) or (b). In either case, you are permitted
to use only the results that precede the result in Euclid’s Book I. What will be
looked for in your proof is coherence — that your statements fit together in a
logical way — and that you justify each step in the proof you choose to give; you
are not expected to use precisely the same language as in the proof given to you.
(a) Proposition I.2: There exists a construction to place at a point P (as an
extremity) a line segment equal in length to a line segment BC.
(b) Proposition I.5 (part): If, in a triangle ABC, |AB| = |AC|, then the interior angles at B and C are equal. (If you select to prove this proposition, you
may use either Euclid’s proof, or the proof of Proclus shown in the solutions
to an assignment question.)
Solution:
(a) Proposition I.2: A proof is given in these notes, 16. Students should remember that, in the operation of drawing a circle with a given centre, we cannot
specify the radius — we can only insist that the circle pass through a given
point; this is all that can be inferred initially from Euclid’s 3rd postulate.
(b) Proposition I.5 (part): A proof of Pons Asinorum is given in these notes,
§3.10.3, page 22. Proclus’s proof is discussed in §A.1.1, page 1002.
Information for Students in Math 348 2003 09
1023
2. Answer the following questions for each of these strip patterns109 :
· · · ∀∀∀∀∀∀ · · ·
· · · ∀∃∀∃∀∃∀∃∀∃ · · ·
· · · ∀¬∀¬∀¬∀¬∀¬ · · ·
· · · ∀I∀I∀I∀I∀I∀I· · ·
· · · ∼ °∼ °∼ °∼ °∼ ° · · ·
· · · 6=∝6=∝6=∝6=∝6=∝ · · ·
· · · 6=÷6=÷6=÷6=÷6=÷ · · ·
· · · `a`a`a`a`a`a · · ·
(a) [5 MARKS] If the pattern has reflective symmetry in some vertical mirror,
describe all locations for vertical mirrors that are possible. For example, for
the pattern ...VVVVV... one would observe that there are two possible
locations for mirrors:
i. between two successive letters V, as ...VV|VVV...
|
ii. along the axis of symmetry of any one letter V, as ...VVVVV...
(b) [5 MARKS] If the pattern has rotational symmetry under a half-turn, describe
the locations of the centres of all half-turn(s).
(c) [5 MARKS] If the pattern has symmetry under translations, describe an interval whose images cover the entire strip, and that is as short as possible;
e.g. for the pattern ...UVUVUVUVUV... one such interval would be UV,
another would be VU; but UVUV would not be acceptable, because it is
not as short as possible.
(d) [5 MARKS] If the pattern has symmetry under a glide reflection, describe the
axis of the translation and a shortest possible interval whose images cover the
entire strip; e.g. in · · · LΓLΓL· · · one such interval would be LΓ.110
(e) [5 MARKS] Determine which line of the table reproduced below gives the
numbers of isometries of the given strip pattern, or argue that none of the
rows of the table has that property.
109
Each version of the test had two of these strip patterns.
The instructions in this part are ambiguous where they refer to a fundamental interval and an axis;
accordingly this question was graded “generously” in those problems where there was a glide reflectional
symmetry: if the part of the problem was otherwise correct, students were accorded full marks for the
part.
110
Information for Students in Math 348 2003 09
1024
Typical pattern
Ref(H) Ref(V) Trans
(i)
...EEEEE...
0
0
∞
(ii)
...bdbdbdbd...
0
0
∞
(iii)
...44444...
0
∞
∞
(iv)
· · · ;;;;; · · ·
0
0
∞
(v) · · · 4 5 4 5 45 · · ·
0
∞
∞
(vi)
· · ·∈∈∈∈∈ · · ·
1
0
∞
(vii)
· · · mmmmm · · ·
1
∞
∞
Symbol Meaning
Ref(H) Reflection in Horizontal mirror
Ref(V) Reflections in Vertical mirrors
Legend:
Trans Translations
HT
Half-Turns
GL
Glide Reflections (with nontrivial
HT
0
0
0
∞
∞
0
∞
GL
0
∞
0
0
∞
∞
∞
Abstract Group
C∞
C∞
D∞
D∞
D∞
C∞ × D1
D∞ × D1
translation)
Solution:
Frieze
· · · ∀∀∀∀∀∀ · · ·
· · · ∀∃∀∃∀∃∀∃∀∃ · · ·
· · · ∀¬∀¬∀¬∀¬∀¬ · · ·
· · · ∀I∀I∀I∀I∀I∀I· · ·
· · · ∼ °∼ °∼ °∼ °∼ ° · · ·
··· =
6 ∝6=∝6=∝6=∝6=∝ · · ·
··· =
6 ÷6=÷6=÷6=÷6=÷ · · ·
· · · `a`a`a`a`a`a · · ·
B.2.2
Axis of
Ref(H)
no
no
no
no
no
no
no
yes
Axes of
Ref(V)
∀, ∀|∀
none
none
∀, I
none
none
none
` | a, a | `
Fund.Interval
(Translation)
∀
∀∃
∀¬
∀I
∼°
6=∝
6= ÷
`a
Centres of
Half-Turns
none
none
none
none
∼, °
none
6=, ÷
`a, a`
Axis
(Glide-Reflection)
none
none
none
none
none
none
none
horizontal lines of `a
Isometry
Group
(iii)
(i)
(i)
(iii)
(iv)
(i)
(iv)
(vii)
The problems that were each on only one test
Version 1, #1 You are given a triangle with vertices A, B, C, in which E is a point
on side AB with the property that |AE| = 2|EB|, and F is a point on side AC
with the property that |AF | = 2|F C|.
1. [13 MARKS] Use vectors — no other method will be accepted — to prove
that EF is parallel to BC.
2. [12 MARKS] Determine, by using information from your vector proof of the
|EF |
previous part, the value of the ratio
.
|BC|
Solution: The information given concerned only the lengths of certain line segments; one should not infer from these statements the equality of certain vectors
until one knows that the line segments involved are parallel; more than that —
one needs to know that they are oriented in the same directions. Students were
not penalized for assuming one or other of the possible locations of the points E
Information for Students in Math 348 2003 09
1025
and F , but were expected to be consistent, and not to reverse directions without
justification.
There are two possible interpretations for the locations of E and F . The multiplicity of interpretations derives from the fact that the hypotheses concerned only
lengths, not directed lengths.
• E, F are respectively in the interiors of sides AB and AC: (This was
the intended interpretation.)
~ = EB
~ + BC
~ + CF
~
EF
1 ~
~ + 1 CA
~
=
AB + BC
3
3
1 ~
~ + BC
~
(AB + CA)
=
3
1 ~
~
CB + BC
=
3
1 ~
~
= − BC
+ BC
3
2 ~
=
BC
3
proving that
1. EF is parallel
¯
¯ to BC.
2
¯
2
~ | = ¯ BC
~ ¯¯ = 2 |BC|;
~ hence |EF | = 3 |BC| = 2 . Note that the quotient
2. |EF
3
3
|BC|
|BC|
3
is between real numbers — one cannot divide one vector by another!
• E, F are respectively in the extensions of sides AB and AC beyond
B and C: (While this was not the intended interpretation, it is still a valid
one.)
~ = EB
~ + BC
~ + CF
~
EF
1 ~
~ + 1 AF
~
=
EA + BC
2
2
1 ~
~ ) + BC
~
=
(EA + AF
2
1 ~
~
EF + BC
=
2
1 ~
~
⇒ EF
= BC
2
~ = 2BC
~
⇒ EF
proving that
Information for Students in Math 348 2003 09
1026
1. EF is parallel to BC.
¯
¯
2
BC|
|EF |
|
~ | = ¯¯2BC
~ ¯¯ = 2|BC|;
~
= |BC|
= 2 . Note that the quotient
2. |EF
hence |BC|
is between real numbers — one cannot divide one vector by another!
• Could the two interpretations be combined? That is, could one of
the points E, F be in the interior of a side, and the other point in
the exterior of the opposite side? Suppose, for example, that E is in the
interior of AB, but F is on AC extended. Then we would have
~ = EA
~ + AC
~ + CF
~
EF
2 ~
~ + CF
~
=
BA + AC
3
2 ~
~
=
BA + 2AC
3
2 ~
~
~ + 4 AC
=
(BA + AC)
3
3
2 ~
4 ~
=
BC + AC
3
3
~ would be parallel to BC only if AC
~ were also in the same direction,
Thus EF
which could happen only if the triangle ABC were “degenerate” — having all
of its vertices on the same line. This was certainly not intended, but could
~ |
EF
make sense, and would give values to the ratio ||BC|
~ , depending on where one
placed the points E, F .
Version 2 #1 The real plane is endowed with the usual orthogonal Cartesian coordinate system (with perpendicular coordinate axes), and A(0, 2), B(−3, 1), C(0, 0),
D(3, 1) are the vertices of a quadrilateral with sides AB, BC, CD, DA.
1. [12 MARKS] List all of the isometries of the quadrilateral. You may use either
of the following forms to specify each isometry:
µ
¶
A B C D
, where U, V, W, X are the
• You may use the form
U V W X
points A, B, C, D in some order.
• Alternatively, you may specify precisely which type of isometry you have
(reflection, rotation, etc.) and describe its action in a precise way so that
the reader can reconstruct the function unambiguously. In this case you
should give the isometry a name, which will be needed in the next part
of the problem.
2. [13 MARKS] Give a complete a Cayley table for the isometries, where the
function applied second is named in the labels for the rows (horizontal lines)
Information for Students in Math 348 2003 09
1027
of the table, and the function applied second is first in the labels for the
columns (vertical lines).
Solution:
1. The distance between vertices A and C is different from that between any
other pair of vertices; under any isometry A and C will either be kept fixed
or interchanged; likewise, B and D will either be both fixed or interchanged.
The isometries are as follows, described both in the “matrix” notation, and
by referring to the geometric properties of the isometry.
(a) The identity mapping is an isometry — itµis a rotation about
¶ the point
A B C D
(0, 1), through a zero angle. It is given by
.
A B C D
(b) The reflection in the y-axis is an isometry. It could also be described by
the formula (x, y)
¶ 7→ (−x, y). In the “matrix” notation it is given by
µ
A B C D
.
A D C B
(c) The reflection in the line y = 1 is an isometry. It could also be described
by µ
the formula (x, y)¶ 7→ (x, 2 − y). In the “matrix” notation it is given
A B C D
by
.
C B A D
(d) The half-turn about the point (0, 1) is an isometry. It could also be
described by the
µ formula (x, y)¶7→ (−x, 2 − y). In the “matrix” notation
A B C D
.
it is given by
C D A B
µ
(b)
µ
ι=
µ
A
µ A
A
µ C
A
C
A
A
B
B
B
D
B
B
B
D
C
C
C
B
C
A
C
A
D
D¶
D
C ¶
D
D ¶
D
B
¶
ι=
µ
A
µ A
A
µ A
A
µ C
A
C
A
A
B
B
B
B
B
D
B
B
B
D
C
C
C
C
C
B
C
A
C
A
D
D
¶
D
D ¶
D
C ¶
D
D ¶
D
B
¶
µ
A
A
B
D
C
C
A
µ A
A
µ A
A
µ C
A
C
B
D
B
B
B
D
B
B
C
C
C
C
C
A
C
A
µ
¶
µ
¶
D
B ¶
D
D ¶
D
B ¶
D
D
µ
D
B
A
C
B
B
C
A
A
µ C
A
µ C
A
µ A
A
A
B
B
B
D
B
B
B
D
C
A
C
A
C
C
C
C
¶
µ
¶
D
D ¶
D
B ¶
D
D ¶
D
B
µ
D
D
A
C
B
D
C
A
A
µ C
A
µ C
A
µ A
A
A
B
D
B
B
B
D
B
B
C
A
C
A
C
C
C
C
D
B
¶
¶
D
B ¶
D
D ¶
D
B ¶
D
D
Version 3 #1 For each of the following statements, indicate whether it is true or false
in general. If it is false, prove that it is false — if possible by describing an
example or a theorem in precise details. You should not expect part marks for
a false statement that you indicate to be true, or for a true statement that you
indicate to be false.
1. [5 MARKS] If n is any positive integer such that n > 6, it is not possible to
arrange n (distinct) points in the plane so that a straight line through any
two of them must pass through a third.
UPDATED TO November 27, 2003
Information for Students in Math 348 2003 09
1028
2. [5 MARKS] Let n be any positive integer greater than 6. If n (distinct)
points in the plane are not all on one straight line, then there exists a line
which contains exactly three of them.
3. [5 MARKS] Let n be any positive integer greater than 6. If n (distinct) points
in the plane are not all on one straight line, then there exists no line which
contains exactly three of them.
4. [5 MARKS] If n > 6 (distinct) points in the plane are all on one straight line,
then there exists a straight line which contains exactly three of them.
5. [5 MARKS] There cannot exist 7 distinct points in the plane with the property
that any three of them lie on a straight line.
Solution:
1. Since Sylvester’s problem was resolved positively [4, §4.7, p. 65], we know that
a set of n points will have the property that any line through 2 of them passes
through a third only if all the points lie in one straight line. The restriction
that n > 6 is irrelevant. But, since the given statement does not exclude the
possibility that the n points lie in one line, that statement is FALSE. (The
counterexample is any set of n distinct points in a line in the plane.)
2. This statement is FALSE for every integer n ≥ 3, in particular, for every
integer n > 6. In providing a sequence of counterexamples we are giving
much more than is needed to disprove the claim: since the given statement
speaks of any integer n — i.e. it is a statement that could be symbolized with
the quantifier ∀n > 6, we have disproved the claim if we can provide one
value of n greater than 6 for which it fails; we will provide a counterexample
for every value of n > 6, and will even observe that our counterexample can
work for n = 3, 4, 5, 6 as well, had the author of the statement been audacious
enough to state it.
Take n distinct points on a circle. Since a line meets the circle in not more
than 2 points, no 3 of these points can be collinear.
But, for n ≤ 2, the claim would be true: it is impossible to place n ≤ 2 points
in the plane in such a way that they are not all on one straight line, and a
proposition of the form p → q, in which p is known to be false, is true.
3. The statement we are considering is of the form
∀n∀configurations¬∃line through 3 points .
This is logically equivalent to
¬∃n∃configuration∃line through 3 points .
Information for Students in Math 348 2003 09
1029
To show that it is FALSE we need only exhibit one value of n > 6 for which
there are n points in some configuration, not all collinear, so that exactly 3
of those points are collinear. For example, take 3 points on one line `1 , and
n − 3 points on another line, `2 , parallel to `1 . Thus the statement is FALSE.
4. Two lines that are not identical meet in a single point. If 2 points on a line
` are to be contained in a line `0 , then `0 = `. Thus, if 3 of n points of ` are
contained in `0 , then all n points are in `0 . Since n > 6, n 6= 3. The claim is
FALSE.
5. Note that no claim is made that any lines need to be distinct. The claim is
FALSE, since we could take 7 distinct points on a single line, and they have
the property that any 3 of them lie on that very line!
If the statement of the problem had required that each line contain exactly 3
of the points, then it would be false, although we haven’t discussed a proof of
that fact; the configuration of 7 points called the Fano geometry, which has
the property that any 3 points lie in a straight line, cannot be “embedded” in
the real plane.
Version 4 #3 For each of the following statements, indicate whether it is true or false.
• If you claim it is false, prove that it is false — if possible by describing an
example or a theorem in precise details.
• If you claim it is true, no proof is required. No part marks need be given for
a false statement that you indicate to be true, or for a true statement that
you indicate to be false.
[To say that a function preserves a set S of points is to say that the
function is an isometry of the plane which maps the set S on to itself.]
1. [5 MARKS] Any rotation which preserves a finite set of points in the plane
must have period 1, 2, 3, 4, or 6.
2. [5 MARKS] Any set of points in the plane must be preserved by some rotation
whose period is 1, 2, 3, 4, or 6.
3. [5 MARKS] The only sets of points in the plane that have an isometry of
period 2 are finite.
4. [5 MARKS] There is no set of points in the plane that has an isometry of
period 7.
5. [5 MARKS] If the symmetry group of a set of points in the plane contains
translations, then the entire set must be contained in a line.
Information for Students in Math 348 2003 09
1030
Solution:
1. This statement is FALSE. For example, consider 5 points equally spaced on
a circle in the plane, forming the vertices of a pentagon. Among the 10
symmetries of this finite configuration (forming the dihedral group, usually
denoted as D5 ), there are 5 rotations about the centre of the circle. One of
these is the identity rotation, through 0 radians. The other 4 rotations all
have period 5.
2. This is TRUE. The identity function has period 1, since 1 is the smallest
positive integer n such that ιn = ι. Also, ι is a rotation about any point one
wishes to choose as centre, through an angle of 0 radians. Any configuration
of points is invariant under ι, hence it is invariant under a rotation of period
1, hence “under some rotation whose periods is 1, 2, 3, 4, or 6.”
3. This is FALSE. Consider the reflection in the y-axis, given, in a plane coordinatized by an orthogonal cartesian system, by (x, y) 7→ (−x, y). The entire
plane is symmetric under this isometry, whose period is 2, since two reflections
yield the identity function. But the set of points in the plane is not finite.
4. This statement is FALSE. Consider the rotation about a point O through an
angle of 2π
radians. This rotation has period 7 — the 7th power is ι, but
7
any smaller power is not. One set of points with this symmetry is the set of
vertices of a regular heptagon (7-gon) centred at O. But, as the problem did
not require that the set be finite, we could also take the set of all points in
the plane, which is surely unchanged under this — or any other — bijection.
5. This statement is FALSE. Consider, for example, the points with integer
coordinates in a plane coordinatized by a rectangular cartesian coordinate
system. This set is not contained in a single line, but is symmetric under
translations. In fact, any function of the form (x, y) 7→ (x + m, y + n) where
m and n are each any integer, is a translation of this set of points.
B.3
Final Examination, December, 2002
1. (a) [5 MARKS] Using homogeneous coordinates, list all the points of the Fano projective plane (the plane constructed using the field F2 of residues
modulo 2).
(b) [5 MARKS] List all the lines of this geometry.
(c) [5 MARKS] List, for each of the points, all the lines with which it is incident.
Information for Students in Math 348 2003 09
1031
2. Let P be the set of even integers — positive, negative, and 0 — and let Q be the
set
{(`, m, n)|((`, m, n) ∈ P × P × P) ∧ ((` < m < n) ∨ (n < m < `))} .
Prove carefully that each of the following axioms (of “ordered geometry”) is satisfied.
(a) [1 MARK] |P| ≥ 2.
(b) [2 MARKS] ∀A∀B((A 6= B) ⇒ (∃C)((A, B, C) ∈ Q)) .
(c) [3 MARKS] ∀A∀B∀C(((A, B, C) ∈ Q) ⇒ (A 6= C))
(d) [3 MARKS] ∀A∀B∀C(((A, B, C) ∈ Q) ⇒ ((C, B, A) ∈ Q)) .
(e) [3 MARKS] ∀A∀B∀C(((A, B, C) ∈ Q) ⇒ ((B, C, A) ∈
/ Q)) .
(f) [3 MARKS] If A and B are distinct points, and C and D are distinct points
in L(A, B), then A ∈ L(C, D). (You are expected to know the meaning of the
notations L(A, B), L(C, D).)
3. [15 MARKS] Let R1 , R2 , R3 , and R4 be reflections, respectively in the x-axis, the
line x − y = 1, the y-axis, and the line x − y = 1. Using the fact that a rotation
can be expressed as a product of reflections, show carefully how one can express
R1 R2 R3 R4 (where R1 is applied first, and R4 last) as a translation.
4. Answer the following questions for each of the following strip patterns composed
of equally spaced characters:
I. · · ·LOΓOLOΓOLOΓOLOΓOLOΓO· · ·
II. · · ·VΛΞVΛΞVΛΞVΛΞVΛΞVΛΞ · · ·
(a) [2 MARKS FOR EACH PATTERN] If the pattern has reflective symmetry
in some vertical mirror, describe all locations for vertical mirrors that are
possible. (Variations in the thicknesses of the letters are not intentional; for
example, V should be read as through it is V.)
(b) [2 MARKS FOR EACH PATTERN] If the pattern has rotational symmetry
under a half-turn, describe the locations of the centres of all such half-turn(s).
(c) [1 MARK FOR EACH PATTERN] If the pattern has symmetry under a
translation which is not the identity, describe an interval of minimum length
whose images under iterations of one translation cover the entire strip; e.g., for
the pattern ...UVUVUVUVUV... one such interval would be UV, another
would be VU; but UVUV would not be acceptable, because it is not as short
as possible.
Information for Students in Math 348 2003 09
1032
(d) [1 MARK FOR EACH PATTERN] Does the pattern have symmetry under a
horizontal reflection? You are expected to prove your claim.
(e) [1 12 MARKS FOR EACH PATTERN] If the pattern has symmetry under a
glide reflection, describe precisely how each type of character in the pattern is
mapped under a glide reflection of your choice; e.g., in · · · VΛVΛV· · · there
is a glide reflection under which any character V is mapped on to a Λ.
5. [10 MARKS] Give in complete detail the proof of Pappus that the interior angles
at the base of an isosceles triangle are equal.
6. For each of the following statements, indicate whether it is true or false.
• If you indicate that a statement is false, you must prove that it is false — if
possible by describing an example or a theorem in precise detail.
• If you indicate that a statement is true, you are not expected to provide a
proof.
• You should not expect part marks when a statement that you indicate to be
true is false; nor when a statement that you indicate to be false is true.
(a) [3 MARKS] For every non-identity element g of every group (G, ∗) there exists
a positive integer n such that the composition of n copies of g, g ∗ g ∗ . . . ∗ g ,
|
{z
}
n copies
is equal to the identity element.
(b) [3 MARKS] There exist sets S of points in the Euclidean plane such that there
are no isometries that preserve S.
(c) [3 MARKS] Let A, B, C be three distinct points in the Euclidean plane, not
lying in the same line. Then the medians of 4ABC all pass through the same
point.
(d) [3 MARKS] There cannot exist 13 distinct points in the Euclidean plane with
the property that any 4 of them lie on a straight line.
(e) [3 MARKS] The product of two reflections of the Euclidean plane is always
either a translation or a half-turn.
7. Describe in careful detail an example of each of the following:
(a) [3 MARKS] A map on the torus which has exactly 18 faces, all of the same
degree.
(b) [3 MARKS] A map on the sphere which has exactly 2 faces, both of degree 5.
Information for Students in Math 348 2003 09
1033
(c) [3 MARKS] A map on the sphere which has exactly 12 faces, all having the
same degree.
(d) [3 MARKS] A map on the sphere, having more than 4 vertices, where all faces
have degree 3.
(e) [3 MARKS] A finite projective plane containing exactly 57 lines.
B.4
Supplemental/Deferred Examination, May, 2003
1. (a) [5 MARKS] Using homogeneous coordinates, list all the points of the projective plane P G(3, 2), (the plane constructed using the field F3 of residues
modulo 3).
(b) [5 MARKS] Then list all the lines of the same geometry.
(c) [5 MARKS] Finally, list, for each of the points, all the lines with which it is
incident.
2. Let P be the set of odd integers — positive or negative — and let Q be the set
{(`, m, n)|((`, m, n) ∈ P × P × P) ∧ ((` < m < n) ∨ (n < m < `))} .
Prove carefully that each of the following axioms (of “ordered geometry”) is satisfied.
(a) [1 MARK] |P| ≥ 2.
(b) [2 MARKS] ∀A∀B((A 6= B) ⇒ (∃C)((A, B, C) ∈ Q)) .
(c) [3 MARKS] ∀A∀B∀C(((A, B, C) ∈ Q) ⇒ (A 6= C))
(d) [3 MARKS] ∀A∀B∀C(((A, B, C) ∈ Q) ⇒ ((C, B, A) ∈ Q)) .
(e) [3 MARKS] ∀A∀B∀C(((A, B, C) ∈ Q) ⇒ ((B, C, A) ∈
/ Q)) .
(f) [3 MARKS] If A and B are distinct points, and C and D are distinct points
in L(A, B), then A ∈ L(C, D). (You are expected to know the meaning of the
notations L(A, B), L(C, D).)
3. [15 MARKS] Let R1 , R2 , R3 , and R4 be reflections, respectively in the x-axis, the
line x + y = 1, the y-axis, and the line x + y = 1. Using the fact that a rotation
can be expressed as a product of reflections, show carefully how one can express
R1 R2 R3 R4 (where R1 is applied first, and R4 last) as a translation.
4. Answer the following questions for each of the following strip patterns composed
of equally spaced characters:
Information for Students in Math 348 2003 09
I.
II.
1034
· · ·??¡¿¿!??¡¿¿!??¡¿¿!??¡¿¿!??¡¿¿!· · ·
· · ·VOΛΞVOΛΞVOΛΞVOΛΞVOΛΞVOΛΞ · · ·
(a) [2 MARKS FOR EACH PATTERN] If the pattern has reflective symmetry
in some vertical mirror, describe all locations for vertical mirrors that are
possible. (Variations in the thicknesses of the letters are not intentional; for
example, V should be read as through it is V.)
(b) [2 MARKS FOR EACH PATTERN] If the pattern has rotational symmetry
under a half-turn, describe the locations of the centres of all such half-turn(s).
(c) [1 MARK FOR EACH PATTERN] If the pattern has symmetry under a
translation which is not the identity, describe an interval of minimum length
whose images under iterations of one translation cover the entire strip; e.g., for
the pattern ...UVUVUVUVUV... one such interval would be UV, another
would be VU; but UVUV would not be acceptable, because it is not as short
as possible.
(d) [1 MARK FOR EACH PATTERN] Does the pattern have symmetry under a
horizontal reflection? You are expected to prove your claim.
(e) [1 12 MARKS FOR EACH PATTERN] If the pattern has symmetry under a
glide reflection, describe precisely how each type of character in the pattern is
mapped under a glide reflection of your choice; e.g., in · · · VΛVΛV· · · there
is a glide reflection under which any character V is mapped on to a Λ.
5. [10 MARKS] Give in complete detail Euclid’s proof of Proposition I.2, which shows
that, given a point P and a line segment AB, it is possible to find a point C
P C = AB.
6. For each of the following statements, indicate whether it is true or false.
• If the statement is false, prove that it is false — if possible by describing an
example or a theorem in precise details. You should not expect part marks
for a false statement that you indicate to be true.
• If the statement is true, and you so indicate, you are not expected to provide
a proof; but there will be no marks if you indicate that a true statement is
false.
(a) [3 MARKS] There exists in every group (G, ∗) containing at least 2 elements
a non-identity element g and a positive integer n such that the composition
of n copies of g, g ∗ g ∗ . . . ∗ g , is equal to the identity element.
|
{z
}
n copies
Information for Students in Math 348 2003 09
1035
(b) [3 MARKS] For some sets S of points in the Euclidean plane every isometry
that preserves S is either a reflection or a half-turn.
(c) [3 MARKS] If T1 , T2 , T3 are three translations in the Euclidean plane, and
if there is a point P such that T1 T2 T3 does not move P , then T1 T2 T3 is the
identity mapping.
(d) [3 MARKS] If there exist 13 points in the Euclidean plane, not all in the
same straight line, which have the property that every line containing 2 of
them contains at least 3 of them, then every isometry of the plane is either a
rotation or a translation.
(e) [3 MARKS] The product of two glide reflections of the Euclidean plane is
always either a translation or a rotation.
7. Describe in careful detail an example of each of the following:
(a) [3 MARKS] A map on the torus which has exactly 1 face.
(b) [3 MARKS] A map on the sphere which has more than 5 faces, none of them
of degree 3 or of degree 5.
(c) [3 MARKS] Three distinct lines in the Real Projective Plane that pass through
the same point at infinity.
(d) [3 MARKS] Three distinct points in the Real Projective Plane that lie on the
line at infinity.
(e) [3 MARKS] A group of isometries of the Euclidean plane which contains
exactly 7 reflections.
Information for Students in Math 348 2003 09
2001
• References to the following sources are often given in the notes for completeness.
Students are not expected to look up sources, but may wish to do so out of curiosity
or a thirst for knowledge.
• The entries in this list may not be in alphabetical order. As the notes are constructed, new entries will be added at the end, so as not to upset the earlier
numbering of references.
C
References
[1] I. Anderson, Combinatorics of Finite Sets. Clarendon Press, (Oxford, 1987). ISBN
0-19-853367-5.
[2] F. Bachmann, Aufbau der Geometrie aus dem Spiegelungsbegriff , 2nd edition
(1973). Springer-Verlag, Berlin. ISBN 3-540-06136-3.
[3] Lewis Carroll [Charles Lutwidge Dodgson] Through the Looking-Glass.
[4] H. S. M. Coxeter, Introduction to Geometry, Second Edition. John Wiley and Sons
(1969). ISBN 0-471-18283-4.
[5] H. S. M. Coxeter, A problem of collinear points. American Mathematical Monthly
55 (1948) 26-28; MR 9, p. 458.
[6] P. Erdős, Gy. Szekeres, A combinatorial problem in geometry, Compositio Math.
2 (1935), 463–470.
[7] L. Euler, Solutio problematis ad geometriam situs pertinentis. Reprinted from Commentarii academiæ scientiarum Petropolitanæ 8 (1736), 1741, pp. 128–140. Printed
as an appendix to [10].
[8] D. W. Farmer, Groups and Symmetry. A Guide to Discovering Mathematics. Mathematical World, Volume 5. American Mathematical Society, Providence, R. I.
(1995), ISBN 0-8218-0450-2.
[9] Heath, T. L.The thirteen books of Euclid’s Elements, translated from the text of
Heiberg, with introduction and commentary, 2nd edition, revised with additions.
Dover Publications, New York, paperbound. Volume 1, xi + 432 pp., ISBN 48660088-2; Volume 2, 436 pp., ISBN 486-60089-0; Volume 3, 564 pp., ISBN 48660090-4.
Information for Students in Math 348 2003 09
2002
[10] D. König, Theorie der endlichen und unendlichen Graphen. Kombinatorische
Topologie der Streckenkomplexe. (Theory of finite and infinite graphs. Combinatorial topology of 1-dimensional complexes.) Akademische Verlagsgesellschaft M. B.
H., Leipzig (1936); Chelsea Publishing Company, New York (1950). A translation
into English has been published by Birkhäuser Verlag (1990), ISBN 0–8176–3389–8.
Another reprint of the original version, published by Teubner Verlagsgesellschaft,
Leipzig (1986), ISBN 3-211-95830-4, has Euler’s paper as an appendix (cf. [7]).
[11] G. R. Morrow, Proclus, A Commentary on the First Book of Euclid’s Elements,
Princeton University Press, Princeton (1970). ISBN 0-691097160-8.
[12] K. H. Rosen, Discrete Mathematics and its Applications, 4th Edition, (McGrawHill, Inc., 1999), ISBN 0–07-289905-0.
[13] S. Stahl, Geometry, from Euclid to Knots Pearson Education Inc,m Upper Saddle
River, NJ, 2003. ISBN 0-13-032927-4.
[14] R. Steinberg, Solution to Problem 4065 , American Mathematical Monthly 51
(1944), 169-171.
[15] W. A. Whitworth, Choice and Chance, (Hafner Publishing Company, New York,
1965). (Enlarged reprint of the original edition, published in 1901)
[16] Ambrose Bierce, The Devil’s Dictionary, 1911.
[17] Marshall Hall, Jr., The Theory of Groups. Macmillan, (New York 1959).