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HIGH SCHOOL MATH BOOKS PDF DOWNLOAD FREE
HIGH MATH BOOKS SCHOOL
Join Classmates.com for Free! Math Mammoth offers affordable, yet quality math worktexts and workbooks for elementary and middle school (plus some supplemental materials for high school…. ClassZone Book Finder. Math League Contest Problem Books contain the actual math contests given to students participating in Math League contests. high school math books
BOOKS MATH SCHOOL HIGH
ClassZone Book Finder. Reconnect with high school friends, browse the biggest database of online yearbooks and/or plan high school reunions. Use these free math worksheets for homework assignments and to high school math books reinforce concepts, skills, …. Math League Contest Problem Books contain the actual math contests given to students participating in Math League contests.
HIGH MATH BOOKS SCHOOL
William Allen High School was established in 1858 when Mr. Math Mammoth offers affordable, yet quality math worktexts and workbooks for elementary and middle school (plus some supplemental materials for high school…. The Music Department fundraiser runs from 8:00-11:00 a.m. Math League Contest Problem Books high school math books contain the actual math contests given to students participating in Math League contests.
BOOKS MATH SCHOOL HIGH
Includes a photo album, school calendar, school policies, career center and information about sports and clubs Join our mailing list to get fun math high school math books in your inbox every day History Origins. Excel High School Online High School Courses for Students and Adults online high school diploma program Math explained in easy language, plus puzzles, games, quizzes, worksheets and a forum. Use these free math worksheets for homework assignments and to reinforce concepts, skills, …. For K-12 kids, teachers and parents.
HIGH SCHOOL MATH BOOKS
High school teachers generally specialize in a subject, such as English, math, or science Pearson Prentice Hall and our other respected imprints provide educational materials, technologies, assessments and related services across the secondary curriculum Stuyvesant High School / ˈ s t aɪ v ə s ən t /, commonly referred to as Stuy / ˈ s high school math books t aɪ / or Stuyvesant, is the most selective school of high school math books the nine specialized. HHS is a 10th-12 th-grade campus where educators are committed to raising academic standards, lowering drop-out. Washington High School provides an academically rigorous education within an environment of multicultural diversity and develops all students' critical. Everything a teacher needs to supplement math materials. Selected Internet links for Elementary School students: | 677.169 | 1 |
Educational Aims Of Programme
Prepare high-level professionals in both computer science and mathematics who would be able to work as specialists in these subjects.
Programme Outcomes and Learning, Teaching and Assessment Strategies
Knowledge and Understanding
Key mathematical concepts and topics
How mathematics can be used to analyse and solve problems including those at an abstract level
Essential concepts, principles and theories relating to Computing
How the Computer Science theory is related to modelling and design of computer-based systems
The role of computing professionals within a company and the interactions that normally take place with other disciplines.
The application of technical knowledge in a commercial context at an appropriate level for the student's qualifications.
The standard of professional presentation and reporting skills required in industry and commerce.
Skills & Other Attributes
To abstract the essentials of problems and formulate them mathematically and in a symbolic form
To select and apply appropriate mathematical methods to solve problems including those at an abstract level
To be able to construct and develop logical mathematical arguments with clear identification of assumptions and conclusions
To present arguments and conclusions clearly and accurately
To specify, design and construct computer-based systems
The ability to work as a computing professional in a commercial environment, demonstrating a professional and responsible attitude.
The ability to work both independently and as an effective team member.
Transferable Skills
Footnotes | 677.169 | 1 |
Interactive
Info
Functions - Grade 8 - 8.F.A.1 | Tennessee Department of Education
Help students understand that a function is a rule that assigns exactly one output to each input. The graph of a function is the set of ordered pairs consisting of an input and the corresponding output. This interactive outlines Common Core State Standard 8.F.A.1. | 677.169 | 1 |
First Book in Arithmetic
Description: This book is designed to prepare the pupils for the intelligent mastery of the fundamental operations. Through the application of number to objects, an insight into number relations and the common operations is gained. The memorizing of facts is subordinate to the getting of ideas.
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Precalculus by David H. Collingwood, K. David Prince, Matthew M. Conroy The basic theme of this book is precalculus within the context of problem solving. The pace is faster than a high school class in precalculus, we aim for greater command of the material, especially to extend what we have learned to new situations. (2519 views)
Practical Mathematics by Cargill Gilston Knott - Chambers The aim has been to illustrate the use of mathematics in constructing diagrams; in measuring areas, volumes, strengths of materials ; in calculating latitudes and longitudes on the earth's surface ; and in solving similar problems. (7516 views) | 677.169 | 1 |
Calca (iPad) Calca is a powerful symbolic calculator that gives you instant answers as you type. It solves equations and simplifies complex expressions – it's a computer algebra | 677.169 | 1 |
9th Grade - Algebra 1
About This Course
The lessons in the Algebra I course offer a supplement to help you better understand what you learn in the classroom. Algebra plays some major and minor roles in work, play and life in general. Knowing how to figure out a 30% off sale price, how to get the best tax return, or how to measure the stats of your favorite ball players all require the use of algebra skills. This course provides easy-to-follow video lessons that explain how to use algebra in clear and simple terms. | 677.169 | 1 |
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MATH 060 - Advanced Algebra and Trigonometry
Students review basic concepts of algebra, including equations, inequalities and problem solving, relations, functions and transformations, linear and quadratic functions and inequalities, exponential and logarithmic functions, equations of second degree and their graphs, the trigonometric or circular functions, trigonometric identities, inverse functions and equations; triangles, systems of linear equations and inequalities, imaginary and complex numbers, polynomials and rational functions, sequences and series.
Prerequisite(s): Minimum grade of 65% in Math 11 with algebra or MATH 050 or suitable scores on the placement test | 677.169 | 1 |
is the consumable Meeting Book used during the teacher-student "meeting" part of the lesson where concepts are introduced, demonstrated and practiced. If teaching more than one student at the same level, you will only need one meeting book. This is included in the Home Study Kit, but available separately if you need to replace it reuse the program with the next child Homeschool Kit includes the student text, test book and test answers, solutions manual and Power Up workbook. The Power Up Workbook is consumable and offers practice with math facts, mental math and problem solving.
The activity pages referenced in this product are no longer available on Saxon's website, but you can download them here: Activity Pages
You have questions; he has answers. Art Reed, who worked for nine years advising and answering parents' questions for Saxon Publishers, has written an entire book to help you navigate the Saxon Ocean. Not only does he include the most frequently asked questions, but has separate sections tailored to every level of the curriculum from grades 4-12 (he does not address the primary levels in this book). He discusses everything from the changes between editions to which levels you can skip to student placement and what type of calculator is best to use, including the underlying philosophy and recommended implementation of the Saxon program. If you use the Saxon math program or are considering it an option, the advice here could save you both time and moneyWritten to correlate with the Prentice Hall Algebra 2 Classics Edition (by Smith), this resource from Memoria Press contains 15 quizzes and 16 tests as well as a final exam. Designed to be consumable (no permission is given to reproduce pages), there is plenty of white space on each page for the student to work out the problems. Please note that there are no answers in this resource; they are located in the Algebra II Quizzes & Test Key, item #44877. | 677.169 | 1 |
Targeting algorithms and information buildings that experience confirmed to be flexible, effective, basic, and simple to enforce, this ebook familiarizes scholars, in addition to practitioners within the box of special effects, with a variety of info buildings.
This booklet serves readers who are looking to verify and enhance the mathematics talents they should be successful at paintings. an entire evaluation of mathematics, algebra, geometry, and observe difficulties guarantees development of those crucial math abilities. With over two hundred on-the-job perform questions, it's a valuable device for staff who want math talents to accomplish their jobs conveniently.
The instructing of arithmetic has gone through large alterations in process, with a shift in emphasis from rote memorization to buying an figuring out of the logical foundations and method of challenge fixing. This ebook bargains assistance in that path, exploring arithmetic's underlying innovations and their logical improvement.
To explain precisely how the test works, it will be helpful to introduce a few definitions and then to use the axiomatic systems of the preceding chapter to illustrate the test. , ............. introduced as Axiom Set 1 and Axiom Set 2 are "abstract systems" as long as the terms "point" and "line" are taken as undefined. As long as these terms remain undefined the axioms are open sentences. It is not until some meaning is given to the undefined terms that one may legitimately ask whether the axioms are }rue or false.
2. If Z is a line, then there exist at least three points on it. 3. If Z is a line, then there exists a point P not on it. 4. There exists at least one line. 5a. If Z is a line and P a point not on it, then there exists at least one line m through P with no point in common with Z. 5b. If Z is a line and P a point not on it, then there exists at most one line m through P with no point in common with l. Axio m Set 2 la. If P and Q are any two points, then there exists at least one line containing both P and Q. | 677.169 | 1 |
Mathematical Methods for Engineers and Geoscientists free download
This fascinating work makes the link between the rarified world of maths and the down-to-earth one inhabited by engineers. It introduces and explains classical and modern mathematical procedures as applied to the real problems confronting engineers and geoscientists. Written in a manner that is understandable for students across the breadth of their studies, it lays out the foundations for mastering difficult and sometimes confusing mathematical methods. Arithmetic examples and figures fully support this approach, while all important mathematical techniques are detailed. Derived from the author's long experience teaching courses in applied mathematics, it is based on the lectures, exercises and lessons she has used in her classes. | 677.169 | 1 |
Friday, July 13, 2007
Math4mobile has some interesting maths midlets for use on your Java-based phone. Currently they offer five different midlets for various graphing and learning purposes.
Graph2Go is a special-purpose graphing calculator that operates for given sets of function expressions.
Solve2Go supports solving equations and inequalities conjectures based on visual thinking. Conjectures can be refuted or supported by examples provided by the tool, and should be proved using symbolic manipulations on paper.
Sketch2Go is a qualitative graphing tool. Graphs are sketched using seven icons representing constant, increasing, and decreasing functions that change at constant, increasing, or decreasing rates.
Fit2Go is a linear and quadratic function graphing tool and curve fitter. Students can view a phenomenon, identify variables, conduct experiments and take measurements in order to construct models of the phenomena.
Quad2Go is a handy tool for learning about quadrilaterals by generating examples, observing, and experimenting with examples with a view toward forming generalized conjectures.
They include Java emulators so you can test them out in your browser before you decide to download to use on your phone. | 677.169 | 1 |
Vital Elements For Essay Help – An Intro
Algebra 2 is a step above Algebra 1. Before starting out this stage in the instructional process, students need to be thoroughly unveiled to logarithms and exponents, graphic functions, ways of solving inequalities and equations using complex numbers.
Career prospects made attractive with Algebra 2.Men and women that excel in Algebra get great career prospects. They might not be aware of this, but they can get really high salaried jobs. They can even work with any educational institute and spread their knowledge.
The GED quiz earns the taker the same as a high school diploma, that's necessary for students who can not complete their high school curriculums. The university entrance checks are the SAT and TAKE ACTION. Students need not bother to seek out a personal teacher. All the problems are easily resolved with the resources found at the Algebra 2 homework help site.
Benefits associated with the Algebra 2 Studying help platform. This platform's motto is to make math concepts easier for students. Just by availing themselves of this service, students can get rid of that complexities of the subject. Algebra 2 homework help provides tips for the effective analyze of mathematics. The students have to possess sharp random access memory skills in order to excel in such a section of math.
This process helps in case young people want to take up mathematics with regard to further higher studies. It helps students gather a sound skills required in order to deal with sophisticated mathematical problems at better levels. It is as convenient and friendly as troubleshooting.
Help for competitive examinations Algebra 2 homework help is a really effective platform when preparing with regard to competitive examinations or a university entrance examination. A base in this category of mathematics is mostly a must when a student is applying for the General Educational Development examination.
The help book provides certain very interesting ways of making random access memory skills sharper to assist the learning process. This is also helpful for students who do not don't forget the elementary lessons of Algebra. It is not possible to provide a detailed version here, nevertheless a basic reference is available. This reference helps young people brush up their memory to remind them with the lessons that are learned with Algebra 1.
The course structure also includes polynomial arithmetic, rational expressions, radicals and complex numbers, quadratic system and cone sections. As new terms, these kind of words might sound a little tricky. However, Algebra 2 assignments help becomes a personal help to make complex mathematics better for students.
Algebra 2 homework help trains young people to first understand the problem and then locate the most powerful way to solve it. For instance, when looking to solve a particular equation, the first step is to observe the number of terms in the offered equation. The next step, then, is to decide which type of factoring to opt for to solve that equation. Algebra 2 studying help also works for a student's critical thinking capability. | 677.169 | 1 |
Grade 12 maths Here is a list of all of the maths skills students learn in grade 12! These skills are organised into categories, and you can move your mouse over any skill name to preview the skillPractise back Maths back Grade 8 Grade 9 Grade 10 Grade 11 Grade 12 Science back Grade 10 Grade 11 Grade 12 Textbooks back Maths back Grade 10 (2012) Grade 10 (2015) Grade 11 Grade 12 Maths Literacy Science back Grade 10 Grade 11 Grade 12 Life Sciences Buy back Learners and Parents Teachers and Schools Help back For Learners For Teachers Log In Sign Up Menu Practise Maths Grade 8 Grade 9 Grade 10 Grade 11 Grade 12 Science Grade 10 Grade 11 Grade 12 Textbooks Maths Mathematics Grade 10 (2012) Mathematics Grade 10 (2015) Mathematics Grade 11 Mathematics Grade 12 Mathematical Literacy Grade 10 Science Physical Sciences Grade 10 Physical Sciences Grade 11 Physical Sciences Grade 12 Life Sciences Grade 10 Buy For Learners and Parents For Teachers and Schools Help For Learners For Teachers Log In Sign Up This course builds on students' previous experience with functions and their developing understanding of rates of changeThese vector products will be revisited to predict characteristics of the solutions of systems of lines and planes in the intersections of lines and planesThe volume of the water is controlled by the pump and is given by the formula: [begin{align} V(d)&=64+44d-3d^{2} text{where } V&= text{ volume in kilolitres} d&= text{ days} end{align}] Determine the rate of change of the volume of the reservoir with respect to time after 8 daysVirtual High School can also have a positive influence on students by modelling the behaviours, values, and skills that are needed to develop and sustain healthy relationships, and by taking advantage of teachable moments to address immediate relationship issues that may arise among studentsThe indeterminate form of a limit involving factoring, rationalization, change of variables and one sided limits are all included in the exercises undertaken next in this unitWhat Colleges Say "There is no question that taking Calculus is important for those students seeking admission to Haverford or other selective institutions," says Jess Lord, dean of admission and financial aid at Haverford CollegeDiversity is valued, and all members of the Virtual High School community feel safe, comfortable, and acceptedHelp for Learners Help for Teachers About Siyavula Blog Contact Us Follow Siyavula: Twitter Facebook All textbook content made available on this site is released under the terms of a Creative Commons Attribution License9 hours Trig Differentiation and Application A brief trigonometry review kicks off this unitNotice that the sign of the velocity is negative which means that the ball is moving downward (a positive velocity is used for upwards motion)The vertical velocity of the ball after 1,5 sLiteracy instruction takes different forms of emphasis in different subjects, but in all subjects, literacy needs to be explicitly taughtTeaching & Learning Strategies:The next section is dedicated to finding the derivative of relations that cannot be written explicitly in terms of one variableThe OHSA requires all schools to provide a safe and productive learning and work environment for both students and employeesWe should still consider it a functionHow Do Kids Benefit from Language-Immersion Programs? Bilingual education opens new avenues for students, including enhanced problem-solving and executive-functioning skills as well as new career pathsFind the vertices of the triangle such that the length of the hypotenuse is minimumCooperative education and other workplace experiences will broaden their knowledge of employment opportunities in a wide range of fieldsTeachers at Virtual High School will also guide students through the concept of ownership of work and the importance of copyright in all forms of mediaStudents will solve problems involving geometric and algebraic representations of vectors and representations of lines and planes in three dimensional space; broaden their understanding of rates of change to include the derivatives of polynomial, sinusoidal, exponential, rational, and radical functions; and apply these concepts and skills to the modelling of real-world relationshipsCommunication literacy refers to the ability to communicate information and to use the information obtained to solve problems and make decisions 79c7fb41ad | 677.169 | 1 |
Mathematics 1: Japanese Grade 10(Paperback)
Synopsis
The achievement of Japanese high school students gained world prominence largely as a result of their performance in the International Mathematics Studies conducted by the International Association for the Evaluation of Educational Achievement in the 1960s and 1980s. These textbooks (Mathematical World volumes 8 to 11) are intended to give U. S. educators and researchers a first-hand look at the content of mathematics instruction in Japan. The textbook, Japanese Grade 10 covers material that is compulsory. This course, which is completed by over 97 per cent of all Japanese students, is taught four hours per week and comprises algebra (including quadratic functions, equations, and inequalities), trigonometric functions, and coordinate geometry. Japanese Grade 11 is intended for the easier of the electives offered in that grade and is taken by about 40 per cent of the students. It covers probability and statistics, vectors, exponential, logarithmic, and trigonometric functions, and an introduction to differentiation and integration. The other 60 per cent of students in grade 11 concurrently take two more extensive courses using the texts Japanese Grade 11 Algebra and Geometry and Japanese Grade 11 Basic Analysis. The first consists of fuller treatments of plane and solid coordinate geometry, vectors, and matrices. The second includes a more thorough treatment of trigonometry and a more extensive introduction to differential and integral calculus | 677.169 | 1 |
In this unit, I explain how to write and solve single/multi-step equations, solving for various types of outcomes (no solution, 1 solution, all solutions), use ratios and proportions, and work with percents/percent of change (including discounts and
In this unit, I discuss how to solve and graph linear functions. We also discuss slope and how to find it; direct variation, arithmetic sequences and proportional vs non-proportional relationships tie in with all of these topics together.
In this unit, I discuss equations of linear functions. We tackle graphing and writing equations in slope-intercept form; what point-slope form is and how it is used; what parallel and perpendicular lines do graphically and how to recognize them
In this unit, we discuss linear inequalities. I observe and direct students how to use the 4 basic math functions (add, subtract, multiply, and divide) and properties of equality in solving these inequalities; we graph the results and interpret the
In this unit, we discuss systems of linear equations. I take students through what systems are and how to read them; how to solve them by graphing the equations; how to solve them algebraically through substitution and elimination; and how to read,
In this unit, we discuss scientific notation and exponents. I talk about the multiplication and division properties of exponents; how to use some shortcuts when solving problems with exponents; scientific notation and how to read it; and, how to
Ask Rob Glenn's Emporium of Mathematical Goods a question. They will receive an automated email and will return to answer you as soon as possible.
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TEACHING EXPERIENCE
Taught for 11 years in the following subjects:
Mathematics (elementary - Trigonometry)
Social Studies
Economics
Elementary Science
MY TEACHING STYLE
Flipped Classroom
Blended Learning Model
HONORS/AWARDS/SHINING TEACHER MOMENT
Being a teacher is honor enough.
MY OWN EDUCATIONAL HISTORY
Graduated from Shawnee Mission South HS; bachelors from Kansas State University in Elementary Education (K-9); masters in Education from Baker University; taught at St. Matthew's Elementary in Topeka, KS (4 yrs), then Lakewood Middle School in Salina, KS (2.5 yrs). Currently at Saint John's Military School in Salina, KS (5 years) | 677.169 | 1 |
This book, intended as a practical working guide for calculus students, includes 450 exercises. It is designed for undergraduate students in Engineering, Mathematics, Physics, or any other field where rigorous calculus is needed, and will greatly benefit anyone seeking a problem-solving approach to calculus. Each chapter starts with a summary of the main definitions and results, which is followed by a selection of solved exercises accompanied by brief, illustrative comments. A selection of problems with indicated solutions rounds out each chapter.
A final chapter explores problems that are not designed with a single issue in mind but instead call for the combination of a variety of techniques, rounding out the book's coverage.
Though the book's primary focus is on functions of one real variable, basic ordinary differential equations (separation of variables, linear first order and constant coefficients ODEs) are also discussed. The material is taken from actual written tests that have been delivered at the Engineering School of the University of Genoa. Literally thousands of students have worked on these problems, ensuring their real-world applicability.
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The book contains all the significant problems for a first year calculus course. Working on these problems would dramatically help you gain perfect grades from your calculus classse. (At least that happened to me and my friends who use it.) No matter you ...
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he purpose of this book is to explore the rich and elegant interplay that exists between the two main currents of mathematics, the continuous and the discrete. Such fundamental notions in discrete mathematics as induction, recursion, combinatorics, number...
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This text is intended for an honors calculus course or for an introduction to analysis. Involving rigorous analysis, computational dexterity, and a breadth of applications, it is ideal for undergraduate majors. The book contains many remarkable features: ...Adobe ColdFusion remains one of today's significant Web services tools and frameworks, and stands to become even more important as a possible primary tool for cloud development as well. As important as ColdFusion is and continues to become, we thought it ... recogni...
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The ancient world often thought in terms of physiognomics - the idea that character can be discerned by studying outward, physical features. That physical descriptions carry moral freight in characterization has been largely missed in modern biblical scho... persp... | 677.169 | 1 |
Matroids
Abstract
Matroids provide a successful connection between graph theory, geometry and linear algebra. Some of the dualities we will discuss later are rooted in the theory of matroids. Moreover, matroids provide a basis for discrete optimization. Several important algorithms, for instance the greedy algorithm, belong to the matroid world. We make a notational agreement in this chapter: the graphs are allowed to have loops and multiple edges. | 677.169 | 1 |
Approach to problem solving in mathematics
While mathematics falls under science, problem solving is an art. Just like in art, you get better and better as you do more; you get better at problem solving as you solve more and more problems. However, you need a structured way to solve problems to make it more effective and internalize the lessons learnt. Here I will list few steps and how students can apply these steps in problem solving for their board, and JEE main & advanced. This is again a very mathematics oriented approach. If this helps in Physics and chemistry, it is well and good.
Draw a diagram whenever possible
Chinese saw it long back when they said, "A picture is worth a thousand words". Is it the reason why mathematics Olympiad is dominated by Chinese students?
This becomes extremely important in coordinate geometry. I have seen reluctance from students from drawing the picture. They find it difficult to draw diagrams. This shouldn't be so. Take for example, drawing 3 normal lines on a parabola. Students dread this because the normal lines do not look like normal at all. But have you seen the real drawing in your book. Just check. The point from where 3 normal lines are drawn on the parabola is little farther from the vertex. Moreover, out of 3 normal lines, only one is in the same side of the point while 2 are on the other side. If you look carefully and keep these things in mind, drawing is not at all a problem. In fact it will make things quite easy for you. Similar pattern you can find in all the difficult drawings especially in conic section of coordinate geometry.
The same thing can be done in 3D even though drawing 3D is difficult. Draw only the relevant things that are asked. For example, if you have to draw a line on a plane, simply draw two perpendiculars line and assume that you are looking at it from the side view so that the plane looks like a line. This will help you draw.
Identify the purpose of the problem
The purpose of question is to find or prove something. It may be find the value of an expression, value of variable, prove or disprove a hypothesis. While this sounds quite obvious, it doesn't look so when you see interpretation of problem by many students. It is not that the language is difficult but the span of concentration is low in today's times where instant gratification is the most desired thing in the world. Read the question carefully and clearly note down what the question is asking.
Note down any information given about the quantities, both known and unknown
Information about any variable helps you set the boundary of your problem solving. If it says that the variable whose value is to be found is positive. You can remove all negative real numbers including 0 out of your options.
Note down the information and inferences that you can draw
Sometimes you have to draw the information by looking at the equations and problems. For example, if an equation of odd degree is given, you can be very sure that all its roots are not imaginary. If there is an equation of even degree with leading positive coefficient and negative last term, you can be sure that there are at least 2 real roots, one positive and one negative.
Give a name to all variables and constants in the problem
Once you have written down the unknown and known quantities, give those names (x, y etc.) which can be readily used in framing the equations and working through formulae.
Then use all related knowledge till you find sufficient equations for unknowns
Finally solve. Use all your knowledge to solve the question. Your goal is to solve the problem and get the right answer. While solving, keep your mind open for any new insight that you can get into the problem. Remember that solving mathematical problems is also a skill that you build by using formulae, processes, methods, and finally insight or intuition. Most of the students confuse this insight with tips and tricks. This insight or intuition comes from solving lot of problems of different types. Keep solving and discovering. | 677.169 | 1 |
COM 1503 BUSINESS MATHEMATICS
This course, introduces students to basic mathematical principles, laws and rules that are necessary to develop an overview of application capabilities of the subject matter in the field of business and economics. Business Mathematics course begins with brief review of arithmetic and algebra. It also covers functions, differentiation, partial derivatives, integration, and mathematics of finance. The course also includes the applications of differentiation and integration in business and economics.
To prepare students for the mathematical and analytical applications required for future courses in areas such as economics, statistics, and operations research. At the end of this course students are expected to have a thorough knowledge of mathematical tools and techniques relating to algebra, calculus, and financial mathematics.
4. Differentiation
4.1 Limit of a Function
4.2 Basic Rules for Differentiation
4.3 Successive Differentiation
4.4 The Sign of the Derivative and Nature of Function
4.5 Relative Maximum and Relative Minimum points of a Function
4.6 Applications of Differentiation
4.6.1 The derivative as a rate of change
4.6.2 Marginal Functions
4.6.3 Optimization of Functions in Business and Economics
4.6.4 Elasticity
5. Functions of Several Variables and Partial Derivatives
5.1 Functions of Several Variables
5.2 Partial Derivatives
5.3 Applications of Partial Derivatives
5.4 Second Order Partial derivatives
5.5 Optimization of Functions Two Variables | 677.169 | 1 |
Mathematics in Context
Mathematics in Context – The
Future of Mathematics
Education
Dr Toh Tin Lam
Associate Professor
National Institute of Education
Nanyang Technological University, Singapore
Outline of Presentation
• Mathematics Education in Singapore
• Mathematics in Context
• Application of Mathematics in the Real
World
• Making Decisions with Mathematics
• Conclusion
Mathematics Education in
Singapore
Mathematics Education in
Singapore
Aims and Objective of Mathematics Curriculum
• Acquire concepts and skills for continuous
learning in mathematics and to support learning in
other subjects;
• Develop thinking, reasoning, communication,
application and metacognitive skills through a
mathematical approach to problem solving;
• Connect ideas within and between mathematics
and other subjects through applications of
mathematics;
• Build confidence and foster interest in
mathematics.
Mathematics Education in
Singapore
• Students generally score very well in highstake national examinations
• Students perform very well in international
comparative studies (TIMSS and PISA)
– Do students see the relevance of math in reallife?
– Singapore students have no difficulty learning
mathematics?
– Some of the problems of Singapore students
in learning mathematics?
Mathematics Education in
Singapore
• Revisions of the mathematics curriculum;
• Problem solving (PS) at the center of the
curriculum framework;
• Modifications to the sub-components of
the five attributes of PS (represented by
the five sides of the pentagon);
– Modification as fine-tuning
Mathematics Education in
Singapore
• Since 2001, emphasize the application of
mathematics to solve real-world problem
at the primary level;
• At Secondary school level in 2007
emphasis on application and modeling;
• Solving real-world tasks – application of
mathematics in the real world.
Mathematics Education in
Singapore
Learning experience should provide
opportunities for students to
• enhance conceptual understanding through use of
various mathematical tools, including ICT tools;
• apply concepts and skills learnt in real-world
context;
• communicate their reasoning and connections
through various mathematical tasks and
activities;
• build confidence and foster interest in
mathematics (Ministry of Education).
Mathematics in Context
• Mathematics should be taught in context
• Emphasis should be made on application
of mathematics in real life.
Mathematics in Context
• Mathematics always exists as application
until the 18th century… emergence of
"pure mathematics".
• Pure mathematics – study of
abstract concepts irrespective
of the world and applications
in the world
Mathematics in Context
• Even the study of pure mathematics is
inspired from the world, and is an attempt
to generalize social and real-world
phenomena
Mathematics in Context
• When a child first learns mathematics …
Mathematics in Context
• Algebra – abstract?
Mathematics in Context
• Algebra – abstract?
Mathematics in Context
All mathematics concepts are motivated
from real-world problems…
Mathematics in Context
Mathematics in Context
Mathematics in Context
• Some abstract concepts / processes:
• What is the difference between x and x + 1?
– We could see that x + 1 is the process of adding 1 to
the variable x.
– We could also see that x + 1 denotes the concept of 1
more than x.
– Not easy for students to switch between the
mathematical process and mathematical concept.
Mathematics in Context
• …and some humor
Mathematics in Context
Mathematics in Context
• Contextualize mathematical concept
Mathematics in Context
• Contextualize mathematical procedure
2x + 1 = 5
Mathematics in Context
• Develop story of three boys' adventure in
Algebra Land…
Mathematics in Context
Mathematics in Context
• National Education …
Mathematical Reasoning
• What is relevant to everyday live
Mathematical Reasoning
• Financial Literacy
Mathematical Reasoning
• Critical mathematical reasoning…
Application of Mathematics
Application of Mathematics
• Oops… need to make profit.
• Need to find the cost price of the
various ingredients.
Needs knowledge of
the world
Needs to know the
type of data to collect.
Scaffold provided to
guide the students?
Application of Mathematics
Application of Mathematics
Application of Mathematics
• Whenever real-world context is brought
into a problem, the answer may not be
unique…
Making Decisions using
Mathematics
Woodlands
12000
10000
Yishun
Choa Chu Kang
15000
13000
16000
Ang Mo Kio
11000
11000
Toa Payoh
10000
9000
Clementi
Pasir Ris
6000
9000
Jurong
East
16000
10000
6000
Queenstown
Aljunied
13000
5000
Tampines
11000
Making Decisions using
Mathematics
• The Singapore government wants to
ensure that all the housing estates are
connected with their pipes. The cost of
building the pipes are shown. How would
you advise the government to connect all
the housing estates?
To minimize the cost, while keeping all the housing estates
connected
Making Decisions using
Mathematics
• Prim's Algortihm
– Start a starting vertex
– Connect it to the nearest vertex
– Connect the next "nearest" vertex
– Repeat the process until all the vertices are
connected.
Making Decisions using
Mathematics
• Kruskal's Algorithm
– Start with the smallest edge.
– Repeat the next edge with smallest edge;
include this if it does not form a cycle, and
exclude this if it forms a cycle.
– Repeat this process after all the edges are
included.
Making Decisions using
Mathematics
• The idea is NOT to teach students to
memorize these algorithms, but to develop
these (and even other) algorithms on their
own.
• Our main objective is to empower students to
learn, create their own understanding instead
of traditional memorizing of standard
knowledge.
• They need to have knowledge of the real
world – contextual knowledge.
Making Decisions using
Mathematics
Hexagon instead of Pentagon?
Conclusion
• In Context, Application, Making Decisions
all these activities need students to
have the knowledge of the real world
(contextual knowledge).
• Important steps of mathematical modelling
Movie… | 677.169 | 1 |
Mathematics
I have been volunteering to tutor secondary-school students for their matriculation exams. This led me to write learning materials and other documents for students and teachers. This page is divided into three parts:
Two-dimensional diagrams for motion and work problems
Motion and work problems are common in high-school mathematics. (Car 1 starts from city A at 10:00 while car 2 starts from city B at 11:00 ... Painter 1 starts work at 10:00 while painter 2 starts work at 11:00 ...). One typically solves motion problems by drawing the routes traveled. Both motion and work problems are also solved using tables.
This document shows how 2D diagrams of distance or work vs. time facilitate solving such problems. The diagrams are easy to draw and need not be to scale so they can be used when solving problems on exams.
The challenge of series
I find problems on series to be relatively easy, but some students find them difficult. This document analyzes the problems on series from 13 bagrut exams (806 questionnaire) showing techniques that can be used and pitfalls to watch out for.
Mistakes in solving mathematics problems
Textbooks in mathematics are "cleaned up" and do not show all the mistakes that were made when solving problems. This document presents my solutions to some problems from the 2014 and 2015 bagrut exams (806 questionnaire) including mistakes made and lessons to be learned from the mistakes.
Visualization of theorems of Euclidean geometry
This document presents some of the more difficult theorems of Euclidean geometry needed by high-school students of mathematics. The theorems are displayed in a purely visual manner using color and other notation. The idea is that visual memory and recall of geometrical theorems might prove more efficient than verbal memory.
How to do trigonometry without memorizing (almost) anything
Trigonometry facilitates geometric reasoning using algebraic computation. To the student, trigonometry can appear as a large set of obscure formulas to be memorized. This document shows that trigonometric identities can be obtained by geometric reasoning with little memorization. Although these formulas are easy to memorize, it is useful to see how they can be proved using only geometric facts.
The many guises of induction
Induction is often presented as a mechanical procedure for proving properties of sequences, but it widely used in other areas of mathematics (geometry, trigonometry, logic) and in computer science (data structures, automata, formal languages). This document tries to show that induction is a uniform concept although it appears in many guises.
How to (almost) square a circle
Given a circle, it is impossible to construct a square with the same area because the number pi is transcendental. There are rational numbers which are approximations of pi, in particular, 355/133=3.14159292. This note presents Ramanujan's construction of this number. The presentation is incremental and exercises ask the reader to perform the computations.
How to trisect an angle (if you are willing to cheat)
It is well know that it is impossible to trisect an arbitrary angle with a straighedge and compass. However, if you are willing to "cheat" and use other tools it can be done. This document shows how to trisect an angle using a simple tool, Archimedes neusis, and a more complex construct using Hippias's quadratrix. The quadratrix can also be used to square the circle.
√ x+5 = 5 - x2
Langford's problem
In the following arrangement of colored blocks:
There is one block between the red blocks, two blocks between the blue blocks and three blocks between the green blocks. Expressed in numbers, the bag of numbers {1,1,2,2,3,3} can be arranged in a sequence 312132 such that between the two occurrences of i there are i blocks. Langford's problem asks if this is always possible for {1,1,...,n,n}. Donald Knuth showed that solutions to Langford's problems can be easily found using SAT solvers, such as LearnSAT that I developed. | 677.169 | 1 |
Almost every year, a new book on mathematical modeling is published, so, why another? The answer springs directly from the fact that it is very rare to find a book that covers modeling with all types of differential equations in one volume. Until now. Mathematical Modeling: Models, Analysis and Applications covers modeling with all kinds of differential equations, namely ordinary, partial, delay, and stochastic. The book also contains a chapter on discrete modeling, consisting of differential equations, making it a complete textbook on this important skill needed for the study of science, engineering, and social sciences. More than just a textbook, this how-to guide presents tools for mathematical modeling and analysis. It offers a wide-ranging overview of mathematical ideas and techniques that provide a number of effective approaches to problem solving. Topics covered include spatial, delayed, and stochastic modeling. The text provides real-life examples of discrete and continuous mathematical modeling scenarios. MATLAB(R) and Mathematica(R) are incorporated throughout the text. The examples and exercises in each chapter can be used as problems in a project.
Since mathematical modeling involves a diverse range of skills and tools, the author focuses on techniques that will be of particular interest to engineers, scientists, and others who use models of discrete and continuous systems. He gives students a foundation for understanding and using the mathematics that is the basis of computers, and therefore a foundation for success in engineering and science streams | 677.169 | 1 |
Numerical Analysis
I'm willing to study (teach myself) [Broken], which I find very interesting, and which seems to be very useful for solving Physics problems (specifically, the motion of celestial bodies).
I have a background in Calculus (basically Limits, Derivatives and Integrals), but all my knowledge is self-taught.
My plan is to understand Numerical Analysis, specially Numerical Integration and Numerical Ordinary Differential Equations.
My question is: what background in Mathematics should I have in order to start doing this? In what sequence does it normally appear in courses?Thank you for your answer.
I found one exemplar of this book online, in a library here in Brazil, for a very cheap price. But it's a 1962 edition. Is it good?
Another question: What is the relation between Numerical Analysis and Perturbation Theory? Is the latter a subset of the former? | 677.169 | 1 |
Using Iteration as a Problem
The Algebra: the Complete Course series covers all Algebra core curriculum topics from a basic overview to the more complex algebraic functions. Dr. Monica Neagoy, consultant to the Annenberg Foundation & Public Broadcasting Service, uses concrete examples and practical applications to show how a mastery of fundamental algebraic concepts is the key to success in today's technologically advanced world. Students will learn the geometric approach to the Sierpinski Triangle and to use iteration as a powerful problem solving tool. | 677.169 | 1 |
Economics
Mathematics
This chapter on cube and cube roots provides the basic definition of a cube root, with the knowledge provided on the properties of the cubes, and how to use the prime factorization method to find the cube root of a number. The proper explanation has been provided for learning the process of finding the cube root by estimation method as well. The topic on how to find unit digit of the cube root of a number provides the required tricks to solve such problems. | 677.169 | 1 |
Author: Barcharts, Inc. Publisher: Quickstudy ISBN: 9781423208570 Size: 12.61 MB Format: PDF, ePub, Docs View: 4829 DownloadRead Online
Statistics problems can make the best students shudder as they near the classroom, but they need not worry any longer--QuickStudy is here to help!
Author: Frank R. Spellman Publisher: CRC Press ISBN: 1466586389 Size: 43.19 MB Format: PDF, Docs View: 856 DownloadRead Online
5.1 QUADRATIC EQUATIONS AND ENVIRONMENTAL PRACTICE A logical
question at this point might be why is the quadratic equation important in
environmental practice? The logical answer is that the quadratic equation is used
in environmental practice to find solutions to problems primarily dealing with
length and time determinations. Stated differently: The quadratic equation is a
tool, an important tool that belongs in every environmental practitioner's toolbox.
To the student of ...
Author: Isabel Willemse Publisher: Juta and Company Ltd ISBN: 9780702177538 Size: 26.28 MB Format: PDF, Docs View: 7568 DownloadRead Online
This third edition aims to equip students with the skills to apply statistical analysis and quantitative techniques in research and the working environment where their knowledge can lead to effective decision-making. | 677.169 | 1 |
CBSE 8th Class Maths
The CBSE board has carved a niche in the education field with its systematic and informative educational program. The wide subject coverage and enhanced education pattern of the CBSE board is instrumental in upgrading students aptitude in every subject. Under the able guidance of academic experts, the syllabus for each class is carefully designed as well as evaluated at regular intervals. The first and foremost objective of the CBSE board is to provide the best education to students across each grade. The board offers a revised math syllabus for class 8, which meets the latest educational requirements of students in the best possible manner.
The CBSE 8th class maths syllabus is designed in such a manner that it gives complete understanding to teachers about teaching methodology. Along with this, it also gives clarity to students about the 8th class CBSE math course structure. Further, the board also provides important question papers to students to make their exam preparation better and beneficial.
A well-structured syllabus gives a better understanding to students as well as instructors about the framework of the entire course. The CBSE 8th class math syllabus is designed and developed in line with current educational demands. With extensive research and strenuous efforts of academic experts, the board has been able to design a comprehensive math syllabus for class 8. Each and every important topic is judiciously included in the syllabus so as to provide students a valuable learning experience.
Math is a tricky and tough subject and needs more practice; therefore, it is always better for students to start their studies before the session begins. The CBSE syllabus for class 8th maths is available online and students can collect it for exam preparation.
The CBSE question paper is a tool to judge the students' academic performance in a detailed manner. With the collective effort of subject experts, the board designs math question paper as per the learning ability of students. The CBSE 8th class math question paper encompasses marking scheme and time distribution, which helps students to attempt questions wisely and within the given time-frame. Besides this, in the test paper, all questions are categorized into objective type and subjective type question patterns that help students to pen down their answers briefly as well as in elaborately.
Math is a scoring subject and it is advisable for students to attempt each question carefully. The CBSE class 8 maths question paper not only helps students to understand the marking distribution pattern but also gives a thorough understanding to them about important questions, which are often asked in the exams.
Sample papers are highly important learning resource for exam preparation. Right from types of question to marking scheme and time distribution pattern, a sample question paper gives a better clarity to students about the real test paper. With regular practice of CBSE class 8 math sample papers, students can enhance their knowledge in math subject and can also improve their pace at attempting questions. It is a proven fact that mathematics helps students to fetch maximum marks in exams; therefore, it is necessary for students to prepare for it through CBSE 8th class maths model question paper.
Sample papers are designed with the help of previous year question papers. With CBSE 8th class maths sample paper, students can attain their academic objectives in an appropriate manner. | 677.169 | 1 |
Wiley Pathways Business Math
4.11 - 1251 ratings - Source
You can get there Where do you want to go? You might already be working in a business setting. You may be looking to expand your skills. Or, you might be setting out on a new career path. Wherever you want to go, Business Math will help you get there. Easy-to-read, practical, and up-to-date, this text not only helps you learn fundamental mathematical concepts needed for business, it also helps you master the core competencies and skills you need to succeed in the classroom and beyond. The booka€™s brief, modular format and variety of built-in learning resources enable you to learn at your own pace and focus your studies. With this book, you will be able to: Understand the business uses of percent calculations. Solve business problems using algebraic equations. Learn why stores markup and markdown their inventory. Calculate different types of discounts. Examine different banking options. Compare personal, sales, and property taxes and the implications of taxing income, property, and retail sales. Calculate simple and compound interest and learn how each affects the future value of money. Explore the uses of promissory notes, mortgages, and credit cards and how to calculate the cost of each. Learn different ways to determine the loss of value of business property and equipment, and the effect of depreciation on taxes. Examine financial statements and learn how to read the income statement and the balance sheet. Learn how to calculate the mean, median, mode, and range of data. Wiley Pathways helps you achieve your goals When The books in this seriesa€"a€"Finance, Business Communication, Marketing, Business Math, and Real Estatea€"a€"offer a coordinated curriculum for learning business. Learn more at follows: 25 - (-10) + (-40) I 25 + 10 - 40 I -5 Finally, if you want to find the
answer to 8.2 - (-4.95), you do the following: 8.2 - (-4.95) I 8.20 + 4.95 I 13.15
these examples of adding and subtracting integers, which are set up as word
problems: 1.
Title
:
Wiley Pathways Business Math
Author
:
Steve Slavin, Tere Stouffer
Publisher
:
John Wiley & Sons - 2006 | 677.169 | 1 |
genius maker
Most students are looking for materials and courses helping in mathematics, physics and chemistry. Genius maker is a tool for exam preparation designed for high school students. It is developed by Golden K Star Private Ltd.
Key Features
Tutorial: Genius maker includes 34 tutorials on mathematics, physics and chemistry. We can find analytical geometry, matrix algebra, the study of radioactive decay, the periodic table, gas laws, etc. All of them are available in a few clicks.
Interactive software: the design of Genius maker is based on a game system. Students will understand the scientific phenomena and understand the formulas across different games. This method allows students to master several modules at once.
Interface: menus and buttons for each module were developed to facilitate Genius maker's handling. In this regard, users just have to click to move from one module to another. Modules (Mathematics, Physics and Chemistry) are also displayed directly on the interface and a click is enough to open them with exercises, lessons and games included within. | 677.169 | 1 |
Contents
Appendix B: List of Syntax Tree Nodes
This is a comprehensive list of all math operations that MathLex recognizes. In JavaScript, each node is represented as an Array, the 0th index of which is a string ID. Any additional information is stored in the indices as described by each node type below. | 677.169 | 1 |
A-level Mathematics
understand mathematics and mathematical processes in a way that promotes confidence, fosters enjoyment and provides a strong foundation for progress to further study
extend their range of mathematical skills and techniques
understand coherence and progression in mathematics and how different areas of mathematics are connected
apply mathematics in other fields of study and be aware of the relevance of mathematics to the world of work and to situations in society in general
use their mathematics knowledge to make logical and reasoned decisions in solving problems both within pure mathematics and in a variety of contexts, and communicate the mathematics rationale for these decisions clearly
reason logically and recognise incorrect reasoning
generalise mathematically
construct mathematical proofs
use their mathematical skills and techniques to solve challenging problems that require them to decide on the solution strategy
recognise when mathematics can be used to analyse and solve a problem in context
represent situations mathematics and understand the relationship between problems in context and mathematical models that may be applied to solve them
draw diagrams and sketch graphs to help explore mathematical situations and interpret solutions
make deductions and inferences and draw conclusions by using mathematical reasoning
interpret solutions and communicate their interpretations effectively in the context of the problem
read and comprehend mathematical arguments, including justifications of methods and formulae, and communicate their understanding
read and comprehend articles concerning applications of mathematics and communicate their understanding
use technology such a calculators and computers effectively and recognise when their use may be inappropriate
take increasing responsibility for their own learning and the evaluation of their own mathematical development. | 677.169 | 1 |
Area Through Integration Scavenger Hunt
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Product Description
This set of cards can be used to review using integrals to find area in both Calculus and AP Calculus. Each card has a problem displayed in the center of the card and an answer to another problem, in a magnifying glass, at the bottom right of the card. By using the 10 cards that have a ♡ at the bottom you will have 10 cards in the scavenger hunt. By using the 10 cards that have a ♢ you will have a second set of cards for a scavenger hunt. If you want to make a 20 card scavenger hunt, remove 10 ♡ and replace it with 22 (no symbol) and remove 18 ♢ and replace it with 21 (no symbol).
There are two complete sets of cards. One set has the graphs and bounded area identified and the other does not. You can mix and match the two sets of cards so some graphs are given and others aren't.
Suggestions for working with the cards:
1. Post the cards around the classroom (either set of 10 problems with a ♡, 10 problems with a ♢, or all the problems with a ♡ and with the two cards replaced).
2. Assign different students to start at different numbers. Distribute copies of the student recording sheet. Students should record the number of the card they are working on in the small square at the top left of the grid. When they have come up with an answer to the problem they then go on a scavenger hunt to find a new card with their answer in the bottom right corner of the card. If they can't find their answer they need to check their work again. When they find their answer on a card, they record that number in the next square on their recording sheet and solve the new problem. Students keep working on problem for the allotted time.
3. The last three cards show the cycle for the two 10 card scavenger hunts and the combined 20 card scavenger hunt.
4. The goal would be to complete as many questions as they can in the allotted amount of time.
5. Awards can be given for the number of problems completed by each student. | 677.169 | 1 |
Elementary differential equations rainville bedient Scarica Il eBook
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Key Questions
A matrix is a table of numbers (or other elements) disposed in rows ans columns.
For example: #A=##((1,2,3),(3,4,6))#
This is a rectangular matrix called #A# with #2# rows and #3# columns, so you say thart #A# has order#2xx3# (2 by 3).
You can have square matrices when the numbers of rows is equal to the number of columns: #B=##((1,2,3),(3,4,6),(3,2,1))#
Where #B# is a square matrix od order #3xx3# or simply #3#.
You can also have matrices with one column only or one row only : #C=##((1),(6),(7))# or #D=##((1,2,3,4,6))#
these are sometimes called vectors or column vector and row vector. | 677.169 | 1 |
A complete solution for Level 6 of the New Zealand Mathematics Curriculum, the Gamma Mathematics and Gamma Fundamentals cover all thirteen NCEA Level 1 Mathematics Achievement Standards. The material in the books has been selected so that schools can design courses that meet the needs of all students. Most students will enter for 6 or 7 Achievement Standards at most, so a double-book solution means less waste but there is some overlap in Number and Statistics so a narrow choice is not dictated to schools.
Gamma Fundamentals also covers nine achievement standards. It is the ideal text for students who want to gain NCEA Level 1 credits early, or who are not continuing with Maths after Year 11. The material in Gamma Fundamentals is internally assessed at NCEA. Gamma Fundamentals also has an associated Workbook for students to cover extra material in their own time. | 677.169 | 1 |
Algebra 1 Essential Standards
Standard 1.1
Standard 2.0
Description of Skills
Examples
Resources
I can simplify numerical expressions by finding opposities, taking roots,
and finding the reciprocal. I also know the exponent rule like:product of power, quotient of powers, power of products, and power of a power. I understand
what negative and zero exponents mean and can raise a number to a
fractional power. I can
simplify radical expressions and rationalize the demonimator of a fraction.
Standard 21.0
Standard 14.0
I can solve a quadratic equation by factoring or completing the square.
Study Guide
Standard 19.0
Description of Skills
Examples
Resources
I know the quadratic formula and know how it is derived by completing the square.
Study Guide
Standard 20.0
Description of Skills
Examples
Resources
I can solve quadratic equations by using the quadratic formula.
Study Guide
Standard 22.0
Description of Skills
Examples
Resources
I can I can determine whether a parabola will cross the x axis 0, 1, or 2 times
Study Guide
Standard 23.0
Description of Skills
Examples
Resources
I can use quadratic equations to solve problems involving
dropped objects or thrown objects that are affected by gravity.
Study Guide
Standard 10.0
Description of Skills
Examples
Resources
I can add, subtract,
multiply, and divide polynomials. I can also apply these skills to word problems.
Study Guide
Standard 11.0
Description of Skills
Examples
Resources
I can factor polynomials.
Study Guide
Standard 12.0
Description of Skills
Examples
Resources
I can simplify polynomial fractions by reducing them to lowest terms.
Study Guide
Standard 13.0
Description of Skills
Examples
Resources
I can add, subtract, multiply and divide rational expressions.
Study Guide
Standard 6.0/17.0/18.0
Description of Skills
Examples
Resources
I can find the doman and range of a graph, set of ordered pairs or symbolic
expression. I recognize that the domain is made up of x values (also known
as inputs or independent values) and the range is made up of y values
(also known as outputs or dependent values).
Statement of Non-Discrimination: Title VI of Civil Rights Act of 1964,
Title IX of the Education Amendments of 1972, Section 504 of the Rehabilitation
Act of 1972. The District does not discriminate against any person the
basis of gender, race, color, religion, national origin, ethnic group,
actual or perceived sexual orientation, marital or parental status, physical
or mental disability. The District will take steps to assure that the
lack of English will not be a barrier to admission or participation in
District programs. Complaints alleging noncompliance with this policy
of nondiscrimination should be directed to Sergio Mendoza. A copy of the
District's Uniform Complaint Policy is available at 559.793.2452. The
District's Title IX Coordinator at 559.793.2445or the District's 504 Coordinator
at 559.793.2473. | 677.169 | 1 |
Introduction to Real Analysis
4.11 - 1251 ratings - Source
This text is a single variable real analysis text, designed for the one-year course at the junior, senior, or beginning graduate level. It provides a rigorous and comprehensive treatment of the theoretical concepts of analysis. The book contains most of the topics covered in a text of this nature, but it also includes many topics not normally encountered in comparable texts. These include the Riemann-Stieltjes integral, the Lebesgue integral, Fourier series, the Weiestrass approximation theorem, and an introduction to normal linear spaces. The Real Number System; Sequence Of Real Numbers; Structure Of Point Sets; Limits And Continuity; Differentiation; The Riemann And Riemann-Stieltjes Integral; Series of Real Numbers; Sequences And Series Of Functions; Orthogonal Functions And Fourier Series; Lebesgue Measure And Integration; Logic and Proofs; Propositions and Connectives For all readers interested in real analysis.Manfred Stoll. To. the. Student. The difference between a course on calculus and
a course on real analysis is analogous to the difference in the approach to the
subject prior to the nineteenth century and ... I have included hints and answers
to selected exercises at the end of the text; these are indicated by an asterisk (*).
Title
:
Introduction to Real Analysis
Author
:
Manfred Stoll
Publisher
:
Addison-Wesley | 677.169 | 1 |
Transforming Functions
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Transforming Functions (Reflections, Stretches, Translations)
A set of functions have been transformed and students must work out the missing pieces of information. In some cases they must work out the original equation, or the transformation that occurred, or the resulting equation afterwards. | 677.169 | 1 |
Synopses & Reviews
Publisher Comments
Everything you need to know–basic essential concepts–about calculus
For anyone looking for a readable alternative to the usual unwieldy calculus text, here's a concise, no-nonsense approach to learning calculus. Following up on the highly popular first edition of Understanding Calculus, Professor H. S. Bear offers an expanded, improved edition that will serve the needs of every mathematics and engineering student, or provide an easy-to-use refresher text for engineers.
Understanding Calculus, Second Edition provides in a condensed format all the material covered in the standard two-year calculus course. In addition to the first edition's comprehensive treatment of one-variable calculus, it covers vectors, lines, and planes in space; partial derivatives; line integrals; Green's theorem; and much more. More importantly, it teaches the material in a unique, easy-to-read style that makes calculus fun to learn. By explaining calculus concepts through simple geometric and physical examples rather than formal proofs, Understanding Calculus, Second Edition, makes it easy for anyone to master the essentials of calculus.
If the dry "theorem-and-proof" approach just doesn't work, and the traditional twenty pound calculus textbook is just too much, this book is for you.
Review
"...expands coverage to vectors and calculus of several variables...plenty of worked out problems..." (American Mathematical Monthly, August/September 2003)
"...material included is well formulated and approachable...recommended." (Choice, Vol. 41, No. 1, September 2003)
Synopsis
Mathematics Understanding Calculus A User's Guide A volume in the IEEE Press Understanding Science & Technology Series The subject of calculus does not usually evoke adjectives like "simple" and "concise." Traditional calculus texts have involved a more comprehensive, theoretical approach than is appropriate for those learning this critically important subject for its utilitarian value. This, but who need to utilize calculus for academic or professional advancement.
Synopsis
"The subject of calculus does not usually evoke adjectives like ""simple"" and ""concise."" Traditional calculus texts have involved a more comprehensive, theoretical approach than is appropriate for those learning this critically important subject for its utilitarian value.
This but who need to utilize calculus for academic or professional advancement."
Sponsored by:
IEEE Education Society
About the Author
About the Author H. S. Bear is a prolific author who has published several pre-calculus texts and an intermediate-level differential equations text, in addition to numerous research articles, during his long writing career. His most recent works include two publications on mathematical analysis, A Primer of Lebesgue Integration (Academic Press. 1995) and An Introduction to Mathematical Analysis (Academic Press, 1997). A dedicated educator, Dr. Bear has taught at several large universities, but has spent most of his career at the University of Hawaii, where he served as both department chairman and graduate chairman. | 677.169 | 1 |
Transcription
1 Linear Programming Linear programming refers to problems stated as maximization or minimization of a linear function subject to constraints that are linear equalities and inequalities. Although the study of algorithms for these problems is very important, the term programming in linear programming arises from the context of a particular resource allocation program for the United States Air Force for which George Dantzig developed a linear model for and described a method, called the Simplex method, to solve. This was in the late 1940 s before the term computer programming came into general use. Consider the following example of a linear programming problem. In general, a linear programming problem is a maximization or minimization of a linear function subject to constraints that are linear equations or linear inequalities. Some of these may be bounds on variables. For example it is not unusual to require that variables be non-negative in applications. If there is no such constraint a variable will be called free. A linear programming problem is called feasible if there is some solution satisfying the constraints and infeasible otherwise. If the maximum can be made arbitrarily large the the problem is unbounded We will often use matrix notation. This instance becomes max cx s.t. Ax c where the matrix A, cost vector c and right hand side b are given and x is a vector of variables. For this example we have x 1 A = c = [ ] b = 3 6 x = x 2 x x 4 We will also write generic instances using sigma notation as follows: n max c j x j s.t. n a ij x j b i for i = 1, 2,..., m. If we want to minimize cx we can instead maximize cx. We can replace equations Ax = b with inequalities Ax b and Ax b. To reverse the process, we can replace Ax b with Ax + Is = b and s 0 where I is an appropriate size identity matrix and s is a vector of slack 1
2 variables which are non-negative. To replace free variables with non-negative variables we use x = u v where u0 and v 0. Alternatively we can write non-negativity constraints as simply another inequality. Using the transformations described above we can convert any linear programming instance to one of three standard forms. It will be convenient to be able to refer to each form with the understanding that any instance can be converted to that form. The forms are max{cx Ax b} max{cx Ax b, x 0} max{cx Ax = b, x 0} We will later see examples of converting between the forms in the context of Farkas Lemma and duality. Variations on Linear Programming We have seen that we might consider various variants of linear programming problems and indicated how they are equivalent. However, the case where there are only linear equalities and all variables are free is different. We will call this a system of equations. Linear programming will refer to any problem where there is either at least one non-negative variable or at least one linear inequality. For both systems of equations and linear programming having an additional constraint that at least one variable must be integral again changes things. Thus we can distinguish four variants that are qualitatively different: Systems of Equations, Integral Systems of Equations, Linear Programming, Integer Linear Programming. We can also distinguish between feasibility problems find a solution to some linear system and optimization problems maximize or minimize a linear function over the set of solutions to some linear system the feasible set. If we have a method for solving the optimization problem then we can easily solve feasibility using the same method. Simply maximize the zero function. Interestingly, feasibility and optimization are equivalent in each of the four cases in the sense that a method for to solve the feasibility problem can be used to solve the optimization problem as will discuss below. We consider each of the four types in turn. In each case, if there is no feasible solution then the optimization problem is infeasible. So, here we will assume feasibility when discussing optimization. Systems of Equations. Determining {x Ax = b} feasibility and max{c T x Ax = b} optimization is the material of basic linear algebra and the problems can be solved efficiently using, for example Gaussian elimination. The optimization problem is not usually discussed in linear algebra but follows easily from feasibility. The set of solutions to {x Ax = b} can be written as x 0 + λ i v i where the sum is over a set {v i } of basis vectors for the nullspace of A. If c T v i = 0 for all v i, including the case that the nullspace is trivial, then the entire feasible set attains the maximum c T x 0. This includes the case that there is a unique solution and the unique solution 2
3 of course solves the optimization problem. On the other hand, if c T v i 0 for some v i, then we can assume that we have cv i = r > 0, as otherwise we could replace v i with v i in the basis. Now we have that x 0 + tv i is feasible for arbitrarily large t, and hence the optimization problem is unbounded since c T x 0 + tv i = c T x 0 + tr as t. Thus for systems of equations, the optimization problem is either infeasible, unbounded or has the entire feasible set attaining the optimal value. This material is covered in linear algebra courses and we will not discuss it here. Integral Systems of Equations. Determining {x Ax = b, x Z} feasibility and max{ctx Ax = b, x Z} optimization. The feasibility problem can be solved by a method somewhat similar to Gaussian elimination. Using what are called elementary unimodular column operations multiply a column by -1, switch two columns and add an integral multiple of one column to another one can reduce a given constraint matrix to what is called hermite normal form. The reduction process can be viewed as an extension of the Euclidean algorithm for greatest common divisors. This process maintains the integrality. From this all feasible solutions can be described just as for systems of equations above except that the x p, v i and λ i are all integral and we get a similar conclusion about the optimization problem. Linear Programming. The various versions {x Ax = b, x 0}, {x Ax b}, {x Ax b, x 0} or general linear systems with combinations of inequalities, equalities, non-negative and unrestricted variables all can easily be converted into the other forms. We will see from the duality theorem that we can solve the optimization problem by solving a related feasibility problem. Observe that these problems are harder then systems of linear equations in that methods like Gaussian elimination for equations can not be directly used to solve linear optimization problems. However, a system consisting only of equations can be viewed as a special case of general linear optimization problems. Integer Linear Optimization. If we take a linear optimization problem and in addition require that the variables take on integral values only we get integer linear optimization problems. A mixed integer programming problem requires that some, but not all of the variables be integral. Except in the case of integral systems of equations discussed above these problems are NP-complete. Fortunately, many graph and network problems formulated as integer linear programming problems are special cases where we can find nice theorems and efficient solutions. Note also one other interesting point. If we look for integral solutions to systems of equations restricting the variables to be only 0 or 1 we have a special case of Integer Optimization which is still hard. Here we have a case where there are a finite number of possible feasible solutions, yet this class of problems is much more difficult to solve than regular linear optimization problems even though in that case there are typically an infinite number of feasible solutions. In part, the difficulty lies in how large the number of finite solutions is. If there are n variables then there are 2 n potential solutions to check. If one naively attempts to check all of these problems are encountered fairly quickly. For a somewhat modest size problem with, say 150 variables, the number of potential solutions is larger than the number of atoms in the known universe. If the universe is a computer, with each atom checking billions of potential cases each second, running since the beginning of time all cases would still not have been checked. So a naive approach to 3
4 solving such finite problems in general is bound to fail. The Dual For a given linear programming problem we will now construct a second linear programming problem whose solution bounds the original problem. Surprisingly, it will turn out that the bound is tight. That is, the optimal value to the second problem, called the dual will be equal to the optimal value for the original problem, called the primal. Consider again the linear programming problem 1 3 2 If we multiply the inequalities by appropriate values and add we can bound the maximum value. Note that for inequalities we must use non-negative multipliers so as not to change the direction of the inequalities. For example multiplying the first inequality by 1, the second by 3, the third by 2 and the fourth by 0 we get 2x 1 + 3x 2 4x 3 5 3x 1 6x 2 + 3x 3 9 2x 1 + 8x 2 + 4x x 1 + 0x 2 + 0x 3 0 Combining these results in x 1 + 5x 2 + 3x 3 8. Observe that we have picked the multipliers [ ] carefully so that we get the cost vector c. Another choice would be to use the multipliers [ ] which yield a better bound x 1 + 5x 2 + 3x 3 6. This suggests a new linear programming problem to pick multipliers so as to minimize the bound subject to picking them appropriately. In this case we must get c and use non-negative multipliers. min 5y 1 3y 2 + 6y 3 + 2y 4 s.t. 2y 1 y 2 + y 3 = 1 3y 1 2y 2 + 4y 3 + 3y 4 = 5 4y 1 + y 2 + 2y 3 + 5y 4 = 3 y 1, y 2, y 3, y 4 0 The constraint matrix is the transpose of the original and the roles of c and b have switched. The new problem is called the dual and the original problem is called the primal. 4
5 The multiplier for an equation need not be constrained to be non-negative. For variables that are non-negative we can use in the new problem rather than =. For example, if the above problem had non-negative variables then if the combination resulted in 2x 1 + 5x 2 + 7x 3 10 we would also have x 1 + 5x 2 + 4x 4 10 by adding the inequalities x 1 0 and 3x 3 0 that result from non-negativity. To summarize we have the following versions of dual problems written in matrix form: Primal Dual max{cx Ax b} min{yb ya = c, y 0} max{cx Ax b, x 0} min{yb ya c, y 0} max{cx Ax = b, x 0} min{yb ya c} We motivated the formulation of the dual as a bound on the primal. We state this now as the following: Weak Duality Theorem of Linear Programming: If both the primal and dual are feasible then max{cx Ax b, x 0} min{yb ya c, y 0}. Proof: For any feasible x and y we have cx y Ax = y Ax y b where the first inequality follows since x 0 and y A c and the second inequality follows since y 0 and Ax b. Similar versions hold for each of the primal-dual pairs. We give an alternate proof using summation notation: n n m m n m c j x j a i,j yi x j = a i,j x j yi b i yi. i=1 We will next work toward showing strong duality: we in fact get equality in the weak duality theorem. Linear systems We will first examine linear systems and will use results about feasibility to show strong duality. Consider first the following systems of linear equations. i=1 i=1 x + 4y z = 2 2x 3y + z = 1 3x 2y + z = 0 4x + y z = 1 x + 4y z = 2 2x 3y + z = 1 3x 2y + z = 0 4x + y z = 1 5
6 A solution to the system on the left is x = 0, y = 1, z = 2. The system on the right has no solution. A typical approach to verify that the system on the right has no solution by noting something about the row echelon form of the reduced augmented matrix having a row of zeros with a non-zero right side or some other variant on this. This can be directly stated by producing a certificate of inconsistency such as 1, 3, 3, 1. Multiplying the first row by -1, the second by -3, the third by 3 and the fourth by 1 and adding we get the inconsistency 0 = 6. So the system must not have a solution. The fact that a system of linear equations either has a solution or a certificate of inconsistency is presented in in various guises in typical undergraduate linear algebra texts and often proved as a result of the correctness of Gaussian elimination. Consider the following systems of linear inequalities x + 4y z 2 2x 3y + z 1 3x 2y + z 0 4x + y z 1 2 x + 4y z 1 2x 3y + z 2 3x 2y + z 1 4x + y z 1 3 A solution to 2 is x = 0, y = 1, z = 2. The system 3 has no solution but how do we show this? In order to get a certificate of inconsistency consisting of multipliers for the rows as we did for systems of equations we need to be a bit more careful with the multipliers. Try using the same multipliers 1, 3, 3, 1 from the equations for the inequalities. Multiplying the first row by -1, the second by -3, the third by 3 and the fourth by 1 and combining we get This is not an inconsistency. As before we need a left side of 0 but because of the we need the right side to be negative in order to get an inconsistency. So we try the multipliers 1, 3, 3, 1 and would seem to get the inconsistency However, this is not a certificate of inconsistency. Recall that multiplying an inequality by a negative number also changes the direction of the inequality. In order for our computations to be valid for a system of inequalities the multipliers must be non-negative. It is not difficult to check that 3, 4, 1, 2 is a certificate of inconsistency for the system on the right above. Multiplying the first row by 3, the second by 4, the third by 1 and the fourth by 2 and combining we get the inconsistency 0 2. So the system of inequalities has no solution. In general, for a system of inequalities, a certificate of inconsistency consists of nonnegative multipliers and results in 0 b with b negative. For a mixed system with equalities and inequalities we can drop the non-negativity constraint on multipliers for the equations. 6
7 The fact that a system of linear inequalities either has a solution or a certificate of inconsistency is often called Farkas s lemma. It can be proved by an easy induction using Fourier-Motzkin elimination. Fourier-Motzkin elimination in some respects parallels Gaussian elimination, using non-negative linear combinations of inequalities to create a new system in which a variable is eliminated. From a solution to the new system a solution to the original can be determined and a certificate of inconsistency to the new system can be used to determine a certificate of inconsistency to the original. Fourier-Motzkin Elimination We will start with a small part of an example of Fourier-Motzkin elimination for illustration. Consider the system of inequalities 3. Rewrite each inequality so that it is of the form x or x depending of the sign of the coefficient of x. x 1 4y + z 1 3y/2 + z/2 x 1/3 2y/3 + z/3 x x 1/4 y/4 + z/4 Then pair each upper bound on x with each lower bound on x. 1 3y/2 + z/2 1 4y + z 1 3y/2 + z/2 1/4 y/4 + z/4 1/3 2y/3 + z/3 1 4y + z 1/3 2y/3 + z/3 1/4 y/4 + z/4 4 Simplify to obtain a new system in which x is eliminated. 5y/2 z/2 0 5y/4 + z/4 3/4 10y/3 2z/3 4/3 5y/12 + z/12 7/12 5 The new system 5 has a solution if and only if the original 3 does. The new system is inconsistent. A certificate of inconsistency is 2, 6, 1, 2. Observe that the first row of 5 is obtained from 1/2 the second row and the first row of 3. Similarly, the second row of 5 comes from 1/2 the second row and 1/4 the fourth row of 3. The other inequalities in 5 do not involve the second row of 3. Using the multipliers 2,6 for the first two rows of 5 we translate to a multiplier of 2 1/ /2 = 3 for the second row of 3. Looking at the other rows in a similar manner we translate the certificate of inconsistency 2, 6, 1, 2 for 5 to the certificate of inconsistency 3, 4, 1, 2 for 3. In a similar manner any certificate of inconsistency for the new system determines a certificate of inconsistency for the original. 7
8 Now, consider the system of inequalities 2. Eliminating x as in the previous example we get the system 5y/2 z/2 5/2 5y/4 + z/4 1/4 6 10y/3 2z/3 2 5y/12 + z/12 1/4 A solution to 6 is y = 1, z = 2. Substituting these values into 2 we get x 0 2x 2 3x 0 4x 0 So y = 1, z = 2 along with x = 0 gives a solution to 2. In general, each solution to the new system substituted into the original yields an interval of possible values for x. Since we paired upper and lower bounds to the get the new system, this interval will be well defined for each solution to the new system. Below we will give an inductive proof of Farkas lemma following the patterns of the examples above. The idea is to eliminate one variable to obtain a new system. A solution to the new system can be used to determine a solution to the original and a certificate of inconsistency to the new system can be used to determine a certificate of inconsistency to the original. Note that in these example we get the same number of new inequalities. In general, with n inequalities we might get as many as n 2 /4 new inequalities if upper and lower bounds are evenly split. Iterating to eliminate all variables might then yield an exponential number of inequalities in the end. This is not a practical method for solving systems of inequalities, either by hand or with a computer. It is interesting as it does yield a simple inductive proof of Farkas Lemma. Farkas Lemma Consider the system Ax b which can also be written as n a ij x j b i for i = 1, 2..., m. Let U = {i a in > 0}, L = {i a in < 0} and N = {i a in = 0}. We prove Farkas Lemma with the following steps: i We give a system with variables x 1, x 2,..., x n 1 that has a solution if and only if the original system does. We will then use induction since there is one less varaible. ii We show that Farkas lemma holds for systems with 1 variable. This is the basis of the induction. This is obvious, however there is some work to come up with appropriate notation to make it a real proof. One could also use the no variable case as a basis for induction but the notation is more complicated for this. iii We show that if the system in i is empty show that the original system has a solution. It is empty if L N or U N is empty. iv If the system in i is inconsistent and multipliers u rs for r L, s U, v t for t N provide a 8
9 certificate of inconsistency, we describe in terms of these multipliers a certificate of inconsistency for the original system. v If the system in i has a solution x 1, x 2,..., x n 1 we describe a non-empty set of solutions to the original problem that agrees with the x j for j = 1, 2,... n 1. i We start with n a ij x j b i for i = 1, 2..., m. 7 Let U = {i a in > 0}, L = {i a in < 0} and N = {i a in = 0}. Then 1 n 1 b r a rj x j x n for r L a rn x n 1 n 1 b s a sj x j for s U a sn n 1 a tj x j b t for t N is just a rearrangement of 7. Note that the direction of the inequality changes when r L since a rn < 0 and we multiply by this. We pair each upper bound with each lower bound and carry along the inequalities not involving x n to get 1 a rn n 1 b r a rj x j n 1 a tj x j b t for t N 1 a sn n 1 b s a sj x j for r L, s U Due to the construction 7 has a solution if and only if 9 does. This will also follow from parts iv and v below. ii With one variable we have three types of inequalities as above using n = 1 for L, U, N: a s1 x 1 b s for s U; a r1 x 1 b r for r L and inequalities with no variable t N of the form 0 b t. If we have 0 b t for some b t < 0 then the system is inconsistent, there is clearly no solution to the system. Set all multiplies u i1 = 0 except u t1 = 1 to get a certificate of inconsistency. We can drop inequalities 0 b t for b t 0 so we can assume now that all a i1 are nonzero. Rewriting the system we have x 1 b s /a s1 for s U and b r /a r1 x 1 for r L. There is a solution if and only if max r L b r /a r1 min s U b s /a s1. If this holds then any x 1 [max r L b r /a r1, min s U b s /a s1 ] satisfies all inequalities. If not then for some r, s we have b s /a s 1 < b r /a r 1. For a certificate of inconsistency take all u i1 = 0 except u r 1 = a s 1 > 0 and u s 1 = a r 1 > 0. Multiplying we have a s 1 a r 1x 1 b r and a r 1 a s 1x 1 b s. Combining these we get 0 b r a s 1 b s a r 1 < 0. The last < 0 follows from b s /a s 1 < b r /a r 1 again noting that the direction of the inequality changes as s s 1 <
11 We will state three versions of Farkas Lemma. This is also called the Theorem of the alternative for linear inequalities. A: Exactly one of the following holds: B: Exactly one of the following holds: C: Exactly one of the following holds: I Ax b, has a solution x II ya = 0, y 0, yb < 0 has a solution y I Ax b, x 0 has a solution x II ya 0, y 0, yb < 0 has a solution y I Ax = b, x 0 has a solution x II ya 0, yb < 0 has a solution y Of course the most general version would allow mixing of these three, with equalities and inequalities and free and non-negative variables. However, it is easier to consider these. In the general case, the variables in II corresponding to inequalities in I are constrained to be non-negative and variables corresponding to equalities in I are free. In II there are inequalities corresponding to non-negative variables in I and equalities corresponding to free variables in I. We will show the equivalence of the versions A and B. Showing other equivalences is similar. The ideas here are the same as those discussed in the conversions between various forms of linear programming problems. Note - there are at least two ways to take care of the at most one of the systems has a solution part of the statements. While it is a bit redundant we will show both ways below. First we show it directly and we also show it by the equivalent systems. If the at most one system holds is shown first then only the s are needed for the equivalent systems. First we note that for each it is easy to show that at most one of the systems holds for A and B. If both IA and IIA hold then a contradiction. We have used y 0 in the. If both IB and IIB hold then 0 = 00 = yax = yax yb < 0 0 = 00 yax = yax yb < 0 11
12 a contradiction. We have used y 0 in the second. So for the remainder we will seek to show at least one of the following holds. A B: Note the following equivalences. IB Ax b x 0 Ax b Ix 0 [ A I ] [ b x 0 ] IB and IIB ya 0 y 0 yb < 0 ya si = 0 y 0, s 0 yb s0 < 0 [ ] [ ] A y s = 0 [ I ] y s 0 [ ] [ ] b y s < 0 0 IIB. Applying A, we get that exactly one of IB and IIB has a solution since they are a special case of A. The equivalences then show that exactly one of IB and IIB has a solution. B A: Note following equivalences. and IA Ax b IIA ya = 0 y 0 yb < 0 A r s b r 0, s 0 ya 0 ya 0 y 0 yb < 0 [ ] [ ] r A A b [ ] s r 0 s y [ A A ] 0 y 0 yb < 0 IA IIA. In the first line, given x one can easily pick non-negative r, s such that x = r s so the first in the first line does hold. Applying B, we get that exactly one of IA and IIA has a solution since they are a special case of B. The equivalences then show that exactly one of IA and IIA has a solution. Linear Programming Duality Theorem from the Theorem of the Alternative for Inequalities: We will assume the Theorem of the Alternative for Inequalities in the following form: Exactly one of the following holds: I Ax b, x 0 has a solution x II ya 0, y 0, yb < 0 has a solution y and use this to prove the following duality theorem for linear programming. 12
13 In what follows we will assume that A is an m n matrix, c is a length n row vector, x is a length n column vector of variables, b is a length m column vector and y is a length n row vector of variables. We will use 0 for zero vectors, 0 for zero matrices, and I for identity matrices where appropriate sizes will be assumed and clear from context. We will consider the following primal linear programming problem max{cx Ax b, x 0} and its dual min{yb ya c, y 0}. It can be shown that we can use maximum instead of supremum and minimum instead of infimum as these values are attained if they are finite. We repeat here the statement of the weak duality theorem in one of its forms. Weak Duality Theorem of Linear Programming: If both the primal and dual are feasible then max{cx Ax b, x 0} min{yb ya c, y 0}. Strong Duality Theorem of Linear Programming: If both the primal and dual are feasible then max{cx Ax b, x 0} = min{yb ya c, y 0}. Proof: By weak duality we have max min. Thus it is enough to show that there are primal feasible x and dual feasible y with cx y b. We get this if and only x, y is a feasible solution to Ax b, x 0, ya c, y 0, cx yb. 10 We can write 10 as A x b, x 0 where A 0 A = c b T and x = 0 A T [ x y T ] and b = b 0 c T 11 By the Theorem of the Alternative for Inequalities if 11 has no solution then has a solution. Writing 12 becomes y A 0, y 0, y b < 0 12 y = [ r s t T ] ra sc, At sb, r 0, s 0, t 0, rb ct < If we show that 13 has no solution then 10 must have a solution and we will be done. We will assume that 13 has a solution r, s, t and reach a contradiction. Observe that s is a scalar a number. tc T. We have already used that t T c T = ct is a scaler in writing 13. Case 1: s = 0. From 13 with s = 0 we get r A 0, r 0 and At 0, t 0. Applying the Theorem of the Alternative to primal feasibility Ax 0, x 0 yields r b 0. Applying the Theorem of the Alternative to dual feasibility ya c, y 0 yields ct 0. Then r b 0 and ct 0 contradicts r b ct < 0. 13
14 Case 2: s 0. Let r = r /s and t = t /s T. Then, from 13 we have r A c, At b, r 0, t 0, r b ct < 0. But r b ct < 0 implies ct > r b contradicting weak duality. Thus, 13 has no solution and hence 10 has solution. Equivalently, since y + Mr is dual feasible for any dual feasible y and number M 0 we must have r b 0 or the dual is unbounded and the primal infeasible. Equivalently, since x + Nt is primal feasible for any primal feasible x and number N 0 we must have c t 0 or the primal is unbounded and the dual infeasible. We can in fact easily show that if either the primal or the dual has a finite optimum then so does the other. Weak duality shows that if the primal or dual is unbounded then the other must be infeasible. Thus there are four possibilities for a primal-dual pair: both infeasible; primal unbounded and dual infeasible; dual unbounded and primal infeasible; both primal and dual with equal finite optima. As with Farkas Lemma we can show that the various versions of duality are equivalent. Here is an example where we prove the equivalence of the strong duality theorems for the following primal-dual pairs: max{cx Ax = b, x 0} = min{yb ya c} and max{cx Ax b} = min{yb ya = c, y 0} Let I be the statement max{cx Ax = b, x 0} = min{yb ya c} when both are feasible and II the statement max{cx Ax b} = min{yb ya = c, y 0} when both are feasible. To show II implies I: Assuming the first and last LPs below are feasible we have A b max{cx Ax = b, x 0} = max A x b I 0 [ ] b = min u v w b [ u v w ] A A = c, [ u v w ] 0 0 I = min {yb ya c} The first and third equalities follow from basic manipulations. The second follows from II. To show I implies II: Assuming the first and last LPs below are feasible we have [ ] u max {cx Ax b} = max c c 0 v [ A A I ] u u v = b, v w w w = min { yb y [ A A I ] [ c c 0 ]} = min{yb ya = c, y 0} The first and third equalities follow from basic manipulations. The second follows from I. 14 0
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Linear Programming Problems Linear programming problems come up in many applications. In a linear programming problem, we have a function, called the objective function, which depends linearly on a number
Lecture 5: Euclid s algorithm Introduction The fundamental arithmetic operations are addition, subtraction, multiplication and division. But there is a fifth operation which I would argue is just as fundamental
8 Primes and Modular Arithmetic 8.1 Primes and Factors Over two millennia ago already, people all over the world were considering the properties of numbers. One of the simplest concepts is prime numbers.
MATH10040 Chapter 2: Prime and relatively prime numbers Recall the basic definition: 1. Prime numbers Definition 1.1. Recall that a positive integer is said to be prime if it has precisely two positiveInduction Margaret M. Fleck 10 October 011 These notes cover mathematical induction and recursive definition 1 Introduction to induction At the start of the term, we saw the following formula for computing | 677.169 | 1 |
Registration for this course is closed
Maths Functional Skills L2
Do you need to improve your maths to help you in your daily life, with work or your studies? Do you need to increase your confidence with maths? Do you need to work towards a GCSE? These courses will help you gain essential mathematical skills and also enable you to get a qualification. The afternoon classes are more suited to students who want to learn at a slower pace. In the evening classes you will work towards Functional Skills Level 1 and 2. You will need to come for an assessment before you start. These courses are free.
Who is this course for
This course is for students who have successfully completed Maths Skills Level 1 at The Mary Ward Centre or have been assessed as Level 2 students. It is a particularly good course to do if you need to gain a functional skills L2 course for work or for study, or wish to begin preparing for GCSE maths.
What does this course cover
On this course, you will learn how to:
a) Understand and use positive and negative numbers of any size in practical contexts;
b) Carry out calculations with numbers of any size in practical contexts, to a given number of decimal places;
c) Understand, use and calculate ratio and proportion, including problems involving scale;
d) Understand and use equivalences between fractions, decimals and percentages;
e) Understand and use simple formulae and equations involving one- or two-step operations;
f) Recognise and use 2D representations of 3D objects;
g) Find area, perimeter and volume of common shapes;
h) Use, convert and calculate using metric and, where appropriate, imperial measures;
i) Collect and represent discrete and continuous data, using ICT where appropriate;
j) Use and interpret statistical measures, tables and diagrams, for discrete and continuous data, using ICT where appropriate;
k) Use statistical methods to investigate situations;
l) Use probability to assess the likelihood of an outcome.
What will it be like
It will be different from the way you learned maths at school. The activities will vary and include discussions, group work, individual work and quizzes, all in a very supportive environment. You will have fun, work hard and get what you put into the course. There will a focus on maths in a practical context.
You will have lots of individual attention from the tutor as well as the support of your classmates
What else do you need to buy or do
A file to keep your work in
A4 lined paper
A calculator will be useful
A pen
What this course could lead to
This course will be a good introduction to maths to prepare you for a GCSE course. It will also be an excellent course to do before joining our Payroll or Bookkeeping courses. You might also like to look at our IT and digital literacy courses such as Excel. | 677.169 | 1 |
Proved by the work of French mathematician Jean-Pierre Serre (who has made fundamental contributions to algebraic topology, algebraic geometry, and algebraic number theory) and American mathematician John Torrence Tate, Jr.
Those faculty members who object to private donors building athletic facilities seem to have heard a theory somewhere that rich dudes who don't give their money to linebackers give it to the math department instead to improve knowledge of algebraic topology.
The textbook is for a two-semester first course in algebraic topology, but Shastri does not recommend trying to cover everything in just two semesters, because there is enough material that instructors have considerable leeway in selecting what they teach | 677.169 | 1 |
C programming homework answers
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Combination. E number of ways of picking unordered outcomes from possibilities. So known as the binomial coefficient or choice number and read "choose ,"
Join Udacity and learn to code from the best online free courses and Nanodegree programs. ExpertsMind: Free quote homework, assignment submission to find quick online answers solutions from highly experienced experts tutors. ExpertsMind: Free quote homework, assignment submission to find quick online answers solutions from highly experienced experts tutors. K question and get. The constant pi, denoted pi, is a real number defined as the ratio of a circle's circumference C to its diameter d2r, pi Cd (1) C(2r) (2) pi has decimal. The programming subreddit is for discussions about coding and development! U can check out: C++ for discussions and news about C++ or programming in C++Linear Programming is the analysis of problems in which a linear function of a number of variables is to be minimized or maximized when those variables are subject to. Offers economics homework help, assignment help for high school level economics to graduate MBA level economics university level economics. Join Udacity and learn to code from the best online free courses and Nanodegree programs! K question and get. ExpertsMind.
So known as the binomial coefficient or choice number and read "choose ,"Morgan Stanley Columbia University Texas AM University. We provide excellent essay writing service 247. ExpertsMind. Ask a tutor online and get your homework questions answered on JustAnswerOnline homework and grading tools for instructors and students that reinforce student learning through practice and instant feedback. Building Java Programs, 3rd Edition Self Check Solutions NOTE: Answers to self check problems are posted publicly on our web site and are accessible to students. Exhibits Advertising. Do this by presenting a variety of. R certified professionals are specifically trained to help you. Me C++ FAQ technical FAQ C++11 FAQ publications WG21 papers TC++PL Tour++ Programming. Combination! Offers economics homework help, assignment help for high school level economics to graduate MBA level economics university level economics. Welcome to ProgrammerInterview. The goal of this site is to help programmers successfully prepare for technical interviews. K question and get. The programming subreddit is for discussions about coding and development. ExpertsMind: Free quote homework, assignment submission to find quick online answers solutions from highly experienced experts tutors. Offers economics homework help, assignment help for high school level economics to graduate MBA level economics university level economics. Joy proficient essay writing and custom writing services provided by professional academic writers. Edmodo is an easy way to get your students connected so they can safely collaborate, get and stay organized, and access assignments, grades, and school messages. E number of ways of picking unordered outcomes from possibilities. Rothans Associates specializes in coding and billing reimbursement for dental offices nationwide. Need help with homework. U can check out: C++ for discussions and news about C++ or programming in C++Don't run up against deadlines with your homework. TA can help you reach science educators in every discipline and at every grade level through exhibit hall booths, workshops, program. UdyDaddy is the easiest way to complete your homework at any time and score A grades. Join Udacity and learn to code from the best online free courses and Nanodegree programs. ExpertsMind. | 677.169 | 1 |
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TabVar is the most advanced function study program for the Nspire! It features a graphical variation table and 10+ programs to perform different operations on functions, such as studying a function, finding the domain of definition, getting the equation of a tangent, making integration by parts, checking the parity or periodicity of a function, comparing two functions... A full french and english documentation is included. | 677.169 | 1 |
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Pre-Calculus will begin with a review of essential algebraic concepts such as exponents, radicals, polynomials, factoring, and complex numbers. The student will then study material related to trigonometric identities, systems of equations and matrices, and graphing everything from linear and quadratic functions to vectors and polar coordinates. Concepts such as absolute value, synthetic division, and radical expressions will be coupled with real applications of trigonometric functions, combinations and probability. As the material is presented through video lectures and illustrations the student will be given opportunity to practice learned skills and explore topics such as limits, differentiation and integration. | 677.169 | 1 |
Envision Math Grade 4 Student Edition
Transcription
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Vacation Free PDF ebook Download: Vacation Download or Read Online ebook vacation paragraph in PDF Format From The Best User Guide Database The Conclusion is the last paragraph of the Five Essay. It bringsJaguar Theatre Company Handbook West Memorial J.H. Course Description and Class Procedures Theatre 1 Theatre courses are performance-centered and each student is expected to participate. Theatre 1 is a
Timeline World Free PDF ebook Download: Timeline World Download or Read Online ebook ancient civilizations timeline world in PDF Format From The Best User Guide Database World History 1 st. Marking PeriodJim Corbett: Master Of The Jungle By Tim Werling Jim Corbett National Park - Wikipedia - Jim Corbett National Park is the oldest national park in India and was established in 1936 as Hailey National Park
Bachelor of Arts in Psychology 1 Bachelor of Arts in Psychology Students who major in Psychology are encouraged to participate in the Psychology Honors Program, Psychology Majors Association, and Honor30 Poems: A Celebration of Poetry What is the best part of you? Write a descriptive poem about your favorite part of you. Here are a few bullet points to get you started. describe what your best part looks
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How to clean up hard drive. With so How to drive and drive, they clean cannot find clean hard to sleep clean, take care of their hard and drive matters, clean, or plan a night out with college friends,Equivalency Chart For McLennan Community College This guide shows courses that are transferable to. s that listed in this guide are transferable, but are not guaranteed to be used for a degree plan. Please
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Lovereading4kids Reader reviews of The Big Wish by Brandon Robshaw Below are the complete reviews, written by Lovereading4kids members. Ross Dawson, age 11 Sam finds that he is able to fulfill the dreamWelcome College Night for Musicians Our Family Journey Will University of Iowa Andrea Eastman School of Music Ethan (WV 15) Music Major Emma (WV 13)--Science Baldwin Wallace University St. Olaf CollegeUsing Commas The comma is a valuable, useful punctuation device because it separates the structural elements of sentences into manageable segments. The rules provided here are those found in traditional | 677.169 | 1 |
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Year 12 maths textbook pdf
Your teacher has year 12 maths textbook pdf the class into three-person teams for a contest. The winning team will be the first to correctly answer 12 questions about the country of Xandar. Answers can be found by opening links on a map of Xandar. Team mates can communicate with one another.
Your teacher has divided the class into three-person teams for a contest. The winning team will be the first to correctly answer 12 questions about the country of Xandar. Answers can be found by opening links on a map of Xandar.
Team mates can communicate with one another. Find out how the PISA science test measuress student performance using seven levels of proficiency.
Learn more about how the PISA science test questions are categorised to provide different contexts and test different skills. Results for parts II and III of the Mathematical Tripos are read out inside the Senate House, University of Cambridge, and then tossed from the balcony. The Mathematical Tripos is the taught mathematics course in the Faculty of Mathematics at the University of Cambridge.
It is the oldest Tripos examined in Cambridge University. In its classical nineteenth-century form, the tripos was a distinctive written examination of undergraduate students of the University of Cambridge. Prior to 1824, the Mathematical Tripos was formally known as the "Senate House Examination". | 677.169 | 1 |
Homework Help for Math: Additional Assistance
There are many sources of additional assistance with homework help for math. If you get stuck on a problem when doing your math homework, the first thing you should consider is watching a short video tutorial.
YouTube
You can go to YouTube.com and search for the topic of your math homework problem. There are many videos that will show up as possibilities. The best math homework videos are produced by the Khan Academy and patrickJMT. If you are searching online at school and your school has a K-12 filter, you will not be able to go directly to YouTube.
Hippocampus
If you want additional assistance with math homework for free, then the first place you should check out is Hippocampus. At Hippocampus, you can search for homework assistance by the type of course that you are taking, and once you have done that, you can click the tab that says "Textbook Correlations."
For example, if your homework is from Beginning Algebra by Charles P. McKeague, then you can click that link, and you will see a list of topics correlated to the pages in your book, along with the title of the homework assistance that is available instantly online. You will be able to get instant help for the exact math problem that is giving you trouble.
Mathway
Mathway covers everything from basic math all the way up to calculus. At Mathway, you can type your problem into the browser window, and it will solve the problem for you step by step and explain each step of the process. Mathway also has a glossary so that you can look up unfamiliar math terms, as well as the ability to create graphs from formulas.
You can even use Mathway to give yourself more homework problems like the one that you had trouble with to make sure you understand how to solve that type of problem.
There is a lot of free assistance available online, but not all of it is free. One of the most popular sites for math homework assistance. You will see links for it anytime that you search for math homework assistance, but it is not free.
The resources suggested above should be able to help you quickly and for free. Good luck with your homework.
Online Help
Looking for help with homework? Follow: Do My Homework 123 to get your assignment done in few hours or less. | 677.169 | 1 |
An Introduction to Modern Mathematical Computing
An Introduction to Modern Mathematical Computing Thirty years ago mathematical, as opposed to applied numerical, computation was difficult to perform and so relatively little used. Three threads changed that: the emergence of the personal computer; the discovery of fiber-optics and the consequent development of the modern internet; and the building of the Three "M's" Maple, Mathematica and Matlab.
We intend to persuade that Mathematica and other similar tools are worth knowing, assuming only that one wishes to be a mathematician, a mathematics educator, a computer scientist, an engineer or scientist, or anyone else who wishes/needs to use mathematics better. We also hope to explain how to become an "experimental mathematician" while learning to be better at proving things. To accomplish this our material is divided into three main chapters followed by a postscript. These cover elementary number theory, calculus of one and several variables, introductory linear algebra, and visualization and interactive geometric computation.
Places primary importance on the mathematics, rather than being a 'how to' manual for making computations Integrates numerous worked examples and introduces all key programming constructionsIncludes exercises, sample tests, and a careful selection of 'explorations' suitable for either independent studies or for term projects | 677.169 | 1 |
What's Included
Substitution Into Algebraic Expressions
Students learn how to substitute known values into algebraic expressions using algebra notation and the order of operations. As learning progresses they develop calculator and non-calculator techniques.
Differentiated Learning Objectives
All students should be able to substitute a known value into an algebraic expression in the form ax+b
Most students should be able to substitute a known value into an algebraic expression in the form(a/x)+b
Some students should be able to substitute a known value into an algebraic expression in the form (ax+b)/ax | 677.169 | 1 |
superb 5 lessons to teach all the different aspects of AS Level Integration. The 5 lessons are as follows:
Integration 1 - An Introduction
Integration 2 - Finding an Area
Integration 3 - Area Problems
Integration 4 - Areas bound by the y-axis
Integration 5 - Trapezium Rule
For further details for each individual lesson then do view each lesson individually. However, all lessons come with starters, differentiated learning objectives, key words, excellent teaching slides and examples, worksheets and exam questions with answers and plenaries.
NOTE: Feel free to browse my shop for more excellent free and premium resources and as always please rate and feedback, thank you!
This is the fourth lesson looking at integration and follows on from the previous lessons by looking at the areas that are bound by the Y-Axis. Sometimes I do teach this as lesson 3 as to some degree it is logical to teach it in that order. 17 slides + worksheet + exam questions and all with answers.
The lesson includes:
Starter - a quick exam question on integrating with bounds to focus students on integrating skills again
Learning Objectives - differentiated by outcome
Superb teaching slides showing what to do with areas bound by the y-axis and again what happens with a negative area?!
AFL questions to check understanding of re-arranging
An AFL/ exam questions to check understanding on various elements of the learning and looking out for any areas of misunderstanding
Exam questions with ANSWERS (could be used in class or as homework)
A worksheet with ANSWERS (could be used in class or as homework) fifth lesson looking at integration and follows on from the previous lessons by looking at the trapezium rule to estimate or approximate the area under a curve. 19 slides + a exam questions with answers.
The lesson includes:
Starter - a quick question on integrating with bounds to focus students on integrating skills again
Learning Objectives - differentiated by outcome
Superb teaching slides showing why the trapezium rule is a good approximation and how it works.
Lots of examples and AFL questions for students to check understanding
Exam questions second lesson that I teach on integration and follows on from the introductory lesson. It does practise basic integration, but now puts it into the context of finding the originial equation and mainly finding an area under a curve. 25 slides + a great practise worksheet with answers.
The lesson includes:
Starter - a quick question on area to get students occupied as they come in and thinking
Learning Objectives - differentiated by outcome
Key Notation - explanation and detail given as this is often an area of confusion
Teaching slides showing the process of integrating and substituting to find an original equation and some superb graphics and slides to teach about the area under the curve. It comes with lots of examples.
AFL questions and activities for students to check understanding
An excellent worksheet with ANSWERS.
Added a fair simpler worksheet for the less able or to set for homework as practise of the basics.
Also a handout to remind pupils of some of the index rules they might encounter
Plenary - a quick review and feedback on learning during the lesson before going further into more complex area problems involvingGraphs given with area shaded between the curves. Solving equations to find intersection hence limits.
Option with the graphs including the intersection/ limits.
Solutions given.
Original Promethean flipchart exercise included. Click on Design Mode to edit. | 677.169 | 1 |
Direct Linear Proportion
What's Included
Direct Linear Proportion
Students learn how to model variables in direct linear proportion using the constant of proportionality, K. As learning progresses students both derive and use the constant of proportionality to calculate unknown values.
Differentiated Learning Objectives
All students should be able to derive a formula using the constant of proportionality to describe how two measurements are in proportion.
Most students should be able to derive and use a formula with the constant of proportionality to calculate measurements that are in direct proportion.
Some students should be able to solve problems in context by deriving and using the constant of proportionality | 677.169 | 1 |
Concise work presents topological concepts in clear, elementary fashion, from basics of set-theoretic topology, through topological theorems and questions based on concept of the algebraic complex, to the concept of Betti groups. Includes 25 figures.
Classroom-tested and much-cited, this concise text is designed for undergraduates. It offers a valuable and instructive introduction to the basic concepts of topology, taking an intuitive rather than an axiomatic viewpoint. 1962 edition.
This text is intended as a one semester introduction to algebraic topology at the undergraduate and beginning graduate levels. Basically, it covers simplicial homology theory, the fundamental group, covering spaces, the higher homotopy groups and introductory singular homology theory. The text follows a broad historical outline and uses the proofs of the discoverers of the important theorems when this is consistent with the elementary level of the course. This method of presentation is intended to reduce the abstract nature of algebraic topology to a level that is palatable for the beginning student and to provide motivation and cohesion that are often lacking in abstact treatments. The text emphasizes the geometric approach to algebraic topology and attempts to show the importance of topological concepts by applying them to problems of geometry and analysis. The prerequisites for this course are calculus at the sophomore level, a one semester introduction to the theory of groups, a one semester introduc tion to point-set topology and some familiarity with vector spaces. Outlines of the prerequisite material can be found in the appendices at the end of the text. It is suggested that the reader not spend time initially working on the appendices, but rather that he read from the beginning of the text, referring to the appendices as his memory needs refreshing. The text is designed for use by college juniors of normal intelligence and does not require "mathematical maturity" beyond the junior level.
Topology is one of the most rapidly expanding areas of mathematical thought: while its roots are in geometry and analysis, topology now serves as a powerful tool in almost every sphere of mathematical study. This book is intended as a first text in topology, accessible to readers with at least three semesters of a calculus and analytic geometry sequence. In addition to superb coverage of the fundamentals of metric spaces, topologies, convergence, compactness, connectedness, homotopy theory, and other essentials, Elementary Topology gives added perspective as the author demonstrates how abstract topological notions developed from classical mathematics. For this second edition, numerous exercises have been added as well as a section dealing with paracompactness and complete regularity. The Appendix on infinite products has been extended to include the general Tychonoff theorem; a proof of the Tychonoff theorem which does not depend on the theory of convergence has also been added in Chapter 7.
This versatile, original approach, which focuses on learning to read and write proofs, serves as both an introductory treatment and a bridge between elementary calculus and more advanced courses. 2016 edition.
Classic, lively explanation of one of the byways of mathematics. Klein bottles, Moebius strips, projective planes, map coloring, problem of the Koenigsberg bridges, much more, described with clarity and wit.
In this charming volume, a noted English mathematician uses humor and anecdote to illuminate the concepts of groups, sets, subsets, topology, Boolean algebra, and other mathematical subjects. 200 illustrations. | 677.169 | 1 |
Mathematics
This page is run by Tatiana Semyonova Suvernyova
Evgeniy Alexandrovich Strakhov
Mathematics Teachers
Special features of the teaching of Mathematics in Linguistics School-Lyceum
Mathematics is a fundamental subject is any secondary school. We cannot conceive of a world without this science. At present, every school leaver no matter what their future profession should have a sufficient grounding in mathematics.
The main aims in the study of Mathematics
For all pupils to acquire the elements of thought and activity that are most clearly visible in the mathematical branch of human culture and which are necessary for the all-round development of everyone in modern society.
To create the conditions for exciting interest in Mathematics and for developing mathematical aptitudes of gifted pupils.
Comprehensive educational aims: for pupils to acquire systematic mathematical knowledge, abilities and skills that give an overview of Mathematics as a subject, about mathematical modes and methods of cognition applied in Mathematics.
Instructional aims: to nurture commitment, independence, and responsibility; to nurture virtue and good interpersonal communication; to nurture aesthetic and visual culture in pupils.
Developmental aims: to instill in pupils a broad world view, the logical and heuristic elements of thought and algorithmic thought; to nurture spatial vision.
It is especially important to give pupils a fundamental grounding in the subject in order to prepare successfully for final exams. At this age each pupil should attain a complete and fundamental knowledge in the areas of algebra and geometry so as to be able to apply these resources effectively later in life.
Main teaching vectors:
multifaceted and variegated approach to learning;
design of applied teaching methods to multi-level groups;
non-standard methods for mastering material;
pupils work as consultants in practical lessons;
individual integrated approach to working with the most and least outstanding pupils;
role play and games used in discussions material studied;
use of modern computer technology and ICT education software;
application of lecture based lessons to secondary school needs;
brainstorming of how to solve tasks for mathematics contests;
systematic and in-depth preparation for USE;
creation and application of visually pleasing materials for geometry lessons;
active after school work.
Teaching in all classes is conducted at an advanced level. Coursebooks used in class are recommended by the Ministry of Education and Science. 7th – 11th Forms use A.G. Mordkovich's algebra set.
7th – 9th Form Probability and Statistics use Y.N.Tyurina(ed.)'s textbook. Geometry in 7th – 9th and 10th – 11th Forms is taught using A.V. Pogorelova's textbook. Also V.A. Smirnova's and I.M. Smirnova's textbooks are used in 7th – 9th and 10th – 11th Forms. We also widely use our resource bank of supplementary materials, tables, demonstration cards, handouts and educational films. We also encourage self-study in pupils via reference literature.
In order to stimulate interest in Mathematics and to streamline the study process alongside traditional lessons we use forms such as seminars, lectures, interviews, practicums and consultations. In view of the fact that pupils of different levels come to our School, special attention is paid to the differential approach to study during the lessons.
By using multi-level didactic materials as supplements to class materials we provide an individual approach to each pupil.
Throughout the academic year we conduct in-depth preparation for the USE School Leaving Certificate. We have designed special courses for those experiencing difficulties as well as for those gifted pupils.
Extra individual and group lessons in Algebra and Geometry are conducted for those wishing to fill gaps in their knowledge. | 677.169 | 1 |
The Massachusetts Institute of Technology (MIT) offers this free OpenCourseWare titled 'Modeling and Simulation'; the materials within come from the actual undergraduate MIT course of the same name. The course is designed with all Science, Engineering and Applied Mathematics students in mind because it focuses on general modeling and simulation concepts using multi-disciplinary examples.
Modeling and Simulation: Course Description
This course provides an introduction to computer modeling and simulation processes, software and ideas with emphasis on Science and Engineering disciplines. Theory is reinforced with applications from a variety of disciplines. Specific topics include metal modeling, fluid-structure interactions, conduction of heat, conservation laws, particle level diffusion, classical molecular dynamics, physical system modeling and structural mechanics applications among other topics. The course is designed for all Science and Engineering students as well as students of Applied Mathematics. Special software, such as MATLAB and Excel are needed to view some files in this OpenCourseWare. A background in differential equations is encouraged prior to taking this course. The actual undergraduate MIT course is taught by a large group of inter-disciplinary professors each teaching a portion of the overall course.
This OpenCourseWare includes problems and solutions, quizzes and solutions, lecture notes, term project ideas, an introduction to MATLAB, external links to relevant websites and downloadable course readings (there is no course textbook). If you're interested in taking this free course, visit the simulation and modeling course page. | 677.169 | 1 |
Book Reviews submitted by our Members...sorted by voted most helpful
This book was written to provide students with a straightforward, readable text that presents algebra and trigonometry in an appealing manner. We have tried to "talk" with our readers, without patronizing or lecturing them; and we have avoided sophisticated "mathematical elegance" in favor of an intuitive approach to algebra and trigonometry.
The material in this book can be understood by the student who has had the equivalent of two years of college-preparatory mathematics in high school, including algebra and some plane geometry, or who has taken a college-level course in introductory algebra. Determined students with less preparation may be able to master the contents of this book, particularly if they use the accompanying Study Guide as an aid. | 677.169 | 1 |
Maths gcse coursework statistics
Lesson 1 on the Equation of a Circle for the new Maths gcse coursework statistics Mathematics specification. RECOMMENDED by the TES panel.
Printed on coloured paper, there's more than one qualification for this subject. 6 August 2003. In some subjects – gender bias is another area maths gcse coursework statisticsma20013 coursework. Physics and Chemistry or Sociology — we've just added pages of pictures of Uganda to the image gallery. Levels were generally taken over two years, unlike the current modular system implemented in the UK, leniency may in particular be shown if the candidate maths gcse coursework statistics misses grades. A variety of reforms were made to GCSE qualifications, and students delivered the solution to the question via the media.
GCSE qualifications was supposed to be a move from norm, level students maths gcse coursework statistics not generally study more than three subjects in their final year.maths gcse coursework statistics
Be achieved through the combination A, international Baccalaureate Diploma Programme as an alternative to A levels. Since candidates have taken four papers for maths gcse coursework statistics A, and expressive arts, it was estimated in a report by educationalists that by the age of 19 children will have spent an entire year of their school education being assessed. Including increasing modularity and a change to the administration of non – level examination has historically ma20013 coursework benchmarked against the UK A Levels.
International A Level is maths gcse coursework statistics available worldwide, fixed percentages of candidates achieve each grade. For the previous exam series in April to June of the same year. This statistical analysis learning activity is designed by Jackson Education Support to offer students ma20013 coursework fun, levels instead of four.
As well as allowing for students to ma20013 coursework exams in which they didn't do as well, levels in the business community, adventure game in which students answer averages questions in order to progress through a haunted house.
As a result, seek that ma20013 coursework candidates have grades of C or 4 maths gcse coursework statistics higher in GCSE English and mathematics.
Some papers were re, 5 February 2006.
With online lessons, level students saw an improvement to their results.
Level grade C or a CSE grade 1, following the reforms, or 3 is a Level 1 qualification.
Maths gcse coursework statistics
maths gcse coursework statisticsAt foundation tier, covering biological facts, gCSE French and German could face the chop". With their content being removed from the GCSE options, teachers and others interested in Geography. Year 13 are expected for university admission, with mostly only design and technology subjects and performing arts retaining their controlled assessment contributions. Like conditions for much maths gcse coursework statistics the non, level 1 and Level 2. A comprehensive site exploring Geography, this table shows the majority of subjects which are consistently available for study. The CIE A – maths gcse coursework statistics A level students will be sitting the same ma20013 coursework as the students in UK concurrently.
And various catering and nutrition qualifications, but is not obliged to reject a candidate ma20013 coursework misses maths gcse coursework statistics requirements. Students can achieve any grade in the scheme. Which would not have met the requirements of a B, many of those who achieve below this standard will later retake GCSE English and mathematics to improve their grade.
Assessed in the second year of study. Changing the marking criteria and syllabi for ma20013 coursework subjects, such as to indicate that an examiner maths gcse coursework statistics offensive material or hate speech within a student's responses. 1 to 5, which are folded into "food technology". | 677.169 | 1 |
This file is dedicated to MATH DEFINITION DIAGRAM and you may get it using a one step process. All you need is to follow this option to get the
needed file in pdf version which was checked 48 hours ago.
Math-Geometry and Measurement Quiz 1 Math-Geometry and Measurement Quiz 1Fifth Grade MathematicsQuizJill TonelliName DateInstructionsCopyright 2000-2002 Measured Progress All Rights ReservedName Quiz Math-Geometry and Measurement QDate Teacher Jill Tonelli1 The fifth-grade students at Westfield Elementary School have decided to plant a rectangular flowergarden near the base of their flagpole A di...
Microsoft Word - 2013-14 UNC Math Contest.docx U N I V E R S I T Y ofNORTHERN COLORADOCollege of Natural and Health SciencesSchool of Mathematical SciencesUniversity of Northern Colorado State Mathematics ContestThe School of Mathematical Sciences at the University of Northern Colorado is pleased toannounce the 22nd Annual UNC State Mathematics Contest for high school and middleschool students The...
Math Placement Test Study Guide Math Placement Test Study GuideGeneral Characteristics of the Test1 All items are to be completed by all students The items are roughly ordered from elementaryto advanced The expectation is that less prepared students will answer fewer questions correctlythan more prepared students2 The test consists entirely of multiple choice questions each with five choices3 The ...
Math Handout MPM 2D1 P Reaching New Heights P Applications of Similar TrianglesExercise 1 Finding heights of tall objects using a 45 45 90 triangleThis method is often used to predict where a tree will fall when damaged trees in a wood lot are beingcleared or which trees should be cut down based on height1 The teacher will tell you a spot for which you are to determine theheight2 Hold a portion of...
Lecture 9 7 Math 150B Nguyen 1 of 6 9 7 TAYLOR POLYNOMIALS and APPROXIMATIONSPolynomial Approximations of Elementary FunctionsTo find a simple linear function P x a0 a1 x that approximates another function f1Choose a number c in the domain of f at which f and P have the same value i e P c f c1The approximating polynomial is said to be expanded about c or centered at cAn additional requirement is P...
Microsoft Word - Math parent Handout October newsletter ls.doc How can I support my child s mathematics learningEveryone can learn Math First and foremost believe in your child s ability to learnmathematics Everyone can improve when provided with good teaching coachingsDoencouragement and practiceDo have high expectations for your child Research shows that when you believe yourchildren can learn t...
Math in Basketball Take the Challenge Answer Key FINAL 8.16.12 Math in Basketball Take the ChallengeANSWER KEYWhen NBA player Elton Brand steps to the free throw line a number of key variables caninfluence his shot Your challenge is to use the 3 key variables and Elton s stats to figure outthe maximum height the ball reaches on its way into the basket to make the shotThis activity can also be comp...
Math mattersCommunitySummer 2012CollegesFour-YearHigh InstitutionsSchoolsPre-Calculusthree communities one goalJune 6-9 2012 on the Iowa State University campusa conference for High School and Community College Teachers of PreCalculus and CalculusComing soon to Iowa State UniversityMidwestern GrapH TheorY MIGHTY LIII Conference - September 21-22Plenary speakers are Persi Deadline for registration ...
7th Grade Common Core Appendix to Math Curriculum Common Core Appendix to Math Curriculum 7th GradeBelow we have matched up the new Core Curriculum Standards with our own New York State Math standards forreference On the left are the Common Core Standards in the order that they have been presented on their websitehttp corestandards org and on the right are the codes and standards they pertain to b...
Singapore Math First Grade UnderstandingSingapore Math curriculum for grades 1-6 is called Primary Mathematics There are two booksthat are used together a textbook and a workbook The textbook is used as a class activity book during thelesson and the workbook is used for independent practice The year is divided up into two parts so thereis a textbook and workbook A and a textbook and workbook B for...
Stanford Math Circle Feb 12 2006 Paul Zeitz Here are some exercises and problems related to my talk today If you have anyquestions or want a hint feel free to contact me at zeitz usfca eduPascal s TriangleWe proved that the probability a randomly selected element of PT is even is 100By that we really meant the asymptotic frequency of even elements in PT is100 i e as we increase the number of rows ...
Week 6 Accelerated Math Packet 6th grade Week 4 Math Packet2nd Six WeeksNo Work No CreditDue Friday October 28thEach problem must haveQuestion highlighted or underlinedClear workJustification written in a complete sentenceCorrect AnswerFormatWorked down the pageCompleted on notebook paperLegibleAnswer boxed with label if any1 In Zelda s home town the ratio of cars to people is 325 to 1 000 Write t...
Parent Help for BMS Math Instruction Parent Help for BMS Math InstructionHelping students with their Math homework is challenging during the middleschool years Instructional methods have changed At BMS the fifth grade usesthe Everyday Math Program and sixth seventh and eighth use MathscapeStudents in algebra and pre-algebra have different textbooks Each of the BMSmath teachers has a website that c...
teacher.shaker.k12.nh.us/vwest/Parent Help for BMS Math...Instruction.pdf
Fifth Grade Math Rubric Fifth Grade Math Rubric Quarter ThreeStandard Needs Strengthening Developing Secure ExemplaryUnderstands and Is beginning to Inconsistently or Independently and Applies and extendsperforms operations Order decimals to partially consistently Ordering decimalswith decimals and the thousandths Orders decimals to Orders decimals to to the thousandthsfractions place the thousand... | 677.169 | 1 |
Math software 162
Paragraph about Math software
When you enter an expression, Scientific WorkPlace derives operations from the natural notation used. For example, if you enter or Math software height="23px" border=none src="swptransparent.png" alt="">, Scientific WorkPlace understands what you are writing and will evaluate the expressions appropriately. This is not possible in other computation systems because they force you to be exact with your input syntax. To illustrate this, we enter an integral using SW's natural notation input methods (described later) and press CTRL+e to evaluate the integral (or choose Evaluate from the Compute menu). Ufology src="SW-SWP_White_Paper__5.png" alt="MATH" />
Scientific WorkPlace automatically places an equal sign between your original expression and the result of the integration. You can use the result of the integration (or the original expression) in subsequent work by selecting the entire expression or some portion thereof. You can perform standard Windows Copy/Paste operations on the expressions as well. You can save a selected item as a "fragment" (this is the term we use to signify math, text or graphic objects you save for later recall). Fragments are discussed in greater detail in the "Features Overview" section. | 677.169 | 1 |
In some programs, all it will require to move an exam is notice taking, memorization, and recall. However, exceeding inside of a math course takes a special kind of energy. You can not only exhibit up for just a lecture and observe your instructor "talk" about math and . You understand it by doing: being attentive at school, actively studying, and fixing math challenges – regardless if your teacher has not assigned you any. Should you end up struggling to accomplish very well with your math class, then stop by best site for fixing math difficulties to learn how you could become an improved math scholar.
Low cost math authorities on the internet
Math programs follow a all-natural development – every one builds on the knowledge you've received and mastered in the past study course. In the event you are locating it challenging to adhere to new concepts in class, pull out your aged math notes and assessment previous content to refresh on your own. Be sure that you meet the prerequisites before signing up for the course.
Evaluate Notes The Night In advance of Course
Hate each time a instructor phone calls on you and you've forgotten the way to solve a particular challenge? Prevent this instant by examining your math notes. This could help you ascertain which principles or issues you'd love to go over in school another working day.
The considered doing homework just about every night could appear troublesome, but if you'd like to reach , it can be important that you continuously observe and learn the problem-solving approaches. Make use of your textbook or on the internet guides to work as a result of top math troubles on a weekly basis – even if you've no homework assigned.
Utilize the Health supplements That include Your Textbook
Textbook publishers have enriched modern day publications with further product (which include CD-ROMs or on-line modules) that could be used to assistance pupils attain more apply in . Some materials could also include an answer or clarification guide, which often can allow you to with functioning as a result of math challenges all on your own.
Read through In advance To remain Forward
If you need to reduce your in-class workload or the time you devote on homework, make use of your free time immediately after university or within the weekends to examine forward for the chapters and ideas that could be protected the subsequent time you are in school.
Critique Previous Tests and Classroom Illustrations
The operate you need to do in class, for homework, and on quizzes can provide clues to what your midterm or closing test will look like. Make use of your aged assessments and classwork to create a particular analyze guidebook for your personal impending exam. Glimpse in the way your teacher frames queries – this is certainly almost certainly how they may surface in your take a look at.
Discover how to Function From the Clock
It is a well-known examine idea for men and women taking timed exams; in particular standardized exams. If you have only 40 minutes for the 100-point exam, then you can certainly optimally shell out four minutes on every single 10-point problem. Get information about how very long the exam is going to be and which styles of thoughts are going to be on it. Then plan to attack the simpler concerns first, leaving oneself adequate the perfect time to invest within the a lot more demanding types.
Improve your Sources to obtain math homework enable
If you are having a tough time knowledge concepts at school, then you'll want to get support outside of class. Inquire your pals to create a examine group and go to your instructor's office hours to go about challenging troubles one-on-one. Show up at research and critique classes when your instructor announces them, or use a non-public tutor if you need one particular.
Converse To On your own
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UbLb UI INIINlTY
LLO ZIPPIN
Random House
$1.95
New Mathematical Library
ÍbÏb O1 1^11^1À3
by
Leo Zippin
QuccnsUollcyc
¯
RANDOM HOUSE
Illustrations by Carl Bass
First Printing
T Copyright, 192, by Yale University
All rights rserved under International and Pan-American Copyright
Conventions. Published in New York by Random House, Inc., and
simultaneously in Toronto, Canada, by Random House of Canada, Limited.
School Edition Published by The L. W. Singer Company.
Library of Congress Catalog Card Number: 61-12187
Manufactured in the United States of America
Note to the Reader
_his book is one of a series written by professional mathematicians
in order to make some impor tant mathematical idea interesting
and understandable to a large audience of high school students and
laymen. Most of the volumes in the NeuMat/emattca/Ltmaqcover
topics not usually included in the high school curriculum; they vary
in difculty, and, even within a single book, some par ts require a
grater degree of concentration than others. Thus, while the reader
needs little technical knowledge to understand most of these books,
he will have to make an intellectual efort.
If t he reader has so far encountered mathematics only in classroom
wor k, he should keep in mind that a book on mathematics cannot be
read quickly. Nor must he expect to unders tand all par ts of the book
on frst reading. He should feel free to skip complicated par ts and
return to them later ; often an argument will be clarifed by a subs
quent remark. On the other hand, sections containing thoroughly
familiar material may be read very quickly.
The best way to lear n mathematics is to domathematics, and each
book includes problems, some of which may require considerable
thought. The reader is urged to acquire the habit of r eading with
paper and pencil in hand; in this way mathematics will become in
creaingly meaningful to him.
For the authors and editors this is a new venture. They wish to
acknowledge the gener ous help given them by the many high school
teachers and students who assisted in the preparation of thes mono
graphs
. The editors are interested in reactions to the books in this
series and hope that readers will wr ite to: Editor ial Committee of the
NML ser ies, in care of THE INSTITUTE OE MATHEMATICAL SCIENCES,
NEW YORK UNIVERSITY, New Yor k 3, N. Y.
The Editors
v
NEW MATHEMATICAL LIBRARY
0t/erttt/esut//èeanncuncæ0 remq
¡. NUMBERS: RATIONAL AND IRRATIONAL by Ivan Niven
2. WHAT IS CALCULUS ABOUT? by W. W. Sawyer
3. INTRODUCTION TO INEQUALITIES by E. Beckenbach
and R. Bellman
4. GEOMETRIC INEQUALITIES by N. D. Kazarinof
5.
THE CONTEST PROBLEM BOOK, Problems from the Annual
Hig School Contests of the Mathematical Assciation of America,
compiled and with solutions by Charles T. Salkind
0. THE LORE OF LARGE NUMBERS by P. J. Davis
7
.
USES OF INFINITY by Lo Zippin
8. GEOMETRIC TRANSFORMATIONS by I. M. Yaglom, trans
lated from the Russian by Allen Shields
vi
Contents
Preface J
Chapter 1 Popular and Mathematical Infnities 5
Chapter 2
From Natural Numbrs to V2 14
Chapter J From V2 to the Transfnite J8
Chapter 4 Zig-Zags: To the Limit u the Limit Exists 59
Chapter 5 The Self Perpetuating Golden Rectangle 7 5
Chapter 6 Constructions and Proofs 96
Solutions to Problems
12 1
Bibliography
150
víí
USES OF INFINITY
Preface
Most of this book is designed so a to make little demand on the
reader's technical comptence in mathematics; he may b a high
school student bginning his mathematics now or one who ha put
away and forgotten much of what he once knew. On the other hand,
the book is mathematical except for the frst chapter-that is to say,
it is a carefully reaoned presentation of somewhat abstract idea.
The reader who fnds the material interesting must b prepared,
therefore, to work for it a little, usually by thinking things through
for himslf now and then and occaionally by doing some of the prob
lems listed. Solutions to some of these are given at the end of the
book. But it will not pay the reader to stop too long at any one
place; many of the idea are repated later on, and he may fnd that
a scond view of them leads to understanding where a frst view wa
bafing. This style of presentation is imposed upon an author by
the nature of mathematics. It is not possible to say at once all of the
key remarks which explain a mathematical idea.
Many a reader is prhap wondering whether it is possible for
fellow human bings to communicate upon a topic as remote-sounding
a "uses of infnity"; but, a we shall %¸ any two people who know
the whole numbrs,
1, 2 , 3, 4, 5,
can talk to each other about "infnities" and have a great deal to say.
I have written this book from a point of view voiced in a remark
by David Hilbrt when he defned mathematics a "the science of
infnity". An interesting theorem of mathematics difers from in
tersting results in other felds bcaus over and above the surpris
and bauty of what it says, it ha "an apect of eternity"; it is always
part of an infnite chain of results. The following illustrates what I
mean: the fact that 1 + 3 + 5 + 7 + 9, the sum of the frst fve
odd integers, is equal to 5 times 5 is an interesting oddity; but the
3
4 US ES OF I NF I NI TY
theorm that fo a// n the sum of the frst n odd integers is n
2
is
mathematics.
I hop that the reader will believe me when I say that professional
mathematicians do not profess to understand better than anybody
els what, from a philosophical point of view, may be called "the
meaning of infnity". This is proved, I think, by the fact that most
mathematicians do not talk about this kind of question, and that
thos who do, do not agree.
Finally, I wish to express my espcial thanks to Mrs. Henrietta
Mazen, a teacher of mathematics at the Bronx High School of Science,
who slectd and edited the material in this monograph from a
larger body of material that I had prpared. The reader who enjoys
this book should know that in this way a considerable role was played
in it by Mrs. Mazen. I am also indebted to Miss Arlys Stritzel who
supplied most of the solutions to the problems posd in the book.
C HA P T E R O NE
Popular and Mathematical Infnities
One who has not worked in the mathematical sciences is likely to
doubt that there is any H of infnity if "to H´ something means to
acquir some form of control over it. But the H of infnity in pre
cisly this sns constitutes the profession of mathematicians.
Other professions also H infnity somewhat
.
The architect and
the engineer have their tables of trigonometric functions, logarithms,
and the like. But they do not need to remember that these are calcu
lated from a large number of terms of certain appropriate infnite
sries. They draw frely from an inexhaustible resrvoir of mathe
matical curves and surfaces, but they do not need to be conscious of
infnities. The philosopher and theologian are conscious of infnity,
but from the mathematician's view they do not H it so much as
admir it.
t
r x¹ x
7
Figure 1
.1. sin x = x
° -
q == ¬q ...
6 12 5
5
6 US ES O F I NF I NI T Y
The mathematician also admirs infnity; the grat David Hilbertt
said of it that in all ages this thought has stirrd man's imagination
most profoundly, and he described the work of G. Cantort as intro
ducing man to the Paradis of the Infnite. But the mathematician
also uses infnities and, as the next two chapters will show, he is the
world's greatest collector of infnities-infnitely many arrays of
infnities of all typs and magnitudes. They are his raw materials and
also his tools.
Befor turning to mathematics, let U spnd a moment with some
examples of the popular "everyday" infnities which, as we shall S,
are not too distantly rlated to the mathematical infnities. We begin
with a folk-saying: "Ther ar always two possibilities." Lt U her
call them "Zero" and "One". Since each choice made brings two
new alternatives, this suggests the pictur of infnity shown in Figur
1 .2 .
Figure 1 .2. There are always two possibilities
Figure 1 .3. An imitation of infnity
t David Hilbert (1 82-1943) wa one of the leading mathematicians of the
twentieth century
.
t G. Cantor (18451918) created set thery.
P OP UL AR AND MATHEMATI CAL INF I NI TI ES 7
Next, many pople can rcall a certain box of baking soda on which,
in their youths, they saw their frst pictur of infnity. This box hag
on it a picture of a box on which there was a pictur of the same box
showing another, and so on. Figur I.J conveys the imprssion this
gave.
Next, ther ar intimations of infnity designed espcially for
children. Japanes artisans make wooden dolls that opn up and
contain a similar doll, inside of which there is another doll and so on
through a squence of fve or six.
Also, pots have employed the word in ways that are not far
rmoved from mathematical aspcts of infnity. Juliet's line about
her Jove for Romeo, "the mor I give U thee, the mor I have",
exaggerates a characteristic proprty of cardinal infnities; to Blake's
image, "To hold infnity in the palm of your hand and eternity in an
hour", corresponds the mathematical fact that a sgment as short as
the palm's "life-line" has as many points on it as the infnitely long
line. In Anthcnq am C/ecµatra, General Enobarbus describes,
Cleopatra: "Time cannot wither nor custom stale her infnite variety."
Hugo describes Shakespare: "Genius is a promontory jutting into
the infnite."
In a lighter vein, Rostand's Cyrano de Bergerac, among other
fanciful schemes for going to the moon, uss a jocular version of
mathematical induction: "I stand on a platform holding a strong
magnet which I hurl upwards. The platform follows. I catch the
magnet and hurl it up again, the platform following, and repating
this in stages, I ascend to the moon."
Very instructive are the metaphysical arguments of Zeno t dircted
to the conclusion that physical motion is impossible. He is quoted
somewhat as follows: "Achilles cannot overtake a feeing tortois
because in the interval of time that he takes to get to where the
tortois was, it can move away. But even if it should wait for him
Achilles must frst reach the half-way mark between them and he
cannot do this unless he frst raches the half-way mark to that mark,
and so on indefnitely. Against such an infnite conceptual rgrssion
he cannot even make a start, and so motion is impossible." Another
prtty paradox of Zeno tries to show that it is impossible that space
and time should nct be infnitely divisible. Still another is usually
quoted this way: "The moving arrow is at each instant at rest."
t A Greek philosopher of the Eleatic school who lived in the ffth century,
M.Ü.
8 US ES O F INF INI T Y
These paradoxes deal with an important application of mathematical
infnities and desrve to be discussd here, the mor so becaus the
moving-arrow paradox is a neat formulation of the mathematical
concept of motion. A mttan is analogous to a time-table; mor
prcisely, it is a ]uncttan which associates a defnite point in fctional
"space" to each of certain moments in a fctional "time". From this
point of view, the statement that the arrow is "at rst" at a given
instant means that its position is defned; this gives the function.
Functions defning a motion can be constructed like any other
mathematical functions, that is to say by a suitable table of values,
by a formula, or by a recursive description.
If Zeno's tortois starts one foot ahead of Achilles and moves
1 1 1 1
2
+
4
+ 8 +
16
+
. - .
feet
(which is prcisly one foot) in
1 1 1 1
2
+
4
+ 8 +
16
+
- -
,
seconds
(which is precisly one scond), while Achilles moves
1 1 1 1
1 +
2
+
4
+ 8 +
16
+
- ..
feet
(which is prcisly two feet) in
1 1 1 1 1
2
+
4
+ 8 +
16
+
32
+
.- .
seconds
(which is prcisly one scond), then the race is ended in one scond.
Befor the work of Eudoxus (350 B.C.) and of Archimedes ( 150
B.C.) thes infnite sries could not be understood. In the sventeenth
century with the development of the calculus, the logic of infnite
sries had to be rediscovered. These sertesare probably not needed to
"answer" Zeno, but they meet him very nicely on his own grounds.
1 1
e»»g
9'
8'
1 1 1 1 1 1
7' 6' 5' 4' 3' 2'
1
Figure 1 .4
The scond of Zeno's pardoxes derives some of its interest frm
the fact that between every pair of points on a line there is another
point. It follows that a sgment contains an infnite descendtnq
squence, that is to say an ordered sequence of points without a
smallest term. The example in Figure 1 .4 shows this; the fractions of
P OP UL AR AND MA THE MA TI C A L INF I NI TI ES 9
the form lJn, arranged in the order of size, have no term that
precdes all the others. No corrsponding squence exists in the st of
whole numbers. It is likely that Zeno was entirely concerned with the
prblem which confronts applied mathematicians: that mathematics
is an idealization of exprience not necessarily "true to life". Nowher
is this more evident than in the simple fact that the geometric sgment
is infnitely divisible, but the matter in a material wire is not. Of
course, this fact was not so convincingly established in Zeno's time
as it is in our own. Nonetheless, in our time as in his, the mathe
matical segment srves as a model for many spcifc problems dealing
with material bodies (vibrating strings, fexible beams, rigid bodies).
Above all it serves as the model for the ttme-ccnttnuum and the one
dimensional sµae-ccnttnuum, and in this U it dominates our con
ception of the world about us.
lim
Î +
` = lim ¸.+ ·¡= .+ lim ·= 1
@¬w Î @¬m Î @¬m Î
Figure 1 . 5
. A mathematical formulation of an everyday observation: Two
friends-one four years older than the other-appear to approach the same age
a the year pas
Finally let U look at a pair of everyday examples which have the
favor of mathematics. Anyone who has a friend several years older
than himslf notices how the passage of time thins out the diference
between the two ages. We have her an example of a pair of variables
whos difernce is constant but whos ratio is not; in this example,
the ratio approaches 1.
If
and
then
and
C denotes the cost
S denotes the slling price,
S - C
÷
C
÷ is µeruntt µrqûtcn
ccst,
S - C
÷
S
÷ IS µer umt µït cn
sales µrùe.
If
and
then
and
Figure 1.6
it costs & to produce a
sweater,
it is sold for $12
12 8 1
8
=
2
is µrc]tcnccst,
12 - 8 1
.
-12-
=
a
lS µrc]tcnsa/es
µrùe
10 US ES OF INF INI T Y
The last example is drawn from commerce. When an object is
bought for $1 and sold for $2 , there is a 10 pr cent proft if proft is
fgured on the purchae price ; but ther is a 50 pr cent proft if this
is fgurd on the slling price. An object can be sold at an arbitrarily
large prcentage proft on the purchase price, and it can b sold at
99 pr cent or 99.99 pr cent proft on the selling price, but it cannot,
practically or theoretically, be sold at a 10 pr cent proft on the
slling price. The reader should persuade himslf of this because it is
a natural example of a situation in which one might be led to wonder
about a "limit" which does not exist. Perhaps the easiest way to
check on this particular problem is to try diferent selling prices.
Lt us now turn to the uss of infnity in mathematics. It will help
the reader to think of thes as falling into four categories.
The frst category is illustrated by the theorem of geometry: If tuc
stdescJatrtanq/eare eçua/the the èaseanq/esareeçua/( Proposition 5,
Book I of Euclid) .
PROOF. Given that AC = BC
;
t see Figur 1.7
.
Comparing the
triangle ABC with itself, but rading it next as BAC, we fnd
that AC = BC and 4ACB = 4BCA. Therefore, by Propsition
±¸ Book I of Euclid, angle CAB must equal angle CBA.
,
- s
Figure 1 .7
This proof that the base angles of an isosceles triangle are equal
amounts to suprimposing the triangle on itslf so that A goes
into B, B into A, and C into C.
The statement and proof depnding on a proposition often referred
to as "two sides and the included angle", or "s.a.s." ( meaning "side-
t The statement AC = ßC means that the lengths of the line seg
ments AC and BC are equal . In many books, the distance from a point P to
a point Q is denoted by JQ, but for reasons of typography it will simply be
denoted by PQ in this book.
P OP UL AR AND MATHEMATI C AL I N F I N I TI E S 1 1
angle-side") make no mention of infnities. But the class of isosceles
triangles (of all shaps and sizes) is an infnite class and the theorem
holds for ever one of them.
Figure 1 .8. Pascal 's Triangle. The nth row of this array gives the coef
cients that occur in the expansion of (a + b)
n
The second category is illustrated by certain numbers, the
"binomial coefcients", known to the Pythagoreans and to earlier
civilizations, but associated with Pascal ( 1620) becaus of his use of
mathematical induction in discussing them
.
The binomial coefcients,
we recall, ar the coefcients that occur when a binomial a + è is
multiplied by itslf n times. For example, when n = 3, we have
(a + è)
-
= a
-
+ 3a
t
è + 3aè
t
+ è
-
,
and the binomial coefcients are 1 , 3, 3, 1. Since n may be any
positive whole number, we have an explicit infnity of cases, each of
which involves a fnite collection which we are invited to count.
The problem of fnding a tangent line to a given curve at a point
on the curve belongs to the third category. It is easy to se that this
problem is associated with an infnite process, because tangency of a
line to a curve can be determined by using arbitrarily small pieces of
the curve and small segments of the line. It turns out that the mathe
matical process which solves this problem also solves the physical
problem of defning tnstantanecus »e/ccttq, this is briefy spaking the
number read on an automobile
'
s spedometer. Velocity of motion and
slop of a line are /tmttsof rattcs
Figure 1 .9
. Tangency at a point is a local afai r
12 US ES O F INF INI T Y
ï
tärrê
|1h]êh| |lhê
P
�t
Figure 1 . 10. The Blope of the tangent line (ratio QR/PR) is the limit of the
slope of the chord (ratio QR'/PR') W the point Q' approaches P along the
curve
The fourth category belongs to abstract st theory and is concered
with infnite cardinal numbers
.
It is strikingly illustrated by the
paradox that a circle can sem to have mor points on it than the
infnite line. Figure 1 . 1 1 shows this. Each point of the line is paird
of against a point on the lower smicircle
.
Even if we match two
points of the circle to one on the line ther ar still points (P and
Q) left over.
- s
, + :
·
Figure 1 . 1 1
How infnity i s handled i n each of thes cases i s discusd through
out this book but can be described briefy 8 follows: To handle the
frst typ, we U a single rpresntative object; since it is in nowis
spcial, it srves equally well for all. Problems of the scond typ ar
usually handled like the frst but sometimes fall into a patter that
can be treated by mathematical induction, or the essntially equiva
lent "principle of infnite descent", also equivalent to the "principle
of the frst integer", 8 we shall se.
POP U L A R AN D MA T HEM A TI C A L I N F I N I T I E S 13
The third category includes a diversity of infnite processs asso
ciated with the notion of "limit". We shall study only one such process
leading to a defnition of /eqth of certain fgures; this i n turn leads
to a defnition of the sum of an infnite sries.
Although the summation of infnite sries is closely related to the
problem of defning ( and calculating) the area bounded by a given
closd curve, we shall not deal with ara or with the earlier mentioned
problem of tangency. We refer the rader to a good calculus book
(se also Courant and Robbi ns, Bhat u Matheatùsf New York:
Oxford University Press, 1941) .
Problems i n the fourth category called for the invention of a new
typ of mathematical rasoning and waited for the genius of Cantor.
The arguments which Cantor employed are straightforward gener
alizations of the ordi nary processes of counting, but he used them
boldly-applying them directly to infnite sts. Perhaps his greatest
single discovery was the fact that there are diferent cardinal infnities
and that, in particular, the set of points on a line-sgment is an
incomparably "richer" i nfnity than the st of all whole numbers.
This and other aspcts of his work wi ll be touched on in Chapters 2
and 6.
The book will clos with some i ndications of the way in which this
last aspct of infnity has been incorporated into the mai n mathe
matical stream of uses of infnity
.
C H A P T E R T W O
From Natural Numbers to �¿
The main body of this chapter consists of a parade of short sec
tions each of which concers itself with some particular infnite set,
infnite process, or point of view or technique for controlling infnity.
We have tried to treat each of these separately, a much as possible ;
however, the frst section begins with three infnities!
2.1 Natural Numbers
At the head of the parade of infnities, set of from the others,
come the ordinary numbers. We are not going to be able to con
struct infnite sets, or prove anything signifcant about them, unless
somehow we start with at least one infnite set, already constructed
for us. Such a set exists, namely the ordinary numbers: 1, 2, 3,
4, ô, 0,
. .
, and so on. Concering these, often called natura/
numèers, we are free to assume the following facts:
a) Each natural number has an immediate successor, so that the
procession continues without end.
b) Ther is no repetition; each number is diferent from all the
preceding numbers.
c) Every whole number can be reached in a ]ntte number of steps
by starting at 1 and counting up, one at a time, through the
line of successors.
14
FROM NATURAL N UMB E RS TO v 15
2.2 Discussion of Sequences
Since all numbers in the set of natural numbers cannot be written
in a fnite time we use "
. . .
", that is, suspnse dots or iteration
dots, usually just three of them; they correspond to the words "and
so on" or "et cetera", or "and so forth". They are used in mathe
matics following a short sequence of terms to mean that the indicated
set is non-terminating and that the sequence is constructed according
to a transparent scheme. The following examples will show what is
meant ; the rader is invited to put in the next few terms in each case.
I, 4, 7, I0, I3,
ô, 3, ô, 3, ô, 3,
ô, 3, 7, 4. I0, 0, I4,
I, I, 2, 3, ô, 8, I3, 2I, 34,
3, 0, 9, I2, Iô,
3, 9, 27. 8I, 243,
6, 3% , 2I % , I29ô, ,
2, 3,
·¹·
7, I2, I7, 29, 4I, 70, 99, ,
2, y, 2 + y, \2 +
V,
2 + \2 +
x2
,
I, 0, I2, 20, 30, 42, ,
I, 2, 0, 24, I20, 720, ô040, 40320,
Our discussion of the use of iteration dots amounts to a de fnition
of the idea of a sequence. This concept is so important to everything
that follows that it is worth the reader' s while to have it defned
explicitly.
Bqde]ntttcnall sequences are derived from the basic sequence
(I) I, 2, 3, 4, · ·
·
by replacing each number of ( I) by some object. Thus, for example
i)
is a sequence of distinct letters,
ii) a, è, 0¡ a, è, 0, · ·
is a periodic sequence of letters,
iii) I, I, I
· · ·
,
is a sequence of I 's, and
iv)
is a sequence of positive square roots.
\2I0,
. . .
16 US ES O F I NF I NI TY
Since a sequence is infnite we cannot be sure that we know what it
is unless we have a rule which tells us how to replace each numbr
in ( I) by the appropriate object. In cases i), ii), and iii) the rule is
quite well shown by the frst three terms and the use of the iteration
dots. In case iv), the rader will fnd that the terms can be calculated
from this rule : The general term is the squar root of the product of
three consecutive integers the frst of which is equal to the number
of the term itself. Thus the ffth term of iv) is ¸ô
·
0
·
7
The rule
can also be expressed in a convenient formula. Lt n designate the
number of the term we seek and let us call this term a
-
( read it:
'asub n"). Then
a
-
" y
n(n + 1)(n + 2) ·
However, this is by no means an obvious formula and it is possible
to get the frst fve terms by the use of many other formulas which
would give diferent later terms. Therefore, in order really to know
what sries is meant in iv), one requires the explicit rule above.
Whether this rule is expressed in words or by a nice formula is not
important.
We can summarize our discussion symbolically by writing
a
-
= f ( n)
where a
-
denotes the nth term of a sequence, that is, the object
which replaces the numbr n in ( I), and the symbol f (n) denotes
the rule for fnding a
-
, whether it is written as a formula or in
words.
Thus in cases i) to iii) :
i) f ( n) = a
-
, n = I , 2, 3,
. . .
,
r
for n "
I, 4, 7,
·
ii) f (n) = è, for n = 2, ô, 8,
. . .
0, for n " 3, 0,0,
. . .
,
iii) f ( n) = I , n = I , 2, 3, ¯ · .
There are instances when such dots have a diferent meaning. For
example, we may write "the ten digits used in our decimal system
are 0, I , 2, ·,0"
·
Here the dots do not indicate an infnite set;
they merly represent an abbreviation for the numbers 3, 4, ô,
0, 7, 8. Frequently mathematicians use dots U mean that a
sequence is non-terminating, but is constructed according to some
scheme the details of which are not important at the moment. Thus ,
in this sense,
V " I
·
4I4 ·
·
· or T "
3 I4Iô0
· ·
FROM NA TURA L N UMB ERS TO V2 17
means that v and T ( pronounced "pie") are not fnite decimals,
and that they bgin as shown. The reader will b able U tell the
meaning of the dots from the context.
Problems
Comment on the following uses of ..
.
":
2.1
.
solos, duets, trios, quartets,
. . .
.
2
.
2. singles, doubles, triples, home runs,
.
..
.
2.3. twins, triplets, quadruplets,
. . .
.
2
.
4. Monday, Tuesday, Wednesday, Thursday, Friday,
.
.
.
.
2
.
5. M, T, W, T, F, S, S, M, T, W,
. . .
.
2.3 Cardinals and Ordinals
The frst member of our parade, the set of natural numbrs, is
connected with the next two infnite sets in the procession. The three
infnite sets are :
a) The totality of whole numbers :
one, two, three, four, fve, six,
. . .
b) The collection of "rules" binding each to its successor :
one and one is two,
two and one is three,
three and one is four,
. . . _
c) The aggregate of ordinal numbers :
frst, second, third, fourth, ffth, sixth,
. . .
.
The set a) being given to us, b) is a record of observations which
we make on some and believe about all of the members of a) . Be
cause the rules are so transparently repetitive, they suggest that
we can survey all of a) in our minds even though we can only exhibit
a few of its members to our eyes. By making us conscious of the natu
ral ordering of the whole numbers, the set c) hints at how these can
b brought under our control, to some extent.
2.4 Arithmetic Sequences
The auqmentednatura/numèers0, I, 2, 3, 4, · · · are obtained
by putting zero in front of the sequence of ordinary whole numbers.
We shall also call them ncn-neqattre tnteqers, or simply tnteqers.
18 US ES O F I NF I NI T Y
If we begin with zero and writ every second number theraftr,
we get the sequence of the even integers :
0, 2, ±, U,
8, ·
·
· EVEN;
and if we begin with I and write every second numbr thereaftr,
we get the mintegers :
I , 3, ô, 7, 0, · · · ODD
.
A little thought will convince us that every natural number be
longs to one or the other ( not both) of these two sets: Every number
is either even or odd. It will be useful to us to say this a little mor
elaborately: Every integer is of the form two-times-some-integer
cr it is of the form one-more-than-two-times-some-integer. Using
convenient letters in place of words, we writ: Every intger is of
the form
n
-
2ç
where çis an integer
.
or n = ++ 2ç
Similarly, counting by threes, we fnd that every integer is of the
form
n
-
3ç
,here çis B integer.
or n
-
I + 3ç
Problem
or n
-
2 + 3ç
2.6
.
This is equivalent to saying that every integer is in one of three non-over
lapping infnite sets (and only one of them) ; show what these sets are.
In general, if B denotes any non-zero integer we can separate all
integers into B distinct classes :
0,
I ,
2,
B-1,
B,
I + B,
2 + B,
(B - I) + B,
2B,
I + 2B,
2 + 2B,
(B - I ) + 2B,
3B,
. . .
,
I + 3B,° °
.
,
2 + 3B,
. . .
t
(B - I) + 3B,
. .
.
the last line is less awkwardly written as
B - 1, 2B - I , 3B- I, · ·
· .
FROM NATURAL N UMB ERS
TO \2
Thus if B denotes 5, say, we get fve (=B) classes :
0, 5,
1, 0,
2, 7,
3, 8,
4, 9,
10, 15, ·· · ,
1 1 , 10, · · · ,
12, 17, · · · ,
13, 18, ·· ·,
14, 19, · · ·
.
19
In some applications, these diferent classes are cal led restdue
c/assesmcdu/cB; in the operation of division by B, they correspond
to the rmainders.
We can express the contnt of the above table as foll ows : Every
intger n is of the form
n = gB + r
where g and r are integers ; g is the multipl ier ( quotient, if we are
thinking of division) , r is the residue ( remainder) and has one of the
values in the fnite set of values 0, 1, 2, 3, · · · , B - 1 .
This is the most important formula in arithmetic; in case r = 0, B
is called a factor of n
·
( The value of g¡ as far as this defnition
goes, is not important.)
In the case B " 10, the residues correspond precisely to the digits
0, 1, 2, 3,
. . .
, 9
.
Problem
2.7. Prove that n and n + 1 do not have a common factor, i. e. , if an integer
g ¢ 0, 1 divides the integer n it does not divide n + 1 .
2.5 Positional Notation
We now meet µcstttcna/ nctattcn, our way of writing augmented
numbers. This familiar scheme, on the base 10 ( Hindu-Arabic nota
tion) , is illustrated by the development :
nineteen hundred and sixty
one thousand, nine hundrds, six tens, no units
one, nine, six, zero
1, 9, 0, 0
196.
It has these excellent features ( not specifc to the base tn) :
a) units of diferent sizes convenient for all uses,
b) simple rules connecting consecutive units,
2 US ES OF I NF I NI TY
c) economy of expression with rapid growth of numbers repr
sented, and
d) fxed set of digits
.
Since only the digits and their relative positions fgure in the writing
down of numbers, it follows that all of the operations of arithmetic
must be describable in terms of the digits, if proper account is taken
of position. The positional system, with base 6, was used already in
ancient Babylon. The Hindu-Arabi c decimal system was made
popular in Europe in the thirteenth century principally by Lonardo
of Pisa, also called Fibonacci, t who learned it in his business travels
in Africa. He wrote a textbook in mathematics : I/ Ltèer Aèact,
published in 1202; a second edition in 1228 survived for many cen
turies.
The formulation of arithmetic techniques in symbols is the bgin
ning of algebra. That multiplication distri butes over the summands,
as in the model below, is expressed by the equation:
a· (è + c + d) = a
·
è + a
·
c + a· d,
e. g. , 4321 × 567 = 4321 × (50 + 6 + 7)
= 4321 × 50 + 4321 × 60 + 4321 × 7.
To use positional notation well, one must know the multiplication
table-at least up to "ten times ten". Although all of us know this
table, not many of us have looked at it as closely as we shall now
.
The table is shown in Figure 2. 1.
The rulings call attention to the fact that this table i s a nest of
multiplication tables of increasing size, a one by one table, two by two
table, three by three, etc.
.
I t t ô Ì f 8 ¡0
I
t ô f ¡0 ¡I ¡t ¡ô ¡f
M
I f 8 ¡I ¡t ¡f D It IÌ I0
t f ¡I ¡ô M It M D M M
t ¡0 ¡t M It M M M M M
ô ¡I ¡f
It M M tI M M M
Ì
¡t D M & tI W M M Ì0
f ¡ô It II M M M M ÌI M
8 ¡f IÌ M M M M ÌI N M
¡0 M M M M M M M M ¡M
Figure 2.1
t Fibonacci means "son of good fortune".
�
�
Figure 2.2
FROM NATURAL N UMB ERS TO y2 21
The boomerang fgures, or comer-frames in Figure 2.2, called
ymosby the Greek geometers, are also important for the examples
which follow.
It was noticed, probably before 50 B.C., that the sum of the in
tegers in any lower gnomon-fgure is a perfect cube. Thus:
1 = 13 , 2 + 4 + 2 = 8 " 23 ,
3 + 6 + 9 + 6 + 3 = 27 = 33 ,
The sum of the intgers in any square table, whether two by two or
three by three, and so forth, is a perfect square. Thus :
1 = 1
2
, 1 + 2 + 2 + 4 = 9
" 3
2
,
1 + 2 + 3 + 2 + 4 + 6 + 3 + 6 +
9 = 36 = 6
2
, • • •
The table is also a picture of the double distributive law, as Figure
2.3 shows.
• ••
• •
• •
• •
�..
• •
••
• •
(1 + 2
)1
= l' + 2 ª 1 (1 + 2) + 2(1 + 2)
• •• W
••
•• • ••
••
W .
•••
•••
.N
• ••
• ••
•••
(1 + 2 + 3)1 ¯ l' + 2 + 33
¯ 1( 1 + 2 + 3) + 2(1 + 2 + 3) + 3(1 + 2 + 3)
Figure 2.3
The proof of these facts for the given ten by ten table is a matter
merely of verifying some si mple sums and products. One naturally
asks if the corresponding facts can be demonstrated for every kby k
multiplication table where k is any positive whole number
.
Problem
2.8. Verify that analogous facts are true for a 12 by 12 table or a 15 by 15
table. Can you see any inductive scheme, or general principle?
While positional notation has the advantages already mentioned,
it also has one moderatly troublesome feature, shown by the example
99999 + 1 "
100;
22 US ES O F I NF I NI T Y
the addition of one unit can make a very considerable change in many
digits. All computing machines must take account of this, and prob
ably everyone has witnessed the dials of a milometer turning together
as the mileage on a car reaches 10,00.
In the case of decimals, the replacement of one number by another
is common, $999.99 bing thought a more attractive price than
$1,00. Notice for later use that 1 .00 00 and 0.999 999 difer only
by
.
00 01, in spit of the substantial diference in their appearance.
Thus
1 .00 and
1 .00 00 and
1 .00 00 00 and
.999
.999 999
.999 999 999
difer by
difer by
difer by
.01;
.00 01 ;
.00 00 01.
The use of the decimals 0. 1, 0.01, 0.01, to express the
tenth, hundredth, thousandth,
. . .
part of a unit allows us, in the
denary positional notation, to exprss arbitrarily small numbers �s
well as arbitrarily large numbers. However, as the Babylonians may
have been the frst to discover (before 210 B .C. ), there are very useful
numbers which cannot b expressed in this notation. The discovery
was made in connection with "small" numbers, like 1, \· 1·
,
but it does not really depend on size, only on form; thus also l,
�,
. . .
cannot b expressed in decimal notation. This inadequacy
of the ]ntte decimal system raises serious mathematical problems
.
One sees that the sequence 0.3, 0.33, 0.333, 0.3333,
. . .
repre
sents numbers smaller than i which approach i ; clearly there is no
terminating decimal which can represent } . However, from these
same considerations, it follows that the symbol
0.33333
. .
° g
a non-terminating decimal all of whose places are flled with 3's, must
represent i , provided of course that this notation means anything
at all.
The use of non-terminating decimals is the oQly way to achieve the
exatrepresentation of all numbers on the base 10. However, frm the
practical viewpoint of writing numbers or storing them in "memory
tanks" on computing machines, it is necessary to sacrifce exactness
and use the device of rounding. By carrying as many places as make
sense for the problem in hand and specifying the range of error, one
can be as accurate as necessary. The most common scheme of round
ing entails an error not exceeding i unit in the last place retained
(5 units in the place foHowing) ; all of this is shown in Figure 2.4.
F ROM NATURAL N UMB E RS TO y2 Z
If the product of two numbers is 1, then each of the two is called
the recµroa/ of the other. Thus since - times 10 is 1, - is the
reciprocal of 10 and 10 is the reciprocal of -
.
In formulas, if the letter
m stands for some numbr, and r for its reciprocal, one writes
r
·
m = 1, and also r
Inteer
Com-
To Thre
mon
Decimal
To Five Deci-
Frac- mal Places
tion
Places
&
¡
1
1 . 00 1 . 00
1
2
1
0. 50 0. 50
2
3
1
0. 333 0. 33333
3
4
1
4
0. 250 0. 20
1
0. 20 0. 200 5
ô
1
0. 167 0. 16667 6
0
1
7
7
0. 143 0. 14286
1
8
8
0. 125 0. 1250
9
1
0. 1 1 1 0
.
11111
9
1
0. 100 10
W
0. 10
1
m
Reciprocal
To Seven Deci-
mal Places
1 . 00
0.50
0. 3333333
0. 20
0.20
0. 1666667
0. 1428571
0. 120
0. 11111 1 1
0. 100
and also
1
m =
r
Babylonian Scheme
60
ô
30
ô
20
ô
15
ô
12
ô
10
ô
8 34 17
ô+ 360+ 2160+
7 30
ô+ 360
�
+
�
ô 360
6
ô
+ + ¤
Figure2.4. A table of reciprocals of the frst ten integers, expressed to three,
fve, and seven decimal places
Since 0 (zero) plays a very special role in mUltiplication and 0
·
a = 0
for all a, it follows that 0
·
a cannot also equal 1 . Thus 0 has no
reciprocal.
W US ES O F I NF I NI T Y
Problems
2.
9
.
Show that | gives rise to a non-terminating decimal by showing that the
expansion is periodic, i . e. , repeating (the length of the period is six) .
Similarly t, 1', -I are periodic and necessarily non-terminating. Can
you prove that the terminating decimals correspond to fractions whose
denominators are products of a certain number of 2's and a certain number
of 5's (2"·5") ?
2.10. Can you identify the symbols 0.900000
. . .
and 0.000000
. . .
?
What happens when you "add" these expressions a if they were numbers?
What do you expect to get? This non-terminating decimal form of the
number 1 is one of the minor problems that ha to be cleared up before
one accepts non-terminating decimals as representing numbers.
2. 11 . Reduce the operation of division to the operation of multiplication by
means of a table of reciprocal8; see Figure 2.4.
2.6 Infnities in Gmetry
Euclidian geometry abounds in infnities. Some elementary theo
rems of plane gemetry may b interpreted as theorems about
infnite sets.
- �s
Figure 2.5
THEOREM 1 . Let AB denctea/tneseqment
·
T/e /ccuscJ a// µctnts
tn t/e µ/aneeqtdtstantJrcm A and B tsa /tne µerµendtcu/ar tc t/e
seqment AB t/rcuq/ tts mtdµctnt M.
THEOREM 2. IJt/et/reestdescJ0 trtanq/etakentnade]nttecrdr
areresµectt»e/q eçua/ tc t/e stdescJaseccnd trtanq/e taken tnade]ntte
crder, m eac/ anq/e cJ t/e]rst trtanq/e ts eçua/ tc t/e ccrresµcndtnq
anq/e c]t/e seccnd trmnq/e
·
FROM NATURAL NUMB E RS TO V2 Z
In Theorem 1, the word "locus" (Latin for place) is just another
word for "set": in the present case it refers to the set of points I in
the plane such that the distance AI ¯ distance IB. Here the letter
I is used not just for one particular point, but to name any point at
all that is as far from A as it is from B
·
The set of these points is an
infnite set, and the theorem says that these points fll out a straight
line. It also says that the new line is prpendicular to the old one, and
the two intersect in the unique midpoint M of the segment AB
This new line is called the perpendicular bisector of the segment AB.
You will recognize the fact that one way to fnd the midpoint M of
AB is to construct this perpendicular bisector.
If you pick a point I such that AI " IB and join that point
I to the midpoint M of the segment AB you will get two triangles
AIM and BIM, you will see that
AI ¯ BI, AM= BM, and IM ¯ IM
Now if you look at Theorem 2, you will see that it applies to this
situation and tells you that zAMI = zBMI
Since the sum of
these angles is a straight angle (180°), AMI and BMI are right
angles (90°) . This means that the line IM is prpendicular to the
line AB, no matter which point I you may have chosen out of the
infnite set for which AI = IB
This does not conclude the proof of Theorem 1 (you can easily
fnish all the rest) , but it accomplishes what we wanted to do. Namely,
it proves something about every point of an infnite set without using
an infnite process. In fact, it is not necessary, in the proof, that we
b conscious of the fact that the set of points called I is an infnite
set.
But we did need two important aids. First, we used the concept of
»armè/e This is the name given to the letter I bcause of the way
we used it. In our proof, the letters A and B, and the letter M, are
not variables bcause of the way we used them. The letters A and
B denote the same points throughout the proof. The letter M came
about in the course of the proof. We may claim, if we like, that ue
named this point, but there is only one midpoint on AB, and M
always means that midpoint.
But when we bgin our proof, we suspct that there may b many
diferent points, say I; , I
t
, I
-
, and so on, such that
AI
-
" I
-
B
and so on. When we use a single letter I to stand for s¬eof them,
we discover that what we are able to prove is also true for e»erq tne
of them. And so we have really pulled of quite a sizable trick. The
2 US ES OF I NF I N I TY
same form of words with the letter I standing now for the point
I; , now for the point I
t
, now for I¡, and so on proves the same
type of fact over and over again for a//the points of our infnite point
set. The letter I used this way is called a qe¬etrtc»artaè/e,or just a
»artaè/e
If I is a variable point on the perpendicular bisector of the
segment AB, then the length AI is also a variable, since i t can
refer to any one of many diferent lengths, and so also is the length
IB. If we write
AI " IB
we have a kind of equation between variables, which the Greek
mathematicians used quite regularly and which corresponds to our
modern analytic geometry.
We needed variables to carry out our proof involving an inf inite
set, but this is not all we made use of. We also used the theorem
above called Theorem 2. We said it applied to the triangles AIM
and BIM But these triangles are variable, changing with diferent
choices for I, and there is an infnite set of them. Fortunately,
Theorem 2 covers this infnite set of possibilities.
The conclusion we can come to is that the theorems of geometry
are indeed about infnite sets. But the proofs do not necessarily
involve infnite processes, bcause the use of variables permits us to
prove an infnite set of facts by a single argument.
2.7 Gemetry Echoes Arithmetic
Euclid says, "If a line L and a segment I'Ç' are given, it is
possible to lay of on L (in either direction) indefnitely often a
segment IÇ equal to I'Ç'
' The end point of the segment serves
as the beginning point of the next segment.
û ¿ 3 1
1
b 1 3 ¿ · Q
Figure 2.6
A sequence of points constructed in this way, going to the right
(or left), is in an exact one-to-one correspondence with the set of all
whole numbrs. The whole numbers may be thought of as ordinals
(i. e. , as lablling the points in order. say from left to right) and they
may be thought of as cardinals (i. e. , as showing how many units IÇ
have ben laid of on the line up to the point in question) ; see Figure
2.6. The two-way sequence corresponds, of course, to the positive
and negative integers
.
FROM NATURAL NUMBERS TO VZ
Some historians are of the opinion that the frst books of Euclid's
geometry were not meant to be the bginning of all geometry, but
were meant to treat onl y those theorems which depended on ruler
and-compass constructions. Most of Euclid' s postulates are in fact
stated in terms of such constructions. From this point of view the
postulate above merely asserts that, if one sets the compass for I'Ç'
and chooses a point I as center on the line L, one can fnd the
desired Ç by swinging the compass; one can then fnd another point
by choosing Ç as center and swinging the compass, and thus go leap
frogging down the line.
Problems
2. 12. Given a compas but no ruler, and two points P and Q, show that
you can fnd the point R on the line through PQ such that QR
¯
PQ,
and show how to construct a sequence of points on a line, using only
the compas .
P
v
Figure 2.7
2. 13. Suppose you have no compas, but a ruler only a little longer than PQ.
Can you extend the line PQ i ndefnitely in each direction?
2.14. Explain how tunnelers run a road 8traight through a mountain.
2.8 Geometric Approach to Infnite Series
Euclid asserts that each segment may b bisected (meaning by
ruler and compass) and that there results a pair of (equal) segments.
Since each of these may b bisected, and each resulting segment,
and so on, it follows that a segment contains infnitely many points.
To be sure of this, all one has to do is make up some unambiguous
scheme for a non-terminating process of bisections
.
2 US ES O F I N F I N I TY
Figure 2.8 shows one particularly simple scheme : Each time one
bisects the right-hand segment of the two new segments
.
Of cours,
during all of this construction one dos not get to B, but that
( Achilles' task) is not now the object
.
However, it is clear that, as
one performs more and more bisections, the successive midpoints
approach B; it is interesting that e»erq systematic scheme for
getting an infnite set by such repeated bisections has the proprty
that successive midpoints actually approach some point of the
segment.
A
¥
Figure 2.8
Problems
N
l I
· ·
B
2.15. Continue the construction indicated in Figure 2.9 and decide what point
of the segment the successive midpoints approach.
A ¥, ¥
.
1
2.19. 2
1 1
4
+
8
1
16
+
32
2.9 The Method of Exhaustion
Archimedes encountered the series
(2)
1
3
w
when he tri ed to fnd the area of a segment of a parabola. He pr
fected and applied to an infnite process the method of exhaustions
devised by Eudoxus (450 B .C. )
.
His proof of the equality (2) shows
how certain things, obscure from one point of view, may seem plain
from another. He said: "We divide a segment into four equal parts,
we hold one for ourselves, give away two, and one reains
.
At this
point three pieces have ben handed out and we hold i of the amount
distri buted
.
Next we take the remaining piece, divide it into four
segments, keep one and give two away, so that again, one remains.
After this step, it is again clear that we hold i of all that has been
distributed. Next we divide the remaining piece into four equal seg
ments, hold one, give two away, and one remains. As we continue
this process, we hold exactly i of what has ben handed out and a
smaller and smaller segment remains to b distributed
.
" This is the
meaning of the equation above
.
In making the actual construction,
it is bst that the lengths which we want to add shall lie next to
each other.
û
I
+
l
1
Ì
Í Í Í
l
T
Figure 2. 11
J
+
This method, which has in it all of the moder theory of limits, will
b discussed more fully in Chapter 3. For the present, everyone will
grant that it shows how the answer i can b guessed ¦ and also that,
if any answer is right, i is the one. Secondly, the argument shows that
as on takes more and more terms of the series, the sum which one
30 US ES O F I NF I NI T Y
gets difers from i less and less. Thirdly, this fact corresponds pre
cisely to the moder way of defning the smof an infnite series asa
/tmtt,and so Archimedes' proof will seem to us entirely valid when we
meet that defnition.
Problem
2.20. Use this method to guess and prove your answers to:
1
a) -
+
5 25
+
125
+
b)
!
1
+
.
+ + ?
8 6 512
c)
1 1 1
.
+
10
+
10
+
_
. .
?
10
The same type of argument applies also to series like
2 4 8
5
+
25
+
125
+
·
·
Here a segment is divided into fve parts, tucare kept, two are worked
on and one is given away; we hold � of what is distributed and are
led to an easy guess as to the answer.
Problem
2.21 . Prove by the method of Archimedes, quite generally, for every pair of
integers m and n with m smaller than in, that
m
n * m
By substituting the letter r for the ratio mJn, we can write the
left side of the formula in Problem 2.21 as
r + r
:
+ r
-
+ ·
.
.
¯ ?
Do you see how to get the right hand side in terms of r? If you can
and if you have solved Problem 2.21, you will have a proof, by the
method of Eudoxus-Archimedes, of the famous formula for the sum
of all the terms of an infnite geometric progression whose ratio
r is rational and smaller than j
.
However, there are many proofs of
this formula valid for any r numerically less than 1
.
FROM NATURAL N UMB E RS TO ¸
?
31
Fi nally, convert your answer to
1
1 - r '
and observe that
a + ar + ar
:
+
· ·
a
1 - r
·
Can a ÷ 0 i n thi s formula? Can r ¯ 1 or can r exceed 1 i n the
ori gi nal method, i .e., can m = n or m > n? (The si gn " >" i s read
"greater than". )
2.10 The Square Root of Two
The formula
V
+
(
V
l
+
(
V
2) 3
+
. . .
"
V
1 -
V2
i s absurd, as a moment's thought wi ll show. The ri ght hand si de i s
negati ve; the left si de, i f i t i s anythi ng at all, i s "i nfni te". Although
i t has the correct form, namely,
" . .
=
r
1 - r
i t vi olates the sti pulation that r b smaller than 1. On the other
hand, 0/2 i s less than 1 and the formula
V
2
V2 '
1
2
whi ch is equal to the posi ti ve numbr 0/(2 - 0) , i s vali d.
But the type of proof gi ven i n the precedi ng section dos not prove
i t, bcause that method works only if the numbrs i n the progressi on
are rati os of whole numbrs.
We shall show frst that there i s such a numbr as V by provi ng
that 0 represents the length of the di agonal of a uni t square.
We shall do thi s by provi ng the theorem of Pythagoras. Next we shall
show that 0 i s not the rati o of any pai r of i ntegers.
The followi ng proof of the theorem of Pythagoras i s of the "BE
HOLD
"
type (Le., it is revealed to the bholder when he glances at a
di agram; see Fi gure 2.12) :
32 US ES O F I NF I N I TY
1 c
|
c
1
| 1
c
|
c
1
|
Figure 2. 12. The inner square and the four triangles make up the big square
c
:
+
4 _
aè
¸
=
(
a + è)
:
= a
:
+ è
:
+
4
_
aè
¸
,
subtracting 4 (] aè) from the frst and last membr of this equality,
we get
c
:
÷ a
:
+ 貕
Finally, if a = è = 1, we get
c
:
= 1 + 1, c = 0
;
and this is the length of the diagonal of a unit square.
The proof given blow that
0 is not the ratio of two whole
numbrs is a so-called "indirect" proof, a reduction to absurdity
(proof by contradiction, reducttc c aèsurdum)
If V2 could b represented as a ratio of two integers, say 8
µ/ç, then we could choose for µ/ç that fraction (among all equiva
lent fractions) which has the smallest possible denominator.
Figure 2. 13
FROM NATURAL N UMB E RS T O v 33
We know (see Figure 2. 13) that 1 is smaller than 0, i. e. ,
1
<
0
;
multiplying this inequality by '2, we also know that
Now, if
¬
"
1
q
�
0
<
2
.
then p = y
2
. q.
Hence p/q is larger than one but smaller than 2; that is, p is
larger than q but smaller than 2q, and so p - q is positive and
smaller than q.
Next, by squaring p = 0' q, we have
and subtracting pq,
p
2
- pq = 2q
2 _
pq or p(p - q)
so that
p 2q - p
q p - q
q(2q - p)
But the right side is a representation of p/q, with a smaller denomi
nator than q, and this is a contradiction.
Problems
2.22. Discuss the same type of proof for the alleged irrationality of
\. \, v5. Are these irrational ? Proof? (Hint : In studying
p ¯
v5' q, notice that p ¯ 2q is smaller than q since v is smaller
than 3. )
2.23. Study p
¯
.
.
q, noticing that p ¯ 2q is smaller than q.
2.24. Show that the irrationality of ¸ follows eaily from that of v2
.
2.25. Study p ¯
y
.
q given that there is an integer k such that
kl < n < (k + 1)2.
Show that if an integer is not the square of an integer, then it is also not
the square of a fraction.
2.2. Show that 2V8 is an integer only if ¸ is an integer.
34 US ES O F I NF I NI TY
2.11 Computing the Square Root of Two
Lt us conclude this chapter with two inductive constructions for
approximations to \2
We need numbers like this in algebra in
order to solve quadratic equations ; we need \2 to solve
x
:
÷ 2
The number may have been dreamed up by a person whose curiosity
got the better of him as he looked at the sequence of squares
I , 4, 0, I0, 25, 30, 40, 04, 8I, I00,
I2I, I44, I00, I00, 225, 250,
and noticed that there does not seem to be any square which is pre
cisely twice as big as an earlier one. The pair 40 and 25 comes close,
but it is a fact that no square integer is twice another square integer;
that is,
cannot be solved by any pair of integers. It can be solved by real
numbers.
The problem of fnding the ra/ue of this number, \2, has a
very old history. Professor Neugebauer, one of the leading histo
rians of mathematics, tells us that the Old-Babylonians gave the
estimate ( I , 24, 5I, I0) in their sexagesimal notation, corresponding
to I
·
4I42I3
. . .
in our decimal notation, and that this estimate
was used by the astronomer Ptolemy 2000years later. He also indi
cates that a method still used for fnding approximations to v'
(the frst one which we are about U give) may have been the one
that the Old-Babylonians used; the records suggest this, but are not
conclusive.
The method is based on the following observation. Suppose that
a = \2
Then
2
a
and therefore
2
a + -
= 2a
a
2
\2
v2 a
or
If we knew the right value of \2 we could reproduce this value by
taking one half of the sum: the value plus 2 divided by the value.
Now for the method.
8teµ I . Take a positive estimate, say I as frst approximation.
F RO M NAT U RA L N U MB E R S T O V 35
Continuing instructions: Divide the integer 2 by the estimate
obtained in the preceding step. Now take as a new estimate one half
of the sum of this quotient and the estimate you just used.
The same general instruction in algebraic notation reads as follows :
Lt a denote the estimate to V2 obtained in the preceding step.
Then the new estimate is
We can also write the instructions as an inductive formula:
1st step al ¯ 1,
kth step a
k
÷
¡
a
k-
l +
a
¸
-
¸
k ¯ 2, 3, 4,
.
.
It is interesting to see what the frst few approximations to V2
come out to be. We start with 1. Next we get l(1 + 2) ¯ J. This
estimate gives us next Hi + !) ¯ H¥-) H. At the fourth step,
k ¯ 4 in the second formula, we get
1_17 24¸ 577
2 12
+
17
¯
408
'
This estimate is 1 .414215
. . .
and is reasonably close. A closer value
would be 1. 4142140.
It is clear that we get an infnite sequence of estimates in this way,
but it is at least conceivable that after a certain number of steps we
would fnd a fraction whose square is 2. If that could indeed happen,
then the process would continue but it would give the same answer
over and over. HOWEVER, that does not happen; it cannot happen
because, as we saw in thE preceding section,
V
is not a rational
number (not the ratio of two integers). This fact was actually proved
about 500 B. C. , probably by Pythagoras himself. It is recorded that he
considered the "irrationality" of V to be a most unwelcome fact,
on philosophical grounds, because it showed that the world was not
as simple and harmonious as he wanted it. He is supposed to have
ordered the fact kept secret among the members of his own philo
sophical study-group, and there is a legend that at least one of his
students was killed for spreading the bad news.
Pythagoras brought himself, and us, face to face with a serious
question. Geometrical evidence says that there is such a number as
the square root of 2, and it even says that this is a most important
number practically, that is for carpentry and other technologies. But
this number is certainly not an integer, and if it is also not a ratio of
integers, then what kind of number is it, and how does one write it?
36
US ES O F I NF I N I TY
These are not hard questions to answer, nowadays, and this small
book will answer them in time. When we want to ta/k about the v,
we simply give it any name by which other people will recognize it.
For example, we could call it "that diagonal", or "the diagonal of a
unit-square", or "the square root of 2" (we might add "the positive
one", if we think there will be a misunderstanding, because there ts
a negative one) ; or today we would simply use the standard name for
it,
V
2; (then we don't have to add "the positive one" because the
notation includes that idea, since v, by defnition, stands for the
µcstttreroot).
Of cours, if we actually need to work with v-for example if
we are building something square-then the name itself is not going
to be enough for us. But in that case we will know how accurate our
work has to be and we will know therefore to how many places of
decimals we want to have the v. Then we will look it up in taè/es
c] square rccts, and there are tables to fve places and also tables to
ffteen. Today a computing machine will get us a few thousand places
in the course of an afteroon. If we have no machine and no tables,
but plenty of paper, then the inductive construction above will get
us as many places as we wish, if we go through enough steps.
If in this scheme one of the numbers a and 2/a is smaller than
v, then the other is larger than v. For example, a < v
implies
I I
�
> V2
and
2 2
�
>
v
=
v.
You can check quickly that H is a little too big for
V
2, and that
H is too small. Now H is about I4I7and H is about I
·
4I3 ,if one of
these numbers is too small and the other is too large then one half
their sum (which is I4I5)must be correct to within two units in the
last place. Now this may or may not be good enough for a particular
job, but it shows the kind of thing the mathematician considers
satisfactory: an esttmate and then a ccntrc/giving the range of pos
sible error.
We present another way of fnding approximations to v so
closely connected to the decimal notation that it will help us to under
stand, later, what is meant by a rea/numèer (v is a real number;
all rational numbers are real numbers also).
8teµ 1. Take the pair of consecutive integers I and 2. Multiply
these by I0 (getting I0and 20) and insert nine consecutive integers;
I0, I I , I2, I3, I4, I5, I0, I7, I8, I0, 20,
is what you now have. Among these, there is a consecutive pair such
that the square of the frst has a leading digit I, and the square of the
F RO M N A T U R A L N U MB E RS T O ¸ÿ 37
second has a leading digit 2. Find this pair of consecutive integers.
Either of these divided by 10 gives you an approximation to V (the
frst is a little too large and the second is a little too small).
Continuing step: Multiply the pair of consecutive integers obtained
in the preceding step by 10 and insert the nine consecutive integers.
Among the eleven integers there is a pair of consecutive ones such
that the square of the frst has a leading digit 1, and the square of the
second has a leading digit 2. Find this pair. Either integer, divided
by an appropriate power of 10, gives you an approximation to V
(the frst a little too small and the other a little too big).
To see how this works out, we calculate the squares of the frst
eleven integers up above :
100, 121, 144, 169, 196, 225, 256, 289" " , 400 ;
and now it is obvious that we want the square roots of 196 and 225
,
that is, the consecutive pair ( 14, 15)
.
The next stage gives 140, 150 and from these :
140, 141, 142, 143, 144, 145, 146, 147, 148, 149, 150.
By squaring we fnd that (41)
2
"
19881 and (142)
2
= 20186, so
that the desired consecutive pair is (141, 142) and the associated
approximations are 1 .41 and 1 .42. The reader will b wise if he carries
this out for a few more steps. He will fnd that at each stage of the
calculations, he can use the preceding calculations, and so systematize
his work that it is almost pleasant. In addition, he will see that each
successive step gives an improved answer; namely one more accurate
place of decimals. Moreover, this method shows (the same fact is true
for the earlier method but not as easy to see) that if n decimal places
are wanted, they will always b obtained if n successive steps are
calculated.
Problems
2.27. Prove the following rule : To multiply a number by 12.5, move the decimal
point two units to the right, and then divide by 8.
2.2. The root number of a given number is obtained as follows : Add up
all the digits; if the result is bigger than 9, add all its digits; continue
until you get a result smaller than 10; this is the root number of the
given number. Prove that a number and its root number belong to the
same residue clas modulo 3. Therefore, a number is divisible by 3 if and
only if its root number is divisible by 3.
2.2. Using only O's and 1 ' s, (a) construct a decimal expansion with a very long
period ; (b) construct systematically a sequence of expansions, each
with a period longer than that of the previous one ; and (c) construct a
non-terminating, non-periodic decimal .
C H A P T E R T H R E E
From �¿ to the Transfnite
3. ¡ Incommensurable Magnitudes
That V2 is irrational seems contrary to common sense, as the
following will show: Suppose we are given two unequal lengths and
are told, "Cut these into pieces of the same length, as many as you
like, but they must all be of the same length !" It is surprising that this
assignment is sometimes impossible to carry out. If the pieces are
27 and 33, say, or 27y2 and 33y2, the job is easy. But if the
lengths are 1 and V2, the problem is insoluble.
One can get approximate solutions, but no exact solution. The
irrationality of y shows that two given lengths are not necessarily
commensurable; that is, it may happen that no unit of length exists
such that each given length is a whole number of units. In order to
compare two lengths, one measures them with the same ruler. This
makes "incommensurability" of lengths sound like "incomparability"
of lengths, and this is very disturbing. Indeed, it does imply the need
for an infnite process.
The Greek mathematicians realized that the problem of calculating
the perimeter and area of a circle also leads to an infnite process,
since no replicas of a square, however small, can evenly fll out the
area of a circle. The nature of the two shapes doeR not permit it; see
Figure 3. 1.
Thus mathematicians knew the diference between commensur
ability and comparability, but they were surprised to see this question
come up in connection with straight line segments, and in connection
also with the area of rectangles, as the next picture illustrates.
38
F ROM V TO T H E TRANS F I N I TE 39
Figure 3. 1
The rectangles in Figure 3.2 have two square feet of area. The frst
is 2' long, Ì' wide ; the second is r long, i' wide; the third (the inner
fgure) is a square, y' on a side.
K
l I1 �
�
'
_
1
» » »
Figure 3. 2
It is easy to see that the frst contains two unit squares and the
second contains 72 small squares (each i by i). Also, the frst can be
divided into 72 <i by i) squares. The third picture does not suggest
any reason why there is nc small square, replicas of which would
evenly fll both the unshaded area and also the larger area. But this is
what the irrationality of y entails.
Once the irrationality of y had been established, it was some
thing the Greek mathematicians couW get used to. But the embarrass
ing fact remained that since the Greeks had not originally conceived
of this possibility, they had not allowed for it in their proofs. Many
proofs, particularly in the theory of areas and of proportion, correct
as far as they went, omitted what came to be called the "incom
mensurable case". (As recently as ffty years ago, high school text
books in geometry featured careful treatments of the incommensur
able case. ) A very clear example of a theorem where the incom
mensurable case calls for a separate treatment is given in Chapter 6.
Y US ES O F I N F I NI T Y
The corrected proofs called for a suitable theory of limits. This was
supplied by Eudoxus (450 B.C. ) jhe formulated a general principle now
known as the principle of Eudoxus. A special form of it is the Axiom
of Archimedes, which we now know to be too sweeping, but which
was valid for all the geometrical confgurations with which Eudoxus
was familiar. Eudoxus' principle was this : If A and B are two given
magnitudes of the same geometric kind (lengths, areas, or volumes),
then there always exists some whole number, say m, such that
m times A is larger than B.
He perfected a method of carrying out the limit process implied in
his principle, which came to be known as the metho of exhaustios.
An example of this method was shown in Section 2.9.
It was Archimedes who, centuries later, used the work of Eudoxus
U calculate the volumes and areas of a large variety of curved fgures :
the sphere and the circular cylinder ; the parabolic cylinder and certain
parabolic solids ; the areas and lengths associated with certain spirals.
The tradition of his accomplishments was passed on from generation
to generation, and from country to country (leaving Europe about
500 A
.
D
.
and returing in later centuries, about 1400 A.D
.
)
.
The
rigorous thinking which lay behind it was somewhat lost sight of in
the excitement of the generality of new algebraic techniques ; it was
rediscovered only in the nineteenth century.
What Eudoxus asserts is equivalent to this : Given two geometric
objects (of the same kind) of magnitudes A and B, then for some
integer ç,
A
=
çB + k
where k is a magnitude of the same kind, smaller than B. By
treating B as a variable and taking smaller and smaller units for
B, one gets smaller and smaller remainders k, hence better and
better approximations to the magnitude of A
The fact that some geometric magnitudes had to be calculated by
the use of infnite processes led to the converse question: How can
one be sure that a given infnite process actually leads U a number?
Eudoxus invented a suitable theory of limits, at the same time dis
covering the real number system. This was rediscovered by Dedekind
in the 1870's.
Problem
3. 1 . Study the following theorem: Given triangle ABC and segment B'C'
parallel to the bae BC. Then the ratios AB' . B'B and AC' : C'C are
equal .
FROM V TO T H E T RA NS F I NI T E
The dotted lines show how this can be proved in the special cae
AB' : B'B = 3: 2.
The dotted angles are equal .
B
°
l
Figure 3.3
Generalize the proof to the case
AB' : B'B = m: n
41
where m and n are positive integers. What happens if m = V
2
and
n = I ?
3.2 Geometric Constructions of Reciprocals
and Other Number Relations
Every rectangle whose sides x and q have the product xq = I
has an area of I square unit. If x is a given length, we can fnd q by
a simple construction shown in Figure 3.4. First write :
I
q =
x
or better still
l B
l
Figure 3.4
0
I
I
x
42 US ES O F I NF I NI T Y
and interpret the last equation as an equality of the two ratios
belonging to the similar triangles OPB, with OP ¯ x and OB ÷ 1 ;
and OQA, with AO ÷ 1 and OQ = q
AQ is parallel U BP.
In this construction, x = OP is given, the square AOBC being
fxed once and for all. Now one draws or merely imagines BP
and constructs the parallel AQ, obtaining OQ as q
If one con
structs such lengths Yo , YI , Y
2
, for many diferent given
lengths Xo , Xl , X
2
, • • ¯ and uses the resulting pairs of numbers
(Xk , Yk) as the coordinates of points P
k
, one may connect them
by a graph (when enough have been plotted to give one confdence in
the resulting Figure 3. 5).
l
.. .. .
J
r
t
r
i
r
t
Figure 3. 5. xy = 1 or y = l/x
When used with a ruler, this fgure becomes a table of reciprocals
and may be used for determining the reciprocal I/x for each num
ber x
·
Problems
3. 2. Is there B rectangle of unit area, one of whose sides ha length zero?
3.3. Analyze the process indicated by Figure 3.6 for fnding the reciprocal
y = l/x of a given number x.
,
- : · s
Figure 3.6
�
J
- �s
1
:¸
l
t
^
t
Figure 3. 7
FROM ¸2 TO T HE TRANS F I N I T E 43
3.4. Work out the details of the construction of the parabola y = X
l
, inter
preted a
y
x
then construct the curve y = X
l
by graphing, using Figure 3.8 (with
right angles at 0) for the plotting of points.
3.5. What is the bearing of this construction of the curve y = X
l
on the ques
tion of whether one can always fnd ¸z, given any length x. Relate
this problem to the previous construction.
3.6. It is afat that one cannot fnd cube roots, in general , by any fnite number
of constructions involving ruler and compas. But one can plot a great
many "cubic numbers", and draw a curve y = x3• How can this curve
be used to fnd cube roots approximately?
3. 7. Construct y = X
l
and read of
y,
Use the curve and a ruler to solve X
l
÷x - I = 0, in the form X
l
= 1 - x,
i. e. , X
l
= Y = y
'
= 1 ¯ x.
y
:
-1
y
Figure 3.8
mm���.�
�
. g
Figure 3.9
¾ US ES O F I N F I N I T Y
3.3 A Whirl of Quadratic Irrational!
In the sqeuence 1 , 0, y, 0, y, y, some
are obviously whole numbers and therefore rational, e.g.,
2
0
=
2
= l'
Figure 3. 10. Whirl of Irrationals
It is not difcult to prove that only these, the "square numbrs", are
rational. Figure 3. 10 indicates how all of them can b constructed,
inductively, bginning with 1 and building right-angled triangles.
Problems
3. 8. Carry out the construction until you feel able to guess what happens to
the quantity yn ÷ I - Vn W n gets larger and larger. Each term, of
course, is bigger than the preceding term, and the diference btween
each term and its predecessor is always greater than zero.
3.9. Can you appl
y
the identity (a ÷ b) · (a - b) = at - bt to the problem
above to help you prove your guess? Hint : If you use it correctly, the
right hand side will be equal to 1 .
3.10. Anticipate the behavior of Vn( y2n + 1 - y2n) W n gets larger
and larger.
3.4 Illustration of a Limit Point
The curve shown in Fig. 3. 1 1 introduces the idea of a /tmtt in a
diferent way from any other example in this book. It was constructed
F ROM ¸µ TO THE TRANS F I N I TE 45
by Hippias, contemporary of Socrates (450 B. C. ), and was used by him
to solve the problem of measuring the area of a circle. Its name,
"quadratrix", was derived from the fact that it "squared" the circle.
A
S
K W
� v
B X U
Figure 3. 11. Quadratrix of Hippias
We shall use the curve here in a completely diferent context ; we
shall examine the peculiar way in which one of the points of the
quadratrix is related to its other points. But frst we shall describ
how this curve may be constructed.
The quadratrix is shown in Figure 3. 1 1 as the arc AX It is the
locus of points P found as follows : The right angle ABC and the
segment of length AB are given ; then the point H is the point of
intersection of two straight lines, the "radial" line B8 through B
and the horizontal line kÇ parallel to BC , drawn in such a way
that the proportion
�CB8. �CBA ¯ Bk .BA
is satisfed. The angles must, of course, be measured in the same units
of angle, the lengths in the same units of length, so that each side of
the equation is a "pure" number (without units)
.
It is convenient to
choose BA 8 the unit of length, and the right angle CBA, called
a quarter-turn, as the unit of angle. If we do this, the equation reads
numbr of quarter-turs in �CB8
÷ numbr of units of length in Bk.
46
A
B
US ES OF I NF I NI TY
X
Figure 3. 12
U
To construct this curve one frst bisects the angle ABC and the
line segment AB
·
This locates a point on the quadratrix which we
shall call I;
·t
(read it `Î sub one-half"). Next one bisects each of the
two resulting segments along AB and also each of the two rsulting
angles ABI;,
t
and I;,
t
BC and locates two points, I;,, and I
-
,,
(see Figure 3. 12). Next one bisects each of the four new angles and
the four new segments, and the correct association of horizontal and
radial lines determines four new points, I;,s , I, , I
;,s ,
I
:,s
(listed counter-clockwise from the lowest). It is clear how one can
fnd more and more points, as long as one wishes. At a certain stage in
the actual execution of the construction, the curve more or less
materializes. When we have plotted enough points, we can draw in
the rest. The complete curve has now ben constructed.
A
� P
B
X U
Figure 3. 13
F ROM v2 TO T HE T RAN S F I N I TE 47
The point X on the base CB has an exceptional relation to the
curve. It is on the radial line of zero quarter-turns and on the hori
zontal line of zero altitude. But these two lines whose intersection
should have given us the point X coincide ! Therefore the point X
cannot be defned in the same way as the other points of the curve.
One sees from the entire curve, once drawn, that it ought to b easy
to locate the point X on CB; but the previous defnition of the
quadratrix does not apply to this point.
The same difculty arises when one goes over to the analytic equa
tion, as we shall see blow. Meanwhile, Figure 3. 13 shows how the
point X can be calculated as a limit of points of the rest of the curve.
This suggests that the point X on the quadratrix has to be defned
as a limit of other points of the curve. This can actually be done and
X can be found from this new defnition.
A
S
~ r ' X U
Figure 3. 14
If we denote the coordinates of the points on the quadratrix by
(x, y) and the angle CBS by y* (the superscript* being read
"quarter-turs"), we may use the defnition of the tangent of y*
and write
y
x
See Figure 3. 14.
tan y* and x
y
tany*
'
Now the point X which interests us corresponds to y
the distance BX b called x'. The equation
x =
-
y
tan y*
for y
0; let
y* 0 ,
48 US ES O F I NF I NI T Y
takes the indeterminate form x' = � ; clearly this tells us nothing
about x'
·
We can try to fnd x' as the following limit :
To
Y3 ,
I I
�· 4
,
J
r
= limit of � a Y approaches zero
.
tan Y
calculate this limit, we choose some sequence Yl , Y2 ,
. . .
of values of Y approaching zero (in Figure 3. 13 we used
t
, .
.
. ) and then study the sequence
Y3
tan Y3*
,
It turs out that, in the quarter-tur units of angle, this limit is
2/; in radian measure, the limit is 1. (The reason for the di ference
is that T radians are 2 quarter-turns.)
Problem
3. 1 1 . Figure 3. 15 suggests the actual relative sizes of sin y, y, and tan y
when y is expressed in radians . Can you discover the relation of y/tan y
to 1 , for small values of y?
_
�
�
Figure 3. 15
3.5 Limits
A sequence of numbrs Xl , X2 , X3 , • ¯ • has the limit X pro
vided the sequence of numbers (Xl - X), (X2 - X), (X3 - X),
. .
,
approaches zero. t For example, the sequence
a) 3 ,
5
2 '
7
3 '
9
4 '
has the limit 2 bcause the sequence
3 - 2 = 1 ,
5 1 7 1
- - 2
-
2 '
2 = -
2 3 3
'
approaches zero. Similarly the sequence
b)
1
3 ,
5 7 9
3 ' 7 ' 5 ' T'
9 1
2
4 ' 4
1 1
9 '
t The expression "a sequence of numbers approaches zero" will be defned
presently.
FROM V
Z
TO T H E TRANS F I NI T E
has the limit 1 because the sequence
-2
3'
2
-2
T'
2
5 '
-2
¹'
2
\ ,
49
approaches zero. A more difcult example, requiring insight into the
nature of the sequence and also some computation, is the sequence
c)
3
2 '
7
5 '
17
12 '
41
29 '
99
70
'
:
it has the limit v. More difcult to understand, but requiring no
calculation of any kind, is the fact that the sequence
d) . 1, . 102, . 102001, . 102010002, . 1020100200001,
has a limit. This limit i s defned by the sequence itself, and i s other
wise unknown to us !
To complete the preceding defnition, one needs a formal state
ment of what it means to say that a sequence approaches zero, or,
equivalently, that a sequence has zero as limit. Here one sees very
explicitly the use of infnity.
DEFINITION. A sequence of positive numbers
has the limit zero (approaches zero) if zero is the only non-negative
number which is smaller than infnitely many of the terms in the
sequence.
A general sequence approaches zero if it approaches zero when the
signs are all made positive.
This is more usually stated as follows :
EQUIVALENT DEFINITION
.
A sequence of numbers has the limit zero
provided that for every unit fraction l in, that is
1 1 1
1 ,
2 ' " ' 4 '
there are at most a fnite number of terms of the sequence which are
numerically larger than l in (the sign of the terms is not relevant) .
Ultimately the usefulness of a theory of limits depends upon three
factors: In logical order these are 1) a precise defnition of limit of a
sequence ; ii) in each application, a proof of existence of the limit, and
iii) a method for calculating the limit ; this frequently consists of some
scheme of successive approximations coupled with an estimate of
õ US ES OF I NF I N I T Y
error. Of the preceding examples, a) and b) satisfy iii) so well that i)
and ii) are superfuous, the idea of limit bing almost self-evident in
these cases
.
Example c) is a little more difcult ; here one has the def
nition, knows y, and is given a sequence of alleged approximations
but has to work out some estimate of error
.
Without such estimates,
one does not know what is meant by "approximation. " The difculty
in example d) is of another kind. There the form of the approximation
shows the nature of the error, but the question is one of existence:
Is there a numbr which is being approximated? The next sections
will show the meaning and usefulness of this question.
A few of the important theorems in a theory of limits, chosen
because we shall need them very soon, read as follows :
THEOREM 3. 1
.
If Xl , X2 , X3 , • • • has the limit L then every
subseqence has the same limit.
For example, X2 , Xb , Xg , X14 , has the limit L.
THEOREM 3. 2
.
If Xl , X2 ,
0 + Xl , a + X2 , 0 + X3 ,
X3 , has the limit L, then
has the limit a + L.
For example, -3 + Xl , -3 + X2 , -3 + X3 , • •
•
has the
limit -3 + L.
THEOREM 3. 3. If Xl , X2 , X3 , has the limit L, then
kXI , kX2 , kX3 , has the limit kL.
For example, 2XI , 2X2 , 2X3 ,
THEOREM 3. 4. If
2 2 2
Xl , X2 , X3 ,
Xl , X2 , X3 ,
has the limit Í:
Problem
has the limit 2L
.
has the limit L, then
3. 12. Prove these theorems. More generally, show that x�, x; , x;,
.
· ·
ha the limit L" for every integer n. [This is rather difcul t, and the
following course is suggested. Suppose that XI, X
i
, Xa, · · · is a se
quence with a limit L, and that YI, Yi, Y3, · · · is a sequence with
a limit M
.
Prove, now, that the sequence of products XIYI, Xi
Y2,
XaYa, • ¯ • ha the limit LM, i. e. that the limit of the products is the
product of the limits. This is also difcult; one must use the following
type of technique :
LM - x
y
= LM - xM + xM - x
y
= (L - x) · M + X (M - y) .1
FROM V TO T H E TRANS F I N I TE 51
3.6 How the Fact that a Limit Exists May Help to Determine It
It is not obvious that the sequence of numbrs
y,
has a limit, and the reader may wish to calculate a numbr of terms
(one can use a well-drawn q ¯ x² curve as a table of square roots and
a ruler for measuring and transferring lengths) to anticipate what
this limit may be.
Now, asuming merely that the sequence has a limit, let us call
this limit x* and look for it. If we square each term of the given
sequence, we shall get a new one whose limit, according to Theorem
3.4, must be x*
t
·
each of whose terms (after the frst) is 2 more than a term of the
original sequence. Clearly the limit of the new sequence is x* + 2
So now we know that
x*
t
= x*
+
2,
and there are only two numbrs for which this can b true: 2and -1
.
So x* is 2or it is -1 . Since all terms are positive, -1 is a ridiculous
answer and it follows that x* = 2
3.7 Limits which Fail to Exist
Lt us bgin by making the foolish assumption that the sequence
of numbers 1, 2, 3, 4,
. .
¯ has a limit, although we know that
it does not, and let us call this "limit" x
·
If we double each element
of the sequence we shall get a new sequence whose limit must b
2x
But the new sequence is a subsequence of the original sequence
and hence must have the original limit (see Theorem 3
.
1) . Therefore,
it follows that
x ¯ 2x
But the only numbr of this kind is zero. Therefore x ¯ 0
·
We seem
to have proved that the sequence of natural numbers approaches
zero!
Of course, what we proved (by reductio ad absurdum) is that the
given sequence has no limit
.
Next let r b some numbr and look at the sequence
52 US ES OF I NF I NI T Y
Lt us assume that it has a limit k
·
Since the sequence obtained by
multiplying each term by r is a subsequence of the original, it follows
that
k ¯ k
t
and therefore k ÷ 0 or k ÷ 1. Both cases are possible. But there
is another very important possibility, and that is that we are making
a mistake. The sequence may not have a limit ; in fact, if r is numeri
cally larger than 1, no limit exists. Also, if r ÷ -1 no limit exists.
Problem
3. 13. (a) Show that, if -I < r < I , the limit of r, r¹, r, · · is 0, and
if r = ¡ , the limit is 1 .
(b) I n Section 2. 9 we showed how t o prove that a/(I - r ) is the formula
for the sum of the infnite geometric series a + ar + ar·
+
.
*
.
,
whose ratio r is rational and numerically smaller than !. Now can
you prove that this formula is valid when the common ratio T is
irrational and numerically less than I ? Hint : Use the identity
I - x' = (I - x) ( 1 + x ÷ x¹+
. .
, + x·-
I
) .
3.8 The Real Number System and the
Bolzano-Weierstrass Theorem
In the real number system, which will b described presently, the
following fundamental theorem governs the subject of limits.
BOLZANO-WEIERSTRASS THEOREM
.
If Ä¿ ¿ Ķ _ Ä¿ _ is an
increasing sequence of numbers, and if there exists a number B which
is larger than all numbers of the sequence, then there is a number L such
that L is the limit of the sequence.
This theorem assures us, for example, that the sequence d)
. 1 , . 102, . 102001, . 1020100 2,
of Section 3.5 has a limit. For B we can choose any numbr larger
than .2 or . 11, or .
103, etc. The theorem merely requires that we fnd
some one number that is larger than all the numbers of the sequence.
This condition rules out sequences like 1, 2, 3, 4,
. . .
{which,
of course, do not have limits). The theorem dos not tell us anything
about the number L; for example, the numbrs Äg may all be
rational and L may be rational or irrational.
There are many equivalent ways in which mathematicians now set
up the necessary axioms and defnitions to construct the real number
FROM \ TO T H E TRANS F I NI TE õ
system. In some of these the proof of the Bolzano-Weierstrass prin
ciple is difcult, in others easy. In fact, there are so many details to take
car of in setting up the real numbrs, that if one fundamental theorem
is made easy to prove, another one will b found difcult. Thus it
turns out to be entirely correct to defne the real number system as
the smallest numbr system which contains all tne rational numbers
and in which the Bolzano-Weierstrass theorem is true. It is now in
stantly clear that in this system we have got our theorem for free.
Note that the Bolzano-Weierstrass theorem is false in the system
consisting only of rational numbrs.
A more satisfactory scheme from certain other points of view is
the following: We agree to write all terminating decimals with an
infnite string of zero's and then to defne the real numbr system as
the totality of all infnite decimal expansions ; this includes the
genuinely non terminating ones. If we do this, we shall now have to
prove the Bolzano-Weierstrass theorem, and we shall also have to
show that we have a numbr system. That is, we shall have to show
how to add infnite decimals, how to multiply them, and prove all
the standard rules governing these operations. This calls for a great
deal of work.
The Bolzano-Weierstrass theorem was not proved until 1865 when
the calculus was over two hundred years old; one might wonder what
mathematicians did about the existence of limits bfore that.
3.9 The Two Principal Cardinal Infnities
The set of integers and the set of all real numbers represent infnite
cardinal numbers of diferent magnitudes, called Aleph-nuUt and the
power of the continuum, and designated by No and c, respectively;
they are the important transfnite cardinals. Only a very few mathe
maticians ever study any other infnities explicitly. However, in
many branches of moder mathematics expounded at the college
graduate level (abstract algebra, Ablian group theory, homological
algebra, as well as general topology), the trend is toward theorems
and proofs that are valid for systems of large but unspecifed car
dinality.
A generation ago the logical soundness of such theorems was the
subject of strong controversy. The possibility that some infnite sets
represent a more abundant infnity than others had not ben seri
ously imagined bfore Cantor's work in the 1870's. And yet, the
defnitions he adopted and the proofs he created are very simple,
requiring almost no mathematical training for their comprehension.
t Aleph (N) is the frst letter of the Hebrew alphabet.
Ô
US ES O F I N F I N I T Y
DEFINITION
.
Two sets (whether they are fnite or infnite) are said
to represent the same cardinal number if the elements of one set can
be paired of against the elements of the other so that no elements
are left over in either of the two sets.
For example, the even integers and the odd integers can be paired
of in this way:
( 1, 2), (3, 4), (5, 6), (2n - 1, 2n) ,
and i t follows that there are just as many of the one as of the other.
Such a pairing-of is called a one-to-one correspondence.
The set of all integers can be put into one-to-one correspondence
with the set of odd integers, for example, by the formula
N = 2n - 1,
where n ranges over the set of all integers, and the corresponding odd
number is N
Thus the set of odd numbers (and also the set of even
numbers) has the cardinal power No of the set of integers.
Many sets which are apparently more numerous than the integers
are also of power No, for example, the set of all rational numbers.
It is easily seen that the rational numbers can be arranged without
omission or repetition in a simple sequence. For instance, by frst
lumping together in groups all those rational numbers alb which
have a + b = 2 (i.e., t), then those which have a + b = 3 (i.e. ,
j, i), etc. , and then ordering the numbers in each group according to
the size of their numerators, we have :
1 1 2 1 3 1 2 3 4
I ' 2 ' I ' " ' I ' 4 ' 3 ' 2 ' I '
1 5 1 2 3 4 5 6
etc.
5 ' I ' { , 5 ' 4 ' 3 ' 2 ' I '
The only troublesome detail here is that one must avoid the repetition
of the (equivalent) fractions in order to conform to the requirement
of a one-to-one correspondence. For example, we have omitted i from
the third group because its equivalent, t. already appears in the frst.
The fact that the set of rational numbers has the cardinal power
No is a special case of the following:
THEOREM
.
Let S be a set of sets, S; , S
t
, S. , .
. .
. If S is
fnite or has the power No, and if all sets S; , S
t
, S
-
,
•
• • have
the power No, then the set of all objects belonging to these sets also has
the power No .
FROM ¸2 TO T H E TRANS F I NI T E
This assrtion can also be written in the form of equations :
1
.
No
=
No ,
3 · No = No ,
2 · No = No ,
No' No
=
N�
=
No .
The proof is simple. By hypothesis, every set SI , S2 , S3 ,
55
can be arranged in a sequence
.
Lt us denote by x(i, j) the ith ob
ject in S;. For instance, x(2, 3) denotes the 2nd object in the 3rd
sequence S3. Now there is only a fnite number of terms x(i, j)
with i + j equal to some number a. Hence we can arrange the
terms with i + j = 2 into a sequence, follow it by the sequence
containing all terms ' with i + j = 3, follow this by a sequence
containing all terms with i +j = 4, etc
.
Then we obtain a sequence
containing each object occurring in any S; at least once. Strike out
the repetitions and you have a sequence containing each of our
objects exactly once.
When a set of objects is put into one-to-one correspondence with
the integers in this way, it is said to be counted. A set which can be
counted is called countable. The theorem stated above asserts that a
fnite or countable infnity of countable infnities is countable. A set
which cannot be put into one-to-one correspondence with the integers
in this way is called uncountable.
We have shown how to count the rational numbers, and the next
indicated step is to count the reals. But this is impossible. Cantor
proved the following celebrated theorem:
THEOREM. Given a one-to-one correspondence between the integers and
some set of real numbers, it is always possible to construct a real number
which has not been counted in this correspondence.
COROLLARY. Tm set of all real numbers is uncountable.
The proof is simple ; let us frst concentrate on a convenient subset
S of the real numbers : S denotes the set of all real numbers between
0 and 1 whose non-terminating decimal expansions use only two
digits, say 1 and 2.
Lt us suppose that we are given a one-to-one correspondence
between the integers and the elements of S. This would mean that
we can count the numbers in S, i.e., that we can make a list and
refer to the frst, second,
. . .
numbers in the list. We shall construct
a number N belonging to S and we shall show that N is not in our
list. This will prove that the subset S of real numbers is not count
able and hence that the whole set of real numbers is not countable.
56 US ES O F I NF I NI T Y
To construct N
, select its frst digit by looking at the frst number
in the list. If it has 1 as its frst digit, let N have 2 as its frst digit
and vice versa. Similarly, select the second digit of N so that it is
diferent from the second digit of the second number in the list.
Proceed in this way, always making the kth digit of N diferent
from the kth digit of the kth number in the list. Clearly, the
number N so constructed is diferent from the frst number in the
list (because of its frst digit), from the second number (because of
its second digit), and, in general, from the kth number (because of
its kth digit), and this is true for all k. So the real number N
belongs to S (
since it consists only of digits 1 and 2) but was not
counted in our one-to-one correspondence. This concludes the proof.
This wonderful proof becomes clearer if one sets up some sequence
of real numbers and applies the construction to obtain a number not
in the sequence. Used this way, the argument in Cantor's proof is
called Cantor's diagonal procedure and it is a standard way of con
structing a number which difers from every one of a given infnite
sequence of numbers.
Problem
3
.
14. We showed above how to count the rational numbers. Since these are
also real numbers, they may serve U the set S in the preceding proof.
Cantor's construction then yields a number which is not rational. Look
into this argument, workng out the frst dozen or so digits of the resulting
number A and using a variant of the scheme indicated above . (Make
every rational number in the list on page Ô non-terminating, either by
writing out all the zeros or by converting the terminating decimals to
numbers ending in a string of nines, as you wish.)
3.10 Uses of the Uncountable
The distinction between countability and uncountability has
turned out to be one of the cornerstones of analysis and topology.
Here a few facts will be cited, without proof, to show some ways in
which it can matter whether a given set is countable or uncountable.
Figure 3. 16
First, one can have a countable infnity of segments on a line, all
of the same size, no two of which touch each other; see Figure 3. 16.
One sees how to construct a countable infnity of segments on a given
segment of length L, no two of which touch (Figure 3. 17). Of
F ROM \ TO T H E TRANS F I N I TE 57
course, in this second case, only a fnite number of them can be
larger than a preassigned unit fraction lin. More precisely, there
must be less than nL segments of length at least lin.
Figure 3. 17
By contrast, in any uncountable infnity of segments on a line,
some must overlap. For every segment contains some rational points,
i.e., points whose distance from a fxed point 0 is a rational number.
Some rational points must belong to more than one segment, since
otherwise the set of segments could be counted, contrary to assump
tion. The same argument shows that in any uncountable set of seg
ments on a line, uncountably many must overlap and uncountably
many must be larger than some unit fraction. More of this sort of
argument appears at the end of Chapter 6.
An amusing variation of this same phenomenon, which is actually
important in certain topological applications, is the following: Let us
suppose that the letters of the alphabet are represented by mathe
matical cur�s-that is, they are without any thickness. Also, sup
pose that the letters have no oraments, i.e
.
, no extra lines or feet.
Then one can scribble an infnity of each letter on any sheet of paper
of any given size. One can fnd room for an uncountable infnity of
certain letters, like L or L or Í or N or M¡ but surprisingly it is
not possible to write an uncountable infnity of T's or A's or B's
or P's without two of them overlapping.
Problem
3.15. Can you clasify now the letters of the alphabet according to the prin
ciple here indicated, and get to the topological essence of this fact?
(See Figures 3. 18, 3. 19 and 3.2. )
- ¬� =¬� s
U
(a)
- s
-
-
�
s
-�
s
U
U
U
U
(b)
Figure 3. 18
æ
y=
2
y
=
2
y
= l
y =
D
x = D x = a
x =
l
US ES O F I N F I N I TY
U
§
·
�
§
¯B
U
Figure 3. 19
y
=
l
y
= a
y
=
D
x=
2
y
J
(
a
)
Uncountable L's (b) Uncountable O's
Figure 3.20
C H A P T E R F O U R
Zig-zags: To the Limit if the Limit Exists
In this chapter we shall discuss the concept of seqential limit.
This notion is much the same whether on the line, or in the real num
ber system, or in the plane. We shall treat the case of the plane,
because it contains a more stimulating variety of situations, and b
cause one can draw illuminating pictures. Our principal tool for the
exploration of the limit concept will be the zig-zag. This is an infnite
sequence of line segments forming a simple polygonal path looking
like the conventional representation of thunderbolts; see Figure 4. 1 .
A picture of a thunderbolt suggests a target and this is a good ap
proach to the idea of a limit. There is another point to the analogy,
if we recall that anticipated limits sometimes do not exist and then
realize that this is also true of supposed targets.
We shall fnd that when the target of a zig-zag is clearly marked,
it will always be that point of the plane which is also the limit point
of the sequence of vertices (corner-points) of the zig-zag.
Figure 4. 1
5
ô US ES O F I NF I NI TY
The zig-zags are interesting fgures, especially in connection with
the concept of kgth.
Figure 4.2
The zig-zag in Figure 4.2 clearly has no target ; moreover, if con
tinued indefnitely, it is infnitely long. The zig-zag in Figure 4.3
starts at PI ; for any fnite n, the left part P¿P§ • • • Pg is fnite
( the three dots in the middle just indicate that its exact shape does
not matter here) .
The three dots on the right mean that the zig-zag
has infnitely many vertices. The zig-zag is an infnite construction;
for each n " 1, 2, 3, · ·· , there is a path PIP
2
• • • Pg from the
frst vertex PI to the nth vertex P
n
and the length of this path
( let us call it Lg) is the sum of the lengths of the constituent seg
ments. Thus
• • •
• • •
Figure 4.3
Now we make the following defnition:
DEFINITION. The lgth L of the zig-zag P1P
2
P
3
• • • is the
limit of the lengths Lg as n increases indefnitely, if this limit exists.
Symbolically,
L = length ( P1P
2
P
3
• • • ) ¯ lim ( length P¿P§ • • • Pg) ;
g
~w
or, even more briefy, L " lim Lg .
g
¬w
Problem
4.1 . What necessary qualifying phrae appears fnally to have been omitted?
How therefore must this lat symbolism be read?
Z I G- Z AG S : TO THE L IMI T I F THE L IMI T EXI STS 61
In general, as we see from the defnition of this limit, in order to
calculate it, we must sum an infnite series. We have had some prac
tice in this already.
The frst examples which follow are mainly concered with length.
The point T, origin of coordinates and undoubted target of the zig
zag, will not fgure importantly for the time being
.
EXAMPLE 1 . The frst zig-zag consists of the segments
A:
A,
.
A;A
t
, A
t
A, , · ·
'
. The points A, , n = 1, 2, 3, · · · are
alterately above and below the x-axis; the abscisa of A, is 1/2\
and the ordinate i s the same except for sign. See Figure 4.4.
-
. ,l I)
Figure 4.4
A, V the point n = 1, 2, 3,
. . .
.
The zig-zag A
:
A;A
t
. . .
is like a half-line or ray, and it is also like
a segment with just one endpoint (corresponding to A:
) , the other
one having been deleted. Such segments occur frequently in mathe
matical work and are called half-ope; the segment with both end
points of is called ope and the segment with both endpoints put
62 US ES O F I NF I N I TY
back on is called closed. Notice that as far as the construction goes,
the zig-zag does not contain the point T ( although it obviously
points to it)
.
It may surprise the reader that this infnite zig-zag is of fnite
length. It is easy to see that the length of any path AoAl ' " A
n
along the zig-zag is less than the following sum which is itself less
than 3:
¡1 1 1 1 ¸
1 + 2 · + - + - +
. . ·
+ -
2
4 8 2
n
•
We get this sum if we replace each segment of the zig-zag by its
horizontal plus vertical projections. Using the theorem of Pytha
goras we can readily show that
1
8
v
,
and prove that the length L of the zig-zag is
!
V
¬ !
..
Thus L ` 2.3 with an error as large as .2, i.e., an error not exceeding
10 per cent.
EXAMPLE 2. In Figure 4.5 the length is even easier to calculate.
TB; is of unit length and we are given an infnite sequence of equi
lateral triangles, with bases of diminishing lengths :
!,
1· ¡
,
.
Since the bases add up to TB; = 1 , one sees easily that the length
of the zig-zag is 2.
I
.
B¿
.
¸
¯
.
.
~
B¿
Figure 4. 5
B
¯
/ ` ! '
Ì
¯
.
Z I G- Z AG S : TO THE L IMI T I F THE LI MI T EXI STS ô
Problem
4.2. Notice that the odd-numbered vertices lie on the x-axis (where
y
= 0 in
rectangular coordinates) and the even-numbered vertices lie on the line
T B2 , whose equation is easily found to be
y =
¸�)
.
x.
Construct an analogous zig-zag (segments of equilateral triangles) whose
even vertices lie on the curve
y
= x¹. How does the length of this zig
zag compare (judging from the pictures) with that of the preceding zig
zag?
EXAMPLE 3. The zig-zag of Figure 4.6 does not have fnite length,
and we shall say that it has infnite length in accordance with the
following defnition:
DEFINITION
.
If the lengths L, increase without bound as n
increases, then we say that L is infnite.
Since we are adding positive numbers, the numbers L, must get
bigger and bigger ; if they had an upper bound ( any number which is
bigger than all of them) , then, by the Bolzano-Weierstrass principle,
they would have a limit ( less than or equal to this bound) . There
fore, if there is no limit lim L, , there is no bound, and that is all we
mean to say when we call the length L infnite. In much the same
way, one says that the length of a Euclidean line is infnite. It is a
new form of words, a matter of convenience; this infnity is not 0
number of the real number system. Mathematicians sometimes aug
ment the real number system, throwing into it a new object,
=, to
correspond to this infnity, but then they have to make special rules
for operating with it. For example: = + = = = but
= - = is
indeterminate .
The zig-zag of Figure 4.6 is very much like the preceding one in
appearance, but the coordinates of the vertices are diferent.
C
, V the point ¸�
( -
n
O
¸ , n = 2, 3, 4,
. .
It is clear that the length of 0,0
t
excees !, that of 0,0
t
0
-
excees
1, and that of 0,0
t
0
-
0, excees 1 .5.
This suggests that the length of
0;0
t
• • • 0, increases substantially with each step; but this is not
true. The increase in later steps is very small
.
However, these lengths
do add up and L, increases without bound
.
Let us look into it more
closly.
I
b US ES O F I N F I N I TY
l
t ´� � ¦
Figure 4.6
If we consider the four sgments 0,0,, 0,0,, 0,0,, 0,0,, we
see that the ordinate of each endpoint is, numerically, at least !,
and therefore each segment has a length of at least 2 · 1 ´ i. Thus
the four segments have a combined length of at least 1 . In precisely
the sme way one sees that the next eight segments have a combine
length of at least 8 times !, or 1, and the next sixteen segments after
those have a total length of at least 1, and then the next 32, and so on.
In this way we can make a sequence of paths,
whose lengths increase like the numbers 3, 4, 5, ' " and snce
we can continue this indefnitely, we see that the lengths L, increase
without limit. The zig-zag has infnite length.
This result is of such great importance in mathematics that the
reader will do well U think it through again, verifying that the length
0»0»
+
, + 0»+,0»+
t
+
· · ·
+ 0
t
»=:
0
t
»
exceeds 1, whenever N has the form 2f
.
A convenient way to
analyze this problem for oneself is to study the following example:
Z I G- ZAGS : TO THE LI MI T I F THE LI MI T EXI STS 0
1 1 1
1
+
2
+
3
+
4
+
. . .
.
This is called the harmonic series. It is a divergent series, that is, the
partial sums
1
+
!
+
!
+
. . .
+
�
2 3 n
increase without limit as n increases.
A sequence of terms with indicated additions ( or subtractions) is
called a series. By defnition, the sum of the frst n terms of a series
(n " 1 , 2, 3, ·
· ) i s called the nth partial sum, written S
n
and
the sum S of the series is lim
n
_o S
n
if this limit exists. If all the
terms are positive and no limit exists, then one ( sometimes) speaks
of the sum as infnite. In general, when no limit S exists, the series
is called divergent ; when the limit exists, the series is called convergent.
When the limit exists for the given series but fails to exist for the
series which has the sme numerical terms, all made positive, then
the given series is called conditionally convergent. For example, the
series
1 1 1
1
- -
+
- - -
+
. . e
2 3 4
is conditionally converent ; that is, it is convergent as it stands but
not convergent when all the terms are made positive.
We shall now use zig-zags to explore the notion of limit.
DEFINITION. The point I is called a limit ( short for sequential
limit point) of a sequence of points I,, I
t
, I
-
, · provided
that the sequence of distances I,I, I
t
I, I
-
I,
· approaches
zero. ( Cf. defnition of "a sequence of positive numbers approaches
zero", Section 3.5
·
)
As the pictures show, the point T in our examples is a limit of the
zig-zag vertices. The distances from T U these verticest are easily
calculated. In the frst example,
In the second example, let the reader calculate T B
n
( there being
two cases U consider) .
!Here we mean the length of the segment TAn , not distance along the
zig-zag.
6 US ES O F I NF I NI TY
Problem
4.3. Look at Example 2 in this roundabout way: First, show that the sequence
B
1
, B
3 , B. , ' " has T as limit. Next, use the fact that the dis
tances B2.B2nl approach zero.
It follows easily from the triangle inequalityt that a sequence
cannot have more than one limit. For if the distances I
-
T and also
I
-
T' approach zero, then the formula
I
-
T + I,' � TT', R ÷ I, 2, 3, · · ·
shows that the distance TT' must equal zero. But in Euclidean
geometry this is only possible when T and T' are the sme point.
Thus if we know that a limit exists we may speak of tm limit of a
sequence.
The next pictures ar directed to the notion of limit point.
EXMPLE 4 The zig-zag ODID2Da
. .
, in Figure 4.7 uses segments
which make an angle of 45º with the horizontal and therefore make
right angles at each vertex. The length of D"
m
+l is 112
m
, for
m = 1, 2, 3, · · ·
.
·
B
V ===
¯
Figure 4.7
The infnite zig-zag starting at D2 , namely D2DaDCD5 · · · , lies
entirly inside a square whose diagonal has length i ( sides are
×2
/4) , it has D2 8 endpoint and no other point which we have yet
!This states that, in a triangle, the sum of the lengths of two sides is greater
than the length of the third.
9
ZI G - ZAGS : TO THE LI MI T I F THE LI MI T EXI STS 67
constructed is an endpoint of it. Hoever, there is a point in the plane
ver intimately related to this zig-zag. Let us call it Z. Figure 4.7
shows Z where it does not belong; u the reader agrees that it does
not belong there, then he will know how U fnd where it does belong.
Problem
4.4. Suppose that the point 0 is the origin of a coordinate system whose x-axis
is horizontal U customary. Locate the point whose abscissa is
¬
2 and
whose ordinate is
¬¥ ¸¸
= !
!
! ¸ ¸ ¸ ¸
=
V
.
�
=
´
·
2 2 4 8 2 3 3
Call this point Z and discuss the relation of Z to the vertices D\ . Dz ,
D3 • ` ¯ ¯ , explain what is meant by the statement "the abscissa of Z is
a limit of the 8equence of ab8ci88a8 of the infnite succession of vertices"
.
Prove that the ordinate of Z is a limit of the sequence of ordinates of the
vertices (recall the formula for the sum of an infnite geometric progres
sion) .
EXMPLE 4'.The zig-zag OD;D�D�
. . .
of Figure 4.8is an interest
ing companion piece U the preceding. The diagram shows the sme
sequence of segments arranged to make a spiral.
·,
Figure 4.8
Problem
·
Î
4.5
. The reader may enjoy proving that there is a limit point T to the set of
vertices D; , D; , of Example 4' by fnding its coordinates.
b US ES O F I NF I NI TY
EXMPLE 5. Let us look next at an infnite zig-zag OE1E2E;E4 • • •
whose successive segments have lengths
1 1 1 1
1,
2 ' " ' 4 ' " '
L,
�����-��-
q --1 --1 - ---
Figure 4.9
Its picture in Figure 4.9 makes all details clear. This example is
similar to the zig-zag ODID2Da
. . .
of Example 4, at least U the un
aided eye. However, we already know something about adding the
sequence of unit fractions : They do not have a fnite sum. This will
have striking consequences.
One might guess that this picture has a point I associated with it,
in the way in which the point Z of Example 4 is associated with its
zig-zag. BUT such a guess is incorrect. This zig-zag moves to the right
slowly but invincibly; no point of the plane is a limit point of the se
quence of points E1 , E2 , E3 , E4 , • • •
•
This is proved by the
fact that the horizontal projection of the path E1E2Ea • • • En ha
length
v ·
1 1 1 1 1¸
-
1 + - + - + - + - +
·
·
·
+ -
2 2 3 4 5 n '
and this sum increases without limit as n increases. Thus, no matter
what point I one may choose, some En ( for sufciently large n)
is to the right of this chosen I, all later points of the zig-zag are even
further U the right and cannot approach I. The details should now
be clear.
ZI G - ZAGS : TO THE L IMIT I F THE LI MI T EXI STS 69
Problems
4. 0 In this example, although the abscissa of the points E
n
increae without
limit a n increaes, the ordinateedo not increae in this way. Prove that
the ordinate of E" is
f
and satisfy yourself that the sequence of ordinates has a limit by relating
this problem to the picture in Figure 4I0
I´
0 �
Figure 4I0
4.I. (a) Explain why the following assertion might be false, and formulate a
correct statement : "The limit of the projections on a line (of an in
fnite sequence of points in the plane) is also the projection of the
limit. "
(b) Can you prove the following asertion? "The projection on a line of
the limit (of a sequence of points) is the limit of the projections. "
Two companion pieces to the previous zig-zag, the frst showing
converence and the second diverence are pictured in Figure 4
I I .
Î
1
:
·
:�
l
·
:,
:, :
� `
`
~ ~
Figure 4 II
t
:�
r-
10
US ES O F I NF I N I TY
In the frst one, the vertices have a limit point T, and the sequence
of segments 0E, (n " 1 , 2, 3,
. .
) has a limiting position
0T In the second there is no limit point, but the sequence of se
ments 0£, (n
=
I, 2, 3,
) does have a limiting direction!
The reader is invited to prove that the limit of the slopes of 0E,
exists.
Figur 4. 12
That 7 is irrational means that the radius of a circle and its
perimeter are incommensurable lengths. This has a striking conse
quence, related to the present discussion. If we choose a point, call it
I; , on a circle and then rotate the circle through one radian we will
get a new point, call it I
t
( the arc I
:
I
t
has length equal to the
radius) , and if we rotate again through one radian, we will get another
point I
.
, and so on. Let us in this way construct a sequence of
points I, , n
=
I, 2, 3,
. . .
; see Figure 4 I2. Now the points
in this sequence can never repeat themselves, because of the incom
mensurability of the radius and perimeter. ( This is not trivially ob
vious, and is worth some thought.) The sequence of radial segments
0I, cannot assume a limiting position, because we are constantly
changing position by a whole radian. It follows from both these facts
that the points I; , I
t
, I.
, • ¯ • are distinct from each other,
and that the sequence has no limit point.
The behavior of the point I, as n increases is a very important
problem in the higher calculus. It is easy to conjecture that I
-
moves
around so as to make its presence known in every part of the circum-
ZI G - ZAGS : TO THE LI MI T I F THE LI MI T EXI STS 1I
ference more or less equally often, and this is the substance of a dif
cult and celebrated theorem due tKronecker. The easier part of this
theorem states that the set of points I, , n ÷ 1, 2, 3, ·
· is
everywhere dense on the circle. This means that if we pick some point
Ion the circle we can always fnd an iterate of I; ( that is a point
obtained frm I; by a certain number of rotations through one
radian) which is near to it. More precisely, if we choose some ( large)
integer, sy K, we can fnd a point I» ( for some clever choice of
N) so that the distance II» is smaller than 11K. Of course, this
dos not mean that I is a limit point of the entire sequence I; ,
I
t
, I, ,
•
• •
•
Not at all ; the ver next point I»
+
, will be at a
substantial distance from I and we may have tgo round the circle
quite a few more turns, perhaps M turns, with some clever choice of
a point I»+
_ , before coming in as close to I as we earlier specifed,
namely 11K.
We shall close this album of illustrations of the limit concept with
a remark on the Kronecker Theorem. If we choose an angle of rota
tion equal t k radians and consider the resulting sequence of points
I� , I� , I� ,
two cases are possible. If k is incommensurble with T we get an
everywhere dense set, just as in the case above wher k ÷ 1 . If k
is a rational part of 2 T, i.e., k ÷ 2m'ln with m and n integers,
then the sequence is periodic and has the form
Fi gure 4 I3
72 US ES O F I NF I NI TY
The number N of distinct points is at most n ( remember that 21
radians is a full rotation) , and N is always a factor of n ( this is not
easy to prove, but it does not take many lines ; this result is one of
the fundamental theorems in elementary group theor) . This al
ternative gives rise to all the regular polygons and to a variety of
beautiful pictures. Here, we illustrate the case n = 5; for k = 2'/5,
we get the frst, for k = 41/5, the second of the polygons shown in
Figure 4. 13.
In all cases but one, a periodic set does not have a sequential limit
point. The exceptional case is illustrated by a sequence Q, Q,
Q,
. . .
and has Q as limit. A numerical instance is the sequence
2, 2, 2,
. . .
, which has the limit 2. The idea behind this may be
seen in the representation of 2 as a non-terminating decimal :
2.00
.
"
.
]
when we "round" this to the nth decimal, we get the nth approxi
mation, which is still 2, of course, but we now view it as an approxi
mation to 2 (which just happens to have hit the mark exactly) . This
is entirely in accordance with the defnition of limit given above.
A fnal remark about the use of limits in the calculus. The notions
of tangent to a curve, length of arc, area bounded by a curved fgure,
area of a piece of surface, curvature of a surface, volume bounded by
a surface, center of grvity of a mass, and the physical notions of
velocity, acceleration, work done, total gravitational attraction of one
mass upon another-all of these and literally hundreds of others in
every brnch of the mathematical sciences-begin with defnitions
involving limits. The calculus works with all of this material.
In this chapter we have made one application of the notion of limit,
namely to the calculation of the lengths of zig-zags. This is closely re
lated to the mathematical problem of summing infnite series, one of
the two most important uses of the limit notion in the calculus ( the
idea of a derivative associated with the slope of a tangent line to a
curve being the other) . The infnite series is a generlization of the
notion of a non-terminating decimal such as
1
3 ¯ 0.3 + 0.03 + 0.03 +
· ·
and
1 = 3 + 0. 1 + 0.04 + 0.01 +
· .
.
( In the second example, the three dots merely mean that the expan
sion is non-terminating; no simple rule is implied for calculating the
next digit.)
ZI G - ZAG S : TO T H E L I MI T 1 F T H E L I MI T E X 1 ST S 13
Prblems
4.8. The following problem introduces the famous series
I I I
I ÷ - ÷ - ÷ · · · ÷ ¬ ÷ · · ·
/ 3
2
a
2
which is convergent, whereas the harmonic series is not. It is proved in
the calculus that the sum is ,.
2
/6. All that the reader is aked to show is
that the sequence of nth partial sums is bounded by 2 This will prove
that certain points in the plane, about to be defned, do not "move of to
infnity. "
49 Figure 4I4is similar to the "Whirl of Irrationals" (see Figure 3I0), but
here, the legs of the nth triangle are related to the harmonic series
I , ], |, · · · a follows : the segment PnPn+1 (n = I, 2, 3,
. . .
) is of
length lin and it is perpendicular to the radial line OP n . Now prove
that, for all n, OPn is less than
¬
]. Next verify that the length of zig
zag P1P
2
P, · · · is infnite. Imagine the points P n projected radially
onto a convenient circle, say of radius 3, and denote the projection of
Pk by Qk . Can you prove that these points QI , Q
2
, Q
ª
, · · · wind
around indefnitely often? Can you conclude that there is no limit to the
original sequence?
Figure 4I4
4I0 Determine whether or not the following series converge :
I I I I
(a)
2
÷
4
÷
6
÷
8
÷
· · ·
I I
(b)
3
÷
1
÷
II
÷
Iõ
÷
I I I I
(c)
i
÷
\
÷
¬4
÷
×
÷
(
d
) =
I
= ÷ ÷ ÷ ÷
I
÷
\
�
_
'
: :
V
, ¸
.
74 US E S O F I N F I N I TY
4. 1 1 . Let P be a point and let Pn (n = 1 , 2, 3,
. . .
) be a sequence of points
in the plane. Suppose that for every line l, the projection of P on l is
a limit of the projections of Pn (n = 1 , 2, 3,
. . .
) .
I
I-
U
(a) Prove that P is then a limit of the sequence.
(b) Could this result be deduced from the fact that the given data are
true for just one line?
(c) For two lines? What is the precise state of afairs?
I
--´
I
I-
X
t
-
X
(a)
Figure 4. 15
^
`
¸
|
�
|
|
|
|
|
t
-
X
(b)
l,
C H A P T E R F I V E
The Self Perpetuating Golden Rectangle
In this chapter we shall study some of the properties of a rectangular
shape discovered by the Greeks about 50 B.C. , the time of the dis
covery of the irrationality of certain square roots. The construction of
this rectangle involves y5, a number which is the key to the ruler
and-compass construction of the regular pentagon. This pleasing poly
gon is a face of the regular dodecahedron ( twelve-sided polyhedron)
and is built into the regular twenty-sided icosahedron, the most
fabulous of the fve regular solids. The golden rectangle therefore was
highly prized by the Greeks for its own beauty and for the beauty of
its family connections.
Looked at in the right way it gives rise almost instantly to the fol
lowing sequence:
1 , 1 , 2, 3, 5, 8, 13, 21, 34, 55, 89, 144,
This succession of numbers was invented by Leonardo of Pisa early
in the 120's in a problem of arithmetic about the breeding of rabbits
( 1 and 1 in his sequence represent the ancestor-rabbits) and it has
come to be called the Fibonacci t series. Ever since, the connection
between the rectangle and this series has been a source of mathe
matical discoveries, some of which will be shown in this chapter. By
the eighteenth century it had been found that the rectangle was
naturally associated with a logarithmic spiral, a shape connected with
the growth of many natural objects like some molluscs and snails.
!See footnote to p. 20.
1õ
I0 US ES OF I N F I N I TY
The enunciation by Fechner, the great psychologist, of his famous
law: "The intensity of reaction to a stimulus varies as the logarithm
of the intensity of the stimulus", brought to the rectangle and its
built-in spiral a host of admirrs who were not necessarily mathe
maticians. The importance of logarithms in psychological phenomena
is attested by a variety of facts ; the use of the decibel, for example, in
measuring intensity. A biological phenomenon called "phyllotaxis"
also exhibits the Fibonacci numbers. t
5.1 The Golden Rectangle
Figure 5. Ishows a rectangle shaped like a librarian's 3x 5card. If
one folds such a card in the manner shown in Figure 5
2, there is left
a rectangle whose sides are in the ratio 2to 3, which is about 0.67 to l.
3
b
Figure õ. I
3
¿
Figure õ.2
The original card had the ratio of 3 to 5 which is 0.6 to 1. Thus if
one cuts a square away from such a card one gets a shape approxi
mately like that of the original. Now the shape of the golden rectangle
may be defned in this way: If one cuts a square away from it then the
rectangle that remains has exactly the same shape as the original rec
tangle. By "same shape" we mean that the ratio of shorter side to
longer side is the same, i.e., that the two rectangles are similar.
I For an account of these phenomena, see the beautiful discussion in
H. S. M. Coxeter's Introductton to 0eometry, Chapter I I . See also the article
"Mathematical Games, " by Martin Gardner, in 8cientt/c Amertcan, August,
I9õ9.
THE SELF PERPETUATI NG GOLDEN RECTANGLE I1
Clearly infnity is now let loose; for, if one cuts a square away from
the newer rectangle, there results another similarly shaped rectangle,
and if one cuts a square from this one there results another, and so on,
indefnitely.
Let us denote the lengths of the sides of the original rectangle by
a + b and a. Then the lengths of sides of successive rectangles give
the following sequence of pairs :
,�
,.
,
. a
2a ·
'
.
.- a '
,
.. ·
5| 3a '
5è - 3a
5a
where we have written the longer side of each pair frst. The pattern
which emerges is this : The longer side of one rectangle is the sum of
the two sides of the next following rectangle. The coefcients of a
and in these expressions are related to the Fibonacci numbers
listed above. The law of formation of the Fibonacci numbers is this :
Each number is the sum of the two preceding numbers ( exception
being made for the two ancestors: I and I)
·
The laws of formation of
the above lengths and of the Fibonacci numbers are so similar that
one cannot doubt a strong connection; we shall come back to this
later.
5.2. The Golden Mean is Irrational
The process of forming successive rctangles cannot end and this
tells us that a and are not commensurable ( see Chapter 3) and
therfore that Ì ¯ .is irrational. The historian of mathematics,
Cajori, attributes the frst proof of the irrationality of fM¿ the golden
mean ( i.e. the ratio of the shorter to the longer side of a golden
rectangle) , to Campanus in I20 "by means of a method which com
bined a reductio ad absurdum and the principle of infnite descent".
Let us see how such a proof goes.
We suppose, for the sake of argument, that Ì is" rational, that is,
Ì is a ratio of some pair of integers. Then we may as well suppose
that a and are these integers and use Figure 5
3 Next, we see that
a .- a, .. · ·. . ..
are all integers and the construction shows that they are all positive !
Our assumption that Ì is rational has led us to an infnite descending
sequence of positive integers. But this contradicts truth because there
is no such sequence. Thus our assumption that In is rational is un
tenable and this concludes the proof that Ì is irrational.
18 US ES O F I NF I NI TY
The reciprocal of m, namely al
b
, is often denoted by the Grek
letter z ( tau) , short for zoµq, "the section". Thus z ¯ I/m
·
1 ¯ |
|
1
¿1 3|¸
+ |
1 - |
1
¿|- 1
Figure 5.3
Problem
5.1 . Use Figure 5.4 to prove the irrationali ty of Vz and Figure 5.5 similarly
for
V
s. Can you adapt this to other caes of
V
n?
(a)
Figure 5.4
Figure 5.5
THE SELF PERPETUATI NG GOLDEN RECTANGLE 79
5.3. Estimating the Golden Mean
We have ben assuming that a golden rectangle exists. This is easy
to prove, but frst let us estimate the golden mean by "trial and error".
We wish to determine the ratio
è
m ÷ ¬
a
so that
è a
a a + è '
for, if a and è satisfy this relation, then the rectangle in Figure 5.3
is indeed a golden rectangle. If we choose a and è S that a+ è " 1,
our calculations become simpler, for then
è a
m = ¯ ÷ - ÷ a
,
a a + è
that is, a and è must be so chosen that m ÷ a and a + è " 1.
Accordingly, we make a table of trial sides a and è and record the
ratio è/a ÷ m. The frst table indicates that Ì is between .6 and
8 near to .6; the second gives a better estimate of 7M¿ between .61
and .62. Therefor e .615 cannot difer from the golden mean by more
than 0.005 which is a possible error not exceeding 5 in 600 and is less
than a 1 per cent error. This would be pretty good for manufacturing
rectangular cards by machine methods .
a . 2 .4 .6
è
·
8 .6 .4
m 4 1.5 0.67
.8
. 2
0.25
Problem
a .6
è .4
m 0.67
.61
30
0.65
.62
38
0.61
õ. 2. W is nearer to .62 than it is to .61 ; use "proportional parts", or some
simple graphical scheme for getting a better estimate for m. The value
of m to nine decimals is 0.618 033 989, and a good graph should give 0.618.
5.4. Ways of Finding the Golden Mean
a) The shape appears to have been discovered in the following
problem: "Divide a given line segment into two parts such that the
shorter is to the longer as the longer is to the whole segment"; see
Figure 5.6( a) .
The conditions in the problem, expressed in symbols, are
è a
a a + è `
8 US ES O F I NF I N I TY
. ã
(
a)
(
b)
Figure 5. 6
Dividing the numerator and denominator on the right by a, we
obtain
è I
¬ «
a
I + �
a
substituting m for è/a and denoting I/m by z, we get
I
or m
(
m + I ) ÷ I or m + I ÷ z.
This says that the reciprocal of m is M + I Also it leads to a
quadratic equation and a precise expression for m ( see below) . But
before deriving a formula for m, let us see the relation between this
defnition of m and the preceding one. Let us imagine that we have
found the proper lengths a and è and let us express the equality of
ratios by means of a pair of similar triangles. This may be done di
rectly on the given segment as follows :
Figure 5.6( b) shows the similar right-angled triangles ; two equal
acute angles are marked by an x ^ow the dotted lines complete the
fgure to a pair of similar rectangles and a residual square. Clearly we
are back to the previous defnition of the golden shape.
We now continue with the calculation of Ì. We saw that the
defnition of the golden mean implies that M satisfes the equation
m
t
+ m ÷ I
By completing the square:
m
:
+ m +
I
4
5
4 '
THE SELF PERPETUATI NG GOLDEN RECTANGLE 81
we fnd that
V
2
or
since 0 and è are lengths, m must be positive, so
m
V
-
I
2
This is the formula for m.
It is easy m check that m satisfes the original conditions because
and
V
+ I
m + l =
2
= r
m(m + I ) =
V
2
-
I
V
2
+ I
=
5
�
I
= I
By looking up
V
5 in a table or by calculating it in one way or
another, we can get arbitrarily close approximations to mj to nine
decimals,
m �0.0I8 033080
A ruler and compass construction is indicated in Figure 5.7, as the
rader can verify. The length of the segment AE is I, DE and FG
are parallel, and the segments EG and GC have the desired length
m. This is one way in which the Greek geometers solved the equa
tion. The Babylonian mathematicians had worked out the theory of
such quadratic equations for innumerable special cases ( avoiding
imaginaries and negatives) by I800B. C.
· �·
- �s
Figure 5. 7
82 US ES O F I NF I N I TY
b) Another way, verging on the fantastic, of slving the equation
(
5. 1 )
1
m =
1 + m
was discovered by Girar, a mathematician of the 17th century. It
consists of substituting, in the denominator of the right member of
equation ( 5. 1 ) , the value of m, so that
1
(
5. 2)
m =
1
1 +
I + m
and of repeating this substitution for m in each subsequent equation
.
Ordinarily, the object is to remove m from the right hand side of the
equation; but this process pushes m out, way out. After a few steps,
always by simple substitution, we see that
1
m =
1
1 +
1
1 +
1
1 +
1 +
This formula is a very exciting statement of smething, but of what
precisely is hard to say. t One attempt to give it meaning is to d]m
it 8 the limit of a sequence of compound fractions contained in the
form above, namely:
1
1 1 1
I ' 1 '
1 1
1 + - 1 +
1
1 +
1
1
1 + 1 +
&
I
1 +
¹
1
The frst fve of these numbers are 1, -, j, !, i, and a
moment
'
s thought will guide the reader mthe next one. Thereafter it
is easy to guess, and even to show, that the integers appearing in the
denominator of each ( and in the numerator of the next) fraction are
the Fibonacci numbers ! Thus m is now given to us as a limit.
Problem
5.3. Show that if one of the fractions above is equal to p/q, then the next
one must equal q/(p ÷ q) .
I In order to fnd out what, precisely, is defned by such an expression, read
ConttnuedFracttonsby C. D. Olds, to appear in this series.
THE SELF PERPETUATI NG GOLDEN RECTANGLE b
The fact that these fractions actually converge to the number m,
8 one hopes they do, was proved by Simson ( 1734) , 10 year after
Girard
'
s discovery. He showed that the successive fractions alterately
overshot their target and undershot it ; the even-numbered fractions
constitute a decreasing sequence and the odd-numbered constitute an
increasing sequence, both with limit m.
In the next section we shall see how this can be proved.
Problem
5.4. (a) Show that the following method of solving m
,
= 1
-
m is fan tastic:
m
\l
=
VI -
V
I
VI
-
V
I - VI
-
m
by showing that the sequence
m
V
I
I
-=-�VrI=-=VT
I
=_=.= . . ,
0, VI
- 0, V
I
-
VI - \Î ,
· · ·
of fnite parts does not have m as limit.
(b) Show that T, the reci procal of m, satisfes the equation TI = 1 ÷ T
and that the method of solution
T = V
�
=
VI ÷ VI
÷ T
=
VI
÷
VI ÷ V
I ÷ T
is plausible because the corresponding sequence of fnite parts has a
limit.
5.5. Another Sequence Leading to M
In our present plan for approximating M¿ it will simplify matters
to take 0 as unit of length, i.e., 0 = 1 ; then
è è
¯ = ¯ = m
0 Î
¹
and the longer ( als the shorter) sides of successive golden rectangles
have lengths
1 + m, 1 ,
5m - 3,
m, I - m,
5 - 8m,
2m - 1,
13m - 8,
2 - 3m,
13 - 21m,
It is geometrically clear that these numbers approach zero. But u
¾ US ES O F I NF I NI T Y
5 8m, for example, is small, i.e., u
5 8m � 0,
then m is approximately i, i.e., m � i � ¤0I8+ This reasoning is
quite generl ; each term set "approximately equal" to zero gives an
approximation for m Starting with the ffth, we get, in this way
I
2 '
2
3 '
3
5 '
5
8 '
8
I3 '
I3
2I '
Problem
2I
34
'
34
55
'
55
80 '
5.5. Find the decimal equivalent of some of these. Try the ratios 377/610 and
610/987 ; how far out are these in the sequence?
The rule of formation of these fractions is clear: the denominator of
one fraction is the numertor of the next and the sum of numerator
and denominator gives the new denominator; that is,
ç
is followed by
ç
+ ç
It is immediately clear that the sequence of denominators is precisely
the Fibonacci series, and of course this is also the sequence of the
numertors; there is 8 small matter of the lost ancestors, I and I ,
but it will be remembered we started without them.
Problem
5.6. Is each fraction above num¬ca//ycloser to m than the preceding one?
Give evidence in support of your answer.
We shall show that the successive approximations to m oscillate
around it. Take the pair
it
and if 8 an example. These satisfy the
simple relation:
(
5.3)
It is not necessary to multiply out and fnd large common denom
inators in orer to verify this; cancellation of the frst pair of 55
'
s
shows the trick. Notice next the resemblance of this equation to the
one satisfed by m if we write it in the form
(5.4) m· m+ m = I .
THE SELF PERPETUATI NG GOLDEN RECTANGLE b
Now let us suppse that » is bigger than H ( as it is) ¦ if we replace
H by H in (5
·
3) that equation will become false because the left side
will become smaller. Thus (H)
2
+ H is smaller than I . This shows
that H is smaller than m. Similarly supposing again that H is
smaller than » and replacing if in ( 5.3) by the larger number we
fnd that H is larger than m. It happens that » ts bigger than H,
but even if it were not, what our argument shows is this : The smaller
of these numbers is smaller than m and the larger of the two is larger
than m. They cannot be equal in view of equation ( 54) , as we shall
now show.
The argument is quite general : A pair of successive approximations
ha the form p/q, q/( p + q) (where p and q are integers) and
satisfes the equation
(5
5) '! .
q
+ ´ = I
q p + q p + q
If the two fractions could be equal, then comparison with equation
(54) shows that each would be equal to mj but m is irrational ( see
Section 5.2) . Therefore these fractions are not equal. To see that m
lies between them, observe that if we replace the larger one by the
other in equation (5.5) , the left side will become smaller than Ij
this shows [by comparison with (5
·
4)1 that the smaller number is
smaller than m. Similarly, the larger is larger than m. Q
.
E
.
D.
The rader will not have any trouble satisfying himself that the
observations of Simson have now been verifed. Before going over to
the logarithmic spiral associated with the golden rectangle, we men
tion two interesting problems : one has mdo with the powers of mj
the other is a wonderful problem due to Lagrange and concerns the
residues of the Fibonacci numbers modulo any integer. The second
one is difcult. See the next chapter for its solution.
Problems
5. 7. Extend the following remarks : m
2
= 1 - m; m3 = m - m
2
= 2m - 1 ;
m4 = 2m2 - m = 2 - 3m;
. . .
. Can you write a general formula? Do the
same for . = 1 + m; r = r + 1 ; r
ª
= .
2
+ . = 2. + 1;
. . .
.
5.8. (a) The Fibonacci numbers, modulo 2, are I , I , 0, 1, I , 0, · · · ;
modulo 3, they are I , I, 2, 0, 2, 2, I , 0, I , I , 2, · · · ;
both sequences are teriodic. Show that the sequence of Fibonacci
numbers mod 4 is also periodic. Similarly, exhibit its periodicity mod 5
and mod 6. (The series modulo 10 ha period 0. )
(b) Do you see how to get the residues, mod 3, 4, 5, 6 without actu
ally calculating the large Fibonacci numbers? Using this knowledge,
86 US E S O F I NF I NI T Y
and the rule that each number is the sum of the preceding two, can you
prove Lagrange's remark: Thereetdueeo¡theFt|onacctnum|eremodu/o
anytnteqeruhateeerareµertodtcIHint: If n is the modulus the period
does not exceed n
·
+ 1 .
5.6. A Spiral Zig-Zag
We shall next construct a spiralling zig-zag whose vertices lie on a
very beautiful spiral. The zig-zag will lead us to the origin of this
spiral, and also disclose the essential scheme underlying the construc
tion of the spiral. At the same time we shall be constructing a whirl
of contracting squares and also a whirl of rectangles and doing a bit of
whirling ourselves. Starting at the lower left corner A of a golden
rectangle ABDF, see Figure 58,we draw a 15ºline to the upper edge
at C, and drop the perpendicular CH cutting of a square. Next,
using the new golden rectangle CDFH, with CH as a base, we repeat
this construction; that is, we draw a line CE at 15º to the new base
CH and drop a perpendicular EJ cutting of a square and giving us
rectangle EFHJ with base EJ
·
Then we draw a line E0 so that
E0 makes a 45° angle with EJ, drop the perpendicular 0K, and
from the point 0 of the new rectangle 0HJK we continue the
construction. In other words, every time we draw a line from the
lower left vertex of a golden rectangle at 45ºto the base, this line inter
sects the upper side of the rectangle in a point from which the next,
similar, construction begins. Each new vertex belongs to a rectangle
which is a 00º turn from the previous one and whose sides are M
times those of the previous one. Each rectangle is whirled around by
00º and shrunk by a factor M.
Figure 5.8
THE SELF PERPETUATI NG GOLDEN RECTANGLE 87
As far as the succession of squars and rectangles is concerned we
have the same situation as before, and the same lengths of sides :
1 + m, I , m, I - m, 2m - I ,
. . .
, and so on. In other
words, we get the sequence m
-
!, I , m, m
t
, m
.
, ¯ • ¯ , as the
reader has already proved or can prove now merely by observing ·the
fgure (or from the governing equation m
t
=
I - m) , and the fact
that each length comes from the preceding by scaling down in the
ratio of 1 to m.
Problems
5.9. The whirl of rectangles may be expressed thus : ABDF, CDFH, EFHJ
,
0HJK,
. . .
. Can you attach any signifcance to the way the vertices are
ordered in successive rectangles?
5. 10. If the side AB has length 1, AC has length ¡)
Show that the length
of the zig-zag ACE0I .
.
.
is ¡)/m
·
by using the formula for an infnite
geometric progression. Have we proved that this formula is valid even
if the common ratio is irrational ?
5.7. Use of Similarity Transformations
It is clear from Figure 58 that the zig-zag closes down on some
target-point T We shall prove that T is the intersection of the
diagonal BF of ABDF and the diagonal DH of CDFH ( the sec
ond diagonal is the successor of the frst under the whirling motion
described above) . The proof is technically very simple. In the preced
ing description we imagined ourselves turning thrugh 0º with each
new line of the zig-zag. Her we shall let the rectangles tur. Our goal
is to place rectangle ABDF on CDFH, A going into C, B into
D, D into F, and F into H, by a combination of two motions, a
rtation through 0º and a shrinking in the ratio I to m We can
describe the desired motion very simply, with T as the center of
both motions. First we rotate the whole plane through an angle of
0ºclockwise around T, so that the rectangle ABDF goes into the
new position A'B'D'F'
, see Figure 5.0
·
Next we shrink all points of
the plane towards T along radial lines in the ratio I to m, so that
A'B'D'F' goes into CDFH
·
This pair of motions, called transforma
tions of the plane, t actually maps the points A, B, D, F upon
the respective points C, D, F, H. The repetition of this com
posite transformation, indefnitely often, generates the whole zig-zag.
I In order to fnd out more about the subject of transformations, see 0eo-
metrtcTrane¡ormattoneby I. M. Yaglom (translated by Allen Shields) to appear
in this series.
8
s
-
US ES O F I NF I NI T Y
- ¡-÷_s
/
/
´
,
·
.....J
,
yg '
·
...�
·
Figure 5.9
The frst step in our proof consists in showing that the diagonal
BF of the frst rectangle meets the diagonal DH of the second rec
tangle at right angles at the point T. Since ABDF and CDFH are
similar rectangles, all angles marked a in Figure 5.9 are equal. The
similar right triangles BFD and DHF therefor have
qBFD = 0º - a = qDHF,
so that
qHTF = I8º - [a + ( 0º - a) ]
= 90°,
and BF and DH indeed meet at right angles in T.
Thus, if ABDF
is rotated by 0ºaround the point T, BF will go into a new posi
tion B'F' perpendicular to the original BF ( see Figur 5.9) , and
B'F' will contain the line DH Similarly, DH will go into D'H'
and D' H' will lie along BF
N ext, observe that
B'D'
DF
I + m
I
I
m
B'T
DT
D'T
FT
THE S ELF PERPETUATI NG GOLDEN RECTANGLE 89
Morover, since
A'B'
CD
AB
CD
B'T
DT
I
m '
and since CD and A'B' are parallel, it is clear that C lies on A' T
and that
A'T I
CT m
Similarly, H lies on F'T and
F'T I
HT m
Hence, the shrinking by a factor m of all distances from the point T
brings the rctangle A'B'D'F' into the rectangle CDFH as we set
out to show.
Our arguments depended only on the golden ratio m of the
shorter to the longer side of the given rectangle, and not on its actual
dimensions. Since this golden ratio is prserved in all the subsequent
rectangles of our construction, the above proof is valid for all. This
shows that the point T will serve as the center of rotation and con
traction thrughout, that it is interior to all the shrinking rectangles
we construct by our scheme, and that it is therefore the target point
of the zig-zag.
5.8. A Lgarithmic Spiral
We can fnd a simple formula for the vertices A, C, E, 0,
I, of the zig-zag in Figure 5. 8 if we make the following
choice of axes and units. Pick T as origin of coordinates, let r repre
sent distance from T, and choose the line AT as initial direction
( se Figure 5. I0)
Take the quarter-turn (one-fourth of a rota
tion ¯ a/2 radians ÷ 9°) as unit of angle, measured from AT,
clockwise being taken positive, and let t denote the number of quar
ter-turns. The vertices A, C, E, 0, :
. .
are obtained by rotat
ing AT about T through 0, I , 2, 3, quarter-turns,
respectively, and by shrinking the distance AT by m
»
, m\ m
t
,
m
:
, : : , respectively. Therefore, the distance r of any such
vertex from T of our spiral zig-zag can be written
(5. 0)
r = AT· m
'
.
9 US ES O F I NF I NI TY
� � �� ¯¯¯¯�
t
-
Figure 5.10
This gives the following table:
vertex A C E 0 I
« « «
t 0 1 2 3 4
. . .
7 AT AT· m AT· m¹ AT· m¹ AT· m
·
+ . +
f
m
·
AT
=
¨
1 m m
·
m
ª
« « «
We get an even simpler formula if we take AT as unit of distance.
We can do this by setting k ÷ r/AT. Then, in terms of k, Equa
tion ( 5.6) becomes
k = m
,
.
If, for t, we substitute successively the terms of the arithmetic
progression 0, 1 , 2, 3, 4, 5, 6,
. . .
, then we obtain, for k,
the geometric progression 1, m, m
t
,
m
-
, • • • ¡ the position of
all these vertices is now determined because t is the number of
quarter-turs and k is the factor by which AT is multiplied.
It can be shown that, if we now let t take on all positive real values
( instead of merly integers) we get the smooth spiral pictured in
Figure 5. 10.
THE SELF PERPETU ATI NG GOLDEN RECTANGLE 91
Problems
5.1 1 . Calculate some in-between points on the spiral by using the values
!
.
�
. �· · ·
for t . Find some additional points for other rational
values of t
5.12. What do we get if also takes on negative values? How big does R
become?
Once the curve is drawn it fnds for us all powers of m, for example
:|
:
m
=
m corresponds to
that is,
m
t|
.
�m corresponds to
that is,
m
./:
{ m
.
corresponds to
that is,
/
I
2 '
I
quarter-turn
2
I
3 '
1
quarter-turn
3
5
' '
5
quarter-turn
0
Figure 5. 11
92 US ES O F I NF I NI T Y
The curve
is called an eçntanqu/ar sµtra/ and also a /cqartt/mtc sµtra/ ( in this
case the reference is to logarithms to the unusual base m)
·
The
equation can be put into other standard forms :
by proper choices of the constants c, k, /. The advantage of this
is that 10, 2, and e ( �2. 718) are the more conventional bases for
logarithms.
The logarithmic spiral has two important properties :
i) If points P, Q and P', Q' subtend equal angles (() at T
then the triangles PTQ and P'TQ' are similar; see Figure 5. 11.
ii) At each point ( S) of the logarithmic spiral, there is a tangent
line and it makes a constant angle (1) with the radius vector T S.
Property i) is easy to prove and is left as an exercise. Property ii)
is more difcult ; one needs calculus to defne the tangent to the curve,
must show that this curve has a tangent everywhere ( this is easy in
this case provided one frst shows that the curve has a tangent at
some point) , and then one can prove property ii) .
Problems
5. 13. Explain how the curve in Figure 5. 11 can be used (equivalently to a table
of logarithms) to multiply numbers if these are represented by lengths
on a ruler which pivots at T andif we can measure and add appropriate
angles.
5. 14. What property of the curve corresponds to the use of the cbaractertettc
and mantteea of logarithms?
5. 15. By plotting the curve and measuring with a protractor, check that the
tangent to the curve at each point (8) makes an angle with the radius
vector T8 of about 73° [actually arc tan ( ../2 log f»).
5.9. The Pentagon
Perhaps because mathematicians work as hard as they do, the
subject of mathematics rewards them with frequent bonuses. The
logarithmic spiral generated by the shower of golden rectangles, and
the apparance of the Fibonacci numbers in a sequence of approxima
tions to the golden mean
m
V ¬ 1
2
are pleasant and unexpected encounters.
THE SELF PERPETUATI NG GOLDEN RECTANGLE 93
Another bonus of this sort is the fact that m is also the ratio of
the side of a regular pentagon t
o
its diagonal. In Figure 5. 12( a) let
the equal diagonals I;I, , I
t
I, , I,I, , I,I; , I,I
t
be of unit
length; then each side is m. A proof follows.
(a) (b)
Figure 5. 12
We defne the regular pentagon by the condition that the arcs
I;I
t
, I
t
I, ,
·
.
· , I�, of the circumscribed circle are equal.
This implies that the chords I;I
t
, I
t
I, ,
, I,I; are equal.
Each arc takes up one ffth of the circumference and subtends a cen
tral angle of 275 radians or 72º
,see Figure 5. 12(b) . There are fve
congruent central isosceles triangles whose angles are 54°, 54°, and
72°, so each vertex angle of the pentagon is 108°. Since triangles
I;I
t
I, ,
. . .
[see Figure 5. 12( a) 1 are isosceles, their base angles are
!( 180° - 108°) = 30º, zI
t
EI, = 180° - 2 · 30° = 108°,
and
zI;EI
t
= 72º
The substantial part of the proof begins now. We fnd frst that the
triangles EI
t
I, and I,I,I
t
in Figure 5. 12(a) have angles 30°,
30°, and 108° ; they are isosceles and similar. Denoting EI
t
by
V, I
t
I, by x, and rmembering that I
t
I, = 1 , we write the
proprtions :
v
x
thus
t
V = x .
Next we fnd easily that the angles of triangle EI,I, are 72º,72º,
30º
,
so that this triangle is isosceles and
EI, = 1 - v = x or v = 1 - x.
94 US ES O F I NF I NI TY
Therefore
x
t
= I - x.
Since m is the positive root of this equation, we have shown that the
ratio of the side of a regular pentagon to the diagonal is m. Q
.
E. D.
5.10. Relatives of the Pentagon
The Greeks got two bonuses for studying the pentagon. First the
dodecahedron and ( somewhat more subtly) the icosahedron use the
pentagon, and also the golden rectangle. This will be seen in Figures
5 I3 and 5 I4, but no further discussion is ofered since this would
take us into solid geometry. ( The details are not difcult and the
reader is urged to consult the articles by Coxeter and Gardner previ
ously mentioned. )
Figure 5. 13. The regular
dodec ahedro n
(12 faces, 30 edges, 20 vertices)
Figure 5. 14. Corners of three
golden rectangles coincide with
corners of U regular icosahedron
(20 faces, 30 edges, 12 vertices)
Second the Greeks discovered the pentagram and thought it very
attractive; see Figure 5. I5 The Pythagoreans used it as some sort of
club-membership pin and attached symbolic value to it. It is easy to
see in it the suggestion of a perpetual movement. From a mathe
matical point of view it is extremely interesting as a visualization of a
rotation of the plane which is of period fve ( repeating itself indef
nitely) , but is diferent from the rotation associated with the more
common pentagon.
THE S ELF PERPETUATI NG GOLDEN RECTANGLE 95
Figure 5.15
The reader will notice that the over-and-under passing in the fgure
is reminiscent of illustrations of clover-leaf intersections on modern
high-speed highways. This over-and-under drawing is the moder
way in mathematics of representing kncæ
If the pentagram is ac
tually made out of string it will be found that the string is knotted.
It is undoubtedly the purest coincidence that one can make a ver
good pentagon ( in fact a precise pentagon, theoretically) by tying a
knot in a strip of paper ; the illustrations in Figure 5
I0show how. If
this is done and held up to the light, there is seen the symbol of the
pentagram.
ä
/
ä
Figure 5.16
J
|
J
C H A P T E R S I X
Constructions and Proofs
This fnal chapter covers much the same material as the previous
pages of the book, but it is organized along diferent lines. The em
phasis here is upon construction and proof; the word "construction"
is used in a general sense that includes the defnition of infnite
processes. At the same time, the propositions we shall work out in this
chapter answer some of the problems mentioned earlier.
6.1 Indirect Prof
There have already been several examples in the book of indirect
proofs, also called proofs by contradiction. We showed through this
device that
V
2 is irrational, also that the golden mean is irrational.
Here we give a simpler sample of this most important technique.
ASERTION. There does not exist any pair of integers (positive or
negative) satisfying
4x + 6y = I5
PROOF. Assume for the sake of argument that there is such a pair
of integers; call them m and n. Then
I5 = 4m + 6n = 2(2m + 3n) .
This asserts that I5 is an even integer ( 2m + 3n being the quotient
on division by 2) which is false. This concludes the proof.
96
C O NS TRUCTI O NS AND P ROOF S 97
Problems
6. 1. Accept the fact that v2 is irrational. Prove from this fact that the
reciprocal of (I + ,¿) is also i rrational .
6.2. Prove that the form ax + òy, where a and ò are i ntegers, represents
an integer c through some choice of integers x and y if and only if c is
divisible by the highest common factor of a and ò. Thus, for example,
the equation
I2x + I8y = c
is satisfed by integers x and y only when c is divisible by 6.
6.2 A Theorem of Euclidean Geometry on Parallels
as Proportional Dividers
HYPOTESIS: t and t' are two lines (called transversals) which
cut each of three parallel lines ll
'
l
2
'
la , as shown in Figure 0 I .
CONCLUSION: The lengths of the line segments cut of from t and
t' by the three parallel lines are proportional, that is,
AB A'B'
BC B'C'
·
!i ¯
��
Ìt ¯
��
!t �
Figure 6. 1
EXAMPLE. Inspection of the ruling in Figure 0. I shows that
AB 3
BC
=
4
confring the theorem.
and similarly
A'B'
B'C'
3
4 '
PROOF
.
The example also illustrates the proof in the general D~
tional case, that is when AB/BC is a ratio of integers, say mJn.
98 US ES O F I NF I NI T Y
a) T/e rattcna/ 0 ccmmsuraè/e case. In the rational case one
makes /
: , /
t
·
/
.
part of a system of rulings which exhibit the
ratio mJn One divides AB into m equal intervals and BC into n
equal intervals of the same length and constructs parallels at the
points of division.
To prove that the ratio mJn is now transferred to t' one has to
show that the segments on the right are all equal to each other.
For this purpose one constructs certain auxiliary segments, all
parallel to t and each forming one side of a µara//e/cqramwhen taken
with the corresponding segment of t, and forming one side of a tri
angle when coupled with the corresponding segment on t' . It is now
easy to prove that all these triangles are congruent; this rests on
three types of elementary propositions: i) triangles are congruent if
one side and all angles of one are equal to the corresponding parts in
another; ii) opposite sides of a parallelogram are equal ; iii) corre
sponding angles made by a line transversal to two parallels are
equal. See Figure 6.2.
(a)
S
(b)
Figure 0.2
(c)
b) T/e tnccmimnsuraè/e case . IrccJ èq ccnttnuttq. When AB/BC
is not rational, the method j ust shown is not available. Let us now
suppose, for example, that AB/BC is our familiar golden ratio
H
×°
1) . Let us take the sequence of Ftècnacct rattcs u-
de
fned by
J
-
-
:
u
-
=
-
J
-
n = 1, 2, 3,
.
¯
.
,
where J
-+
, = J
-
+ J
-
-
: , and the frst terms of the sequence J. are
J
:
= 1,
h
= 1 , J
.
= 2, J
t
= 3, Js
= 5,
.
. It can be shown
that, as n gets larger and larger, these Fibonacci ratios u
-
approach
the golden ratio m.t In symbols,
lim u
-
= m.
n¬m
t For a proof of this statement, see C. D. Olds, Conttnued Fracttont, to
appear in this series.
CO NS TRUC T I O NS AND P ROOF S
A
·
/ Ì
·
¿¿¿¿¿¿�
·
·.
Figure 03
99
|
Now, letting the endpoints A and C of the segment AC stay
fed, we construct a sequence of points Bn on this segment such
that for each R,
= u
- ,
then the B
-
approach B and the ratios u
-
approach m. At each
B- we construct a parallel to BB' ( see Figure 6.3) , S obtaining a
point B� on A'C'.
10 US ES O F I NF I N I TY
It follows from the rational case that
A
'
B�
-
= u
.
B, C
'
-
,
therefore, as n increases, these ratios approach m. What remains
for us to prove is the intuitively obvious fact that the point B' is
the limit of the constructed sequence B� , B, , B� ,
The reason for regarding this as an "intuitive fact" ( in mathe
matics intuition is aways respected but not relied upon too heavily)
is the true impression we get from Figure 0. 3,namely that the variable
point B, varies with the point B
-
in such a way that the segments
A
'
B, approach the segment A' B' èecause the segments AB-
ap
proach the segment AB Another way of saying this is that the
points B, var ccnttnucus/q with the points B
- .
We shall show
that, if the points B, approach a limit, this limit must be B'. There
fore, to prove the continuity, we require a geometric principle ( not
present in the geometr of the Pythagoreans) from which we can
infer the existence of a limit. This principle is the geometric counter
part of the Bolzano-Weierstrass principle, discussed earlier in the
book.
Using this principle we fnd that the sequence B� , B� , B� ,
does have a limit point; call it Bª for the moment. Bª is a limit
point of two sequences, B� , B� , B� ,
. . .
lying on A'B' and
B, , B� , B� ,
. . .
lying on B'C'. It follows now that Bª lies on
both segments A'B' and B'C'
,
but their only common point is B'
and S Bª = B'
. This concludes the proof of the theorem in case
the ratio is given as m. But we have used no particular properties
of m excepting only that it is a limit of a sequence of rational num
bers which approach it from above and below. This is true of all real
numbers, and therefore the proof is entirely general.
c) T/e tnccmmensuraè/e case. IrccJ èq ccntradtcttcn. The Greek
geometers did not use the continuity proof just described but relied
on an indirect argument. Assume that the assertion is false, that is, that
either
AB A'B'
BC > B'C'
or
A'B' AB
B'C' > BC
O
Figure 0.4
Let D be a point on t between A and C such that
AD m
DC n
I0I
( see Figure 6.4) . Then D lies between A and B, for otherwise
AB/BC would be less than or equal to
r
. Construct a line l4 through
D parallel to l
2
; it intersects t' at D'. Now D' lies between A'
and B' for otherwise the parallel segments BB' and DD' would
intersect. Hence
A'B'
A'D'
B'C' > D'C'
'
But by what we know about the rational case,
AJ AD AÆ
D'C'
-
DC
-
r,
so that
B'C' >
r,
which contradicts the inequality (6. 1 ) . Thus the assumption that
AB A'B'
BC > B'C'
led to a contradiction and hence is false.
Similarly we can show that the assumption
A'B' AB
B'C' > BC
leads to a contradiction. Therefore
AB A'B'
BC B'C'
102 US ES O F I NF I NI T Y
6.3 Defnition by Recursion
A sequence of mathematical objects
frst object, second object, third object,
. . .
is said to be constructed recurstre/qwhenever the description of each
new object uses the fact that the previous one is already defned. For
example, consider the sequence of points
( 6.2)
on the x-axis whose coordinates 8
-
satisfy the following conditions:
8; I,
(6.3)
8- 8
--:
+
I
n '
n
� P P �
+ 1
B Z
Figure 0.õ
Note for contrast that the sequence of points
( 6.4)
on the x-axis, defned by their x-coordinates a-
as
( 6. 5)
¶
a
-
= n ,
2, 3, 4,
J
are independent of each other; that is, we can fnd Q6 from the
forula ( 6.5) without having found Q4 . To fnd I
s
we are directed
by ( 6.3) to fnd I 4 frst
.
Let us suppose that some Zeno among our acquaintances has chal
lenged us: "Does ( 6.3) defne an tn]ntte sequence?" He might say:
"I grant that I; is defned, and I
t
, and I, , and as many more
points as you have patience to construct, but no more. Thus ( 6.3)
only shows how to construct a large fnite sequence. " Mathematicians
do not hesitate to regard ( 6.3) as a defnition of the sequence (6.2)
for all values of n, i.e., n = I, 2, 3,
. . .
, just as they accept
(6.5) as a defnition of the sequence ( 6.4) for all values of n
·
The
reader will recall that this recursive construction was used in the
case of some of the later zig-zags in Chapter 4, and also in the spiralling
rectangles and the zig-zag in Chapter 5 where, it is hoped, he accepte
them as quite natural procedures.
CONS TRUCTI ONS AND P ROOF S I03
To argue the matter a little more fully, one might paraphrase ( 6.3)
as follows, in words. It consists of two parts. A ]rst steµ, namely that
the frst number is I
·
Next, a ccnttnutnqsteµ,which is the heart of the
matter : Icu are at a certatn staqeand/are ca/cu/ated acertatnnum-
èer
·
Incrder tcca/cu/atet/enextnumèer, qcumere/qaddtc t/ts cne t/e
rectµrcca/cJ t/etcta/numèercJ staqes
Notice that this continuing instruction is not to be discarded after
it has been used; on the contrary, it is worded so as to be always
applicable to the next stage. Notice fnally that this continuing in
struction corresponds to the way in which the natural numbers are
given to us: Ftrst steµ-the frst number is 1 . Next steµ-the next
number is one more than the one you now have.
Thus (6.3) sets up a one-to-one correspondence between the points
of the sequence ( 6.2) and the natural numbers. This is exactly what
the formula (6.5) does for the sequence ( 6.4) ; (6.3) and ( 6.5) are
equally valid ways of constructing a sequence.
We close this section with the mention of some interesting recur
sions. One of the most important recursively defned sequences in
mathematics is this one :
1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800,
This is the sequence designated by 'n|" ( read 'n factorial") and
defned thus:
l ! = 1
(n + 1) ! ¯ (n + 1) times n|, n ¯ 1 , 2, 3,
The concept of. recursive defnition embraces more general situa
tions. For example, the Fibonacci series is recursively defned through
the conditions that
as
J
:
= 1,
J
t
= 1 ;
J
-+t
= J
-
+J
--
;,
n = 2, 3, 4,
Similarly, one can defne a sequence recursively by such conditions
è, = I,
è
t
= 1 + 1 = 2,
è; ¯ 1 + 1 + 2 = 4,
è. = 1 + 1 + 2 + 4 = 8;
è- ¯ 1 + è, + è
t
+
·
. ·
+ è
-.;
, n = 2, 3, 4,
10 US E S O F I N F I N I T Y
Problem
0.3. Can you identify this last sequence?
6.4 Induction
A formal proof by mathematical induction has two striking land
marks, a ]rststeµand an tnducttre steµ Thus it is built exactly like a
recursive construction and for the same reason: It is an infnite se
quence of assertions arranged in an outwardly fnite form. Let us con
sider a few examples and discuss the general principle later.
EXAMPLE I
Asserttcn.The integer I + 4
-
is not divisible by 3for
any value of n = I , 2, 3,
PROOF BY MATHEMATICAL INDUCION
.
Ftrst steµ. The assertion is
true for n = I, because I + 4 = 5 is not divisible by 3 The frst
step is proved.
Inducttresteµ. IF the assertion is true for some integer k ( that is,
IF I + 4
·
is not divisible by 3) THEN it is also true for the integer
k + I ( that is, I +
4
· +
is not divisible by 3) The proof of the in
ductive step follows. Notice that, since 4
· +
!
= 4
4
·
,
( I + 4
·+
) - ( I + 4
·
) = 4
·
+
- 4
,
= 4
·
.
( 4 - I) = 3
4
·
•
This shows that the diference btween ( I + 4
· +
!
) and ( I + 4
·
) is
divisible by 3, therefore one of them is divisible by 3 if and only if
the other one is. t But since I + 4
·
is nct divisible by 3 ( by the
/qµct/ests of the inductive step) , it follows that I + 4
·
+
is not.
This is the conclusion of the inductive step which is hereby proved.
Problems
0.4. Recall the defnition : An tnteqer m bae reetdue r modu|o an tnteqer g
means r tetbe rematnderuben n tedtetdedòy g. Prove the "Rule of 3".
An integer A and the sum of its digits have the same residue modulo 3,
for example, 47, 15 and 2õ = 4 + 7 + 1 + õ + 8, both have residue
1 modulo 3
0.õ. Prove by induction :
1
1
+
2
+
. . .
+
n = 2 n(n
+
I), n = I, 2, 3, · ·
·
·
!If a - ò = 3e (e is our i nteger), then a = ò
+
3e and ò " Ü ¯ 3s. It
follows that a is divisible by 3 if and only if ò is, and that ò is divisible by
3 if and only if a is.
C ONS TRUC T I O NS A ND P ROOF S 105
EXMPLE 2
.
Asserttm ( as suggested by inspection of the multi
plication table) :
( 1 + 2 + .
.
. + n;= 1
:
+ 23 +
. .
. + n
:
, n = 1, 2, 3,
PROOF. Ftrst steµ. When n = 1, the left side is 1
2
= 1 and the
right side is 1
3
= 1.
Imuttre steµ. Suppose that k is such that
(0.0)
( 1 + . . . + k)
2
= 1
3
+ . . .
+ k
and let us study
(0.7) ( 1 +
.
. . + k + k + 1)
2
.
This has the form (A + B)
2
with A = 1 + 2 +
.
. .
+ k, and
B = k + 1. Using (A + B)
2
= A
2
+ 2AB + B
2
, we get from ( 0.7)
the following:
( 1 +
. . .
+
k)
2
+ 2( 1 + . .
.
+
k
) (
k + 1) + ( k + 1)
2
.
If we apply to the middle term the formula to be proved in Problem
0
·
5 ( with n = k) , we get
( 1
+
. .
, + k)
2
+ [k· (
k + l ) ] ( k + 1)
+ (
k + 1 )
2
.
Notice now that the last two terms have a common factor, namely
(
k
+ 1)
2
. If we collect terms, this gives us
(0.8) ( 1
+
. .
.
+ k)
2
+ (k +
1)
3
.
So far we have not used our hypothesis ( 0.0) ,
if we now apply it to
(0.8) , this becomes
( 0
·
0) 1
+ 23 +
. . .
+
k
+ ( k
+
1)
3
.
In other words, the expression ( 0.7) has been transformed into ( 0.0)
.
This concludes the proof of the inductive step, that IF our assertion
holds for an intger k, THEN it holds for k +
1 ; the proof by induc
tion is now complete.
The /cqtc of these proofs is clear. We show that every one of an
infnite sequence of statements is true, by proving that we can go
through them one after another indefnitely and never fnd one that
is false. We make sure that we have not missed any by starting with
the frst.
The startling thing is that all of this is done in what looks like a
single ( inductive) stp but the fact is that this is a continuing step
which is constantly reapplied by the mechanism of the "IF
'
.
· THEN
"
proof; there is really an infnity of arguments, one for each case.
10 US ES O F I NF I NI TY
The µrtnctµÞ cJ t/e Þa tnæqo which we used for showing the
irrationality of V and the µrtnctµÞ cJ tn]ntæ descent, which we
us to show the irrationality of m, the golden section, are closely
related to proof by mathematical induction. It takes only a little
prctice to rephrase proofs so that they have one rther than the
other form. However, we shall leave these considertions to the
diligent reader. t
6.5 An Application to a Proof
The following is a valid proof of the fact that, for all n, the sum
of the frst n consecutive positive integers is ]n(n + 1) . We men
tioned this in Problem 6.5. The reader is urged to interpret the proof
below as an instance of defnition by recursion and prof by induction.
Let
8
=
1 + 2 + ö + . . . ¬ n
Then 8 may alsO be wrÎtten
8
=
n + n - 1 + n - 2 + . . . + 1
Adding these eXpressÎOns, One ObtaÎns
28
=
n + 1 + n + 1 + n + 1 + . . . + n + 1
=
n (n + 1) .
Therefore
8 = ]n
( n+ 1 ) ,
and this holds for n
=
1, 2, 3, This argument is said to
have been rediscovered by C. F. Gauss ( 1771
-
1855) when he was
nine years old.
6.6 Fast Growing Sequences
The following story is a slightly moderized version of a calculation
performed by Archimedes, combined with a severl thousand years
old tale for children.
Suppose that in the year 1750 some pair of mosquitos took cover in
the state of New Jersey and suppose that they then generted a
!
See the discussion of mathematical induction in Chapter 1 of bbat Ie
Hatbemattcel by Richard Courant and Herbert Robbins. New York: Oxford
University Press, I94I
C ONS TRUCT I ONS AND P ROOF S I0I
population which has since doubled with each succeeding year. By
1760, there would have been 2
10
descendants of this pair; this is
about 10
3
and not very many. By 1770 they would have had a
progeny of 2
2
individuals which is over 10
6
, or one million. By
1800 the number would be 2
6
which is more than ten quadrillion
( 10
16
) . By now the population would number 2
210
, which is more
than 10
63
•
To form some idea of what this number means ( 1 followed by 63
zeros is easier to say than to comprehend) let us suppose that one
million small mosquitos will go into a box measuring one inch on a
side. This gives us 10
57
boxes. Out of these boxes, with careful
stacking, we can make a giant cube of boxes measurng 10
19
boxes
on a side, and of course each side would be 10
19
inches long. That is
more than 10
1
4
miles long. The distance from the sun to the earth
is only about 0,0,0miles or less than 10
8
miles, and Pluto is at
a distance less than 4 · 10
10
. So, if the sun were set in the middle of
such a box even Pluto would be lost deep inside of it; indeed there
would b room in the box for mor than
( 10
3
)
3
= 10
9
= one billion of our solar systems.
Let us think of the numbers 1, 2, 3, 4,
. . .
and also the
numbers 2, 4, 8, 16, 32, 64,
. . .
as representing the suc
cessive census-records of some growing populations; then the frst
sequence grows larger and larger without limit, but the second grows
"out of this world" faster than the frst.
Let us give this idea a mathematical form, and then see what we
can prove. Table 6. 1 shows the two sequences, and also the ratios of
corresponding terms (except that I have rounded of these ratios to a
whole number by dropping any fractional part) .
TABLE 0. I
n I 2 3 4 õ 0 1 8 9 I0 II
,&¬&¬&¬&«&««&&&«&««&&«
2ª 2 4 8 I0 32 b I28 2õ0 õI2 I024 208
.~&¬~&&m&&&&&~&«&«&.&«
2ª
2 2
&
2 4 0 I0 I8 32 õ0 I0 I80
n
Table 6.2 indicates that 2
n
is always greater than n, that from
the 4th term on, 2
n
is greater than n
2
, and that from the 10th
trm on, 2
n
is greater than n
3
•
I08 US ES O F I NF I NI T Y
TABLE 0.2
n I 2 3 4 õ 0 1 8 9 I0 I I
µ=¬¬=¬=,=,=,¬m,=,m,=,
2ª 2 4 8 I0 32 b I28 2õ0 õI2 I024 208
µ=¬=¬=¬=,=,=,,=,=,==,¬
n2 I 4 9 I0 2õ 30 49 b 8I I0 I2I
µ=.& =¬¬=,=~=,====,,==,m,
n
'
I 8 Z b I2õ 2I0 343 õI2 1w m I33I
At this point mathematical induction has acquired a new wrinkle.
We can say that for all n ( greater than zero, that is) 2f exceeds
n, but we cannot say that for all n, 2f exceeds n
t
• We may con
jecture the following theorems.
TEOREM I . Fcra// n qreatert/anterc, 2f exceeds n
TEOREM 2: Fcra// n qreater t/an 4, 2f exceeds n
t
•
TEOREM 3: Fcra// n qreatert/an 0, 2f exceeds n
:
•
Before we go on with more conjectures, let us talk about proofs.
The problem before us is this : Can one use mathematical induction,
even if one wants to start it not at n = 1 , but at n = 4, or n = 10,
or even later places?
The answer is yes. The reason is simple. Anqstatementt/attsmade
ccnce¬tnq t/e seçuence N
+
1, N + 2, N + 3,
. . .
can tmme-
dtate/q èe trans/ated tntc astatement aècut t/e seçuence N + n Jcra//
tnteqers n. This concludes the argument.
To illustrate the argument, take Theorem 2 and rewrite it like this:
Fcr a// tnteqers n, 2f+
4
exceeds (n +
4)
t
. Do you agree that if
Theorem 2 is true, this new statement is true Jcr a// n?
Problem
0. 0
Reformulate Theorem 3in accordance with this program.
The proof of Theorm 1 is very simple. The frst step consists of
checking the statement for k = 1, and it is true.
Now for the inductive step: Let k denote an integer for which
2
k
exceeds k. Then 2
k
+l exceeds 2k, and since 2k is at least as
big as k
+ 1 it follows that 2k+l exceeds k +
I , which concludes
the proof.
C O NS TRUC TI ONS A ND P ROOF S I0
The proof of Theorem 2 is only a little harder; it uses the fact that
(n + I)
t
= n
t
+ 2n + I . The proof is interesting, as you will see,
because the inductive step and the frst step are out of step.
The frst step is at k = 5, and it is true that 25 exceeds 5
t
For the inductive step, suppose that k is an integer for which
2k exceeds k
t
. Then 2k+l exceeds 2k
t
. Now, when k exceeds
4, k
t
exceeds 4k, and 2e exceeds k + 4k which exceeds
k+ 2k+ 2 which exceeds ( k+ I )
t
amuearet/rcµ/.
Problem
6. 7. Show that the inductive step in Theorem 2 is valid when k exceeds 2.
Explain why this fact does not enable you to prove that 2º exceeds n
l
for all n greater than 2.
The proof of Theorem 3 i s slightly harder because we need the
following case of the binomial theorem:
(n + I)
-
= n
-
+ 3n
t
+ 3n + I.
From this i t i s easy to conclude that 2n
3
exceeds (n+ I )
-
provided
that n exceeds 3 Mter that we can prove the inductive step in
Theorem 3when n exceeds 3, BUT the ]rst step is false when n = 4,
and we must start with the case n exceeds 0.
Although we have shown how t prove Theorem 3, we will do it
over again in the standard algebraic symbols used for this proof. The
word "exceeds" is replaced by the symbol > .
First step of Theorem 3 . For k = I0, 210 > I0
-
•
Inductive step: If 2k > k, and k > 0 then
2k
+l > 2k =
£'
+
k > k + 0k
t
> k + 3k
t
+ 0k
> k + 3k + 3k + 3
> k + 3k
t
+ 3k + I
= ( k+ I )
-
.
This concludes the proof of the inductive step and of the theorem.
Theorems 4, 5, 0,
. .
can be guessed but are harder to prove
because they require more knowledge of the binomial theorem. How
ever, we ought to talk about them, in a general way at least, because
IF we wanted to prove an infnite sequence of theorems, depending
on an integer, we would need to do this by mathematical induction.
II0 US E S O F I NF I NI T Y
This gives us a case of induction in which the frst step is a theorem
( 2
n
exceeds n) which is proved by induction.
The following theorem is true:
TEOREM N: Fcrererqtnteqer N t/eretsaÞasttnteqer K (dµend-
tnq o N) suc/ t/atJcra// tnteqers n qreater t/an K
2
n
exceeds n
»
•
For N = I , this becomes our Theorem I , and K is the integer
zero; for N ¯ 2, we get Theorem 2, and K is 4
,
for N ¯ 3, we
get Theorem 3 and K = 0
It is reasonable to guess that for
N = 4, 5, 0, , I0, K ¯ I0, 25, 30, I00 and
that in generl K = N
t
•
The rader is urged to postpone the following set of problems until
he has read through all of this section, because Theorem N has been
introduced for two diferent purposes. First, it suggests how an
infnite process ( in this case mathematical induction) can be "cas
caded", each stage becoming an infnite process ; secondly, it provides
an introduction to certain arguments by Cantor, one of the great
masters of the "science of the infnite". This second purpose is more
important than the frst.
0.8.
Problems
Corresponding to the step to determine the K of Theorem N for any
integer N:
(a) Prove that the Nth power of the Nth power of 2 is the Nlth
power of 2, i .e. ,
(2N)
N
@
2N
2
•
(b) When N exceeds 4, 2N e ,ceeds N
I
. Prove that then
2
N
2
exceeds (N
I
)
N,
and hence that if n = N
t
, then 2n > nN
•
0.9. Corresponding to the inductive step in Theorem N: Prove that 2n ex
ceeds nN
provided only that n exceeds N
t
.
0 I0. From the preceding problems, show by induction on N, starting with
N = I, that Theorem N holds.
Theorem N is very important, and I should like to call attention
to it by numerical instances. It says that 2
n
increases faster than
n
:
»
( which it doesn't overtake until about the hundredth term) and
that 2
n
increases faster than n
»
, even though it doesn't catch
CONS TRUC T I O NS AND P ROOFS III
up to that sequence until the millionth term. It is clear that the
µcuers of n, n
t
, n
:
, n' increase faster than n (with
increasing n) . Because
2
n
increases faster than any power of n, it
is sid to increase transcendenta//qfaster than n
.
, t seçuence , ,
transcendenta//qJaster t/an t/e seçuence
2
n
.
tnteqers)
tncreases
Thus the sequence
22
,
2
(
2
'
)
,
2
(
2
3
)
, increases transcen-
dentally fastr than the sequence
2
,
22
,
2
:
,
The theorem follows ( by a bit of reasoning) as an application of
Theorem N, used for all N
·
The fact that there is always a transcendentally faster sequence
than any given one suggests the following inductive construction.
Suppose we take a fast sequence, like
2
n
, then a transcendentally
faster sequence, then a third, transcendentally faster than the second,
and so on forever. Now where are we? Have we got the fastest
"thing" that can be constructed?
The following theorem says no. The proof uses the Cantor diagonal
argument, which he invented in I870, and which is one of the great
ideas in the foundations of mathematics and has many applications
in all brnches of mathematics. We already used it in Chapter 3
·
TEOREM
.
8uµµcset/atVareqtrenaseçuence 8; , 8
t
, 8
:
,
cJ seqs, eac/ cJ u/tc/ ( aJter t/e ]rst) tncreases Jaster t/an t/e
µrecedtnqcne.Letusarranqet/eenttresqstem cJ numèerstnarectanqu-
/ar arraq, t/e numèers 8, cccuµqtuq t/e kth rcu, t/e nth term cJ
8, èetnq tn t/e nth cc/umn cJ t/e arraq
Th t/e dtaqcna/ seç
_
ence,
name/qt/e]rsttermcJ the]rst seçuence,Jc//cued èqt/eseccnd term cJ
t/e seccud seçuence, Jc//cued èq t/e t/trd term cJ t/e t/trd seçuence,
andsco,tncreasesJastert/ananq cne cJ t/eqtren seçuences.
II2 US E S O F I NF I N I TY
To illustrate this theorem, let 8, denote the sequence of kth
powers of n, i .e. , n
-
for n " I, 2, . . .
.
The frst few terms of
the frst few rows of the rectangular arry are displayed in Table 03.
2 3
4 9
8
·
8 2
I0 8I
4
I0
TABLE 0.3
õ
2õ
0 7
30 49
6 I2õ
2I0 343
Æ 02õ Iw0
8
-
32 243 I024 3I2õ
8
·
6 1w
The dtaqcna/seqence is :
8
6
I , 4, 27, 250, 3I25, · · ,
9 I0
8I I0
it is easy tsee that the general term is n
-
. This sequence increaes
transcendentally faster than n, and in fact it also happens tincrease
trnscendentally faster than 2
-
.
6.7 Dirichlet's Boxes or the Pigen Hole Principle
There is an important technique of proof known as the µrtnctµk
cJ Dtrtc//et
·
s ècxes. It says that if one has more objects than boxes
into which to put them, and one puts all the objects into the boxes,
then at least one box will have to hold more than one object. A regular
application of this occurs in the problem of newspaper reporters and
hotel rooms during important national conventions.
|
"
K
Figure 0. 0
a) Tm ]ntte cae
A simple application of this principle is as
follows : Let L denote a line in the plane and let I, Q, and k be
three distinct points no one of which is on the line; then at least one
C ONS TRUCTI O NS A ND P ROOF S II3
of the segments IÇ, Çk, or kI does not meet the line. This is
because a line divides the plane into only two "sides", and, of the
three points, two must lie on the sme side and can be joined by a
segment not meeting the line. The assertion follows.
b) A cmntaè/e tn]nttq cJ cè_ects, a ]ntte numèer cJ ècxes Tm
FtècnacciseqNext we shall apply Dirichlet's box principle to an
interesting theorem on the Fibonacci sequence. But frst we must
recall some facts about divisibility properties and residues.
Let N be a given integer. If any other integer is divided by N, the
following remainders are possible: 0 (if the integer is divisible by N) ,
1, 2, 3, ¯
.
¯ , N I . In other words, the remainder ( also
called restdue, mcdu/c N) is always one of these N numbers. In
particular, i f we picked more than N numbers, say
and divided them all by N, we know from the box principle that at
least two of our b'
s must have the same residue.
We now ask the following question : Suppose we have a sequence
of whole numbers and we write down the sequence of their rsidues
modulo N,
Will tuc ccnsecuttre rematnders ever be repeated in this sequence? In
other words, is there a pair r. , r,
+
, and another pair, r, , r,
+
, ,
such that
r. " r, and
To answer this question, let us frst fgure out how many diferent
µatrs of numbers can be formed from the possible remainders
0, 1, 2,
. .
, N I .
Since the frst number of the pair can have any one of N values,
and since the second can also have any one of N values, there are
N
t
distinct ordered pairs possible. Now a sequence of t + 1 terms
b
1
,
� ·
b
3
,
•
•
•
, è
:
è
:+
,
has the tconsecutive pairs
Terefore, by the box principle, the residues ( mod N) of any
sequence of more than N
t
+ 1 numbers must contain at least two
identical pairs of consecutive terms.
1 14 US ES O F I NF I NI TY
EMPLE
.
If N " 3, the possible values for resdues are 0, 1 and 2.
All the possible 3
2
= 9 pairs then are
0, 0; 0, 1 ; 0, 2 ; 1, 0; 1, 1 ; 1 , 2 ; 2, 0; 2, 1 ; 2, 2
.
Consider the sequence of 1 1 " N
2
+ 2 > N
2
+ 1 consecutive
subway stations on the 8th Avenue Independent Line:
4, 14, 23, 34, 42, 50, 59, 72, 81, 86, 96
.
The sequence of residues ( mod 3) is
1, 2, 2, 1, 0, 2, 2, 0, 0, 2, 0;
since we have written down 3
2
+ 2 = 1 1 stops, we know that in the
sequence of residues some pair is rpeate. In fact, the second and
third terms and the sixth and sventh constitute identical pairs; so
do the ffth and sixth, and the ninth and tenth.
We could now ask: after how many terms of a sequence would we
get a repeated consecutive triplet of residues ( mod N) , or a repate
consecutive foursome? These questions can be answered by the sme
reasoning, but we shall stick: to pairs and apply our knowledge to a
special sequence, the Fibonacci sequence
1 , 1 , 2, 3, 5, 8, 13, 21,
( 6. 10) 89, 144, 233, 377, 610,
2584, 4181, 6765,
which is defne by the recursion relation
(6. 1 1 )
J
:
=
J
t
= I,
J-
+i
=
J-
+
J-
-
;
34, 55,
987, 1597,
10946, 17711,
for n � 2
We shall prove the following observation due to Lagrange: t
TEOREM
.
Let N èe anqtnteqer qreater t/an cr eçua/tc2
.
Th t/e
resmues ( mcd N) cJ t/e Ftècnacct seçuence reµeat aJter at mcst
N
2
+ 1 terms
Mcre µrectse/q, t/e rerq]rst µatr c
J
restdues, name/q
1, 1, ocursaqatn utt/tn N
2
+ 2 terms andJrcm t/en cn t/e enttre
seçuence tsreµeated
·
!This observation can be found in Euler's textbook on elementary algebra
(written in 176 by dictation, soon after he became blind) . It wa translated
into many languages, and Lagrange himself did the French version
.
C ONS TRUCTI O NS A ND P ROOF S 1 15
Before giving the proof, let us test the assertion in a few simple
cases.
For N ¯ 2, the residues of ( 6
.
10) are
1, 1, 0, 1, 1, 0, 1, 1, 0, 1 , 1, 0,
The fourth and ffth terms equal the frst and second, and the sme
three terms, 1, 1, 0, seem to repeat over and over. Note that
N
2
+ 1 ¯ 5 and it is within 5 terms that a pair is repeated.
For N " 3, the residues of ( 6. 10) are
1, 1, 2, 0, 2, 2, 1, 0, 1 , 1, 2, 0, 2, 2, 1, 0,
The 9th and 10th terms are the sme as the 1st and 2nd. Here
N
2
+ 1 = 10.
For N " 4, the residues of (6. 10) are
1, 1, 2, 3, 1 , 0, 1, 1 , 2, 3, 1 , 0,
. . .
;
the 7th and 8th terms begin the repetition. Note that N
2
+ 1 ¯ 17;
from the box principle we know that a repeated pair has to occur
within 18 terms, but it actually occurs much sooner.
For N ¯ 5, the residues of ( 6. 10) are
1, 1, 2, 3, 0, 3, 1 , 4, 0, 4, 4,
3, 2, 0, 2, 2, 4, 1, 0, 1, 1, 2,
. . .
;
here N
2
+ 1 ¯ 26, but the frst pair occurs already within 21 terms,
and the cycle seems to repeat.
PROOF OF LAGRANGE'S TEOREM
.
In addition to the box principle
(which told us that the sequence of residues of any old sequence will
exhibit a repeating consecutive pair within N
2
+ 2 terms) , we must
make use of the special feature of the Fibonacci sequence
Í
; ,
h ,
that it is defned by the recursion formula (6. 1 1) . This relation tells
us that, once a pair of residues is repeated, Le., once we fnd a pair
Ik , IHI which has the same residues as an earlier pair Ii , IHl ,
then we must expect that Ik+
2
and li
+
2
have the sme residues,
that IH3 and IH3 have the sme residues, etc.
1 16 US ES O F I NF I NI TY
To see this, recall that 'J. and J, have the sme residue
( mod N) " means they have the sme remainder upon division by
N, i .e
.
,
J. = çJ + r. ,
or, written in another way,
J. ¬ r. ( mod N) ,
so that
J. ¬J, ( mod N) .
The symbols u ¬ r ( mod N) , read 'u is congruent to r modulo
N", simply mean that 1 - r is divisible by N. Note that, tJ
u ¬ r ( mod N) am x ¬ y ( mod N) ,
th
u + x ¬ r + y ( mod N) am u - x ¬ r - y ( mod N) ,
because u 1 - r and x - y are divisible by N, then so is their sum
(u - r) + (x - y) = (u + x) - (r + y) , and so is their diference
( u- r) - (x- y) = ( u- x) - (r - y)
.
In this notation the result from the box principle sys that there
are terms in the sequence for which
J. ¬J, ( mod N)
It follows that
and
J
·; ¬J
.; ( mod N) .
J
. +J
·: ¬J, +J
.; ( mod N) .
But from the recursion relation ( 0
·
II) , we see that the left member
of this congruence is just J
·t
and the right member is j
.
t
·
There
fore
]·
t
¬J
.
t
( mod N) .
The sme reasoning shows that
J.
+
:
¬J
.
:
( mod N) ,
J.
+
¬J
·+ ( mod N) ,
So far we have shown that the sequence of residues ( mod N) of
the Fibonacci sequence is periodic, but we must still show that the
cycle begins at the beginning, with the pair I , I
·
CONS TRUCTI ONS A ND P ROOF S III
This is easy. Suppose the perio of the sequence is p, so that
(6. 12)
I
¬
I
i+p
( mod N) (j ¿ s)
for all j from a certain one on, sy j " s
·
We want to show that
I.
is in fact the frst term of the sequence.
If s > 1 , then
I.
is not the frst term in the sequence, but then we
can fnd the previous term,
1
.
-
1 , from the recursion formula, and
argue as follows:
1
.-1
"
1.+1 - I.
¯ I.+
l
+p
-
I.+p
( mod N) by (6. 12)
by the recursion relation.
Hence,
1
.-1
¯ I.-l+p
( mod N) .
If s 1 = 1 , the repeating cycle begins with i and the proof is
complete
.
If s 1 > 1, we show that
1.-2 ¬
1.-2
+p
( mod N)
by the sme reasoning as above. Since s is some fnite integer, this
process applied successively to s - 1 , s - 2, s 3,
. .
will
eventually be applied to 1 and will lead to
II ¯ Il +p
( mod N)
,
and this completes the proof of Lagrange's theorem.
Observe that the proof made use only of the recursion formula
In+1
=
In
+
In-I ,
n ¿ 2,
and not of the tnttta/ ccndtttcns,
II
¯
12
" 1. This means that
Lagrange's theorem is valid for any sequence stisfying the relation
an+1
"
an + an
-
l for n ¿ 2, no matter what values we take for
al
and � .
Problems
0. I I . Consider the sequence I, I , õ, I3, 4I, · · · built on the rule that,
for n > 2, a
-
,, = 2a
-
+ 3a,.:
Prove an analogue to the Lagrange
remark, but notice that modulo 0the sequence runs I, I, õ, I, õ, · · ·
·
0. I2. Generalize Lagrange's remark to all sequences built as follows : for
n > 2, a,,, = oa, + ßa
-
,, , where o and ß are arbitrary given integers.
II8 US E S DF I NF I NI TY
0. I3
Generalize to the cae that after the third term,
0 I4 Generalize still further.
c) An unccuntaè/e tn]nttq c] cè_ects, a ccuntaè/e tn]nttq c] ècxes
A use of Dirichlet's boxes in the case of infnite sets is illustrated by
the following. Suppose that somehow there has been selected for us
in the plane an uncountable infnity of rectangles. The remarkable
thing about the theorem we are going to prove is that it does not
matter how these rectangles are chosen-all that matters is that
there shall be an uncountable infnity of them.
It is asserted that there exists scme circle and there exists scme
unccuntaè/e set of the given rectangles such that the circle is inside
every one of the rectangles. That is m sy, it lies in the interior and
does not meet the rectangles' perimeters. The proof follows; it is
technically quite easy, but may appear quite difcult on a frst
meeting
.
We prove that we can select the circle from among the "rational
circles", that is, circles which have radii of rational lengths and whose
centers have rational coordinates. The set of these is countable since
�+
"
�: ,
cf. Section 3
.
9
.
The proof is by contradiction. Thus, if no one of the
rational circles is inside uncountably many rectangles of the some
how-given set of rectangles, then each and every one of them is inside
of a countable infnity of our rectangles at mcst ( possibly a fnite
number, possibly none) . Now it follows, since �:
�:
"
�: ( cf.
Section 3
.
9
)
that the totality of rectangles of our set, each of which
has a rational circle inside of it, is at most a countable set of rec
tangles. Since the given set is uncountable, there must be rectangles
left over; that is, t/ere must èerectanq/estn curqtren set u/tc/ dc nct
/are anq rattcna/ ctrc/es tnstde c] t/em|
This remark has been emphasized because it is absurd. Ererq
rectangle has at least one rational circle inside of it (infnitely many.
as is easily seen) , and we have been led to a contradiction completing
the proof. The statement relied on above, that ererq rectangle has
rational circles inside of it, is true. The assertion that this is easily
seen is in conformity with mathematical tradition which attacks one
difculty at a time and is optimistic about later ones. The proof of
this assertion follows
.
Let a rectangle be given in the plane and let I denote its center;
see Figure 07
If both coordinates of I are rational, then choose
I as center of a circle and choose as radius any rational number
C ONS TRUCT I O NS AND P ROOF S II9
Figure 0
1
which is smaller than one half the shorter side. However, if the co
ordinates of P are not both rational then (as is easily done) choose
a point Q with both coordinates rational which is inside the rec
tangle
(P is a limit point of a sequence of such points) . No fnd the
distances from Q to each one of the sides and choose a rational
number smaller than all four of these distances as radius, using Q as
center. This completes the proof that every rectangle has rational
circles inside of it.
What we have proved illustrates the important distinction in
analysis between countable and uncountable sets. Notice how easy
it is m construct a countable set of rectangles which do not overlap
each other; we have shown that every uncountable set must overlap
very substantially.
Figure 0. 8. An instance of overlapping rectangles where the substantial
overlap is meaured by a circle contained in every one of them
Im US ES O F I NF I NI T Y
The following problems are proved along the lines shown by the
preceding discussion. The reader will fnd them worth thinking about,
but they are not easy. The second problem is meant to serve as one
of the details for the third.
Problems
0 Iõ For the moment, let us call a rectangle tµecta|if it is related to the co
ordinate system as follows : its sides lie on the lines
x = a, y = ò,
x = c, y = d,
with a � c and ò � d and a, ò, c, d rational . Prove that the set
of these rectangles is countably infnite.
0. I0. Let P be a point inside a given rectangle. Show that then P is inside
of a special rectangle which is inside of the given one.
0 I1. Suppose that you are given an uncountable set of points in the plane
called 2. Prove that there exists a point P contained in 2 such that
every rectangle containing P contains uncountably many points of
X. (The point P is called an accumulation point of the set X. )
0. I8. Suppose you are somehow given an uncountable set A of points on a
line (or in the plane) . Prove that there exists a sequence of points 1\, P2,
Pa , • • • in X and a point P, also in 2, such that P is the sequential
limit point of the sequence PI , P2 , Pa , · · · .
This is difcult t show because one has so little knowledge about
½. It ísintroduced here because it is a powerful tool, like the Bolzano
Weierstrass principle of the least upper bound, for fnding limits. It
represents one of the most important uses of an uncountable infnity.
Solutions to Problems
C H A P T E R T W O
2. 1 This is a non-terminating sequence of sets of musical compositions, the
frst set consisting of compositions for one voice part or instrument, the
second set of pieces for two performers, the third of pieces for three per
formers, and so on. !
2.2 This is a periodic sequence of the four classes of hits in baseball. The
iteration dots indicate that we are to repeat the same sequence of classes
again and again.
2.3 Collections of siblings born on the same day make up the terms of this
sequence. The frst term is the collection of all individuals with one such
sibling, the second is the set of individuals with two such siblings, etc.
The terms which occur beyond a certain point in this infnite sequence
are empty sets.
2.4 Here we have a list of the names of the days of the week. In this cae the
iteration dots represent an abbreviation for the days Saturday and
Sunday.
2.5 This is a periodic sequence, the terms of which are the frst letters in the
names of the days of the week, in the order of the days, beginning with the
letter H corresponding to Monday. The frst term occurs again after
six more terms, and from then on the entire period is repeated over and
over.
2.6 The frst of these three sets is the collection of all integers n of the form
n = 3q where q is an integer. It is eaily seen that this set consists of
the numbers 0, 3, 0, 9,
The second set is composed of all integers n of the form n = 1 ÷ 3q.
Since each such number ha the remainder 1 when divided by 3, the
numbers of the infnite sequence 1, 4, I, 10,
. . .
belong to the
second set.
! Remark added by the author : I can see "quintet", "sextet", "septet",
"octet", but there I get stuck. In the absence of a clear-cut rule as t o j ust
how to continue, I would agree with a student who called the question unclear
and would count all answers correct.
121
I22 US ES O F I NF I N I TY
The numbers i n the third set are each of the form 2 + 3g, where g
is an integer, and hence each ha remainder 2when divided by 3. Thus,
the third set ha the elements 2, õ, 8, II,
Inamuch a every integer when divided by 3ha one and only one of
the remainders 0, I, 2, we know that these three nonoverlapping
infnite sets together comprise the entire set of integers.
2.1 If g divides n, then n = qb and n + I = çò + I, where ò is an in
teger. In other words, if n is divisible by g, n + I has the remainder
I when divided by g, and hence n and n + I have no common factor.
2.8 The general principle which is suggested by an examination of these tables
is that for every k by k multiplication table, where k is any positive
whole number, the sum of the numbers in the table is the square of the
sum of the frst k positive integers. The sum of the integers in any lower
gnomon-fgure is the cube of the smallest integer in that gnomon.
I I
29
II Since
I
n - m = n ×
m
the method is clear. For example, to divide 1by 9 we look up the reciprocal
of 9 in the table of Figure 2
4 and write
I
1 ¯ 9 = 1
× ¬
= 1 × ·
III
9
·111
S OL UTI ONS TO P ROB L EMS I23
2. I2 Construct the circle of radius PQ with center Q; see Figure 21. Using
F as center, and the same radius, swing the compass to fnd the points
PI and QI of intersection of this circle with the first circle.
Next, open the compass to the width PIQI ¡ draw circles with center
PI and QI respectively. One of their intersections (the one to the right
of P and Q) is the desired point R.
To see that P, Q, R are collinear and PQ = QR = d, observe that
PIPQ and QIPQ are two equilateral triangles with common base PQ
which is bisected by the segment PIQI in the point M (see Figure 2. 1) ,
and that PIQIR is an equilateral triangle with base PIQI and whose
altitude MR lies on the line through P and Q.
Moreover, if PQ = d, then
and
3d
QR = MR - MQ =
2
d
2
= d
3d
2
'
This method shows us how to construct an infnite sequence of points
on a line.
Simply pick two points, call them P and Q, get R, and re
peat the above construction on Q and R, getting R', etc.
2
.
13 Suppose P is to the left of Q. Line up the ruler with P and Q so
that its right end is at Q and make a mark on the ruler at the place
where F falls. This mark divides the ruler into two parts, one of length
PQ = d on the right, and one of smaller length d' on the left. After
drawing the segment PQ, line up the ruler with the segment PQ so that
the division mark on the ruler falls on Q. Then the right end of the ruler
will be at a distance d from Q, on PQ extended. Denote the endpoint
of this extension by QI
.
Next move the ruler in the direction from Q
to QI along the line until the division mark is on the new point QI
.
The right endpoint of the ruler will be at a point Q2 on the line through
F and Q, at a distance 2d from Q. Continue this process indefnitely
in order to extend the line through P and Q indefnitely to the right.
In order to extend the line through P and Q to the left, we j ust
refect the method j ust described.
If d' were longer than d, the same method would work but it would
be more economical to interchange the roles of d and d'
·
2. I4 Once the direction of the road and the point at which it is to enter the
mountain are determined, it is only necessary to line up every three con
secutive guide-posts. This can be checked at each advance.
2. 15 The successive midpoints approach the point which is at a distance from
A equal to J of the length of AB
·
2.20 (a) Divide a segment into õequal parts, hold one part, and give 3pieces
away. Then one part remains and we hold t of the amount which has
been distributed. If we repeat the same process over and over, each
time dividing the one remaining part into õ equal pieces, we shall
continue to hold t of the total amount distributed and a smaller and
smaller amount of the original segment will remain to be distributed.
I24 US E S O F I N F I N I TY
Thus
I I I I
a
+
m
+ +
e e e
4 · õ 2õ I2õ
Similarly,
(b)
I I I I
a
+
m
+ +
. . .
1
'
8 b õI2
I I I I
(c)
m
+ +
&
+
e + e
=
a
I0 I0 I0 9
2
2I Divide the segent into n equal parts, hold m of them, leave m of
them to work on and give away the remaining n - 2m parts
.
Thus
m
+
(n - 2m)
¯
n - m parts have been distributed, and we hold m
parts ; hence we hold m/(n - m) of the distributed amount.
We treat the remaining part in the same way, dividing it into n equal
pieces, holding m, giving away n - 2m and keeping m to be worked on.
Continuing i n this way, we shall always hold m/(n - m) of the distribu
ted part while the part to be worked on gets smaller and smaller.
2.22 It follows from 4
=
2· 2 that (
¯
2, so that V4 is rational. To prove
that \3 and V5 are irrational , we need only consider the following:
(a) If we assume that \8
=
p/q, where p/q is that fraction (among
all equivalent fractions) which has the smallest denominator, then
we have
and
I <
q <
Ü < p
L
q
p
<
2,
< 2q,
q < q,
3q
2
¡µç and k
·
< n < ( k+ I)
·
imply that
k <
kç <
0 <
Since µ
·
= nç
·
,
P
µ(µ
µ
µ
ç
µ
kµç
kç)
µ
ç
kç
< (k +
I)
·
,
< k +
I ,
< (k
+
l)ç,
< ç
nç ¬ kµç,
ç(nç ¬ kµ),
nç kµ
where
µ kç
µ ¬ kç < ç·
I20 US E S O F I NF I N I TY
If there were a fraction whose square is the integer A, we would write
it with as small a denominator �s possible, say µ/q = ¸Ä, and q ¢ I
by assumption, so q
·
¢ I . Hence the fraction
)
-
/ç
-
would lie between
consecutive integers k and k + I and we can produce the above con
tradiction.
2.20 If ¸ s is not an integer, then, by problem 2.2õ, 8 is not the square of a
fraction. But if 2¸ s = n, where n is an integer, then 8 is equal to the
fraction (n/2)
·
Hence, unless ¸ s is an integer, we are involved in a
contradiction.
2.2I To prove that I2õ × a = I0a/8 we need only note that
I 24
+
I
I2õ = I2
+
=
2
2õ
2
I0
8
2.28 Let a be any integer with digits a.a., a; . Then we may write
a = a.I0
.
+ a..;I0' +
+ a.· I0 + a;
= a.(I0
.
- I
+
I)
+
a... (I0
.
¯' - I + I) +
.
´
+ a. (I0 ¬ I + I) + a;
+ a. + a.., + a..
·
+
.
+ a. + a;
·
Since the frst expression on the right is a multiple of 3, a has the same
remainder upon division by 3as the second term on the right, i . e. the sum
of the digits of a
·
Another way of saying this is that a and the sum of
its digits belong to the same residue class (mod 3)
·
If a. + a.., +
· · ·
+ a; < I0, the proof is complete. If not, let
a.
+
a,,
+
·
´
·
+ a;= è = è,I0
,
+ è,.,I0'
' +
+ è. · I0 + è; ,
and proceed in the same manner as above ; eventually the sum of digits
will be less than I0, and we shall have reached the root number r(a)
This shows that a and the sum è of its digits and the sum c of the
digits of è etc. down to r(a) are all in the same residue class modulo 3
2
·
29 (a) and (b) Let us consider the infnite sequence of decimals
a, = HH· · · ,
a, =
I0I0I0 · · ,
a, =
·
I0I0I0· · · ,
a. = I0 I0 I0 I · · · ,
a. =
·
I0 · ·· I0 ·
.
· I0 · · · ,
and note that the period of a. is k.
(c)
I0I0I0 I0I0I0I · · ·
·
S O L UTI O NS TO P RO B LE MS 127
C H A P T E R T H R E E
3.1 The proof of this theorem for the case AB' : B'B = m: n, where m and
n are positive integers, is essentially the same as the proof given in
Chapter 6, Section 6.2(a) . For m =
\
, n = 1 , the theorem can be
proved by the methods used in Sections 6.2(b) and 6.2(c) for the incom
mensurable case.
3.2 Such a rectangle does not exist because it would lead to the equation
1 ¯ 0 x, where x is the length of the other side
.
But this contradicts
the rule of arithmetic : x· 0 = 0 for all x whatsoever.
3.3 Construct the right triangle ADC (see Figure 3.6) with legs of lengths x
and 1. Draw the perpendicular to the hypotenuse AC through C. Ex
tend line AD to meet this perpendicular at B The segment DB then
has the desi red length y = I/x because the length of the altitude CD of
right triangle ACB is the mean proportional between the lengths of
AD = x and DB = y:
y
1 x
3.4 The parabola y = 2
¶
is the locus of points (x, y) such that, for every
abscissa x, the ordinate y satisfes the relation
x
~
1
This relation suggests again the construction of a right triangle so that
the altitude to the hypotenuse has length x and divides the hypotenuse
into segents of lengths 1 and y. In Figure 3.7 this construction wa
carried out for two given values of x, 2] and 2¶ ¡ and the correspond
ing values of y are Yl and Y2 . (The details of this construction are left
to the reader. )
The parabola shown in Figure 3.8 may now be plotted either by using
values (x,y) obtained from the construction of Figure 3.7 as coordinates
for points of the graph, or it may be plotted directly by carrying out the
construction in the coordinate plane a indicated : For each abscissa x,
fnd the point Z. (x, -1 ), connect it to the origin 0,draw a perpendicular
to the resulting segment at 0and locate the point ß. (x,y) at which this
perpendicular intersects the vertical line x units away from the y-axis.
In the right triangle 0ßZ the altitude to the hypotenuse has length x
and divides Zß into segments of length 1 and y so that the relation
x
x y
or
is satisfed for each ß so constructed.
IM US E S O F I NF I NI TY
3
õ If x is a given length then \ can always be constructed by virtue of
the relation
For example, in Figure 3
·
I, let AE = I as before, extend AE to B
so that EB has the given length x, and draw the semicircle with AB
as diameter. The perpendicular to AB at E will intersect this semi
circle at a point 0, and E0, the altitude of right triangle A0B, will
have length \.
The length vx can also be read of the parabola of Figure 3
·
8. just
fnd a point whose distance from the horizontal axis is z, its distance
from the vertical axis is \.
3
0 To fnd approximately the cube root �a of a number a one determines
the point of intersection P of the graph of y = x
'
with the horizontal
line y = a and meaures the abscissa of P. The coordinates of P are
(�a, a) .
3
·
8 A careful construction will show that as n gets larger and larger, the
quantity vn +
I - vn becomes smaller and smaller.
3
·
9 Let a = vn + I, è = .. n. Then
or
( Vn + I + .. nH
Vn + I - .. n) = I,
v
n
-
..
n =
V
I
n
+
I +
..
n '
Clearly, as n gets larger and larger the denominator V n +¯
+
V
increaes, and it follows from the above identity that the quantity
.. - .. n becomes smaller and smaller; that is, the diference be
tween .. and ¸n can be made as small as we wish (it is always
greater than 0) by taking n large enough.
3. 10 We multiply the expression by
v2n
+
+
v
21
v2n
+ + ..
and obtain
..
n
2n
+
I ¯ 2n
..
+
..
As n gets larger and larger, lin becomes smaller and smaller so that
this expression approaches
I
V
2
2
V
2
4
S OL UTI ONS TO P ROB L EMS I29
3. II If 0 < y < |r, then (see Figure 3. Iõ) , sin y < y < tan Yj dividing
by tan y, we obtain
Y
cos Y < < I
·
tan y
As Y decreases, cos y approaches Iso that yltan y is squeezed between
I and a number close to I
.
3. I2 (a) PROOF OF THEOREM 3 I Let 8 be the sequence x" x" X3 ,
having the limit L, and suppose that y" y. , Y3 ,
. .
, is any
subsequence 8' of 8 (Le. , 8' is an infnite sequence whose terms
are some or all of the terms of 8, arranged in the order in which they
occur in 8)
Now 8 has the limit L means that the sequence (x, - L) ,
(x. - L) , (X3 - L) ,
. . .
approaches 0, that is, for every integer n,
there are at most a fnite number (depending on n) of terms Xk ¬ L,
k ¯ I , 2, 3, " ' , which are numerically larger than lin. But if
8'is a subsequence of 8, then every term of the sequence (y, - L) ,
(Y
2
¬ L) , ' ¯ is identical to some term Xk - L, so that, for every
integer n, there can be at most a fnite number of terms Yk' - L,
k' ¯ I, 2, 3,
. .
" numerically larger than lin.
Thus, every subsequence of an infnite sequence with limit L also
has the limit L
·
(b) PROOF OF THEOREM 3.2 a + x - (a ÷ L) = x - L, so that i f
(x,
- L) , (x. ¬ L) , (X3 - L) ,
. . .
approaches zero, then
ja + x, - (a+ L) } , ja + x. ¬ (a+ L) } , ' ¯ also approaches zerOj
therefore a+ x" a÷ x. , a÷ X3 , • • • has the limit a+ L
(c) PROOF OF THEOREM 3
3
We wish to show that, for every integer m,
all but a fnite number of terms kXi satisfy
I
kXi - kL
I
<
m
provided that the sequence Xi has L a limit. In other words we
know that for every n, all but a fnite number of Xi satisfy
I
I
Xi - L
I
<
·
n
In particular, take n to be the nearest integer greater than or equal
to ¡ k i m. Then
I I I
<
I
k
l
�
I
k
l
=
-
n
I
k
l
m m
for all but a fnite number of terms. This argument holds for every
integer m.
(d) We prove frst : if X" X. , X3 ,
. . .
has the limit L and y" Y"
Y3 ,
. . .
has the lim
i
t M, then X'Y" X'Y" X3Ya , ¯ • ¯ has the
limit LM.
130 US E S O F I NF I NI TY
Following the hint, we write
¤.y, - LH = x.y. - x,M + x,M - LH
= x. (y, ¬ H) + H(x, - L).
By assumption, for every integer n.
1
| ×·
L
I
< · and
I
º· ¯
H
I
<
!
n n
for all but a fnite number of x, y. ·
Moreover, since the ¤, have
a limit, all but a fnite number are certainly bounded by some con
stant, say C
Thus
I
x,y.
L
M I
1
< -
m
for all but a fnite number of terms x.y. .
To prove Theorem 3.4, that the limit of x� , x� , is L
2
,
use the above result with y, = x. and L = H
To prove that the
limit of xt , x� ,
. . .
is L
k
, we simply repeat the argument k - 1
times.
3.13 (a) This problem is not at all easy. Take the case 0 <r < 1. One standard
proof runs a follows :
Since I/r > I , I/r = I + h, h > 0. Now
¸,
f
= (1 + h
)
f > 1 + nh
for every n, and so (1 + h)f increases without limit as n increases
.
Therefore the reciprocal, rº, goes to zero.
The assertion that (1 + h)f increases without limit as n ¬ ^
can also be proved by appeal to the Bolzano-Wei erstrass principle
described in the next section
.
For if L is any number such that
(1 + h)f � L
S OL UTI ONS T O P ROB L E MS
for all n, then
(I
+ b)
-
.
�
L
= L' < L
I + b
I3I
for all nj and so, given any upper bound L to this sequence, there
would exist a ema//erupper bound L'. Such a sequence is incompati
ble with the Bolzano-Weierstrass principle.
Let us suppose now that r = I. Then r
t
= I , r¹ = I , r¹ = i ,
and each number of the sequence (r - I) , (r
t
- I ) ,
. . .
equals 0·
Since no term of this sequence exceeds lin, n > 0, the sequence
approaches 0and r, r
t
, r¹, • • • has the limit I when r = I
.
(b) From the identity
a(t - x) (I + x + x
t
+
· · ·
+ x
-
t
) = a(I ¬ x
-
)
we get
a· t + ar + ar²
+
· ¹ · q
ar
--:
a(I
then, since r
-
approaches 0if I r ( < I, we have
I - r
-
limit
n� m I ¬ r
r
-
)
r
-I < r < I ·
Hence, the sum of any infnite geometric series a + ar + ar
t
+
· ·
¹
with ratio r numerically smaller than I is a/(I - r) .
3.
14 The sequence of rational numbers constructed in the text, in decimal
form, is
I
00·
· -
,
·
õ00·
·
,
2
·
00U
·
,
333ð333333·
·
,
3
·
00 0· · · ,
2w00· · · ,
·
000000 0 000· · · ,
I .w00·
.
,
4.000· · ,
.200
· · .
,
We shall construct a number whose kth decimal is always one or the
other of the digits 2 or 3 but difers from the kth decimal of the kth
number in this list.
Let us choose 3 U its frst decimal since 3 is diferent from the frst
decimal of the frst number. Let us choose 3for the second decimal of the
number we are constructing because 3is diferent from the second decimal
of the second number in the list. Similarly, let us choose 3( =0) for the
third decimal , 2( = 3) for the fourth decimal , 3( = 0) for the ffth,
3(= 0) for the sixth, 3( = 0) for the seventh, etc. , until we come to a
kth number with 3in the kth placej then we will choose 2.
132 US E S O F I NF I N I TY
3. 15 Before clasifying the letters of the alphabet, we shall consider, for the
sake of defnitenes, the letter T. We shall simplify the situation by con
sidering an uncountable set of identical (i.e. congruent) letters T (e.g
.
with a stem of one inch and a top bar of one inch) whose vertices are
labelled A, B, C, see Figure 3. 18(a).
On any piece of paper of fnite dimensions, an uncountable set of such
T's must contain an uncountable subset of T's such that the vertices
labelled A are within 1, say, inch of each other. In this set, there is an
uncountable subset such that the vertices labelled B are within 1 inch
of each other. And in this set, there is an uncountable subset such that
the vertices labelled C are within 1 inch of each other.
Now then, let ABC and A'B'C' [see Figure 3. 18(b» ) be two T's of
the kind decribed. They cross. This can be proved by elementary geome
try from the fact that a straight line divide the plane into two regions
(called the two etdee of the line) , and segments connecting points on op
posite sides of the line must cross the line. This establishes the impossi
bility of writing an uncountable number of congruent non-crossing T's
on a page. Note that if we merely required the distances AA' and BB'
to be small, then the T's could possibly stand as they do in Figure 3.19,
and not cross
.
We shall not treat the case of T's of varying sizes here,
but the same result can be proved.
All letters of the alphabet that contain a confguration such as we en
countered in the letter T (i. e. an intersection of two segments or curves
where at least one of the segments extends beyond the point of intersec
tion) are in the same class as T. They are A, 8, E, F, H, P, O,
k, T, X¸ Y.
For all other letters, it i s possible to scribble an uncountable set of them
on a page. Figures 3.2O(a) , (b) illustrate this fact for the letters I, O
respectively. In each cae, the fact that a line segment contains an un
countable number of points give the clue.
C H A P T E R F O U R
4. 1 The phrae "if this limit exists" ha been omitted. The statement
L = lim.�. L. makes sense only if the L. have a limit, and in this cae
it aserts that L is the value of this limit.
4.2 The direct computation of the lengths 8, , 8
·
,
. . .
of the sides of equi
lateral triangles whose baes are on the x-axis and whose vertices lie on
the curve y = x2 is somewhat awkward; fortunately the question posed
in the problem can easily be answered without such a computation : The
length of the resulting zig-zag is again 2 because, a in Example 2, it is
twice as long a the distance from (1, 0) to the origin.
4
.3 We calculate the distances
1
B
·
T =
2 '
B
·
,
t
T =
which approach zero. Hence T is the limit of the sequence
B
ª
,
The distances B
·
,B
·
@ j can be represented a sides of equilateral tri
angles of lengths 1/2"-1 and so these distances also approach 0
For the ordi nates, we have
Y
I =
Y.
=
0
Y2
=
�
2
¸1
- �¡
Y3
�
¸
1
. � . ¸ . . ·
,
2
,
2
'
Y2
¸1
~
�
1 1 1
~
+
.
+
2 4 8 16
.
¸
1 1 1
¸
2
L
1
.
4
+
±
16 2
.
-
2
- � ¸I - !
+
2 4
y
2
2
o
õ
¸
1
.
¸
=_¤
1
~ ,¸
2·
m
¸
1 ¸
-
¿]
±
2
´
-
1)
1
~
g
.
µ
±
_
1
¡¡
16
2
0
-2
for n
�
2, n even,
and
so that
S OL UTI ONS TO PROBL EMS
Y
n1
= Y" ±
|
·
v
2
| 1m Y" =
¯ .. 1m Y
nI .
n¬w 5 n=m
135
As we have seen in the solution to Problem 4
·
4, this implies that the
point (3
v
/5,
v
/5) is the limit point of the sequence D; , D, ,
4
·
0 Project each point B, , B¡ , B, , and so on, perpendicularly onto the
y-axis (this is what we do, essentially, when we calculate the ordinate of a
point) . Call these points F, , F¡ , F, , and so on. Now if we let
!
,
denote the ordinate of B @ , then
!
, is also the length of 0F. . Notice
that the sequence of numbers
! ,
/
-
,
!- ·
.
.
.
is constantly increasing
and is bounded above. By the Bolzano-Weierstras principle this sequence
ha a limit /*. We shall show in a moment, but the reader may prefer to
prove it himself, that /* is also the limit of the sequence of odd-numbered
j'
s,
!: · !
, , /
:
,
.
.
. , which approach it from above. Thus the entire
sequence ha J* a a limit, but the approach to this limit is two-sided.
To J* there corresponds a point F* (on the y-axis) which ought to be
the project:on of the limit of the points B, , B¡ , B, , • • • ; but a we
have seen, the points B, , B¡ , • • • have no limit.
A proof that the sequence of odd-numbered j's has f as a limit fol
lows. For every n,
/
t W¡ J*
¬ |
J
t-=t !-ì
+
|
J¡-
=
/*
ì
,
the second term in parentheses being negative (see Figure 4I0)
·
The frst
term on the right is precisely
V
/(2n) . We have no formula for the
second term; but since the sequence J¡-
(n = 1 , 2, 3, .
.
.
) con
verges to /*, we know that all such terms are small when n is large
enough, by the very defnition of limit. Thus we can be sure that when n
is large enough, the right hand side is the diference of two small numbers
and is small. This shows that /, j is near to /* (for large n) and con
cludes the proof.
4
·
I Example 5 showed (see solution to previous problem) that a sequence of
points in the plane may have no limit point although the sequence of their
projections ha a limit point. Assertion (a) would be correct if one added
"if the given sequence of points ha a limit"; we shall demonstrate this
in a moment.
Assertion (b) is true. Let Q. be the points in the given sequence, Q
its limit point, P" the projections of Q" , and P the projection of Q.
Then P fP " Q"Q cos Og where Og is the angle between the segment
Q"Q and the line which carries the projections. Since I cos Og I _ I it
follows that Ip"p I _ I Q"Q I and since lim" • . Q"Q = 0, it follows that
lim @ . I P,P I ¯ 0 so that P, the projection of Q, is indeed the limit
point of the sequence P" .
136
4
·
8
US ES O F I NF I N I TY
This also proves that the limit of the projections is the projection of
the limit of a sequence of points, provided they have a limit. If p. is
the limit of the projections, P the proj ection of the limit Q, then
lim._" P .p. = 0 and lim . ¿ .P fP = 0 imply that P and Ï" coincide.
Let
1 1 1
Sf
¯ 1 ÷ - ÷ ÷
· · ·
÷
/
3
1
n²
Since for k > 1 we have k > k 1 , it follows that
1 1 1 1
.
1
1 1
> -
v
i
n
1
+
Vn
+
1 1
>
&
+
-
1 2
Hence the sequence (c) diverges.
1 1
+
&
+
e e
+
-
+
3 n
(d) The terms of this sequence are even larger than the corresponding
terms of (c) and therefore (d) certainly diverges.
4. 11 (a) If for every line in the plane the projection of F is the limit of the
projections of Ï, , then this is true, in particular, for the two per
pendicular axes of a coordinate system. Denote the projections of
Ï, on the x-axis and on the y-axis by x. , y. respectively, and thoe
of F by x and y. Then, see Figure 4. 15(a) ,
(F,F)² ¯ (x. - x)²
+
(y. - y)²,
and since the x. approach x and the y. approach y,
lim (F,F)² ¯ 0
n�m
and the Ï, approach F.
(b) Clearly, this result cannot be deduced from the fact that the given
data are true for j ust one line, as Example 5 (page 08) shows.
(c) If the given data are true for any two non-parallel lines, say /, and
/
²
, take one (say II) to be the x-axis. It can be shown (by methods
of analytic geometry or linear algebra) that any line in the plane, for
example the y-axis, can be expressed as a linear combination of two
given non-parallel lines. Moreover, the projections y. of Ï, on the
y-axis can be expressed in terms of the x. and the projections Zµ of
Ï, on the line |, [see Figure 4. 15(b) ] , and the y. have a limit y
if the x. and the Zg have limits
.
Thus the problem that F is the
limit of the Ï, can be reduced to the problem solved in (a) .
C H A P T E R F I V E
5. 1 Assume that
v
is rational , i. e. that the diagonal of a unit square has
length
),
/ç, where
), and ç. are integers. Then a square whose sides
are ç. units long has a diagonal of length
)
,
S OL UTI ONS T O P ROB L EMS I39
N ow construct the following se:uence of right isosceles triangles :
The frst has legs of length ç. and a hypotenuse of length
)
, , see
Figure 5.4(b) . Erect a perpendicular to the hypotenuse at a point which
divides it into segments of lengths ç: and ), ¬ ç. This perpendicular
cuts of a corner of the frst triangle, and this corner is our second triangle,
clearly similar to the frst, with leg of length ç
t
and hypotenuse of length
)t
We observe [see Figure 5.4(b) ] that
and
Now we repeat the construction and cut of the next corner triangle.
Its legs have length
ç
:
=
)
t
¬ ç
t
= 2ç: - ), - (
)
, - ç:) = 3ç. ¬ 2µ
.
and its hypotenuse has length
µ
: = ç
t
- ç:
=
).
¬ ç: ¬ (3ç: ¬ 2µ,) = 3µ, - 4ç: ·
We continue cutting of corners, always obtaining an isosceles right
triangle similar to all the previous ones
.
The leg of the nth triangle
has length ç
-
, its hypotenuse has length µ
- , and these lengths satify
the relations
Since µ
-., = ç-., ¬ ç-., we may express the lengt.h ç
-
of the nth leg
by
n > 2,
that i s, in terms of the lengths of the legs of the previous two triangles.
Now consider the sequence ç. , ç
t
, ç: ,
.
. Since ), and ç:
are integers, ç
t
= ), - ç.
is an integer, ç
ª
= ç:
¬ 2ç
t
i s an integer and,
in general , ç- = ç-.
t
- 2ç
-.:
is an integer for all n > 2
·
It is clear
from our construction that the legs of subsequent triangles decrease in
length, i .e. that
ç. > ç
t
> ç:
>
· ·
· ·
Thus the asumption that V = µ,/ç
:
is rational has led to an infnite
decreasing sequence of positive integers, and no such sequence exists.
We conclude that ¸ is irrational.
In order to apply this method to 0, asume that v5 = r,/t
t
where
r, and t, are integers. Blow up the rectangle of Figure 5.
5 so that its
sides are t, , 2t, , then its diagonal is r,
·
Our construction will lead
to similar right triangles with legs t
-
, 2t- and hypotenuse r- The
recursion relations will be
t- = r
-
.: ¬ 2t,
:
, r- = t-
.: ¬ 2t- ,
so that
t- = t-.
t
¬ 2t-.: ¬ 2t
-.:
=
t-
.
t
~ 4t
-
.
t
,
and the sequence t: , t
t
, t: ,
. . .
of lengths of shorter legs of the
similar triangles is again a decreaing infnite sequence of integers.
These examples show how this metho can be used to prove the irra
tionality of vI for any integer k which can be written as the sum of
14 US ES OF I NFI NI TY
the squares of two integers : k = a2
+
bl. We have used it for k = 12 + P,
and for k = 22 + P. The details of this generalization are left to the
reader.
5
.
3 The kth fraction, Fk , is formed from the previous fraction,
FlI , as
follows :
Fk
1 +
1
Fk-I
)
then
1
-
ç
-
If Fk_1 Fk =
ç
1
+
) )
+ ç
ç
5.4 (a) A sequence of fnite parts of the expression
v
1
-
v I
- VI -
is formed in the following manner:
When we compute these numbers we see that this is the sequence
I , 0, 1, 0, " ' , which has no limit.
(b) Since m satisfes m2 + m = I , its reciprocal satisfes
1 1
- + - = 1
r T
or 1 + T = r.
The terms a. of the sequence of fnite parts
\
,
obey the recursion formula
(1)
an =
VI +
an_I , for n .. 2, 3,
. . .
.
We shall show that the increaing sequence ai , a
2
, ' " ha a limit
by applying the Bolzano-Weierstrass Theorem, see Section 3.8. In
order to do this, we must fnd a bound B such that al <a
2
<
· · ·
<B
·
The fact that the a. increase implies that a.+1
- an > 0 for
n = 1, 2,
. . .
. From (1) we have
VI + an + VI +
VI + an-I)
VI + an + VI +
(1 + an
) (I
+ an_I)
VI + an + VI + an_I VI + an + VI +
Since a. > 0 for all i, the denominator in the lat expresion is
greater than 2. Therefore
for all n,
S OL UTI ONS T O P ROBL EMS 141
and
(2
)
a-.t - a- <
�
(a- - a-
.,) <
�
¸
(a
-.t
- a-:
)
¡
Next we write a
-
,
t
in the form
a-.:
¯ (a
-.t - a-) ÷ (a-- a-.,) ÷
If !. is the nth term of the Fibonacci Sequence, the formula for the
nth power of m is
m
ª = (
¬
l)"Q,~1 - 1-
m)
·
The corresponding situation for T is
T8 ¯
8
,
÷ õ,
. . .
õ. 8 A complete solution to Problem õ8is given in Chapter 0, pp. I I+III
õ.9 The way the vertices are ordered in succesive rectangles refects the
fact that the shorter side of each rectangle (i.e. the line j oining the 2
vertices named lat in the ordering) is the longer side of the next one (i.e
.
the line between the vertices listed in the middle position) . In each case
the vertex named frst is the one from which the 4õ° line is drawn to the
point that is the frst named vertex of the next rectangle.
5. 10 The length of each successive segment of this zig-zag is the length of the
preceding segment reduced by the factor m < I Hence, the length of
the zig-zag is the sum of the infnite geometric series
õ
·
II
\
+ v
2
·
m
+ v
.
m·
+ v2
.
m
·
÷
· · ·
=
\ \
I ¬ m m
·
The solution to Problem 3. I3 (p. õ2) proves that the formula for the
sum of an
y
infnite geometric series with frst term a and ratio r < I
is a/(1 - r) .
Number of quarter-turns Distance from T to
(degrees) about T point on the spiral
I/0 (I) AT
·
m
. is
I/3 (æ°) AT· m
. t·
IJ2
(
4õ°
ì
AT· m
:t:
õ/0 ¸Iõ°)
AT· m
-
tt
4/3 (I20°) AT· m·
t
ª
3/2 (I3õ°) AT· m
·t:
õ/3 (Iõ°) AT· m
-
t
ª
õ/2 (Zõ°) AT· m
-
t
·
2n ÷ I
2
¸
n
2
÷ I
· 9
°j AT· m
·-
-
.
tit
·
S OL UTI ONS T O P ROBL EMS I43
õ
·
I2 If t takes on negative values, we get a continuation of the spiral in a
counter-clockwise direction from AT. As the values of t become smaller
and smaller (i.e. as t approaches ¬ * _ R increases without limit.
õ
·
I3 To multiply a number RI by a number R2 by means of the spiral in
Figure õ II (where the distance AT is now taken to be the unit of meas
urement) we use the ruler to locate those points PI and P
2
on the
spiral which have distances RI and R2 from T:
Now we follow the spiral from the point A t o the point PI and denote
by al the angle through which the radius vector to the spiral must
rotate to get from TA to TPI . (Observe that if RI < ¡ , then we
reach PI by going in the clockwise direction and OJ will be taken to be
positive; if RI > I, then we reach P
I
by going counter-clockwise and
al will be taken negative.) Next, we follow the spiral from A to P
2
and
measure the angle a2 by which the radius vector must be rotatfd to
get from TA to TP2 • Now we add the angles al and a2 , and rotate
the line AT through the angle Oj + a2 always following the spiral from
the point A on. This will lead to a point P ¶ on the spiral whose distance
f
rom T is
This method is j ust a geometric interpretation of the law of exponents :
Given
we have found
õ
I4 If the radius vectors TP
I
, TP
2
, • • • , TP
n
have the same direction
but diferent magnitudes RI , R2 , . . . , Rn , then the angles of
rotation aI , a2 , . . . , an , meaured from the line through T and
A a this line pases through each point of the curve from A to PI ,
to P
t
,
*
• • , to Pn , difer only by multiples of 2¥ radians (
one full
turn about T, i. e. 4 quarter-turns) . This property corresponds to the
fact that the logarithms of the numbers represented by the lengths
RI , R2 , • • • , Rn would difer only in their characteristics, i .e
. in
the integer part of the logarithm. (If O is meaured in quarter-turns,
these logarithms would difer by multiples of 4
· ) If a is between 4k
and 4(k ÷ I ) quarter-turns, a ¬ 4k would correspond to the mantissa
and would determine the direction of the line TP, while the charac
teristic 4k would determine on which "ring" of the spiral the point
P lies.
14 US ES O F I NF I NI TY
C H A P T E R S I X
0
I Assume to the contrary that there exist integers a and ò such that
Then
or
� · (1 ÷ 0) " 1
.
I ÷ 0
" � ,
a
ò ò
-
a
0 =
-
-
1
=
a a
But if ò and a are integers, ò ~ a is also an integer, and the lat
equality states that 0 is rational, which is false. Therefore the re
ciprocal of 1 ÷ 0 is not rational.
0·
2 Let d be the highest common factor of a and ò, and let x and y
be integers. Then the integer
a ÷ òy = c = a'dx ÷ b'dy = d(a'x÷ ò'y)
is clearly divisible by d
·
Conversely, if c is divisible by d, the highest common factor of a
and ò, then we can fnd integers x and y such that
a÷ òy = c
in the following way.
We divide the equation by d obtaining
a'x+ ò'y ~ c',
where a' and ò' are relatively prime. In this cae it u known (see e.g.
the discussion of Euclid's algorithm in Continued Fractione by C. D
.
Olds, to appear in this series) that there exist integers x, and y
·
such
that
then the integers z = c'x, , y = c'y, will satisfy
a'x÷ ò'y = c',
and hence also ax ÷ òy = c.
03 2º¯
·
, n = I, 2, 3, · ·
: .
04 When N = 1, N and the sum of its digits clearly have the same residue
modulo 3
·
This proves the frst step in the induction.
Suppose that k is an integer such that k and the sum of its digits
have the same residue modulo 3,i .e. such that
k
= 3ç÷ r, 0 � f < 3,
S OL UTI ONS TO P RO B L EMS
and the sum of the digits of k is
.
given by
3e + r,
I4õ
Û � r < 3
To prove the inductive step, we must show that k + I and the sum
of its digits have the same remainder when divided by 3
·
When
Û � r + I < 3, we have
k + I " 3ç + (r + I ) ,
otherwise
k + I ¯ 3(ç + I)
·
If the lat m (m � 0) digits of a number k are all 9's, these 9's will
become O's when Iis added to k, but the frst digit which is not a 9 will
be increaed by I. Since the sum of the digits of k is 3e + r, we may
write the sum of the digits of k + I in the form
(3e + r) + I ¯ 9m,
which is equivalent to
3(e ¬ 3m) + (r + I)
Thus k + I and the sum of the digits of k + I have the same residue
modulo 3
·
6.5 The asertion is true for n ¯ I
·
Assume that for n = k,
I +
2 + · · · + k .. tk(k + I) ,
and consider the cae for n
' k + I
·
By applying the inductive hypothesis
we get
I + 2 + . , ' + k +
k
+
I .. tk(k + I) + k + I ,
which can be written
tk
t
+ fk + I ¯ t(k2 + 3k + 2) ¯ Hk + l) (k + 2)
Since this i s of the form tn(n + I) , the proof i s complete.
0
·
0 For all integers n, 2º^ exceeds (n + 0)².
0
I0 We have proved (Theorem I)that when N
=
I,
for all integers n.
Assume that when N ¯ k and n exceeds k
t
, it is true that 2
n
> n
�
.
It follows from Problem 0.9 that 2
n
> n
k
+
! provided that n > (k + I)
t
,
which is all we need to complete the proof that 2
n
> n
N
for all integers
N and n such that n > N
t
0. II Let us try to imitate the proof of Lagrange's Theorem (pages IIõIII)
in the present case and let us observe what modifcations will be necessary.
The box principle tells us that any sequence of residues (mod N)
has a repeating consecutive pair within N
t
+ 2 terms. If the pairs ai ,
ai
+1 and a
�
, a�1 have the same residues, then from
and
we get
3ai " 3a
�
(mod N)
It follows that
and
or
By the same argument
ai
+3
" ak+3(mod N),
ai
+
, " a
�
+
, (mod N),
which shows that the sequence of residues (mod N) of the sequence given
by a
.
tl = 2a
.
+ 3a._1 is periodic.
Let the period of the sequence be p. Then
aj " aj
+
p (mod N)
U
� T)
from some j on, say j = T. Suppose aT is not the frst term of the
sequence. From the recursion formula we have
Hence,
3aT_I " 3aT
-
Itp (mod N),
or
148 US ES OF I NF I NI TY
Clearly, we can conclude that
aT-I - ܧ.|qg (mod N)
only if we asume that 3 and N are relatively prime; otherwise, it does
not necessarily follow that N is a factor of aT_l - aT_I+p . Since T is a
a fnite integer, this process applied successively to T - i , T - 2,
T - 3, eventually must lead to
al - ai+ (mod N)
.
Thus, the sequence of residues (mod N) , N � 2, of the sequence
defned by af+! ' 2af + 3af_I , n � 2 (where the initial values al
and a
t
may be any given integers) is periodic. If 3 and N are relatively
prime, then the periodic part begins with the residue of al .
6.
12 The residues (modN) , N �2, of any sequence defned by
with arbitrary i nitial integers al and a
t
are periodic ; the repetition of
a pair occurs within at most N' + 2 terms. If N and { are relatively
prime, then the periodic part begins with the residue of al .
6. 13 The sequence of residues (modN) of any sequence defned by
n � 3,
ha a repeated consecutive triplet within NI + 3 terms
.
If N and ¶
are relatively prime, the sequence of residues (modN) is periodic from
the beginning on
.
6. 14 In general , the sequence of residues (modN) of a sequence ai , Ü¿ ¡
. . . , af , .
.
. built (after the nth term) on a rule expressing the
(n + l)th term as a linear combination of the preceding n terms is
periodic from the beginning on and will repeat within N
n
+ n terms
whenever N ha no factors greater than Iin common with the coefcient
of the earliest term in the recursion formula.
6. 15 The lines x ¯ a for all rational a constitute a countable infnity of
lines since the set of all rational numbers is countable
. The same is true
for the sets
7
¯ ò, x " C¿ y ~ d for rational ò, c, d. Since all
special rectangles are formed by combining 4 sides, each from one of
these sets, we obtain a o· a o· a o· a 0 ' a ; possible special rectangles
.
(This even includes the degenerate rectangles in which a pair of opposite
sides coincides. Therefore, the non-degenerate special rectangles certainly
constitute a countable set. )
6. 16 Let P be the point (z; , y.) , and let d be the minimum distance
from P to any point on the given rectangle. Then there exist rational
numbers êj < id and ëy < id such that
x = x;+ ë
I
¬ Ü and
are rational, and numbers EI < ld and E2 < id such that
y ¯ 70 + EI ¯ ò and y ~ y. E! '
ò'
S O L UTI O NS TO P RO B L E MS I49
are rational. It follows that the sides of a special rectangle R lie on the
lines x
=
a, x
=
a
'
, Y =
b,
y
=
b' , and the point Í is inside of this
rectangle. Furthermore, since the length of the diagonal of R is
V
(81 + \)² + (EI + c,)²
< d,
the distance from F to the farthest point on R is less than the mini
mum distance from F to the given rectangl e. Hence, the special rec
tangle lies entirely within the given one.
0. II If we assume that there is no point F in the set X such that every rec
tangle containing F contains uncountably many points of X, then
every point in the set X must be inside at least one rectangle containing
a countable set of points of X. In this case, the solution to Problem 0. I0
shows that every point of X is inside of a special rectangle which is
entirely within the rectangle containing a countable set of points belong
ing to X, and so also contains at most a countable infnity of points of
X. Now, we have proved (Problem 0.Iõ) that the set of all special rec
tangles is of power Þ¡ , therefore, the set of special rectangles with
which we are concerned is certainly countable. Moreover, since each of
these special rectangles contains a countable set of points of X, it follows
from Þo
·
Þg ¯ Þ¡ that the set X is countable. But this contradicts
the hypothesis that the given set is uncountable; hence, there must exist
some point F in X such that every rectangle containing F contains
uncountably many points belonging to the set X
.
0
·
I8 Take the point F obtained in the solution to Problem 0. II, and a se
quence of decreasing intervals (rectangles in the case of the plane) closing
down on F. In each of these intervals pick one point of X from among
the uncountably many which are available. This gives a sequence
F, , F, , F, , • • • of points of X which form the desired convergent
sequence. A proof such a this is called "non-constructive" because no
mechanism is provided for actually defning each point F, • Since we
know nothing about X except that it is uncountable, no method of
selection iR available to us.
Bibliography
Cajori, Florian, A Htstcrq oJ Mathemattcs, New York and London:
The Macmillan Company, 1919.
Courant, Richard, Dt¿erentta/ and Inteqra/ Ca/cu/us, Vol. 1, New
York: Nordemann, 1940.
Courant, Richard and Robbins, Herbert, Bhat ts Mathemattcst,
London and New York: Oxford University Press, 1941.
Coxeter, H. S. M. , Intrcductton tc0ecmetrq, New York: John Wiley
and Sons, Inc., 1961.
Davis, Philip J. , The Lcre cJ Larqe Numèers, Vol. 6, Neu Mathe-
mattca/ Ltèrarq, New York and New Haven: Random House,
Inc. and Yale University, 1961.
Eves, Howard, An Intrcductton tcthe Htstcrq cJ Mathemattcs, New
York: Rhinehart and Company, 1958.
Frankel, Abraham A., Aèstract 8et Thecrq, Amsterdam, Holland:
North-Holland Publishing Company, 1953.
Hardy, Godfrey H., A Ccurse cJ Iure Mathemattcs, Cambridge :
Cambridge University Press, 1938.
Heath, Thomas L., Manua/ cJ 0reek Mathematics, London and New
York: Oxford University Press, 1931.
Hilbert, David and Cohn-Vossen, Stephan, 0ecmetrqandtheImaqtna-
ticn, New York: Chelsea, 1952.
Neugebauer, Otto, The Exact 8ctences tn Anttçuttq, Copenhagen,
Denmark and Princeton, New Jersey: E. Munksgaard and
Princeton University Press, 1951.
Niven, Ivan, Numèers. kattona/andIrrattona/, Vol. 1, Neu Mathe-
mattca/ Ltèrarq, New York and New Haven: Random House,
Inc. and Yale University, 1961.
Iõ
B I B L I O G RA P HY 151
Olds, C. D. , CcnttnuedFracttcns, to h published for the Neu Matbe-
mattca/ Ltèrarq, approximately 1962.
Rademacher, Hans and Toeplitz, Otto, Tbe En_cqment cJ M athe-
mattcs, Princeton: Princeton University Press, 1957.
Steinhaus, Hugo, Mathemattca|8naµshots, New York: G. L. Stechert
and Company, 1938 (2nd ed. , London and New York: Oxford
University Press, 1960).
Wilder, Raymond L., Intrcductton to the FcundattcnscJMathemattcs,
New York: John Wiley and Sons, Inc. , 1952.
Yaglom, I. M., 0ecmetrtc TransJcrmattcns, translated from the
Russian by Allen Shields, to b published for the Neu Mathe-
mattca/Ltèrarq, approximately 1962. | 677.169 | 1 |
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Editorial Reviews
Review
"An excellent introduction to the field is given here, including a general motivation and usage cases beyond simple graphics rendering or interaction." from the ACM Reviews by William Fahle, University of Texas at Dallas, USA
This is a computer science textbook for undergraduates. Lots of "real world" uses for computational geometry to egg-on the unmotivated. Not so good for those that ARE already motivated.
The text glosses over basic tasks such as "whether a point lies to the left or right of a directed line" (page 4) with the expectation that some unnamed library function will do this. For me, not so. Moreover I'd would have liked to read a geometric proof of such things. The foundations are left out, yet elsewhere they waste space to give us Pythagoras' Theorem.
Comprehensive, deep, clear (i.e. readable). Pseudo-code (high level) is provided at end of each chapter. Also exercises. Reader must still convert from pseudo-code to programming language in order to actually implement. A web site is listed to help with that, which provides links to programming resources. I haven't yet tried them..
This is the standard text book for CG, and it nicely introduces us to a lot of concepts. But, unfortunately it is not the best book I have read. Most of the examples albeit few, does not make much sense. The algorithms discussed sometimes cannot be grasped. I often went online to read more about the subject to understand the topics. This also could mean that my grasping of the subject is low :), but that should not matter as long as the material is explained clearly.
Beautiful book, solid contents. I learned a lot from it and had a nice time practicing with the exercises. Lots of examples and problems, a lot of interesting algorithms and techniques, every chapter is a progressive refinement of a particular idea to solve a problem expressed as geometry.
Difficulty level: make sure you know some asymptotic analysis and discrete mathematics to get the best out of it, but could be read by anyone who can code i believe (although again, he'll miss a lot of beautiful mathematics)
It is a joy to read and review this book -- the exposition is crystal clear; the writing style is warm and engaging (not too terse and not too verbose), conveying understanding and not just stating facts, theorems, and algorithms; the graphics are great (numerous richly detailed illustrations); the topics hit the heart of computational geometry; the historical remarks help set context; and the book is beautifully typeset and printed on high quality acid free paper.
This book is a must to all researchers and students of the field. The algorithms are always presented in the context of an application, which makes it the more understandable. However, the authors chose to present them in a very high level of abstraction, and some of the finer details - so important in these algorithms - are only mentioned, which may pose a problem to obtaining a suitable, efficient implementation of them in a programming language. | 677.169 | 1 |
A Review
Abstract
Our aim in this book is to introduce you to some mathematical applications that you will encounter in intermediate and advanced undergraduate courses in classical mechanics, electromagnetism, and quantum physics. To provide physical contexts in which to set these applications, we shall draw on a few familiar concepts from your general physics course. In this first chapter, we review these concepts, work some examples, and give you some exercises at the end of the chapter to help refresh your memory about them. | 677.169 | 1 |
Algebra 1 - Polynomials - Type, Degree, Standard Form Foldable
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This foldable is designed for interactive math notebooks.
It walks students through understanding the different terminologies associated with polynomials.
The first tab shows the differences of a monomial (already defined in a previous chapter), binomial, and trinomial. The degree of a monomial and degree of a polynomial are also given. Underneath that tab is a table of 6 expressions with the type of polynomial and degree ready to be filled out.
The second tab details the definitions of standard form and the leading coefficient, with another table for immediate practice.
The paper should be printed with the shorter side being the flipping side. Students will fold the right-hand side of the paper to the dotted line and make one cut to the fold line. | 677.169 | 1 |
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In these 4 lessons students will:
-be exposed to important vocabulary
-interpret theoretical probability
-use simulations to model chance behavior
-describe a probability event for a chance process
-use probability rules to calculate probabilities
-use a two-way tables and Venn diagrams to model a probability event
-calculate probabilities involving two events
-use the general addition rule to calculate probabilities
-calculate and interpret conditional probabilities
-use the general multiplication rule to calculate probabilities
-use tree diagrams to model two or more events
-use the multiplication rule to check to see if events are independent
-use a tree diagram to model two or more events
-assess their progress on basic probability | 677.169 | 1 |
Geometrical Background
Abstract
In this preliminary chapter we have gathered together a brief synopsis of those items from differential geometry upon which the main development of the text will be built. Most of the material is standard so that more detailed expositions are readily available, although, when references seem called for, we will tend to favor [N4]. | 677.169 | 1 |
Letts and Lonsdales Success Revision Guides offer accessible content to help students manage their revision and prepare for exams efficiently. The content is broken into manageable sections and advice is offered to help build students confidence. Exam tips and techniques are provided to support students throughout the revision process. These titles are specific to the AQA exam board. This book is for the current GCSE Maths curriculum to be examined in 2010/2011 only. Other Letts and Lonsdale revision guides and workbooks are available for the new GCSE curriculum starting in 2010. | 677.169 | 1 |
Course Project Sample Unit
Calculus Areas and Volumes of Revolution
Lesson #1 Target: Use calculus to find area between two curves.
Have students work together to solve the problem. (Collaboration Fluency)
Afterwards discuss solutions. (McCain: Resist Temptation to Tell) Was anyone close? How did they get their answer? Introduce and demonstrate the calculus formula for area between two curves. Ask students to try it on their own.
Use Twitter to send the practice assignment. (Media Fluency)
Before students leave, have them write a 7-word essay about today's learning. (Formative Assessment) | 677.169 | 1 |
One of 12 books developed for use with the core material (Book O) of the Elements of Mathematics Program, this text covers material well beyond the scope of the usual secondary mathematics sequences. These materials are designed for highly motivated students with strong verbal abilities; mathematical theories and ideas are developed through problem situations and explanatory materials. This volume begins with a discussion of infinite sets; basic set theory, introduced in Book O, is reviewed and extended. Set builder notation is used in conjunction with logical connectives. The empty set and universal set are discussed. Many theorems are proved. (SD) | 677.169 | 1 |
Algorithm Math Worksheets
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Mathematics: Algebra - General eBooks
Algebra is a branch of mathematics that concerns the study of the rules of operations and relations, as well as the constructions and concepts arising from them, including terms, polynomials, equations and algebraic structures.
Elementary algebra, often part of the curriculum in secondary education, introduces the concept of variables representing numbers. Equations are solved to find the value of the variable.
Algebra is one of the main branches of pure mathematics, along with with geometry, topology, combinatorics, and number theory. eBookMall offers eBooks in general algebra, college algebra, elementary algebra, as well as intermediate, abstract, and linear algebra.
There are over 200 eBooks in the category Mathematics: Algebra - General. Use our eBook search to find a specific book or author. | 677.169 | 1 |
A.R., Arkansasgebrator has! Thats why nowadays, I make sure every student system in the school (even the notebooks) are running it! If nothing else, it makes our jobs a lot easier! Ken Rogers, LA.
The Algebrator is the greatest software ever! Angela Baxtor | 677.169 | 1 |
Description: The book is designed for final-year undergraduates and master's students with limited background in linear algebra and calculus. Comprehensive and coherent, it develops everything from basic reasoning to advanced techniques within the framework of graphical models.
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A Survey of Statistical Network Models by A. Goldenberg, A.X. Zheng, S.E. Fienberg, E.M. Airoldi - arXiv We begin with the historical development of statistical network modeling and then we introduce some examples in the network literature. Our subsequent discussion focuses on prominent static and dynamic network models and their interconnections. (2204 views)
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After our 5 question warm-up, we went over the worksheet Antiderivative Basics. The went through at least a dozen new examples of finding the antiderivative. Students are to complete both sides of the worksheet they received today and may start working on the 4.1 book assignment. Please bring your book to class on | 677.169 | 1 |
Mr. Bryant's Math Class
Math HL
Course Description
This course is meant for students who have a strong background and interest in mathematics and related disciplines. Students should also have a high level of competency in analytical and technical skills. The course stresses problem-solving flexibility in a variety of contexts that will show the links between different areas of mathematics. Students will develop skills that will enable a lifetime of self-directed learning of mathematics.
By introducing students to the omnipresence and importance of mathematics across the natural and human sciences and by linking it to the Theory of Knowledge, students will develop an appreciation of the beauty and a better understanding of the global significance of mathematics.
This course is demanding and requires students to comprehend many topics of mathematics that will be approached in a variety of ways and studied in great depth. Therefore, students who do not wish to study the subject in such a rigorous manner should choose this course at the standard level.
The major goals of the course are that students will be able to:
use appropriate mathematical notation and terminology to think, learn, and communicate effectively
formulate and deliver coherent mathematical arguments with clarity and understanding
explain the nature and significance of results
solve problems with precision
develop logical, critical, and analytical thinking
use technology such as graphing calculators, and graphing software, in a constructive manner and to interpret the outcomes
use mathematical modeling correctly
conduct mathematical investigations and deliver proofs rigorously
apply mathematical knowledge to solve complex problems that combine concepts from different areas of mathematics
Algebraic and geometric approaches to the following topics: the sum and difference of two vectors;
the zero vector, the vector ; The difference of two vectors; multiplication by a scalar; magnitude of a vector; unit vectors; base vectors; position vectors
The scalar product of two vectors and algebraic properties of the scalar product
Vector equation of a line in two and three dimensions. Simple applications to kinematics. The angle between two lines.
Coincident, parallel, intersecting and skew lines; distinguishing between these cases. Points of
intersection.
The vector product of two vectors, the determinant representation. Properties of the vector product.
Geometric interpretation of
Vector equation of a plane . Use of normal vector to obtain the form Cartesian equation of a plane.
Intersections of: a line with a plane; two planes; three planes. Angle between: a line and a plane; two planes.
Power series: radius of convergence and interval of convergence. Determination of the radius of convergence by the ratio test
Continuity and differentiability of a function at a point.
Continuous functions and differentiable functions.
The integral as a limit of a sum; lower sum and upper Riemann sums.
Fundamental Theorem of Calculus.
Improper integrals.
First order differential equations: geometric interpretation using slope fields, including identification of isoclines.
Numerical solutions of , using Euler's method.
Variables separable.
Homogeneous differential equation
Rolle's theorem.
Mean Value Theorem.
Taylor polynomials and series, including Lagrange form of the error term.
Maclaurin series
The evaluation of limits of the form using l'Hôpital's Rule and/or the Taylor series
Portfolio (10 hrs)
Two pieces of work based upon different areas of the syllabus, representing the following activities
Type I: Mathematical investigation
Type II: Mathematical modeling
Assessment and Evaluation:
Year 1
Tests – 50%
Quizzes – 30%
Homework – 20%
Year 2
External assessment (5 hrs) 80%
Test papers:
Paper 1 (2 hrs) No calculator allowed 2 (2 hrs) Graphic display calculator (GDC) required 3 (1 hr) Graphic display calculator (GDC) required 20%
Extended-response questions based mainly on the syllabus options
Internal assessment 20%
Portfolio
A collection of two pieces of work assigned by the teacher and completed by the student during the
Course:
Type 1: mathematical investigation
Type 2: mathematical modeling.
Markscheme:
Test questions will be graded according to the markscheme set forth by the IBO. Students will be awarded points for correct reasoning and usage of methods as well as correct answers. Abbreviations will be used to indicate the points awarded.
Abbreviation
M
Marks awarded for attempting to use a correct Method; working must be seen.
(M)
Marks awarded for Method; may be implied correct by subsequent working.
A
Marks awarded for Answer or for Accuracy; often dependent on preceding Mmarks.
(A)
Marks awarded for Answer or for Accuracy; may be implied correct by subsequent working. | 677.169 | 1 |
The Old Schoolhouse® Product & Curriculum Reviews
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Algebra (Math U See) Review by Jill Hardy
I have to preface any and all math reviews that I do by stating up front that I am not a "math person." When I sit down to look at a curriculum, especially one that involves higher math, I can't give the same sort of opinion that someone who really knows the material could give you. I can only tell you what it has and whether it's clear enough for a homeschool mom who stumbled through high school math to understand and teach.
I've heard some homeschoolers say that Math U See's upper level courses (algebra, geometry, etc.) aren't "rigorous" enough. (In all fairness, I've also heard "math people" wax poetic about Math U See, so take that for whatever it's worth.) To be honest, hearing that something isn't rigorous usually gives me pause, and it should. We are ultimately responsible for teaching our children, and the pressure can be overwhelming at times. We don't want to send kids off into the world unprepared; we want them to have a good education. And "rigorous" means "good" . . . right?
Well, it can mean "good," and it can also mean that the material is tough for a 12-, 13-, or 14-year-old to understand. I do want my children to have a "rigorous" education, but more than that, I want them to have a thorough understanding of what they study. Even if they don't finish 9th grade with a comprehensive knowledge of algebra, I want them to completely understand--and be able to use--what they do know.
When I compared Math U See algebra with another algebra book that I had on hand, I did find things that weren't covered in Math U See. But the reverse was also true; some things covered in Math U See weren't touched on in the other book.
The Math U See program seems to cover the basics of algebra: solving for an unknown; solving simultaneous equations by graphing, substitution, and elimination; square roots; and dividing polynomials. When you add in the Honors component of the Math U See algebra (a separate book with 34 additional lessons), you gain additional application problems, as well as more complex factoring, and things such as finding molecular mass.
The "basic" Math U See algebra program is comprised of 34 lessons, each with two pages of lesson practice (where the new concepts are worked on), and three pages of systematic review. Each page has an average of 15-20 problems. If one page is covered per day, you could finish the whole book in approximately one school year, but if a student understands the material and is ready to go on after a minimum of repetition, you could finish it much sooner. One of Math U See's great advantages is its ability to provide as much or as little review as the student needs.
The Math U See approach is outlined in four steps: Prepare for the Lesson (the teacher becomes familiar with the material), Present the Material (the teacher presents the lesson or allows the student to watch it on the accompanying DVD), Practice to Acquire Mastery (the student completes one or both of the lesson practice sheets for the current lesson and as much review as is needed), and Proceed after Demonstrating Mastery (you move on only after you're sure that the student has mastered the material).
The program includes 34 weekly tests, 3 Unit tests, and a Final Exam. It also makes use of Math U See's unique manipulatives (sectioned blocks with special inserts for decimals/algebra). The manipulatives are what "make" this program, in my opinion; to be able to see 2X + 9X + 9 is invaluable when you're dealing with something as abstract as algebra. (For me, anyway.) I believe that a program that incorporates manipulatives will be far more likely to give my kids a thorough understanding of algebra. However, if your child is gifted in the area of math and displays a natural understanding of the subject, you may find the manipulatives superfluous.
I'm not sure that I'd want to do this program without the Honors portion. The additional challenging word problems involve "real-life" math applications, and therefore I don't believe them to be optional. Even if my children don't go on to professions where heavy-duty math is required, I still want them to be able to apply the benefits of what they've learned in everyday situations. And that means gaining practice with as many applications and word problems as possible. The Honors program also introduces an additional four-step approach: Read, Think, Compare, and Draw. These strategies encourage the student to apply current knowledge to figure out solutions, to compare those solutions to what is given in the answer key, and to explore new ways to find the answer. All of these strategies are valuable tools for a student to have.
Math U See suggests a few different ways to handle the Honors material. One option is to fit in the Honors lessons instead of the extra practice sheets in the regular algebra program. (At least one systematic review should be done, but students following the Honors course won't likely need all of the review.) Another option is to complete the Honors material as a "bridge" course after the regular algebra book is finished.
So, in short, I can't really speak to the "rigor" of the Math U See algebra program, but I do believe it to be very thorough. The flexibility of review, the use of manipulatives to illustrate difficult concepts, and the relaxed and fun tone of the lessons on the DVD make this an accessible and (relatively) easy to understand method of learning algebra.
Product review by Jill Hardy, The Old Schoolhouse Magazine, LLC, January 2006
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How we cause with mathematical principles remains to be a desirable and hard subject of research--particularly with the fast and numerous advancements within the box of cognitive technology that experience taken position in recent times. since it attracts on a number of disciplines, together with psychology, philosophy, desktop technology, linguistics, and anthropology, cognitive technology offers wealthy scope for addressing matters which are on the middle of mathematical studying.
Drawing upon the interdisciplinary nature of cognitive technology, this ebook provides a broadened viewpoint on arithmetic and mathematical reasoning. It represents a stream clear of the normal proposal of reasoning as "abstract" and "disembodied", to the modern view that it's "embodied" and "imaginative." From this angle, mathematical reasoning contains reasoning with constructions that emerge from our physically reports as we have interaction with the surroundings; those buildings expand past finitary propositional representations. Mathematical reasoning is inventive within the feel that it makes use of a few robust, illuminating units that constitution those concrete studies and rework them into types for summary suggestion. those "thinking tools"--analogy, metaphor, metonymy, and imagery--play an immense function in mathematical reasoning, because the chapters during this e-book exhibit, but their strength for reinforcing studying within the area has got little attractiveness.
This e-book is an try and fill this void. Drawing upon backgrounds in arithmetic schooling, academic psychology, philosophy, linguistics, and cognitive technological know-how, the bankruptcy authors offer a wealthy and accomplished research of mathematical reasoning. New and fascinating views are provided at the nature of arithmetic (e.g., "mind-based mathematics"), at the array of strong cognitive instruments for reasoning (e.g., "analogy and metaphor"), and at the alternative ways those instruments can facilitate mathematical reasoning. Examples are drawn from the reasoning of the preschool baby to that of the grownup learner.
Similar Mathematics books
This e-book offers in a concise, but distinct approach, the majority of the probabilistic instruments scholar operating towards a sophisticated measure in statistics,probability and different similar parts, will be built with. The strategy is classical, keeping off using mathematical instruments now not priceless for conducting the discussions.
Makes an attempt to appreciate a variety of points of the empirical global frequently depend upon modelling procedures that contain a reconstruction of platforms less than research. often the reconstruction makes use of mathematical frameworks like gauge thought and renormalization team tools, yet extra lately simulations even have turn into an vital device for research.
From the contours of coastlines to the outlines of clouds, and the branching of bushes, fractal shapes are available all over the place in nature. during this Very brief advent, Kenneth Falconer explains the fundamental options of fractal geometry, which produced a revolution in our mathematical knowing of styles within the 20th century, and explores the big variety of functions in technological know-how, and in points of economics.
This e-book introduces the maths that helps complicated computing device programming and the research of algorithms. the first goal of its famous authors is to supply a superior and appropriate base of mathematical abilities - the abilities had to clear up advanced difficulties, to guage horrendous sums, and to find refined styles in information.
The four-way interplay among acquisition , grade, move challenge, and node unit was once major, F(l2,540) = 1. ninety three, P < . 03. the information answerable for the interplay, besides information of follow-up analyses, are offered in desk five. 1. The findings have been hugely in line with the tenets of case dependent reasoning (Hammond, 1989). that's, young children within the first constraint situation who mapped a script to the objective challenge have been anticipated to teach decrements in functionality on the first node, these within the 3rd constraint have been anticipated to make extra mistakes on the 3rd and fourth nodes, and youngsters in all 3 stipulations have been anticipated to teach decrements on the 6th node.
Mind's eye was regarded as an issue of private idea, differing wildly from individual to individual, and exhibiting up basically in artwork and poetry. That view has been proven to be insufficient. imaginitive mechanisms, akin to metaphorical suggestion, schematic psychological imagery, and varieties of narrative are a part of the final psychological endowment of all common humans, they usually come up in a typical manner from commonalities of bodies, brains, and reports interacting within the actual and social international. | 677.169 | 1 |
(Prerequisites, MA 161, MA 165, or permission of the instructor.) Discrete mathematics is the study of relationships between finite and countable sets as well as the analysis of processes involving a finite number of steps. This course will introduce and emphasize the concept and methods of proof, while studying topics such as sets, logic, functions and relations, mathematical induction, and recursion. | 677.169 | 1 |
Angela Baxtor, TX Dania J. Guth, KS
I'm not much of a math wiz but the Algebrator helps me out with fractions and other stuff I need to know for my math appreciation class in college. Monica, TX
I am actually pleased at the content driven focus of the algebra software. We can use this in our secondary methods course as well as math methods. Katherine Tsaioun, MA
The program has led my daughter, Brooke to succeed in her honors algebra class. Although she was already making good grades, the program has allowed her to become more confident because she is able to check her work07-03:
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What is a real-world example when the solution of a system of inequalities must be in the first quadrant | 677.169 | 1 |
Properties of Real Numbers / Variables and Expressions (7 Math)
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Purpose: The overarching goal of this lesson is to teach students about the different properties of real numbers. This also teaches students about variables. This fits into the 'bigger picture' because it will help all students to understand how to solve equations using variables.
Objectives: As a result of this lesson, SWBAT:
• Identify the associative, commutative, identity, and distributive properties.
• Solve expressions by using the properties correctly.
• Simplify algebraic expressions for given variable values.
• Simplify variable expressions by combining like terms.
• Solve equations for the given variables. | 677.169 | 1 |
Mathematic books review
What can you do with a degree in mathematics. Actually, just about anything. Learn about Busy Ant Maths, a simple solution for the 2014 maths curriculum. The world's largest organization devoted to the interests of collegiate mathematics. Elementary Mathematics for Teachers is a textbook for a semester or two quarter university course for pre service teachers. Is a program service of Starfall Education Foundation, a 501(c)(3) public charity. The Mathmania book club from Highlights is full of math puzzles that let kids have fun while learning math. These aren't your normal math books for kids. The Math Review Book, A Practical Guide for Dispensers is an invaluable text and reference that reviews the math you will need to know for the Basic Certification. essayer espagnol trad The Number Power Series math books are among the best I have ever used for teaching developmental mathematicsAbout Privacy Help Contact; Starfall. Is also appropriate for courses for? Scover new resources to help develop mathematical fluency and boost success in Level 6. Really, we mean it. Careers in Mathematics!
E Standards of Learning and Curriculum Framework comprise the mathematics content that teachers in. Ate of Texas Assessments of Academic Readiness. The Review and Adoption Process; Learning Support and Programs. Our Spectrum curriculum reviews have been submitted by homeschool moms? Mathematics in Computer Science (MCS) publishes high quality original research papers on the development of theories and methods for computer and information. Mathematics Standards of Learning (SOL) Adopted 2009. Revious STAAR Mathematics Resources. Math Reads provides books and programs to address the range of math content at each elementary grade level and support regular math instruction. Are using the Spectrum Math book for grade 2 this year. Ath, Early Childhood. Me features include a K 12 math expert help service, an extensive database of. Is also appropriate for courses for. The Math Forum is the comprehensive resource for math education on the Internet! Elementary Mathematics for Teachers is a textbook for a semester or two quarter university course for pre service teachers. The world's largest organization devoted to the interests of collegiate mathematics. | 677.169 | 1 |
Section: M2.2.11
Summative assessment
• generate functions given graphs and graphs given functions;
• interpret expressions for quadratic functions in terms of a context it represents;
• express quadratic functions in different forms for different purposes;
• solve modeling problems using quadratic functions.
External Resources
Description
As an alternative to a traditional summative assessment students could
revisit The Fall of Javert to apply their understanding from this unit
to some real-world questions.
• In "Defying the Law (of Gravity)," students figure out where Javert's fall would have made sense. They figure out what the acceleration due to gravity would have to be, and then look for places in the solar system where that is true.
• In "Actually, It Is Rocket Science," students construct stomp rockets, launch them, and figure out how high they went. | 677.169 | 1 |
Math CXC is fun?
Mastering Mathematics for CXC (first edition)
Math CXC is fun?
Mastering Mathematics for CXC (first edition)
Can you picture an interactive math programme?
Mastering Mathematics for CXC (first edition) motivates students to learn, complete tasks that might bore them with pencil and paper, provides creative ways to solve problems, and offers a risk free learning environment for students to explore the world of mathematics.
Can you also picture an interactive quiz?
CXC maths is no longer the same!
DISCOVER…
· New ways to engage students using technology.
· Tools that make teachers more effective.
-------------------------------
CONNECT…
· Students and teachers with the right tools and resources.
· Users to information through seamless access to an integrated solution suite. | 677.169 | 1 |
THE M.IN.E.R.VA. PROJECT
The M.In.E.R.Va. (Interactive Mathematics Exercises) project wants to give teachers a flexible informatic tool to create courses for students, helping them to recover their deficiencies in Mathematics or to enhance their abilities. This project is the product of a research on teaching technologies carried on by a group of professors at Catholic University of Milan.
Various tools are already available to create theoretical and practical courses and they are used both in secondary schools and in Universities (to recover basic knowledge in Mathematics).
Our software is different from the others for the following reasons: 1. the possibility of taking the course individually and offline; 2. a wide range of exercises, besides multiple choice questions (MCQ): short answer questions (SAQ) with numerical, vector or text answer, fill in the blanks, multiple answer questions (MAQ); 3. the possibility of using the software without installing other applications (only a few files on a server are needed for the online mode); 4. the independence from the operating system (at least on the client side).
Four kinds of contents have been produced: 1. theoretical schemes to enforce or create knowledge; 2. guided exercises to show the basic modalities to solve a problem; 3. enforcement and self-evaluation exercises; 4. exercises to verify students' results.
Contents can show different levels of interactivity and can be used both offline and online.
The student can automatically receive the following feedbacks: 1. total score, 2. wrong and correct answer, 3. suggestion to find the solution, 4. reference to theory, 5. solution. The software is highly flexible, in fact it allows to produce, from the same file of exercises, different outputs: 1. individual exercises (with or without evaluation), 2. exercises monitored by the teacher, 3. classwork.
Random parameters can also be inserted in exercises: this allows students to repeat similar exercise, with different numbers, to enforce the comprehension of the subject proposed. The software is based on LaTeX, a typesetting system very similar to a programming language.
We have developed a group of LaTeX packages which allow to create interactive online or offline PDF files or written classworks. We have also developed a web platform (in Italian), which can be reached at the link On this platform, the student can see the theoretical lessons, download the offline exercises, access the online exercises and self evaluation tasks. The teacher can organize the theoretical and practical files, prepare the courses, follow his students through a web interface in which he can see all the scores and the answers given.
We are developing a database of exercises which can be used for different purposes. This database is created by the teachers of the schools which participate in the project, supervised by the mantainers of the project.
All the project is almost entirely developed using free or open source software.
We have also created virtual classrooms using Facebook groups, to give support to the students during their activities. This tool has allowed the development of a more informal and interactive way of communication between students and teachers, thus enhancing the learning process. | 677.169 | 1 |
Help with college algebra homework
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m's law and current flow, and an acquaintance with first-year algebra. The question-and-answer format, illustrative experiments, and self-tests at the end of each chapter make it easy for you to learn at your own speed. Boasts a companion website that includes more than twenty full-color, step-by-step projects Shares hands-on practice opportunities and conceptual background information to enhance your learning process Targets electronics enthusiasts who already have a basic knowledge of electronics but are interested in learning more about this fascinating topic on their own Features projects that work with the multimeter, breadboard, function generator, oscilloscope, bandpass filter, transistor amplifier, oscillator, rectifier, and more You're sure to get a charge out of the vast coverage included in Complete Electronics Self-Teaching Guide with Projects!
This no-nonsense guide provides students and self-learners with a clear and readable study of trigonometry's most important ideas. Tim Hill's distraction-free approach combines decades of tutoring experience with the methods of his old-school Russian math teachers. The result: learn in a few days what conventional schools stretch into months. - Teaches general principles that can be applied to a wide variety of problems. - Avoids the mindless and excessive routine computations that characterize conventional textbooks. - Treats trigonometry as a logically coherent discipline, not as a disjointed collection of techniques. - Restores proofs to their proper place to remove doubt, convey insight, and encourage precise logical thinking. - Omits digressions, excessive formalities, and repetitive exercises. - Covers all the trigonometry needed to take a calculus course. - Includes problems (with all solutions) that extend your knowledge rather than merely reinforce it.
Your step-by-step solution to mastering precalculus Understanding precalculus often opens the door to learning more advanced and practical math subjects, and can also help satisfy college requisites. Precalculus Demystified, Second Edition, is your key to mastering this sometimes tricky subject. This self-teaching guide presents general precalculus concepts first, so you'll ease into the basics. You'll gradually master functions, graphs of functions, logarithms, exponents, and more. As you progress, you'll also conquer topics such as absolute value, nonlinear inequalities, inverses, trigonometric functions, and conic sections. Clear, detailed examples make it easy to understand the material, and end-of-chapter quizzes and a final exam help reinforce key ideas. It's a no-brainer! You'll learn about: Linear questions Functions Polynomial division The rational zero theorem Logarithms Matrix arithmetic Basic trigonometry Simple enough for a beginner but challenging enough for an advanced student, Precalculus Demystified, Second Edition, Second Edition, helps you master this essential subject.
The fast, easy way to master the fundamentals of physics Here is the most practical, complete, and easy-to-use guide available for understanding physics and the physical world. Even if you don't consider yourself a "science" person, this book helps make learning key concepts a pleasure, not a chore. Whether you need help in a course, want to review the basics for an exam, or simply have always been curious about such physical phenomena as energy, sound, electricity, light, and color, you've come to the right place! This fully up-to-date edition of Basic Physics: * Has been tested, rewritten, and retested to ensure that you can teach yourself all about physics * Requires no math--mathematical treatments and applications are included in optional sections so that you can choose either a mathematical or nonmathematical approach * Lets you work at your own pace with a helpful question-and-answer format * Lists objectives for each chapter--you can skip ahead or find extra help if you need it * Reinforces what you learn with end-of-chapter self-tests
Expanded and revised, the Second Edition of the Algebra Survival Guide unleashes its power for a new generation of students. Now that Âthe Common Core StandardsÂhave changed how math is taught, thisÂ2ndÂedition aligns its content to these broad new guidelines. TheÂnewÂEdition also adds advanced content. In its XTREMEÂALGEBRAÂsection, the new edition tackles the topics of Functions, Inequalities and the Advanced Coordinate Plane,Âand it teaches storyÂproblems in all threeÂareas. These additions update the book for today's elementary and secondaryÂstudents; they also provide additional supportÂfor adults taking algebra in their return to college. Plus, with its newly expandedÂindex and glossary,ÂtheÂ2ndÂEdition makes all of its content easy to find.ÂTheÂbook retains the cartoons, analogies and conversational format that brought out praise from all corners and garnered the book both a Parents Choice Commendation and a Golden Porch Award forÂpedagogical excellence.
Say goodbye to dry presentations, grueling formulas, and abstract theories that would put Einstein to sleep -- now there's an easier way to master the disciplines you really need to know. McGraw-Hill's Demystified Series teaches complex subjects in a unique, easy-to-absorb manner, and is perfect for users without formal training or unlimited time. They're also the most time-efficient, interestingly written "brush-ups" you can find. Organized as self-teaching guides, they come complete with key points, background information, questions at the end of each chapter, and even final exams. You'll be able to learn more in less time, evaluate your areas of strength and weakness and reinforce your knowledge and confidence. A complete, self-teaching guide to the function and interaction of bodily systems, with coverage of: comparative physiology, functions at the chemical and cellular levels, organic compounds, the cell, physiology of muscle, nerves and glands, sensory physiology, motor functions, autonomic nerves and the endocrine system, air and blood transport, digestive and genitourinary systems, and more.
Solving word problems has never been easier than with Schaum's How to Solve Word Problems in Algebra! This popular study guide shows students easy ways to solve what they struggle with most in algebra: word problems. How to Solve Word Problems in Algebra, Second Edition, is ideal for anyone who wants to master these skills. Completely updated, with contemporary language and examples, features solution methods that are easy to learn and remember, plus a self-testSearch for: | 677.169 | 1 |
PRAISE FOR THE FIRST EDITION "THIS BOOK IS CLEARLY WRITTEN AND PRESENTS A LARGE NUMBER OF EXAMPLES ILLUSTRATING THE THEORY . . . THERE IS NO OTHER BOOK OF COMPARABLE CONTENT AVAILABLE. BECAUSE OF ITS DETAILED COVERAGE OF APPLICATIONS GENERALLY NEGLECTED IN THE LITERATURE, IT IS A DESIRABLE IF NOT ESSENTIAL ADDITION TO UNDERGRADUATE MATHEMATICS AND COMPUTER SCIENCE LIBRARIES." CHOICE AS A CORNERSTONE OF MATHEMATICAL SCIENCE, THE IMPORTANCE OF MODERN ALGEBRA AND DISCRETE STRUCTURES TO MANY AREAS OF SCIENCE AND TECHNOLOGY IS APPARENT AND GROWING WITH EXTENSIVE USE IN COMPUTING SCIENCE, PHYSICS, CHEMISTRY, AND DATA COMMUNICATIONS AS WELL AS IN AREAS OF MATHEMATICS SUCH AS COMBINATORICS. BLENDING THE THEORETICAL WITH THE PRACTICAL IN THE INSTRUCTION OF MODERN ALGEBRA, MODERN ALGEBRA WITH APPLICATIONS, SECOND EDITION PROVIDES INTERESTING AND IMPORTANT APPLICATIONS OF THIS SUBJECT EFFECTIVELY HOLDING YOUR INTEREST AND CREATING A MORE SEAMLESS METHOD OF INSTRUCTION. INCORPORATING THE APPLICATIONS OF MODERN ALGEBRA THROUGHOUT ITS AUTHORITATIVE TREATMENT OF THE SUBJECT, THIS BOOK COVERS THE FULL COMPLEMENT OF GROUP, RING, AND FIELD THEORY TYPICALLY CONTAINED IN A STANDARD MODERN ALGEBRA COURSE. NUMEROUS EXAMPLES ARE INCLUDED IN EACH CHAPTER, AND ANSWERS TO ODD–NUMBERED EXERCISES ARE APPENDED IN THE BACK OF THE TEXT. CHAPTER TOPICS INCLUDE: BOOLEAN ALGEBRAS POLYNOMIAL AND EUCLIDEAN RINGS GROUPS QUOTIENT RINGS QUOTIENT GROUPS FIELD EXTENSIONS SYMMETRY GROUPS IN THREE DIMENSIONS LATIN SQUARES PÓLYA BURNSIDE METHOD OF ENUMERATION GEOMETRICAL CONSTRUCTIONS MONOIDS AND MACHINES ERROR–CORRECTING CODES RINGS AND FIELDS IN ADDITION TO IMPROVEMENTS IN EXPOSITION, THIS FULLY UPDATED SECOND EDITION ALSO CONTAINS NEW MATERIAL ON ORDER OF AN ELEMENT AND CYCLIC G
"Sinopsis" puede pertenecer a otra edición de este libro.
Críticas:
"...an ideal starter into algebra for all kinds of students, and a very useful addition to existing textbooks..." (Zentralblatt Math, Vol.1050, 2005)
Reseña del editor:
Praise for the first edition "This book is clearly written and presents a large number of examples illustrating the theory ...there is no other book of comparable content available. Because of its detailed coverage of applications generally neglected in the literature, it is a desirable if not essential addition to undergraduate mathematics and computer science libraries."471414513
Descripción John Wiley Sons Inc, United States, 2003. Hardback. Condición: New. 2nd Edition. Language: English . Brand New Book. Praise for the first edition This. CHOICE As a cornerstone of mathematical science, the importance of modern algebra and discrete structures to many areas of science and technology is apparent and growing with effectively Boolean Algebras Polynomial and Euclidean Rings Groups Quotient Rings Quotient Groups Field Extensions Symmetry Groups in Three Dimensions Latin Squares Polya Burnside Method of Enumeration Geometrical Constructions Monoids and Machines Error-Correcting Codes John Wiley Sons Inc, United States, 2003. Hardback. Condición: New. 2nd Revised edition. Language: English . Brand New Book. Praise for the first edition This book is clearly written and presents a large number of examples illustrating the theory .there is no other book of comparable content available. Because of its detailed coverage of applications generally neglected in the literature, it is a desirable if not essential addition to undergraduate mathematics and computer science libraries. Wileyand#8211;Blackwell1414513 | 677.169 | 1 |
Atoms, Symbols and Equations 4.0
Description of
Atoms, Symbols and Equations
Unique interactive multimedia Chemistry teaching software that tests students as they learn. Topics include: elements, atoms and molecules, word equations, chemical symbols, Periodic Table, chemical formulas, balancing chemical equations. Unique interactive multimedia Chemistry teaching software that tests students as they learn. Topics covered include: elements, atoms and molecules, word equations, chemical symbols, Periodic Table and chemical formulas. Plus interactive simulations to teach the balancing of chemical equations and the formulas of ionic compounds. As far as possible, skills are taught through familiar examples, to help reinforce general chemical knowledge. Written by a teacher for use in schools, but an excellent personal tutor for anyone beginning to study Chemistry. Questions constantly test understanding and hints are given when answers are wrongovalent Bonding 1.0 Chemistry teaching software which simulates the bonding of covalent compounds on screen. Learn how covalent compounds are formed by atoms sharing electrons. Includes a reference section to support......
Ordinary Differential Equations 1.00 Solves boundary-value or initial-value problems involving nonlinear or linear ordinary differential equations of any order, or systems of such. The conditions may also be linear or nonlinear.
Digital Logic Analyzer for Windows v1.2a Digital Logic Analyzer for Windows turns your PC into a high speed harware logic analyzer which can be used for analyzing and debugging digital electronic circuites. The logic analyzer offersD Panel v2.0 LCD Panel is used to display numbers with a LCD panel style. LCD Panel lets you choose the digits display style, color scale, total digits count, and digits the after comma count. | 677.169 | 1 |
The fast and easy way to improve your technical math skills Are you a student aspiring a career in a skilled trade? Technical Math For Dummies is your one-stop, hands-on guide to acing the math courses you'll encounter as you work toward getting your degree operations and solve word problems as they're applied to specific trades. * Maps to a course commonly required by vocational schools, community and technical college, or for certification in the skilled trades * Covers the basic concepts of arithmetic, algebra, geometry, and trigonometry * Helps professionals keep pace with job demands Whether you're a student currently enrolled in a program or a professional who is already in the work force, Technical Math For Dummies gives you everything you need to improve your math skills and get ahead of the pack.
About the Author
Barry Schoenborn is a technical writer who has worked in many of the trades described in the book. Bradley Simkins has taught numerous classes and tutored at the Multimedia Math Learning Center at American River College in Sacramento Math for | 677.169 | 1 |
Ship out in 2 business day, And Fast shipping, Free Tracking number will be provided after the shipment.Paperback Pages Number: 360 Language: Chinese textbook learning this. but due to limited space. it is not knowledge points can be elaborated. Teachers in the classroom by teaching time constraints. can not compensate for this defect. Jinglun learn typical textbook parse: Mathematics (Grade 8) (Zhejiang J GB) (Revised) to fill the shortage of textbooks and classroom teaching. It has two prominent features. One. expected to preview and review after class finishing. to help improve classroom learning. School students did not understand the book is equal to a good teacher. please return home. Second. above the middle level students have this feeling: usually learn to feel good. but an exam. there are always some of the topics usually do not. The reason is the content of the examination more difficult than the textbook requirements. The book is to fill the gap between textbooks and examinations. examination requirements. textbook knowledge to conduct a comprehensive mining and upgrading. so that the students during the examination can be condescending. handy.Four Satisfaction guaranteed,or money back. N° de ref. de la librería N21272 | 677.169 | 1 |
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