text stringlengths 6 976k | token_count float64 677 677 | cluster_id int64 1 1 |
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Fun Self-Discovery Tools
Introduction to Matrices
Introduction to Matrices defines Matrices and their elements, and demonstrate how they are used to organize data. Matrices are much more powerful than simple organizational tools. You can perform operations with matrices just like you can with numbers. You will learn the steps involved in Adding and Subtracting Matrices and even learn about Matrix Multiplication. You will explore the Commutative Property of Matrices and Associative Properties of Matrices, which can help you when Solving basic Matrix Equations | 677.169 | 1 |
GeoGebra
GeoGebra is a mathematical tool that can solve equations, graph functions, create constructions, analyze data, and create 3D plots. It has a simple interface which lets you construct and analyze polygons, conic sections, and other shapes using your mouse. Support for touch-screen interface makes it easier to work with graphs.
More about GeoGebra
Get GeoGebra v5.0.360; you will need 50.03 MB of free space on the hard drive; GeoGebra Geometry and GeoGebra Pre-Release are its known alternative names. This program is provided at no cost by its creator GeoGebra Inc. It is in the Education category, Science with .ggb formats support. Versions 5.0 and 4.9 are still the most popular ones and can be run on Windows XP/7/8/8.1/10. The most popular installers for the software are GeoGebra.exe and GeoGebra-JOGL2.exe. A number of antivirus services checked the file and found it safe to use. | 677.169 | 1 |
Grades 10,11,12
Prerequisite: Either a high grade in Algebra II, or have taken or are currently enrolled in Honors Algebra II
AP Statistics is very similar to an introductory statistics course in many colleges. In this year long course students will be introduced to the major concepts and tools for collecting, displaying, analyzing, and drawing conclusions from data. Computers and calculators will aid in exploring the data and displaying it, while the Internet will be utilized to discover existing sets of data and studies. Certain distributions of data will be examined and characteristics identified. A well-developed design for collecting data will be studied and implemented throughout the course. This class will prepare students for the optional Advanced Placement exam for possible college credit. A graphing calculator is required for the course, and an Internet license is useful. The EHS Mathematics Department recommends the TI-83 Plus. Note: This is a year-long course. Students should register for all three courses.
AP® and Advanced Placement® are registered trademarks of the College Board. Used with permission. | 677.169 | 1 |
Books
Geometry & Topology
Highly regarded for its exceptional clarity, imaginative and instructive exercises, and fine writing style, this concise book offers an ideal introduction to the fundamentals of topology. Originally conceived as a text for a one-semester course, it is directed to undergraduate students whose studies of calculus sequence have included definitions and proofs of theorems. The book's principal aim is to provide a simple, thorough survey of elementary topics in the study of collections of objects, or sets, that possess a mathematical structure.
The author begins with an informal discussion of set theory in Chapter 1, reserving coverage of countability for Chapter 5, where it appears in the context of compactness. In the second chapter Professor Mendelson discusses metric spaces, paying particular attention to various distance functions which may be defined on Euclidean n-space and which lead to the ordinary topology.
Chapter 3 takes up the concept of topological space, presenting it as a generalization of the concept of a metric space. Chapters 4 and 5 are devoted to a discussion of the two most important topological properties: connectedness and compactness. Throughout the text, Dr. Mendelson, a former Professor of Mathematics at Smith College, has included many challenging and stimulating exercises to help students develop a solid grasp of the material presented.
The rules are simple . . . The math is easy . . .The puzzles get harder and harder!
Once you match wits with area mazes, you'll be hooked! Your quest is to navigate a network of rectangles to find a missing value.
Just Remember:
Area = length × width
Use spatial reasoning to find helpful relationships
Whole numbers are all you need. You can always get the answer without using fractions!
Originally invented for gifted students, area mazes (menseki meiro), have taken all of Japan by storm. Are you a sudoku fanatic? Do you play brain games to stay sharp? Did you love geometry . . . or would you like to finally show it who's boss? Feed your brain some area mazes—they could be just what you're craving! diversions in the Orient. The tangram is not entirely new to America, since Yankee sea captains brought books of tangram puzzles back from Canton and Shanghai. Edgar Allan Poe was a devotee, while on the other side of the world Napoleon is said to have whiled away his time with them. Around the turn of the twentieth century, the two great modern puzzlers, H. E. Dudeney and Sam Loyd, greatly enlarged the traditional field of tangram situations. This collection gathers together nearly 330 tangrams, the best creations of both Chinese and Occidental puzzle devisers. Included are some of the most striking Oriental puzzles, carefully selected from rare nineteenth-century books and some of the most inventive and imaginative inventions of Loyd and Dudeney. Although individual tangrams have always been favorite members of most collections of mathematical amusements, to our knowledge this is the first book devoted entirely to this popular form in many decades. Some of the tangrams in this collection are relatively easy and can be solved without too much brain-straining. Others are difficult, and many demand quite a bit of mental sweat before they are resolved. All, however, are delightful concoctions in recreational form-perception and will provide many hours of pleasure.
Practice makes perfect! Get perfect with a thousand and one practice problems!
1,001 Geometry Practice Problems For Dummies gives you 1,001 opportunities to practice solving problems that deal with core geometry topics, such as points, lines, angles, and planes, as well as area and volume of shapes. You'll also find practice problems on more advanced topics, such as proofs, theorems, and postulates. The companion website gives you free online access to 500 practice problems and solutions. You can track your progress and ID where you should focus your study time. The online component works in conjunction with the book to help you polish your skills and build confidence.
As the perfect companion to Geometry For Dummies or a stand-alone practice tool for students, this book & website will help you put your geometry skills into practice, encouraging deeper understanding and retention. The companion website includes:
Hundreds of practice problems
Customizable practice sets for self-directed study
Problems ranked as easy, medium, and hard
Free one-year access to the online questions bank
With 1,001 Geometry Practice Problems For Dummies, you'll get the practice you need to master geometry and gain confidence in the classroom.
Each page in Common Core Math Workouts for grade 6Hit the geometry wall? Get up and running with this no-nonsense guide!
Does the thought of geometry make you jittery? You're not alone. Fortunately, this down-to-earth guide helps you approach it from a new angle, making it easier than ever to conquer your fears and score your highest in geometry. From getting started with geometry basics to making friends with lines and angles, you'll be proving triangles congruent, calculating circumference, using formulas, and serving up pi in no time.
Geometry is a subject full of mathematical richness and beauty. But it's a subject that bewilders many students because it's so unlike the math they've done before—it requires the use of deductive logic in formal proofs. If you're having a hard time wrapping your mind around what that even means, you've come to the right place! Inside, you'll find out how a proof's chain of logic works and even discover some secrets for getting past rough spots along the way. You don't have to be a math genius to grasp geometry, and this book helps you get un-stumped in a hurry!
Find out how to decode complex geometry proofs
Learn to reason deductively and inductively
Make sense of angles, arcs, area, and more
Improve your chances of scoring higher in your geometry class
There's no reason to let your nerves get jangled over geometry—your understanding will take new shape with the help of Geometry For Dummies.
The classic Heath translation, in a completely new layout with plenty of space and generous margins. An affordable but sturdy student and teacher sewn softcover edition in one volume, with minimal notes and a new index/glossary.
Tough Test Questions? Missed Lectures? Not Enough Time?
Fortunately, there's Schaum's. This all-in-one-package includes more than 1,100 build confidence, skills, and knowledge for the highest score possible105 fully solved problems
Concise explanations of all calculus concepts
Expert tips on using the graphing calculator
Fully compatible with your classroom text, Schaum's highlights all the important facts you need to know. Use Schaum's to shorten your study time--and get your best test scores!
Three-time New York Times bestselling author Danica McKellar and popular "Dancing With The Stars" contestant now makes it a breeze to excel in… Geometry!
Hollywood actress and math whiz Danica McKellar has completely shattered the "math nerd" stereotype. For years, she's been showing girls how to feel confident and ace their math classes – with style! With Girls Get Curves, she applies her winning techniques to high school geometry, giving readers the tools they need to feel great and totally "get" everything from congruent triangles to theorems, and more. Inside you'll find: | 677.169 | 1 |
Introduction to Scientific Computing A Matrix-Vector Approach Using Matlab
by Charles F. Van LoanUnique in content and approach, this book covers all the topics that are usually covered in an introduction to scientific computing -- but folds in graphics and matrix-vector manipulation in a way that gets readers to appreciate the connection between continuous mathematics and computing. MATLAB 5 is used throughout to encourage experimentation, and each chapter focuses on a different important theorem -- allowing readers to appreciate the rigorous side of scientific computing. In addition to standard topical coverage, each chapter includes 1) a sketch of a "hard" problem that involves ill-conditioning, high dimension, etc.; 2) at least one theorem with both a rigorous proof and a "proof by MATLAB" experiment to bolster intuition; 3) at least one recursive algorithm; and 4) at least one connection to a real-world application. The book revolves around examples that are packaged in 200+ M-files, which, collectively, communicate all the key mathematical ideas and an appreciation for the subtleties of numerical computing | 677.169 | 1 |
FREE MATH LESSON - "Geometry Terms: Student Reference Set"
The sources in this book contain terms, pictures and definitions of geometry concepts for the elementary classroom. Simply copy a set for each student in your class. Then place them in their math journals. Now they have a quick, ready reference for geometry. | 677.169 | 1 |
A-APR. Arithmetic with Polynomials and Rational Expressions
A-APR.A. Perform arithmetic operations on polynomials.
A-APR.A.1. Understand that polynomials form a system analogous to the integers, namely, they are closed under the operations of addition, subtraction, and multiplication; add, subtract, and multiply polynomials.
A-APR.C.4. Prove polynomial identities and use them to describe numerical relationships. For example, the polynomial identity $(x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2$ can be used to generate Pythagorean triples.
A-APR.C.5. Know and apply the Binomial Theorem for the expansion of $(x + y)^n$ in powers of $x$ and $y$ for a positive integer $n$, where $x$ and $y$ are any numbers, with coefficients determined for example by Pascal's Triangle.The Binomial Theorem can be proved by mathematical induction or by a com- binatorial argument.
A-APR.D. Rewrite rational expressions.
A-APR.D.6. Rewrite simple rational expressions in different forms; write $\frac{a(x)}{b(x)}$ in the form $q(x) + \frac{r(x)}{b(x)}$, where $a(x)$, $b(x)$, $q(x)$, and $r(x)$ are polynomials with the degree of $r(x)$ less than the degree of $b(x)$, using inspection, long division, or, for the more complicated examples, a computer algebra system.
A-APR.D.7. Understand that rational expressions form a system analogous to the rational numbers, closed under addition, subtraction, multiplication, and division by a nonzero rational expression; add, subtract, multiply, and divide rational expressions. | 677.169 | 1 |
Seller
Description
The book contains detailed GUIDELINES for the challenges of integration.
Compiled by the textbook: SG Lukinova, NE Lepp. MATHEMATICS. Integral calculus. Differential equations. Rows.
The book discussed in detail an example of the control solutions №6. "Integral calculus. Application of the economy." Objective 2.
The book is a separate EXE-file and does not require any additional software | 677.169 | 1 |
MathMagic Pro Edition for Adobe InDesign is an equation editor mainly for use with Adobe InDesign software in editing any mathematical expressions and symbols with WYSIWYG interface and various powerful featuresThe use of computation and simulation has become an essential part of the scientific process. Being able to transform a theory into an algorithm requires significant theoretical insight, detailed physical and mathematical understanding, and a working level of competency in programming. | 677.169 | 1 |
Description: Matrices, vectors, vector spaces, transformations. Covers all topics in a first year college linear algebra course. This is an advanced course normally taken by science or engineering majors after taking at least two semesters of calculus (although calculus really isn't a prereq) so don't confuse this with regular high school algebra. Linear algebra is a branch of mathematics that studies vector spaces, also called linear spaces, along with linear functions that input one vector and output another. Such functions are called linear maps (or linear transformations or linear operators) and can be represented by matrices if a basis is given. Thus matrix theory is often considered as a part of linear algebra. Linear algebra is commonly restricted to the case of finite dimensional vector spaces, while the peculiarities of the infinite dimensional case are traditionally covered in linear functional analysis.
Resources: OpenCourseware from Khan Academy, MIT, UC Berkeley, Stanford along with many of the World's finest University's. | 677.169 | 1 |
Although differential geometry courses at Brown University have
been using computer-generated media such as color films, slides,
and video for a long time, recently computers have taken an
even larger role in facilitating the understanding of geometry.
Over the past year, students have been participating in interactive
computer labs that allow them to investigate curves and surfaces
with the aid of three-dimensional computer graphics. This paper
is aimed at reporting on these experiences. | 677.169 | 1 |
MAT 1150 Advice
Showing 1 to 3 of 8
This course offers a basic understanding of algebra in a way that is easy for beginners and people who are not good at math to understand. Mrs. Walker provides plenty of guidance for students who need more help, while allowing more advanced students the freedom that they desire.
Course highlights:
graphing, radicals, logarithms, exponents
Hours per week:
3-5 hours
Advice for students:
Do the practice problems in the book and pay attention to what will be on the tests!
The best advice i give to all students is study hard, and always attend class.
Course Term:Fall 2016
Professor:ms. heardway
Course Required?Yes
Nov 30, 2016
| Would recommend.
Pretty easy, overall.
Course Overview:
I'd recommend this as a bridge course for student that may have taken accelerated math classes. It helps fully prepare you for college mathematics.
Course highlights:
The highlights of the curse would definitely be the teaching style. For a class that was around 4 hours long, he teacher had no problem making the material more interesting and easier to pay attention. Her animated speaking allowed me to stay focused on the material for long periods of time.
Hours per week:
3-5 hours
Advice for students:
Don't take the class for granted. Take in everything you can from the class and don't take it too lightly. | 677.169 | 1 |
Introduction to Statistics Advice
Showing 1 to 2 of 2
Broad with a lot of practice is needed together algebraic problem solving skills and arithmetic's previous knowledge is required.
Course highlights:
provides critical thinking and logic, theoretic operations and computations of statistical data. I learnt how to interpret graphs, charts, histograms organize and represent data. how to apply and compute probability and connect statistical data or survey . Able to understand role of umbers prediction analysis approximation, quantities, distributions, estimations errors, precision and accurately interpreting results. acquired, synthesized information and creatively use that information to solving problems. Able to connect linear algebra and binomial probability applications in real world.
Hours per week:
9-11 hours
Advice for students:
learn how to learn from different sources. start early on homework sets. don't get stuck on one idea see the big picture. Time sacrifice practice on own during free time. try learn things well the first time around and build solid foundation. find out something you like about mathematics and focus on that it will give interest and more liking. always understand concepts instead of memorizing. The point is nobody was born mathematician so be positive and put the intention and passion like it and you will succeed in this class.
Course Term:Spring 2017
Professor:gabresselassie daniel
Course Required?Yes
Course Tags:Math-heavyBackground Knowledge ExpectedGreat Discussions
Mar 29, 2016
| Would not recommend.
Not too easy. Not too difficult.
Course Overview:
While he thought he was being very helpful, he was hard to understand and would get extremely frustrated with students when they said they didn't understand what he was explaining.
Course highlights:
I feel like I learned a lot about statistics and it helped me work on my fraction and graph skills. | 677.169 | 1 |
Georgia Georgia.
This set of books is unlike any other textbooks. Each text is written in the style of a novel with a humorous story line. Each section tells part of the life of Fred Gauss and how, in the course of his life, he encounters the need for the math and then learns the methods. Tons of solved examples. Each hardcover textbook contains ALL of the material – more than most instructors cover in traditional classroom settings. Includes tons of proofs.
Written by Dr. Stanley Schmidt with the intent to make math come alive with lots of humour, clear explanations, and silly illustrations that stick in the mind. The student will learn to think mathematically.
Completion of this series prepares student for third year college math.
Singapore Math books are clear, logical, and sequential. There is a strong focus on mental math. Word problems and geometry are integrated throughtout the series. Singapore Math® books are a popular choice of homeschool families and for parents looking for math activity books to support what their child is learning in school. There are also titles to help home educators understand the foundations of Singapore Math® methodology, giving you tools to help your student be successful and have fun.They offer math texts from pre-K to 12th grades. This series challenges children to think through and understand mathematical concepts instead of simply memorizing facts and algorithms. One of the benefits of using this program is its affordability. The textbooks are inexpensive and are reusable. The consumable workbooks are priced so that even families with multiple children using this program will find it affordableRightStart Mathematics uses the AL Abacus to provide a visual, auditory, and kinesthetic experience. The elementary and intermediate program lessons guide the teacher day-by-day and year-by-year, helping children understand, apply, and enjoy mathematics. The RishtStart Mathematics homeschool program is set up with levels, rather than grades, so that your child can begin at the proper level and advance at their own pace. | 677.169 | 1 |
Martin Gardner starts off Riddles with questions on splitting up polygons into prescribed shapes and he ends this publication with a suggestion of a prize of $100 for the 1st individual to ship him a three x# magic sq. inclusive of consecutive primes. purely Gardner may healthy such a lot of varied and tantalizing difficulties into one booklet.
Get the grade you will have in algebra with Gustafson and Frisk's starting AND INTERMEDIATE ALGEBRA! Written with you in brain, the authors offer transparent, no-nonsense factors to help you research tough ideas comfortably. organize for assessments with various assets positioned on-line and in the course of the textual content similar to on-line tutoring, bankruptcy Summaries, Self-Checks, preparing workouts, and Vocabulary and inspiration difficulties.
Undemanding ALGEBRA bargains a realistic method of the learn of starting algebra options, in step with the wishes of cutting-edge pupil. The authors position specific emphasis at the labored examples in every one part, treating them because the basic technique of guideline, considering that scholars depend so seriously on examples to accomplish assignments.
Additional resources for Understanding elementary algebra with geometry : a course for college students
Sample text
Two examples may look the same, but the instructions may be asking you to do two different things. For example: Identify the following property: a ϩ (b ϩ c) ϭ (a ϩ b) ϩ c versus On the other hand, two different examples may have the same instructions but require you to do different things. For example: Evaluate: 2(3 Ϫ 8) versus Evaluate: 2 ϩ (3 Ϫ 8) You are asked to evaluate both expressions, but the solutions require different steps. It is a good idea to familiarize yourself with the various ways in which the same basic instructions can be worded.
In the denominator, ؊10 ؊ 10 means ؊10 ؉ (؊10). 100 ϭ ᎏᎏ Ϫ20 ϭ Ϫ5 As the example shows, the location of parentheses can make a great difference in what an example means. We will frequently emphasize the importance of reading an example carefully so that you clearly understand what it is saying and what it is asking. Properties of 0 Finally, we conclude this section with some special properties of the number 0. We are familiar with the numerical facts that 5 и 0 ϭ 0 and 11 и 0 ϭ 0. In fact, 0 times any whole number is equal to 0, and this fact extends to the negative integers as well.
Reviewing the material before doing exercises makes each solution you go through more meaningful. The better you understand the concepts underlying the exercise, the easier the material becomes and the less likely you are to confuse examples or forget steps. When reviewing the material, take the time to think about what you are reading. Try not to get frustrated if it takes you an hour to read and understand a few pages of a math text—that time will be well spent. As you read your text and your notes, think about the concepts being discussed: (a) how they relate to previous concepts covered and (b) how the examples illustrate the concepts being discussed. | 677.169 | 1 |
it is hardly in the embryoni
stage of development. very new. Yet. Compared to its older sister |mathemati
s| whi
h is thousands of years
old.Chapter 1
Introduction
In histori
al terms. The s
ien
e of
omputing is
yet newer.
Be
ause of the
hallenges of programming (whi
h means instru
ting a dumb ma
hine
how to solve ea
h instan
e of a problem) and the unpre
edented s
ale of programming
problems.
omputing s
ientists have had to hone their problem-solving skills to a very
.
omputing s
ien
e is already
having a major in
uen
e on our problem-solving skills. the digital
omputer is very. amounting to a revolution in the
art of ee
tive reasoning.
1. and to
hanges in the way that mathemati
s
is pra
tised. This has led to advan
es in logi
.ne degree.1
Algorithms
Solutions to programming problems are formulated as so-
alled algorithms. These le
tures form an introdu
tion to problem-solving using the insights
that have been gained in
omputing s
ien
e. An algorithm is a well-de.
The four people take dierent amounts
of time to
ross the bridge. The . but they only have one tor
h between them. Consider the following problem. that are
exe
uted in turn in order to solve the given problem. It is dark. whi
h is typi
al of some of
the exer
ises we dis
uss.
onsisting of a number of instru
tions.ned pro
edure.
A
on
rete example may help to understand better the nature of algorithms and their
relation to problem solving. and it is ne
essary to use a tor
h when
rossing the bridge. when two
ross together they must pro
eed at the speed of
the slowest.
Four people wish to
ross a bridge. The bridge is narrow
and only two people
an be on it at any one time. You may want to ta
kle the problem before reading further.
instead. 2008
. an algorithm might input four numbers. is a mu
h greater understanding of the solution. so that it is always
arried when the bridge is
rossed.
most of the problems are quite diÆ
ult. with only elementary mathemati
al knowledge.
The key to ee
tive problem-solving is e
onomy of thought and of expression | the
avoidan
e of unne
essary detail and
omplexity.
May 28. like the
bridge-
rossing problem above.
This book aims to impart these new skills and insights to a broad audien
e. It aims to demonstrate the importan
e of mathemati
al
al
ulation. the algorithm should
ompute an output whi
h is related
to the input by a
ertain so-
alled input-output relation.
however. and output the total time needed to get all four a
ross the bridge. but the
hosen examples are typi
ally not mathemati
al. the output should be 30 and if the input
is the numbers 1 .
The sequen
e of instru
tions solves the problem if the total time taken to exe
ute the
instru
tions is (no more than) 17 minutes. the bridge-
rossing problem
an be generalised by allowing the number of people to be variable. A typi
al instru
tion will be: \persons x and y
ross the bridge" or \person z
rosses the bridge". be
ause it is ne
essary to formulate very
learly and pre
isely the pro
edure for solving the problem. whereby a single error
an
ause an entire system to abruptly \
rash". Introdu
tion
the bridge. 20 . 6 the output should be 17 . 5 . For example. the harder it gets. 4 . 19 . an algorithm will have
ertain inputs . 3 . it is perhaps not so surprising that the
hallenges of algorithm design have had an immense impa
t on our problem-solving
skills. The book also aims to
hallenge.
An algorithm is typi
ally more general than this. they are problems that are readily understood by a lay
person. the
rossing time for ea
h
person. The
more general the problem.
Coupled with the unforgiving nature of digital
omputers.
Formulating an algorithm makes problem-solving de
idedly harder. Normally.2
1. thousands or even millions lines of
ode.
The solution to this problem |whi
h we won't dis
lose just yet!| is
learly a sequen
e of instru
tions about how to get all four people a
ross the bridge. for ea
h input.
Algorithmi
Problem Solving
Roland Ba
khouse. at least to the untrained or poorly trained
pra
titioner.
if the input is the numbers 1 . (For instan
e. The mastery of
omplexity is espe
ially
important in
omputing s
ien
e be
ause of the unpre
edented size of
omputer programs:
a typi
al
omputer program will have hundreds. using
an example-driven approa
h.) The advantage.
Show that all four
an
ross the bridge within 17 minutes. The pro
ess of formulating an
algorithm demands a full understanding of why the algorithm is
orre
t. The input values are
alled the
parameters of the algorithm. In the
ase of the bridge
rossing problem.
2.3
1.2
Bibliographic Remarks
I . Bibliographi
Remarks
1.
The problem is also known as the \
ashlight" problem and the \U2" problem. Rote [Rot02℄ gives a
omprehensive bibliography.rst found the bridge problem in [Lev03℄. 2008
.
it is reputed to be used by at least one major software
ompany in interviews for new
employees.
Algorithmi
Problem Solving
Roland Ba
khouse.
May 28.
Chapter 2
Invariants
\Invariant" means \not
hanging".
\law" and \pattern". An invariant of some pro
ess is thus some attribute or
property of the pro
ess that doesn't
hange. possibly the most
important.
The re
ognition of invariants is an important problem-solving skill.
We begin as we mean to go on. This
hapter introdu
es the notion of an invariant. We . and dis
usses a number
of examples of its use. Other names for \invariant" are \
onstant".
rst present a number of problems for you to
ta
kle. Some you may .
nd easy. but others you may .
nd diÆ
ult or even impossible to
solve. If you
an't solve one, move on to the next. To gain full bene
t, however, it is
important that you try the problems
rst, before reading further.
We then return to ea
h of the problems individually. The
is now straightforward. it is your turn again. whi
h we dis
uss in detail. This pro
ess is
repeated as the problems get harder. we suggest a sequen
e of steps whi
h lead dire
tly to the solution. The se
ond problem.
Then.rst problem we dis
uss
in detail. you should be able to solve the next
ouple of problems.
The third problem is quite easy. and
then leave you to solve some more. From a proper understanding of the solution to these initial
problems. and how to reason about assignments. but. we
introdu
e some basi
skills related to
omputer programming | the use of assignment
statements. we demonstrate how to solve one problem. whi
h otherwise
would be quite hard. We leave it to you to solve. be
ause the
te
hniques are new. You should . Along the way. but involves a new
on
ept. showing how an invariant is used to solve the problem.
(Thus ea
h
ut splits
one pie
e into two pie
es. in the usual way.
1. A
ut is made by
taking a single pie
e and
utting along one of the grooves. Chocolate Bars.
5
May 28. 2008
.
A re
tangular
ho
olate bar is divided into squares by horizontal and verti
al
grooves. It is to be
ut into individual squares.)
Algorithmi
Problem Solving
Roland Ba
khouse.nd them mu
h easier to solve.
Invariants
Figure 2.1 shows a 4×3
ho
olate bar that has been
ut into .6
2.
Tumblers.ve pie
es.
Eleven large empty boxes are pla
ed on a table.1: Cho
olate-Bar Problem. An unknown number of
the medium boxes is sele
ted and. eight small boxes are pla
ed.
How many
uts in total are needed to
ompletely
ut the
ho
olate into all its
pie
es?
2. (See .
Figure 2. eight medium boxes are pla
ed. Empty Boxes. How many boxes are there
in total?
3. The
uts
are indi
ated by solid lines.
Several tumblers are pla
ed in a line on a table. into ea
h.
some are the right way up. Some tumblers are upside down. An unknown number of the boxes
is sele
ted and. into ea
h.
At the end of this pro
ess there are 102 empty boxes.
g. 2.2.) It is required to turn all the tumblers the
right way up. However, the tumblers may not be turned individually; an allowed
move is to turn any two tumblers simultaneously.
Figure 2.2: Tumbler Problem.
From whi
h initial states of the tumblers is it possible to turn all the tumblers the
right way up?
4. Black and White Balls
Consider an urn
lled with a number of balls ea
h of whi
h is either bla
k or white.
There are also enough balls outside the urn to play the following game. We want
Algorithmi
Problem Solving
Roland Ba
khouse.
May 28, 2008
7
to redu
e the number of balls in the urn to one by repeating the following pro
ess
as often as ne
essary.
Take any two balls out of the urn. If both have the same
olour, throw them away,
but put another bla
k ball into the urn; if they have dierent
olours then return
the white one to the urn and throw the bla
k one away.
Ea
h exe
ution of the above pro
ess redu
es the number of balls in the urn by one;
when only one ball is left the game is over. What, if anything,
an be said about
the
olour of the
(See .nal ball in the urn in relation to the original number of bla
k
balls and white balls?
5. Dominoes
A
hess board has had its top-right and bottom-left squares removed so that there
are 62 squares remaining.
3: Mutilated Chess Board
been provided. Is
it possible to
over all 62 squares of the
hessboard with the dominoes without
any domino overlapping another domino or sti
king out beyond the edges of the
board?
6.3. ea
h domino will
over exa
tly two squares of the
hessboard. 2.) An unlimited supply of dominoes has
Figure 2. Tetrominoes
A tetromino is a .g.
Show that at least
one side of the re
tangle has an even number of squares. Overlapping tetrominoes or tetrominoes that sti
k out
from the sides of the board are not allowed.g.
May 28. 2. 2008
.
Algorithmi
Problem Solving
Roland Ba
khouse.4.
Assume that the board is made up of squares of the same size as the ones used
to make the tetrominoes.)
The following exer
ises all
on
ern
overing a re
tangular board with tetrominoes.
(a) Suppose a re
tangular board is
overed with tetrominoes.
Show that
the number of squares is a multiple of 8 . L-.and I-tetromino
(b) Suppose a re
tangular board
an be
overed with T-tetrominoes.8
2.
(d) An 8×8 board
annot be
overed with one O-tetromino and . T. Invariants
Figure 2.
(
) Suppose a re
tangular board
an be
overed with L-tetrominoes. Z-.4: O-. Show that
the number of squares is a multiple of 8 .
we begin with one pie
e and zero
uts. 2008
. What doesn't
hange.
Why not?
2. or a \
onstant". It is to be
ut into individual squares.
May 28. in the usual way. This is an
\invariant". however. It being a
onstant means that
it will always be one. Thus. Equivalently. and the number of pie
es in
reases by one.fteen L-tetrominoes. of the pro
ess of
utting the
ho
olate bar.1
The Solution
The solution to the
ho
olate-bar problem is as follows.1. the
number of
uts in
reases by one.
is the dieren
e between the number of
uts and the number of pie
es. So.1
Chocolate Bars
Re
all the problem statement:
A re
tangular
ho
olate bar is divided into squares by horizontal and
verti
al grooves. is one. the number of
uts will always be one less than the number of pie
es. the dieren
e between the number
of pie
es and the number of
uts. at the outset. A
ut is made by taking a single pie
e and
utting along one of the grooves. the
number of
uts and the number of pie
es both
hange.
Now.
Algorithmi
Problem Solving
Roland Ba
khouse. Whenever a
ut is made.
(Thus ea
h
ut splits one pie
e into two pie
es. That is. the number
of pie
es will always be one more than the number of
uts.)
How many
uts in total are needed to
ompletely
ut the
ho
olate into
all its pie
es?
2. no matter how many
uts have been made.
We let
variable p
ount the number of pie
es. For more
omplex problems.1.
2.
Abstraction The mathemati
al solution begins by introdu
ing two variables. this is an easy problem to solve. the number
of
uts needed is one less than the number of pie
es. we have used English to des
ribe the solution. rather than formulate the
solution in a mathemati
al notation. be
ause it is more su
in
t and more pre
ise. mathemati
al notation
helps
onsiderably. and we let variable c
ount the number of
uts.1.
This .
The values of these variables des
ribe the state of the
ho
olate bar.2
The Mathematical Solution
On
e the skill of identifying invariants has been mastered.9
2. Let us use this problem
to illustrate what we mean.
For this reason. Cho
olate Bars
We
on
lude that to
ut the
ho
olate bar into all its individual pie
es.
a
ordingly.
The problem has be
ome a \mathemati
al" problem. That is. the variables p
and c do not
ompletely
hara
terise the state of the
ho
olate bar. say. We \abstra
t" from the problem a
olle
tion
of variables (or \parameters") that
ompletely
hara
terise the essential elements of the
problem.
One of the inessential details is that the problem has anything to do with
ho
olate
bars! This is totally irrelevant and. rather than a \real-world" problem.
Other inessential details that have been eliminated are the sequen
e of
uts that have
been made. Real-world problems are very hard. in
ontrast. In this step. and the shapes and sizes of the resulting pie
es. inessential details are eliminated. The problem
ould equally well have been about
utting postage stamps from a sheet of stamps. has been eliminated. making . to solve. four
uts have been
made. problems that su
umb to mathemati
al analysis are
relatively easy. if not
impossible. be
ause it is about properties of
numbers.rst step is
alled abstra
tion. Knowing that. or the sequen
e of
uts that have been made to rea
h that state.
ve pie
es, does not allow us to re
onstru
t the sizes of the individual
pie
es. That is irrelevant to solving the problem.
The abstra
tion step is often the hardest step to make. It is very easy to fall into
the trap of in
luding unne
essary detail, making the problem and its solution over
ompli
ated. Conversely, de
iding what is essential is far from easy |there is no algorithm for doing this!| . The best problem-solvers are probably the ones most skilled in
abstra
tion.
(Texts on problem-solving often advise drawing a
gure. This may help to
larify
the problem statement |for example, we in
luded
g. 2.1 in order to
larify what is
meant by a
ut| but it
an also be a handi
ap! There are two reasons. The
rst is that
extreme
ases are often diÆ
ult to
apture in a .
This is something we return to
Algorithmi
Problem Solving
Roland Ba
khouse.
May 28.gure. 2008
.
10
2. The se
ond is that . Invariants
later.
gures often
ontain mu
h unne
essary detail. as exempli.
ed
by .
g. Our advi
e is to use . 2.1.
whi
h are two quite dierent things. is used instead of the assignment symbol. c := p+1 . c+1 ). c+1 .)
The next step in the problem's solution is to model the pro
ess of
utting
the
ho
olate bar. pronoun
ed \be
omes". Java being again
an example. This is a major problem be
ause it
auses
onfusion between assignments and equalities. This is a nuisan
e. The two sides are
separated by the assignment symbol \ := ". For example. p \be
omes" p+1 . leading to diÆ
ult-to-. In words. ea
h expression on the right side. Some programming languages do not allow simultaneous assignments. \ = ". c+1 . but only that. respe
tively.
An assignment ee
ts a
hange of state.gures with the utmost
aution. and the value of
E after repla
ing all variables as pres
ribed by the assignment. (For
example. mathemati
al formulae
are most often far more ee
tive.
An invariant of an assignment is some fun
tion of the state whose value remains
onstant under exe
ution of the assignment. Java is an example. Instead of a
simultaneous assignment. and then repla
ing the
values of p and c by these values. p−c is an invariant of the
assignment p .) We
an
he
k that E
is an invariant simply by
he
king for equality between the value of E . An assignment is exe
uted by evaluating. the equality
1A
word of warning (for those who have already learnt to program in a language like Java or C):
The assignment statements we will be using are often
alled simultaneous assignments be
ause several
variables are allowed on the left side. and c
\be
omes" c+1 . The state is then
hanged by
repla
ing the value of ea
h variable on the left side by the value of the
orresponding
expression on the right side. p−c is an expression depending on variables p and c . The right side is a
omma-separated list of expressions (in this
ase. restri
ting the
programmer to a single variable on the left side in all assignments. c ). This is how the assignment statement models the pro
ess of making a
single
ut of the
ho
olate bar1 . No variable may o
ur more than
on
e in the left side.
in the
urrent state. a left side and a right side. one has to write a sequen
e of assignments. For example.
Mu
h worse is that the equality symbol. the state |the number of pie
es and
the number of
uts| is
hanged by evaluating p+1 and c+1 . p .
Suppose E is an expression depending on the values of the state variables. their values being updated simultaneously on
e the right side has
been evaluated.
p+1 . The left side is a
omma-separated list of variables (in this
ase. The list must have length equal to the number of variables on the left side. In our example. c := p+1 .
An assignment statement has two sides. Most novi
e programmers frequently make the mistake of
onfusing
the two. and even experien
ed programmers sometimes do. We do so by means of the assignment statement
Assignments
p .
even if your tea
hers do not do so.
Algorithmi
Problem Solving
Roland Ba
khouse. 2008
. always remember to pronoun
e an assignment as \left side becomes right side".
and not \left side equals right side". If you do
write Java or C programs.
May 28.nd errors.
2008
.
n. c := p+1 .
ls := rs .
does not
hange the value of
Given an expression.
m
simulaneously de
reasing
m + 3×n . c+1 . The left side of this equality is the expression E
is the expression E after repla
ing all variables as pres
ribed by the
holds whatever the values of
assignment
p. n−1] = (3×n) + (m+3) + (n−1)
The invariant rule for assignments is then the following.
Algorithmi
Problem Solving
Roland Ba
khouse. n−1] = (m+3) + 3×(n−1)
(m+n+p)[m .
and we
onsider the
assignment
m .
Here are
some examples:
(p−c)[p . n := m+3 .1.
E[ls := rs]
is used to denote the expression obtained by repla
ing all o
urren
es of the variables in
E
listed in
ls
by the
orresponding expression in the list of expressions
rs . p := 3×n .
May 28.
p and c .
As another example. n := m+3 . So. n−1
We
he
k that
m + 3×n
is invariant by
he
king that
m + 3×n = (m+3) + 3×(n−1) .
Simple algebra shows that this holds.
E
is an invariant of the assignment
ls := rs
if.
and an assignment. in
reasing
n
by
1. for all instan
es of the variables in
E. Cho
olate Bars
11
p−c = (p+1) − (c+1) . n .c
and the right side
assignment.2. m+3 . This
he
ks that p−c is an invariant of the
:= p+1 .
E. suppose we have two variables
m
and
by
3.
E[ls := rs] = E . c+1] = (p+1) − (c+1)
(m + 3×n)[m .
Invariants
Induction The .12
2.
So. p = 1 and c = 0 . So. p−c = 1 . So.
p = s . p−c is invariant. c . at that time.
Initially. the number of
uts. When the bar has been
ut into all its squares. initially.nal step in the solution of the
ho
olate problem is to exploit the
invarian
e of p−c . But. p−c = 1
no matter how many
uts have been made. satis. where s is the number of squares.
doesn't play anymore).
the value of the expression is un
hanged (obviously. if the value of an expression is un
hanged by some assignment to its variables. the value of the expression remains
un
hanged. for three times. Both times. the value will be un
hanged no matter how
many times the assignment is applied. An unknown number
of the boxes is sele
ted and into ea
h eight medium boxes are pla
ed. Two players
ompete in
ea
h game. and seek an
invariant.1
2
2. by assumption.es
s−c = 1 . so the end result is also no
hange. It is that.
alled the prin
iple of mathemati
al
indu
tion.
Note that the
ase of zero times is in
luded here. That is. Applying the assignment twi
e means
applying it on
e and then on
e again. An
Algorithmi
Problem Solving
Roland Ba
khouse. the value
of the expression is un
hanged. And so on. the loser is kno
ked out (i.
May 28.
The winner of the tournament is the player that is left after all other players have been
kno
ked out. four times. The prin
iple is very simple. How many games are played before the tournament winner is de
ided? (Hint:
hoose suitable variables. if the assignment is applied zero times. A number of
important problem-solving prin
iples have been introdu
ed | abstra
tion.
This
ompletes our dis
ussion of the
ho
olate-bar problem.
et
. be
ause applying the assignment
zero times means doing nothing).2
Empty Boxes
Try ta
kling the empty-box problem.
An important prin
iple is being used here. Re
all its statement.
Summary
A kno
kout tournament is a series of games. We will see these prin
iples again and again throughout these le
tures. c = s−1 . invariants
and indu
tion. the winner
arries on. it is vital to solving the problem in the
ase that the
ho
olate bar has exa
tly one square (in whi
h
ase zero
uts are required). 2008
. It is very important not to forget
zero. The number of
uts is one less than the number of squares.e. In the
ase of the
ho
olate-bar problem.)
Exercise 2.
Eleven large empty boxes are pla
ed on a table. If the assignment is applied exa
tly on
e.
Suppose there are 1234 players in a tournament. That is.
The Tumbler Problem
unknown number of the medium boxes is sele
ted and into ea
h eight small
boxes are pla
ed.13
2. Identify the initial values of e and f . How many boxes
are there in total?
The following steps should help in determining the solution.
At the end of this pro
ess there are 102 empty boxes. Introdu
e the variables e and f for the number of empty and the number of full
boxes.
2.
1.3. respe
tively. Identify the .
5.nal value of e . Combine the previous steps to dedu
e the .
4.
3. Identify an invariant of the assignment. Model the pro
ess of putting eight boxes inside a box as an assignment to e and
f.
Hen
e dedu
e the .nal value of f .
May 28. an
allowed move is to turn any two tumblers simultaneously.nal
value of e+f .
Several tumblers are pla
ed in a line on a table. From whi
h initial
states of the tumblers is it possible to turn all the tumblers the right way up?
The problem suggests that we introdu
e just one variable that
ounts the number of
tumblers that are upside down.
There are three possible ee
ts of turning two of the tumblers. All of these are irrelevant. 2008
. some are the right way up. Some tumblers are upside
down.
This is a key to ee
tive problem-solving: keep it simple!
2.3
The Tumbler Problem
Let us now look at how to solve the tumbler problem. or whi
h are full and whi
h are empty. However. It is required to turn all the tumblers the
right way up.
and a solution that introdu
es variables representing these quantities is grossly over
ompli
ated. Re
all the statement of the
problem. Let us
all it u . or the
number of small boxes.
Note that this solution does not try to
ount the number of medium boxes. Two tumblers that
are both the right way up are turned upside down. the tumblers may not be turned individually. This is modelled by the assignment
Algorithmi
Problem Solving
Roland Ba
khouse.
Finally.
Turning two tumblers that are both upside down has the opposite ee
t | u de
reases
by two. the third possibility is to exe
ute
skip .
but it is better to have a name for the statement that does not depend on any variables.
The
hoi
e of whi
h of these three statements is exe
uted is left unspe
i. Invariants
u := u+2 . one upside down. In this
example. In programming terms. \Skip" means \do nothing" or \having no ee
t".
the other the right way up) has no ee
t on u .
We use the name skip . it is equivalent to the assignment
u := u .14
2. This is modelled by the assignment
u := u−2 . this is modelled
by a so-
alled skip statement. So. turning two tumblers that are the opposite way up (that is.
The parity of u is a boolean value: it is
either true or false . we
an dis
ount skip . three. four. An invariant
of the turning pro
ess must therefore be an invariant of ea
h of the three.) and it is false if
u is odd (one. It is true if u is even (zero.ed.
Everything is an invariant of skip . We therefore seek an
invariant of the two assignments u := u+2 and u := u−2 . So. What does not
hange
if we add or subtra
t two from u ?
The answer is: the so-
alled parity of u . two. eight et
. .
(even.(u+2) = even.
Zero is an even number.
That is. Let us write even.ve. even. If there is an even number at
the outset.
Then. seven.).
May 28. the parity
of the number of upside-down tumblers will not
hange.u)[u := u+2] = even. no matter how many times we turn two tumblers over. if there is an odd number at the outset.u .
(even.
That is. so the answer to the question is that there must be an even
number of upside-down tumblers at the outset.(u−2) = even. there will always be an even number.u is an invariant of the assignment u := u+2 .u .u for this Boolean quantity.
there will always be an odd number. even.
The goal is to repeat the turning pro
ess until there are zero upside-down tumblers.u)[u := u−2] = even.u is also an invariant of the assignment u := u−2 .
Algorithmi
Problem Solving
Roland Ba
khouse. Also. 2008
. et
.
We
on
lude that.
Suppose c denotes the
number of
overed squares. the number of squares
along one side is m and the number along the other side is n . the solution to
problem 6(a): if a re
tangular board is
overed by tetrominoes. 7 mod 4 is 3 . c = m×n and. It's a preliminary to solving
6(b). Have a peek if you want
to. In words. and
we say \ c is a multiple of 4 ". the words \is an invariant property" are omitted.
May 28. The pro
ess of solving more diÆ
ult problems typi
ally
involves formulating and solving simpler subproblems. Show that
the number of squares is a multiple of 8 . For
example. the argument has been about tetrominoes in general.) Then. (That is.2. m×n is a multiple of 4 . The
hessboard problem is a
little harder. 2008
. Apply the method of introdu
ing appropriate
variables to des
ribe the state of the balls in the urn. What we have just shown is. In fa
t. pla
ing a tetromino on the board is modelled by
c := c+4 .
Suppose a re
tangular board
an be
overed with T-tetrominoes. Then. (Hint: use the
olouring of the squares
on the
hessboard.) Problem 6(a) should be a bit easier.
Identify an invariant. ( c mod 4 is the remainder after dividing c by 4 .4
Tetrominoes
In this se
tion. it must be the
ase that either m or n (or both) is a multiple of 2 . but
an be solved in the same way.
Thus.
Note that. suppose the tetrominoes
over an m×n board. More often. illustrates a general phenomenon in solving problems . and draw the appropriate
on
lusion.
Re
all the problem.
A brief analysis of this problem reveals an obvious invariant. c mod 4 is always 0 . and not
parti
ularly about T-tetrominoes. one
ould say that a
Algorithmi
Problem Solving
Roland Ba
khouse. whi
h we do |together with 6(a)| in the next se
tion.4. So. This gives us the opportunity
to introdu
e a style of mathemati
al
al
ulation that improves
larity. so far.
Now. and 16 mod 4 is 0 . we say \ c is a multiple of 4 is an
invariant property".
2.) Initially c is 0 .
The dis
overy of a solution to problem 6(a). Then express the pro
ess of
removing and/or repla
ing balls by a
hoi
e among a number of assignment statements. we present the solution of problem 6(b). in fa
t. in this way. at least one of the sides
of the re
tangle must have even length. For the produ
t m×n of two numbers m and n to be a
multiple of 4 . c mod 4 is invariant. Tetrominoes
15
You should now be in a position to solve the problem of the bla
k and white balls
(problem 4 in the introdu
tory se
tion).
so. so c mod 4 is 0 mod 4 .
whi
h is 0 .
This is a two-step
al
ulation. Invariants
\diÆ
ult" problem is one that involves putting together the solution to several simple
problems. we want to introdu
e a style for presenting
al
ulations that is
learer
than the normal mixture of text with interspersed mathemati
al expresssions. Just
keep on solving simple problems until you have rea
hed your goal!
At this point. \diÆ
ult" problems be
ome a lot more manageable. To introdu
e the style we repeat the argument just given.16
2. Here it is in the new style:
an m×n board is
overed with tetrominoes
⇒
{
invariant: c is a multiple of 4 . The . Looked at this way.
c = m×n
}
m×n is a multiple of 4
⇒
{
property of multiples
}
m is a multiple of 2 ∨ n is a multiple of 2 .
m is a multiple of 2 or n is a multiple of 2 . that the number
of
overed squares is always a multiple of 4 (whatever the shape of the area
overed)
together with the fa
t that. \an m×n board is
overed with tetrominoes implies m×n is a multiple
of 4 " or \an m×n board is
overed with tetrominoes only if m×n is a multiple of
4 . if an m×n board has been
overed.rst step is a so-
alled \impli
ation" step. as indi
ated
by the \ ⇒ " symbol. following the \ ⇒ " symbol is a hint why the statement is true.
The se
ond step is read as:
If m×n is a multiple of 4 . proved earlier.
(Alternatively. The step is read as
If
an m×n board is
overed with tetrominoes.
Again. the \ ⇒ " symbol signi.")
The text between
urly bra
kets. the number of
overed
squares is m×n . m×n is a multiple of 4 . Here the hint is the
ombination of the fa
t.
The symbol \ ∨ " means \or". is less spe
i. but not both. Note
that by \or" we mean so-
alled \in
lusive or" | the possibility that both m and n are
multiples of 2 is in
luded.
The hint.e.es an impli
ation. i. in this
ase. A so-
alled \ex
lusive or" would mean that m is a multiple
of 2 or n is a multiple of 2 . it would ex
lude this possibility.
The
on
lusion of the
al
ulation is also an \if" statement. but you should have suÆ
ient knowledge of multiplying
numbers by 4 to a
ept that the step is valid. You may or may not be familiar with the general theorem.
May 28. 2008
.
. It is:
Algorithmi
Problem Solving
Roland Ba
khouse. The property that is being alluded to has to do
with expressing numbers as multiples of prime numbers.
Clearly. In
luding the symbol
\ ⇒ " makes
lear the relation between the expressions it
onne
ts. Colouring this one square dierently from the other three suggests
olouring the squares of the re
tangle in the way a
hessboard is
oloured. we present
al
ulations in whi
h \ ⇐ " is the
onne
ting symbol. and 4 is not a multiple of 8 .
one with three bla
k squares and one white square. This gives us two types. m is a multiple of 2 or n is
a multiple of 2 . In
luding hints within
urly bra
kets between two expressions means that the hints may
be as long as we like. Tetrominoes
If an m×n board is
overed with tetrominoes. (See . Later. More importantly.
Suppose we indeed
olour the re
tangle with bla
k and white squares. whi
h is often the most ee
tive way to reason.
Let us now ta
kle problem 6(b) head on.)
What distinguishes a T-tetronimo is that it has one square that is adja
ent to the
other three squares. We
all them dark and light T-tetrominoes. A 4×1 board
an be
overed with 1 I-tetronimo. they may even in
lude other sub
al
ulations. The T-tetrominoes should be
oloured in the same way. and one with three white squares and
one bla
k square. the solution must take a
ount
of the shape of a T-tetronomo.
This style of presenting a mathemati
al
al
ulation reverses the normal style: mathemati
al expressions are interspersed with text.4. as on a
hessboard.17
2. rather than the other way around. (It isn't true for I-tetronimoes.
it allows us to use other relations. Su
h
al
ulations work ba
kwards from a goal to what has been
given.
d
ounts the number of dark T-tetrominoes that have been used.
Pla
ing a dark tetromino on the board is modelled by the assignment
d . 2008
. w+3 . b+1 .5.
and l
ounts the number of light tetrominoes.
May 28. w := l+1 . In addition. The variable b
ounts
the number of
overed bla
k squares. 2.
An invariant of both assignments is
Algorithmi
Problem Solving
Roland Ba
khouse.) Pla
ing
the tetrominoes on the board now involves
hoosing the appropriate type so that the
olours of the
overed squares mat
h the
olours of the tetrominoes. w := d+1 . whilst w
ounts the number of
overed white
squares.g. b . w+1 . b+3 . b .
Figure 2.5: Dark and light T-tetrominoes
We introdu
e four variables to des
ribe the state of the board.
Pla
ing a light tetromino on the board is modelled by the assignment
l .
no matter how
many T-tetrominoes are pla
ed on the board.nition of substitution
}
(b+1) − 3×d − (l+1)
=
{
arithmeti
}
b − 3×d − l .
Now. the initial value of b − 3×d − l is zero. another invariant of both assignments is
w − 3×l − d .
We
an now solve the given problem.
whi
h means that the number of bla
k squares
equals the number of white squares
}
b=w
⇒
{
b − 3×d − l = 0
w − 3×l − d = 0
}
(b = w) ∧ (3×d + l = 3×l + d)
Algorithmi
Problem Solving
Roland Ba
khouse. it will always be zero. So.
May 28. 2008
. Similarly. the value of w − 3×l − d will
always be zero.
Similarly.
a re
tangular board is
overed by T-tetrominoes
⇒
{
from problem 6(a) we know that at least one
side of the board has an even number of squares.
(How easily you
an adapt the solution to one problem in order to solve another is a
good measure of the ee
tiveness of your solution method. It shouldn't be too diÆ
ult
to solve 6(
) be
ause the solution to 6(b). you should be able to repeat the
same argument as above.5.19
2. Similarly.
We
on
lude that
If a re
tangular board is
overed by T-tetrominoes. Indeed. Look at other ways of
olouring the
squares bla
k and white. whi
h suggests that
it
an be solved in a similar way. above. takes
are to
learly identify those steps
where a property or properties of T-tetrominoes are used. the number of
overed
squares is divisible by 8 .
You
an now ta
kle 6(
). Be
areful to
he
k that all steps remain valid. the solution also
learly identi. it
an. The problem looks very mu
h like 6(b). Additional Exer
ises
{
⇒
arithmeti
}
(b = w) ∧ (l = d)
b − 3×d − l = 0
{
⇒
w − 3×l − d = 0
}
b = w = 4×d = 4×l
{
⇒
arithmeti
}
b+w = 8×d
{
⇒
b+w is the number of
overed squares
}
the number of
overed squares is a multiple of 8 . Having found a suitable way.
2008
.
Exercise 2. on
e 6(
) has been solved.
they are diÆ
ult to adapt to new
ir
umstan
es.
May 28. The total number of obje
ts is
redu
ed by repeatedly removing two obje
ts of dierent kind. Good lu
k!
2. and repla
ing them by an
obje
t of the third kind.5
Additional Exercises
Given is a bag of three kinds of obje
ts.
Identify exa
t
onditions in whi
h it is possible to remove all the obje
ts ex
ept
one.2
2
Algorithmi
Problem Solving
Roland Ba
khouse.)
Problem 6(d) is relatively easy. As a result.es where the fa
t that the area
overed is re
tangular is exploited. Badly
presented
al
ulations do not make
lear whi
h properties are being used.
The problems of the bla
k and
white balls is from [Gri81℄.2 was posed to me by Dmitri Chubarov. both new and old. The author of the problem is apparently not stated. It was posed (in a slightly dierent
form) in the Russian national Mathemati
s Olympiad in 1975 and appears in a book
by Vasiliev entitled \Zada
hi Vsesoyuzynykh Matemti
heskikh Olympiad" published in
Mos
ow. Their publi
ations
ontain many examples of mathemati
al
puzzles. I do not know their origin. The domino and tumbler
problems are old
hestnuts.) Vierkant
Voor Wiskunde |foursquare for mathemati
s| is a foundation that promotes mathemati
s in Dut
h s
hools. (See
2.
Algorithmi
Problem Solving
Roland Ba
khouse.6
2. The tetromino problems I found in the 1999 Vierkant Voor
Wiskunde
alendar. I have made grateful use of them throughout this text.
Exer
ise 2. some of whi
h I have made use of.
Thanks go to Jeremy Weissman for suggestions on how to improve the presentation of
the tetronimo problems. Invariants
Bibliographic Remarks
The empty-box problem was given to me by Wim Feijen. 2008
.nl/puzzels/.vierkantvoorwiskunde. 1988.
May 28.
so we
an't dispense with it. it is often the
ase that. they have startling. elsewhere. On
e a problem has been broken down in this way. The amount of work involved explodes as the problem size gets bigger. be
ause it helps to understand the nature of problem-solving. We
shall see how inappropriate or unne
essary naming
an in
rease the
omplexity of a
Algorithmi
Problem Solving
Roland Ba
khouse. We use them as simple illustrations of \brute-for
e"
sear
h and problem de
omposition.
Brute-for
e sear
h means systemati
ally trying all possibilities. postponing the use of a brute-for
e sear
h for as long
as possible. brute
for
e
an be applied. However.
An important issue that emerges in this
hapter is naming the elements of a problem. Problem de
omposition involves exploiting the stru
ture of a problem to break it down into smaller. ultimately. lots of
areful.
21
May 28.
All river-
rossing problems have an obvious stru
tural property. hidden beauty. but is often overlooked. it is mu
h better to spend more
eort in de
omposing a problem. namely the symmetry
between the two banks of the river. The exploitation of symmetry is a very important
problem-solving te
hnique.
making it impra
ti
al for all but toy problems. Indeed. but does require a lot of
areful and a
urate work. brute for
e isn't even pra
ti
al for implementation
on a
omputer.
You may already have seen the problems.
ombined with problem
de
omposition. brute for
e is the only
solution method. it is useful to know what
brute for
e entails.
Problem de
omposition is something we humans are mu
h better at. But. Using brute
for
e is not something human beings are good at. or similar ones.Chapter 3
Crossing a River
The examples in this
hapter all involve getting a number of people or things a
ross
a river under
ertain
onstraints. 2008
. As illustrations of
brute-for
e sear
h |whi
h is how their solutions are often presented| they are extremely
uninteresting! However.
more manageable problems.
De
iding on what and how names should be introdu
ed
an be
ru
ial to su
ess. a
urate work is something more suited to
omputers. as illustrations of the use of symmetry. It's a te
hnique that
doesn't require any skill. Nevertheless. parti
ularly when using brute for
e.
Crossing a River
problem.
Overweight
Ann. the goat should not be left alone with the
abbage
(otherwise.
3.
How
an all three
ouples get a
ross the river?
3.1
Brute Force
Goat. a
abbage and a wolf a
ross a river.
How
an the farmer a
hieve the task?
2. They have one boat that
is only big enough to a
ommodate one adult or two
hildren. the goat would eat the
abbage). Cabbage and Wolf
The goat-.
However. Our
main purpose in showing the brute-for
e solution is to illustrate the pitfalls of
Algorithmi
Problem Solving
Roland Ba
khouse. Ann is
Dee is
100
46
kilos.2. Col and Dee are on one side of a river.
Problems
Goat.
How
an all the adults and all the
hildren
ross the river? Make
lear any assumptions you are obliged to make. Bob is
49
kilos.
Adults and Children. Also.
A farmer wishes to ferry a goat.
his boat is only large enough to take one of them at a time. They have one rowing boat that
an
arry at most 100 kilos.
A group of adults and
hildren are on one side of a river.22
3.
4. 2008
. Bob.2
3. Bob
an't row. The
husbands are so jealous of ea
h other that none is willing to allow their wife to be
with another man.1
1.
poor
May 28. the wolf would eat the goat). if they are not themselves present. and the wolf should not be left alone
with the goat (otherwise.
Three
ouples (husband and wife) wish to
ross a river. Col is
52
kilos and
kilos. making several trips a
ross the river ne
essary. They have one boat that
an
arry at most two people. making several trips
a
ross the river ne
essary.
The Jealous Couples. Cabbage and Wolf. making it impossible to solve even with the aid of a very powerful
omputer.and wolf-problem is often used to illustrate brute-for
e sear
h.
abbage.
How
an they all get to the other side?
3.
and the wolf
should not be left alone with the goat (otherwise. a
abbage and a wolf a
ross a river. and we
all their possible values L (for left) and R (for right).
This means that we
an represent a state by four variables. we introdu
e some terminology that is useful when
dis
ussing the eÆ
ien
y of a parti
ular solution to a problem. g (for goat). making
several trips a
ross the river ne
essary.
A farmer wishes to ferry a goat.
How
an the farmer a
hieve the task?
The problem involves four individuals. Brute For
e
problem-solving skills. ea
h of whi
h has one of
two values. his boat is only large enough to take one of them at a time.2. the wolf would eat the
goat). c (for
abbage) and w
(for wolf).
In the goat-. so we do not need to introdu
e a variable to represent its
position. a state des
ribes on whi
h bank ea
h of the four
individuals
an be found. and ways of
hanging from one state to another | the state transitions.
However. the goat should not be left alone
with the
abbage (otherwise. Note that the boat
is always where the farmer is.
A brute-for
e sear
h involves
onstru
ting a state-transition graph that models all
possible states. the goat would eat the
abbage).
abbage-. We
all the variables f (for farmer). Additionally. Also. wolf-problem.23
3. A value of L means at the left bank. A state transition is a
hange of state that is allowed by
the problem spe
i. A value of R
means \at the right bank". and ea
h is at one of the two river banks.
whilst the
abbage and wolf are at the
left bank.
ation. a diagram
an be drawn depi
ting a state-transition
graph. The farmer and goat are at the right bank. However. two states between whi
h there is a valid state
transition are:
1. The lines have arrows on them if some state transitions are
not reversible. For example.
2.e. The requirement that the goat
annot be left alone with the
abbage is expressed
by the system invariant
Algorithmi
Problem Solving
Roland Ba
khouse. there are 24 (i. if so. and the state transitions are drawn as lines
onne
ting the
ir
les. sixteen) dierent
ombinations of values. 2008
. If all state transitions are
reversible. in this problem some of these
ombinations are ex
luded. the arrows are not ne
essary and the diagram is
alled an undire
ted graph. All four are at the left bank. The states are drawn as
ir
les.
For the very simplest problems. the diagram is
alled a dire
ted graph.
We are going to draw a state-transition graph to demonstrate the brute-for
e solution to
this problem.
If four variables
an ea
h have one of two values.
May 28.
eliminating the ones that are not allowed.24
3.
That is.
or the goat and
abbage are on dierent banks ( g 6= c ). Similarly. c and w
are required to be equal.
The table below shows the ten dierent
ombinations. the total redu
es to ten. but dierent from f . The graph in
. (Noti
e that when f and g are
equal all
ombinations of c and w are allowed. Crossing a River
f = g = c ∨ g 6= c . we enumerate all the possible transitions between these states. when f and g are dierent.)
f
g
c
w
L
L
L
L
L
R
R
R
R
R
L
L
L
L
R
L
R
R
R
R
L
L
R
R
L
R
L
L
R
R
L
R
L
R
L
R
L
R
L
R
Now. either the farmer. This ex
ludes
ases where g
and c are equal.
If we list all states. the requirement that the goat
annot
be left alone with the wolf is expressed by the system invariant
f = g = w ∨ g 6= w . the goat and the
abbage are all on the same bank ( f = g = c ).
Ea
h solution
is given by a path through the graph from \LLLL" box to the \RRRR" box.
May 28.1 does just this. and returns alone. it is
lear that there are two solutions to the problem.g. 3. This is the path
from LLLL to LRLL. There are
no arrows on the edges be
ause ea
h transition
an be reversed. The upper
path gives the following solution:
1. and the
abbage and wolf are at the left bank. 2008
. and the
edges of the graph |the lines
onne
ting the boxes| represent transitions.
At the very left. The farmer takes the goat to the right bank. The nodes of the graph |the boxes| represent states.
Algorithmi
Problem Solving
Roland Ba
khouse. This is
represented by the line
onne
ting the \LLLL" box to the \RRLL" box. the box labelled \LLLL" represents the state where all four are on
the left bank.
From the graph. The only allowed transition from this state is to the state where the farmer
and goat are at the right bank.
and wolf-problem is not representative.
We
an see how qui
kly the sear
h spa
e
an grow by analysing what is involved in
using brute for
e to solve the remaining problems in se
tion 3. In the jargon used by
omputing s
ientists.1: Goat-. For other problems.2
State-Space Explosion
There is often a tenden
y to apply brute for
e without thinking when fa
ed with a new
problem.
4. and returns with the goat. there are at least two transitions. Also.and wolf-problem.
3. given by the lower path. However. brute for
e does not \s
ale
up" to larger problems. So. So. ea
h of whom
an be on one side or other of the
Algorithmi
Problem Solving
Roland Ba
khouse. Wolf-Problem
2. Cabbage-.
In the \overweight" problem. and a manageable
number of transitions.
unlike in the goat-. The farmer takes the goat to the right bank. there are 16 possible states. The farmer takes the wolf to the right bank.2. from the initial state there are four dierent transitions. from
most other states. it should only be used where it is unavoidable.1.25
3. too large even for the most diligent problem-solvers. Brute For
e
RRRL
LLLL
RRLL
LLRL
LRLL
RLRR
RRLR
LLRR
RRRR
LLLR
Figure 3. This is the path
from LLRL to LLRR.
abbage.
3. there are four named individuals and no restri
tions
on their being together on the same side of the bank.
Here. 2008
. inter
hanges \
abbage" and\wolf"
in the se
ond and third steps. The goat-. there are six individuals involved. This is the path from LLRR to RRRR.
The alternative solution.
May 28. and returns alone. Brute for
e is only
useful for very simple problems. no restri
tion on the size of the state
spa
e is possible. the total number of transitions
is large. The farmer takes the
abbage to the right bank. the sear
h spa
e qui
kly be
omes
mu
h too large.
abbage. This
is the path from LRLL to LLRL.
The situation in an unskilled solution of the \jealous-
ouples" problem is even worse. the
above |thoughtless!| solution has a manageable number of states.2.
Crossing a River
river bank. the number of states is 26 .
64 ! That's an impossible number for any human being to
ope with.26
3. there are .e. i. If we give ea
h individual a distin
t name. and we haven't even
begun to
ount the number of transitions. In another variation on the jealous-
ouples
problem.
namely
that it
an only be applied in spe
i. 1024 . and the boat
an take three people at a time. and a yet larger
number of transitions. That
means.e. not real problems. there are are 210 . i.ve
ouples. if all are named. Take note: these are \toy" problems. dierent states.
The \adults-and-
hildren" problem illustrates another failing of brute for
e.
and not in the general
ase.
ases. The number of
adults and
hildren is not spe
i.
Yet. The goat-
abbageand-wolf problem is a good example. But. We speak of an \exponential" growth in the
number of states ( n is the exponent in 2n ). diagrams qui
kly be
ome a
problem in themselves | apart from the size of paper needed.
Drawing state-transition diagrams is equally inee
tive. even for
quite small n . do we really need to distinguish between all
four? In the dis
ussion of the state spa
e. well-
hosen problem.
The use of a
omputer to perform a brute-for
e sear
h shifts the meaning of what
is a \small" problem and what is a \large" problem.
In the goat-
abbage-and-wolf problem. there are. how are the nodes to be
pla
ed on the paper so that the diagram be
omes readable?
3. Instead. in prin
iple. 2n is a very large number. the \
abbage" and the \wolf". Spe
i. Whenever the state spa
e of a
lass of
problems grows exponentially. But. But
onstru
ting
a diagram is rarely helpful in problem-solving. A diagram
an o
asionally
be used to illustrate the solution of a simple.3
Abstraction
The state-spa
e explosion is often
aused by a failure to properly analyse a problem. the
\goat".2. If there are n individuals in su
h
a problem. The so-
alled \state-spa
e explosion problem" gets in the way.ed in this problem. it is in fa
t the easiest of all to
solve. The river-
rossing
problems illustrate \state-spa
e explosion" very well. it means that even the largest and fastest super
omputers
an only ta
kle quite small instan
es. distin
t names are given to the \farmer". 2n dierent states to be
onsidered. but not as mu
h as one might
expe
t. we remarked on a \similarity" between the
wolf and the
abbage. a
parti
ularly frequent
ause is unne
essary or inappropriate naming.
This \similarity" also emerged in the solution: two solutions were obtained. In the restated problem. then. Why. we
all
Algorithmi
Problem Solving
Roland Ba
khouse. this time with a naming
onvention that omits the unne
essary distin
tion between the wolf and the
abbage. are the \wolf" and
the \
abbage" distinguished by giving them dierent names?
Let us restate the problem. 2008
.
May 28. the goat
annot be left with either the wolf or the
abbage.
symmetri
al in the inter
hange of \wolf" and \
abbage".
ally.
making it de
idedly boring.
May 28. Indeed. it is
easy to dis
over. and adds to the size of the state spa
e. and redu
ing a problem to its essentials is
alled abstra
tion. Their solutions are then put together to form a solution to the original problem.
A farmer wishes to ferry an alpha and two betas a
ross a river.
How
an the farmer a
hieve the task?
Now the problem be
omes mu
h easier to solve. an alpha should not be left alone with
a beta. making use of
general problem-solving prin
iples. making the problem
more
ompli
ated than it really is. We en
ounter the importan
e of appropriate naming
time and again in the
oming
hapters. However.
3. Also. and then one beta a
ross. a problem has an inherent stru
ture that fa
ilitates de
omposing the problem
into smaller problems.
his boat is only large enough to take one of them at a time. But. They have one
boat that
an
arry at most two people. The smaller problems
an then be further de
omposed until they
be
ome suÆ
iently manageable to be solvable by other means. 2008
. Be
ause there is only one solution.
The problem-solving prin
iple that we learn from this example is very important. Bear it in mind as you read. The husbands are so jealous of ea
h other that none is willing to
allow their wife to be with another man if they are not themselves present. making several trips a
ross the river
ne
essary.
Re
all its statement:
Three
ouples (husband and wife) wish to
ross a river.3
Jealous Couples
Very often.
How
an all three
ouples get a
ross the river?
Algorithmi
Problem Solving
Roland Ba
khouse. there is only one solution:
Take the alpha a
ross. returning with the alpha. and it is unne
essary to
onstru
t a state-transition diagram for the
problem. The pro
ess of omitting
unne
essary detail. followed by the alpha. Then take
the se
ond beta a
ross.
Avoid unnecessary or inappropriate naming.
When elements of a problem are given individual names. perhaps even by brute
for
e. making several
trips a
ross the river ne
essary. it distinguishes them from other
elements of the problem.27
3. It
an be solved by brute for
e.3.
The jealous-
ouples problem is an ex
ellent example. Poor
solutions to problems are ones that fail to \abstra
t" adequately. Jealous Couples
the goat an \alpha" and the
abbage and the wolf \betas". it
an be solved mu
h more ee
tively.
it is possible to get any number of \things" a
ross the river: repeat
the pro
ess of letting two
ross from left to right. this
means that a boat that
an
arry four people is suÆ
ient to ferry an arbitrary number
of
ouples a
ross the river.
learly. and we have just shown that
it is at most four. Then.
Now. If the
apa
ity is one
(or less) the maximum number of
ouples is zero.
ouples are not jealous of ea
h other! ) Sin
e a
ouple is two people. Rather than ta
kle
the problem as stated. and then . In the original problem. by repla
ing \thing" by \
ouple". the question is how many
ouples
an be ferried with a boat of
apa
ity
two. there is no
maximum. But. we
an spe
ify the
apa
ity of the boat and ask what is the maximum
number of
ouples that
an be ferried a
ross with that
apa
ity. the minimum
apa
ity is at least two (sin
e it is not possible to ferry more than one person a
ross a
river in a boat that
an only
arry one person at a time).
Alternatively. a boat with
apa
ity two is needed|
and we are required to give a
onstru
tive proof that this is the
ase. what is the
minimum
apa
ity needed to ferry n
ouples a
ross the river? Obviously. followed by one returning from right
to left.
Suppose there is one boat that
an
arry two \things". and if the
apa
ity is four.28
3.
This simple analysis gives us a dierent slant on the problem. Crossing a River
What's The Problem?
Before we ta
kle this parti
ular problem. So.1
3. let us try to determine what the essen
e of the
problem is. and there are no other restri
tions. we
are given the answer |in the
ase of three
ouples.
The new problems look more diÆ
ult than the original. there is often
an advantage in not knowing the answer | be
ause we
an sometimes gain insight by
generalising. we
an ta
kle a related problem. and how many
ouples
an be ferried with a boat of
apa
ity three.3. until at most two remain on the left bank. namely. (After all. we infer that a boat that
an
arry two
ouples at one
rossing
an be used to ferry an arbitrary number of
ouples a
ross the
river. what is the minimum
apa
ity needed to ferry three
ouples a
ross the river? More generally.
That is.
First.3.
3. This suggests seeking a solution that gets ea
h
ouple
a
ross in turn.rst solving simpler instan
es of the general problem. there are three
ouples. we de
ompose the problem into three subproblems: get the .2
Problem Structure
The stru
ture of this problem suggests several ways in whi
h it might be de
omposed.
get the se
ond
ouple a
ross. A
ording to the maxim \ladies
before gentlemen". we
ould try .
Another de
omposition is into husbands and wives. and get the third
ouple a
ross.rst
ouple a
ross.
followed by all the
Algorithmi
Problem Solving
Roland Ba
khouse.rst getting all the wives a
ross.
May 28. 2008
.
there is no diÆ
ulty
in transferring the wives away from them. Getting all the husbands a
ross . The reason is that.
Getting all the wives a
ross.rst getting all the
husbands a
ross. followed by all the wives. whilst their husbands remain at the left bank turns out
to be easy. if the husbands all stay in one pla
e.
abbage-. wolf-problem exhibits the leftright symmetry very well.)
3. seems mu
h harder. this is a
ru
ial step in . The
introdu
tion of notation involves naming the elements of the problem that we want to
distinguish. The diagram also illustrates the symmetry between the
abbage and wolf. the result is a pro
ess for
getting the same group of people from right to left. whilst
their wives stay at the left bank.3
Denoting States and Transitions
We begin by introdu
ing some lo
al notation to make the solution strategy pre
ise. getting the
husbands to join their wives may prove to be harder than getting the wives to join their
husbands. there doesn't seem mu
h to
hoose between them. As dis
ussed earlier. by
ignoring them and using brute-for
e.
(The state-transition diagram for the goat-. The pro
ess of getting
a group of people from left to right
an always be reversed. This is indeed what we do.3. On the other hand. Ladies before gentleman. Perhaps a symmetri
solution is
possible! If that is the
ase. one key stru
tural property of the problem that we have not yet
onsidered.rst.
There is. and that is a major
saving. or age before beauty. however. It is the symmetry between the left and right banks. we only need to do half the work. we lost the opportunity of a redu
tion in eort. Both symmetries were to be expe
ted from the problem statement.
2008
.nding a solution.2H || 2W . whilst 1C means a husband and wife who do form
a
ouple. Number is an extremely important mathemati
al abstra
tion. for example. wives and
ouples that is relevant to the
problem's solution. A state is denoted by two sequen
es separated by bars. we use letters H . wife and
ouple. respe
tively.
May 28.
Note that we do not name the individual people as in. 1H. and all three
wives are at the right bank.
Clare et
.2H means one
ouple and two husbands. for example.
These are pre
eded by a number.1W means a husband
and wife who do not form a
ouple. Bob. W and C to mean husband. A se
ond example of a state is 1C. An example is 3H || 3W . Ann. whi
h denotes
Algorithmi
Problem Solving
Roland Ba
khouse.
Here.
A state des
ribes a situation when ea
h individual (husband or wife) is at one of the
banks. It is only the number of husbands. 2H means two husbands. We exploit the notation
to distinguish between
ouples and individuals.
We distinguish between states and a
tions. 3C means
three
ouples and 1C. for example.
whi
h denotes the state in whi
h all three husbands are at the left bank.
The starting state is thus 3C || and the required . Crossing a River
the state in whi
h one
ouple and two husbands are at the left bank and two wives are at
the right bank.30
3.
leaving three husbands at the left bank and one wife at the right bank. An
example is 3H |2W| 1W . this denotes the a
tion of transporting two wives a
ross the
river. and. Sin
e the position of the
boat must alternate between the left bank and the right bank. this ambiguity is easily
resolved.
The notation allows valid and invalid states/a
tions to be easily identi.
Note that the notation for states and a
tions does not spe
ify the position or dire
tion
of the boat.
ould be ambiguous.nishing state is || 3C .
An a
tion is when some individuals are being transported a
ross the river. taken out of
ontext.
1W || 1C. whilst all three wives are at the right bank. beginning in the state where two
ouples and one husband are at
the left bank. beginning
with the state 3C || and ending with the state || 3C . 1C. and S denotes a sequen
e of a
tions. who is on the other side of the river).
3H |3W| is invalid be
ause the boat
an only
arry at most two people. 2008
.
In general. If p and q denote states. that alternates between states and
a
tions.
Of
ourse. but the
he
ks are straightforward. it
will result in state q . for example. if the sequen
e of a
tions S is performed beginning in state p .
Algorithmi
Problem Solving
Roland Ba
khouse. So. Care is needed. an a
tion ee
ts a transition between states. Also. detailed solution to the problem is a sequen
e. It is easy to make
a mistake and make an invalid
laim.
{
p }
S
{
q }
is the property that.
May 28.1H || 1W
}
3H |2W| 1W
{
3H || 3W
}
is the property that.1H is invalid (be
ause there is a wife who is on the same side of
the river as a man other than her husband.) Additional notation helps to express
the result of a
tions. a
omplete.
{
2C. (In the terminology of state-transition diagrams.
An a
tion results in a
hange of state. letting two wives
ross will result in a state in whi
h all three husbands
are at the left bank. we should always
he
k the validity of su
h properties.ed. For example.
This strategy is realised by de
omposing S0 into three sequen
es S1 .
The sequen
e S1
hanges the state from the start state to the state where all the wives are
at the right bank and all the husbands are at the left bank.
Our strategy
an be summarised as exploiting two properties of the problem.4
Problem Decomposition
Using this notation we
an express our strategy for de
omposing the problem. Jealous Couples
3.
{ 3H || 3W } S2 { 3W || 3H } .
The fa
t that it is easy to get the wives from one side to the other whilst their
husbands remain on one bank.
Finally. doing S1 followed by S2 followed by S3 .
{ 3W || 3H } S3 { || 3C } .3. whi
h we denote by
S1 . will a
hieve the obje
tive of
hanging the state from the start state (everyone
is at the left bank) to the . S3 .31
3. S2 and S3 su
h
that
{ 3C || } S1 { 3H || 3W } .
The left-right symmetry. S2 . the sequen
e S3
hanges the end state of S2 to the state where everyone is
at the right bank. The goal
is to
onstru
t a sequen
e of a
tions S0 satisfying
{ 3C || }
S0 { || 3C } .3. So. The sequen
e S2
hanges the
end state of S1 to the state where the positions of the wives and husbands are reversed.
but. getting all the wives a
ross the river. its task is to transfer all the wives from the right bank to the left bank. if we
onstru
t S1 .
Algorithmi
Problem Solving
Roland Ba
khouse. 2008
. Symmetry is
aptured by
making the fun
tion of S3 entirely symmetri
al to the fun
tion of S1 . an odd number of
rossings will be ne
essary.nal state (everyone is at the right bank).
We now have to ta
kle the problem of
onstru
ting S1 and S2 .
learly.
The de
omposition is into three
omponents be
ause we want to exploit symmetry. (It is a problem that
an be solved by brute for
e. If we
onsider the
reverse of S3 .
As mentioned earlier.
May 28.)
Here is how it is a
hieved. it is a simple task to
onstru
t S3 dire
tly from it.
So. leaving their husbands at
the left bank is easy. if ne
essary.
The length of S2 must also be odd.
T2
{
3W || 3H
} .nishing
position is the left bank. This is a requirement for S2 to follow S1 and be followed by
S3 .
Again. If the solution is to
remain symmetri
.
1C |1C| 1C
.
Algorithmi
Problem Solving
Roland Ba
khouse. 2008
. it must surely take the following form:
{
3H || 3W
}
T1
.
May 28. we look for a de
omposition into three subsequen
es.
so that
you
an fully appre
iate how mu
h more ee
tive it is than brute-for
e sear
h.34
3. not all intermediate states are shown. but not every step. by re
ording the main steps.
(In this solution. Crossing a River
.2H . |2W| 1C.5
A Review
Pause awhile to review the method used to solve the jealous-
ouples problem. This helps to do
ument the
solution. ea
h de
omposition is into three subsequen
es. 1W |1W| 1C.2H
{
|| 3C
}
.
The
onstru
tion seeks at ea
h stage to exploit the symmetry between the left and
right banks.3. Sin
e the number of
rossings will inevitably be odd.)
3. Too mu
h detail in program
do
umentation
an be a hindran
e. where the .
{
3W || 3H
}
1W |2W| 3H .
Naming the unknown sequen
es.
The . and to avoid error.rst and last are \mirror images" in some
sense. and formally spe
ifying their fun
tion using the
{ p } S { q } notation helps to
larify what has to be a
hieved.
But. the solution to the
problem
annot be used in other
ontexts. solve the following related problem:
There are . For the proof
of the pudding. it is easy to remember.nal solution involves eleven
rossings. Moreover.
making a re
onstru
tion of the solution very simple. be
ause the solution method is well stru
tured. That's too many for anyone to
ommit
to memory. but the solution method
an.
Determine how to transport all the
ouples
a
ross the river. Determine a solution to this
problem. The
ase that there are four jealous
ouples. and their boat
an
arry a maximum of three.1 (Five-couple Problem)
2
Unfortunately. and their boat
an
arry a maximum of three individuals. but it is not symmetri
. the symmetry between the left
and right banks does not guarantee that every river-
rossing problem has a symmetri
solution. has a solution. Four is less than .
The following hint may be helpful.ve jealous
ouples.
Exercise 3.
ve. you will have
solved the problem of transporting . by now. and.
So.1). try to
modify the solution for .ve
ouples a
ross the river (exer
ise 3.
You should be able to
.ve
ouples to obtain a solution for four.
nd two solutions in this way. one being obtained from the other by reversing left and
right.2 (Four-couple Problem)
Algorithmi
Problem Solving
Roland Ba
khouse.
May 28. 2008
.
Exercise 3.
A solution is symmetri
if this transformation maps the solution
to itself. it is impossible to
transport six or more
ouples a
ross the river. It is
alled a Hoare
triple.
Hint: Both problems
an be handled together. if the boat
an hold a maximum of two people.
Show that.
2
3.3
At most half of the husbands
an
ross together.
The boat
an only hold one
ouple. there is a transformation from solutions to solutions based on
reversing left and right.4. if the boat
an hold a maximum of three people. it is impossible
to transport four or more
ouples a
ross the river. individual solutions need not be symmetri
.4
Rule of Sequential Composition
The { p } S { q } notation we used for solving the jealous-
ouples problem is the
notation used for spe
ifying and
onstru
ting
omputer programs. That is.)
2
Show that.3. he was one of the . (Sir Tony Hoare is a British
omputing s
ientist who pioneered te
hniques for
formally verifying the
orre
tness of
omputer programs. The
ru
ial properties are:
Exercise 3. but the set of solutions
is symmetri
. Rule of Sequential Composition
35
(In general.
rst to use
the notation.)
A
omputer program is spe
i.
The allowed input values are spe
i.ed by a relation between the input values and the output
values.
and the
output values are spe
i.ed by a so-
alled pre
ondition. p .
a program to
ompute the remainder r and dividend
d after dividing number M by number N would have pre
ondition
N 6= 0 .
Algorithmi
Problem Solving
Roland Ba
khouse.
If S is a program. For example.
May 28. the program variables will
satisfy property q . Pre
onditions and post
onditions
are properties of the program variables. and p and q are properties of the program variables. if the program variables satisfy property p before exe
ution of statement
S . q .ed by a post
ondition .
fpgSfqg
means that. afterwards. 2008
. exe
ution of S is guaranteed to terminate and.
Crossing a River
(sin
e dividing by 0 is not allowed) and post
ondition
M = N×d + r ∧ 0 ≤ r < N .36
3. the spe
i.
If the program is S .
ation of the program is thus
{ N 6= 0 } S { M = N×d + r ∧ 0 ≤ r < N } . S1 . S2 . if S1 . Thus. A semi
olon is usually used to denote sequen
ing. S2 and
S3 are programs. the individual
omponents are exe
uted one
after the other.
Programs are often
omposed by sequen
ing. S3 denotes the program that is exe
uted by .
This is
alled the sequential
omposition
of S1 . and then exe
uting S3 . an intermediate
ondition.
then exe
uting S2 . S2 and S3 .
A sequential
omposition is introdu
ed into a program when the problem it solves
is de
omposed into subproblems. is
invented.
given a pre
ondition p and a post
ondition q . The problem of
onstru
ting a program S satisfying the spe
i. In the
ase of a de
omposition into two
omponents. r say.rst exe
uting S1 .
ation
fpgSfqg
is then resolved by letting S be S1 . S2 and
onstru
ting S1 and S2 to satisfy the
spe
i.
This is what we did in solving the jealous-
ouples problem. The intermediate
onditions
3H || 3W and 3W || 3H were then introdu
ed in order to make the .
If the problem is de
omposed into three subproblems. The initial problem
statement has pre
ondition 3C || and post
ondition || 3C .
ations
fpgS frg
1
and
frgS fqg
2
. two intermediate
onditions are
needed.
The intermediate
ondition r thus a
ts as post
ondition for S1 and pre
ondition for S2 .
There are dierent ways of using the rule of sequential
omposition. The stru
ture of
the given problem may suggest an appropriate intermediate
ondition. the task is then to identify
the intermediate
ondition and the .rst de
omposition. Alternatively. the
problem may suggest an appropriate initial
omputation S1 .
then the task be
omes one of identifying
the intermediate
ondition r and the initial
omputation S1 .
May 28. Re
all its statement:
Algorithmi
Problem Solving
Roland Ba
khouse.nal
omputation S2 . 2008
.
A
on
rete illustration is the bridge problem posed in
hapter 1.
when two
ross
together they must pro
eed at the speed of the slowest.4. Rule of Sequential Composition
Four people wish to
ross a bridge.
The bridge is narrow and only two people
an be on it at any one time. It is dark. The
four people take dierent amounts of time to
ross the bridge. and it is ne
essary to use a
tor
h when
rossing the bridge.37
3. The . but they only have one tor
h between them.
It is easy to see that .rst person takes
1 minute to
ross. the third 5 minutes and the fourth
10 minutes. the se
ond 2 minutes. The tor
h must be ferried ba
k and forth a
ross the bridge. so
that it is always
arried when the bridge is
rossed.
Show that all four
an
ross the bridge within 17 minutes.
ve trips are needed to get all four people a
ross the bridge.
The .
The times have been
hosen in this example so that the shortest time is a
hieved
when the two slowest
ross together. Sin
e there are only two return trips.) Using the times as identi. The question is
whether they should
ross together or seperately. See exer
ise 3. to a
hieve
the shortest time. the two slowest should not make a return journey. at least two people never make a return trip. Clearly. (It isn't always the
ase that the best strategy is
to let the two slowest
ross together.ve trips
omprise three trips where two people
ross together interleaved with two
trips where one person returns with the tor
h.4. and
there are four people.
5.5. the task be
omes one of determining sequen
es
of
rossings S1 and S2 .10 || q } p |5.10 || q }
{ p.10 } .10| q
as one step.5.10 }
and
{ p || q.10 || } S1 { p. the solution will therefore in
lude.5.ers for
the individuals.
Consider the tor
h problem in the
ase that the
rossing times of the four individuals form the input parameters. Seeking to exploit symmetry.5. Suppose the .
We leave you to
omplete the rest.5. for some p and q .10 } S2 { || 1. and p and q su
h that
{ 1. a
rossing of the
form
p |5.2.10| q { p || q.2.
in an optimal solution.
Apply your solution to the following two
ases:
Exercise 3. 2008
.rst person
takes t1 minutes to
ross. You may assume that. every
forward trip involves two people and every return trip involves one person. the third t3 minutes and the fourth
t4 minutes. Find a method of getting all four a
ross that
mimimises the total
rossing time.4 (The Torch Problem)
Algorithmi
Problem Solving
Roland Ba
khouse.
May 28. Assume that t1 ≤ t2 ≤ t3 ≤ t4 . the se
ond t2 minutes.
3 minutes and 3 minutes. 4 minutes.
Hint: In the spe
i. 1 minute.
(b) The times taken are 1 minute. Crossing a River
(a) The times taken are 1 minute.38
3. 4 minutes and 5 minutes.
this isn't always the best strategy. 5 and
10 .
Note that this exer
ise is mu
h harder than the spe
i. An alternative strategy is to let the fastest
person a
ompany ea
h of the others a
ross in turn. You will need to evaluate the time
taken for both strategies and
hoose between them on the basis of the times. t3 and t4 are 1 .)
(a) How many times must the tor
h be
arried a
ross the bridge in order to get all four
people a
ross? (In
lude
rossings in both dire
tions in the
ount. In order
to derive the solution methodi
ally. What are the dierent strategies? Evaluate the time taken for ea
h. Use a
onditional statement to de
ide whi
h
strategy to use. Hen
e.
(d) Give a solution to the general problem. what
an you say about the number of people who do not make a return
trip? Whi
h of the four people would you
hoose not to make a return trip? (Give
a
onvin
ing argument to support your
hoi
e. (Note that steps (a)
and (b) were already dis
ussed above. we suggest the following steps. t2 . 2 .) How many of
these are return trips?
(b) Comparing the number of times a return journey must be made with the number of
people.)
(c) Now fo
us on how to get the people who do not make a return trip a
ross the
bridge.
ase dis
ussed above (where t1 .
onstru
t a formula for the minimum time needed to get all four people a
ross in
the general
ase. the shortest time is a
hieved by letting the two slowest
ross together.
However. respe
tively).
ase dis
ussed above.
May 28. The original problem was
arefully
worded so that this was not required. This would mean enumerating all the dierent ways of getting four people
a
ross the bridge.4 above). 2008
. How many ways are there?
Exercise 3.
2
Suppose a brute-for
e sear
h is used to solve the tor
h problem (exer
ise
3.5
2
Algorithmi
Problem Solving
Roland Ba
khouse. This is not
just be
ause the input times are parameters but primarily be
ause you must establish
without doubt that your solution
annot be bettered.
problems.39
3. Summary
3. This suggests ta
kling the problems by de
omposing them into three
omponents. Complexity theory. or similar. lends greater for
e to this argument. we have
ontrasted brute-for
e sear
h with problem de
omposition. no matter how mu
h
bigger and faster
omputers be
ome. there is a symmetry between the left and right
banks.
Brute-for
e sear
h should only be used as a last resort.5. Modern
omputer te
hnology
means that some problems that are too large for human beings to solve do be
ome solvable.5
Summary
In this
hapter.
even with the most powerful
omputers. whi
h you will study in a
ourse on algorithm design.
In all river-
rossing. they
an never
ompete with the in
rease in size of
the state spa
e
aused by modest in
reases in problem size. but the state-spa
e explosion makes the method impra
ti
al for realisti
problems.
Problem de
omposition seeks to exploit the inherent stru
ture of the problem domain.
the .
Along the way. they be
ome more
manageable. Unfortunately. in
ontrast to the
solutions. if the problems are ta
kled in this way.
Most importantly. and often have
lear. whi
h are only relevant to one parti
ular problem. easily remembered and easily reprodu
ed solutions. the issue of de
iding what to name (and what not to name) has
emerged as an important problem-solving skill that
an have signi.rst and last being symmetri
al in some way. the solution method
an be applied repeatedly. but. this strategy has no
guarantee of su
ess.
we abstra
t the few that
are relevant.
ant impa
t on the
omplexity of the problem. The pro
ess is
alled abstra
tion | from the myriad of
details that surround any real-world des
ription of a problem. introdu
ing appropriate.
learly de.
(Chess. The game ends when it is no longer possible to make
a move.1
Matchstick Games
A mat
hsti
k game is played with one or more piles of mat
hes. it introdu
es the importan
e of algebra in problem
solving. A winning strategy is then what we
all \maintaining an invariant".
The key to winning is the re
ognition of invariants. a
ording to a given rule. it will mean ensuring that the opponent is always pla
ed
in a position from whi
h losing is inevitable. Games provide very
good examples of algorithmi
problem solving be
ause playing games is all about winning. \Maintaining an invariant" is an important te
hnique in
algorithm development. in essen
e. The
hapter is also about trying to identify and exploit
stru
ture in problems. The goal is to have some method (i. in order to give
the
avour of the games that we
onsider.
41
May 28.
A mat
hsti
k game is an example of an impartial. Following it. for example.
The next se
tion introdu
es a number of games with mat
hsti
ks.e. and
bla
k
an only move bla
k pie
es. Moves involve removing one or more mat
hes from one of the
piles. be
ause white
an only move white pie
es. and the other player is the
winner. this
hapter is
a
ontinuation of
hapter 2. Two players take it
in turn to make a move.
4.Chapter 4
Games
This
hapter is about how to win some simple two-person games.) \Complete information" means that both players
Algorithmi
Problem Solving
Roland Ba
khouse. 2008
. two-person game with
omplete
information. In this sense. is not impartial. we develop a method of systemati
ally identifying winning and losing positions in a game (assuming a number of
simplifying
onstraints on the rules of the game). \algorithm") for de
iding what to do so that
the eventual out
ome is a win. \Impartial" means that rules for moving apply equally to both players. So. Here. The player whose turn it is to move is the loser.
e. 9 et
. If m is the number of mat
hes in su
h a position
(so. 6 . The losing positions are the positions where the number of mat
hes
is a multiple of 3 (that is. in
ard games like poker.). it is always possible to
hara
terise the positions as either winning or losing positions. Games
know the
omplete state of the game. This means that there are either 0 mat
hes
left. the number of mat
hes is 0 . The opponent is then put in a position where
the number of mat
hes is a multiple of 3 . In
ontrast. This is
either 1 or 2 .
In an impartial game that is guaranteed to terminate no matter how the players
hoose their moves (i. and an allowed move is to remove
1 or 2 mat
hes. The remaining
positions are the winning positions. or any move they make will result in there again
being a number of mat
hes remaining that is not a multiple of 3 .
that guarantees a win. The following exer
ises
ask you to do this in spe
i.42
4.
A winning position is one from whi
h a perfe
t player is always assured of a win.
As an example. it is usual
that ea
h player does not know the
ards held by the other player. the strategy is to remove m mod 3 mat
hes1. A
losing position is one from whi
h a player
an never win. 3 . and so the move is valid. when playing against a perfe
t
player. A winning strategy is an algorithm for
hoosing moves from winning positions. in whi
h
ase the opponent loses. the players have
in
omplete information about the state of the game. suppose there is one pile of mat
hes. the possibility of stalemate is ex
luded). m is not a multiple of 3 ).
Ea
h player is allowed to remove 1 mat
h.
ases.
an you
see how to win a game in whi
h an allowed move is to remove at least one and at
most N mat
hes. There is one pile of mat
hes. Ea
h player is allowed to remove 0 mat
hes. What are
the winning positions?
2. There is one pile of mat
hes.
1. What
are the winning positions?
3. Can you see a pattern in the last two problems and the example dis
ussed above
(in whi
h a player is allowed to remove 1 or 2 mat
hes)? In other words. where N is some number .
your
opponent removes 1 mat
h.
What are the winning positions and what is the winning strategy?
5. 3 or 4 mat
hes. There is one pile of mat
hes.
ex
ept that it is not allowed to repeat the opponent's last move.)
What are the winning positions and what is the winning strategy?
1 Re
all
that
m mod 3
denotes the remainder after dividing
Algorithmi
Problem Solving
Roland Ba
khouse.xed in advan
e?
4. (So. Ea
h player is allowed to remove 1 . your next move must be to remove 3 or 4 mat
hes. 3 or 4 mat
hes.
m
by
3. 2008
.
May 28. if. Ea
h player is allowed to remove 1 . say. There is one pile of mat
hes.
We begin with the
simple mat
hsti
k game where a move is to remove one or two mat
hes from a single
pile of mat
hes. from the right pile.2. the algorithm does give a
better understanding of what is involved. What are the winning positions and what is the winning strategy?
8.1
Assumptions
We make a number of assumptions about the game. and thus not pra
ti
al for more
ompli
ated games. There are two piles of mat
hes. 2 or 3 mat
hes. What are the winning positions and what is the winning strategy?
4. A move is to
hoose one pile.
4. 1 . we show how to sear
h systemati
ally through all the positions of the
game.
remove 1 . labelling ea
h as either a winning or a losing position. from the left pile. in order that the sear
h will work.43
4. from that pile. and from the right pile 1 thru2 7 mat
hes may
be removed. 1 or 2 mat
hes may
be removed. A move is to
hoose one pile. and.
2 or 3 mat
hes may be removed. we formulate what is required of a winning strategy. Although a brute-for
e
sear
h. A move is to
hoose one pile and. from the left pile 1 .2. There are two piles of mat
hes.
We assume that the number of positions is .
3 or 4 mat
hes may be removed. Winning Strategies
6. What are the winning positions and what is the
winning strategy?
7. There are two piles of mat
hes. and
an be used as a basis for developing more
eÆ
ient solutions in parti
ular
ases.2
Winning Strategies
In this se
tion.
The .nite.
We assume that the game is guaranteed to terminate no matter how the players
hoose their moves.
rst assumption is ne
essary be
ause a one-by-one sear
h of the positions
an
never be
omplete if the number of positions is in.
The se
ond assumption is
ne
essary be
ause the algorithm relies on being able to
hara
terise all positions as
either losing or winning.nite.
Stalemate positions are ones from whi
h the players
an
ontinue the game inde. we ex
lude the possibility that there are stalemate positions.
2 We
use \thru" when we want to spe
ify an in
lusive range of numbers.
Algorithmi
Problem Solving
Roland Ba
khouse.
May 28.nitely.
so that neither wins or loses. 3 and 4 . 2 . The English expression \ 1 to 4 " is ambiguous about whether the
number 4 is in
luded or not. \ 1 thru 4 "
means the numbers 1 . 2008
. For example.
2
Labelling Positions
The .44
4. Games
4.2.
Ea
h edge is from one node
to another node.1.rst step is to draw a dire
ted graph depi
ting all positions.1 is a graph of the mat
hsti
k game des
ribed at the beginning of se
tion
4.
0
1
2
3
4
5
6
7
8
Figure 4.
The nodes in .
A dire
ted graph has a set of nodes and a set of edges. Fig. When graphs are drawn.1: Mat
hsti
k Game. and all moves in the
game. and edges are
depi
ted by arrows pointing from the from node to the to node. 4. Players may take one or two mat
hes at ea
h turn. nodes are depi
ted by
ir
les.
Having drawn the graph. It is impossible to move from
the position in whi
h no mat
hes remain. there is
only one move that
an be made.g. From the node labelled n . one is allowed to
remove one or two mat
hes. there is
an edge to the node labelled n−1 and an edge to the node labelled n−2 . there is exa
tly one
edge. From all other
nodes. there are two edges. 4. From the position in whi
h one mat
h remains. namely to remove the remaining mat
h. That is.1 are labelled by a number. A player who . where n is at least 2 . there are no edges. from
a position in whi
h the number of remaining mat
hes is at least 2 . the number of mat
hes remaining in
the pile. From the node labelled 0 . to the node labelled 0 . we
an begin labelling the nodes as either losing positions
or winning positions. From the node labelled 1 .
A player who . if playing against a perfe
t opponent.nds themself in a losing position will inevitably
lose.
provided the right
hoi
e of move is made at ea
h turn. one for losing positions. the other for winning
positions:
A node is labelled losing if every edge from the node is to a winning position.nds themself in a winning
position is guaranteed to win.
At .
The labelling rule has two parts.
A node is labelled winning if there is an edge from the node to a losing position.
after
all.rst sight. it may seem that it is impossible to begin to apply these rules. the .
rst rule de.
This is be
ause. the statement \every edge from the node is to a winning position" is
true.nes losing positions in terms of winning positions.
May 28. It is indeed the
ase that all of the (non-existent) edges is to a winning position.
Algorithmi
Problem Solving
Roland Ba
khouse. 2008
. if there are no
edges from a node. whilst the se
ond
rule does the reverse. we
an begin by labelling
as losing positions all the nodes with no outgoing edges. It seems like a vi
ious
ir
le! However.
2. A statement of the form \every x has
property p " is what is
alled a for-all quanti. Winning Strategies
This is an instan
e of a general rule of logi
.45
4.
ation. or a universal quanti.
ation.
Su
h a statement is said to be va
uously true when there are no instan
es of the \ x "
in the quanti.
In a sense. empty) be
ause it is a
statement about nothing.e.
Returning to .
ation. the statement is \va
uous" (i.
nodes 1 and 2 are labelled \winning".
Next. It is indeed a losing position.1. whi
h we know to be a losing position. be
ause the rules of the game spe
ify that a player
who
annot make a move loses. be
ause there are no edges
from it. from ea
h. be
ause.g. there is an edge to
0 . 4. the node 0 is labelled \losing". Note that the edges we have identi.
every move is to a position starting from whi
h a win is guaranteed. node 3 is labelled \losing".
Now.ed di
tate
the move that should be made from these positions if the game is to be won. be
ause both edges from node 3 are to nodes ( 1
and 2 ) that we have already labelled \winning". From a position in whi
h there are 3
mat
hes remaining.
A player that .
a pattern is emerging from this pro
ess. The winning
positions are the remaining positions.
The pro
ess we have des
ribed repeats itself until all nodes have been labelled.2 shows the state of the labelling pro
ess at the point that node 7 has been
labelled but not node 8 . The
ir
les depi
ting losing positions are drawn with thi
k
lines.nds themself in this position will eventually lose.2.3
Formulating Requirements
The terminology we use to des
ribe the winning strategy is to \maintain invariant"
the property that the number of mat
hes is a multiple of 3 .
Clearly. In programming terms. the
ir
les depi
ting winning positions are the ones from whi
h there is an edge
drawn with a thi
k line.
Algorithmi
Problem Solving
Roland Ba
khouse.
Fig. then nodes 7 and 8
are labelled \winning". 2008
.
0
1
2
3
4
5
6
7
8
Figure 4. Nodes
4 and 5 are labelled \winning". the winning strategy is to remove one or two
mat
hes so as to leave the opponent in a position where the number of mat
hes is on
e
again a multiple of 3 .
May 28. and so on. Winning edges are indi
ated by thi
k edges. 4.2: Labelling Positions. The pattern is that the losing
positions are the ones where the number of mat
hes is a multiple of 3 . then node 6 is labelled \losing".
4. These edges depi
t the winning move from that position.
the
orre
tness of the winning strategy is expressed by the following
annotated program segment:
{
n is a multiple of 3 .46
4. Games
we express this property using Hoare triples. Let n denote the number of mat
hes in
the pile.
{
2
}
2 ≤ n → n := n−2 fi
n is not a multiple of 3 }
n := n − (n mod 3)
{
n is a multiple of 3 }
There are . and n 6= 0
if 1 ≤ n → n := n−1
. Then.
ve
omponents of this program segment. The . ea
h on a separate line.
Within these bra
kets is a non-deterministi
hoi
e |
indi
ated by the \ 2 " symbol| among a number of so-
alled guarded
ommands. The triple. hen
e. This expresses the assumption that we begin from a position in
whi
h the number of mat
hes is a multiple of 3 . Starting in a given state.
omprising the . Removing one mat
h is only allowed if 1 ≤ n .fi statement in the se
ond line models an arbitrary move. At least one of these guards. where b is a boolean-valued expression
alled
the guard.
In this way. A
guarded
ommand has the form b → S . If none of the guards evaluates to true . If several guards evaluate to true .
and then exe
uting its body. the if . is true be
ause of the assumption that n 6= 0 . and
possibly both. removing two mat
hes |modelled by the assignment
n := n−2 | is \guarded" by the
ondition 2 ≤ n . and S is a statement
alled the body. Conditional statements are
re
ognised by \ if . the statement n := n−1 is \guarded"
by this
ondition. a
onditional
statement is exe
uted by
hoosing a guarded
ommand whose guard evaluates to true .
The se
ond line is a so-
alled
onditional statement.
The post
ondition of the guarded
ommand is the assertion \ n is not a multiple
of 3 ". Similarly. an arbitrary
hoi
e of
ommand is made.fi " bra
kets. and non-zero. exe
ution is aborted3 .rst
line is the pre
ondition.
spe
ifi
ally. thus asserts that. and a valid move is made that redu
es the number of mat
hes
by one or two. then.
The fourth line of the sequen
e is the implementation of the winning strategy.rst three lines. on
ompletion of the move. The . remove n mod 3 mat
hes. the number of mat
hes will not be a
multiple of 3 . if the number of
mat
hes is a multiple of 3 .
fth line is the .
2008
. the number of mat
hes will again be a
multiple of 3 .
Algorithmi
Problem Solving
Roland Ba
khouse.
3 If
you are already familiar with a
onventional programming language. as used here. In su
h statements. the
hoi
e
of whi
h of the optional statements should be exe
uted is
ompletely determined by the state of the
program variables.nal post
ondition. whi
h asserts
that. after exe
ution of the winning strategy. In a non-deterministi
hoi
e.
May 28. the
hoi
e is not
ompletely determined. you will be familiar with
deterministi
onditional statements | so-
alled if-then-else statements.
47
4. For ea
h winning position. one has to identify a way of
al
ulating
a losing position to whi
h to move. results in a state in whi
h n is not a multiple of 3. the losing and winning positions. beginning from a state in whi
h n is a multiple of 3 . Winning Strategies
In summary. a winning strategy is obtained by
hara
terising the losing positions by
some property.
removing n mod 3 mat
hes results in a state in whi
h n is again a multiple of 3 . losing say. and the winning strategy must satisfy
the following spe
i. Subsequently. The winning positions are then the positions that do
not satisfy losing .2. the algorithm that is used is the winning strategy. and making
an arbitrary move. The end positions (the positions where the game is over)
must satisfy the property losing .
In general.
More formally.
{ losing position. If the
starting position is a losing position.
From a losing position that is not an end position. and not an end position }
make an arbitrary (legal) move
. if the starting position is a winning position. i.
ation. in su
h a way that the
following three properties hold:
End positions are losing positions. the . the losing positions and the winning positions. the se
ond player is guaranteed to win.
From a winning position.
If both players are perfe
t. not a losing position }
apply winning strategy
{ losing position }
In summary. resulting in a losing position. every move is to a winning
position. the winner is de
ided by the starting position.
{ winning position. Vi
eversa. it is always possible to apply the winning strategy. a winning strategy is a way of
hoosing moves that divides the positions
into two types.e.
Starting from a losing position.
Algorithmi
Problem Solving
Roland Ba
khouse. by assumption. and positions where the number
of moves remaining is zero. this quantity will always be true
in a game played by perfe
t players.
May 28. a winning strategy maintains invariant the boolean quantity
(the number of moves remaining before the game ends is even) equals (the
position is a losing position).
Formally.
and will make a mistake. one
an only hope that one's opponent is not perfe
t. whi
h is an even number.rst player is guaranteed to win. 2008
.
Sin
e end positions are losing positions.
48
4. The
number M is a natural number. Games
We re
ommend that you now try to solve the mat
hsti
k-game problem when the
rule is that any number of mat
hes from 1 thru M may be removed at ea
h turn. .
xed in advan
e. We re
ommend that you try to solve
this general problem by .
rst
onsidering the
ase that M is 0 . Next.
Constru
t a table instead. try working out the
ase that M is 3 . This
ase also has an easy solution. whi
h is the
ase we have just
onsidered.
onsider the
ase that M
is 1 . As a last resort. Now. but slightly more
ompli
ated. This
ase has a very easy
solution.
Do you see a pattern in the solutions? If you don't see a pattern immediately. A diagram is mu
h too
ompli
ated. 1 and 2 (in parti
ular. try a bit
harder. the extreme
ases 0 and 1 ) in order to
he
k the pattern you have identi. return to the
ases that M is 0 . although it is a
ase that is very often negle
ted.
ombine
these two
ases with the
ase that M is 2 . (Don't draw a diagram.) Then.
you have to determine for ea
h of 365 days (or 366 in the
ase of a
leap year) whether naming the day results in losing against a perfe
t player. Finally. For ea
h.
Two players alternately name dates.
Ea
h part of this exer
ise uses a dierent rule for the dates that a player is allowed to
name. stating whi
h player should win. State also
if it depends on whether the year is a leap year or not. The
winner is the player who names 31st De
ember. In pra
ti
e.
Hint: in prin
iple.
Exercise 4. or in
rease the day of the month
by an arbitrary amount. and the starting date is 1st January. Begin by identifying the days in De
ember that one should
avoid naming. the . devise a winning strategy. formulate the
orre
tness of the strategy
by a sequen
e of assertions and statements.ed. (For example.1 (31st December Game)
a) (Easy) A player
an name the 1st of the next month.
a pattern soon be
omes evident and the days in ea
h month
an be grouped together
into winning and losing days. as we did above for the
ase that M is 2 .
3
Subtraction-Set Games
A
lass of mat
hsti
k games is based on a single pile of mat
hes and a (.
or a date in January other than the 1st.rst player begins by naming 1st February.)
b) (Harder) A player
an in
rease the day by one. leaving the month un
hanged.
2
4. or name
the 1st of the next month.
nite) set of
numbers. a move is to remove m mat
hes. A
game in this
lass is
alled a subtra
tion-set game. and the set of numbers is
alled the
subtra
tion set.
May 28.
Algorithmi
Problem Solving
Roland Ba
khouse. 2008
. where m is an element of the given set.
If r is 0 or 2 .
Beginning with position 0 . Subtra
tion-Set Games
49
The games we have just dis
ussed are examples of mat
hsti
k games.2. For example. 3 . the position is a losing
Algorithmi
Problem Solving
Roland Ba
khouse.
Continuing this pro
ess. Table 4. and
a position is a winning position if there is a move from it to a losing position. We may therefore
on
lude that. or remove 1 mat
h to move to position 2 . we noti
e that the pattern of winning and losing
positions repeats itself.
In other words.
More interesting examples are obtained by
hoosing a subtra
tion set with less regular
stru
ture.1 shows the entries when the size of the pile
is at most 6 .
remove four mat
hes. \position 1 " means the position in whi
h there is just
one mat
h in the pile. the subtra
tion set is {1 .2.M} . the subtra
tion set is {1. 4} .4. the winning and losing positions
an always be
omputed. after dividing the number of mat
hes by 7 . positions 3 and 4 are winning positions be
ause from both a
move
an be made to 0 . 2008
. 4} . it will
ontinue to
do so forever. We refer to
the positions using this number. In the
ase that the position is a winning position. 3 .
The results are entered in a table. if the rule is
that 1 thru M mat
hes may be removed at ea
h turn. we get the next seven entries in the table: see table 4. We exemplify the pro
ess in this se
tion by
al
ulating the winning and losing
positions when the allowed moves are:
remove one mat
h. For example.1 and 4.. So. \position 0 " means the position in whi
h there are
no mat
hes remaining in the pile.
r say. Note that there may be a
hoi
e of winning move. and the middle row shows whether or not it is
a winning (W) or losing position (L). for the subtra
tion set {1 .
Positions in the game are given by the number of mat
hes in the pile.
Comparing tables 4. It suÆ
es to enter just one move in the
bottom row of the table.
from position 3 there are two winning moves | remove 3 mat
hes to move to position
0 . whether
or not the position is a winning position
an be determined by
omputing the remainder. we identify
whether ea
h position is a winning position using the rules that
a position is a losing position if every move from it is to a winning position.
May 28.
the bottom row shows the number of mat
hes that should be removed in order to move
from the position to a losing position.
For any given subtra
tion set. and working one-by-one through the positions. On
e the pattern begins repeating in this way.
remove three mat
hes. and so on.3. 2 is a losing position be
ause the
only move from 2 is to 1 . The top row is the position.
Table 4.2: Winning (W) and Losing (L) Positions for subtra
tion set {1 , 3 , 4}
position. Otherwise, it is a winning position. The winning strategy is to remove 1 mat
h
if r is 1 , remove 3 mat
hes if r is 3 or 5 , and remove 4 mat
hes if r is 4 or 6 .
The repetition in the pattern of winning and losing positions that is evident in this
example is a general property of subtra
tion-set games, with the
onsequen
e that, for a
given subtra
tion set, it is always possible to determine for an arbitrary position whether
or not it is a winning position (and, for the winning positions, a winning move). The
following argument gives the reason why.
Suppose a subtra
tion set is given. Sin
e the set is assumed to be
nite, it must
have a largest element. Let this be M . Then, from ea
h position, there are at most
M moves. For ea
h position k , let W.k be true if k is a winning position, and false
otherwise. When k is at least M , W.k is
ompletely determined by the sequen
e
W.(k−1) , W.(k−2) , . . . , W.(k−M) . Call this sequen
e s.k . Now, there are only 2M
dierent sequen
es of booleans of length M . As a
onsequen
e, the sequen
e s.(M+1) ,
s.(M+2) , s.(M+3) , . . . must eventually repeat, and it must do so within at most 2M
steps. That is, for some j and k , with M ≤ j < k < M+2M , we must have s.j = s.k . It
follows that W.j = W.k and the sequen
e W repeats from the k th position onwards.
For the example above, this analysis predi
ts that the W-L pattern will repeat from
the 20 th position onwards. In fa
t, it begins repeating mu
h earlier. Generally, we
an say that the pattern of win-lose positions will repeat at position 2M+M , or before.
To determine whether an arbitrary position is a winning or losing position involves
omputing the status of ea
h position k , for su
essive values of k , until a repetition in
s.k is observed. If the repetition o
urs at position R , then, for an arbitrary position
k , W.k equals W.(k mod R) .
Suppose there is one pile of mat
hes. In ea
h move, 2 , 5 or 6 mat
hes
may be removed. (That is, the subtra
tion set is {2 , 5 , 6} .)
Exercise 4.2
(b) Identify a pattern in the winning and losing positions. Spe
ify the pattern by giving
pre
ise details of a boolean fun
tion of n that determines whether a pile of n
mat
hes is a winning position or not.
Verify the pattern by
onstru
ting a table showing how the fun
tion's value
hanges
when a move is made.
2
This exer
ise is
hallenging; its solution involves thinking beyond the
material presented in the rest of the
hapter.
Figure 4.3 shows a variant of snakes and ladders. In this game, there is just one
ounter. The two players take it in turn to move the
ounter at most four spa
es forward.
The start is square 1 and the
nish is square 25 . the winner is the .
rst to rea
h the
.
)
(b) Identify the winning and losing positions. (These are not the same as the squares. it falls down to the tail of the snake.
(c) Some of the positions
annot be identi. if the
ounter lands at the foot of a ladder. Use the rule that a losing position is one
from whi
h every move is to a winning position.nish. Think
arefully about squares linked by a snake or a ladder. if the
ounter lands on the head of a
snake.
Exercise 4. As in the usual game of snakes and ladders.3
(a) List the positions in this game. and a winning position is one from
whi
h there is a move to a losing position.
it
limbs to the top of the ladder.
we look at how to exploit the stru
ture of a game in order to
ompute a
winning strategy more ee
tively.1 have more than one pile of
mat
hes.4
Sums of Games
In this se
tion.ed as winning or losing in this way.
2
4.
The later examples of mat
hsti
k games in se
tion 4. one of the piles must . Explain
why. When a move is made.
Algorithmi
Problem Solving
Roland Ba
khouse. 2008
. The game is thus a
ombination of two games. mat
hes
may be removed from the
hosen pile a
ording to some pres
ribed rule. this parti
ular way of
ombining games is
alled summing the games.
May 28. whi
h may dier
from pile to pile. then.rst be
hosen.
A move in the sum game is a move in one of
the games. and the moves are represented by the
edges.3: Snakes and Ladders. a position in the \sum" of the games is given by a pair
Xx where \X" names a node in the left graph. A move is then to
hoose one of the
oins. given two games ea
h with its own rules for making a move. two
oins are used. Ea
h graph represents a game.
In the \sum" of the games.
Both the left and right games in . the sum of
the games is the game des
ribed as follows. The nodes in the left graph and right graphs are named
by
apital letters and small letters. Players take it in turn to move the
ounter at most
four spa
es forward. Imagine a
oin pla
ed on a node. A position in the sum game is the
ombination of a position in the
left game and a position in the right game. and displa
e it along
an edge to another node.4 is an example of the sum of two games.
Figure 4. and \x" names a node in the right graph. one
oin being pla
ed over a node in
ea
h of the two graphs.
In general.
a move has the ee
t of
hanging exa
tly one of \X" or \x".
where the positions are represented by the nodes. so that we
an refer to them later. Games
21
22
23
24
25
20
19
18
17
16
11
12
13
14
15
10
9
8
7
6
1
2
3
4
5
Figure 4. we
all the two games the left
and the right game. respe
tively. Thus. A move is then to displa
e the
oin along one
of the edges to another node. For
larity.52
4.
2. Sums of Games
O
M
N
J
K
F
L
G
H
D
A
k
h
I
B
f
C
j
g
c
E
i
d
a
e
b
Figure 4. The left and right games are represented by the two graphs.53
4. and \x" is
the name of a node in the right graph.2 is unavoidable when determining their
winning and losing positions. However. A move
hanges exa
tly one of X or x.4.
A position is a pair Xx where \X" is the name of a node in the left graph. the left game in .
for
e sear
h pro
edure des
ribed in se
tion 4.4: A Sum Game.
4 has 15 dierent postions. and for sums of games in general. thus. we study how to
ompute a winning strategy for the sum of two games. We
.g.
For this game.
In this se
tion.
and the right game has 11 . the sum of the two games has 15×11 dierent positions. 4. a brute-for
e sear
h is highly undesirable.
We .nd that the
omputational eort is the sum (in the usual sense of addition of numbers)
of the eort required to
ompute winning and losing positions for the
omponent games.
rather than the produ
t.
nd.
May 28. 2008
. Deriving a suitable generalisation forms the
ore of the analysis. however.
Algorithmi
Problem Solving
Roland Ba
khouse. that it is not suÆ
ient to know just the
winning strategy for the individual games.
May 28. An allowed move is to
hoose any one of the piles and remove at least
one mat
h from the
hosen pile. m = n = 0 . game: given
a (single) pile of mat
hes. Suppose there are two
piles of mat
hes. there is an equal number of mat
hes in both piles.
This is indeed the
ase.
{
if
m 6= n
}
m < n → n := n − (n−m)
2 n < m → m := m − (m−n)
fi
Algorithmi
Problem Solving
Roland Ba
khouse. any move will be to a position where m 6= n . As always.
Formally. applies the winning strategy for the individual game| .
This game is the \sum" of two instan
es of the same. In this
simple game. In the end
position.54
4.
hoosing the pile with the larger number of mat
hes. the opponent
wins by removing the remaining mat
hes in the other pile. in the
end position. 2008
.4.
The symmetry between left and right allows us to easily spot a winning strategy. the winning positions are. If a player removes all the mat
hes from
one pile |that is. Otherwise. Subsequently. very.
{
if
m = n ∧ (m 6= 0 ∨ n 6= 0)
1≤m →
2 1≤n →
}
redu
e m
redu
e n
fi
. will restore the property that m = n . That is. the positions in whi
h the pile has at
least one mat
h. The position in
whi
h there are no mat
hes remaining is the only losing position. there is no restri
tion on the number of
mat
hes that may be removed. the game is lost when a player
annot make
a move. either 1 ≤ m or 1 ≤ n ). and the winning strategy is to remove all the mat
hes. the
orre
tness of the winning strategy is expressed by the following sequen
e of assertions and program statements. very simple. obviously.
It qui
kly be
omes
lear that knowing the winning strategy for the individual games
is insuÆ
ient to win the sum of the games. and removing the ex
ess mat
hes
from this pile.1
Symmetry
A Simple Sum Game
We begin with a very simple example of the sum of two games. From a position in whi
h m = n . This suggests that the losing positions are given by m = n . Games
4.
Suppose we let m and n denote the number of mat
hes in the two piles. and a move is possible (that
is. namely 0 . a move is to remove at least one mat
h from the pile.
Sin
e. we infer that 1 ≤ n−m ≤ n . In the
ase that m < n . in the
ase that 1 ≤ n .
The property m 6= n is the pre
ondition for the winning strategy to be applied. or n
hanges in value. and redu
ing
n . but not both. so
that n−m mat
hes
an be removed from the pile with n mat
hes. in the
ase that 1 ≤ m . The
fa
t that either m
hanges in value. n−(n−m)
simpli. m < n or n < m .4.
Equivalently.55
4. Sums of Games
{
m=n
}
The non-deterministi
hoi
e between redu
ing m . guarantees
m 6= n after
ompletion of the move. models an arbitrary
hoi
e of move in the sum game.
Who should win and what
is the winning strategy? Generalise your solution to the
ase that there are initially n
petals and a move
onsists of removing between 1 and M adja
ent petals (where M is
.4.
{
m 6= n
{
m<n ∨ n<m
if
}
}
m < n → { 1 ≤ n−m ≤ n } n := n − (n−m) { m = n }
2 n < m → { 1 ≤ m−n ≤ m } m := m − (m−n) { m = n }
fi
{
4. The pre
ondition expresses the legitima
y of the move. We see shortly that this is how to win
all sum-games. Note how the two assignments
have been annotated with a pre
ondition and a post
ondition.
The Daisy Problem Suppose a daisy has 16 petals arranged symmetri
ally around
its
entre. Here is a
ouple. after the assignment n := n−(n−m) . no matter what the individual
omponents are.1 is another example of the importan
e of symmetry. the property
m = n will hold. it is
lear that.
The following sequen
e of assertions and program statements summarises the argument just given for the validity of the winning strategy. the winning
strategy is to ensure that the opponent is always left in a position of symmetry between
the two individual
omponents of the sum-game. The solutions
an be found at the end of the book. The
ase n < m is symmetri
. A move involves removing one petal or two adja
ent
petals.4.es to m . The winner is the one who removes the last petal.
There are many examples of games where symmetry is the key to winning.2
m=n
}
Maintain Symmetry!
The game in se
tion 4. the post
ondition is the losing property that the
strategy is required to establish. There are two players.
56
4. Games
Figure 4.3
More Simple Sums
Let us return to our mat
hsti
k games. do you
assume about the
oins in order to justify your answer?
4.4. su
h that it does not overlap any
oin already on the table.5: A 16 -petal daisy
The Coin Problem Two players are seated at a re
tangular table whi
h initially is
bare. They ea
h have an unlimited supply of
ir
ular
oins of varying diameter. The winner is the one who puts the last
oin on the table.1 is
to restri
t the number of mat
hes that
an be removed. A variation on the sum game in se
tion 4. The
players take it in turns to pla
e a
oin on the table.4.
Who should win and what is the winning strategy? (Harder ) What. Suppose the restri
tion is that
at most K mat
hes
an be removed from either pile (where K is . if anything.
If. m and
n denote the number of mat
hes in the two piles. in advan
e). Consequently.
The ee
t of the restri
tion is to disallow some winning moves. as before. For example. if K is .xed. it is not allowed to remove m−n
mat
hes when K < m−n . the property m = n no longer
hara
terises the
losing positions.
the opponent
an
then remove the mat
h to win the game.
A more signi. the position in whi
h one pile has two
mat
hes whilst the se
ond pile has no mat
hes is a losing position: in this position a
player is for
ed to move to a position in whi
h one mat
h remains.xed at 1 .
Worse is if we break symmetry further by imposing
dierent restri
tions on the two piles: suppose. we impose the limit M on
the number of mat
hes that may be removed from the left pile. where M 6= N .
ant ee
t of the restri
tion seems to be that the strategy of establishing
symmetry is no longer appli
able. for example.
May 28. suppose
Algorithmi
Problem Solving
Roland Ba
khouse. Alternatively. and N on the number of
mat
hes that may be removed from the right pile. 2008
.
for a pile of
m mat
hes. if one is a mat
hsti
k game
and the other is the daisy game. the property is satis. for example. This suggests that. is to
ontinually establish the property that
the remainder after dividing the number of mat
hes by M+1 is 0 .
( M is the maximum number of mat
hes that
an be removed from the left pile. \symmetry" between the piles
is formulated as the property that
m mod (M+1) = n mod (N+1) .)
This. a form of \symmetry" is a key to the winning strategy: symmetry
is too important to abandon so easily!
We saw. the number m mod (M+1) determines whether the position is a winning
position or not. in the two-pile game.4. is the
orre
t solution. and N
is the maximum number that
an be removed from the right pile. Sums of Games
the left and right games are
ompletely dierent. indeed.57
4. If this is the
ase.2. In the end position. that the way to win the one-pile game. Thus. how is it possible to maintain symmetry? Nevertheless. where both piles have 0
mat
hes. with the restri
tion
that at most M mat
hes
an be removed. in se
tion 4.
58
4. so.
The idea is to de. Games
Consider a game that is the sum of two games. A position in the sum game is a pair
( l.r ) where l is a position in the left game. A
move ae
ts just one
omponent. a move is modelled by either a (guarded) assignment
l := l ′ (for some l ′ ) to the left
omponent or a (guarded) assignment r := r ′ (for some
r ′ ) to the right
omponent. and r is a position in the right game.
on left and right positions.r . how do
we spe
ify the fun
tions L and R ?
The analysis given earlier of a winning strategy allows us to distill the spe
i.r ) is a losing position exa
tly when L. respe
tively.l = R.ne two fun
tions L and R . in su
h a way that a position ( l. say.
The question is: what properties should these fun
tions satisfy? In other words.
l = R.r ) satisfying L.
First.l = R. should result in a winning position |a position ( l.l 6= R.r }
.l 6= R. it must be the
ase that L
and R have equal values on end positions.
We
an satisfy the .
Third.
{
L.r | should result in a losing position |a position ( l. applying the winning strategy. from a winning position |a position ( l. that is not an end position.r | .
{
if
L.l = R.
Se
ond.r ∧ (l is not an end position ∨ r is not an end position ) }
l is not an end position →
hange l
2 r is not an end position →
hange r
fi
{
L.l 6= R.r ) is an end position of the sum game exa
tly when l is an end position
of the left game and r is an end position of the right game.
That is.r |.r }
.l 6= R.r ) satisfying L.r ) satisfying
L.r | .r ) satisfying L. sin
e ( l.r }
apply winning strategy
{
L. every allowed move from a losing position |a position ( l.
ation.l = R. That is.
rst and se
ond requirements if we de.
L.
′
′
′
Similarly. and require that:
For end positions l and r of the respe
tive games.
Algorithmi
Problem Solving
Roland Ba
khouse.l = 0 = R. 2008
.r 6= R.r ′ .ne L and R to be fun
tions
with range the set of natural numbers.l 6= L.r .
For every l su
h that there is a move from l to l in the left game.
R. for every r ′ su
h that there is a move from r to r ′ in the right game.l . L.
May 28.
l and R.l < R.r < L. and the
hoi
e
of 0 as the fun
tions' value at end positions is quite arbitrary.l .r or R. If L.59
4.r are dierent natural numbers. The advantage of this
hoi
e arises from the third requirement.
either L.4. Sums of Games
Note that the
hoi
e of the natural numbers as range of the fun
tions. This allows us to re.
(See below.
For this to work. Similarly.r .r < L.l ′ = n . we require that:
For any number m less than R.r →
hange r
2 R.ne the pro
ess of applying the winning
strategy.l < R.r }
.l .r < L.l →
hange l
fi
{
L.r }
{
if
L. it is possible to move from r to a position r
′
su
h that R.l 6= R.l < R.
The bulleted requirements are satis. it must be possible
to move from l to a position l ′ su
h that L. for any number n less than L.l .l = R. by
hoosing to move in the right game when L.r and
hoosing to move in
the left game when R.r ′ = m .)
L.
ed if we de.
The pre
ise de.ne the fun
tions L and R to be the
so-
alled \mex" fun
tion.
The mex value of p . denoted mexG.nition of this fun
tion is as follows.p .
Let p be a position in a game G .
is de.
q = n .
For every natural number m less than n . there is a legal move in the
game G from p to a position q satisfying mexG. su
h that
There is no legal move in the game G from p to a position q satisfying
mexG. A brief.4.
4. informal des
ription of the mex
number of a position p is the minimum number that is ex
luded from the mex numbers
of positions q to whi
h a move
an be made from p . n .q = m .5
Using the MEX Function
We use the game depi
ted in .ned to be the smallest natural number.
\Mex" is short for \minimal ex
ludant".
4 to illustrate the
al
ulation of mex numbers.
onsequently.g. the only way to
ompute the mex numbers is by a brute-for
e sear
h of all positions. 4. This is easily done
Algorithmi
Problem Solving
Roland Ba
khouse.
May 28.
The graphs do not have any systemati
stru
ture. Figure
4.6 shows the mex numbers of ea
h of the nodes in their respe
tive games. 2008
.
by de. Subsequently. The mex number of a node is the smallest natural number
not in
luded among the mex numbers of its su
essors. Games
2
1
2
0
0
0
1
1
0
0
0
3
1
0
0
2
1
0
0
0
0
1
1
2
1
0
Figure 4.6: Mex Numbers.) The
number is.60
4.
(A su
essor of a node p is a node q su
h that there is an edge from p to q . The end positions are ea
h given mex number 0 . a mex number
an be given to a node when all its su
essors have already been given a mex number.
by hand.
nition.7 shows a typi
al situation. Fig. 4. The node at the top of the . the smallest number that is not in
luded in the mex numbers
of its su
essors.
in either the left or right graph. The latter is larger ( 3 against 2 ). suppose we play this game. Let us suppose the starting position is \Ok". This
is a winning position be
ause the mex number of \O" is dierent from the mex number
of \k". but there are su
essors with the mex numbers 0 and 1 .
Now. whi
h has the same mex number as \O". the winning strategy is to move in the
right graph to the node \i". In the
situation shown. The . to a node with mex number
dierent from 2 . The opponent
is then obliged to move.gure
is given a mex number when all its su
essors have been given mex numbers. So. the mex number given to it is 2 be
ause none of its su
essors have
been given this number.
rst player then repeats the strategy of ensuring that the mex
numbers are equal. until eventually the opponent
an move no further.
May 28. 2008
.
Algorithmi
Problem Solving
Roland Ba
khouse.
the saving grows as the size of the
omponent games
in
reases.
Note that. give the winning move in the form X m where \X" is one of \L" (for
Algorithmi
Problem Solving
Roland Ba
khouse.
May 28.4
Left Game Right Game \losing" or winning move
10
20
?
20
20
?
15
5
?
6
9
?
37
43
?
Table 4.
In the sum game. therefore. For winning positions. this is just the sum (in the usual sense of the word)
of 15 and 11 . and is mu
h less than their produ
t.61
4.4. 5 or 6 mat
hes may be removed. 5 or 6 mat
hes may be removed. we have to sear
h
through 26 dierent positions. 165 . But. state whether it is a winning or a losing position. 2 . In the left game.
The table below shows a number of dierent positions in this game.
a) Consider the subtra
tion-set game where there is one pile of mat
hes
from whi
h at most 2 .7: Computing mex numbers.3: Fill in entries marked \?"
For ea
h position. A position is
given by a pair of numbers: the number of mat
hes in the left pile. Cal
ulate the mex number for
ea
h position until you spot a pattern. In the right game. 1 or 2 mat
hes
may be removed at ea
h turn. Moreover.
Exercise 4. The unlabelled node is given the mex number 2 . be
ause of the la
k of stru
ture of the individual games. Sums of Games
0
1
4
5
Figure 4. in order
to
al
ulate the mex numbers of ea
h position. This is a substantial saving
in
omputational eort. a move is made by
hoosing to play in the left game. and the number of
mat
hes in the right pile. or
hoosing to
play in the right game. we have to sear
h
through all 15 positions of the left game and all 11 positions of the right game. In total.
b) Consider a game whi
h is the sum of two games. 2008
.
making a total of m×n
squares.
2
A re
tangular board is divided into m horizontal rows and n verti
al
olumns.62
4. if the board has 4×4
squares. The number of squares is
alled the area of the board. where m and n are both stri
tly positive integers.) For example. Games
\left game") or \R" (for right game). and m is the number of mat
hes to be removed. when a
ut has been made a part whose area is at
most the area of the other part is dis
arded. (This means that the part with the smaller
area is dis
arded if the two parts have dierent areas. Also. if the board has 4×5 squares. and one of the two parts is
hosen
arbitrarily if the two areas are equal. 4×3 . or 4×4 squares. Ea
h of the two players takes it in turn
to
ut the board either horizontally or verti
ally along one of the dividing lines. (Boards with 3×4 and
4×3 squares are ee
tively the same. a
single move redu
es it to 2×5 . 3×5 . A
ut
divides the board into two parts. the orientation of the board is not signi. a single move redu
es it to either 2×4 or 3×4 squares.
A game is played on the board as follows.
if the board has 5
olumns. for n less than
3 .
al
ulate whi
h positions are winning and whi
h positions
are losing for the . 5 ≤ n . at whi
h point the player whose turn it is to play loses. No other moves are possible be
ause.)
The game ends when the board has been redu
ed to a 1×1 board.
Exercise 4. At this point. The
omponent games are
opies of the same game. and for n greater than 4 .5
(a) For the
omponent game. A move
in the game is to repla
e m by a number n su
h that n < m ≤ 2n . 2n < 5 . at ea
h move. This question
is about
al
ulating the mex numbers of the
omponent games in order to determine a
winning move even when the board
annot be made square. A position in the game is given by a stri
tly positive integer m .
ant. the
player whose turn it is to play loses. This game
is as follows. the number of
olumns
an be redu
ed to 3 or 4
be
ause 3 < 5 ≤ 6 and 4 < 5 ≤ 8 .
This game is a sum game be
ause. a
hoi
e is made between
utting
horizontally or verti
ally. the game ends when
m has been redu
ed to 1 . For
example.
The game is easy to win if it is possible to make the board square.
rst 15 positions.
Algorithmi
Problem Solving
Roland Ba
khouse.
May 28. A winning position is a position from whi
h there is a move to a losing
position. Make a general
onje
ture about the winning
and losing positions in the
omponent game and prove your
onje
ture. A losing position is a position from whi
h every move is to a winning
position. position 1 . The end position.
Base your proof on the following fa
ts. 2008
. is a losing
position.
4.
al
ulate the mex number of ea
h of the . Sums of Games
(b) For the
omponent game.63
4.
rst 15 positions. The
.
Give the results of your
al
ulation in the form of a table with two rows.
Split the table into four parts. Part i gives the mex numbers of positions 2i thru
2i+1−1 (where i begins at 0 ) as shown below.rst row is a number m and the se
ond row is the mex number of position m . (The .
ed
in part (a)) is 0 . You should also be able to observe a pattern in the way entries
are .
the .)
(c) Table 4. The pattern is based on
whether the position is an even number or an odd number.lled in for part i+1 knowing the entries for part i .3 shows a position in the board game.
. Using your table of mex
numbers.rst
olumn shows the number
of
olumns and the se
ond
olumn the number of rows. or otherwise.
ll in \losing" if the position is a losing position. If the
position is not a losing position. .
and n is the number whi
h
it should be
ome.
May 28.
2
Algorithmi
Problem Solving
Roland Ba
khouse. \C" or \R" indi
ates whether the move is to redu
e
the number of \C"olumns or the number of \R"ows.ll in a winning move either in the form \C n " or
\R n ". 2008
. where n is an integer.
The analysis
of the \sum" of two games exempli.
Brute-for
e sear
h is only advisable for small.
ation enabled us to
formulate a brute-for
e sear
h pro
edure to determine whi
h positions are whi
h. unstru
tured problems.
Again.es the way stru
ture is exploited in problem solving. the fo
us was on problem spe
i.
ation. By formulating a notion of \symmetry"
between the left and right games. we were able to determine a spe
i.
and mex numbers are sometimes
alled \Sprague-Grundy" numbers.
ation of the \mex"
fun
tion on game positions. and sums of more
than two games.
ompared
to a brute-for
e sear
h. well-explored area of Mathemati
s. It is a theory that is be
oming in
reasingly important in Computing S
ien
e. One reason for this is that problems that beset software design. are solved using mex numbers. it is easy to formulate games having very simple rules but for whi
h
no eÆ
ient algorithm implementing the winning strategy is known. to show how Nim. We have not
developed the theory suÆ
iently.
May 28. (What is missing is how to
ompute the
mex number of the sum of two games. 2008
. su
h as the
se
urity of a system. with the user of the software as the
adversary.
Mex numbers were introdu
ed by Sprague and Grundy to solve the \Nim" problem. whi
h we have only tou
hed
upon in this
hapter. in this
hapter.)
Algorithmi
Problem Solving
Roland Ba
khouse. after their originators. are often modelled as a game. Another reason is that games often provide ex
ellent examples of \
omputational
omplexity". Nim is a well-known mat
hsti
k game involving three piles of mat
hes. The use of mex fun
tions substantially redu
es the eort
needed to determine winning and losing positions in the \sum" of two games.
Game theory is a ri
h.
2008
.5 was
suggested by Atheer Aheer.65
4.
Algorithmi
Problem Solving
Roland Ba
khouse.6. Exer
ise 4.
May 28.
The 31st De
ember game (exer
ise 4.1) is adapted from [DW00℄. Conway and Guy is the
bible of game theory.6
Bibliographic Remarks
The two-volume book \Winning Ways" [BCG82℄ by Berlekamp. Bibliographi
Remarks
4.
" The problem is
(a) Can it be determined whether the native is a knight or a knave?
(b) Can it be determined whether there is gold on the island?
2. Suppose you
ome a
ross two of the natives. are easy exer
ises in the use of
al
ulational
logi
. and similar logi
puzzles. B and C.
1. these. It is rumoured that there is gold buried on the island. The island has two types of natives. 2008
. In
ontrast.
67
May 28."
What
an be inferred about the number of knights?
Algorithmi
Problem Solving
Roland Ba
khouse. Case analysis is a
lumsy way of ta
kling the problems.
The temptation is to solve su
h problems by
ase analysis |in a problem involving n
natives. Will you get the same answer in ea
h
ase?
3.
onsider the 2n dierent
ases obtained by assuming that the individual natives
are knights or knaves| . You ask both of them whether the
other one is a knight.
tional island that is often used to test students'
ability to reason logi
ally. \knights" who always
tell the truth. and \knaves" who always lie. You ask one of the natives
whether there is gold on the island. Logi
puzzles involve dedu
ing fa
ts about
the island from statements made by its natives without knowing whether or not the
statements are made by a knight or a knave. The native replies: \There is gold on this
island is the same as I am a knight. whi
h we introdu
e in this
hapter. There are three natives A. Suppose A says \B and C are the same type.1
Logic Puzzles
Here is a typi
al
olle
tion of knights-and-knaves puzzles.
5.
Knights and Knaves
4. We say that m2−n2 and (m+n)×(m−n) are equal. 2008
. What question should you ask A to determine whether A and B are the same type
(i. What question should
you pose to A to determine whether or not C is telling the truth?
5. You would like to determine whether an odd number of A. both knights or both knaves)?
8. Laws are typi
ally primitive. What is the question you
should ask?
9. Formulate a single
yes/no question that the tourist
an ask su
h that the answer will be yes if the left
fork leads to the restaurant. We learn. and otherwise the answer will be no. for example. They are \primitive" in the sense that they
annot be
broken down into simpler laws. where one bran
h leads to a restaurant and
one doesn't.
equalities between expressions.2. and write
m2−n2 = (m+n)×(m−n) . both involving zero are:
n+0 = n . Devise a question that allows you to determine whether a native is a knight. We
all them
axioms. and they are \general" in the sense that they hold
independently of the values of any variables in the
onstituent expressions.
May 28. Two examples of axioms. A tourist
omes to a fork in the road. independently of the
values of m and n .
and
n−n = 0 .
5. Suppose C says \A and B are as like as two peas in a pod".
You may ask one yes/no question to any one of them.2
5.
Algorithmi
Problem Solving
Roland Ba
khouse. A native of the island is standing at the fork.
6.
The basis for these
al
ulations is a set of laws. B and C is a knight.1
Calculational Logic
Propositions
The algebra we learn at s
hool is about
al
ulating with expressions whose values are
numbers.e. how to manipulate an expression like m2−n2 in order
to show that its value is the same as the value of (m+n)×(m−n) . What question should you ask A to determine whether B is a knight?
7. but general.68
5.
(Cal
ulational) logi
is about
al
ulating with expressions whose values are so-
alled
\booleans" | that is.
m < n < p . in
ontrast to other
axiomatisations of logi
. . Equality
is the most basi
on
ept of logi
|a fa
t . The
three examples above are all atomi
. either true or false .
n = 0 (whi
h is either true or false depending on the value of n ). for example. For
example. Cal
ulational Logi
both of whi
h are true whatever the value of the variable n . n and p . and \if" that are used to
ombine atomi
propositions. then . ). \or". . \asso
iativity of addition" is the name given to the equality:
(m+n)+p = m+(n+p) . .
whi
h is true for all m . The laws are often given names so that we
an remember them more easily. whi
h emphasise logi
al impli
ation (if . . Logi
is about rules for manipulating
the logi
al
onne
tives | the operators like \and".2. that is the
on
ern of the problem domain being dis
ussed.
Cal
ulational logi
pla
es emphasis on equality of propositions. whi
h
an be broken down into the so-
alled
onjun
tion of m < n and
n<p.
Logi
is not
on
erned with the truth or otherwise of atomi
propositions. Atomi
propositions are propositions that
annot be broken down into simpler propositions. and n < n+1 (whi
h
is true for all numbers n ). We say they are true \for
all n ". Boolean-valued expressions are
alled propositions . A non-atomi
proposition would be.69
5. Examples of su
h expressions are \it is
sunny" (whi
h is either true or false depending on to when and where \it" refers) .
who lived from 1646 to 1716 and who was the .rst re
ognised by Gottfried Wilhelm Leibniz.
Then. the
statements made by the natives are propositions. Also.rst to try to formulate logi
al reasoning|
and equality of propositions is no ex
eption. Suppose A denotes the proposition \A is a knight". That is. and
suppose native A makes a statement S. and a knave always lies.2
Knights and Knaves
Equality of propositions is
entral to solving puzzles about knights and knaves. We see shortly that equality of propositions
is parti
ularly spe
ial. and so is a proposition.
5.2. 2008
.
May 28. If A is a native of the island. the
ru
ial observation is that the values
of these two propositions are the same.
A=S . Re
all
that a knight always tells the truth.
the statement \A is a knight" is either true or false . re
ognition of whi
h
onsiderably enhan
es the beauty and power
of reasoning with propositions.
Algorithmi
Problem Solving
Roland Ba
khouse. A statement like \the restaurant is to
the left" is either true or false.
This doesn't tell us anything! A moment's thought
on. or A is not a knight and the
restaurant is not to the left. if A says \the restaurant is to the left".
Using this rule.
where L denotes the truth value of the statement \the restaurant is to the left". Knights and Knaves
For example. then
A=L .70
5. A is a knight and the restaurant is to the left. if A says \I am a knight". In
words. we dedu
e
A=A .
If native A is asked a yes/no question Q . Asked the question \is B a knight?" A will respond \yes" if they are both the
same type (i. For example. . A simple. Otherwise the response will be
\no". That is. the response will be \yes" if A is a knight and the answer is
really yes. or A is a knave and the answer is really no. That is.
Be
ause these rules are equalities. the response to the question is the truth
value of A = Q . Both knights and knaves would
laim that they are knights. A's response is \yes" or \no" depending
on the truth or falsity of A = B .
as A = A .e.rms that this is what one would
expe
t. A = B ). asked the question \are you a knight" all natives will answer \yes". otherwise \no". the algebrai
properties of equality play a
entral
role in the solution of logi
puzzles formulated about the island.
x = y is the same as y = x . That is x = x whatever the value (or type) of x . B's response is B = A . if x = y and f is any fun
tion then f. it
is re
exive .
But. As dis
ussed
above. Third. Finally. First. therefore.x = f. it is transitive .rst example
is if A is asked whether B is a knight.
5.
That is. That is. Reversing the roles of A and B. the two responses will always be the same. Se
ond. A's response is A = B . Note
that this argument does not involve any
ase analysis on the four dierent values of A
and B .2. it is symmetri
. equality is symmetri
.y (where the in.3
Boolean Equality
Equality |on any domain of values| has a number of
hara
teristi
properties.
The
al
ulational properties of equality of booleans are dis
ussed in the next se
tion
before we return again to the knights and knaves. if x = y and y = z
then x = z . and B is asked whether A is a knight.
Equality is a binary relation. This last rule is
alled substitution of equals for equals
or Leibniz's rule . re
exivity. When studying relations. the
Algorithmi
Problem Solving
Roland Ba
khouse. It is a
fun
tion with range the boolean values true and false . 2008
. symmetry and
transitivity are properties that we look out for.
May 28. When we study fun
tions.x dot
denotes fun
tion appli
ation). Equality is also a fun
tion.
or
hara
ters. But. That is.
The re
exivity of equality is expressed by the rule
(p = p) = true . perhaps surprisingly. (p = q) = r is also false . in all but one
ase. viewed as a fun
tion. In other words. q and r are numbers. what about asso
iativity of equality? Is equality an asso
iative
operator?
The answer is that. (p = q) = r is a meaningful boolean value. or sequen
es. the question doesn't make sense. an elementary example is the
following. Cal
ulational Logi
sort of properties we look out for are asso
iativity and symmetry. q and r are all booleans it makes sense to
ompare the
boolean p = q with r for equality.
You should
he
k this property by
onstru
ting truth tables for (p = q) = r and for
p = (q = r) and
omparing the entries. y and z .71
5. q and r .1)
[Associativity]
((p = q) = r) = (p = (q = r)) . so too is p = (q = r) . for one be
ause it enhan
es
e
onomy of expression. That is. 2008
. and the third is false . it makes sense to ask whether equality of boolean values is
asso
iative | and. When p .
Symmetry of equality.
x+y = y+x
and
x×y = y×x . for all booleans p . You should observe that the entries for whi
h
(p = q) = r is true are those for whi
h an odd number of p . is just the same as symmetry of equality. et
. it is. Asso
iativity
of a binary fun
tion only makes sense if the domains of its two arguments and the range
of its result are all the same.
Algorithmi
Problem Solving
Roland Ba
khouse.
May 28.2.
The asso
iativity of equality is a very powerful property. For example.
viewed as a relation. The one ex
eption is equality
of boolean values.
Similarly.
They are also both symmetri
: for all x and y .
x + (y + z) = (x + y) + z
and
x × (y × z) = (x × y) × z . We will see several examples. It also makes sense to
ompare these two values for
equality. If two of
the three are true.
(5. q and r is true . addition
and multipli
ation are both asso
iative: for all x . The expression (p = q) = r just doesn't make sense when
p .
et
. We use it several times below. string.72
5. for boolean
p .4
Hidden Treasures
We
an now return to the island of knights and knaves.). whatever its type (number. and dis
over the hidden treasures. But.2. Knights and Knaves
This holds for all p . boolean.
This rule is most
ommonly used to simplify expressions by eliminating \ true " from an
expression of the form p = true .
5.
Let us
onsider the . we
an apply the asso
iativity of equality to get:
p = (p = true) .
We
on
lude that there is gold on the island.
Suppose. that the native is at a fork in the road. but it is not possible to determine whether
the native is a knight or a knave.rst problem posed in se
tion 5. What
an we dedu
e if a native
says \I am a knight equals there is gold on the island"? Let A stand for \the native is
a knight" and G stand for \there is gold on the island". So. now. and we dedu
e that
A = (A = G)
is true.1.
substitution of equals for equals
}
true = G
=
{
equality is symmetri
}
G = true
=
{
G = (G = true)
}
G .
true
=
{
A's statement
}
A = (A = G)
=
{
equality of booleans is asso
iative
}
(A = A) = G
=
{
(A = A) = true . 2008
. Then the native's statement is
A = G .
May 28. You want to formulate
Algorithmi
Problem Solving
Roland Ba
khouse. and you want to determine
whether the gold
an be found by following the left or right fork.
Note that this analysis is valid independently of what L denotes. Then. Cal
ulational Logi
a question su
h that the reply will be \yes" if the left fork should be followed. the question to be posed is P = A .73
5. Let Q be the question to be posed. Let L denote \the gold
an be found by following the left fork. or whether there are
any knaves on the island." The requirement is that L is the same as the
response to the question.
As usual. the question Q to be posed is L = A . we give the unknown a name. and \no"
if the right fork should be followed. the question may be simpli. we require that L = (A = Q) . That is.
L = (A = Q)
=
{
equality is asso
iative
}
(L = A) = Q .
So.2. It might be that
you want to determine whether there is a restaurant on the island. the response to the question will be A = Q . In general. That is. or whatever. But. ask the question \Is the truth value
of `the gold
an be found by following the left fork' equal to the truth value of `you are
a knight' ". if it is required to determine whether
some proposition P is true or false. In the
ase of
more
omplex propositions P .
as we saw earlier.
But. B and C denote the propositions A (respe
tively.
B and C) is a knight. one expression
an be substituted for the other. determines whether C is telling the truth. Q = (A = C) . Here.
To solve this problem. Formulate a question that. So we know that C = (A = B) . A = (A = B) simpli.
5.2.
The response we want is C . But. B and C.
Suppose there are three natives of the island.ed. we let A . We also let Q be the unknown question.5
Equals for Equals
Equality is distinguished from other logi
al
onne
tives by Leibniz's rule: if two expressions are equal.
C's statement is A = B .4.2. when posed to
A. we
onsider one
simple example of the use of Leibniz's rule. A. Substituting equals for equals.
Q = (A = (A = B)) . by the analysis in se
tion 5. and C says \A
and B are both the same type". So.
but set out as a
al
ulation of Q .
Q
=
{
rule for formulating questions
Algorithmi
Problem Solving
Roland Ba
khouse. 2008
.es to B .
}
May 28. with
hints showing the steps taken at ea
h stage. the question to be posed is
\is B a knight?". Here is this argument again. So.
y and z ). (x⊕y)⊕z = x⊕(y⊕z) for
all x . More importantly. we
an
hoose to simplify u⊕w for any pair of subexpressions u and
w. The expression
be
omes more
ompa
t be
ause of the omission of parentheses.t in
al
ulations
is immense. x⊕y = y⊕x for all x
and y ) the gain is even bigger. we
an write x⊕y⊕z without fear of ambiguity. If the operator is also symmetri
(that is. If a binary operator ⊕ is asso
iative (that is. be
ause then. the
expression is unbiased.
In. we may
hoose to simplify x⊕y or y⊕z depending on whi
h is
the most
onvenient. if the operator is used to
ombine several
subexpressions.
Do we understand
in. the formula is more
ompa
t (sin
e m is not written twi
e). we are guided to the inferen
e that 0 ≤ n . 0 ≤ m ≤ n . But. we have a dilemma.x notation is also often used for binary relations.
Here. The algebrai
property that is being
hidden here is the transitivity of the at-most relation. If the relation between m and n
is m < n rather than m ≤ n and we write 0 ≤ m < n . Here.
In this way. More importantly. the operators are being used
onjun
tionally : the meaning is 0 ≤ m and m ≤ n . We write. it is an inferen
e
that is so fundamental that the notation is designed to fa
ilitate its re
ognition.
In the
ase of equality of boolean values. for example. the
inferen
e is more
omplex sin
e there are two relations involved. we may infer that 0 < n .
x operator is a symbol used to denote a fun
tion of two arguments that is written between the
two arguments. The symbols \ + " and \ × " are both in.
in just the same way as we would read x+y+z ? The two readings are.3. as
x = (y = z) . equally. unfortunately.
not the same (for example true = false = false is false a
ording to the . y and z ? Or do we read it \asso
iatively" as
(x = y) = z .
or. Equivalen
e and Continued Equalities
equality as a relation and read a
ontinued expression of the form
x=y=z
as asserting the equality of all of x .75
5.
pn are equal.
we write both p = q and p ≡ q . A
ordingly. these both mean the same.e as p = q and q = r | . There are advantages in both readings. When p and q are expressions denoting boolean
values. whi
hever is the most
onvenient| whereas a
ontinued expression
p=q=r
is to be evaluated
onjun
tionally |i.
May 28. . as (p ≡ q) ≡ r or p ≡ (q ≡ r) .
It would be very
onfusing and. More generally. . = pn
means that all of p1 . But a
ontinued expression
p≡q≡r . ≡ pn
Algorithmi
Problem Solving
Roland Ba
khouse. 2008
. a
ontinued equality of the form
p1 = p2 = . . . indeed. is to
be evaluated asso
iatively |i. dangerous to read x = y = z in any other way
than x = y and y = z .rst reading but
true a
ording to the se
ond and third readings).e. whilst a
ontinued equivalen
e of the
form
p1 ≡ p2 ≡ .
The solution to this dilemma is to use two dierent symbols to denote equality of
boolean values | the symbol \ = " when the transitivity of the equality relation is to be
emphasised and the symbol \ ≡ " when its asso
iativity is to be exploited. . the
onjun
tional
reading (for other types) is so universally a
epted |for good reasons| that it would
be quite una
eptable to try to impose a dierent
onvention.
omprising more than two boolean expressions
onne
ted by the \ ≡ " symbol. the meaning of a sequen
e of expressions separated by
equality symbols would depend on the type of the expressions. . Also. . p2 .
and it is a major drawba
k to have to
hoose one in favour of the other. otherwise. .
Shortly.
Moreover. They invariably
involve a
ontninued equivalen
e. A .76
5. being
an unfamiliar word. we re
ommend that the \ ≡ " symbol is pronoun
ed as \equivales".
sin
e the out
ome is not ae
ted) and then evaluating the expression as indi
ated by the
hosen parenthesisation. its use will help to avoid misunderstanding. we introdu
e a number of laws governing boolean equality. Knights and Knaves
has the meaning given by fully parenthesising the expression (in any way whatsover.
(5.2)
[Reflexivity]
5.1
true ≡ p ≡ p .3.rst example is its re
exivity.
Even and Odd Numbers The .
Examples of the Associativity of Equivalence
This se
tion
ontains a
ouple of beautiful examples illustrating the ee
tiveness of the
asso
iativity of equivalen
e.
in general.
May 28. the equivalen
e p ≡ q ≡ r is true exa
tly when an odd number of p . if we parenthesise the statement as
m+n is even ≡ (m is even ≡ n is even) .
Another way of reading the statement is to use the fa
t that. Parenthesising it as
(m+n is even ≡ m is even) ≡ n is even .rst example is the following property of the predi
ate
even on numbers. So the
property
aptures four dierent
ases:
or
or
or
(( m+n
(( m+n
(( m+n
(( m+n
is even)
is odd)
is odd)
is even)
and
and
and
and
(m
(m
(m
(m
is even)
is odd)
is even)
is odd)
and
and
and
and
Algorithmi
Problem Solving
Roland Ba
khouse. (A number is even exa
tly when it is a multiple of two.
(n
(n
(n
(n
is even))
is even))
is odd))
is odd)) . Then. q and r is true.
it states that the number m+n is even exa
tly when the parities of m and n are the
same.
it states that the operation of adding a number n to a number m does not
hange the
parity of m exa
tly when n is even.
It will help if we refer to whether or not a number is even or odd as the parity of the
number. 2008
.)
m+n is even ≡ m is even ≡ n is even .
the produ
t x×y is positive if the signs of
x and y are equal. For non-zero numbers x and y . the produ
t x×y is negative.
Assuming that x and y are non-zero. our justi. even though no
ase analysis is involved.
Sign of Non-Zero Numbers The sign of a number says whether or not the number
is positive. Avoidan
e of
ase analysis is vital to ee
tive reasoning. this one statement neatly
aptures a number of dierent
ases. If the signs of x and y are dierent. Using the
asso
iativity of equivalen
e the value of \ m+n is even" is expressed in one simple
formula. this rule is expressed as
x×y is positive ≡ x is positive ≡ y is positive .3. There are four
distin
t
ombinations of the two booleans \ m is even" and \ n is even". without any repetition of the
omponent expressions.5.
Just as for the predi
ate even. Indeed. Equivalen
e and Continued Equalities
77
The beauty of this example lies in the avoidan
e of
ase analysis. rather than as a list of
dierent
ases.
It is akin
to introdu
ing equality of numbers by .3. At the
present time.2
On Natural Language
Many mathemati
ians and logi
ians are not aware that equality of booleans is asso
iative.
5.
ation of the rule is
the statement
x×y is positive ≡ (x is positive ≡ y is positive) . there is
onsiderable resistan
e to a shift in fo
us from impli
ation to
equality.
those that do are often unaware or dismissive of how ee
tive its use
an be.
The other parenthesisation |whi
h states that the sign of x is un
hanged when it is
multiplied by y exa
tly when y is positive| is obtained \for free" from the asso
iativity
of boolean equality. Most
ourses on logi
introdu
e boolean equality as \if and only if".
(\If it is raining.
The most probable explanation lies in the fa
t that many logi
ians view the purpose
of logi
as formalising \natural" or \intuitive" reasoning. and our \natural" tenden
y
is not to reason in terms of equalities.ning an \at-most and at-least" operator. I will
take my umbrella.") The equality symbol was . but in
ausal terms.
the
ontinued
equivalen
e \a blind man
an see through two eyes equivales a blind man
an see through
one eye equivales a blind man
an see through no eyes" may seem very odd. To take a
on
rete example. Natural language has
no
ounterpart to a
ontinued equivalen
e.
May 28. if not
nonsensi
al. even though it is a
tually true!
Algorithmi
Problem Solving
Roland Ba
khouse. in the history of mathemati
s.rst introdu
ed into mathemati
s by
Robert Re
orde in 1557. 2008
. is quite re
ent. whi
h. were
equality \natural" it would have been introdu
ed mu
h earlier.
nowadays.78
5. people sometimes say. the most
ommon way to express time is
in words: like \quarter to ten" or \ten past eleven". for example. Knights and Knaves
This fa
t should not be a deterrent to the use of
ontinued equivalen
e. But. a very long time ago) there was probably similar resistan
e to the
introdu
tion of
ontinued additions and multipli
ations. Cal
ulational requirements (eg.
wanting to determine how long is it before the train is due to arrive) have in
uen
ed
natural language so that. For example. 9:45 or 11:10 in
everyday spee
h. At one
time (admittedly. The eviden
e is still present
in the language we use today. we still don't .
Negation is a unary
operator (meaning that it is a fun
tion with exa
tly one argument) mapping a boolean
to a boolean. but will always lag a long way behind. The goal is not to mimi
\natural" reasoning.
Native A says. Changes in natural language have o
urred. but to provide a more ee
tive
alternative. \B is a knight equals I am not a knight". In
fa
t. are fundamental to
the
al
ulation. and is denoted by the symbol \ ¬ ". and will
ontinue to o
ur. What
an you determine about
A and B?
This problem involves a so-
alled negation : the use of \not". this is what we
a
tually use when we want to
al
ulate the time dieren
e between 9.
as a result of progress in mathemati
s. There are two natives. The
language of mathemati
s has developed in order to over
ome the limitations of natural
language.
5. several laws of arithmeti
. A and B. in
luding asso
iativity of addition.nd it a
eptable to say 10:70! Yet. written as a pre.45 and 11:10.4
Negation
Consider the following knights-and-knaves problem.
) The goal is to
simplify this expression. we know that A ≡ S . if A makes a statement S .
for this problem:
A ≡ B ≡ ¬A .
Reading this as
¬p = (p ≡ false) .
In order to ta
kle this problem.
May 28.x to its argument. For arbitrary proposition p .
(We swit
h from \ = " to \ ≡ " here in order to exploit asso
iativity. \ ¬p " is pronoun
ed \not p ".
Using the general rule that. If
p is a boolean expression. the law governing ¬p is:
(5.3)
[Negation]
¬p ≡ p ≡ false .
Algorithmi
Problem Solving
Roland Ba
khouse. we get. it is ne
essary to begin by formulating
al
ulational
rules for negation. 2008
.
79
5. Negation
it fun
tions as a de.4.
So.
This simpli. we are given that:
A ≡ B ≡ ¬A . In addition. Reading it the other way:
(¬p ≡ p) = false
it provides a way of simplifying propositional expressions.
Returning to the knights-and-knaves problem.nition of negation. we also get the property:
p = (¬p ≡ false) . the symmetry
of equivalen
e means that we
an rearrange the terms in a
ontinued equivalen
e in any
order we like.
Suppose.3) with p := A
}
false ≡ B
=
{
law (5.es to ¬B as follows:
A ≡ B ≡ ¬A
=
{
rearranging terms }
¬A ≡ A ≡ B
=
{
law (5.
May 28. we want to simplify
¬p ≡ p ≡ q ≡ ¬p ≡ r ≡ ¬q . Note how (5.2) and (5.
We begin by rearranging all the terms so that repeated o
urren
es of \ p " and \ q " are
grouped together. in
onjun
tion with the symmetry and asso
iativity of equivalen
e. but A
ould be a knight or a knave.3) to redu
e the number of o
urren
es of \ p " and \ q " to
at most one (possibly negated).3) is used in two
dierent ways.3) with p := B and rearranging }
¬B . B is a knave.
provides a way of simplifying
ontinued equivalen
es in whi
h one or more terms are
repeated and/or negated. for example.3). In this parti
ular example we obtain
true ≡ p ≡ false ≡ r . 2008
.
So.
Now we
an use (5.
Algorithmi
Problem Solving
Roland Ba
khouse.
The law (5. Thus we get
¬p ≡ ¬p ≡ p ≡ q ≡ ¬q ≡ r .
we use (5. The result is that the original formula is simpli.2) and (5.3) again.80
5. Knights and Knaves
Finally.
ed
to
¬p ≡ r . this pro
ess
an be
ompared with the simpli.
Just as before.
q and r .3) are all that is needed to de.ed to
q + 2r
by
ounting all the o
urren
es of p . Again.
The two laws (5. the details are dierent but the pro
ess is essentially identi
al.2) and (5. an o
urren
e of −p
an
elling out an
o
urren
e of p .
(See
hapter 3. If we let n denote the number of
rossings.
The name refers to the use of the rule in the form (p ≡ q) = (¬p ≡ ¬q) .4)
[Contraposition]
p ≡ q ≡ ¬p ≡ ¬q .
We used the rule of
ontraposition impli
itly in the river-
rossing problems. but whi
h is surprisingly useful.) Re
all that ea
h problem involves getting a group of people from one side of
a river to another. 2008
.
May 28.ne the way that negation
intera
ts with equivalen
e. and l
2 Other
tion.
(5. is the rule we
all
ontraposition 2 . A
simple example of how these two laws are
ombined is a proof that ¬false = true :
¬false
=
{
law ¬p ≡ p ≡ false with p := false
}
false ≡ false
=
{
law true ≡ p ≡ p with p := false
}
true .
authors use the name \
ontraposition" for a less general rule
ombining negation with impli
a-
Algorithmi
Problem Solving
Roland Ba
khouse.
5. using these two laws we
an derive several other laws. using one boat.5
Contraposition
A rule that should now be obvious.
We are given that.n)
}
¬(even. and the boat
hanges side. initially.
Another example is the following.n ≡ l is invariantly true .n ≡ l .n ≡ l)[n . the number of
rossings in
reases by one.
we
on
lude that even. l := n+1 . the boat is on the left side
equivales the number of
rossings is even.
In words. Sin
e zero is an even number. a
rossing of the river is
modelled by the assignment:
n . Suppose it is required to move a square arm
hair
sideways by a distan
e equal to its own width.81
5. The rule
of
ontraposition tells us that
even. (See .n) ≡ ¬l
=
{
ontraposition
}
even. ¬l . ¬l]
=
{
rule of substitution
}
even. l := n+1 .n ≡ l
is invariant under this assignment. This is be
ause
(even.(n+1) ≡ ¬(even. Contraposition
denote the boolean \the boat is on the left side of the river". In words. the boat is on the left side.5.(n+1) ≡ ¬l
=
{
even.
) However. like a
hess board. Is it possible to move the
hair as desired? If so. Suppose the arm
hair is initially positioned along
a north-south axis.1. the
hair is
so heavy that it
an only be moved by rotating it through 90◦ around one of its four
orners. how? If not. that the
oor is painted alternately with bla
k and
white squares.1: Moving a heavy arm
hair.2.
Algorithmi
Problem Solving
Roland Ba
khouse. 2008
.
The answer is that it is impossible.gure 5. why not?
Figure 5.
May 28. (See Figure 5. The
requirement is to move the arm
hair from a north-south position on a bla
k square to a
north-south position on a white square. also.) Suppose the arm
hair is initially on a bla
k square. Suppose. with ea
h of the squares being the same size as the
arm
hair.
rotating the arm
hair
Now. moving the arm
hair sideways one square
hanges
the
olour but does not
hange the dire
tion.
Roland Ba
khouse. That is. A
hessboard is an
Algorithmi
Problem Solving
8×8
grid of squares. ¬dir
The rule of
ontraposition states that an invariant of this assignment is
col ≡ dir .
May 28. Then. and dir represent the dire
tion that the arm
hair is
fa
ing (say. two pla
es left or right and one
pla
e up or down. no matter how many
times the arm
hair is rotated. the value of this expression will remain equal to its initial value. Knights and Knaves
col represent the
olour of the square that the arm
hair is on (say. an invariant of rotating the arm
hair through
the
hair is on a bla
k square
≡
90◦
around a
orner point is
the
hair is fa
ing north-south
whi
h is false when the
hair is on a white square and fa
ing north-south.82
5. a knight's move is two pla
es
up or down and one pla
e left or right. vi
e versa.2: Invariant when moving a heavy arm
hair. let boolean
about any
orner is represented by the assignment:
col .
But. 2008
.5 (Knight's Move)
In the game of
hess. true for north-south and false for east-west).
90◦
Figure 5.
In words.
Exercise 5. dir := ¬col .
and is impossible to a
hieve by
ontinually rotating the arm
hair as pres
ribed. or. it
hanges the value of
col ≡ dir .
So.
true for bla
k and false for white).
Suppose ea
h person
ounts the number of times
they shake hands. means that.6.
There are n numbers in the range 0 to n−1 . We
on
lude that everyone shakes hands with between 0 and n−1 people.
Suppose there are n people. Handshake Problems
Show that it is impossible to move a knight from the bottom-left
orner of a
hessboard to the top-right
orner in su
h a way that every square on the board is visited
exa
tly on
e. In this way. for all
x and y . xSy is read as \person x shakes hands with person y ".
We are required to show that (at least) two people shake hands the same number
of times.
Hint: How many moves have to be made? Model a move in terms of the ee
t on the
number of moves and the
olour of the square on whi
h the knight is standing. identify
a relation between the two that is invariant under a move. for any two people |Ja
k and Jill. and suppose we use x and y to refer
to people.
May 28. say| Ja
k shakes hands with
Jill equivales Jill shakes hands with Ja
k. These are
that it is a binary relation .) It being a \symmetri
"
relation means that. a relation is any boolean-valued
fun
tion.
Algorithmi
Problem Solving
Roland Ba
khouse. someone shakes hands 0 times and someone shakes hands n−1 times. Finally.
2
5. 2008
. (In general.
In parti
ular.
Suppose we abbreviate \shake hands" to S .
Cru
ial to how we solve this problem are the properties of shaking hands. or just
x shakes hands with y .
xSy ≡ ySx .
The simplest example of a handshake problem is this: Suppose that at a party. everyone shakes hands with at most n people. Then. and it is anti-re
exive .
The symmetry of the shake-hands relation makes this impossible.
However. Then the symmetry of \shakes hands" gives us the rule. the anti-re
exivity property is that noone shakes hands with themselves. for any two people |Ja
k and Jill. say| Ja
k shakes
hands with Jill is either true or false. Show that at least two people have the same
ount. it being \anti-re
exive" means that
no-one shakes hands with themselves. Let us explore the
onsequen
es of the properties of the shake-hands relation
with respe
t to the number of times that ea
h person shakes hands.83
5. Binary means that it is a fun
tion of two arguments. some
people shake hands and some don't.6
Handshake Problems
Logi
al properties of negation are fundamental to solving so-
alled handshake problems . The negation of \two people shake
hands the same number of times" is \everyone shake hands a distin
t number of times". it is symmetri
. It being a \binary
relation" on people.
Suppose a number of
ouples (husband and wife) attend a party.) However.4. This statement is formulated as B 6= A . whereas the
Queen never does. and b shakes hands
with a . the
out
ome might be dierent. x doesn't shake hands with y equivales y doesn't shake hands with x .
Then. Like \shake hands". the same as saying \B is a knight equals I am not a knight". Were we to
onsider a similar problem involving a dierent relation. but the greeting is
not returned.
In words. substituting equals for equals. if we repla
e \shake hands" by some other
form of greeting like \bows or
urtsies". and one bows to the other.
¬(xSy) ≡ ¬(ySx) . in fa
t. and gets a different answer every time.7
Inequivalence
In the knights-and-knaves problem mentioned at the beginning of se
tion 5.
Note
arefully how the symmetry and anti-re
exivity of the shakes-hands relation
are
ru
ial. and so we
on
lude that two people must shake hands the same
number of times. the property need not
hold3.
in order to exploit asso
iativity. Some people
shake hands. \rub noses" is a symmetri
and anti-re
exive relation. whi
h is not symmetri
.
as is \don't shake hands". anyone meeting the British Queen is required to bow or
urtsey. Note that we swit
h from \ = " to \ ≡ " on
e again. others do not. But then. How many times did the host and the host's partner shake
hands?
Exercise 5.
suppose person a shakes hands with noone and person b shakes hands with everyone. Now.
May 28. the
\host". A might
have said \B is dierent from myself". Husband and wife never shake hands. i.
3 At
the time of writing. 2008
. The relation is not symmetri
.e. in parti
ular. for all x and y . or
¬(B = A) This is. It's a bit more diÆ
ult to solve. ¬(aSb) . we have both ¬(aSb) and
aSb .
as the following
al
ulation shows. asks everyone else how many times they have shaken hands. Knights and Knaves
The
ontrapositive of this rule is that.e. One person.
The assumption that everyone shakes hands with a distin
t number of people has
led to a
ontradi
tion. the property does
hold.
Here is another handshaking problem.84
5.6
2
5.
but the essen
e of the problem remains the same: \shake hands" is a symmetri
relation. For example. whi
h is false. a does not shake hands with b .
Algorithmi
Problem Solving
Roland Ba
khouse. (Suppose there are two people. if \shake hands" is repla
ed by \rub noses". bSa . i.
(5.7)
[Inequivalence]
¬(p ≡ q) ≡ p ≡ ¬q . The . for all propositions p and q .
Note how asso
iativity of equivalen
e has been used silently in this
al
ulation. Note
also how asso
iativity of equivalen
e in the summary of the
al
ulation gives us two
properties for the pri
e of one.7.
We have thus proved. Inequivalen
e
¬(B ≡ A)
=
{
the law ¬p ≡ p ≡ false with p := (B ≡ A)
}
B ≡ A ≡ false
=
{
the law ¬p ≡ p ≡ false with p := A
}
B ≡ ¬A .85
5.
86
5. we have shown that p 6≡ q 6≡ r and p ≡ q ≡ r are
equal. we
an write the
ontinued inequivalen
e p 6≡ q 6≡ r without fear of ambiguity4.
As a . as a byprodu
t. Note that. Knights and Knaves
As a result.
Exploitation of the asso
iativity of equivalen
e eliminates the tedious and
inee
tive
ase analysis that is often seen in solutions to logi
puzzles. Show that.
Inequivalen
e
an be repla
ed by equivalen
e in the
But.9
Prove that ¬true = false
2
Exercise 5. and this possibility is never exploited!
Algorithmi
Problem Solving
Roland Ba
khouse.11 (Encryption)
(p 6≡ (q 6≡ r)) ≡ ((p 6≡ q) 6≡ r) . That is. in addition to the standard properties
of equality.
en
ryption and de
ryption pro
ess. the result is b
independently of the key a . that is
Exercise 5.
is used to en
rypt data.
2
On the island of knights and knaves.8
Summary
In this
hapter. A
and B. we have used simple logi
puzzles to introdu
e logi
al equivalen
e |
the equality of boolean values| .8. To en
rypt a single bit b of data. Equivalen
e
has the remarkable property of being asso
iative. the most fundamental logi
al operator. the re
eiver uses the same key a to
ompute
a 6≡ c .
5 This
operation is usually
alled \ex
lusive-or" in texts on data en
ryption. c . What question should you ask A to determine whether A and B are dierent
types?
Exercise 5. The re
eiver de
rypts the re
eived
bit. if bit b is en
rypted and then de
rypted in this way.10 (Double Negation)
Prove the rule of double negation
¬¬p = p .87
5. Summary
2
Exercise 5.
May 28. a key a is
hosen and the
en
rypted form of b that is transmitted is a 6≡ b .12
2
5. very few s
ientists and engineers are aware of the algebrai
properties of equivalen
e. you en
ounter two natives. 2008
. using the same operation5.
2
The fa
t that inequivalen
e is asso
iative. it is not
ommonly known
that ex
lusive-or and inequivalen
e are the same.
parti
ularly if one
tries to express its properties in natural language.W. This is an entertaining book whi
h leads on from simple logi
puzzles to a dis
ussion of the logi
al paradoxes and Godel's unde
idability theorem. having been mentioned by Alfred Tarski in his PhD thesis. where its dis
overy is
attributed to J. Dijkstra in his work on program semanti
s and mathemati
al
method.g. 2008
.) Nevertheless.
see [Ba
03℄ or [GS93℄.
May 28. whi
h in
ludes dis
ussion of
onjun
tion (\and").
Tarski is a famous logi
ian. [DS90℄. provides ample eviden
e that the
adheren
e to \natural" modes of reasoning is a major impediment to ee
tive reasoning. but to
provide a more powerful alternative. disjun
tion (\or"). and introdu
ing the symbol \ 0 " to denote it.
enturies-long pro
ess of a
epting zero as a
number.
The fa
t that equality of boolean values is asso
iative has been known sin
e at least the
1920's . The painful. Lukasiewi
z. (See e. For a
omplete a
ount of
al
ulational logi
.
The purpose of a
al
ulus is not to mimi
\natural" or \intuitive" reasoning.
But Smullyan's proofs invariably involve detailed
ase analyses.88
5. (See the paper \On the primitive term of logisti
" [Tar56℄.)
The origin of the logi
puzzles is Raymond Smullyan's book \What Is The Name Of
This Book?" [Smu78℄. follows-from (\if") and impli
ation (\only if").
Algorithmi
Problem Solving
Roland Ba
khouse. its usefulness was never re
ognised until
brought to the fore by E. The exploitation of the
asso
iativity of equivalen
e in the solution of su
h puzzles is due to Gerard Wiltink
[Wil87℄. Knights and Knaves
The asso
iativity of equivalen
e
an be diÆ
ult to get used to. this should not be used
as an ex
use for ignoring it. However.
su
h problems are very easy to solve. we then solve the problem in two steps. First. where the number of mat
hes
is a parameter. we apply the indu
tion step to
onstru
t a solution to problems of size 1 from
the known solution to problems of size 0 .1
Example Problems
All the following problems
an be solved by indu
tion. the problem might involve a pile of mat
hsti
ks. et
.
6. (They are often dismissed as \trivial". given
a solution to a problem of size n . a problem of size n+1 . 3 .
The idea is that we somehow measure the \size" of instan
es of a problem. then problems of size 4 . In the . we apply the indu
tion step again to
onstru
t a solution to problems of size 2 from
the known solution to problems of size 1 . We
an now solve problems
of size 3 . This is
alled the basis of the indu
tion. And so it goes on. we know how to solve problems of size
2 . we
onsider problems of size 0 .Chapter 6
Induction
\Indu
tion" is the name given to a problem-solving te
hnique based on using the solution
to small instan
es of a problem to solve larger instan
es of the problem.
whole number | thus 0 . Usually. We also know how to solve problems
of size 1 . an instan
e of the problem is then a parti
ular pile of mat
hes. whole number1. 2 . how the size of an instan
e of a problem is measured
is quite obvious from the problem des
ription.) Se
ond.
By this pro
ess. This is
alled the indu
tion step. for an arbitrary natural number n . et
.
we show how to solve. and its
size is the number of mat
hes in the pile. we
an solve problems of size 0 . A requirement is that the size is a non-negative. For
example. We use the term natural number for a nonnegative.
Having de
ided how to measure size. Almost invariably. 1 . Then.
There are.
89
May 28. 2008
. very good reasons why
0
0
from the natural numbers. and
1 Warning :
Mathemati
ians often ex
lude the number
however. making a break with tradition imperative. the size is the number
of lines.
should always be in
luded. it is expli
tly given by the parameter n .
Algorithmi
Problem Solving
Roland Ba
khouse.rst. In the se
ond and third problems.
it is the number of disks. See . You then
have to solve the indu
tion step.90
6. the basis should be easy. We dis
uss ea
h problem in turn in
oming se
tions. Cutting the Plane
A number of straight lines are drawn a
ross a sheet of paper.
1. ea
h line extending
all the way from from one border to another. In ea
h
ase. Indu
tion
in the fourth.
Show that it is possible to
olour ea
h region
bla
k or white so that no two adja
ent regions have the same
olour (that is. so
that the two regions on opposite sides of any line segment have dierent
olours). and the others are left un
overed.g. Trapeziums
An equilateral triangle. where n is a natural
number2.)
NB: The
ase n = 0 should be in
luded in your solution. with side of length 2n for some natural number n .2 shows. the paper
is divided into a number of regions. In this way. See .
Figure 6.
Show that it is possible to
over the remaining squares with (non-overlapping)
triominoes.
On the right is a triomino. 6. A triomino is an L-shape made of three
grid squares. is
made of smaller equilateral triangles. (Fig.
2.
A bu
ket-shaped trapezium is made from three equilateral triangles.
3.1: Bla
k and White Colouring. Figure 6. One grid square is
overed.1. The topmost equilateral triangle is
overed. an 8×8 grid with one square
overed. 6. on the left. The individual squares are
alled grid squares. Triominoes
A square pie
e of paper is divided into a grid of size 2n×2n .3 shows a solution in one
ase.
g.
Show that it is possible to
over the remaining triangles with (non-overlapping)
trapeziums.2. 6. See .
5 for the solution in the
ase that n is 2 .
4. Towers of Hanoi
The Towers of Hanoi problem
omes from a puzzle marketed in 1883 by the Fren
h
douard Lu
as. under the pseudonym M. In
lude the
ase n = 0 in your solution. Claus. 6.
NB.
mathemati
ian E
The puzzle is based on a legend a
ording to whi
h there is a temple in Bramah
where there are three giant poles .g.
On the .xed in the ground.
rst of these poles. in de
reasing order of size. ea
h of
dierent size.
at the time of the world's
reation. (See . God pla
ed sixty four golden disks.
6. one per day.g. 6. The monks' task will
be
omplete when they have su
eeded in moving all the disks from the . from one pole to another a
ording
to the rule that no disk may ever be above a smaller disk.) The Brahmin monks were
given the task of moving the disks.
)
Algorithmi
Problem Solving
Roland Ba
khouse.g. it does provide the basis for
onstru
ting a so-
alled iterative
solution to the problem.
May 28. on the day that they
omplete their task.
poles to the se
ond and. the world will
ome to an end!
Constru
t an indu
tive solution to this problem. The base
ase is when there are
no disks to be moved.
(We see later that the indu
tive solution is
ertainly not the one that the Brahmin
monks use.4. However. 2008
. 6.
ea
h line
extending all the way from from one border to another.93
6. See .6: Towers of Hanoi Problem
6.2
Cutting The Plane
Re
all the statement of the problem:
A number of straight lines are drawn a
ross a sheet of paper.2. Cutting The Plane
Figure 6.
the number of lines is an obvious measure of the \size" of the
problem.g.
For this problem. thus.
For brevity. The goal is. we
all a
olouring of the regions with the property that no two adja
ent
regions have the same
olour a satisfa
tory
olouring.1. to solve the problem \by indu
tion on the number of lines".
Algorithmi
Problem Solving
Roland Ba
khouse. assuming that we
an solve
the problem when there are n lines |this is the indu
tion step| .
May 28. so that the two regions on opposite sides of any line
segment have dierent
olours). 2008
. Show that it is possible
to
olour ea
h region bla
k or white so that no two adja
ent regions have the
same
olour (that is. 6. the paper is divided into a number of regions. In this
way.
This means that we have to show how to solve the problem when there are zero lines
|this is the \basis" of the indu
tion| and we have to show how to solve the problem
when there are n+1 lines. where n is an arbitrary number.
6.94
6. The sheet of paper is divided into one
region. either
olouring meeting the
onditions
of a solution (be
ause there is no pair of adja
ent regions). su
h pairs of regions will have the same
olour.
For the indu
tion step. and so the existing
olouring is not satisfa
tory. We now suppose that an additional line is drawn on the paper. and this
an be
oloured bla
k or white. Indu
tion
The
ase where there are zero lines is easy. Fig. The
plane has been divided into twelve regions by .7 is an example. and the dierent regions have been
oloured bla
k or white so that no
two adja
ent regions have the same
olour. we assume that a number of lines have been drawn on the
sheet of paper. This assumption is
alled the indu
tion
hypothesis. This will
divide some of the existing regions into two.
either side of the additional line. 2008
.
Figure 6.
easily remembered. Let us
all these regions the left and right regions. Additional line shown in red. In order to guarantee
that. shown in red for
larity. and invert all the
olours in that region. It is just a
onvenient. Now. the regions have the same
olour. This gives a satisfa
tory
olouring
of the left region (be
ause inverting the
olours of a satisfa
tory
olouring gives a satisfa
tory
olouring). as required. It also gives a satisfa
tory
olouring of the right region (be
ause the
Algorithmi
Problem Solving
Roland Ba
khouse.
May 28. This
has had the ee
t of dividing four of the regions into two. and vi
e-versa) also gives a satisfa
tory
olouring.
The key to a solution is to note that inverting the
olours of any satisfa
tory
olouring (that is. way of naming the regions. we do not imply
that the additional line must be from top to bottom of the page.
hoose. the
left region. (By this
hoi
e of names. has been added.) Note that the assumed
olouring is a
satisfa
tory
olouring of the left region and of the right region.7: Cutting the Plane.
hanging a bla
k region to white. all regions have opposite
olour.ve lines. say. On either side of the red line. adja
ent regions have dierent
olours. the additional line divides the sheet of paper into two regions.
Elsewhere. The task is to show how to modify
the
olouring so that it does indeed be
ome a satisfa
tory solution. thus in
reasing the number
of regions by four. An additional line. and the regions
oloured bla
k
and white.
6.95
6.
Figure 6. In order to apply the
onstru
tion to an instan
e
of the problem with. Blue has been used instead of bla
k in
order to make the inversion of the
olours more evident.8: Cutting the Plane. Cutting The Plane
olouring hasn't
hanged.2. say. The
olours are inverted to one side of the additional line
(bla
k is shown as blue to make
lear whi
h
olours have
hanged). we begin by
olouring the whole sheet of paper.8 shows the ee
t on our example.
This
ompletes the indu
tion step. Also. be
ause they have
hanged from being the same to being dierent. seven lines. the
olouring of adja
ent
regions at the boundary of the left and right regions is satisfa
tory.
Then the lines are added one-by-one. Ea
h time a line is added.
Fig. and was satisfa
tory already). the existing
olouring
is modi.
until all seven lines have been added.
The algorithm is non-deterministi
in several ways. The initial
olouring of the sheet
of paper (bla
k or white) is unspe
i.ed as pres
ribed in the indu
tion step.
The ordering of the lines (whi
h to add .ed.
) is also unspe
i. et
.rst.
whi
h to add next.
ed. and whi
h the \right" region is unspe
i. whi
h region is
hosen as the \left"
region. Finally.
ed. This means that the .
But that doesn't matter. The .nal
olouring
may be a
hieved in lots of dierent ways.
nal
olouring
is guaranteed to be \satisfa
tory". as required in the problem spe
i.
The problem assumes the lines are drawn on a pie
e of paper.
Che
k your understanding by
onsidering variations on the problem.
May 28. or on the surfa
e of a doughnut?
We remarked that the algorithm for
olouring the plane is non-deterministi
.1
2
Algorithmi
Problem Solving
Roland Ba
khouse.
ation. 2008
. Is the solution still
valid if the lines are drawn on the surfa
e of a ball. How
many dierent
olourings does it
onstru
t?
Exercise 6.
Why is it required that the lines are straight? How might this assumption be relaxed
without invalidating the solution.
is the
number n . This one square is.
The obvious measure of the \size" of instan
es of the problem. Re
all the statement of the problem. the one that is
overed. We solve the problem by indu
tion on n. i. suppose we
onsider a grid of size 2n+1×2n+1 . and the others are left un
overed. We make the indu
tion hypothesis that it is possible to
over any grid of size 2n×2n with triominoes if. Indu
tion
Triominoes
As a se
ond example of an indu
tive
onstru
tion. A triomino is an L-shape
made of three grid squares.
A square pie
e of paper is divided into a grid of size 2n×2n . . where n is
a natural number.3
6. there is exa
tly one square. Show that it is possible to
over the remaining
squares with (non-overlapping) triominoes. The individual squares are
alled grid squares. in this
ase. is
how the base
ase is solved.1. One grid
square is
overed. That
is.
Now. then.
leaving no squares un
overed. The grid then has size 20×20 . let us
onsider the grid problem posed
in se
tion 6.96
6. It takes 0 triominoes to
over no squares! This. 1×1 . inevitably.e.
The base
ase is when n equals 0 .
bottom-right. We
an apply the indu
tion
hypothesis to them if just one square in ea
h of the three is
overed. We have to show how to exploit this hypothesis
in order to
over a grid of size 2n+1×2n+1 of whi
h one square has been
overed. an
arbitrary grid square has been
overed. Let
us
all the four grids the bottom-left.)
The bottom-left grid is thus a grid of size 2n×2n of whi
h one square has been
overed. the remaining squares in the bottom-left grid
an
be
overed with triominoes.rst. One
grid square is already
overed. top-left. (If not.
A grid of size 2n+1×2n+1
an be subdivided into 4 grids ea
h of size 2n×2n . as yet. This is done by
pla
ing a triomino at the jun
tion of the three grids. and top-right grids. simply
by drawing horizontal and verti
al dividing lines through the middle of ea
h side. This leaves us with the task of
overing the bottom-right. the entire grid
an be rotated about
the
entre so that it does be
ome the
ase.
top-left and top-right grids with triominoes. This square will be in one of the four sub-grids. We may
assume that it is in the bottom-left grid. By the indu
tion hypothesis. as shown in .
None of the squares in these three grids is
overed.
May 28. 6. the indu
tive hypothesis is applied to
over the remaining squares of the bottomright. top-left and top-right grids with triominoes.g.
Now.9.
2
Algorithmi
Problem Solving
Roland Ba
khouse. On
ompletion of this pro
ess. the
entire 2n+1×2n+1 grid has been
overed with triominoes.
Exercise 6. 2008
.2
Solve the trapezium problem given in se
tion 6.1.
4
Looking For Patterns
In se
tions 6. and then testing whether the
onje
tures
an be dedu
ed from existing knowledge. In simple terms. whereby the dedu
tions
made are guaranteed to be true provided that the laws on whi
h they are based are true.
6.
Guess-and-verify is a brief way of summarising mathemati
al indu
tion. as it is normally understood. simply reformulations of existing knowledge. In
ontrast.es one sub-grid.3. the experimental s
ien
es. for example. It's a pro
ess of looking for patterns in a set of observations. and may have to be dis
arded if the predi
tions turn out to be false. The indu
tion hypothesis is then used
to
over all four sub-grids with triominoes. the pro
ess we des
ribed is
alled \mathemati
al indu
tion". indu
tion is about looking for patterns. refers to a pro
ess of
reasoning whereby general laws are inferred from a
olle
tion of observations. (Guessing is the
formulation of a
onje
ture. Te
hni
ally.
is pla
ed at the jun
tion of the other three grids.
Mathemati
al indu
tion is a
ombination of indu
tion and dedu
tion. based on his observations of plant and animal life in
remote parts of the world.
Laws formulated by a pro
ess of indu
tion go beyond the knowledge on whi
h they
are based. shown in blue. at best.
however.
\Indu
tion".2 and 6. su
h laws are only probably true. A triomino.
In a sense. we have seen how indu
tion is used to solve problems of a
given \size". laws dedu
ed from existing laws add nothing to our sto
k of knowledge sin
e
they are. A famous
example of indu
tion is the pro
ess that led Charles Darwin to formulate his theory
of evolution by natural sele
tion. veri. formulating the patterns as
onje
tures. they are tested by the predi
tions they
make.
\indu
tion". is more general. as used in. In the experimental s
ien
es.
dedu
tion is the pro
ess of inferring laws from existing laws. thus introdu
ing inherently new knowledge.
1 or 2 mat
hes is 0 .98
6. 4 or 5 mat
hes is 1 . by de. So. 7 and 8 mat
hes are winning positions. However. 4 .
The basis for the indu
tion is when n equals 0 . A pile of 0 mat
hes is. the \size"
of a pile of 3 .)
Several of the mat
hsti
k games studied in
hapter 4 provide good examples of mathemati
al indu
tion. the game dis
ussed in se
tion 4. and
piles with 1 . Exploring
this game. a
losing position be
ause. and so on. This is a
onje
ture about all positions made from observations on just nine positions.2: there is
one pile of mat
hes from whi
h it is allowed to remove one or two mat
hes.
we
an verify that the
onje
ture is true by using mathemati
al indu
tion to
onstru
t
a winning strategy. Indu
tion
orre
t. we measure the \size" of a pile of mat
hes not by the
number of mat
hes but by the number of mat
hes divided by 3 . The indu
tion hypothesis is that a pile
of 3n mat
hes is a losing position. indeed. 2 . the \size" of a pile of 0 .
In order to use indu
tion. rounded down to the
nearest natural number. 3 or 6 mat
hes is a losing position. we dis
overed that a pile with 0 . and a pile of 3n + 1 or 3n + 2 mat
hes is a winning
position. and winning positions are the remaining positions. There seems to be
a pattern in these numbers: losing positions are the positions in whi
h the number of
mat
hes are a multiple of 3 . Re
all. for example.2. 5 .
the opponent is
left with either 3n + 2 or 3n + 1 mat
hes. A pile of 1 or 2 mat
hes is a winning position be
ause the player
an remove
the mat
hes. whose turn it is. and a pile of 3(n+1) + 1 or 3(n+1) + 2
mat
hes is a winning position. by the indu
tion hypothesis. We have to show that
a pile of 3(n+1) mat
hes is a losing position.
and a pile of 3n + 1 or 3n + 2 mat
hes is a winning position.
Now. But. By taking 1 mat
h in
the . the game is lost when it is no longer possible to
move. this leaves
the opponent in a winning position. we assume that a pile of 3n mat
hes is a losing position.nition. The player. suppose there are 3(n+1) + 1 or 3(n+1) + 2 mat
hes. That is. must remove 1 or
2 mat
hes.
Suppose there are 3(n+1) mat
hes. for the indu
tion step. leaving the opponent in a losing position. Hen
e. the position in whi
h there are 3(n+1) is a
losing position. leaving either 3(n+1) − 1 or 3(n+1) − 2 behind.
Now.
the player leaves the opponent in a
position where there are 3(n+1) mat
hes. and thus veri.rst
ase. the positions in whi
h there are 3(n+1) + 1 or 3(n+1) + 2 are both winning
positions. and 2 mat
hes in the se
ond
ase.
Hen
e. This we now know to be a losing position.
This
ompletes the indu
tive
onstru
tion of the winning moves.
es the
onje
ture that a position is a losing position exa
tly when the number of mat
hes is a
multiple of 3 .
Algorithmi
Problem Solving
Roland Ba
khouse. 2008
.
May 28.
5.99
6. it is vital that any
onje
ture is properly veri. The Need For Proof
6.5
The Need For Proof
When using indu
tion.
The points are
hosen in su
h a way that all
interse
tion points of pairs of
hords are distin
t. This suggests that the
number of portions. It is too easy to
extrapolate from a few
ases to a more general
laim that is not true. for arbitrary n . Indeed. This se
tion is about a non-trivial example of a false
onje
ture. 2 . Many
onje
tures
turn out to be false.
Suppose n points are marked on the
ir
umferen
e of a
ir
ular
ake and then the
ake is
ut along the
hords joining them. 4 and 8 .10 shows the
ase when n is 1 . this
onje
ture is supported by
the
ase that n = 5 . is 2n−1 . 3 or 4 . in how many portions
does this
ut the
ake?
Figure 6. (We leave the reader to draw the .10: Cutting the
ake
The number of portions is su
essively 1 .ed. only by subje
ting them to the rigours of proof
an we be sure of
their validity. 2 .
Figure 6. The question is.
) In this
ase. for n = 6 .gure. whi
h is 25−1 . However. the number of portions is 31 !
(See . the number
of portions is 16 .
) Note that n = 6 is the .11.g. 6.
The derivation of the
orre
t formula for the number of portions is too
ompli
ated to
dis
uss here. and impose the requirement that n is dierent from 0 . is to dismiss this
ase. suspi
ions about the
onje
ture
would already have been raised | it doesn't make sense to say that there are 20−1 portions. but inadequate
way out.rst
ase in whi
h the points are not allowed to
be pla
ed at equal distan
es around the perimeter.6
From Verification to Construction
In mathemati
al texts. indu
tion is often used to verify known formulae. Veri. even though
utting the
ake as stated does make sense! The easy. but it does in
lude the
ase that n equals 0 !
6.
Had we begun by
onsidering the
ase that n = 0 .
ation is
important but has a major drawba
k | it seems that a substantial amount of
lairvoyan
e
Algorithmi
Problem Solving
Roland Ba
khouse. 2008
.
May 28.
The
ase n = 6 .11: Cutting the
ake.100
6. not 26−1
is needed to
ome up with the formula that is to be veri. Indu
tion
Figure 6. The number of portions is 31 .
And. if one's
onje
ture is
wrong.ed. veri.
Indu
tion is not important in
omputing s
ien
e as a veri.
ation gives little help in determining the
orre
t formula.
ation prin
iple but be
ause it is a fundamental prin
iple in the
onstru
tion of
omputer programs. This
se
tion introdu
es the use of indu
tion in the
onstru
tion of mathemati
al formulae.
The problem we
onsider is how to determine a
losed formula for the sum of the
k th powers of the .
for example. but what happens if the answer is not already known! Suppose.
that you are now asked to determine a
losed formula for the sum of the 4 th powers of
the .
ation: the te
hnique works if the answer
is known.
Redu
ing the guesswork to a minimum. and trial-and-error
annot be
ompletely
eliminated. 2 and
3 .
does not seem to be appli
able unless we already know the right side of the equation. the sum of the m th powers of the .
ation. for m equal to 1 . The key to su
ess is simpli
ity.
A simple pattern in the formulae displayed above is that. Don't be over-ambitious. repla
ing it by mathemati
al
al
ulation is the key to su
ess. say. Leave the
work to mathemati
al
al
ulation.
Can you guess what the right side would be in this
ase? Can you guess what the right
side would be in the
ase that the term being summed is.
not!
Constru
ting solutions to non-trivial problems involves a
reative pro
ess. k27 ? Almost
ertainly. using the prin
iple of mathemati
al indu
tion. formulate the pattern in pre
ise mathemati
al terms and then
verify the pattern. This means
that a
ertain amount of guesswork is ne
essary. The general
idea is to seek a pattern.
Indu
tion
an be used to
onstru
t
losed formulae for su
h summations.
(The sum of the .rst n numbers is a polynomial in n of degree
m+1 .
the sum of the
.rst n numbers is a quadrati
fun
tion of n .
and the sum of the .rst n squares is a
ubi
fun
tion of n .
rst n
ubes is a quarti
fun
tion of n .) This pattern is also
on.
A strategy for determining a
losed formula for. + n0 = n . . say. the sum of the fourth powers is thus
to guess that it is a . .rmed in the (oft-forgotten)
ase that m is 0 :
10 + 20 + .
so let us illustrate the pro
ess
by showing how to
onstru
t a
losed formula for 1 + 2 + . The
al
ulation in this
ase is quite long.. b and c .
Then.)
We
onje
ture that the required formula is a se
ond degree polynomial in n . + n . let us use S. The method des
ribed here is more general.fth degree polynomial in n and then
. If this is the
ase.n to denote the proposition
S. .0
=
{
de..n = a + bn + cn2 .
For brevity.
P.n to denote
use indu
tion to
al
ulate
the
oeÆ
ients
1+2+ . please bear with us. Here is how the
al
ulation
goes. say
a + bn + cn2 and then
al
ulate the
oeÆ
ients a . +n . . (Some readers will
already know a simpler way of deriving the formula in this parti
ular
ase.
We also use P.
(n+1)
=
de. Then
P. the
oeÆ
ient of n0 . To do so.102
6.
So the basis of the indu
tion has allowed us to dedu
e that a .n is true. we
al
ulate b and c . Indu
tion
S.0 = 0 (the sum of an empty set of numbers
{
is zero) and arithmeti
}
0=a . we make the indu
tion hypothesis that 0 ≤ n
and P.0 = a + b×0 + c×02
=
S. Now. is
0 .
+ nm as a polynomial fun
tion for any given natural number m .n + (n+1)m (where S.. ..
2
2
Extrapolating from this
al
ulation. The steps
in the algorithm are: postulate that the summation is a polynomial in n with degree
m+1 .(n+1) is S. we have
thus
al
ulated that
1+2+ .0 is 0
and S. 2008
.n denotes 1m + 2m + . . + nm ) to determine
Algorithmi
Problem Solving
Roland Ba
khouse. one
an see that it embodies an algorithm to express
1m + 2m + .
May 28. . Use the prin
iple of mathemati
al indu
tion together with the fa
ts that S. .rst n numbers is a quadrati
in n . +n =
1
1
n + n2 .
Try to identify a simple pattern in the way
winning and losing positions are grouped. .
Algorithmi
Problem Solving
Roland Ba
khouse. .4
n
1
n
2
6. .
From the fa
t that there are n o
urren
es of n+1 we
on
lude that the sum is
1
n(n+1) .
n+1
2
Consider a mat
hsti
k game with one pile of mat
hes from whi
h m
thru mat
hes
an be removed.6
an be summarised as \Don't guess! Cal
ulate. However. . .. . and what the winning
strategy is.
Remark : In the
ase of the sum 1 + 2 + .3 Use the te
hnique just demonstrated to
onstru
t
losed formulae for
10 + 20 + ." We put this
into pra
ti
e in this se
tion. + n+1 . solve the system of
equations. + n there is an easier way to derive the
orre
t formula. . or m is 2 and n is 3 ). . . By
onsidering a few simple examples (for example.
Exercise 6. + n2 ..
Then add the two rows together:
+ n+1 + .103
6. . m
is and is arbitrary.
+
1
.
May 28.7
Fake-Coin Detection
The motto of se
tion 6. End of remark.
Exercise 6. Fake-Coin Dete
tion
a system of simultaneous equations in the
oeÆ
ients. formulate a general rule for determining
whi
h are the winning positions and whi
h are the losing positions.. + nm
2
for m greater than 1 . . Simply write down the required sum
1
+ 2 + . 2008
. Introdu
e variables to represent the grouping. this method
annot be used for determining 1m + 2m + .
and
al
ulate the values of the variables.
and immediately below it
n
+
n−1
+
. + n .7. Finally.
Avoid guessing the
omplete solution. + n0 and 12 + 22 + .
We are
told that among them there is at most one \fake"
oin.
All \genuine"
oins have the same weight. Indu
tion
Suppose we are given a number of
oins. The problem is how to use the pair of s
ales optimally in order to
. and all the rest are \genuine".104
6. whereas a \fake"
oin has a dierent weight
to a \genuine"
oin. ea
h of the same size and shape.
and needs revision. More pre
isely. if there are no other
oins to
ompare it with? Our
onje
ture has
broken down. in this
ase. if the
number. Then there are 1 + 2m dierent possibilities that
an be observed with a pair
of s
ales: \ 1 " possibility is that all
oins are genuine.
Algorithmi
Problem Solving
Roland Ba
khouse. Several
stages are needed. however.
We have almost rea
hed the point at whi
h we
an state our problem pre
isely. or they may tip to the right.
Now. there are
at most 3n dierent out
omes3. we are given (3n−1)/2
oins about whi
h we know nothing ex
ept that at most one is fake. Often. and we are also given
3 Note
the impli
it use of indu
tion here. they may
balan
e. of
oins is greater than (3n−1)/2 .1
Problem Formulation
When we use a pair of s
ales to
ompare two weights |an operation that we
all a
omparison | there are 3 possible out
omes: the s
ales may tip to the left. For n equal to 1 . of whi
h at most one is fake and the rest are
genuine. it is impossible to guarantee that a fake
oin
an be found with n
omparisons. if there is one. given (3n−1)/2
oins of whi
h at most one is fake. otherwise. the
onje
ture is
learly true. we run into a problem. it is possible to
establish that all are genuine or identify the fake
oin (and whether it is lighter or
heavier than a genuine
oin) using at most n
omparisons.
6. an essential element of problem
solving is to
learly identify the problem itself. Our formulation of the problem and its
eventual solution illustrates several other aspe
ts of \real" problem solving. the number of
oins among whi
h at most
one fake
oin
an be dete
ted is at most m . This means that.7. This gives an upper bound on what
an be a
hieved
using a pair of s
ales. This is deliberate. there are \ 2 " ways
that ea
h of the \ m "
oins may be fake (by being lighter or heavier than a genuine
oin). This means that with n
omparisons.
For n equal to 0 . We
onje
ture that. suppose we are given m
oins.
Note the element of vagueness in this problem statement. The assumption
is that there is one
oin (sin
e (31−1)/2 = 1 ). But how
an we tell whether this one
oin
is fake or genuine. m . in
luding some \ba
ktra
king" and revision.
May 28. with n
omparisons. 2008
. all of
whi
h are genuine. where 1 + 2m = 3n . we don't say what we mean
by using the s
ales \optimally". Thus. there are no
oins.nd the fake
oin.
We propose to modify the
onje
ture by assuming that we have at our disposal at
least one additional
oin that is known to be genuine.
using
at most n
omparisons.
6. if it exists. in its solution. We have
to show how to . For
brevity. The base
ase.
an be found among c. or determine that all
oins are genuine.
The Basis With zero
omparisons. The problem is to
onstru
t an algorithm
that will identify the fake
oin.
Induction Step Now.7. Suppose n is at least zero. we ta
kle the indu
tion step. n equal to 0 . By indu
tion. if it exists. is thus solved. let us use c. Fake-Coin Dete
tion
at least one
oin that is known to be genuine.2
Problem Solution
Our formulation of the problem begs the use of indu
tion on the number of
omparisons.105
6.
n . we
an report immediately that all
oins in a
olle
tion of (30−1)/2 are genuine.n
oins using at most n
omparisons. we may assume that a fake
oin.n to denote (3n−1)/2 .7.
nd a fake
oin. if it exists. using at most n+1
omparisons.(n+1)
oins. among c.
Consider the .
from whi
h one
infers that none of the
oins on the s
ales is fake.rst
omparison. the number of
oins on the two s
ales must be equal. The algorithm would then pro
eed to
try to . and leaving some on the table. To be able to draw any
on
lusion from the
omparison. It involves putting some number of
oins on the left
s
ale. some on the right s
ale.
One possible
onsequen
e of the
omparison is that the s
ales balan
e.
Combined with the indu
tion hypothesis.) We
on
lude
that in the .n + 1 = 3n .nd a fake
oin among the
oins left on the table.
c.n + 1 .(n+1) = (3n+1−1)/2 = 3×((3n−1)/2) + 1 = 3 × c.(n+1) − c. Now.n
oins must be left
on the table.
It also di
tates how many
oins should be put on the s
ales | this is the dieren
e
between c.(n+1) and c.n is the maximum number of
oins among whi
h a fake
oin
an be found with n
omparisons. this di
tates that c. This is be
ause c.n .(n+1)
oins whose kind we must determine. it
an be made even by using one of the
oins we know to be
genuine. in addition to the c. (Re
all the assumption that we have at least one
oin that is known to be
genuine.
This is an odd number.n = 2 × c.
So
c.
n + 1
oins should be put on ea
h of the two s
ales.rst
omparison.
The next step is to determine what to do after the . c.
There are
three possible out
omes.rst
omparison is made. of whi
h we have already dis
ussed one. 2008
.
Algorithmi
Problem Solving
Roland Ba
khouse.
May 28. If the s
ales balan
e.
the base
ase is easy. n )
omparisons are needed to determine this fa
t.
The Marked Coin Problem In this way.e. Exa
tly one of the
oins is fake. there is one
oin. And
ru
ially.106
6. Indu
tion
the fake
oin should be sought among the c.
and
an be eliminated from
onsideration. we
an
on
lude that all the
oins left on the table are genuine. If the s
ales tip
to one side. The new problem
is this. But we are still left with 3n
oins none of
whi
h we know to be genuine. It is too weak! If
the s
ales tip to one side. That is.
For the indu
tion step. Suppose we are supplied
with 3n+1 marked
oins. 3n is greater than c.
the problem we started with has been redu
ed to a dierent problem.
Again. 0 (i. By
\possibly lighter" we mean genuine. and all the
oins on the other side are possibly lighter than a genuine
oin. By \possibly heavier" we mean
genuine. We are unable to
apply the indu
tion hypothesis to this number of
oins.
Constru
t an algorithm that will determine. or fake and heavier. or fake and lighter. we know that all the
oins on that side are possibly heavier than a genuine
oin. and all the rest are genuine.
The
omparison does tell us something about the
oins on the s
ales. The problem is
what to do if the s
ales tip either to the left or to the right.n
oins left on the table. we
an mark all the
oins on the
s
ales one way or the other.
At this point. we realise that the indu
tion hypothesis doesn't help. In the . the fake
oin
among 3n marked
oins. ea
h of whi
h is marked either \possibly
light" or \possibly heavy". Suppose a number of
oins are supplied.n . in the
ase that the s
ales do not balan
e. with at most n
omparisons. After the
omparison. we pro
eed as for the earlier problem. whi
h must be the fake
oin. If n equals 0 .
as already determined.
To draw any
on
lusion from the
omparison.rst
omparison.
The
oins are marked in two dierent ways. Combining this with the markings. we need to determine how to pla
e
the
oins a
ording to their markings. we
on
lude that all
oins
on the left s
ale are possibly heavy.
Now. Similarly. We
al
ulate the numbers as follows. l1+h1 and l2+h2 should be
equal. Furthermore. they should equal 3n .
Suppose l1 possibly light
oins are pla
ed on the left s
ale and l2 possibly light
oins on the right s
ale. the
oins must be divided equally: 3n
oins must be left on the table. 2008
. In order to apply the indu
tion
hypothesis in the
ase that the s
ales balan
e. and some are left on the table. So.
May 28. suppose h1 possibly heavy
oins are pla
ed on the
left s
ale and h2 possibly heavy
oins on the right s
ale. That is. if the
omparison
auses the s
ales to tip to the left. we
on
lude that the l1 possibly light
oins
Algorithmi
Problem Solving
Roland Ba
khouse. some
oins are put on the left s
ale. and thus 3n put on the left s
ale and 3n on the right
s
ale. and all the
oins on the right s
ale are possibly
light.
some on the right. we require that the number of
oins
on the left s
ale equals the number on the right.
The fake
oin is identi.
The Complete Solution This
ompletes the solution to the marked-
oin problem. until ea
h s
ale has its full
omplement of 3n
oins. we require that the number of
oins not eliminated be equal to 3n . this leaves h1+l2
oins to be
investigated further. whatever the out
ome of the
omparison.
leaving l1+h2
oins to be investigated further.
Again.7.
one on the left and one on the right. if the s
ale tips to the right. in order to apply the indu
tion hypothesis.107
6. Simply pla
e the
oins on the s
ales two at a time. the h1 possibly heavy
oins on the left s
ale and the l2 possibly heavy
oins on the right s
ale are genuine. at least two of them will have the same marking. We must arrange the
oins so that ea
h s
ale
ontains
equal numbers of
oins of the same kind. we
infer that l1 = l2 and h1 = h2 . by
hoosing any three
oins. Fake-Coin Dete
tion
on the left s
ale and the h2 possibly heavy
oins on the right s
ale are in fa
t genuine
(sin
e possibly heavy and possibly light equals genuine). Conversely. The
hoi
e
an always be made
be
ause there are always at least three
oins from whi
h to
hoose.
This requirement
an be met. This
imposes the requirement that h1+l2 = l1+h2 = 3n . Together with l1+h1 = l2+h2 .
and thus to the unmarked-
oin problem. always
hoosing two
oins with the same marking.
Pro
eed with the possibly heavy
oins on the left s
ale and the possibly light
oins on the right s
ale. A
ording to the out
ome of
the
omparison.
So too are the
possibly light
oins on the right s
ale and the possibly heavy
oins on the left s
ale.
If the s
ales tip to the right. in su
h a way that there is an
equal number of possibly light
oins on ea
h of the s
ale. (3n−1)/2 + 1 and (3n−1)/2 .
So too are the
possibly light
oins on the left s
ale and the possibly heavy
oins on the right
s
ale.
If the s
ales balan
e.
Divide the
oins into three groups of sizes (3n−1)/2 .
The solution to the unmarked-
oin problem when the number of
oins is (3n+1−1)/2
is as follows. one of the following is exe
uted. all the
oins on the s
ales are genuine. Pro
eed with the
oins
left on the table.
Pro
eed with the possibly heavy
oins on the right s
ale and the possibly light
oins
on the left s
ale.
If the s
ales tip to the left.
Pla
e the . the
oins on the table are genuine.ed from a
olle
tion
of 3n+1 marked
oins by pla
ing 3n
oins on ea
h s
ale. the
oins on the table are genuine.
May 28. and leave the third group on the table. Determine the
out
ome of the
omparison. 2008
. and pro
eed as follows:
Algorithmi
Problem Solving
Roland Ba
khouse.rst group on the left s
ale together with the supplied genuine
oin. Pla
e the
se
ond group on the right s
ale.
with the ex
eption of the supplied genuine
oin as \possibly heavy".108
6.5
2
Given are n obje
ts.
If the s
ales tip to the left. Mark all the
oins on
the left s
ale.
alled the unique obje
t.
Suppose you are given a number of obje
ts. Apply the solution
to the unmarked-
oin problem (indu
tively) to the
oins on the table. you are provided with a pair of s
ales. with the ex
eption of one. In all other respe
ts. even though the development ne
essitates
several su
h hypotheses. all the
oins on the balan
e are genuine. Indu
tion
If the s
ales balan
e. You are required to determine
whi
h is the unique obje
t. the
oins on the table are genuine.
If the s
ales tip to the right. This is the epitome of the art of ee
tive reasoning. with the ex
eption of the supplied genuine
oin as \possibly light". Apply the solution to the
marked-
oin problem to the 3n marked
oins.
3m+1 obje
ts
an be split into 3 groups of 3m obje
ts. whi
h is
the lighter and whi
h is the heavier. A pair of
s
ales is provided so that it is possible to determine.
Mark the
oins on the right s
ale as \possibly light". the obje
ts are identi
al. whi
h has a dierent
weight. Mark all the
oins
on the left s
ale.
Mark the
oins on the right s
ale as \possibly heavy". Apply the solution to the
marked-
oin problem to the 3n marked
oins. for any two of the obje
ts. All the obje
ts have the
same weight. (Hint: for the indu
tion step. Quite the opposite: ea
h hypothesis is systemati
ally
al
ulated
from the available information. For this purpose. where 1 ≤ n .
a) How many
omparisons are needed to .
Show. Note that at no stage
is a guess made at an indu
tive hypothesis.)
Can you identify whether the unique obje
t is lighter or heavier than all the other
obje
ts?
Exercise 6. that at most 2×m
omparisons are needed to identify
the unique obje
t when the total number of obje
ts is 3m . by indu
tion on m .
We ask the reader to review the development of this algorithm. ea
h of dierent weight. the
oins on the table are genuine.
that it is possible to determine whi
h is the lightest and
whi
h is the heaviest obje
t using 2n − 3
omparisons. B . C and D . Show how to .
) Suppose there are 4 obje
ts with weights A . by indu
tion on n .nd the lightest obje
t?
b) Show. Assume that 2 ≤ n . and suppose A < B
and C < D .
nd the lightest and heaviest of all four with two additional
omparisons. Use this to show how to .
Summary
d) Suppose there are 2m obje
ts.109
6. Show. that it
is possible to .8. where 1 ≤ m . by indu
tion on m .
and using
mathemati
al
al
ulation to . (Hint:
make use of (
). The prin
iple of
mathemati
al indu
tion is that instan
es of a problem of arbitrary \size"
an be solved
for all \sizes" if
(a) instan
es of \size" 0
an be solved. The pro
ess may involve some
reative
guesswork. by striving for simpli
ity. it is possible to
adapt the method to solve instan
es of \size" n+1 . whi
h is then subje
ted to the rigours of mathemati
al dedu
tion.)
2
6. for arbitrary n .
Using indu
tion means looking for patterns.
(b) given a method of solving instan
es of \size" n .nd the lightest and heaviest obje
ts using 3m − 2
omparisons. The key
to su
ess is to redu
e the guesswork to a minimum.8
Summary
Indu
tion is one of the most important problem-solving prin
iples.
Chapter 7
The Towers of Hanoi
This
hapter is about the Towers of Hanoi problem. and is often used in
omputing s
ien
e and arti. The problem is dis
ussed in many
mathemati
al texts.
ial intelligen
e as
an illustration of \re
ursion" as a problem-solving strategy.
The Towers of Hanoi problem is a puzzle that is quite diÆ
ult to solve without a
systemati
problem-solving strategy. Indu
tion gives a systemati
way of
onstru
ting
a .
A better solution is obtained by
observing an invariant of the indu
tive solution. Another
reason is to illustrate how diÆ
ult it
an be to understand why a
orre
t solution has
been found if no information about the solution method is provided. However. In this way.rst solution. under the pseudonym M. but
also the properties of logi
al equivalen
e. the Towers of Hanoi problem is one that is not
solved in one go.1
7.
For this problem. this
hapter brings together
a number of the te
hniques dis
ussed earlier: prin
ipally indu
tion and invariants. we begin with the solution of the problem.1. this solution is undesirable.
7.1
Specification and Solution
The End of the World!
The Towers of Hanoi problem
omes from a puzzle marketed in 1883 by the Fren
h
douard Lu
as. Claus. several steps are needed before a satisfa
tory solution is found.
mathemati
ian E
The puzzle is based on a legend a
ording to whi
h there is a temple in Bramah
where there are three giant poles . One reason for doing
so is to make
lear where we are headed.
xed in the ground. On the .
God pla
ed sixty-four golden disks. (See . at the
time of the world's
reation. ea
h of dierent size.
in de
reasing order of size.rst of these poles.
from one pole to another a
ording to the rule that no
disk may ever be above a smaller disk. 7.1.) The Brahmin monks were given the task of
moving the disks. one per day. 2008
.g.
111
May 28. The monks' task will be
omplete when they
Algorithmi
Problem Solving
Roland Ba
khouse.
The Towers of Hanoi
have su
eeded in moving all the disks from the .112
7.
On every alternate day. the world will
ome to an end!
Figure 7. On day 0 . With these assumptions. we assume that the poles are arranged at
the three
orners of a triangle.rst of the poles to the se
ond and. To formulate the solution. the solution
is the following. Movements of the disks
an then be su
in
tly des
ribed
as either
lo
kwise or anti
lo
kwise movements.2
Iterative Solution
There is a very easy solution to the Towers of Hanoi problem that is easy to remember
and easy to exe
ute. We assume that the problem is to move
all the disks from one pole to the next in a
lo
kwise dire
tion. beginning on the . on
the day that they
omplete their task. We also assume that days
are numbered from 0 onwards. the disks are pla
ed in their initial position
and the monks begin moving the disks on day 1 .1: Towers of Hanoi Problem
7.1.
If the total number of
disks is odd.
Algorithmi
Problem Solving
Roland Ba
khouse. The
dire
tion of rotation depends on the total number of disks. Otherwise. The
rule for moving the smallest disk is that it should
y
le around the poles. the smallest disk is moved. the smallest disk should
y
le in a
lo
kwise dire
tion.rst day.
May 28. it
should
y
le in an anti
lo
kwise dire
tion. 2008
.
that is. a 4 -disk puzzle.
The algorithm terminates when no further moves are possible. Take
are to
y
le the
smallest disk on the odd-numbered moves and to obey the rule not to pla
e a disk on
top of a disk smaller than itself on the even-numbered moves. It is easy to see that be
ause
of this rule there is exa
tly one move possible so long as not all the disks are on
one pole. on an evennumbered day when all the disks are on one-and-the-same pole. you will . a disk other than the smallest disk is moved | subje
t to the
rule that no disk may ever be above a smaller disk. If you do. say.2.113
7.
Try exe
uting this algorithm yourself on. Indu
tive Solution
On every other day.
1. and so on.nd
that the algorithm works. you
an exe
ute
the algorithm on larger and larger problems | 5 -disk. provides no help whatsoever in understanding how the solution is
onstru
ted. Matters would be made even worse
if we now pro
eeded to give a formal mathemati
al veri. it only serves to impress |look
at how
lever I am!| but in a reprehensible way.3
WHY?
Presenting the problem and its solution. 6 -disk. Depending on how mu
h patien
e you have. like this.
7. If anything.
ation of the
orre
tness of the
algorithm. we . This is not how we intend to pro
eed! Instead.
Suppose that
the task is to move M disks from one spe
i.
7.2
Inductive Solution
Constru
ting a solution by indu
tion on the number of disks is an obvious strategy.rst present an indu
tive
solution of the problem. Then. we show how to
derive the algorithm above from the indu
tive solution. by observing a number of invariants.
Let us begin with an attempt at a simple-minded indu
tive solution.
pole to another spe
i.
After doing this.
Algorithmi
Problem Solving
Roland Ba
khouse. the basis is easy.
Here.)
As often happens. For the indu
tive step. we see that naming the poles is
inadvisable. 2008
.
May 28. and the problem is to show how to move n+1 disks from A to B. and the third pole C. Let us
all these poles A and B. we soon get stu
k! There is only a
ouple of ways that the indu
tion hypothesis
an be used. and
we have no hypothesis about moving disks from this pole. we have exhausted all possibilities of using the indu
tion hypothesis be
ause n disks are now on pole B. (Later. When the number of disks is 0 no steps are
needed to
omplete the task. but these lead nowhere:
1.
pole. Move the top n disks from A to B. we assume that we
an move n
disks from A to B.
On
e again. and we have no hypothesis about moving disks
from this pole. Move the smallest disk from A to C. we have exhausted all possibilities of using the indu
tion hypothesis. The Towers of Hanoi
2.
be
ause n disks are now on pole B.114
7.
The mistake we have made is to be too spe
i. Then move the remaining n disks from A to
B.
about the indu
tion hypothesis. The
way out is to generalise by introdu
ing one or more parameters to model the start and
.
we make a
ru
ial de
ision. we observe that the problem exhibits a rotational symmetry.
as we did in . The rotational
symmetry is obvious when the poles are pla
ed at the
orners of an equilateral triangle. B and
C.
At this point. Rather than name the poles (A. say).nish positions of the disks.
namely. as is often done.g. we need only say whether it is to be moved
lo
kwise or
anti
lo
kwise from its
urrent position.1. the generalisation of the Towers of Hanoi
problem be
omes how to move n disks from one pole to the next in the dire
tion d . Also.
where d is either
lo
kwise or anti
lo
kwise.
The importan
e of this observation is that only one additional parameter needs to
be introdu
ed. The alternative of naming the poles leads
to the introdu
tion of two additional parameters. That is. the start and . (This rotational symmetry is obs
ured by pla
ing the poles in a
line. in order to spe
ify how a
parti
ular disk is to be moved. the dire
tion of movement.) The problem does not
hange when we rotate the poles and disks
about the
entre of the triangle. 7.
It is not suÆ
ient to simply take the problem spe
i.
Now. we
an return to the indu
tive solution again.nish positions of the
disks. This is mu
h more
ompli
ated sin
e it involves unne
essary additional detail. We need to take
are in formulating the indu
tion hypothesis.
This is be
ause the problem spe
i.
ation
as indu
tion hypothesis.
We must show how to move n+1 disks from one
pole to its neighbour in dire
tion d . We need a stronger indu
tion hypothesis. we assume that the disks are numbered from 1 upwards. where d is either
lo
kwise or anti
lo
kwise. with the smallest
disk being given number 1 .
The indu
tion hypothesis we use is that it is possible to move the n smallest disks.
Algorithmi
Problem Solving
Roland Ba
khouse. the requirement that a larger disk may not be pla
ed on
top of a smaller disk may be violated. 2008
. it is ne
essary to move
n disks in the presen
e of M−n other disks.
ation assumes that there
are exa
tly M disks that are to be moved. but no disk is on top of a disk smaller than itself). If some of these M−n disks are smaller
than the n disks being moved.
May 28.
In the
ase that n is 0 . the sequen
e of moves is the empty sequen
e. In the
ase of
n+1 disks we assume that we have a method of moving the n smallest disks from one
pole to either of its two neighbours. beginning from any valid starting
position (that is. For
onvenien
e.
from one pole to its neighbour in the dire
tion d . a starting position in whi
h the disks are distributed arbitrarily over
the poles. When using indu
tion.
the move is valid be
ause the n disks smaller than disk n+1 are not on
the pole to whi
h disk n+1 is moved. However. (This a
tion may pla
e disk n+1 on top of another
disk.2. We
an begin by moving the n smallest disks in the dire
tion d .
and all n+1 smallest disks have now been moved from their original position to the
neighbouring pole in dire
tion d .) Finally. This pla
es them above disk n+1 . Any other initial
hoi
e of move would pre
lude the use of the indu
tion hypothesis.
Some further thought (preferably assisted by a physi
al model of the problem) reveals
that the solution is to move the n smallest disks in the dire
tion ¬d . Indu
tive Solution
Given the goal of exploiting the indu
tion hypothesis. there is little
hoi
e of what to
do. Then disk n+1
an be moved in the dire
tion d. The
ode
de.115
7.
The following
ode summarises this indu
tive solution to the problem. we use the indu
tion hypothesis again
to move the n smallest disks in the dire
tion ¬d . or in the dire
tion
¬d .
The semi
olon operator
on
atenates sequen
es together.d pres
ribes how to move the n smallest disks one-by-one from one pole
to its neighbour in the dire
tion d . d ′ i where n is the number of disks. following the rule of never pla
ing a larger disk on
top of a smaller disk. meaning
lo
kwise and anti
lo
kwise.d := 1. n. 2008
.)
H2. di] .nes Hn. For example. Taking the pairs in order from left to right.d to be a sequen
e of pairs hk .d . The pair hk . but unrestri
ted use of re
ursion
an be
unreliable.
This indu
tive pro
edure gives us a way to generate the solution to the Towers of
Hanoi problem for any given value of n | we simply use the rules as left-to-right
rewrite rules until all o
urren
es of H have been eliminated.cw
}
Algorithmi
Problem Solving
Roland Ba
khouse. [ ] denotes an empty sequen
e and [x] is a sequen
e
with exa
tly one element x .
May 28.d = [ ]
Hn+1. rather
than true and false in order to improve readability.
H0. d ′ i means move the
disk numbered k from its
urrent position in the dire
tion d ′ .d = Hn . Dire
tions are boolean values. here is how
we determine H2. k is
a disk number and d and d ′ are dire
tions. The form of re
ursion used here is limited. the
omplete
sequen
e Hn. (We use cw and aw .cw . true representing a
lo
kwise
movement and false an anti-
lo
kwise movement. Re
ursion
is a very powerful problem-solving te
hnique. ¬d
Note that the pro
edure name H re
urs on the right side of the equation for Hn+1. des
ribing the solution as an
\indu
tive" solution makes
lear the limitation on the use of re
ursion. Hn . [hn+1 .
Be
ause of this we have what is
alled a re
ursive solution to the problem. ¬d .
disk 1 being the smallest.cw
=
{
2nd equation. Disks are numbered from 1 onwards.
[ ] . The implementation of the
indu
tive solution.aw
}
H0. H0.1.awi] .aw
=
{
2nd equation. [ ]
=
{
on
atenation of sequen
es
}
[h1.cw . and so on.1 is an iterative solution to
the problem.cw . H1.d := 0.
As an exer
ise you should determine H3.awi] . you will
qui
kly dis
over that this indu
tive solution to the problem takes a lot of eort to put
into pra
ti
e.e. H0. The
omplete expansion of the equations in the
ase of n = 3 takes 16
steps. [h1.aw in the same way. H0.cwi .d denote the length of the sequen
e Hn. it is a solution that involves iteratively (i. on the other hand.awi] . [ ] . [h1. If you do.cwi] .cw
=
{
1st equation }
[ ] . Let Tn.awi] . [h2. h1.cwi] . The memory of Bramin monks is unlikely to be large enough
to do that!
The number of days the monks need to
omplete their task is the length
of the sequen
e H64.awi . [h2. in the
ase of n = 4 takes 32 steps.cw .aw .d . Derive
an indu
tive de. h2. That is.116
7.awi] . involves maintaining a sta
k of the sequen
e of
moves yet to be exe
uted. The Towers of Hanoi
H1. n.cw .cwi] . [h1. This is not the easy solution that the
Bramin monks are using! The solution given in se
tion 7. [h1. [h2. repeatedly) exe
uting
a simple pro
edure dependent only on the
urrent state.
nition of T from the indu
tive de.
(You should .nition of H .
) Use this de.d is independent of d .nd that
Tn.
2008
. T1 and T2 .
Use indu
tion to show how to
onstru
t a state-transition diagram that shows all
possible states of n disks on the poles. Hen
e.1
2
Use indu
tion to derive a formula for the number of dierent states in
the Towers of Hanoi problem.
Use the
onstru
tion to show that the above solution optimises the number of times
that disks are moved.2
2
Algorithmi
Problem Solving
Roland Ba
khouse.
Exercise 7. and the allowed moves between states.
May 28. or
otherwise.
Exercise 7. Prove
your
onje
ture by indu
tion on n .nition to evaluate To . formulate a
onje
ture expressing Tn as an arithmeti
fun
tion of n .
The Iterative Solution
7.3.1.2. It has two main
elements: the .3
The Iterative Solution
Re
all the iterative solution to the problem.117
7. presented in se
tion 7.
the se
ond is that
the disk to be moved alternates between the smallest disk and some other disk. the value of even. we show that the smallest disk always
y
les around the poles.d the boolean value
even.
This formula allows us to determine the dire
tion of movement d of disk k . Spe
i.
The key is that. for all pairs hk .k ≡ d ′ is invariant (that is always true or always false). the parameter \ n+1 " is repla
ed by \ n " and \ d " is repla
ed by
\ ¬d ". its dire
tion
of movement is invariantly
lo
kwise or invariantly anti
lo
kwise).(n+1) ≡ d remains
onstant
under this assignment. d ′ i in the sequen
e Hn+1. and we
al
ulate
the dire
tion of movement of ea
h.
for all moves hk .
Whether even. In this
se
tion.d )
will depend on the initial values of n and d .d is applied. we
do more than this.
Cyclic Movement of the Disks
In this se
tion. Then.k ≡ d ≡ even. This is a simple
onsequen
e of the rule of
ontraposition dis
ussed in se
tion 5. When the formula for
Hn+1. d ′ i in the sequen
e Hn+1. Sin
e even. we show how these properties are derived from the indu
tive solution.rst is that the smallest disk
y
les around the poles (that is. Let us suppose these are N and D .(n+1) ≡ ¬(even. In fa
t.5. We show that all the disks
y
le around the poles.N ≡ D .n) . di . we have
even.k ≡ d ′ is true or false (for all pairs hk .
and all odd-numbered disks should
y
le in a
lo
kwise dire
tion. and all odd-numbered disks should
y
le in an anti
lo
kwise dire
tion. At the time of his dis
overy. so that the disks were arranged on all three poles. all evennumbered disks should
y
le in a
lo
kwise dire
tion. However. all even-numbered disks should
y
le in an anti
lo
kwise
dire
tion. and the same dire
tion as D if N is odd.
the smallest disk (whi
h is odd-numbered) should
y
le in a dire
tion opposite to D if
N is even. In parti
ular.
The poles were arranged in a line and not at the
orners of the triangle so he wasn't
sure whi
h dire
tion was
lo
kwise and whi
h anti
lo
kwise. Vi
e-versa. the monks had
got some way to
ompleting their task. if it is required to move an odd number of
disks in a
lo
kwise dire
tion.
ally. on the day of his
arrival he was able to observe the monks move the smallest disk from the middle pole
Exercise 7.
May 28.3
Algorithmi
Problem Solving
Roland Ba
khouse. 2008
.
An explorer on
e dis
overed the Bramin temple and was able to se
retly
observe the monks performing their task.
if it is required to move an even number of disks in a
lo
kwise dire
tion.
118
7. We now want to give a formal proof that the sequen
e
Hn. On the next day. The Towers of Hanoi
to the rightmost pole. And. two
onse
utive moves of a disk other than the smallest have no
ee
t on the state of the puzzle.d satis. he saw the monks move a disk from the middle
pole to the leftmost pole. two
onse
utive moves of the smallest disk are wasteful as they
an always be
ombined into one.
By examining the puzzle itself. namely that the disk that is
moved alternates between the smallest disk and some other disk. Did the disk being moved have an even number or an odd
number?
2
Alternate Disks
We now turn to the se
ond major element of the solution.
After all. it is not diÆ
ult to see that this must be the
ase.
Let us
all a sequen
e of numbers alternating if it has two properties.es this property. The .
whi
h we denote by diskn.
The sequen
e of disks moved on su
essive days.d .rst
property is that
onse
utive elements alternate between one and a value greater than
one. the se
ond property is that if the sequen
e is non-empty then it begins and ends
with the value one. We write alt.ks if the sequen
e ks has these two properties. is
obtained by taking the .
d and ignoring the
se
ond.rst
omponent of ea
h of the pairs in Hn. Then.
from the de.d . Let the sequen
e that is obtained in this way be denoted by diskn.
The explorer left the area and did not return until several years later.
On his return.
straightforward property of diskn for the se
ond. It transpired that
one of the monks had made a mistake shortly after the explorer's . he dis
overed the monks in a state of great despair.rst
onjun
t.
}
true .
The
hapter began
Algorithmi
Problem Solving
Roland Ba
khouse.4
even. they had got into a never-ending loop!
Fortunately.
Exercise 7.rst visit but it had
taken the intervening time before they had dis
overed the mistake. maintaining
invariant the relationship
Exercise 7. the explorer was able to tell the monks how to pro
eed in order to return
all the disks to one-and-the-same pole whilst still obeying the rules laid down to them
on the day of the world's
reation.
May 28. and all the blue disks are on the third pole. Make use of the
fa
t that. They would then be able to begin their task afresh.5 (Coloured Disks)
2
7. 2008
. initially.)
2
Suppose ea
h disk is
oloured.
Devise an algorithm that will sort the disks so that all the red disks are on one pole. beginning in a state in whi
h n disks are all on the same pole. The state of the disks
was still valid but the monks had dis
overed that they were no longer making progress
towards their goal. You
may assume that.n ≡ d ≡ even.4
Summary
In this
hapter we have seen how to use indu
tion to
onstru
t a solution to the Towers of
Hanoi problem. (Hint: The disk being moved will still alternate between the smallest and
some other disk. dierent disks may be
oloured dierently. Be
ause of the monks' mistake this will not be
onstant. red white or blue. Several indu
tive
onstru
tions have been dis
ussed. You only have to de
ide in whi
h dire
tion the smallest disk should
be moved.k ≡ d ′
for the dire
tion d ′ moved by disk k will move n disks in the dire
tion d .
The
olouring of disks is random.
What was the algorithm the explorer gave to the monks? Say why the algorithm
is
orre
t.
all the white disks are on another pole. all disks are on one pole.
indu
tive
proofs of properties of the sequen
e of movements of the disks being used to establish
the
orre
tness of the iterative solutions. 2008
. The Towers of Hanoi
with an indu
tive
onstru
tion of a graph representing all possible moves of the disks
in the general
n -disk
problem.5
Bibliographic Remarks
Information on the history of the Towers of Hanoi problem is taken from [Ste97℄.
May 28.
7. The formulation and
proof presented here is based on [BF01℄.
The
hapter has also illustrated two important design
onsiderations: the in
lusion
of the
0 -disk problem as the basis for the
onstru
tion (rather than the 1 -disk problem)
and the avoidan
e of unne
essary detail by not naming the poles and referring to the
dire
tion of movement of the disks (
lo
kwise or anti
lo
kwise) instead. This solution was then tranformed to an iterative solution.120
7. The graph was used to justify an indu
tive solution to
the problem itself.
Algorithmi
Problem Solving
Roland Ba
khouse. A proof
of the
orre
tness of the iterative solution was published in [BL80℄.
The generalisation is to
onsider an arbitrary number of people.
Spe
i.Chapter 8
The Torch Problem
In this
hapter. the task is
to get all the people a
ross a bridge in the optimal time. we present a solution to a more general version of the tor
h problem in
exer
ise 3.4.
the problem we dis
uss is the following.j whenever i < j . we
an always
onsider pairs (t.i . where i ranges over people. where i < j . we assume that t.i < t.
For simpli
ity. ordered lexi
ographi
ally. parti
ularly in
omparison to the . If the given times are su
h that t. we obtain a total ordering on times with
the desired property. Assuming that the
rossing times are distin
t makes the arguments simpler.
N people wish to
ross a bridge.
Renaming the
rossing \times" to be su
h pairs. Person i takes time t.1
Lower and Upper Bounds
The derivation that follows is quite long and surprisingly diÆ
ult.
The bridge is narrow and at most 2 people
an be on it at any one time. (This means that we assume
the people are ordered a
ording to
rossing time and that their
rossing times are
distin
t. but
is not essential.
ally. when two
ross together they must pro
eed at the speed of the
slowest. i) .i = t.)
8.i to
ross
the bridge.
The people are numbered from 1 thru N.j for some i and j . It is dark. and it is ne
essary to use a
tor
h when
rossing the bridge. but they only have one tor
h between them.
Constru
t an algorithm that will get all N people a
ross in the shortest time.
whi
h is quite simple. This has to do with the dieren
e between establishing an
\upper bound" and a \lower bound" on the
rossing times.nal algorithm. It's important to appre
iate where
pre
isely the diÆ
ulties lie.
121
May 28. 2008
.
Algorithmi
Problem Solving
Roland Ba
khouse.
In other words. The Tor
h Problem
In the original problem given in
hapter 1 .
We
an use the same instan
e of the tor
h problem to further illustrate the dieren
e
between lower and upper bounds.
A mu
h harder problem is to show that 17 minutes is a lower bound on the time
taken. the question asked
was to show that all four
an
ross the bridge within 17 minutes. an upper bound
is established by exhibiting a sequen
e of
rossings that takes the required time. the
question asks for a so-
alled upper bound on the time taken. Most of us. 2 minutes. 5 minutes and 10 minutes. when
onfronted with the tor
h problem
above. there are four people with
rossing times
of 1 minute. will .122
8. Cru
ially. Showing that it is a lower bound means showing that the time
an never be
bettered. In general.
that the optimal
solution is to let one person a
ompany all the others one-by-one a
ross the bridge.
8. The
more eÆ
ient algorithm assumes a knowledge of algorithm development that goes beyond
the material in this book. It may be. The greatest eort goes into showing
that the algorithm simultaneously establishes a lower bound on the
rossing time.) Similarly. by exhibiting the
rossing sequen
e that gets all four people a
ross in
17 minutes does not prove that this time
annot be bettered.e. (Indeed. ea
h
time returning with the tor
h for the next person.
The way to avoid unne
essary detail is to fo
us on what we
all the \forward trips". The algorithm we derive is quite simple but. it
is not. Su
h a solution takes a total time of 2+1+5+1+10 . i. on its own. it only establishes
an upper bound on the optimal
rossing time. we present two algorithms for
onstru
ting an optimal sequen
e. then it is extremely inee
tive. for example.
By exhibiting the
rossing sequen
e. If our solution method requires that
we detail in what order the people
ross.rst explore the solution in whi
h the fastest person a
ompanies the others
a
ross the bridge.
In this
hapter. we have not established that it is a lower bound. the goal is to
onstru
t an algorithm for s
heduling N people to
ross
the bridge.
Algorithmi
Problem Solving
Roland Ba
khouse. The
ombination of equal lower and upper bounds is
alled an exa
t bound. whi
h is a very large number even for quite small values
of N . we have established that 19 minutes is an upper
bound on the
rossing time.
In se
tion 8. 2008
.6. Doing so is mu
h harder
than just
onstru
ting the sequen
e. The number of
dierent orderings is (N−1)! . 19 minutes.
The problem asks for a sequen
e of
rossings but there is an enormous amount of freedom
in the order in whi
h
rossings are s
heduled.2
Outline Strategy
On
e again. the main issue we have to over
ome is the avoidan
e of unne
essary detail. this is what is
meant by an optimal solution.
May 28.
the tor
h must always be
arried. if we
an
ompute the optimal
olle
tion of forward trips. and a return trip is a
rossing in the opposite dire
tion.123
8. we mean a \bag" of sets of people.
The idea is that.
First. the return
trips needed to sequen
e them
orre
tly
an be easily dedu
ed.
In order to turn this idea into an ee
tive solution. we need to pro
eed more formally.2.
Informally. when
rossing the bridge. by the \
olle
tion" of forward trips. This means
that
rossings alternate between \forward"and \return" trips. whereas a set is de. where a forward trip is a
rossing in the desired dire
tion. The
mathemati
al notion of a \bag" (or \multiset" as it is sometimes
alled) is similar to a
set but. the forward trips do the work whilst the return trips servi
e the forward trips. Outline Strategy
Re
all that.
a bag is de.ned solely by whether or not a value is an element of the
set.
For example. a
bag of
oloured marbles would be spe
i.ned by the number of times ea
h value o
urs in the set.
b s and c s in whi
h a o
urs on
e.ed by saying how many red marbles are in the
bag. For brevity. for example). b o
urs twi
e and c o
urs
no times. how many blue marbles. and so on. for example. 2∗b . we also write {1∗a . 2∗b . Even though when
we write down an expression denoting a bag we are for
ed to list the elements in a
ertain
order (alphabeti
al order in {1∗a . the order has no signi.
It is important to stress that a bag is dierent from a sequen
e. We will write. 2∗b} to denote the same bag. 0∗c} to
denote a bag of a s. 0∗c} . {1∗a .
If we are obliged to distinguish between forward
and return trips. for example.
The expressions {1∗a .
A trip is given by the set of people involved in the trip. 2∗b .3} is a
trip in whi
h persons 1 and 3
ross. {1. 0∗c} and {2∗b . 0∗c} both denote the same bag. we pre. So.
an
e. 1∗a .
As we said above.x the trip with either \ + " (for forward) or \ − " (for return).3} denotes a forward trip made by persons 1 and 3 and −{2} denotes a return
trip made by person 2 .
We
all a sequen
e of trips that gets everyone a
ross in a
ordan
e with the rules a
valid sequen
e. our fo
us will be on
omputing the bag of forward trips in an
optimal sequen
e of trips. We begin by establishing a number of properties of sequen
es
of trips that allow us to do this. We will say that one valid sequen
e subsumes another valid sequen
e
if the time taken by the .
So +{1.
The problem is to . Note that the
subsumes relation is re
exive (every valid sequen
e subsumes itself) and transitive (if
valid sequen
e a subsumes valid sequen
e b and valid sequen
e b subsumes valid
sequen
e c then valid sequen
e a subsumes valid sequen
e c ).rst is at most the time taken for the se
ond.
2008
.
Formally.
Algorithmi
Problem Solving
Roland Ba
khouse.nd
a valid sequen
e that subsumes all valid sequen
es. a valid sequen
e is a set of numbered trips with the following two properties:
The trips are sets.
May 28. ea
h set has one or two elements. and the number given to a
trip is its position in the sequen
e (where numbering begins from 1 ).
(A trip T is made by person i if
i∈T . The Tor
h Problem
Odd-numbered trips in the sequen
e are
alled forward trips. beginning and ending with forward trips. even-numbered trips
are
alled return trips.
The trips made by ea
h individual person alternate between forward and return
trips.)
Immediate
onsequen
es of this de.124
8. The length of the sequen
e is odd.
nition whi
h play a
ru
ial role in .
A regular sequen
e is a
valid sequen
e that
onsists entirely of regular forward and return trips.nding an
optimal sequen
e are:
The number of forward trips is one more than the number of return trips. and a regular
return trip means a return trip made by exa
tly one person.
The number of forward trips made by ea
h individual person is one more than the
number of return trips made by that person.
A regular forward trip means a forward trip made by two people.
The .
1) is to show that every valid sequen
e is subsumed by one
in whi
h all trips are regular.rst step (lemma 8. The signi.
)
Most importantly.
(Knowing the bag of forward trips. not even the order in whi
h they are made needs to be known. we
an thus repla
e the problem of . This is be
ause ea
h person makes one fewer return
trips than forward trips.
The number of forward trips is
N−1 and the number of return trips is N−2 .
an
e of this is threefold. knowing just the bag of forward trips in a regular sequen
e is
suÆ
ient to re
onstru
t a valid regular sequen
e.
(Re
all that N is the number of people. In this way.)
The time taken by a regular sequen
e
an be evaluated knowing only whi
h forward
trips are made. the time taken for the return trips
an be
al
ulated. it is easy to determine how many times ea
h
person makes a return trip. Sin
e all su
h sequen
es take the
same total time.
nding an optimal sequen
e of
forward and return trips by the problem of .
We prove the obvious property that. 2008
.
May 28. We
an then use indu
tion to determine the
omplete solution.
the two slowest people do not return.
Algorithmi
Problem Solving
Roland Ba
khouse.nding an optimal bag of forward trips
Finding an optimal bag of forward trips is then a
hieved by fo
using on whi
h people
do not make a return trip. in an optimal solution.
Every valid sequen
e
ontaining irregular trips is subsumed by a stri
tly
faster valid sequen
e without irregular trips. Regular Sequen
es
8.125
8.
Lemma 8.1
Proof Suppose a given valid sequen
e
ontains irregular trips.3.3
Regular Sequences
Re
all that a \regular" sequen
e is a sequen
e in whi
h ea
h forward trip involves two
people and ea
h return trip involves one person. We
an always restri
t attention to
regular sequen
es be
ause of the following lemma. We
onsider two
ases:
the .
rst irregular trip is forward and the .
rst irregular trip is ba
kward.
If the .
q} v −{p.rst irregular trip is ba
kward. p say. u .
(Note that the forward trip made by p involves two people be
ause it is assumed that
the . and remove p
from both trips. More formally.
hoose an arbitrary person. suppose the sequen
e has the form
u +{p. Identify the forward trip made by p prior to the ba
kward trip. v and w are subsequen
es and p o
urs nowhere in v . making the
trip.r} w
where q and r are people.
) The number of irregular
rossings is not redu
ed.
Now suppose the . The time taken is no greater sin
e.
and it is true for p be
ause the trips made by p have
hanged by the removal of
onse
utive forward and return trips. we have to
he
k that the trips
made by ea
h individual
ontinue to alternate between forward and return. This is true
for individuals other than p be
ause the points at whi
h they
ross remain un
hanged. for any x and
y . but the total number of person-trips is
redu
ed. sin
e a
new irregular forward trip has been introdu
ed. t.) Repla
e the sequen
e by
u +{q} v −{r} w
This results in a valid sequen
e.rst irregular trip is ba
kward.p ↑ y ≥ x+y . (To
he
k that the sequen
e remains valid.p ↑ x + t. the time for whi
h is no greater than the original
sequen
e.
rst irregular trip is forward. There are two
ases to
onsider: the
irregular trip is the very .
rst in the sequen
e. and it is not the very .
rst.
If the .
Clearly. Also.
If the . the total time taken is redu
ed. sin
e times are positive.) These two
rossings
an
be removed. the
number of person-trips is redu
ed. it means that one person
rosses and
then immediately returns. (We assume that N is at least 2 .rst trip in the sequen
e is not regular.
rst irregular
rossing is a forward trip but not the very .
2008
.rst. let us suppose it
is person p who
rosses. and suppose q is the person who returns immediately before
Algorithmi
Problem Solving
Roland Ba
khouse.
May 28.
126
8. The Tor
h Problem
this forward trip. (There is only one su
h person be
ause of the assumption that p 's
forward trip is the .
r} w −{q} +{p} v
(where w does not involve p or q ) to
u +{p. transform
u +{q.p > t. (This is be
ause we assume
that p ′ s forward trip is the . it must involve q and not p . the total
rossing time is redu
ed (sin
e.r} w v .p ↑ x ). There
are two
ases: it is a forward trip or it is a return trip. Moreover. That is. for
any x . and the number of person-trips is also redu
ed. Swap p with q in this trip and
remove q 's return trip and p 's irregular
rossing.q ↑ x + t.
The result is a valid sequen
e.
If it is a return trip.q + t.rst irregular trip. suppose the sequen
e has the form
u −{q} +{p} v
Consider the latest
rossing that pre
edes q ′ s return trip and involves p or q. it is made by one person only. t.
If it is a forward trip.) That is.
That is.) That person must be
p .rst irregular trip in the sequen
e.p + t. and the number of person-trips is also redu
ed. the total
rossing time is redu
ed (sin
e. transform
u −{p} w −{q} +{p} v
(where w does not involve p or q ) to
u −{q} w v .
t.
2
8. Swap p with q in this return trip.
The result is a valid sequen
e.p > t.q + t. subsumes the given valid sequen
e and
has a smaller person-trip
ount.
We have now des
ribed how to transform a valid sequen
e that has at least one irregular
rossing. Moreover.
Repeating this pro
ess whilst there are still irregular
rossings is therefore guaranteed to
terminate with a valid sequen
e that is regular. and remove q 's return trip and p 's irregular
rossing.4
Sequencing Forward Trips
Lemma 8. the transformation has the ee
t of stri
tly de
reasing the total time taken.1 has three signi.q ).
May 28. it means that the number of forward
trips in an optimal sequen
e is N−1 and the number of return trips is N−2 . This is
be
ause every subsequen
e
omprising a forward trip followed by a return trip in
reases
Algorithmi
Problem Solving
Roland Ba
khouse.
ant
orollaries. First. 2008
.
127
8. after the . Sequen
ing Forward Trips
the number of people that have
rossed by one.4. and the last trip in
reases the number
by two. Thus.
1∗{7.2} means that persons 1 and 2
make 3 forward trips together.
This is a non-obvious property of the forward trips and to prove that it is indeed the
ase we need to make some
ru
ial observations. whi
h we identify with numbers
in the range 1 thru N . 7 and 8 ) all make 1
forward trip and.
so. 1∗{1. 6 .3↑t.1 is that. after 2×(N−2) + 1 trips everyone has
rossed. similarly. The
number of times a person o
urs is the number of forward trips made by that person. This is be
ause the bag of forward trips enables us
to determine how many times ea
h individual makes a forward trip. given just the bag of forward
trips
orresponding to a regular sequen
e. suppose the forward trips in a regular sequen
e are as follows:
3∗{1. The remaining people ( 4 . it is possible to
onstru
t a regular sequen
e
to get everyone a
ross with the same
olle
tion of forward trips. it means that the total time taken to
omplete any regular sequen
e
an be
evaluated if only the bag of forward trips in the sequen
e is known. not even the order in
whi
h the trips are made is needed.6) + 1 × (t. N−2 people have
rossed and 2 have not. person
2 makes 3 forward trips and hen
e 2 return trips. 1∗{3. re
all that 3∗{1. 1∗{3. The elements of the sets in F are people. and hen
e 3 return trips.8)
+
3 × t. The total time taken is thus:
3 × (t.2} . Division of people into these two
Algorithmi
Problem Solving
Roland Ba
khouse. et
. whilst person 3 makes 2 forward
trips and hen
e 1 return trip.7↑t. and the bottom line gives the
time taken by the return trips. ea
h with exa
tly two elements and ea
h having a
ertain
multipli
ity.1↑t. 5 .
For example.5} . 1∗{1.
The third important
orollary of lemma 8. and ea
h number in the range must o
ur at least on
e. Hen
e. Note that no indi
ation is given of the order in whi
h the forward
trips o
ur in the sequen
e.8}
(The trips are seperated by
ommas. person 1 makes 4 forward trips.) Then. hen
e.6} .2) + 1 × (t.
We will
all a person a settler if they make only one forward trip. no return trips.5) + 1 × (t.3↑t.
May 28. That
is.1↑t.2 + 1 × t.
ounting the number of o
urren
es of ea
h person
in the bag. 2008
. and the total number of return trips is 6 .
Se
ond. from whi
h the total time for the
return trips
an be
omputed.rst 2×(N−2) trips. we
all a person
a nomad if they make more than one forward trip.3
(The top line gives the time taken by the forward trips.) Note that the total number of forward trips is 7 (one
less than the number of people).6} means that persons 1 and 6 make one forward
trip together.4} .4) + 1 × (t.1 + 2 × t. F is a
olle
tion of sets.
Suppose F is a bag of forward trips
orresponding to some regular sequen
e. the number
of times ea
h individual returns
an be
omputed.
128
8. If both people in a trip are settlers we say the trip is hard . The Tor
h Problem
types
auses the trips in F to be divided into three types depending on the number of
settlers in the trip. if one
of the people is a settler and the other a nomad. we say the trip is .
rm . .
the number of .nally. the number of hard trips. #firm and #soft to denote the number
of nomads. if both
people are nomads we say the trip is soft. #hard .
Now. suppose we use #nomad .
in the
olle
tion F . respe
tively. sin
e ea
h nomad makes one more forward trip than
return trip. Sin
e ea
h soft trip
ontributes 2 to the number of forward trips made by
nomads.
The number of return trips equals the total number of forward trips made by individual
nomads less the number of nomads. Then. and ea
h . the number of trips in F is
#hard + #firm + #soft .rm trips and the number of soft
trips.
Equivalently.rm trip
ontributes 1 .
But the number of forward trips is one more than the number of return trips. That is. the number of return trips is thus
2 × #soft + #firm − #nomad .
#hard + #firm + #soft = 2 × #soft + #firm − #nomad + 1 .
#hard + #nomad = #soft + 1 .
We summarise these properties in the following de.
F and #nomad.nition.F denote. respe
tively. then
(8. every person is an element of at least one trip in F . ea
h trip in F involves two people.F + 1 .F + #nomad. the number of trips in
F that involve two settlers. the number of trips in F that involve no settlers.
Definition 8.2 (Regular Bag)
{i | 1 ≤ i ≤ N} is
alled a regular N
Suppose N is at least 2 .)
h∀i : 1 ≤ i ≤ N : h∃T : T ∈F : i∈T ii
(Informally. 2008
. #soft.
May 28. A bag of subsets of
bag if it has the following properties:
Ea
h element of F has size two.F = #soft.)
If #hard. and
the number of nomads in F .3)
2
#hard. (Informally.
Algorithmi
Problem Solving
Roland Ba
khouse.
Formally. what we have proved is the following.F .
Then.i × rF.ii
where
(8.F .F = 0 ⇒ #hard. Moreover.4.6)
rF.i is the number of times that person i returns. by de.F = 1 ∧ #firm.ii × #FT i + hΣi :: t.
If #nomad.5)
hΣT : T ∈F : h⇑i : i∈T : t.F = 0 = #soft. Sequen
ing Forward Trips
129
Lemma 8.7)
#nomad. the bag of forward trips F that is obtained from the sequen
e by forgetting
the numbering of the trips is a regular N bag. there are no nomads and hen
e.)
2
Immediate
onsequen
es of (8. the
time taken is
(8. Let #FT denote the multipli
ity of T in F.8.
( rF.4 Given a valid regular sequen
e of trips to get N people a
ross.F is zero.i = hΣT : T ∈F ∧ i∈T : #FT i − 1 .3) are:
(8. the time taken by the sequen
e
of trips
an be evaluated from F . where N
is at least 2 .
no soft or .nition.
The
onverse is immediate
from (8.F ≥ 1 ⇒ #soft. It is the
ase that #nomad.
If there is only one nomad. a valid regular sequen
e of
trips
an be
onstru
ted from F to get N people a
ross. It follows from (8.F is at least 1 .
Now we
an show how to
onstru
t a regular sequen
e from F . not all
an
ross at on
e and. The easiest
ase is when F
onsists of exa
tly one
trip (with multipli
ity one).F = 0 = #soft.3) that #hard.F .
G iven N (at least 2 ) and a regular N bag.9)
N > 2 ∧ #hard. by (8.F is at least 1 .5).F ≥ 1 .F = 1 .8)
#nomad.3) that #soft. That is #soft.
#nomad.
We need to
onsider three
ases.F
is zero.3). The time taken by the sequen
e
is given by (8. The sequen
e is then just this one trip.
(8.F is 1 .
The se
ond
ase is also easy. It follows from (8.F also equals zero.
(8.F = 1 ≡ #hard. every
trip in F is . In this
ase. So. #hard. there are no trips involving two nomads.rm
trips. so.3).
Lemma 8.
If N (the number of people) is greater than 2 .10
Proof The proof is by indu
tion on the size of the bag F .
2008
.rm. That is.
May 28. The sequen
e is simply obtained by listing all the elements of F in an arbitrary
Algorithmi
Problem Solving
Roland Ba
khouse.s} where n is the nomad and s is a
settler. ea
h trip has the form {n.
The Tor
h Problem
order and inserting a return trip by the nomad n in between ea
h pair of forward trips
of whi
h the .130
8.
#soft. It follows.F is at least 1 .9). In this
ase. by (8.s} for some s .rst is the trip {n.F is non-zero and F has more than one trip. by de.
The third
ase is that #hard.
2
Optimisation Problem Lemmas 8. there is a regular sequen
e
orresponding to F ′ whi
h gets
the remaining people a
ross.m} where n and m are both
nomads.4 and 8. (That is. Choose any soft trip in F .nition of soft that #nomad. Redu
e the
multipli
ity of the
hosen hard trip and the
hosen soft trip in F by 1 .) We get a new bag F ′ in whi
h the number of trips
made by ea
h of n and m has been redu
ed by 1 and the number of people has been
redu
ed by 2 .10 have a signi.m} and is followed by
the return of n . remove
one o
urren
e of ea
h from F . Suppose it is {n. By indu
tion.F
is at least 2 . then an arbitrary hard trip and then the return of m . Constru
t the sequen
e whi
h begins with the trip {n.
we seek a regular bag as de. Instead of seeking a sequen
e of
rossings
of optimal duration.
ant impa
t on how to
solve the general
ase of the bridge problem.
ned in de.
5).2 that optimises the
time given by (8.
In solving this problem. The value of
the . it is useful to introdu
e some terminology when dis
ussing
the time taken as given by (8.5).nition 8. It is this problem that we now solve. There are two summands in this formula.
we
all h⇑i : i∈T : t.i the return time of person i
(or person i 's return time).
8.rst summand we
all F 's total forward time and the value of the se
ond summand
F 's total return time. Given a bag F and a trip T in F . More spe
i.5
Choosing Settlers and Nomads
This se
tion is about how to
hoose settlers and nomads. We establish that the settlers
are the slowest people and the nomads are the fastest.) For ea
h person i .i × rF.ii × #FT
the forward time of T in F . (Sometimes \in F " is omitted if this is
lear from the
ontext. we
all the value of t.
q) an inversion
if. within F . Call a pair of people (p.
ally. additionally.
Every regular bag is subsumed by a regular bag for whi
h all settlers
are slower than all nomads.
Lemma 8. q is a nomad and p is faster than q . person 2 may also be a nomad. 2008
.
Proof
Algorithmi
Problem Solving
Roland Ba
khouse. we establish
that in an optimal solution there are at most 2 nomads.
May 28. p is a settler. Person 1 is always a nomad if
N is greater than 2 .11
Suppose the regular N bag F is given.
Moreover.
The forward times for the trips not involving p or q are. Choosing Settlers and Nomads
Choose any inversion (p.p .q − t. un
hanged. Suppose q and p are inter
hanged everywhere in
F .q ).p . the return time is
learly redu
ed by at least
t. We get a regular N bag. This is veri.131
8. of
ourse.q − t.p < t. The
forward time for the trips originally involving q are not in
reased (be
ause t.5.
The forward time for the one trip originally involving p is in
reased by an amount that
is at most t. q) .
ed by
onsidering two
ases. The .
it suÆ
es to observe that
t. the number of inversions is de
reased by at least
one. In this
ase. arithmeti
}
t.
Finally.r} where
r 6= q . monotoni
ity of max
}
t.p)
=
{
distributivity of sum over max. swapping p and q has no ee
t on the trip.
The net ee
t is that the total time taken does not in
rease.r .q − t.q ↑ (t. a bag F
is obtained that subsumes the given bag. Also.p ≤ t.r + (t. the transformed
bag subsumes the original bag. the times for all other forward trips are un
hanged. by repeating the pro
ess of identifying and eliminating inversions.p ↑ t. In this
ase.p))
≥
{
t.rst
ase is when the
trip involving p is {p.q .q} . Thus.q ↑ t.q − t. the trip involving p is {p. That is.12
In any . In the se
ond
ase.r + (t. and
the in
rease in time taken is 0 .
2
Every regular N bag is subsumed by a regular N bag F with the
following properties:
Corollary 8.
the nomad is person 1 . (Note that this is the empty
set when j equals 1 .)
The multipli
ity of {1.
we may assume that the nomads are slower than the settlers.
Every soft trip in F is {1.rm trip in F .2} . and the hard
trips are {k: 0 ≤ k < j−1: {N − 2×k . (Note: the multipli
ity of this trip in the bag
an be
an arbitrary number.11. From 8. in
luding 0 .)
Suppose F is a regular N bag that optimises the total travel time.
Suppose there is a .2} in F is j . N − 2×k − 1}} . for some j where 1 ≤ j ≤ N÷2 .
This has no ee
t on the forward
Proof
Algorithmi
Problem Solving
Roland Ba
khouse. 2008
.
Repla
e i in one o
urren
e of the trip by person 1 .
May 28.rm trip in F in whi
h the nomad is person i where i is not 1 .
The Tor
h Problem
time. in the
ase of person 1 .132
8. property (8. whi
h
ontradi
ts F being optimal.3). The number of trips made by
i de
reases by one.
in
reases. We
laim that this results in a regular N bag.1 ). So it remains to
he
k property (8. the total return
time is redu
ed (by t. If person i is still a nomad after the
repla
ement. but remains positive be
ause i is a nomad in F .3) is maintained be
ause the type (hard.4 for the properties required of a
regular bag.i − t. (Please refer to lemma 8. The number of
trips made by all other persons either remains un
hanged or. . sin
e i is slower than the other person in the trip. the size of ea
h set in F is un
hanged.)
Of
ourse. However.
person i may be
ome a settler. If so. The se
ond trip is either a . That is. person i is an element of two
trips in F. However. the number
of nomads may be de
reased by the repla
ement.rm or soft) of ea
h
trip remains un
hanged.
rm trip in F and be
omes a hard trip. or it
is a soft trip in F and be
omes .
rm. for some j where j is at least 2 . and there are j−1 hard trips.2} .
Now suppose there is a soft trip in F dierent from {1.2} results in a regular bag with a stri
tly
smaller total travel time. it is easy to
he
k that (8.
ontradi
ting F being optimal. In both
ases.
We may now
on
lude from (8.3) is
maintained. When
there are no soft trips. A similar argument to the
one above shows that repla
ing the trip by {1. all the trips are .2} in F is j .3) that either there are no soft trips or the multipli
ity
of {1.
rm or hard. as we have shown. person 1
is the only nomad in . But.
it follows that
person 1 is the only nomad in F and. Thus
persons 1 and 2 are the elements of one (. from (8.rm trips and there are no nomads in hard trips. that there are no hard trips.3).
N − 2×k − 1}} .
(The multipli
ity of ea
h hard trip is 1 . when j is at least 2 .rm) trip in F . Then the settlers
are persons 3 thru N . so we
an ignore the distin
tion between bags
and sets.
It remains to show that. the hard trips form the set
{k: 0 ≤ k < j−1: {N − 2×k .)
Assume that the number of soft trips is j where j is at least two. the remaining
N − 2×(j−2) being elements of . and 2×(j−1) of them are elements of hard trips.
Any regular bad
learly remains regular under
any permutation of the settlers.rm trips. So we have to show that
hoosing the settlers so that
the hard trips are .
2
Algorithmi
Problem Solving
Roland Ba
khouse. This is done
by indu
tion on the number of settlers in the hard trips.lled in order of slowness gives the optimal arrangement.
May 28. 2008
.
and the last two terms give the return
times.1 + 1 × t.2 + (N−1−1) × t.ii + (N−2) × t. the value of j is 1 . If there are no soft trips.rst three terms give the forward times.
But
HF.)
There are two
ases to
onsider.1 + (1−1) × t.1 + FF.2
=
{
de.
In this
ase.1 . the total time taken is
Proof
hΣi : 2 ≤ i ≤ N : t.
1
=
{
arithmeti
}
hΣi : 2 ≤ i ≤ N : t. In this
ase. HF.1 .2 + (N−2) × t.ii + t. the value of j is equal to the number of soft trips and is at
least 2 .ii + (N−2) × t.nition of HF and FF .j is the forward time for the hard trips in F and FF.
If there are soft trips. arithmeti
}
0 + hΣi : 3 ≤ i ≤ N : t.j is the
forward time for the .
de.1 and person 2 's return time is
(j−1) × t. Finally.
whi
h is what we expe
t the number of return trips to be.2 is the forward time for the j soft
trips. (Person 2 is an element of j forward trips. where j is at least 2 .rm trips in F . person 1 's return time is (N−j−1) × t.2 . and person 1 is an element of
j + (N − 2×(j−1) − 3 + 1) forward trips. Note that the sum of j−1 and N−j−1 is N−2 . Also.)
2
For all j . j × t.
2008
.
May 28.
Now. OT . we observe that
Algorithmi
Problem Solving
Roland Ba
khouse.j to be the optimal time taken by a regular
N bag where the multipli
ity of {1.2} in the bag is j .j is given by (8.ne OT .14). That is.
(N−2j+2) + t.j = −(t.j
an be determined by a sear
h for the point at whi
h the dieren
e fun
tion
hanges from being negative to being positive.(N−2j+1) + 2 × t. If the test evaluates to false.j = t.1 ≤ −t.
the remaining N−2j+2 people are s
heduled to
ross in N−2j+1 . this is followed by the return of person 1 .j = −t.134
8.
The simplest way to do this and simultaneously
onstru
t a regular sequen
e to get
all N people a
ross is to use a linear sear
h. At ea
h
iteration the test 2 × t.
That is.(N−2j+2) .(j+1) − HF.
Note that
OT .2 − t.2 − t. OT .(N−2k+1) ≤ t.2 − t.
OT .1 + t.j in
reases as j in
reases. beginning with j assigned to 1.(N−2j+1)
=
t is in
reasing
{
}
j ≤ k . If it evaluates to true . N−2j+1} and then the return of person 2 .(k+1) − OT .(N−2j+1)) .j ≤ OT .k
=
above
{
}
−t. the
soft trip {1.(j+1) − FF.(N−2k+1) + 2 × t.1
=
arithmeti
}
{
t. A
onsequen
e is that the minimum
value of OT .2 ≤ t.(j+1) − OT .(j+1) − OT .(N−2j+1) is performed.(j+1) − OT . The Tor
h Problem
HF.1 .(N−2j+1) + 2 × t.
As a
onsequen
e.2} is s
heduled.
and
FF. then the hard
trip {N−2j+2 .
es how they are s
heduled.
The remaining N−2j+2 people are then s
heduled to
ross in N−2j+1 .
and the
\nomads" (the people who do make return trips) are the fastest. this
property
ould have been very diÆ
ult to establish.
8. The greatest
hallenge in
an optimisation problem of this nature is to establish without doubt that the algorithm
onstru
ts a solution that
an not be bettered. A major step in solving the problem
was to eliminate the need to
onsider sequen
es of
rossings and to fo
us on the bag
of forward trips.
Even though these properties may seem unsurprising and the . The proof
of the fa
t that there are at most two nomads is made parti
ularly easy by the fo
us
on the bag of forward trips. Via a number of lemmas.7
Conclusion
In this
hapter. An optimal sequen
e is \regular"
| that is.
Many of the properties we proved are not surprising. the \settlers" (the people who never make a return trip) are the slowest. ea
h forward trip is made by two people and ea
h return trip is made by
one. Less obvious is that
there are at most two nomads and the number of \hard" trips (trips made by two settlers)
is one less than the number of trips that the two fastest people make together. if we had had to reason about the sequen
e of trips. we have presented an algorithm to solve the tor
h problem for an arbitrary number of people and arbitrary individual
rossing times. we established a number of properties of an
optimum bag of forward trips whi
h then enabled us to
onstru
t the required algorithm.rm trips as
des
ribed above.
it is also sometimes
alled the \bridge problem". this
hapter is about the
apa
ity 2
problem.136
8.8
Bibliographic Remarks
In Ameri
an-English.
8. In terms of this more general problem. This dieren
e is super. the tor
h problem is
alled the \
ashlight problem".
Rote des
ribes the solution in terms of \multigraphs" rather than bags.
Always beware of
laims that something is \obvious". The solution presented here is based on a solution to
the yet-more-general tor
h problem in whi
h the
apa
ity of the bridge is a parameter
[Ba
07℄. the solution is essentially the same as a solution given by Rote [Rot02℄. The Tor
h Problem
| the interest in the tor
h problem is that the most \obvious" solution (letting the
fastest person a
ompany everyone else a
ross the bridge) is not always the best solution.
The main dieren
e is lemma 8. the
apa
ity of the bridge is 3 and the travel
times are 1 . (If N = 5 . Rote does not prove that any
regular bag
an be
onverted into a sequen
e of
rossings.
Algorithmi
Problem Solving
Roland Ba
khouse. Many \obvious" properties of an optimal solution turn
out to be false.10. of
ourse. his a
ount is in
orre
t. in
ontrast to the use of bags
whi
h does. 1 . . . is just
a set of two people. 2t2 − t1 might not o
ur in the list | the standard
1 . (Stri
tly. Ea
h edge of a \multigraph"
onne
ts two people and.
an be determined easily by lo
ating the value 2t2 − t1 in
the sorted list of ti 's" but. Rote's formulation of the solution
in terms of \multigraphs" does not appear to generalise. 2 . and
doesn't suggest using binary sear
h. the shortest time is 8 . whi
h is a
hieved using 5
rossings. 4 . 2008
. it is no longer the
ase that an optimal solution uses a
minimum number of
rossings. but only proves that it is possible for the regular bags
orresponding to
optimal
rossings. He does
laim that this is
indeed the
ase.) The notion of \regularity" of
rossings has to
be generalised in a way that allows for some forward trips not to be \full" (in the sense
that the full
apa
ity of the bridge is not utilised). the
problem is mu
h harder to solve. hen
e. at one point.
ial. For example.
The shortest time using 3
rossings is 9 .
Rote des
ribes the linear-sear
h method of determining the optimal solution. 5 and 10 minute problem is an example!)
Rote gives a
omprehensive bibliography in
luding pointing out one publi
ation where
the algorithm is in
orre
t. When the
apa
ity of the bridge is also a parameter. 4 and 4 . He
says \the optimal value .
May 28.
by suitably de
omposing the problem.Chapter 9
Knight's Circuit
The problem ta
kled in this
hapter is a parti
ularly hard one. the problem be
omes
solvable.
ombined with ee
tive reasoning skills. Yet.
The problem is to .
.nd a Knight's
ir
uit of a
hessboard. That is.
From ea
h of the
orner
squares. one for ea
h square. However. visiting
ea
h square exa
tly on
e. in prin
iple. but from ea
h of the 16
entral squares.nd a sequen
e of
moves that will take a Knight in a
ir
uit around all the squares of a
hessboard. there are 64 squares on a
hessboard. it
an be solved by a
systemati
.
that means 64 moves have to be
hosen. exhaustive examination of all the paths a Knight
an follow around a
hessboard | a so-
alled brute-for
e sear
h. there is a
hoi
e of just 2 moves. there
is a
hoi
e of 8 moves (see . and ending at the square at whi
h the
ir
uit began.
The problem is an instan
e of a sear
h problem.
Lots
of
hoi
e is usually not a problem but.
But. The Knight's
ir
uit problem is hard be
ause of this
riti
al
ombination of
an explosion with an implosion of
hoi
e. when
ombined with the very restri
tive requirements that the path forms a
ir
uit that visits every square exa
tly on
e.
and the Knight's moves. 9. all is not lost. although many.g. This gives a massive amount of
hoi
e in the paths that
an be followed. are spe
i. it does be
ome
a problem.1). The squares on a
hessboard are arranged in a very simple pattern. from the remaining squares either 4 or 6 moves are
possible.
137
May 28.ed by one simple rule (two squares
horizontally or verti
ally. There is a great
deal of stru
ture. whi
h we must endeavour to exploit.1
Straight-Move Circuits
Finding a Knight's
ir
uit is too diÆ
ult to ta
kle head on. Some experien
e of ta
kling
simpler
ir
uit problems is demanded.
9. 2008
.
Algorithmi
Problem Solving
Roland Ba
khouse. and one square in the opposite dire
tion).
Suppose you want to make a
ir
uit of a
hessboard and you are allowed to
hoose a set of moves that you are allowed to make.138
9.
Let's turn the problem on its head.1: Knight's Moves. What
sort of moves would you
hoose?
The obvious . Knight's Cir
uit
Figure 9.
then return to the starting square. although it isn't immediately obvious.rst answer is to allow moves from any square to any other square. but ex
luding diagonal moves. (These are the moves that
a King
an make. In
that
ase. Let's
onsider
hoosing from a restri
ted set of moves. But that is just too
easy.) We
all these moves straight moves.
hoose a move to a square that has not yet been visited until
all the squares are exhausted. You may be able to . whatever its size |
starting from any square. it's always possible to
onstru
t a
ir
uit of any board.
The simplest move is one square horizontally or verti
ally.
Is it possible to make a
ir
uit of a
hessboard just with straight moves?
The answer is yes.
but let us try to .nd a
straight-move
ir
uit by trial and error.
to the left or right. rather than restri
t the
problem to an 8×8 board.
May 28.1
(a) What is the relation between the number of moves needed to
omplete a
ir
uit of
the board and the number of squares? Use your answer to show that it is impossible
to
omplete a
ir
uit of the board if both sides have odd length. please ta
kle the following exer
ise. beginning and ending at the same square. or up or down. is it possible to visit every
square exa
tly on
e. is it possible to
omplete a straight-move
ir
uit of the entire board? That is.
Exercise 9.
As is often the
ase.
Its solution is relatively straightforward.nd one more systemati
ally. (Hint:
ru
ial is
Algorithmi
Problem Solving
Roland Ba
khouse. let us
onsider an arbitrary re
tangular board. Assuming
ea
h move is by one square only. making \straight" moves
at ea
h step?
In order to gain some familiarity with the problem. it is easier to solve a more general problem. 2008
.
We may take as the
indu
tive hypothesis that a straight-move
ir
uit of both boards
an be
onstru
ted. (Both m and n must be non-zero be
ause the problem
assumes the existen
e of at least one starting square. a 2m × n board
an be split into two boards of sizes 2p × n and 2q × n . (The
onvention
we use is that the .1 is that a straight-move
ir
uit is only possible if the
board has size 2m×n . ea
h row having n squares.139
9. This suggests that
we now try to
onstru
t a straight-move
ir
uit of a 2m×n board. for m at least one
and n greater than one.)
(c) Show that it is possible to
omplete a straight-move
ir
uit of a 2×n board for all
(positive) values of n . by indu
tion on m . Otherwise. the
answer does not depend on the sort of moves that are allowed.)
2
The
on
lusion of exer
ise 9. There are two straight moves from
ea
h of the
orner squares.
The key to the
ombination is the
orner squares. (A 2×n board has two rows. We
just need to
ombine the two
onstru
tions. imagine that a 2m × n board is divided into a
2p × n board and a 2q × n board.
where m is greater than 1 . Now. when m is greater
than 1 .1.) Also. the number of squares
is 2m . the 2×n
ase providing the basis for the
indu
tive
onstru
tion. say. a straight-move
ir
uit
an
always be
ompleted when the board has size 2×n . That is. for positive numbers m and n . with the former above the latter. for positive n .
To
omplete the indu
tive
onstru
tion. we need to
onsider a board of size 2m × n . and any straight-move
ir
uit must use both. In parti
ular.
it must use the horizontal moves. Su
h a
onstru
tion is hopeful be
ause.
where both p and q are smaller than m and p+q equals m . Straight-Move Cir
uits
that ea
h move is from one square to a dierent
oloured square. one side has length 2m
and the other has length n .)
(b) For what values of m is it possible to
omplete a straight-move
ir
uit of a board
of size 2m×1 ? (A 2m×1 board has one
olumn of squares.
Figure 9.)
Now.rst number gives the number of rows and the se
ond the nunber
of
olumns of the board. the shape of the dotted lines gives no indi
ation of the shape of the
ir
uit that
is
onstru
ted. The two horizontal. red
lines at the middle-left of the diagram depi
t the horizontal moves that we know must
form part of the two
ir
uits. (Of
ourse. The blue dotted lines depi
t the rest of the
ir
uits.) The bottom-left
orner of the 2p × n board is immediately
above the top-left
orner of the 2q × n board. Constru
t straight-move
ir
uits of these
two boards. repla
e the
horizontal moves from the bottom-left and top-left
orners by verti
al moves.2 shows the result diagrammati
ally. as shown
by the verti
al green lines in . to
ombine the
ir
uits to form one
ir
uit of the entire board.
It is easy to
onstru
t a straight-move
ir
uit of all its squares
but the middle square. 9. (See .
Consider a 3×3 board.
As mentioned above. no
ir
uit is possible. when a board has an odd number of squares.4.g.
and when it is not possible. a straight-move
ir
uit
of all but one of the other four squares |the squares adja
ent to a
orner square. the middle-left square|
annot be
onstru
ted. However.5.2
Algorithmi
Problem Solving
Roland Ba
khouse.
Exercise 9.
May 28. for
example.g. 2008
.) It is also possible to
onstru
t a straight-move
ir
uit of all its squares but one of the
orner squares. 9. to
onstru
t a straight-move
ir
uit of all the squares but one in a board of odd size.
Explore when it is possible.
2
9.2. omitting one of the
squares.2
Supersquares
Let us now return to the Knight's-
ir
uit problem.141
9.4: A straight-move
ir
uit for a 6 × 8 board. as shown in .5: Straight-move
ir
uits (shown in red) of a 3 × 3 board. Supersquares
Figure 9.
Figure 9. The way this is done is to imagine that the 8 × 8
hessboard is divided into a 4 × 4 board by grouping together 2 × 2 squares into \supersquares". The key to a solution is to exploit
what we know about straight moves.
9. the Knight's moves
an be
lassi.g.
If this is done.6.
a Knight's move
is straight if it takes it from one supersquare to another supersquare either verti
ally
above or below.ed into two types: Straight moves
are moves that are \straight" with respe
t to the supersquares. that is. or horizontally to the left or to the right. Diagonal moves are moves
that are not straight with respe
t to the supersquares. 2008
.
May 28. a move is diagonal if it takes the
Algorithmi
Problem Solving
Roland Ba
khouse.
In . Knight's Cir
uit
Figure 9.6: Chessboard divided into a 4 × 4 board of supersquares.142
9.
Knight from one supersquare to another along one of the diagonals through the starting
supersquare.
by de. That is. let
us use c (short for \
entre") as its name.
Both are equivalent to rotating the 2 × 2 square through 1800 about its
entre.v . Flipping the
olumns and then
ipping the
rows is the same as
ipping the rows and then the
olumns.rst
ipping the
olumns
and then
ipping the rows of the square.h = h. So.
v. That is.
3)
v.
(9.nition of c .
We have now identi.h = c = h.v .
whi
h is the do-nothing operation. all
amount to doing nothing.3).c = n . for greater brevity we use n (short for \no
hange"). doing one operation x followed by two operations
y and z in turn is the same as doing .
Also. That is. together with the fa
t that doing one operation
after another is asso
iative (that is.5)
n.ed three operations on a 2 × 2 square.5).
(9. Flipping twi
e
verti
ally. or twi
e horizontally. we have used skip to name su
h an
operation. (9.4)
v.
The three equations (9. doing nothing before or after any operation is the same as doing the operation.
(9. There is a fourth operation. Elsewhere.n = x . Here. or rotating twi
e through 1800 about the
entre.v = h.x = x.h = c.4) and (9.
)
Use the table to verify that. h and c . for x and y in the set {n.
(Avoid
onstru
ting the
omplete table be
ause it is quite large!)
Exercise 9. v .
Identify a
omplete set of operations on a 2 × 2 square and extend your solution to exer
ise 9. z) = (x .v.
(In prin
iple.6
(9.e. r .c} .
ea
h labelled by one of n . i.x operator for \followed
by" fa
ilitates this all-important
al
ulational te
hnique. y) . z .c} .v.y = y. you need to
onsider 43 .
(9. in one
ase
lo
kwise and in the other anti
lo
kwise.
Exercise 9. 2008
. for x and y in the set {n. dierent
ombinations. Let
r denote the
lo
kwise rotation and let a denote the anti
lo
kwise rotation. (y .8)
x .9
2
Algorithmi
Problem Solving
Roland Ba
khouse. Constru
t
a table that shows the ee
t of performing any two of the operations n .)
2
Two other operations that
an be done on a 2 × 2 square are to rotate it
about the
entre through 90◦ .x . The table should have four rows and four
olumns. a or c in
sequen
e.
Che
k also that.6 so that it is possible to determine the ee
t of
omposing any pair of operations. (Use the physi
al pro
ess of
ipping squares to
onstru
t the entries.h.)
Constru
t a two-dimensional table that shows the ee
t of exe
uting
two operations x and y in turn.7)
x. Think of ways to
redu
e the amount of work.
May 28. 64 .h.
Partitioning the Board
9.3.3
Partitioning the Board
The identi.145
9.
ation of the four operations on supersquares is a signi.
Then the remaining . Suppose one of the supersquares is labelled \ n ".
ant step towards
solving the Knight's-
ir
uit problem.
v . depending on their position relative to the starting square. all 64 squares of the
hessboard are split into four disjoint sets.fteen supersquares
an be uniquely labelled as n . h or c
squares. to the right of it is an \ h " square. and diagonally
adja
ent to it is a \ c " square.
As a
onsequen
e. Then
immediately above it is a \ v " square. Figure 9.9 shows
how this is done. Supersquares further away are labelled using the rules
for
omposing the operations. Suppose we agree that the bottom-left square is an \ n " square. In
.
we
an
onstru
t a straight-move
ir
uit of a 4 × 4 board.
Now. That means we
an
onstru
t straight-move
ir
uits of ea
h of the four sets
of squares on the
hessboard. Knight's Cir
uit
the simple fa
t that it is possible to
onstru
t a straight-move
ir
uit of a board of whi
h
one side has even length and the other side has length at least two. a \straight" Knight's move is \straight" with respe
t to the supersquares of a
hessboard. In .146
9. In parti
ular.
and a
ir
uit
of all the yellow squares. a
ir
uit of all the blue squares. a
ir
uit of all the green squares. The . 9.9.
We now have four disjoint
ir
uits that together visit all the squares of the
hessboard.g. this means
onstru
ting a
ir
uit of all the red
squares.
The way to do this is to exploit
the \diagonal" Knight's moves.nal step is to
ombine the
ir
uits into one. (Refer ba
k to .
Figure 9.
Figure 9. by
ombining the red-blue and green-yellow
ir
uits.11 shows one way of
hoosing the straight and diagonal moves in order to
ombine red and blue
ir
uits.
Figure 9.
in ea
h
ase. These straight moves are repla
ed by the diagonal moves.g. 9. we
an
ombine the green and yellow
ir
uits into a \green-yellow"
ir
uit.10 shows s
hemati
ally how red-blue and green-yellow
ir
uits are formed. and
green and yellow
ir
uits | in ea
h
ase. it is easy to . shown as solid
lines. a single move in ea
h being depi
ted by
a dotted line. The straightmove
ir
uits are depi
ted as
oloured
ir
les. For example.)
A simple way of
ombining the four
ir
uits is to
ombine them in pairs. similarly.10: S
hema for forming \red-blue" and \green-yellow"
ir
uits. Exploiting symmetry. we
an
ombine the red and blue
ir
uits into
a \red-blue"
ir
uit of half the board.7 for the meaning of \diagonal" in
this
ontext. and then
ombine the two pairs. a
omplete
ir
uit of the board is obtained. Finally. two \parallel" straight moves are repla
ed by
two \parallel" diagonal moves. two straight moves (depi
ted by dotted lines) in the respe
tive
ir
uits are
repla
ed by diagonal moves (depi
ted by straight lines).
9.
A slight diÆ
ulty of this method is that it
onstrains the straight-move
ir
uits that
an be made. To
omplete a
ir
uit of the whole board. The method of
ombination is indi
ated by the dotted and solid lines: the
straight moves (dotted lines) are repla
ed by diagonal moves (solid lines). a green-yellow
ir
uit has
to be
onstru
ted. This is left as an
exer
ise.
onsidering the suggested method for
ombining red and blue
ir
uits in . and this
ir
uit
ombined with the red-blue
ir
uit. For example. with this red-blue
ir
uit as basis.12 to form a \redblue"
ir
uit.g.
9. the straight-move
ir
uit of the red squares is
onstrained
by the requirement that it make use of the move shown as a dotted line.g. However. no
onstraint is pla
ed on the blue
ir
uit (be
ause there is only
one way a straight-move
ir
uit of the blue squares
an enter and leave the bottom-left
orner of the board).11. The diÆ
ulty is
resolved by .
\Parallel" red and blue moves. Indi
ate
learly how the individual
ir
uits have been
ombined to form the entire
ir
uit.
are repla
ed by diagonal moves. but not for smaller boards.13 illustrates another way that the
ir
uits
an be
ombined. shown as thi
ker solid lines. the result is a
ir
uit of the
omplete board.
The four straight-move
ir
uits are depi
ted as
ir
les. This is possible in
the
ase of an 8 × 6 board. Knight's Cir
uit
Figure 9.148
9.
To
arry out this plan. the four diagonal moves in .
Exercise 9. thus
ombining the two
ir
uits. one segment of whi
h has been
attened and repla
ed by a dotted line. shown as dotted lines.
ir
uits appropriately.
In order to
onstru
t a Knight's
ir
uit of smaller size boards. the dierent pairs of
ombining moves need to be positioned as
lose as possible together.10
2
Figure 9.
Constru
t a Knight's
ir
uit of an 8 × 8 board using the s
heme dis
ussed above.12: A \red-blue"
ir
uit. If these are repla
ed by diagonal moves (represented in the
diagram by solid bla
k lines). Do the same for an 8 × 6 board. The dotted lines represent straight moves between
onse
utive points.
one
Algorithmi
Problem Solving
Roland Ba
khouse. no straight-move
ir
uit is possible. more
are has to be taken in the
onstru
tion of the straightmove
ir
uits. try turning the board through 90◦ whilst
maintaining the orientation of the
ombining moves. whenever. However. by exploiting exer
ise
9. and the strategy used in exer
ise
9.)
2
Exercise 9.nd.2 for how this is done.
Constru
t a knight's
ir
uit of an
s
heme. Be
ause this superboard has an odd number of
(super) squares. If you en
ounter diÆ
ulties. for ea
h
ir
uit.
The strategy is to
onstru
t four straight-move
ir
uits of the board omitting one of
the supersquares.
(The
onstru
tion of the
ir
uit is easier for an 8 × 8 board than for a 6 × 8 board
be
ause. it is possible to
onstru
t Knight's
ir
uits for boards of
size (4m + 2) × (4n + 2) . (Re
all exer
ise 9. 2008
.10 is not appli
able.12 Division of a board of size (4m + 2) × (4n + 2) into supersquares yields
a (2m + 1) × (2n + 1) \super" board. in the latter
ase.
for any
8×8
and a
6×8
board using the above
Explain how to extend your
onstru
tion to any board of size
m
and
n
su
h that
m≥2
and
4m × 2n
n≥3.2.
May 28.) Then. both m and n are at least 1 .
150
9. Knight's Cir
uit
move is repla
ed by two moves |a straight move and a diagonal move| both with end
points in the omitted supersquare. This s
heme is illustrated in .
14.
Complete the details of this strategy for a 6 × 6 board.14: Strategy for Constru
ting a Knight's Cir
uit of (4m + 2) × (4n + 2) boards.
One supersquare is
hosen. and four straight-move
ir
uits are
onstru
ted around the
remaining squares.
11
00
00
11
000
111
000
111
Figure 9.g. These are then
onne
ted as shown. In order to
onstru
t the twelve
ombining moves depi
ted
in . 9. Make full advantage of the
symmetry of a 6 × 6 board.
g. whenever.14 . the remaining nine
an be found by
rotating the moves through a right angle. 9. whi
h means it is a very good example of dis
iplined problem-solving
Algorithmi
Problem Solving
Roland Ba
khouse.
2
9. it suÆ
es to
onstru
t just three. the Knight's
ir
uit problem is a truly diÆ
ult
problem to solve.4
Discussion
In the absen
e of a systemati
strategy. 2008
. both m and n are at least
1.
May 28.
Explain how to use your solution for the 6 × 6 board to
onstru
t Knight's
ir
uits of any board of size (4m + 2) × (4n + 2) .
This is indeed the
ase for the method we have used to solve the
Knight's
ir
uit problem. Dis
ussion
skills. The method has been applied to
onstru
t a
ir
uit of an
8 × 8
hessboard.
The key ingredients are
the
lassi.4.
The key
riterion for a good method is whether or not it
an be extended to other
related problems.151
9.
and
ombining these together. but the method
an
learly be applied to mu
h larger
hessboards. The method we have used to solve the problem is essentially problem de
omposition | redu
ing the Knight's
ir
uit problem to
onstru
ting straight-move
ir
uits.
A drawba
k of the method is that it
an only be applied to boards that
an be divided
into supersquares. it is not possible to
onstru
t a Knight's
ir
uit of a
board with an odd number of squares.
ation of moves as \straight" or \diagonal". a
omplete
hara
terisation of the sizes of
board for whi
h a Knight's
ir
uit exists is given in se
tion 9. for some m and n . whi
h is obviously highly impra
ti
al. (That is. That leaves open the
ases where the board has
size (2m) × (2n + 1) .) For those interested. As we have seen.5. and
using diagonal moves to
ombine straight-move
ir
uits . Contrast this with remembering the
ir
uit
itself.
straight-move
ir
uits of supersquares. if not impossible.
and reprodu
e a Knight's
ir
uit on demand. it is easy to remember these ingredients. one side has even length and the
other has odd length.
The Knight's-
ir
uit problem exempli.
On
e the method has been fully understood.
es a number of mathemati
al
on
epts whi
h
you will probably en
ounter elsewhere. The n . In general. and the fa
t that this relation partitions the
squares into four disjoint sets is an example of a general theorem about \equivalen
e
relations". h and c operations together form
an example of a \group". the relation on squares of being
onne
ted by straight moves
is an example of an \equivalen
e relation".
The Knight's-
ir
uit problem is an instan
e of a general
lass of problems
alled
\Hamiltonian-Cir
uit Problems". v . the input to a Hamiltonian-
ir
uit problem
is a so-
alled \graph" (a network of \nodes" with \edges"
onne
ting the nodes) and the
requirement is to .
before
returning to the starting node.nd a path through the graph that visits ea
h node exa
tly on
e. instan
es of a
lass of problems
alled \NP-
omplete" problems. in turn. Hamiltonian-
ir
uit problems are. NP-
omplete problems are problems
hara
terised by ease of veri.
May 28. it is easy to
he
k whether or not it is
orre
t (for example. no eÆ
ient methods
Algorithmi
Problem Solving
Roland Ba
khouse. for the
lass of NP-
omplete problems. however. given a
putative solution. That is. 2008
.
ation but diÆ
ulty of
onstru
tion. it is easy to
he
k whether or not it is a
Knight's
ir
uit). given
any sequen
ing of the squares on a
hessboard.
A
ording to one of these [Ma
Quarrie.
9. St.℄ the .5
Boards of Other Sizes
To be written. Andrews Univ.6
Bibliographic Remarks
Solutions and histori
al information on the Knight's
ir
uit problem
an easily be found
on the internet. \Complexity theory" is the name given
to the area of
omputing s
ien
e devoted to trying to quantify how diÆ
ult algorithmi
problems really are.
9. Knight's Cir
uit
are known at this time for
onstru
ting solutions.152
9.
The solution to exer
ise 9. Both
solutions involve sear
hing for \parallel moves". Mi
haelis's solution is slightly preferable
be
ause just two pairs of \parallel moves" have to be found at ea
h stage.11.
Dijkstra [Dij92℄. His solution is for a standard-sized
hessboard and uses the method of
ombining straight-move
ir
uits des
ribed in exer
ise 9. The pairwise
ombination
of straight-move
ir
uits is due to Diethard Mi
haelis [private
ommuni
ation℄.
long before
hess was invented. making it a
little bit harder to do.rst
Knight's tour is believed to have been found by al-Adli ar-Rumni in the ninth
entury. 2008
.12 was
onstru
ted by the author.
May 28.
The use of supersquares in solving the Knight's
ir
uit problem is due to Edsger W.
Algorithmi
Problem Solving
Roland Ba
khouse. with
useful feedba
k from Mi
haelis. Dijkstra's
solution involves sear
hing for four pairs (twi
e as many) at the same time.
Initially.p := m+1 .
In general. 1234−1 players must be kno
ked
out. Hen
e. n−1 .
The algorithm to remove obje
ts maintains the invariant that all obje
ts are of the
same kind equivales there is only one obje
t remaining. k and g are
both equal to 0 .3H |3W| . (In ea
h
ase.
2
3.
2
2. 5H |3W| 2W
. 2008
.1 1233 games must be played. either zero or two). and redu
e
the number of obje
ts. p−m . k
and g are always equal. if there are p players.
Consider the dieren
es m−n .1
{
5C ||
}
2C.3H |1W| 2W . Let k be the number of players that have been kno
ked
out. To de
ide the tournament.
The repla
ement pro
ess is modelled by the assignment
m. we
on
lude that the goal is impossible to rea
h if the starting
state has zero odd dieren
es. the number of odd dieren
es of any three numbers
is always even (i. p−1 .Solutions to Exercises
2. Every time a game is played.
153
May 28.e.
{
5H || 5W
}
Algorithmi
Problem Solving
Roland Ba
khouse.2 Let m . the tournament
onsists of p−1 games. n and p denote the number of obje
ts of ea
h kind. this number of games must be played. one more player is kno
ked out. and let g be the number of games that have been played. 2C. The goal is also impossible to rea
h if the obje
ts are all
of the same kind and there is more than one of them. Choosing to in
rease the number
of obje
ts of the kind that o
urs least frequently will maintain this invariant. If there is more than one obje
t
remaining.n. It is easily
he
ked that the parity of
ea
h is un
hanged by the assignment.) Also. there must be two obje
ts of dierent kind. the dieren
e either is un
hanged
or in
reases or de
reases by 2 . So. n−p . Sin
e the goal is to rea
h a state in whi
h there
are two odd dieren
es.
or all husbands are on one
of the two banks.
lH = 0
Algorithmi
Problem Solving
Roland Ba
khouse. (Re
all the assumptions about
M
and
N .
Property (a) is
learly true initially. (If
not. |3W| 1C. under the given assumptions on
(a)
M
and
N. 2W |1W| 1C. and let
. either there are no single individuals on either bank. The number of wives on the right bank. In both
ases. the wives outnumber the husbands
on the left bank or on the right bank.
After the
rossing.
assumption that
M < lH . and a boat
ouples. The number of husbands
on the right bank.
We note that an invariant is
(lH = lW) ∨ (lH = 0) ∨ (lH = N) . It is a requirement of any solution that this property is an invariant.)
denote the number of husbands on the left bank. we
laim that.
is at most the minimum of
properties are
ommon to the
ases of a boat of
apa
ity
of
apa
ity
Let
lH
3
and
6
2
and
4
3
and
N/2 . Note that
lH 6= 0
both before and after the
rossing. (Before the
rossing.
(These
ouples.Solutions to Exer
ises
.
either
The boat is on the left bank and
(14)
M < lH . either
0 < lH < lW
or
0 < rH < rW .
(13)
That is. denoted
rH . 2W |2W| 4H
. a
rossing is made from left to right. 2008
.
2
3. let
lW
denote the number
of wives on the left bank. suppose (a) is true and.)
Now.
Similarly.
is then
N−lW .3
Let
We assume that
N
is at least
2
and
M
N
denote the number of
ouples. the solution is invalid.
or
(b)
the boat is on the right bank and
(15)
M ≤ lH . then.
{
155
2C || 2C }
2W |2H| 2C .3H .3H
{
|| 4C
}
M
denote the
apa
ity of the boat.)
Now.
{
4W || 4H
}
2W |2W| 4H .
In words.
is then
N−lH . denoted
rW .
lH = 0
is ex
luded by the
is impossible be
ause at most
M
May 28.
lH is
un
hanged. (Stri
tly. (b) be
omes true. of
ourse.) That is. (We omit the justi. and (lH = lW) ∧ (lH 6= N) . (b) is true after the
rossing. we have established that (a) is true after the
rossing. either (a) or (b) is true. the
ase that the boat has
apa
ity
0
an be ex
luded from
onsideration!)
Again. it is impossible to maintain that lH = lW ).
In both
ases. and M < lH both before and after the
rossing. property (a) is true initially. we have established that. A right-to-left
rossing
auses lH and/or lW to in
rease. we need to assume that M is
non-zero in order to assert that lH 6= 0 . and some husbands
ross. after the
rossing M < lH and
(of
ourse) the boat is at the left bank. and lH
does not de
rease. (This is be
ause (16) must be maintained and the
rossing
annot
in
rease lH . Thus. if (b) is true and a right-to-left
rossing
is made. no husbands
ross. the
rossing is invalid.4 As stated in the hint. on the left side of the bank. at all times. we
onsider two
ases: this time. it must be the
ase that lH = N (be
ause (16) must be maintained
and.
If only wives
ross.
If (lH = lW) ∧ (lH 6= N) before the
rossing.Solutions to Exer
ises
156
husbands
an
ross together.
We
onsider two
ases before the
rossing: lH = N . the property lH = lW must be maintained by the
rossing.
But (14) is assumed to be true before the
rossing. That is. the invariant (13)
an be strengthened: the
rossing
is made starting in a state in whi
h
(16)
(lH = lW) ∨ (lH = N) . Also.
Sin
e only one
ouple
an
ross at one time. lH 6= 0 after the
rossing.
If husbands
ross. (a) is true after the
rossing.
2
3. an equal number of wives and husbands must make the
rossing. Obviously. if lH = N before the
rossing. vi
e-versa. Thus. after the
rossing. (b) is true after the
rossing. we assume that every forward trip involves two people and
every return trip involves one person.
Sin
e at most N/2 husbands
an
ross together. if only wives
ross. (b) is true. (a) is true after the
rossing. Sin
e lH 6= 0 before the
rossing. if (a) is true and a left-to-right
rossing is made. at
least N/2 are left at the left bank. and the boat is at the right bank. the a
t of
rossing in
reases lH . (a) be
omes true.
In summary.
Now suppose (b) is true and a
rossing is made from right to left.) So.
In both
ases. so. it
an never
be the
ase that all husbands are at the right bank. So. After the
rossing.
and must result in a state satisfying this property. Otherwise. Thus. the value of lH is de
reased by at most 1 . That is. That is.
Note also that a left-to-right
rossing
auses lH and/or lW to de
rease.
onsequently. (15) is true after the
rossing. the boat
is.
May 28.)
A
onsequen
e is that an optimal solution
onsists of exa
tly 5
rossings. Sin
e ea
h return trip is made by just one person. 2 people never make
Algorithmi
Problem Solving
Roland Ba
khouse. of whi
h 2
are return trips.
See
hapter 8 for a general argument why this assumption may be made. 2008
.
ation of this assumption here.
an optimal solution is found when persons 3 and 4 |the two
slowest| do not return. with person 1 returning. Clearly. (Consider any sequen
e of
rossings that gets all four a
ross
the bridge.
Then persons 3 and 4
ross.
(The in.Solutions to Exer
ises
157
a return trip. then one or both of
the two fastest does not make a return journey. The total time taken using this strategy is
t1↑t2 + t1 + t3↑t4 + t2 + t1↑t2 . a new sequen
e is found whi
h is at least as fast as the
original sequen
e | sin
e the time for at least one return trip is redu
ed and the time
for at most one forward trip is in
reased by the same amount. By inter
hanging a slower person with
a faster person in the sequen
e. and person 2 returns. The strategy \let the two slowest
ross together" is implemented by letting persons 1 and 2 (the two fastest)
ross. Finally. If one or both of the two slowest makes a return journey.)
There are two strategies for getting the two slowest a
ross: let them
ross together
or let them
ross seperately. persons 1 and 2
ross
again.
and . then person 1 returning.
(This
orresponds to person 1 and person 2
rossing.) If the two slowest
ross
seperately. then the best strategy is to let person 1
ross with them and then return for
the other. Implementing this strategy takes a total time of
t1↑t2 + t1 + t1↑t4 + t1 + t1↑t3 .es to t4 . then
person 1 and person 4
rossing. However. then person 1 returning. not simplifying the formula
just yet makes it is easy to identify the people in ea
h
rossing.
The
order
hosen here fa
ilitates the
omparison of the times.)
Comparing the total times. immaterial. 3 and 4
ross is.nally persons 1 and 3
rossing. the . of
ourse. The order in whi
h persons 2 .
)
Applying this solution to the two spe
i. (There is a small element of nondeterminism
in this solution: when t2 + t2 = t1 + t3 an arbitrary
hoi
e may be made between the two
strategies.rst strategy should be used when t2+t2 ≤ t1+t3 and
the se
ond strategy when t1+t3 ≤ t2+t2 .
e.
the two slowest should
ross together. 18 minutes. 4 minutes. 1 minute. 3 minutes and 3 minutes: Sin
e 1+1 ≤ 1+3 . we get:
(a) The times taken are 1 minute. The shortest time is 4+1+5+1+4 .)
Algorithmi
Problem Solving
Roland Ba
khouse. 2008
.e.e.
i.
May 28.
(b) The times taken are 1 minute. 15 minutes.
ases.
i. 4 minutes and 5 minutes. the two slowest should
ross seperately. (The shortest time if the two slowest
ross together is 4+1+5+4+4 . Sin
e
1+5 ≤ 4+4 . i. 7
minutes. The shortest time is 1+1+3+1+1 .
in August. whi
h has an even number of days. the last day of November).2
The .
The pattern
hanges in June. The
opponent is then for
ed to name the 1st of the next month. This is for
ed by the opponent naming 31st
O
tober.
We
on
lude that the se
ond player is guaranteed to win.Solutions to Exer
ises
158
2
3. This means that 31st Mar
h is a losing day. the winning strategy is to name the last day of the month. i. the pattern we saw for De
ember and November
re
urs. This is in line with the terminology of losing and winning
positions. and any day in September is a winning position.) That is. Similarly. Otherwise. The player who
names 1st July loses. and any day in July is a winning position. Every odd-numbered day in Mar
h is a losing day. also. (Take
are: The \losing positions" are the days that
the winning player names.
2
4.
This means that every even-numbered day in May is a winning day. like De
ember.e. the strategy is to name the next day of the year. Whether the year is a leap
year or not makes no dieren
e. the
odd-numbered days are losing positions. September.
Again. This
is for
ed by the opponent naming 30th November (that is.
2
4.5 6×2×3×3×1 . any even-numbered day in June is a losing day. the odd-numbered days in January are losing days. other than 31st De
ember results in losing. In general. .
b) In De
ember. In O
tober.
Sin
e Mar
h has an odd number of days. the losing positions are the odd-numbered days and the winning positions are the even-numbered days. the odd-numbered days are losing
positions. the player who names 1st De
ember wins. Any day in November is
thus a winning position. the player whose turn
it is must name an even-numbered day and. 108 .
July or February.
In parti
ular. so. and every day in February is
a winning day. be
ause the
opponent
an then name 30th November. if the last-named day is an odd-numbered day.1 a) Naming any day in De
ember.
onsequently. The strategy is to name
the 1st day of the following month when the last-named day is in November.
Similarly. will eventually lose. Finally. every evennumbered day in April is a winning day. it does not matter if it is a leap-year. naming any day other than 30th November results in losing.
2
The squares that are not positions are the ones at the foot of a ladder or at the
head of a snake.
The pattern repeats in the se
ond eleven positions.rst eleven positions are shown in table 1. See table 2. Positions that
annot be identi.
The mex
Algorithmi
Problem Solving
Roland Ba
khouse. winning
positions or stalemate positions. Any other move from 4 would be to a winning
position.
Position
Type
Move to square
1 2 4 5 7 13 14 16 18 21 22 23 24 25
WWS WW L WWL WWWWL
3 3 6 9 9
18 18
25 25 25 25
Table 3: Snakes and Ladders. this is
learly the wrong thing
to do. Labelling of winning and losing positions assumes that every
game is guaranteed to terminate no matter how either player moves. the positions are
hara
terised as losing positions. From this position.
When a game has
y
les. If
y
les o
ur this
assumption is not valid.
May 28. a winning position is one from whi
h there is a move to a losing
position. Losing (L) and Stalemate (S) positions
2
4.Solutions to Exer
ises
Position 0
Type
L
Move
159
1
2
3
4
5
6
7
8
9
10
L
W
W
L
W
W
W
L
W
W
2
2
5
6
6
5
6
Table 1: Winning (W) and Losing (L) Positions for subtra
tion set {2 . and a stalemate position is one from whi
h there is a move to a stalemate
position and there are no moves to losing positions. Position
4 is the only stalemate position.4
a) See table 4 for the mex numbers up to and in
luding position 10. 6}
attributable to
y
les in the positions. 5 . a
y
le is a sequen
e of moves that begins and
ends at the same position. The opponent will then use the same strategy and the game will
ontinue forever.
From a stalemate position the best strategy is to move to a stalemate position sin
e.
Table 3 shows all positions and the winning move from winning positions. A losing position is one from whi
h every move is to
a winning position. 6}
Position 11 12 13 14 15 16 17 18 19 20 21
Type
L L W W L W W W L W W
2
2
5
6
6
5
6
Move
Table 2: Winning (W) and Losing (L) positions for subtra
tion set {2 . Winning (W). 5 . 2008
.
if there is an alternative of moving to a winning position. a move to square 6 has the ee
t
of returning the
ounter to position 4 .
for all i . But
n < 2i+1−1 ≤ 2×n
=
{
meaning of
ontinued equalities
}
n < 2i+1−1 ∧ 2i+1−1 ≤ 2×n
=
{
integer inequalities .
When i equals 0 .
The proof is in two parts: we show that. 5 . all other
positions are winning positions. 2008
.
}
May 28. (Other answers
an
be given for the winning moves. also. 2i+1−1 equals 1 . for all i . 6} . the mex number of position m is m mod 3 . from a position n where.)
2
4. Together with the mex
numbers for the right game given above. Position 1 is an end position and thus a losing
position.5 (a) The losing positions are positions 2i+1−1 where i is a natural number.
0
Position
Type
L
Mex Number 0
1
L
0
2
W
1
3
W
1
4
L
0
5
W
2
6
W
1
7
W
3
8
L
0
9
W
2
10
W
1
Table 4: Mex numbers for subtra
tion set {2 . symmetry of ∧
Algorithmi
Problem Solving
Roland Ba
khouse. every move from position 2i+1−1
is to a position n where 2i−1 < n < 2i+1−1 . we
an
omplete table 5.
Left Game Right Game \losing" or winning move
10
20
R14
20
20
losing
15
5
R0
6
9
R4
37
43
losing
Table 5: Winning moves
In the left game. When i is greater than 0 .Solutions to Exer
ises
160
numbers repeat from here on.
n 6= 2i+1−1 we show that we
an
hoose i so that there is a move from n to position
2i−1 . the mex number for position m is equal to the
mex number for position m mod 11 . every move from position 2i+1−1 is to a position
n where n < 2i+1−1 ≤ 2×n . that is.
in order to move the knight as required.
2
5.6 Suppose the number of
ouples is n . there are at least two people
other than the host and the host's partner.n . Be
ause neither is the host. It follows that the two people who shake hands 0 and 2n − 2 times
are husband and wife.5
then
Let col be the
olour of the square. it also follows that neither is the
host's partner. there must be someone who
shakes hands with k people for ea
h k between 0 and 2n − 2 (in
lusive). and n be the number of moves. the only person other than the host is the host's partner.
An invariant of this assignment is
col ≡ even.
Algorithmi
Problem Solving
Roland Ba
khouse. If 2n − 1 of them |everyone but
the host| shake hands with a dierent number of people. but the
olour of the square doesn't
hange.
May 28. 2008
.n . it is thus the
ase that everyone ex
ept that person's partner shakes hands with
at least one person.Solutions to Exer
ises
162
No. of Rows
2
15
4
11
4
14
13
6
21
19
\losing" or winning move
C1 (or R11)
C2 (or R9)
R9
losing
C19 (or R10)
Table 6: Solution to re
tangle game
In general. of
ourse). Sin
e
ouples
do not shake hands. in
luding the host.(2×n) equals n and mex. By the symmetry of the shake-hands
relation.n . of Columns No. In this
ase. who
ea
h shake hands with between 0 and 2n − 2 people.
So. A move is
col . both shake hands 0 times. The person who shakes hands 2n − 2 times does so with everyone
ex
ept their partner (and themself. n := ¬col . n+1 .
whi
h is impossible. There are 2n people. a
hange has to be made to col ≡ even.
If n is 1 .(2×n + 1) equals mex.
An odd number of moves ( 63 ) is needed.
Now suppose that n is greater than 1.
2
5. Consider the two people who shake hands
0 and 2n − 2 times. mex.
9
¬true
=
{
law
¬p ≡ p ≡ false
law
true ≡ p ≡ p
law
¬p ≡ p ≡ false
with
p := true
}
true ≡ false
=
{
with
p := false
}
false . The number of times ea
h person shakes hands is then redu
ed
by one.
Ea
h time.10
¬¬p
=
{
Algorithmi
Problem Solving
with
p := ¬p
Roland Ba
khouse. The host and the host's
partner must therefore have both shaken hands
n−1
times. the number of times
the host and the host's partner shake hands is redu
ed by one.
the other
n−1
That is. 2008
.
2
5.Solutions to Exer
ises
163
Now suppose we dis
ount this
ouple. we
onsider the party
onsisting of
ouples.
Repeating this pro
ess. we eliminate all the
ouples one by one until the party has
been redu
ed to just the host and the host's partner. again. So.
2
5. all but the host have shaken hands a distin
t number of times.8
(a) false
(b) false
(c) false
(d) p
(e) false
(f) q 6≡ r
(g) p
(h) true
2
5.
}
May 28.
2
5.nition of negation: (5. Assume indu
tively that.
there are exa
tly two
olourings. When there are n+1 lines.
May 28. A line on a
ball does this. If the lines are not
straight and they interse
t in a segment of a line. This is
learly the
ase when there are no lines.
The solution remains valid provided every line
uts the surfa
e in two. This is be
ause. the
olours
of the adja
ent regions are not the same before the inversion takes pla
e
ontrary to the
assertion made above. inverting the
olours of one of the two
regions does not guarantee that the
olouring of adja
ent regions at the boundary of the
left and right regions is satisfa
tory.
2
6. Cut the paper along the
hosen line. for ea
h half. i. along the line segment. Combining these gives four dierent ways of
olouring
the entire sheet of paper. However.3) }
b .1 It is required that any two lines interse
t in a single point.
The number of
olourings is always two no matter how many lines there are.
hoose any one of
the lines. In words. Q ≡ ¬B . whereas a line on a doughnut need not. ask A
whether B is a knave.12 Let Q be the question. 2008
.e. Q ≡ A ≡ A 6≡ B . two of these are unsatisfa
tory be
ause the
olours
Algorithmi
Problem Solving
Roland Ba
khouse. Then.
there is just one obje
t.5 When m is 0 .Solutions to Exer
ises
165
of regions adja
ent at the
hosen line must be dierent. Split the 3m obje
ts into 3 groups ea
h of
3m−1 obje
ts. at most a further 2×(m−1)
omparisons are required
to . At most 2
omparisons are needed to determine whi
h of the
groups it is. This leaves exa
tly two ways of
olouring the paper with n+1 lines. One of these 3 groups will have a dierent weight to the other two. This is the unique obje
t and 0 (whi
h
equals 2×0 )
omparisons are needed to dis
over that fa
t.
Suppose now that m is greater than 0 .
2
6. Then. by indu
tion. whi
h
will be of equal weight.
This gives a total of 2×(m−1) + 2 .
It is possible to determine whether the unique obje
t is lighter or heavier than the
others (although.
omparisons as required by the indu
tion hypothesis. the answer is that it is both
lighter and heavier than all the rest). It
an be de
ided in the . 2×m . in the
ase that there is just one obje
t. i.e.nd the unique obje
t in that group.
assume that n−1
omparisons are needed to . it is
lear that 0
omparisons are needed.
2
a) For n = 1 .rst two
omparisons. For the indu
tion step.
nd the lightest of n obje
ts. To .
nd the
lightest of n+1 obje
ts. use n−1
omparisons to .
one extra
omparison has been made. it is
lear that 1
omparison is needed.nd the lightest of n obje
ts. The lightest of the two is the lightest of
them all.
b) For n = 2 . For the indu
tion step. assume that
2n − 3
omparisons are needed to . Also. then
ompare this obje
t with the ( n+1 )th obje
t. making n in total.
To .nd the lightest and heaviest of n obje
ts.
nd the
lightest and heaviest of n+1 obje
ts. use 2n − 3
omparisons to .
making (2n − 3) + 2 . The heaviest of the two is the heaviest of the four. Call the ( n+1 )th obje
t N .e. The lightest
of L and N is the lightest of them all.nd the lightest and
heaviest of n obje
ts. i. Compare B
and D . and the heaviest of H and N is the heaviest of
them all. .
To weigh four obje
ts. This requires two extra
omparisons. 2(n+1) − 3 in
total. Call these L and H . The lightest of the two is the lightest of the four.
) Compare A and C .
Call the lighter one A and the heavier
one B .rst
ompare two.
ompare the remaining two obje
ts and
all the lighter one C and the
heavier one D . Then pro
eed as above.
d) For m = 1 . Likewise. it is
lear that 1
omparison is needed to .
Sele
t and
ompare any two of the obje
ts.
Suppose there are 2(m+1) obje
ts. And.
Let the lightest be A and the heaviest B . we
an . 1 = 3×1 − 2 . By indu
tion.nd the lightest and heaviest
of 2 obje
ts.
su
h that A < B and
6. C and D . A . We now have four obje
ts.
May 28. 2008
. respe
tively.6
Algorithmi
Problem Solving
Roland Ba
khouse.nd the lightest and
heaviest of the remaining 2m obje
ts in 3m − 2
omparisons. B . Let these be C and
D .
And.
2
7. the lightest and heaviest of these four
an be found in 2 further
omparisons.1 Formally we have
To.d
=
{
de. By part (
). These are then the lightest and heaviest of all 2(m+1) obje
ts.Solutions to Exer
ises
166
C < D . the
total number of
omparisons is 1 + (3m − 2) + 2 whi
h equals 3(m+1) − 2 .
May 28. 2 .¬d . .d = 0
Tn+1. 2008
.d = 2 × Tn.d (in parti
ular the repeated multipli
ation by 2 )
suggest that Tn.d is 2n−1 .d is 1 and T2. just as we did for the equations for
H .¬d + 1 .d is 3 (in ea
h
ase for all d ).
2
Algorithmi
Problem Solving
Roland Ba
khouse. T1.¬d + 1 + Tn.
That is. . The simple indu
tive proof is omitted. This
and the form of the equation for Tn+1.nition of T (twi
e) and length
}
Tn. .d is 0 . .
To. 1 .
If we expand these equations for n = 0 . we dis
over that To.
So. For example.
suppose there are . In any
state. the disks on any one pole are in order of de
reasing size.Solutions to Exer
ises
167
`
Figure 15: State-transition diagram for 0-disk problem.2 We begin by
onsidering the permissible states that the puzzle may be in. if we want to spe
ify
the state of the puzzle we only need to spe
ify whi
h pole ea
h disk is on.
7.
disk 5 must be on the bottom of pole B and disk 2 must be on top of it.
No other arrangement of the disks satis. and disk 1 must be on top of
disk 3 . disks 3 and 4 are on pole A and disk 5 is on pole B . Then disk 4 must
be on the bottom of pole A .ve disks and suppose we spe
ify that disk 1 is on pole A . disk 3 must be on top of it. disk 2 is
on pole B . Also.
es the rule that no disk is above a disk smaller
than itself.
The state of an n -disk puzzle
an thus be spe
i.
ed by a sequen
e of n pole names.
The .
Now we
onsider the transitions between states.rst name in the sequen
e is the lo
ation of disk 1 . We
onsider . and so on. the k th name in the sequen
e is the lo
ation (pole name)
of disk k . That is. Sin
e ea
h disk may be on one of three poles we
on
lude that there are 3n
dierent states in the n -disk problem. the se
ond is the lo
ation of
disk 2 .
This is shown in . then the 2 -disk problem. and then we
onsider the general n -disk problem. then the 1 -disk problem.rst the problem where
there are no disks.
When there are no disks there is exa
tly one state: the state when there are no disks
on any of the poles.
g. (You may have diÆ
ulty seeing the . 15.
gure. (See .
It
onsists of a single dot!)
We now explain how to
onstru
t the state-transition diagram for the ( n+1 )-disk
problem. for an arbitrary n . given that we have
onstru
ted the diagram for the n -disk
problem.
es the lo
ation of
the largest disk.
May 28. 2008
. Thus. ea
h state in the state-transition diagram for the n -disk problem
Algorithmi
Problem Solving
Roland Ba
khouse.
Solutions to Exer
ises
168
gives rise to 3 states in the state-transition diagram for the ( n+1 )-disk problem. That
is. a state in the state-transition diagram for the ( n+1 )-disk problem is spe
i.
B or C .
Consider .ed by
a sequen
e of n pole numbers followed by the pole name A . We split the
permissible moves into two sets: those where the largest disk (the disk numbered n+1 )
is moved and those where a disk other than the largest disk is moved.
But its position doesn't ae
t the permissibility or
otherwise of a move of a smaller disk. The . where the pole name p is either A . That means that every transition from state s to
state t in the n -disk problem is also a valid transition from state sp to state tp in the
( n+1 )-disk problem.rst moving a disk other than the largest disk. When doing so. B or C . the largest
disk may be on pole A . B or C .
rst step in
the
onstru
tion of the state-transition diagram for the ( n+1 )-disk problem given the
state-transition diagram for the n -disk problem is to make three
opies of the latter.
The p th
opy is then modi.
Being the largest
disk it may only be moved if all the other disks are on one and the same pole dierent
to the pole that the largest disk is on.
or three edges in the undire
ted state-transition diagram: an edge
onne
ting the states
AnB and AnC . The
onstru
tion is shown s
hemati
ally in . the disk numbered n+1 . This gives six possibilities for moving disk n+1 .
Now
onsider moving the largest disk. an edge
onne
ting the states BnC and BnA and an edge
onne
ting
the states CnA and CnB .ed by simply adding p at the end of ea
h sequen
e of pole
numbers labelling the nodes.
e. be
ause the dire
tion of movement is opposite to that of the smallest disk
(whi
h has an odd number).
2
7.k ≡ d ′ .g.) Set d to odd. when it is no longer possible to determine k in step 1). 16.
2
7.
2.4 The algorithm is to repeatedly exe
ute the following pro
edure until it
an no longer
be exe
uted (i. the three
inner triangles representing the set of all moves that do not move disk n+1 . (Re
all
that disk 1 is the smallest disk.3 Even. of
ourse). Suppose it is possible to move disk k in the dire
tion d ′ .
1.
3.
The
orre
tness is justi. Move disk k (in the dire
tion d ′ . Move the smallest disk in the dire
tion d . where k > 1 .
we know that the
.ed as follows. When step 1 is exe
uted.
To do this.rst k−1 disks are all on the pole in dire
tion ¬d ′ from disk k . it is ne
essary
to move the k−1 smallest disks in the dire
tion ¬d ′ .
May 28. Progress is made if
these k smallest disks
an be transferred to the same pole. 2008
. The dire
tion that disk 1 has to
be moved is thus d where
Algorithmi
Problem Solving
Roland Ba
khouse.
starting with the largest and ending
with the smallest. Otherwise.Solutions to Exer
ises
170
even. moving one-by-one to all the
squares in the same row. suppose it needs
to be moved in dire
tion d from its
urrent position. we get that d = (odd.1
(a) The number of moves that have to be made equals the number of squares. During
this sequen
e of moves the smallest disk will
ontinue to move in the same dire
tion.) The
orre
tness of the Towers of Hanoi program then guarantees that this a
tion will initiate
a sequen
e of moves after whi
h all k−1 disks will have been moved onto disk k . but. Finally.
Simplifying. initially
k is N and we are done when k is 0 . as required.(k−1) ≡ ¬d ′ ≡ even. no straight-move
ir
uit is possible. the dire
tion of the smallest disk may or may not be reversed. the dire
tion that the smallest
disk is moved should be the same as the dire
tion that disk k is moved. a straight-move
ir
uit
of a 2×n board is
ompleted by starting at a
orner.
The only time that step 1
annot be exe
uted is when all the disks are on the same
pole.
(c) See (b) for a
ir
uit of a 2×1 board. otherwise the smallest disk is moved in the opposite dire
tion to disk k . Move the smallest k−1 disks in
the dire
tion ¬d . (In words. we ensure
that the k smallest disks are on the same pole. 2008
. at least one
square is two moves from the starting square. then move disk k to its rightful position. de
rease k by 1 . On
ompletion. the
olour of the
urrent square is dierent from the
olour
of the starting square. it is impossible to visit su
h a square
and return to the starting square without visiting the in-between square more than
on
e. if k is also
odd. then returning via the se
ond row. after an odd number of moves. de
rease k by 1 .
2
Algorithmi
Problem Solving
Roland Ba
khouse.5
2
9. After an
odd number of moves. move to the other square and then move ba
k again| .k ≡ d ′ ) . so. Ea
h time the value of k is reassigned.
If the k th disk is on the right pole.1 ≡ d . For n greater than 1 .
otherwise.
2
The solution is to pla
e the disks in order. however. it is impossible to return
to the starting square. If m is greater than 1 .
(b) It's easy to see that a straight-move
ir
uit of a 2×1 board is possible |starting
at one of the squares.
Continue this pro
ess until k is 0. Let k denote the number of the disks still to be repla
ed.
May 28. So.
7.
(It doesn't matter whether numbering
starts at zero or one. the same
ondition applies. Then.2 For the 3 × 3 board. n) . with the 3 × 3 board as the base
ase| a
ir
uit of the board with the
omitted square.Solutions to Exer
ises
171
9. Constru
t
ir
uits of these three boards. The
onstru
tion is to split the board into four re
tangular
boards in su
h a way that the to-be-omitted square is at a
orner of a board with an odd
number of squares. For larger boards.m = even. a
ir
uit
an be
onstru
ted exa
tly when the omitted square is
not adja
ent to a
orner square. The other three boards ea
h have an even number of squares.
onne
t the
ir
uits together as shown in .n .) Then a
ir
uit
an be
onstru
ted of the remaining squares exa
tly
when even. and
|indu
tively. and at
least one of them has at least one square. Suppose
the
oordinates of the omitted square are (m.
Veri.ed by observing that the table is symmetri
about the top-left
to bottom-right diagonal.
The
ase that x . 2008
.
May 28. y or z is n
an be dealt with simply.
ation of the asso
iativity property is mu
h more tedious. This leaves 27 other
ases to
Algorithmi
Problem Solving
Roland Ba
khouse.
but straightforward" proof!
2
9. it suÆ
es to understand how all the entries are generated by a small set of
primitive transformations. (To be
ompleted.
n
r
a
c
n
r
a
c
n r
r c
a n
c a
a
n
c
r
c
a
r
n
Table 8: Sequential Composition of Rotation Operations
There are 24 dierent ways to assign a dierent
olour to ea
h of the squares in
a 2 × 2 board. so that the size of the full table is 24 × 24 ! Rather than
omplete the
full table. This is an example of a \tedious.Solutions to Exer
ises
172
onsider.9
.)
2
See .
18. all that is required is to \shorten" the straight-move
ir
uits in order to
a
ommodate the smaller board. In the
ir
uit shown.
Note how the
hoi
e of parallel moves
onstrains the
hoi
e of red
ir
uit. a substantial number of
ir
uits
an be found all
based on the same set of
ombining parallel moves.10
2
9. and solid bla
k lines. the
red moves are entirely di
tated by this
hoi
e. depi
ting the diagonal moves. Moves
indi
ated by dotted lines are repla
ed by the diagonal moves indi
ated by solid bla
k
lines.
Figure 20 shows the
ir
uits obtained in this way. In this way. (But note that they
annot be used to
onstru
t a
ir
uit of a 6 × 8 board. the green.
The parallel moves used to
onne
t
ir
uits of dierent
olours are indi
ated by dotted
lines depi
ting the straight moves.g. There is some freedom in
hoosing a green
ir
uit. it suÆ
es to use the te
hnique detailed in . In
ontrast.
In order to
onstru
t a
ir
uit for any board of size 4m × 2n . In fa
t. blue
and yellow
ir
uits were
onstru
ted by \
opying" the red
ir
uit. there is
omplete freedom
in
hoosing a blue or yellow
ir
uit.)
9. these are the moves that are repla
ed.
but not
omplete freedom. The dotted lines are not part of
the
ir
uit.11 Figure 19 shows details of how the straight-move
ir
uits are
ombined.
The same set of
ombining parallel moves
an be used to
onstru
t a
ir
uit of an
8 × 6 board. where m is at least
2 and n is at least 3 .
on
e for ea
h of the straight-move
ir
uits in the solution to the
8 × 6 -board problem.Solutions to Exer
ises
173
Figure 18: A Knight's Cir
uit.12 We begin by identifying the moves shown in . The dotted lines depi
t
straight moves that are repla
ed. The diagonal moves that repla
e them are depi
ted by
solid bla
k lines. This
onstru
tion has to
be applied four times. Solid lines indi
ate the
ir
uit.
2
9.
extending straight-move
ir
uits to boards of arbitrary size.
14.g. See . 9.
)
Now it is easy to . (Note the
symmetry. 21.g.
See
.ll in the straight-move
ir
uits around the remaining squares.
it is easy to extend the straight-move
ir
uits.
May 28.
For the general problem. 2008
.g. 22.
2
Algorithmi
Problem Solving
Roland Ba
khouse.
(Dotted lines are not part
of the
ir
uit. as detailed in
. these are the moves that are repla
ed by diagonal moves.Solutions to Exer
ises
175
Figure 20: Knight's Cir
uit of an 8 × 6 and an 8 × 8 board. | 677.169 | 1 |
NUMERACY EXAM INFORMATION
More information regarding this matter to follow.
Foundations and Pre-Calculus 10 Honours
This course is designed to provide students with mathematical understandings and critical thinking skills identified for post secondary studies in both the arts and the sciences. Topics include trigonometry, irrational numbers, powers involving integral and rational exponents, polynomials, relations and functions, system of linear equations, sequences and series, and financial literacy.. Students will write a Provincial Numeracy Exam in June. At the end of this course, students are prepared for Apprenticeship and Workplace 11, Foundations of Mathematics 11, and/or Pre-Calculus 11. | 677.169 | 1 |
Calculators: You are welcome to use a calculator on your homework, but there
will be NO calculators allowed on any of the exams!
Grading: The grades will be calculated as follows:
Weekly Homework 10%
Midterm 25%
Midterm 25%
Midterm 10%
Final Exam 30%
(Note: There will be 3 midterms. Your lowest midterm score will count for
10% of your grade and your top two midterm scores will each count for 25% of
your final grade.)
Teaching Philosophy: I believe strongly that mathematics, at its
core, is the art/experience/science of problem solving and pattern recognition.
It is inherently a creative process, one to be struggled with, repeated, and
enjoyed. The process requires imagination, persistence, courage, processing
time, and ultimately produces experiential, mathematical skill. It is from
this perspective that I teach. I'm not as concerned with the destination, i.e.
the answer, as I am about the journey of problem-solving and mathematical
exploration since it is exactly the entirety of the journey that creates
the answer. And, self-confidence and mastery are then natural by-products
of the mathematical journey.
Weekly Homework:
I will collect homework during the first class of each week.
All of the homework assigned the previous week is due at that time.
Each section of homework will be worth ten points. For example,
if you have three sections of homework assigned for one week, then
that homework set is worth a total of 30 points.
You will get the full credit if you DO every problem. I will
NOT be grading for correctness, so it is your responsibility to
make sure you understand the problems and their solutions. This
is basically motivation for you to do the homework because that is
the only way to survive a math class. (Please notice that there
is no way to get an A in this course if you choose not to do any of
the homework. On the other hand, turning in all of the homework
can help your grade substantially.)
The homework is to be turned in according to the following
instructions:
The homework set MUST be stapled together with the
corresponding cover sheet as the first page. (I will
provide the cover sheets in class.) A homework set turned
in without being stapled together or without a cover sheet
will not be counted!! It will be returned to you without
being graded. You can staple it with the cover sheet and
turn it in the following week to be graded...but it will then
count as a late section! Please do not come to class hoping
that I or someone else will have a stapler. Be prepared
when you arrive.
To get full credit for the homework, you MUST show all your work!
If you turn in just the answers, with no work shown, you will get
a -1 score for that section. If this happens three times, you will
no longer be allowed to turn in homework for any credit!!!
I will accept 10 late homework sections, up
to two weeks late, throughout the semester for full credit. I will not accept homework more than two weeks late. I
accept these late homework sections to allow for illness,
oversleeping, hectic schedules, etc. Do not ask for special
favors with regard to the homework policy, unless the
circumstances are extraordinarily severe, because my response
will not be in your favor. Also, it is not necessary nor
recommended that you tell me why your homework is late.
This policy is meant to be flexible enough to cover all reasons
and so I don't get calls and emails every week with your personal
drama stories.
I only collect homework during the first class of the
week!! If you need to turn in homework late, you will have
to do so on the first class of the week (one or two weeks after
its due date). This is the only time and location that I
will accept homework. If you slide homework underneath my office
door, I will throw it away without looking at it.
You are responsible for knowing these policies. Please take it
seriously because the flexibility listed here is literally all there is.
WebCT: I will put your grades online on WebCT.
You can get there easily from the main University of Utah website
(There's also a link from my website.) To log in, you use the same student id
and password that you use for Campus Information System. I do my best to
update the grades on a regular basis and keep everything accurate.
However, I would advise you to check your grades often to make sure there
were no data entry mistakes. I'm always happy to correct any mistakes I've
made. You just need to let me know about them.
Grading Scale: The grade scale will be the usual:
A (93-100), A- (90-92), B+ (87-89), B (83-86), B- (80-82), C+ (77-79),
C (73-76), C- (70-72), D+ (67-69), D (63-66), D- (60-62), E (0-59).
I will not curve the grades, as there will be no need, since I've
taught this class many times and know what to expect.
Other Policies: Due to experience, I have decided to make some
additional policies regarding my classroom administration and grading.
There will be no retakes of exams…ever. Your score is
what you get.
There are no alternate exams, period. Make sure you make your
travel plans around the exam dates, which I will keep fixed. If
your parents decide you need to come home early for winter break,
then you will simply miss the final exam for this class and therefore
fail. The only exceptions I will ever make to this policy is for
extreme cases of illness, in which case you will need to provide
thorough documentation of said illness.
I will demand respectful behavior in my classroom. Examples of
disrespect include reading a newspaper or magazine in class, social
chatting with your friend in class, text-messaging your buddies during
class or cuddling with your girl/boyfriend in class. If you choose to
be disrespectful during my class, I can guarantee I will take action
to terminate your disruptive behavior.
There will be no cursing nor negative ranting (for example,
"math sucks") on any written work turned in. The penalty
for such things on your written work will be a zero score on that
assignment or test!
You need to have a valid email address registered with Campus
Information System. I will regularly send emails to the class and hold
you accountable for receiving that information.
If you have crisis-level extenuating circumstances which require
flexibility, it is completely your responsibility to communicate with me
as soon as possible. The longer you wait to communicate with me, the
less I can and am willing to do to help.
If you have questions about any exam grade, or you want
to appeal the grading of the exam, you must bring it to me within one
week of the exam. After that, you will have to live with whatever
grade you got.
Please make sure you do your best throughout the semester and come
talk to me if you need further study strategies. I will NOT offer
any extra credit at the end of the semester or any other way for you
to improve your grade at that time. If you ask me toward the end of
the semester if I'll make special arrangements for you to improve
your grade by some means, I will automatically deduct one percentage
point from your overall grade, just for asking the question!! | 677.169 | 1 |
The quiz: On Tuesday we had a little `quiz', covering
some of the preliminaries for the course. The quiz was returned via
crowdmark. The median and mean scores were 11 and 10.9 out of 15,
respectively. I leave it up to you to draw conclusions. Just keep in
mind that last year, only 50 percent of initially enrolled students
managed to finish and pass the course. The `course change date' for changing from MAT240 to MAT223, is October 3 . See this link for details.
Tutorials: The tutorials started this
week. Although the rooms are booked for two hours, the actual tutorial
won't go much over one hour. Following the `formal' part of the
tutorial, the TA will be available for additional questions. Please note that all our three TA's (Debanjana, Jeff and Ivan) also hold office hours, I
included the updated information (location, email addresses) on the course syllabus . (You may have to hit `refresh' if you downloaded this before.)
In class, we covered: Meaning of the notation $P \Rightarrow Q$ and $P \Leftrightarrow Q$. More properties of fields, for example
$ab=0 \Rightarrow\ a=0 \mbox{ or } b=0$.
The field of complex numbers. Real and imaginary part, complex conjugate, absolute values, and their geometric interpretation in the complex plane. Basic properties (some of this will have to wait until next week): how to take the inverse of a complex number, geometry of multiplication of complex numbers, polar coordinates, triangle inequality $|z+w|\le |z|+|w|$, fundamental theorem of algebra (without proof).
Homework: Assignment #1 is due Friday 11:00pm. No late work is
accepted. Assignment #2 will be released before Friday 11:00 pm.
Additional homework (not to be handed in):
Do examples with complex numbers! Problems with solutions can be found
on the web, for example
here
and many other places.
What's wrong with the calculation
$$ 1=\sqrt{1}=\sqrt{(-1)(-1)}=\sqrt{-1}\sqrt{-1}=i\cdot i=-1\ \ \ \ \ ??\ \ \ $$
Well, it assumes that $\sqrt{z w}=\sqrt{z}\sqrt{w}$ for complex numbers $z,w$, which is something one has to be careful about. It's true for positive real numbers, using the positive square root. The square root of a complex number is only given up to sign, and there's no consistent convention `fixing the sign' for which this formula would always be true. | 677.169 | 1 |
Hey guys ,I was wondering if someone could help me with book mathcad pdf? I have a major project to complete in a couple of months and for that I need a good understanding of problem solving in topics such as hyperbolas, mixed numbers and side-side-side similarity. I can't start my assignment until I have a clear understanding of book mathcad pdf since most of the calculations involved will be directly related to it in one way or the other. I have a question set , which if somebody could help me solve, would help me a lot.
You can try out Algebrator. This software literally helps you solve questions in algebra very fast. You can plug in the questions and this program will go through it with you step by step so you can understand easily as you solve them. There are some demos available so you can also take a look and see how incredibly helpful the program is. I am sure your book mathcad pdf can be solved faster here.
A extraordinary piece of math software is Algebrator. Even I faced similar problems while solving matrices, factoring expressions and proportions. Just by typing in the problem from homework and clicking on Solve – and step by step solution to my math homework would be ready. I have used it through several math classes - Basic Math, Algebra 2 and Remedial Algebra. I highly recommend the program. | 677.169 | 1 |
Algebra 2 Curriculum
This bundle contains warm-ups, notes, homework assignments, quizzes, unit tests, a midterm test, end of year review materials, and a final exam for Algebra 2. This bundle does not contain activities.
Nothing like a good criminal investigation to liven up math class!
Based off my popular CSI projects, I have created Whodunnits? The Whodunnits plan to focus on single content topics as opposed to units and will focus more on skill building than application. Every day can't be a project day, so Wh
Inequality problems are made slightly novel with the element of hangman. Students practice solving multi-step inequalities with this hangman activity/worksheet. A built-in message makes grading the practice (and also giving students feedback as they work) a snap!
Different levels with the same fo
This activity is designed to review solving systems of equations by graphing (or writing systems of equation by the graph). This activity also gets students up and about.
Place the 10 cards on the wall around your room. Students pick any card to begin with. They should graph the function on the bot
Algebra Warm Ups
Great for a "Back to School Algebra Review" or
"Summer Algebra Review" too!
Perfect as a review over the summer. With just 4 questions a day for 76 days, students will be fully prepared for the new school year!!!
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This activity combines the skill of writing algebraic expressions with the classic board game CLUE. Students will use the clues they gather from correctly writing algebraic expressions to solve the mystery of Who Killed Mr. X. Pression.
This kit offers two ways to use this activity.
Use as a Sc
Students match 20 cards of graphs, scatter plots, sets of coordinates, mappings and tables to cutouts of each relation's correct domain and range in this activity. Once all cards are matched, students can also sort their cards into "function" or "not a function" and record their answers on the incl
Parent Functions and Transformations (Algebra 2 Curriculum - Unit 3)
The purpose of this unit is to provide the foundation for the parent functions, with a particular focus on the linear, absolute value, and quadratic function families. A chart is provided with all the parent functions that can be | 677.169 | 1 |
Synopsis
This is a book where you will find many solved exercises about Algebra, Linear Algebra, Calculus and Mathematical Analysis. It is very useful for all those students needing help and support. The exercises are solved step by step, explaining how and why the exercise is solved by this way. It's not a theory book. And you will find exercises about: equations, numbers satisfying any conditions, non linear system equations, vectors and algebraic geometry, integral domains, Diophantine equations, linear algebra, absolute value, inequalities (any of them using absolute value), limits, continuity, function domain, derivable functions, study of a function (maximum, minimum, increasing and decreasing intervals, concavity), series and integrals (any of them using Beta and Gamma functions).
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$54.99 USD
Your price
$43.99 USD
You save $11.00 (20%) and
You'll see how many points you'll earn before checking out. We'll award them after completing your purchase. | 677.169 | 1 |
Microsoft Mathematics 4.0 Download FREE!
Microsoft Mathematics provides a set of mathematical tools that help students get school work done quickly and easily. With Microsoft Mathematics, students can learn to solve equations step-by-step while gaining a better understanding of fundamental concepts in pre-algebra, algebra, trigonometry, physics, chemistry, and calculus.
Microsoft Mathematics 4.0
From basic math to pre-calculus, Microsoft Mathematics includes features which makes it easy to create graphs in 2D and 3D, calculate numerical results, solve equations or inequalities, and simplify algebraic expressions.
Microsoft Mathematics also includes a full-featured graphing calculator that's designed to work just like a handheld calculator. Additional math tools help you evaluate triangles, convert from one system of units to another, and solve systems of equations.
The new Version 4 is now available for free!
Known Issues:
Original show and plot functions have changed to show2d and plot2d In Microsoft Math 3.0, the show and plot functions were used to plot 2D graphs. In this version of the program, you must specify show2d and plot2d instead. If you open a worksheet file from a previous version of the product, any expressions that use show or plot will need to be updated.
Unable to type or edit expression in worksheet input pane after pressing the left arrow key or the backspace key repeatedly You can use the backspace key or the left arrow key to move left so you can edit an expression in the worksheet input pane. However, if you press these keys repeatedly, the input pane may stop accepting input in some cases. To fix this, press the right arrow key once, and then you can type again.
Changing the color scheme causes problems with high contrast displays If your computer display is set to High Contrast mode, some areas of the workspace may become unreadable when you change color schemes in Microsoft Mathematics. To fix this problem, turn High Contrast off and then turn it back on again.
Screen reader does not read inline error messages In many cases, Microsoft Mathematics displays inline error messages to explain syntax errors or to describe a problem. To give screen readers access to this information, you can change your options to show these messages as popup dialogs instead. On the File tab, click Options, and then click Display inline error messages as popup messages.
The appearance of 3D graphs is not optimal. Installation of Microsoft DirectX is recommended because it takes advantage of hardware acceleration options, such as video cards. By default, Microsoft Mathematics opens the DirectX Setup wizard when the Mathematics Setup Wizard is complete. However, if your computer does not have DirectX installed or cannot use the technology, Microsoft Mathematics uses an alternative rendering format for 3D graphs | 677.169 | 1 |
Math: 8th Grade Expressions and Equations 8.EE.A.2 Five Worksheets
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There are five worksheets for the Common Core standard 8.EE.2.A, starting off with a mixed level worksheet, then moving through levels 1 - 4, with 4 being the hardest. There are 10 questions per worksheet, for a total of 50 questions. Answer key included. | 677.169 | 1 |
2.0 Students solve systems of linear equations and inequalities (in two or three variables) by substitution, with graphs, or with matrices.
Algebra ll - Chapter 9
3.0 Students are adept at operations on polynomials, including long division.
Algebra ll - 2.2, 2.4, 2.5, 2.7
4.0 Students factor polynomials representing the difference of squares, perfect square trinomials, and the sum and difference of two cubes.
Algebra ll - Chapter 3
5.0 Students demonstrate knowledge of how real and complex numbers are related both arithmetically and graphically. In particular, they can plot complex numbers as points in the plane.
Algebra ll - 6.2, 6.7
6.0 Students add, subtract, multiply, and divide complex numbers.
Algebra ll - 6.2
7.0 Students add, subtract, multiply, divide, reduce, and evaluate rational expressions with monomial and polynomial denominators and simplify complicated rational expressions, including those with negative exponents in the denominator.
Algebra ll - Chapter 4
8.0 Students solve and graph quadratic equations by factoring, completing the square, or using the quadratic formula. Students apply these techniques in solving word problems. They also solve quadratic equations in the complex number system.
Algebra ll - 6.4 - 6.7, 8.5, 3.6
9.0 Students demonstrate and explain the effect that changing a coefficient has on the graph of quadratic functions; that is, students can determine how the graph of a parabola changes as a, b, and c vary in the equation y = a(x-b)2 + c.
Algebra ll - 8.5
10.0 Students graph quadratic functions and determine the maxima, minima, and zeros of the function.
Algebra ll - 8.5
11.0 Students prove simple laws of logarithms.
Algebra ll - 11.5
11.1Students understand the inverse relationship between exponents and logarithms and use this relationship to solve problems involving logarithms and exponents.
Algebra ll - 11.1, 11.6
11.2 Students judge the validity of an argument according to whether the properties of real numbers, exponents, and logarithms have been applied correctly at each step.
Algebra ll - 11.6
12.0 Students know the laws of fractional exponents, understand exponential functions, and use these functions in problems involving exponential growth and decay.
Algebra ll - 2.3, 5.4, Chapter 11
13.0 Students use the definition of logarithms to translate between logarithms in any base.
Algebra ll - 11.5
14.0 Students understand and use the properties of logarithms to simplify logarithmic numeric expressions and to identify their approximate values.
16.0 Students demonstrate and explain how the geometry of the graph of a conic section (e.g., asymptotes, foci, eccentricity) depends on the coefficients of the quadratic equation representing it.
Algebra ll - 8.5(quadratic)
17.0 Given a quadratic equation of the form ax2 + by2 + cx + dy + e = 0, students can use the method for completing the square to put the equation into standard form and can recognize whether the graph of the equation is a circle, ellipse, parabola, or hyperbola. Students can then graph the equation.
Chapter 8
18.0 Students use fundamental counting principles to compute combinations and permutations.
Algebra ll - 12.1, 12.2
19.0 Students use combinations and permutations to compute probabilities.
Algebra ll - 12.4, 12.5, 12.6
20.0 Students know the binomial theorem and use it to expand binomial expressions that are raised to positive integer powers.
Algebra ll - 12.2
21.0 Students apply the method of mathematical induction to prove general statements about the positive integers.
Algebra ll - 12.2
22.0 Students find the general term and the sums of arithmetic series and of both finite and infinite geometric series.
Algebra ll - 12.2
23.0 Students derive the summation formulas for arithmetic series and for both finite and infinite geometric series.
Algebra ll - 12.2
24.0 Students solve problems involving functional concepts, such as composition, defining the inverse function and performing arithmetic operations on functions.
Algebra ll - Chapter 10
25.0 Students use properties from number systems to justify steps in combining and simplifying functions. | 677.169 | 1 |
Synopsis
Mathematics is, by its very nature, an abstract discipline. However, many students learn best by thinking in terms of tangible constructs. Enhancing Mathematics Understanding through Visualization: The Role of Dynamical Software brings these conflicting viewpoints together by offering visual representations as a method of mathematics instruction. The book explores the role of technology in providing access to multiple representations of concepts, using software applications to create a rich environment in which a student's understanding of mathematical concepts can flourish. Both students and instructors of mathematics at the university level will use this book to implement various novel techniques for the delivery of mathematical concepts in their classrooms. This book is part of the Research Essential collection.
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$165.00 USD
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You save $16.51 (10%) and
You'll see how many points you'll earn before checking out. We'll award them after completing your purchase. | 677.169 | 1 |
Barnett
ISBN: 9780070045767
This new edition aims to bridge the students first experience manipulating algebraic expressions to the more abstract skills they will develop in later courses. Students begin with a review of topics from elementary algebra. Inequalities take on a greater role and complex numbers are introduced. After the review and extension of the quadratic equation, the course takes on a more abstract and graphing direction, dealing with two variables, relations and functions and systems of equations and inequalities. Each concept is illustrated with an example, followed by a parallel problem wth an answer. The book contains many realistic applications from the physical sciences, business and economics and social sciences, and chapter review sections include a summary of the important terms and symbols. | 677.169 | 1 |
This ebook is available for the following devices:
iPad
Windows
Mac
Sony Reader
Cool-er Reader
Nook
Kobo Reader
iRiver Story
more
Design Theory, Second Edition presents some of the most important techniques used for constructing combinatorial designs. It augments the descriptions of the constructions with many figures to help students understand and enjoy this branch of mathematics.
This edition now offers a thorough development of the embedding of Latin squares and combinatorial designs. It also presents some pure mathematical ideas, including connections between universal algebra and graph designs.
The authors focus on several basic designs, including Steiner triple systems, Latin squares, and finite projective and affine planes. They produce these designs using flexible constructions and then add interesting properties that may be required, such as resolvability, embeddings, and orthogonality. The authors also construct more complicated structures, such as Steiner quadruple systems.
By providing both classical and state-of-the-art construction techniques, this book enables students to produce many other types of designs. | 677.169 | 1 |
Mathematics Extension 1
Online Mathematics Extension 1 Tutoring
HSC Mathematics Extension 1 is an interactive and engaging course. The course aims to help you on your way to HSC and your future by teaching the fundamental knowledge of the methods and theorems used in the NSW curriculum.
Our highly qualified e-tut tutors cover an array of theories, solutions and examination techniques that will help you to achieve your desired mathematical scores.
We offer both Preliminary and HSC NSW Subjects
Our online interactive and enjoyable virtual classroom will help you to really grasp Mathematics Extension 1.
The preliminary course includes the following subjects:
Other inequalities
Circle geometry
Further trigonometry (sums and differences, t formulae, identities and equations)
Angles between two lines
Internal and external division of lines into given ratios
Parametric representation
Permutations and combinations
Polynomials
Harder applications of the Preliminary 2 Unit course
Our Year 12 HSC Course includes these fascinating subjects:
Methods of integration
Primitive of sin 2x and cos 2x
Equation dN/dt = k(N – P)
Velocity and acceleration as a function of x
Projectile motion
Simple harmonic motion
Inverse functions and inverse trigonometric functions
Induction
Binomial theorem
Further probability
Iterative methods for numerical estimation of the roots of a polynomial equation
Harder applications of HSC 2 Unit topics
Both preliminary and HSC NSW courses aim to help you appreciate the role of mathematics in solving practical problems as well as appreciating the interrelationships between ideas that are drawn from various areas of the subject.
We help you to develop the skills to apply mathematical techniques to solve practical problems using a variety of strategies that will investigate the mathematical models of numerous situations such as projectiles, exponential growth and decay and binomial probability.
We also help you to develop the ability to use mathematical language comprehensively as well as diagrams that will help you to communicate in an array of situations. Our tutors will guide learners to evaluate solutions to problems and communicate then appropriately.
What Career Can You Follow With Mathematics Extension 1?
Many lucrative professions favour employees who have truly mastered NSW mathematics as well as complimentary subjects such as Chemistry and Physics. Such careers include:
Mathematics professor
Economist
Environmental mathematician
Robotics engineer
Geophysical mathematician
Designer using computer graphics and mathematical modelling
Ecologist
Geometrics engineer
Why You Should Choose Online Tutoring For Mathematics Extension 1
We specialise in Year 11 and 12 subject selection for a number of subjects. e-tut online tutoring offers a safe and dynamic environment for you to learn in with the help of individual attention from well-trained tutors who are master of their subjects.
Our online Mathematics Extension 1 classes comprise one-on-one sessions or small group sessions of up to 5 learners. Our classes last for about 60 minutes and you are afforded the flexibility of studying for as long or little as you like each week.
Our Online Classroom
Our top-notch and technologically advanced virtual classroom makes for interactive and engaging learning sessions. From the comfort of your PC you are able to communicate via two-way audio functionality with both tutors and peers. You can also engage in texting, video sharing, easy file uploading and even remote computer functions.
We keep track of tests that you have taken, your attendance and the amount of time you spend in lessons too so that we can confidently help you to achieve success with Mathematics Extension1.
We also provide an array of examples, theories, tests and examination preparation techniques to help you on your way to your dream job and desirable HTC ATAR.
Online tutoring with e-tut greatly eliminates the costs that are typically involved with having to hire a private tutor or purchasing study material or even attending a learning establishment. It also saves on travel time – which can be spent in front of the PC mastering your subjects! | 677.169 | 1 |
In this course we will review Algebra 1 Content and cover the following units: Linear Equations and Inequalities, Systems of Equations and Inequalities, Quadratics, Sequences and Series, Polynomial Functions, and ACT Prep. This course will prepare the student for Algebra II. | 677.169 | 1 |
June 26 – June 30 Online work and discussion
Long-term Assignment – Interview a student to assess their understanding of the Order of Operations. This is a 'pre-test' more or less, of your abilities as well as a fact finding assessment mission. Engage in the formative assessment of a student's understanding of the order of operations. Student should be in 4th-7th grade. This is an open-ended assignment. You do whatever you think is best to gather the information you need regarding this student's thinking and the Order of Operations. You can do this at any time before the start of class and up to the due date of July 10th. Hold on to your notes from the assessment. You will need them for Q&DWrite-up 3–Formative Assessment of Order of Operations Due July 10. – Write up a one page report on your assessment of student's understanding of the Order of Operations
Note: You will need to meet with this student again at some point between July 21 and July 28th. You must have watched and completed the July 22nd on line class before you meet with student for second time. Further instruction posted for that part of assignment after July 10th due date for this first part of assignment.
Topic: What is Algebra in the Elementary classroom?
Welcome to ELM 558! I look forward to working with you all Face to Face.
elm-558-syllabus-summer 14. Take a look and feel free to shoot me an email if you have any questions or concerns about anything, including scheduling issues.
For now, let's get started thinking about Algebra and what it means to the elementary classroom teacher.
Intro to Faulkner. Watch this video (18 minutes) of me presenting ideas about the brain and mathematics at a TEDxNCSU lecture. If you have watched it once before, go ahead and watch it again. If you have it memorized, you can skip it. Watching this video will give you an idea about my ideas about mathematics and a sense for my study habits in high school : ). I hope you will find it entertaining.
Before you begin your reading address the following questions on your own. Play around with each and consider as many possible approaches as you can. At least two. Both pictures from Suh, 2007.
Discussion Board Post due by June 30th.
#1
Consider the pattern above. It is a staircase pattern and we see it here at step 1, step 2, step 3 and step 5. How would you describe this pattern to someone else? You can use words and/or symbols. Find at least 2 ways to describe the pattern.
How would you tell someone to figure out how many total blocks there are, for the whole staircase, if you have a staircase that has its highest height at 100 steps? If you had to add up all those total blocks, what would be the quickest way to do so, that you can think of?
REMEMBER – play with this on your own for some time before you post or take a look at other posts.
Post your thinking on
Here is a square number pattern. There are many wonderful patterns built into square numbers. How many can you find? In order to discover more patterns, consider using colored tiles or colored pencil and paper to play with the growth from square to square. Describe the patterns you see, at least 2, using words and/or symbols. One way to think about what you are doing is to figure out a way to tell someone how to get from any perfect square to the next perfect square. If I said, 'add 11 to 25' that would tell me how to get from one particular perfect square (25 or 5 squared) to the next perfect square (36 or 6 squared), but would not help me get from ANY perfect square to the next perfect square. How could I tell someone to get from ANY perfect square to the next? Or, perhaps you can describe a pattern that you can find, a way to decompose any given square number other than in the basic form of a square, NxN.
REMEMBER – play with this on your own for some time before you post or take a look at other posts.
Post your thinking on
Once you are done posting your ideas, read Suh, 2007. You can post more after you have read the article if you have more ideas you want to share! Also, go back and read at least a few other people's postings
* Be prepared for class – be ready to describe and explain the solution(s) that someone other than you generated.
That's it for this on-line session. You might want to get ahead by going to next week's on-line session. There are 5 chapters of reading there, along with a discussion paper and discussion board posting due. | 677.169 | 1 |
Anatomy of Quadratic Function Mini-Bundle
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Students become familiar with using the terminology of a quadratic function along with practicing how to use the standard form of a quadratic equation to find the axis of symmetry and vertex. A 3-day lesson plan is included for how I use the materials. I break the notes down into two days worth of instruction using the student and teacher notes. Two different skill sheets are included to assess student progress. A differentiated task card activity is also a part of this bundle. Additional notes are provided for how I differentiate the activity into three different levels to reach the learning needs of all my students.
Check out the preview for a more detailed look at the lesson plans and materials provided! | 677.169 | 1 |
What is the best calculator for me?
For a school student it's always better to use basic calculators as during exams or class many teachers are not comfortable with students using scientific calculators. Basic calculators are use for simple calculations. These calculators are also used in daily household necessities.
Scientific calculator is a must have gadget for mathematics, science and engineering students. These calculators are required to do complex calculations which normal calculators can not perform. These calculators can solve harder problems in an easy way. However, the functions and features of Scientific calculators vary from device to device and brand to brand.
Choosing the right device for one's needs is important. Best Scientific calculators list given below:
Financial calculators are used by students of Finance, accounting & management. These calculators are also necessary for the people exploring financial sectors. Professionals & businessman also uses Financial calculators for simplify calculations.
Professionals, businesses and other financial employees uses printing calculator. These calculators do have normal business and finance functions, also they can print the calculation in a roll of printing paper. | 677.169 | 1 |
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THIS book falls naturally into two parts. In Chapters 1-5 the basic ideas and techniques of partial differentiation, and of line, multiple and surface integrals are discussed. Chapters 6 and 7 give the elements of vector field theory, taking the integral definitions of the divergence and curl of a vector field as their starting points; the last chapter surveys very briefly some of the immediate applications of vector field theory to five branches of applied mathematics. Throughout I have given numerous worked examples. In these I have paid particular attention to those points which in my own experience I have found to give most difficulty to students. In the text I have denoted spherical polar coordinates by (/-, 0, ψ)9 and cylindrical polar coordinates by (p, ψ, ζ), so that ψ measures the same angle in both systems. Since there is no one standard notation for these systems, the reader will meet different notations in the course of his reading, and in quoting examination questions in the exercises I have kept to the notation of the originals. The Exercises at the end of each section are intended to give practice in the basic techniques just discussed. The Miscellaneous Exercises are more varied, and contain many examination questions. | 677.169 | 1 |
New Scientific Applications of Geometry and Topology(Hardback)
Synopsis
Geometry and topology are subjects generally considered to be 'pure' mathematics. Recently, however, some of the methods and results in these two areas have found new utility in both wet-lab science (biology and chemistry) and theoretical physics. Conversely, science is influencing mathematics, from posing questions that call for the construction of mathematical models to exporting theoretical methods of attack on long-standing problems of mathematical interest. Based on an AMS Short Course held in January 1992, this book contains six introductory articles on these intriguing new connections. There are articles by a chemist and a biologist about mathematics, and four articles by mathematicians writing about science.All are expository and require no specific knowledge of the science and mathematics involved. Because this book communicates the excitement and utility of mathematics research at an elementary level, it is an excellent textbook in an advanced undergraduate mathematics course | 677.169 | 1 |
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Further Mathematics : A Level at Gateway College
Course description
This course is delivered using a blend of learning methods including use of online resources to supplement classroom taught lessons. Students will need strong independent learning skills. The units FP1 and FP2 include work on Matrices and Complex Numbers,
Further Trigonometry and Calculus. There is some flexibility over the choice of the other units.
Course content
This course is aimed at students who enjoy Mathematics and feel confident with the subject. It is especially recommended for students intending to study Maths (on its own or as part of a combined degree) at University. This may be a good choice of a fourth
AS level, if you have already chosen Maths and two others.
Entry requirements
6 GCSEs at grade 9-5 including English Language. A minimum of 4 GCSEs at grade 6.
Plus GCSE Maths grade 8. Students must also be studying or have completed AS Level Maths.
Gateway Sixth Form College was rated as being a 'Good' college by Ofsted when they visited in January 2014. They said, the College "provides a harmonious, welcoming atmosphere for a highly diverse student population and offers a distinctive mix of advanced-level subjects and vocational courses at foundation, intermediate and advanced level."
We celebrate our diversity throughout the year by organising themed events, fundraising activities and inviting visiting speakers to work with our students in helping to raise knowledge and awareness of our many different cultures.
We hope that you wil join our growing Gateway Community as you take the next important step in your journey to success. | 677.169 | 1 |
Comment: Has been read, but is in excellent condition. Pages are intact and not marred by any notes or highlighting. The spine remains undamaged. All orders are dispatched within 1 working day from our UK warehouse. Established in 2004. No quibble refund if not completely satisfied.
This text prepares students for future courses that use analytic ideas, such as real and complex analysis, partial and ordinary differential equations, numerical analysis, fluid mechanics, and differential geometry. This book is designed to challenge advanced students while encouraging and helping weaker students. Offering readability, practicality and flexibility, Wade presents fundamental theorems and ideas from a practical viewpoint, showing students the motivation behind the mathematics and enabling them to construct their own proofsWilliam Wade received his PhD in harmonic analysis from the University of California―Riverside. He has been a professor of the Department of Mathematics at the University of Tennessee for more than forty years. During that time, he has received multiple awards including two Fulbright Scholarships, the Chancellor's Award for Research and Creative Achievements, the Dean's Award for Extraordinary Service, and the National Alumni Association Outstanding Teaching Award.
Wade's research interests include problems of uniqueness, growth and dyadic harmonic analysis, on which he has published numerous papers, two books and given multiple presentations on three continents. His current publication, An Introduction to Analysis,is now in its fourth edition.
In his spare time, Wade loves to travel and take photographs to document his trips. He is also musically inclined, and enjoys playing classical music, mainly baroque on the trumpet, recorder, and piano.
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I admit, it's quite a while since i used this book. Also, I don't remember mistakes, typos, or non-sense. However, I thought it to be rather heavy on theorems and lacking clear examples a bit. I didn't find any better substitute at the time, so maybe that just the nature of the beast. I feel I lacked some more graphic, explanatory sections, describing the nature and implications of the theorems however. I don't know how others feel about it, but this book wasn't really for me.
It bridges the gap between Calculus and Analysis. There are probably better books for the topic, but this one gets the job done. Exercises are fairly thorough and standard. The sections are slow and proofs are a bit wordy, but this is a good thing for new comers anyway. With that said, it can get a little boring reading this book. Boring or not, if you tried to read Rudin and found it hard to understand, it is probably better to start by thoroughly going over something like this (let each proof simmer in your mind) before pursuing elegance.
Most of the reviews praising this book are done by doctoral students, graduate students or mathematicians themselves. The problem is that i am an undergraduate student, using this book and i don't have the expertise of a doctoral student praising this book. This book is horrible. In some cases, the positive reviews for this book have a date that is prior to when this book was published --- which really shows how incompetent amazon is approving the reviews -- previous edition reviews are merged together with current edition review -- which in no way is the same thing. Positive reviews dated 1999, 2000, 2001 etc should all be deleted because this is a 2003 edition book. Therefore the average customer review ratings of this book are fictitious and heavily overrated. This book is basically way too advanced and beyond the scope of comprehension for an undergraduate student. In my class, 11 out of 20 students got F on the first test, 3 got D's and only 5 got C or higher. That is what a typical undergraduate class using this book scores on tests. The exercises don't seem to relate to the chapter. It seems the proof exercises are brought from another universe even after you've studied the chapter thoroughly. The book is completely vague and the author is presuming way too much as far as the student's background is concerned. I think this kind of book belongs more in graduate and doctoral school. I have to use other books to learn the material. Its pure torture and is killing every student in my class. | 677.169 | 1 |
Free eBooks Download, ePub book, college books
Mathematics for Electricity & Electronics
With its fresh reader-friendly design, MATHEMATICS FOR ELECTRICITY AND ELECTRONICS, 4E equips learners with a thorough understanding of essential algebra and trigonometry for electricity and electronics technology. Well-illustrated information sharpens the reader's ability to think quantitatively, predict results, and troubleshoot effectively, while problem sets for drill and practice… | 677.169 | 1 |
Date and Time
About Lenin Kumar Gandhi
I have 5 years of experience in teaching IB curriculum and with overall experience of 15 years. Also, I am an IB Examiner. I use innovative methods of teaching Math for easy and effective understanding along with exciting Math games and activities brought into my pedagogy.
About the Course
The session is about 'Asymptotes' which is explained through my recorded video clip which includes explaining through the basic definition and thus through varied examples connecting the content and hence would be very effective in the learning process. A headphone with a good internet connectivity is recommended for the online session.
Topics Covered
Students learn about 'Vertical Asymptote'.
Who should attend
All enthusiastic learners are part of the joining.
Pre-requisites
Quadratic factorisation.
What you need to bring
Laptop or a System, Headphones, Internet connection (min. 1 MBPS)
Key Takeaways
Students learn about Asymptotic behavior and more importantly on how to solve the problems involving 'Vertical Asymptote'. | 677.169 | 1 |
FUNCTIONS in Multiple Representations
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5 VIEWS of a FUNCTION
Tables-Equations-Graphs-Illustration-Scenarios
This resource provides 25 UNIQUE scenarios that students will interpret and represent in different ways. It has been a great activity for my students to see the connections between the different representations of a function.
(Illustration may not be considered a mathematical representation, however, being able to visualize the situation is a skill lacking in many students that encourages mathematical thought)
Students are given any one of the five representations (views) and are asked to create the other four.
Included
teacher notes
a template for presenting the five views of a function
25 SCENARIOS
5 tables
5 equations
5 graphs
5 Illustrations
5 scenarios
solutions for all 25 situations (student answers for scenarios will be unique)
This activity can be used to introduce each functional representation or it can be used after the content is taught to make the connections between each. The functions are all linear, first quadrant functions will real applications.
"5 Views of a Function" engages students as they remember, understand, apply, analyze, evaluate and create.
Although this is NOT a matching game. However, the solution pages could be cut apart and adapted to create a matching game. I love it when one resource can be utilized in different ways! Please leave feedback to let me know how you use this activity. | 677.169 | 1 |
Critical Juncture Ahead! Proceed with Caution to Introduce the Concept of Function
Gregorio Ponce
Algebra teachers find themselves overwhelmed by the sheer
volume of information that they have to teach their students. One way to
address this issue is to have teachers teach for understanding by focusing on
concepts. The author shares how language can help uncover the core ideas that
define the concept of function and provides activities that have been
successfully used with students to introduce the core ideas and illustrate the
usefulness of a function outside the classroom. Teaching functions is the
vehicle for this process.
This is available to members of NCTM. If you are interested in a NCTM membership, join now. You may also purchase this article now for online access. | 677.169 | 1 |
With faster pacing and less review, this is the perfect text for those students in precalculus who are going on to calculus-clarity without compromise. A hallmark of this text is its focus on problem solving. Students are provided with numerous opportunities to reason and think their way through various problems. The author's goal is to get the students to see that the mathematics in this book is interesting, useful, and worth studying. Emphasis is placed on using graphic, numeric, and analytic points of view in discussing and in solving problems throughout the text.
"synopsis" may belong to another edition of this title.
About the Author:
Mark Dugopolski was born in Menominee, Michigan. After receiving a B.S. from Michigan State University, he taught high school for four years, and then went on to receive an M.S. in mathematics from Northern Illinois University. He also received a Ph.D. in the area of topology from the University of Illinois at Champaign-Urbana. Mark has been teaching at Southeastern Louisiana in Hammond, LA, ever since.
Mark has been writing textbooks for about fifteen years. He is married and has two daughters, and enjoys playing tennis, jogging, and riding his bicycle in his spare time. Despite the presence of sailboats both on and in his precalculus textbooks, Mark does not sail-- it's his passion for showing students how mathematics is used in the real world that led him to the world of sailing. | 677.169 | 1 |
Calc 3D is a collection of mathematical tools for highschool and university. The calculator can do statistics, best fits, function plotting, integration. It handles vectors, matrices, complex numbers, coordinates, regular polygons and intersections. For objects ( like point, line, plane and sphere) distances and intersections are calculated. Cartesian, spherical and cylindrical coordinates can be transformed into each other. Carthesian plot | 677.169 | 1 |
Grade 11 Platinum Mathematics 2013 Teachers Guide
nanomaterials free full text electrostatic assembly of
Grade 11 Platinum Mathematics 2013 Teachers Guide involve some pictures that related one another. Find out the most recent pictures of Grade 11 Platinum Mathematics 2013 Teachers Guide here, so you can obtain the picture here simply. Grade 11 Platinum Mathematics 2013 Teachers Guide picture uploaded ang published by Admin that saved inside our collection. Grade 11 Platinum Mathematics 2013 Teachers Guide have a graphic associated with the other. Grade 11 Platinum Mathematics 2013 Teachers Guide It also will include a picture of a sort that might be observed in the gallery of Grade 11 Platinum Mathematics 2013 Teachers Guide. The collection that consisting of chosen picture and the best among others. They are so many great picture list that may become your ideas and informational purpose ofGrade 11 Platinum Mathematics 2013 Teachers GuideGrade 11 Platinum Mathematics 2013 Teachers Guide picture. We provide image Grade 11 Platinum Mathematics 2013 Teachers Guide Grade 11 Platinum Mathematics 2013 Teachers Grade 11 Platinum Mathematics 2013 Teachers Guide. We hope you enjoy and find one of the best collection of pictures and get inspired to decorate your residence. If the hyperlink is busted or the image not entirely onGrade 11 Platinum Mathematics 2013 Teachers Guideyou can contact us to get pictures that look for We offer imageGrade 11 Platinum Mathematics 2013 Teachers Guide | 677.169 | 1 |
New to math, I need help!It would be helpful to have an idea of what level of Calculus and Algebra you need. I can however refer you to a few resources that I find helpful to this day (2nd year Engineering Student).
The type of math you can expect:
1) Calculus 1: Differentiation, Basic introduction to Integration, Types of functions.
2) Calculus 2: Integration Techniques, Parametrisation of curves ,etc.
3) Calculus 3: Line Integrals, Integrals in 3D, etc.
4) Linear Algebra: Solving systems of equations using matrices, Complex Numbers, etc.
The content might not exactly be the same but I'm guessing that a lot of the stuff is covered in a similar way.
Videos:
PatrickJMT - You've probably already heard of him, if not just google and you can access his videos from either youtube or his website. The most helpful math videos in my opinion. Also he gets the ideas across nicely and quickly.
KhanAcademy - I need not even be mentioning this, by far the most popular learning resource on the internet as of this moment.
MathTutorDVD - You have to buy the DVDs but I found the method of teaching here to be very simple and helpful.
Textbooks:
Stewart Caculus: Early Transcendentals (Currently using this textbook for Calc 3 and used it for Calc 1 and 2, really good textbook).I just looked up community colleges in my area, but none seem to have mathematically focused programs.
Look at the specific mathematics courses which the local community college offers. You do not need a highly focused program at the start, but you could move to one when your academic mathematical knowledge puts you ready.
The website shows a Physics degree program and lists Cacluluses one through three, and a linear algebra. Being a community college, the Math department MUST have the less advanced courses too. Check directly with the college. Nothing seemed shown in the website.
$10000 per year even for a full-time course load seems too high for a c.c. Is that what the costs are like in that geographic area?
According to the table, a resident of Allegheny Co. can take up to 19 credits at a flat rate of about $1600 a semester. Out of state students would pay about $4800.
Twelve credits or more is equivalent to being a full-time student, according to the school.
Staff: Mentor
The website shows a Physics degree program and lists Cacluluses one through three, and a linear algebra. Being a community college, the Math department MUST have the less advanced courses too. | 677.169 | 1 |
The purpose of this mathematical paper is to discuss the applications of ordinary mathematical problems concerning rates of change that result in becoming variables in a problem.
Extracts from this document...
Introduction
Introduction The purpose of this mathematical paper is to discuss the applications of ordinary mathematical problems concerning rates of change that result in becoming variables in a problem. This paper will also analyze the practical uses of calculating rates, as well as focus on rates in terms of differential calculus. Background Information To be quite honest, the field of mathematics had never been my forte; I have never seemed to be able to grasp concepts that come so easily to others. However, the area of science fascinates me. Last year, in Biology class, we were appointed to complete an assignment in which we recorded the rates of enzyme reactions when different variables acted upon the substrate. I found the entire experiment intriguing, and didn't realize, until recently, that it was in any way related to Calculus. A rate of change of an object is the speed at which a variable can change over a specific period of time. ...read more.
Middle
Mathematical Applications In order to find the rate of change of an object, one would, essentially, be required to find the slope of a graph. Differential Calculus can be applied in this situation, in that by finding the derivative. For example, consider a man who is 6 feet tall and walking toward a lamppost 20 feet high at a rate of 5 feet per second. The light at the top of the lamppost (20 feet above the ground) is casting a shadow of the man. At what rate is the tip of his shadow moving and at what rate is the length of his shadow changing when he is 10 feet from the base of the lamppost? If z were to equal the distance from the tip of the shadow to the base of the lamppost, y equaled the length of the shadow, and x equaled the distance from the man to the base of the lamppost, we would also know that dx dt = -5 ft/sec From that, we could ...read more.
Conclusion
time was measured. Nine beakers with different concentrations of enzymes were utilized. Small paper discs were soaked in each of the different enzyme solutions of different concentrations, after which they were placed on the bottom of a hydrogen peroxide solution (the substrate) and timed in order to see how long it would take for the enzyme to react with the substrate and cause the paper disc to float to the top of the hydrogen peroxide solution. Concentration of Substrate Time (sec.) 0 0 0.10 10.63 0.20 5.44 0.30 4.9 0.50 4.3 0.80 2.19 1 % 1.69 2 % 1.47 3 % .97 Conclusion Differential calculus has many practical uses outside of the classroom in the 'real' world, and is a valuable tool in determining the rates of changes that are occurring, whether they be chemical reactions, such as in the case of the enzyme and substrate concentration experiment, or in other cases, such as rates of changes in the acceleration or deceleration of a car, or in the rate a rock falls off of a cliff. 2 ...read more.
Arefin Khan Aim: Investigate the effect of temperature on the activity of catalase (from potato). Introduction: Enzymes are biological catalysts. They speed up metabolic reactions in the body but remain chemically unchanged themselves. Enzymes contain an active site. This is a region, normally a depression or cleft, to which another molecule may bind. | 677.169 | 1 |
Chapter 15 - Real Number Systems
"This chapter begins by reviewing the real number system and then move to introducing the imaginary and ultimately the complex number system. Using the powers of exponent rules, students discover the necessity of the number i. This discovery leads to students exploring whether quadratic functions have one, two, or no real roots" | 677.169 | 1 |
MATH 103R - Technical Mathematics I with Review - 5.00 credits
Prerequisite: MATH 20 or MATH 20L with a C grade or higher OR appropriate score on placement test. A review of basic math operations including decimals, fractions, percents, and order of operations. Algebraic expressions, linear equations and systems of linear equations, functions, exponents, graphical analysis, quadratic equations, factoring common factors and difference of squares, unit conversions, percents, and tolerances, clearance, interference, mean, median and mode presented through problem solving activities. The focus of the course is on application of the math in real life situations and industrial disciplines through the use of critical thinking skills.
MATH 105 - Algebra and Trigonometry for Land Surveyors - 4.00 credits
Prerequisite: MATH 40 or MATH 40L or appropriate placement test score. Review of order of operations, scientific notation, rounding and significant digits. Review of basic area and volume formulas with applications to more general shapes. Quadratic and linear functions including piecewise definitions. Distance formula, midpoint formula, equations of circles. Map reading, contours and elevation. Classification of angles and triangles. Right triangle trigonometry. Conversions between radians and degrees/minutes/seconds and decimal degrees. Law of sines, law of cosines, arc length, vectors, and bearing. For all topics there will be an emphasis on applications appropriate to the study of land surveying.
MATH 110 - Intermediate Algebra - 3.00 credits
Prerequisite: MATH 31 with a grade of C or better, or satisfactory score on the math placement test.. The study of simplification of complex fractions, solutions to rational equations, solution of linear equations and inequalities with applications, solutions of absolute value equations and inequalities, quadratic functions and equations with applications, ratios and proportions, solutions of linear systems of equations with applications, rational exponents and radicals, introduction to functions and graphs, and graphing linear equations in 2 variables.
MATH 110R - Intermediate Algebra with Review - 5.00 credits
Prerequisite: MATH 40 or 40L with a grade of C or better, or satisfactory score on the math placement test. A combination of the topics in MATH 40 and MATH 110. The study of operations with polynomials, operations with rational expressions, properties of exponents, solution of linear equations and inequalities with applications, solution of absolute value equations and inequalities, solution of quadratic equations with applications, solution of linear systems of equations with applications, rational exponents and radicals, introduction to functions and graphs, and graphing linear equations in 2 variables.
MATH 119 - Mathematical Reasoning and Modeling - 3.00 credits
Prerequisite: MATH 85 with a grade of C or better or appropriate placement test score. The purpose of this course is to provide a comprehensive overview of the skills required to navigate the mathematical demands of modern life and prepare students for a deeper understanding of information presented in mathematical terms. Emphasis is placed on drawing conclusions, making decisions, and communicating effectively in mathematical situations that depend upon multiple factors. To that end, students will develop critical thinking and problem solving skills to solve ill-defined problems with multiple solutions.
MATH 120 - College Algebra - 3.00 credits
Prerequisite: MATH 95 with a grade of C or better or appropriate placement test score. A study of various types of equations and inequalities, functions and their inverses, theory of higher degree equations, systems of equations, determinants, logarithms and exponentials, and applications.
MATH 120R - College Algebra with Review - 5.00 credits
Prerequisite: MATH 110 or appropriate placement test score. A combination of topics in Intermediate Algebra and College Algebra. A study of various types of equations and inequalities, functions and their graphs, inverse functions, systems of equations, determinants, logarithms and exponential applications.
MATH 150 - PreCalculus - 5.00 credits
Prerequisite: MATH 95 with a grade of C or better or appropriate placement test score. A study of various types of algebraic equations and inequalities, functions and their inverses, theory of higher degree polynomial equations, systems of equations and inequalities, logarithms, exponentials, and applications. A study of trigonometric functions and their inverses, formulas and identities, conditional equations, radian measure, arc length, angular velocity, function graphing and solution of triangles.
MATH 175 - Calculus for Business and Social Science - 3.00 credits
Prerequisite: MATH 120 or higher or appropriate placement test score. Quadratic, polynomial, rational exponential, and logarithmic functions used in differential and integral calculus application in business, economic and social science.
MATH 180 - Analytic Geometry and Calculus I - 5.00 credits
Prerequisite: MATH 130 or 150. A study of plane analytic geometry, limits, continuity, the derivative for functions of a single variable, differentials, indefinite and definite integrals, the Fundamental Theorem of Calculus, and applications of the derivative and integral.
MATH 190 - Analytic Geometry and Calculus II - 5.00 credits
Prerequisite: MATH 180. A study of the calculus of elementary transcendental functions; integration by parts, by trigonometric substitution, by partial fraction and by miscellaneous substitutions; improper integrals; L' Hospital's Rule; conic sections; the transformation of axes, infinite series, parametric and polar equations and their derivatives; and graphs, area, and arc length in polar coordinates.
MATH 20L - Basic Mathematics/Lab - 3.00 credits
MATH 210 - Analytic Geometry and Calculus III - 5.00 credits
Prerequisite: MATH 190. A study of analytic geometry in three dimensions, functions of more than one variable and their calculus, directional and partial derivatives, vector functions and their calculus, two- and three-dimensional applications, multiple integrals, and line integrals.
MATH 230 - Differential Equations - 3.00 credits
Prerequisite: MATH 190. Solution and application of ordinary differential equations including the nth order non-homogeneous linear cases. Laplace transform, and power series methods.
MATH 31 - Pre-College Mathematics - 3.00 credits
Review of all basic mathematical operations. Fractions, decimals, proportions, percentages and real numbers. Elementary geometry (perimeter, area and volume). Review of all operations in real numbers. Solutions of linear equations in one variable, using and manipulating formulas. Properties of exponential numbers, definition and basic operations with polynomials. Graphing linear equations in two variables.
MATH 32 - Pre-College Mathematics II - 3.00 credits
Prerequisites: MATH 31. Continuation of Math 31 topics including a review of all basic mathematical operations. Fractions, decimals, proportions, percentages and real numbers. Elementary geometry (perimeter, area and volume). Review of all operations in real numbers. Solutions of linear equations and inequalities in one variable, using and manipulating formulas. Properties of exponential numbers, definition and basic operations with polynomials and solutions of polynomial equations by factoring. Basic operations and simplification of rational expressions. Graphing linear equations in two variables. Self-paced based on initial diagnostic assessment.
MATH 40 - Introductory Algebra - 3.00 credits
Prerequisite: MATH 20 or MATH 20L or appropriate placement test score. Review of all operations and properties of real numbers with special attention
MATH 40L - Introductory Co-Laboratory Algebra - 3.00 credits
Prerequisite: MATH 20 or MATH 20L or appropriate placement test score. Review of operations and properties of the Real Number System. Operations on polynomials, exponents, and rational expressions. Solving and graphing linear equations. Applications are emphasized throughout the course.
MATH 85 - Mathematical Literacy - 3.00 credits
Prerequisites: MATH 20, 31, or 40 with a satisfactory grade or satisfactory score on the math placement test. Math Literacy for College Students is a one-semester preparatory course for Statistics or Mathematical Reasoning and Modeling intended for students whose programs do not require Precalculus, College Algebra, or Calculus. The emphasis is on active learning, applications, and context. Topics include utilizing a spreadsheet tool, numeracy, notation, formula manipulation, data analysis, pattern recognition, mathematical reasoning, linear and exponential models, and basic statistics. The successful student will be well-prepared for MATH 115 and MATH 119, and can also continue into MTH 95 if a switch to a science, technology, engineering, or math related field is desired.
MATH 91 - Elements of Algebra - 3.00 credits
Review of all basic mathematical operations. Fractions, decimals, proportions 92 - Elements of Algebra - 3.00 credits
Prerequisite: Math 91 93 - Elements of Algebra - 3.00 credits
Prerequisite: MATH 92 95 - Algebra Principles - 5.00 credits
Prerequisite: MATH 31 with a grade of C or better, or satisfactory score on the math placement test. The study of Algebraic principles including: Operations with polynomials, rational expressions, properties of exponents; solutions of linear equations and inequalities and solutions of absolute value equations and inequalities with applications; solutions of quadratics by factoring, completing the square, and quadratic formula; ratios and proportions; solutions of linear systems of equations with applications; rational exponents and radicals; introduction to functions and graphs; graphing linear equations in two variables. | 677.169 | 1 |
You will use it for some classes after those classes, and then you will learn (especially for diff eq) that there are tables that you can reference for solutions to most problems. The point is that you have to learn to crawl before you can walk. They want you to know where the "easy way" came from so that you can think more abstractly in the future. Thinking is why engineers are on the 2nd floor and the technicians are on the 1st floor.
I think I agree with Ben. There's a computer program available to figure out almost anything these days, and most people I know only understand that data goes in, and an answer comes out. They either can't or don't want to understand the underlying theory or process that is involved in any problem. I think the fact that you're a member of this site and thus probably a DIYer means that you give a **** about how stuff works. College calculus blows and, no I've never had to use it in 20 years as an engineer. But I still more or less understand how it works, and I think the mental discipline required to work through **** like 2nd derivatives of transcendental equations has helped me get through everyday work ****.
Those are pretty much the replies I was expecting. I think I wouldn't mind the class so much if we were able to use reference tables like neogenesis2004 talked about. My prof, an Indian woman who writes faster than I can think, insists that we memorize everything in this class, and that meant 32 different new equations for the aformentioned test. Maybe I can retake the class over the summer at a local tech school.
Quote:
Originally Posted by Ben
Don't quit. Push through it. The path of most resistance can be rewarding.
Hey, don't surrender on passing yet - chances are your classmates aren't exactly dancing through either. I had a raw test average of 54(yes, out of 100) in Physical Chem, and ended up with a B on the curve!
I bought the book Calculus for Dummies a few weeks ago, and it actually helped with understanding the concepts behind the stuff we were doing, but it didn't really have examples for me to follow to actually figure out how to do the problems. The book required for the class just makes my head hurt.
My deal is I can do the problems in the book just fine most the time. If i can't the solutions manual helps me of course. We have online quizes and they destroy me though. There is nothing in the book like those questions and the exams are based off those quizes. If I didn't have a rifle teammate that was willing to help me I'd fail for sure.
I hate to sound like the stickler, but I really don't like the idea of so many professionals not using calculus. It's extraordinarily fundamental. All those computer programs mentioned use calculus, albeit numerical approximations, whether you like it or not. And those programs were all written by people who use calculus. On top of that, all of those pieces of software have limitations, and they will all choke if given the correct problem. The scariest thing, though, is when you DO get an answer, but it is wrong due to the user's lack of understanding (and I don't mean input error), but use the spat-out answer anyway. That to me is terrifying.
I hate to sound like the stickler, but I really don't like the idea of so many professionals not using calculus.
My circuits professor made a good joke about mathematicians vs engineers. If you ask a mathematician to calculate how long it will take for a capacitor to discharge he will tell you that it is never fully discharged (infinity). When you ask an engineer to calculate how long he will tell you that once its close to zero it is close enough. (RC)
I think it's pretty much a given that one way or another, we all use calculus. My point was that the grunt work of taking the courses lets you understand just what it is the computer is doing for you, so that the numbers or drawing or whatever has a real context. It sucks while you're going through it, but you understand later on when you see the practical applications.
Im a MET major, and I have to take tech calc I & II. Will I ever use it, probably not..., I'm just about to finish up Trig this semester, and move on to physics, statics, and maybe calc I. I would really give MET a second look, it is looked at very similarly as an ME, you get lots of hands on, and there isn't as much theory and math.
-Michael-
Nothing is harder than calc 2. 2nd order diff equations are hard but are only about 25% of that class. The rest is...well not easy but not calc 2 hard.
I do use infinite series but only as numerical analysis short cuts. I also use the basic position/velocity/accel derivative but that's not exactly taxing. I use integration in the real-time apps I write but it's nothing like what I did in that class. Still, it's nice to know the theory when doing it. Hell, I guess I do use it. | 677.169 | 1 |
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2 AUTHOR Bettye C. Hall is the former Director of Mathematics in the Houston Unified School District in Houston, Texas. Ms. Hall, admired as a teacher s teacher because of her practical perspective on students and teachers needs, is active as a mathematics consultant, speaker, and workshop leader throughout the United States. REVIEWERS JAMES GATES, ED.D. Executive Director Emeritus, National Council of Teachers of Mathematics YVONNE S. GENTZLER, PH.D. Associate Professor, College of Education, University of Idaho Copyright 1999 by Prentice-Hall, Inc., Upper Saddle River, New Jersey All rights reserved. Worksheets and tests may be duplicated for classroom use, the number not to exceed the number of students in each class. Notice of copyright must appear on all copies. No other part of this book may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or by any information storage or retrieval systems, without permission in writing from the publisher. Printed in the United States of America.
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6 iv Using Algebra Tiles Effectively
7 Using Algebra Tiles Effectively v
8 Algebra Tiles: Tools for Understanding Purposes of the Professional Development Workshop Today s mathematics teachers face tremendous challenges, including changes in curricula, teaching techniques, and assessment practices. To help teachers rise to these teaching challenges and opportunities, Prentice Hall presents this self-contained train the trainer workshop, developed by nationally-recognized and respected mathematics educators. This workshop is designed to aid educators in meeting the diverse needs and expectations of their students. The introduction of algebra tiles and other manipulatives into the classroom provides mathematics teachers with exciting opportunities to empower students of all learning styles. Through hands-on activities, this workshop will help teachers become familiar with the uses and applications of algebra tiles. New users will become comfortable using algebra tiles in their classrooms. Experienced users will learn new applications. All users will more fully appreciate the ways that these manipulative tools can help them reach a greater percentage of students. Components of This Workshop This workshop contains activities and visual aid masters that are suitable for use by staff developers, department chairpersons, curriculum specialists, or lead teachers in conducting a workshop on using algebra tiles effectively. This workshop also includes a statement regarding the purpose of the workshop, the intended audience, a list of necessary pre-workshop materials and audio-visual requirements, as well as suggestions for room preparation. Because of the wide variety of activities provided, middle school, high school, firstyear, and veteran teachers will all benefit from this program. The activities can be customized to accommodate different needs and levels of understanding. Trainers can select the activities that best meet their specific time constraints and professional requirements. You can introduce the session using any of a variety of ideas that are presented at the beginning. Next, there is a variety of activities designed to address the workshop goals and objectives. At the end of the workshop are summary ideas and suggestions for closing activities. The workshop contains masters for making transparencies or black line copies of the visual aids. vi Using Algebra Tiles Effectively
9 Facilitator s Roles and Responsibilities As facilitator, you should read through the workshop to become familiar with the objectives and activities. If any of the material is new to you, explore it thoroughly. Select the activities that best suit the needs of the educators attending your workshop. Each section and activity includes the suggested amount of time deemed necessary to accomplish the goal of the activity. Create a reasonable schedule for your workshop, being sure to allow time for exploration and questions. Suggestions for Getting Started Introduce yourself. Workshop participants may or may not know each other. If the latter is true, set aside a few minutes for members of the group to get acquainted with one another. Encourage them to talk about their own experiences and their students experiences using algebra tiles. This should ease participant anxiety. Customizing the Workshop As is always the case, knowledge is best understood and interpreted when it meets the needs of the learner. Knowing the needs of the group to be served, the facilitator will be in the best position to customize the workshop. This workshop, Using Algebra Tiles Effectively, contains a sufficient variety of activities so that you can use the workshop with middle grade teachers or high school teachers, educators new to using algebra tiles or educators experienced at teaching with manipulatives and algebra tiles or any combination of these groups. On page 2 of the workshop is a table containing the activities and suggested amount of time necessary to accomplish each activity. Use this table to determine which activities you will use in this workshop. Using Algebra Tiles Effectively vii
10 Using Algebra Tiles Effectively (x + 3)(x 2) = x 2 + x 6 When I listen, I hear. When I see, I remember. But when I do, then I understand. This workshop helps participants use algebra tiles effectively in their classrooms. Purpose Middle School and/or High School Teachers Materials Algebra tiles (1 set for every two participants) Flip charts or large pieces of newsprint for recording drawings and number sentences. Crayons or colored markers (for each group) Participants Pre-Workshop Preparation Audio-Visual Requirements Overhead projector and screen Algebra tiles for the overhead projector Blank transparency film and pens for transparencies Transparencies of Visual Aids 1, 2, 3, 4, 5A, 5B, 6A, 6B, 7A, 7B, 8A, 8B, 9A, 9B Room Preparation 1. Select desks or tables with flat tops as workspaces. 2. Arrange tables to accommodate groups of two or four. 3. Provide sufficient wall space for posting summaries of group discussions on flip charts or newsprint. USING ALGEBRA TILES EFFECTIVELY 1
12 In this workshop, the Facilitator acts as the teacher and models the lessons on an overhead projector. The participants act as students and experience the benefits of hands-on activities in math classrooms. Welcome Participants (5 minutes) 1. After you introduce yourself, discuss the purpose of the workshop: Say: The purpose of this workshop is to help you help your students learn how to use algebra tiles effectively. Say: Algebra tiles help to develop concepts related to integers, algebraic expressions, equations, and polynomials. Workshop Opening: Examining the Algebra Tiles The Facilitator works with Visual Aid 1 and algebra tiles on the overhead projector while participants work in pairs, or in groups of 4, at their desks. Materials: Visual Aid 1 Algebra tiles for the overhead projector Algebra tiles for each group Newsprint or large paper sheets 2. Explain the approach you will use in the workshop. Say: In this workshop, I, as the Facilitator, will act as the teacher. I ll use algebra tiles and Visual Aid transparencies to model the activities on an overhead projector. You, the participants, will act as students and experience the benefits of hands-on activities in math classrooms. 3. Tell participants which activities you will use in the workshop. When appropriate, pictures of models will appear in this column. Choose at least two activities in addition to the workshop opening and closing. USING ALGEBRA TILES EFFECTIVELY 3
13 Examine the Algebra Tiles (5 minutes) 4. Have participants spread the algebra tiles on their worktables and examine them. Note Use a transparency of Visual Aid 1. Place algebra tiles on the transparency. 5. Discuss the colors and shapes of the different algebra tiles x x x 2 x 2 Ask questions such as: Ask: What do you notice about all the negative tiles, 1, x, and x 2? Answer: All the negative tiles are red. 6. Have participants store the tiles in a corner of their worktables so they will have room to make models in the center of their tables. 7. Inform participants that they will use the newsprint to record their findings by drawing the models and writing the number sentences and equations they will create in the workshop activities. Classroom Applications Encourage participants to consult the Teacher s Editions for Prentice Hall Middle Grades Math, Prentice Hall Algebra, or Prentice Hall Advanced Algebra for classroom activities related to algebra tiles. 4 USING ALGEBRA TILES EFFECTIVELY
14 Adding Integers (10 minutes) Activity 1: Adding Integers The Facilitator works with algebra tiles on the overhead projector while participants work at their desks in pairs or in groups of 4. Materials: +1 tile 1 tile Visual Aid 2 Newsprint 1. Use your algebra tiles on the overhead projector in the steps below to illustrate the addition of two positive numbers: Have participants show two groups of positive tiles. In one group, model +5. In the other group, model +3. Ask How can we model with these tiles? Elicit the fact that to add the two groups, they should be moved together. Ask: What number sentence describes the model? Answer: = 8 2. Since the sum of a number and its opposite is zero, together, a positive tile and a negative tile represent zero and are called a zero pair. Use your algebra tiles to model a zero pair on the overhead projector. Ask: What number sentence describes this zero pair? Answer: 1 + ( 1) = 0 3. Have participants model 3 + ( 3) at their desks while you model the expression on the overhead projector. Model 3 with 3 yellow tiles and model 3 with 3 red tiles. Ask: What number sentence describes the model? Answer: 3 + ( 3) = 0 USING ALGEBRA TILES EFFECTIVELY 5
15 4. Have participants model 2 + ( 7) at their desks while you model the expression on the overhead projector. Model 2 with 2 positive tiles and 7 with 7 negative. Ask: What expression does this model represent? Answer: 2 + ( 7) Have participants join the two groups of tiles. Match pairs of positive and negative tiles and remove them. Elicit the fact that you can remove zero pairs because their value is zero. Elicit the fact that you cannot form any more zero pairs because all yellow tiles have been used. Ask: After removing the zero pairs, what tiles are left? Answer: 5 red tiles, representing 5 Ask: What number sentence describes the model? Answer: = 5 5. Challenge participants to find another sum. Ask: What is the sum of 6 + 4? Have participants model the expression and find the sum. Their models should show 4 zero pairs and 2 red tiles, similar to what is shown at the right. On a piece of paper, have teams record their findings by drawing and coloring the algebra tiles and writing the number sentence. 6. Place a transparency of Visual Aid 2 on the overhead projector. Cover the Model/Answer column. Have participants use algebra tiles to find the following sums. After each group has recorded its models and number sentences, uncover the Model/Answer column and have participants compare their models to those on the transparency. a. 4 + ( 9) b. 3 + ( 8) c. 9 + ( 3) Answer: = 2 Use a transparency of Visual Aid 2 (5 minutes) Classroom Applications Encourage participants to consult the Teacher s Editions for Prentice Hall Middle Grades Math, Prentice Hall Algebra, or Prentice Hall Advanced Algebra for more activities related to integers. 6 USING ALGEBRA TILES EFFECTIVELY
16 Subtracting Integers (10 minutes) Activity 2: Subtracting Integers The Facilitator uses algebra tiles on the overhead projector while participants work at their desks in pairs or in groups of 4. Materials: +1 tile 1 tile 1. Use algebra tiles on the overhead projector in the steps below to illustrate the subtraction of two integers. Visual Aid 3 Newsprint Say: We can model 6 ( 2). Start with 6 negative tiles. To subtract 2, remove 2 negative tiles. Ask: What number sentence describes the model? Answer: 6 ( 2) = 4 2. Say: We can model 4 7. Have each group place four positive tiles in a group. Ask them to add zero pairs until they have seven positive tiles in the set. Have the participants remove seven positive tiles. Say: This is one form of subtraction. Ask: How many tiles are left? Answer: Three red tiles, representing 3. Ask: What number sentence describes the model? Answer: 4 7 = 3 3. Place a transparency of Visual Aid 3 on the overhead projector. Cover the Model/Answer column. Have participants use algebra tiles for the following subtractions. After each group has recorded its models and number sentences, uncover the Model/Answer column. a. 2 6 b. 2 + ( 6) c. 3 8 d. 3 + ( 8) 4. Elicit discussion about the differences and sums in Visual Aid 3. Say: Complete the following statement. To subtract an integer, you can add. Use a transparency of Visual Aid 3 (5 minutes) Answer: its opposite USING ALGEBRA TILES EFFECTIVELY 7
17 One pre-service teacher commented about the class in which he was a substitute teacher: They just don t get it unless they use the tiles. I ve tried without them, and it just doesn t work. Simplifying Algebraic Expressions (15 minutes) Activity 3: Simplifying Algebraic Expressions The Facilitator works with algebra tiles on the overhead projector while participants work at their desks in pairs or in groups of 4. Materials: +1 tile 1 tile 1. Describe the algebra tiles the participants will use. Point out that negative tiles are red. x tile x tile 2. Write this expression on the chalkboard or on a transparency: Write: 5 + 4x Ask: How can we model this expression? Give participants an opportunity to respond. Then model the expression with the group. 3. Write this expression on the chalkboard or on a transparency: Write: 3x + 2 4x 5 Say: Before we model this expression, remember that subtracting is the same as adding the opposite, so we can write the expression as 3x ( 4x) + ( 5). Ask: How can we model this expression? Give participants an opportunity to respond. Then model the expression with the group. Visual Aid 4 Newsprint 5 + 4x 3x + 2 4x 5 8 USING ALGEBRA TILES EFFECTIVELY
18 4. Ask: How can we simplify the expression? Elicit the fact that simplifying means collecting like terms (like tiles) by using zero pairs. Remove zero pairs of x tiles and zero pairs of integer tiles. Say: After we move aside the zero pairs, the simplified expression is left. 5. Write this expression on the chalkboard or on a transparency: Write: 2x + 5 4x 5 Ask: How can we model this expression? Give participants an opportunity to respond. Then model the expression with the group. Ask: How can we simplify the expression? Repeat that simplifying means collecting like terms (like tiles) by using zero pairs. Remove the integer zero pairs. Ask: What is the simplified expression? 6. Place a transparency of Visual Aid 4 on the overhead projector. Cover the Model/Answer column. Have participants use algebra tiles to model and simplify the following expressions. After each group has recorded its models and expressions, uncover the Model/Answer column. a. 4x + 8 3x b. 5x 9 2 3x c. 3x x 6 Answer: 3x + 2 4x 5 = x 3 2x + 5 4x 5 Answer: 6x Use a transparency of Visual Aid 4 (5 minutes) Classroom Applications Encourage participants to consult the Teacher s Editions for Prentice Hall Middle Grades Math, Prentice Hall Algebra, or Prentice Hall Advanced Algebra for further classroom activities related to integers. USING ALGEBRA TILES EFFECTIVELY 9
19 Algebra tiles can help students develop concepts related to integers, algebraic expressions, equations, and polynomials. Activity 4: Solving Linear Equations Solving Linear Equations (15 minutes) The Facilitator works with algebra tiles on the overhead projector while participants work at their desks in pairs or in groups of 4. Materials: +1 tile 1 tile 1. Describe the algebra tiles, shown at the right, that the participants will use. Point out that negative tiles are red. x tile x tile 2. Write this expression on the chalkboard or on a transparency: Write: x 2 = 7 Ask: How can we model this equation? Give participants an opportunity to respond. Then model the equation. Point out that just as the two sides of an equation are separated by an equals sign, the two parts of the equation are separated by the bar in the model. Visual Aid 5 Newsprint x 2 = 7 Ask: How can we use algebra tiles to solve this equation? Give participants an opportunity to respond. Say: To isolate the variable, that is, to get the x tile alone, add 2 positive tiles to each side of the equation. Remove the zero pairs to show the solution. Ask: What is the solution? Answer: x = 9 10 USING ALGEBRA TILES EFFECTIVELY
20 3. Write this expression on the chalkboard or on a transparency: Write: 2x + 3 = 9 Ask: How can we model and solve this equation? Give participants an opportunity to respond. Then manipulate the tiles while you explain each step. 2x + 3 = 9 Add three negative tiles to each side to create zero pairs on the side with the x-tiles. Remove zero pairs to show 2x = 12. Say: We want to get x alone for a solution. First, we can make two groups of equal numbers of tiles on each side of the bar. Then we can remove one set of the tiles from each side of the bar. Remember that whatever we do to one side of an equation, we must do to the other side. 2x = 12 2x 2 = 12 2 Ask: What is the solution? Answer: x = 6 4. Place a transparency of Visual Aid 5A on the overhead projector. Cover the Model/Answer column. Have participants use algebra tiles to solve the following equation. After each group has recorded its model and solution, uncover the Model/Answer column. Say: Solve this equation: 3x 2 = 4 5. Place a transparency of Visual Aid 5B on the overhead projector. Cover the Model/Answer column. Have participants use algebra tiles to solve the following equations. After each group has recorded its models and solutions, uncover the Model/Answer column. Say: Solve this equation: 5x + 6 = 4 Use a transparency of Visual Aid 5A. Answer: x = 2 Use a transparency of Visual Aid 5B. Answer: x = 2 USING ALGEBRA TILES EFFECTIVELY 11
21 In this activity you will challenge participants to work in cooperative groups to analyze a word problem, determine how they can use algebra tiles to model the problem, and how they can use algebra tiles to solve the problem. Using Algebra Tiles to Solve a Word Problem (10 minutes) 1. Read this problem to participants or display a transparency of Visual Aid 6A 2. Tell participants they will work together in their groups for 5 minutes to: Write an equation for the problem. Model the problem with algebra tiles. Solve the problem. Activity 5: Solving Word Problems with Algebra Tiles Materials: Each group of 4 participants needs enough algebra tiles to represent 51 unit tiles and 5 x-tiles. Visual Aids 6A, 6B Use a transparency of Visual Aid 6A to present the problem. Allow 5 minutes. Remind students to record their models and equations. Equation: 3x x + 7 = After 5 minutes display a transparency of Visual Aid 6B on the overhead projector. Discuss the equation, the models, and the solution with participants. Uncover the transparency for Visual Aid 6B row by row. Classroom Applications Encourage participants to consult the Teacher s Editions for Prentice Hall Middle Grades Math, Prentice Hall Algebra, or Prentice Hall Advanced Algebra for further classroom activities related to problem solving. 12 USING ALGEBRA TILES EFFECTIVELY
22 There are many more activities that can be done with algebra tiles beyond representing algebraic expressions and linear equations. Algebra tiles can be used to represent polynomials and operations with polynomials. The tiles can also be used to factor trinomials and to solve quadratic equations. Modeling Polynomials (5 minutes) 1. Review the meaning of each algebra tile with participants. Activity 6: Adding and Subtracting Polynomials Materials: +1 1 x x x 2 x 2 Visual Aids 1, 7A, 7B Newsprint 2. Place the tiles to the right on the overhead projector. Ask: What expression does this model represent? Answer: 2x 2 3x 4 3. Challenge participants to build a model of the expression: Say: Build a model for this expression: 2x 2 2x 3. 2x 2 2x 3 Adding Polynomials (5 minutes) 1. Remind participants that they can model addition of polynomials by modeling the two polynomials, joining them, and removing zero pairs. Ask: How can we model this addition? 2x 2 + 3x + 5 and x 2 2x 3 Accept all reasonable answers. Then model the addition with tiles on the overhead projector. Call attention to the zero pairs. Ask: What is the sum? Answer: 3x 2 + x + 2 USING ALGEBRA TILES EFFECTIVELY 13
23 2. Place a transparency of Visual Aid 7A on the overhead projector. Cover the Model/Answer section. Have participants use algebra tiles to add the following polynomials. After each group has recorded its models and solutions, uncover the Model/Answer section. Say: Find the sum: (3x 2 + 2x 4) + ( 2x 2 + x 3) Subtracting Polynomials (5 minutes) 1. Remind participants that they can model subtraction of a polynomial by adding the model of the inverse of the polynomial to be subtracted to the model of the first polynomial. Ask: How can we model this subtraction? 2x 2 + 4x 5 (x 2 + 2x 3) Accept all reasonable answers. Then model the addition with tiles on the overhead projector. Use a transparency of Visual Aid 7A. Answer: x 2 + 3x 7 By the time students get to polynomials, they should know that subtraction is adding the inverse, or opposite, number. 2. Build a model for 2x 2 + 4x Subtract x 2 + 2x 3 by adding its opposite, that is x 2 2x + 3. Model this expression. Combine models and remove zero pairs to model the result. Ask: What expression does the resulting model represent? Answer: x 2 + 2x 2 4. Place a transparency of Visual Aid 7B on the overhead projector. Cover the Model/Answer section. Have participants use algebra tiles to subtract the following polynomials. After each group has recorded its models and solutions, uncover the Model/Answer section. Say: Find the difference: (2x 2 + 2x 1) (x 2 x + 3) Use a transparency of Visual Aid 7B. Answer: x 2 + 3x 4 14 USING ALGEBRA TILES EFFECTIVELY
24 In this activity participants will use rectangular arrays to model the product of two polynomials. You can relate the multiplication of polynomials to a basic whole-number multiplication table. Place the first factor vertically under the multiplication dot. Place the second factor horizontally to the right of the multiplication dot. Multiplying Polynomials (15 minutes) 1. Review the meaning of each algebra tile with participants. Activity 7: Multiplying Polynomials Materials: +1 1 x x x 2 x 2 Visual Aids 1, 8A, 8B Newsprint 2. Ask participants how the table on Visual Aid 8A shows basic multiplication facts. Ask: How can we find the product 3 4 in this table? Ask: What is the product of 3 4? Use Visual Aid 8A. Answer: Find the entry to the right of 3 and under = Help participants relate a basic multiplication table to a rectangular array to model the multiplication of polynom (x + 3) (x 2) USING ALGEBRA TILES EFFECTIVELY 15
25 Ask: How can we model (x + 3)(x 2)? Give participants a chance to respond and model the multiplication with algebra tiles on the overhead projector. (x + 3)(x 2) Ask: What is the product of (x + 3)(x 2)? 4. Place a transparency of Visual Aid 8Ax 1)(x 4) 5. Place a transparency of Visual Aid 8B 2x + 2)(x 3) Answer: x 2 + 3x 2x 6 = x 2 + x 6 Use a transparency of Visual Aid 8A. Answer: x 2 5x + 4 Use a transparency of Visual Aid 8B. Answer: 2x 2 + 8x 6 Classroom Applications Encourage participants to consult the Teacher s Editions for Prentice Hall Middle Grades Math, Prentice Hall Algebra, or Prentice Hall Advanced Algebra for further classroom activities related to polynomials. 16 USING ALGEBRA TILES EFFECTIVELY
26 In this activity participants will use rectangular arrays to model factoring polynomials. Relate factoring polynomials to Activity 7, Multiplying Polynomials. The first factor is placed vertically under the multiplication dot. The second factor is placed horizontally to the right of the multiplication dot. Factoring Polynomials (15 minutes) 1. Review the meaning of each algebra tile with participants. Activity 8: Factoring Polynomials Materials: +1 1 x x x 2 x 2 Visual Aids 1, 9A, 9B Newsprint 2. Briefly review Multiplying Polynom Give participants a chance to respond. Then model the multiplication with algebra tiles on the overhead projector. 3. Show the model at the right on the overhead projector. (x + 3)(x 2) = x 2 + x 6 Use a transparency of Visual Aid 8A or model the expression with algebra tiles on the overhead projector. Call attention to the fact that the model shows a rectangular array with the tiles arranged in descending order. Ask: What expression does this model represent? Answer: x 2 + 4x 3x 12 USING ALGEBRA TILES EFFECTIVELY 17
27 4. Challenge participants to suggest how the expression can be factored by using algebra tiles. Give participants a chance to respond and then demonstrate on the overhead projector how to factor a polynomial. Build an axis around the rectangle. To factor the polynomial, find the dimensions of the rectangle. Determine which tiles should be placed on the horizontal axis and which tiles should be placed on the vertical axis. Call attention to the fact that all negative tiles should be placed on the same axis. Ask: What expression does the resulting model represent? 5. Place a transparency of Visual Aid 9A + 5x Place a transparency of Visual Aid 9B 7x Answer: (x 3)(x + 4) Use a transparency of Visual Aid 9A. Answer: x 2 + 5x + 6 = (x + 2)(x + 3) Use a transparency of Visual Aid 9B. Answer: x 2 7x + 12 = (x 4)(x + 3) Classroom Applications Encourage participants to consult the Teacher s Editions for Prentice Hall Middle Grades Math, Prentice Hall Algebra, or Prentice Hall Advanced Algebra for further classroom activities related to factoring polynomials. 18 USING ALGEBRA TILES EFFECTIVELY
28 While a representative from each group displays the pages with the group s recordings, you may wish to collect the algebra tiles and other materials that must be returned. Closing Activities (10 minutes) Workshop Closing: Recapping the Activities 1. Have a representative from each group display around the room the recordings (drawings, number sentences, equations) from that group. 2. Recap the activities that have been presented and end by having participants discuss what they have learned and how they can use it. Ask: How can the activities that we worked on today have an impact on students in your classroom from now on? 3. Stress with participants the importance of continuing to experiment with algebra tiles in their classrooms. Say: Remember, algebra tiles can help students develop concepts related to integers, algebraic expressions, equations, and polynomials. 4. You may wish to refer your group to any worthwhile articles, books, or Web sites with which you are familiar. 5. Take any questions that remain. 6. Collect all algebra tiles and other materials that have been distributed. For more professional development information and training materials, visit the Prentice Hall site: 7. You may wish to close with the quotation that introduced this professional development workshop: Say: Remember: When I listen, I hear. When I see, I remember. But when I do, then I understand. USING ALGEBRA TILES EFFECTIVELY 19
35 VISUAL AID 6A SOLVING WORD PROBLEMS Model the word problem with algebra tiles. Use the model to solve the problem. Record your findings 26 USING ALGEBRA TILES EFFECTIVELY 1999 Prentice-Hall, Inc.
36 VISUAL AID 6B SOLVING WORD PROBLEMS Use the model for Visual Aid 6A. Solve the word problem. Record your findings. 3x x + 7 = 38 Simplify the left side of the equation. Add 13 to each side of the equation to form zero pairs. Remove zero pairs. 5x = 25 x = 5 Five students rode in each car. USING ALGEBRA TILES EFFECTIVELY 1999 Prentice-Hall, Inc. 27
A Concrete Introduction to the Abstract Concepts of Integers and Algebra using Algebra Tiles Table of Contents Introduction... 1 page Integers 1: Introduction to Integers... 3 2: Working with Algebra Tiles...MATH 110 College Algebra Online Families of Functions Transformations Functions are important in mathematics. Being able to tell what family a function comes from, its domain and range and finding a function
INTRODUCTION Algebra for All and No Child Left Behind are phrases in the education community that suggest, and in many cases require, action. They give impetus for mathematics teachers at all levels to
Multiplying Binomials Standard: Algebra 10.0 Time: 55 mins. Multiplying Binomials and Factoring Trinomials Using Algebra Tiles and s Materials: Class set of Algebra Tiles or access to a computer for eachSolving Systems of Linear Equations Substitutions Outcome (learning objective) Students will accurately solve a system of equations algebraically using substitution. Student/Class Goal Students thinking
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UC Davis, School and University Partnerships CAHSEE on Target Mathematics Curriculum Published by The University of California, Davis, School/University Partnerships Program 006 Director Sarah R. Martinez,The High School Math Project Focus on Algebra Rhinos and M&M s (Exponential Models) Objective The objectives of this lesson are for students to explore the patterns of exponential models in tables, graphs,Graphing Equations with Color Activity Students must re-write equations into slope intercept form and then graph them on a coordinate plane. 2011 Lindsay Perro Name Date Between The Lines Re-write each
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Multiplication Fact Power and Shortcuts Objectives To discuss multiplication facts and the importance of fact power; and to review fact shortcuts. epresentations etoolkit Algorithms
Title Polynomial Functions, Expressions, and Equations Big Ideas/Enduring Understandings Applying the processes of solving equations and simplifying expressions to problems with variables of varying degrees.
Session 5 Integrating Technology in the Classroom Copyright 2012 SEDL Connecting Kids to Mathematics and Science 5.1 Acknowledgments Connecting Kids to Mathematics and Science was made possible through
UNIT TWO POLYNOMIALS MATH 421A 22 HOURS Revised May 2, 00 38 UNIT 2: POLYNOMIALS Previous Knowledge: With the implementation of APEF Mathematics at the intermediate level, students should be able to: -
Lines, Lines, Lines!!! Slope-Intercept Form ~ Lesson Plan I. Topic: Slope-Intercept Form II. III. Goals and Objectives: A. The student will write an equation of a line given information about its graph.
Creating A Grade Sheet With Microsoft Excel Microsoft Excel serves as an excellent tool for tracking grades in your course. But its power is not limited to its ability to organize information in rows and
visit us at Using the Area Model to Teach Multiplying, Factoring and Division of Polynomials For more information about the materials presented, contact Chris Mikles mikles@cpm.org From CCA
Core Florida Math for College Readiness Florida Math for College Readiness provides a fourth-year math curriculum focused on developing the mastery of skills identified as critical to postsecondary readiness
4th Grade Math Homework Based on the Utah State Core Standards and Objective 4th Grade Math Homework (Standards Based) created by Lara Dean Over the years I have recognized the need for a homework programOhio Standards Connection: Number, Number Sense and Operations Benchmark C Apply properties of operations and the real number system, and justify when they hold for a set of numbers. Indicator 1 Identify
Solving Systems of Linear Equations Graphing Outcome (learning objective) Students will accurately solve a system of equations by graphing. Student/Class Goal Students thinking about continuing their academic
2.3. Finding polynomial functions. An Introduction: As is usually the case when learning a new concept in mathematics, the new concept is the reverse of the previous one. Remember how you first learned
ACCUPLACER Arithmetic & Elementary Algebra Study Guide Acknowledgments We would like to thank Aims Community College for allowing us to use their ACCUPLACER Study Guides as well as Aims Community College
LESSON 0 addition and multiplication of polynomials LESSON 0 Addition and Multiplication of Polynomials Base 0 and Base - Recall the factors of each of the pieces in base 0. The unit block (green) is x.The following pages include many problems to practice factoring skills. There are also several activities with examples to help you with factoring if you feel like you are not proficient with it. ThereMath Board Games For School or Home Education by Teresa Evans Copyright 2005 Teresa Evans. All rights reserved. Permission is given for the making of copies for use in the home or classroom of the purchaser
Grade 4 Level Math Common Core Sampler Test This test sample is made to be used give students and teachers a basic overview of key Grade 4 Common Core grade level work. All questions are aligned to the
CUTTING EXPENSES Outcome (lesson objective) Students will apply strategies for reducing expenses. They will practice operations with whole numbers and will construct bar graphs. Student/Class Goal Students
Algebra 2 PreAP Name Period IMPORTANT INSTRUCTIONS FOR STUDENTS!!! We understand that students come to Algebra II with different strengths and needs. For this reason, students have options for completing
Ohio Standards Connection Patterns, Functions and Algebra Benchmark E Solve open sentences and explain strategies. Indicator 4 Solve open sentences by representing an expression in more than one way using
Unit 10 Area Model Factoring Research-based National Science Foundation-funded Learning transforms lives. Dear Student, When we multiply two factors, we get their product. If we start with the product,Title: Integers: Quick, Fun and Easy To Learn Brief Overview Students will identify and understand positive and negative integers. Using this understanding, the students will use the number line, studyTeacher OBJECTIVES MATERIALS DIRECTIONS Students will learn how the ancient Maya counted, and will practice addition and subtraction using Maya number glyphs. Provided by Classroom Teacher: Pens or pencils | 677.169 | 1 |
In Brief;Reviews;Mathematics;Books
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The EdExcel Certificate of Achievement in Mathematics is designed to reward students who have difficulty in progressing beyond level 3 of the national curriculum. Divided into three parts, the book is tailored to this certificate but it could be used with any similar course or as a supplement to other schemes. The content is clearly laid out on full colour pages. The explanations are good, but some of the exercises are too short. An activities and assessment pack provides more material to support teaching the course | 677.169 | 1 |
MATH 253 - Introduction to Mathematical Reasoning
Credits:3
Introduction to Mathematical Reasoning is a transition to advanced mathematics, with a focus on precise mathematical reasoning and communication. Introductory topics in mathematical logic and proof-writing are discussed, including direct proof, proof by contradiction and contraposition, and mathematical induction. These ideas are applied to problems in elementary number theory, set theory, functions and relations, and cardinality. Further topics may be discussed if time allows.
USI Core 39: Ways of Knowing-Scientific and Mathematical Inquiry.
Prerequisite(s): Satisfactory placement score or MATH 111. May be taken concurrently with MATH 230. | 677.169 | 1 |
Shape is expounded to operate. An aircraft wing has the shape it does as a result of its lifting functionality. The pillars of the Parthenon and the girders of a skyscraper are formed to the aim of helping their enormous buildings. equally, the shape of an algebraic expression or equation displays its functionality. Algebra: shape and serve as initial version introduces each one function--linear, energy, quadratic, exponential, polynomial--and offers a learn of the elemental kind of expressions for that functionality. Readers are inspired to ascertain the elemental kinds, see how they're developed, and view the position of every part. in the course of the textual content, there are instruments sections put on the ends of chapters to assist readers collect the talents they should practice simple algebraic manipulations.
This quantity presents an entire realizing of the basic explanations of routing congestion in present-day and next-generation VLSI circuits, deals recommendations for estimating and relieving congestion, and gives a serious research of the accuracy and effectiveness of those strategies. The e-book comprises metrics and optimization concepts for routing congestion at a variety of phases of the VLSI layout move.
Introduce your scholars to the newest that Microsoft workplace has to provide with the recent new release of Shelly Cashman sequence books! For the prior 3 a long time, the Shelly Cashman sequence has successfully brought computing device talents to thousands of scholars. With Microsoft Excel 2013, we're carrying on with our background of innovation via bettering our confirmed pedagogy to mirror the educational kinds of today's scholars.
Simply because such a lot of first-year writing scholars lack the fundamental abilities the path calls for, studying professional McWhorter supplies them regular information in the course of the demanding situations they face in educational paintings. winning collage Writing deals broad guide in energetic and significant interpreting, useful suggestion on research and faculty survival abilities, step by step suggestions for writing and study, exact insurance of the 9 rhetorical styles of improvement, and sixty one readings that offer robust rhetorical types, in addition to an easy-to-use instruction manual within the whole version.
Submit 12 months be aware: First released in 2009
------------------------
This most generally used textbook within the box has been completely revised and up-to-date to mirror alterations within the future health care and the renewed concentrate on future health care details expertise tasks. new chapters hide Federal efforts to reinforce caliber of sufferer care by using wellbeing and fitness care info know-how and method concerns. also, reflecting the elevated specialise in international overall healthiness, the e-book good points a global point of view on health and wellbeing care details expertise.
Case reviews of organisations experiencing management-related info process demanding situations were up to date and several other new circumstances were extra. those reality-based instances are designed to stimulate dialogue between scholars and allow them to use techniques within the publication to real-life eventualities.
1 REORDERING AND REGROUPING Solution 31 Since the bike shop sells q bicycles during a normal month, and since that number triples during a sale month, it sells 3q bicycles during a sale month. Similarly, the price is (1/2)p during a sale month. Thus, 1 Revenue during sale month = (3q) p . 2 By regrouping and reordering the factors in this product we get Revenue during sale month = (3q) 1 p 2 1 3 = 3 · qp = qp. 5, times the original revenue. Example 3 In a population of 100 prairie dogs there are b births and d deaths over a one-year period.
If xyz = 100, find the value of (3x)(2y)(5z). x 22. If xyz = 20, find the value of (2z)( )(6y). 4 23. Rewrite the expression a + 2(b − a) − 3(c + b) without using parentheses. Simplify your answer. 24. A car travels 200 miles in t hours at a speed of r mph. If the car travels half as fast but three times as long, how far does it travel? ) (a) If in January 2009 the amount (in billions of gallons) flowing into the reservoir is A and the amount flowing out is B, write an expression for the amount of water in the reservoir at the end of January 2009.
B) Is the expression 2x + 6 equivalent to the expression x + 3? Solution (a) If we divide both sides of the first equation by 2, we have 2x + 6 = 10 2x + 6 10 = 2 2 x + 3 = 5. dividing both sides by 2 Therefore, the two equations are equivalent. Dividing both sides of an equation by 2 produces an equivalent equation. You can verify that x = 2 is the solution to both equations. (b) The expression 2x + 6 is not equivalent to x + 3. This can be seen by substituting x = 0 into each expression. The first expression becomes 2x + 6 = 2(0) + 6 = 6, but the second expression becomes x + 3 = 0 + 3 = 3. | 677.169 | 1 |
1. Prerequisites *
At the start of this course the student should have acquired the following competences: an active knowledge of
English
general knowledge of the use of a PC and the Internet
general notion of the basic concepts of
The student knows the basic concepts of set theory and of linear algebra.
specific prerequisites for this course
Content of the course `Groepen en Ringen' ('Groups and Rings'). The student should be familiar with concepts of vector spaces, fields, groups. The student has some basic knowledge on rings.
2. Learning outcomes *
The student knows the most fundamental concepts and theorems from field theory and classical Galois theory.
The student is familiar with diverse properties of field extensions. The student can compute the Galois group for several classical situations.
The student understands the connection between the concept of field extensions and Galois theory, on the one hand, and the problems of solving polynomial equations and of ruler and compass constructions, on the other hand.
3. Course contents *
Rings of polynomials, the procedure of joining a root of a polynomial to the field.
Algebraic field extensions, normal extensions.
Definition and construction of an algebraic closure of a field.
Basis of Galois theory, Galois correspondence between subextensions and subgroups. | 677.169 | 1 |
August 9, 2017
Homeschool Review Crew: Our Review of No-Nonsense Algebra
I don't know about you but some of my children have struggled with high school math classes. When we were given the opportunity to review No-Nonsense Algebra from Math Essentials, we were more than happy to try it out.
This is a physical book but comes with free online classes taught by Richard W. Fisher. What a wonderful combo for such a great price!!! The online classes are short yet very informative. His approach is to teach a step by step process so the student can fully understand how to do the equations in each lesson.
There is a section right in front of the book that explains how to use this book. I was impressed with the steps that each lesson follows. You begin with an Introduction, this step explains each new topic and important terms that will be used in the lesson. Next, we have the Helpful Hints section that gives your student tips and shortcuts to use as they complete the lesson. After reading the hints carefully, you move onto the Examples. The examples will show you how to efficiently and correctly complete each problem in the lesson. The author suggests that you copy each example onto a piece of paper. This will help you obtain an in depth understanding of the process. Step 4 is to do the Exercises in the lesson. After you complete these, you work on the Review section. This section will assist you in remembering what you have been taught. Finally, after you are finished with all the exercises and review questions, you will go to the back of the book, find the lesson you are doing, and correct your work.
It is this series of actions that set this math book apart from others. My son, who has struggled with math in the past, has a better grasp on the topics being taught. He is not sitting there with a blank look on his face. He is able to comprehend what he has wrestled with in the past.
The book has 10 Chapters that include:
Chapter 1- Necessary Tools for Algebra
Chapter 2- Solving Equations
Chapter 3- Graphing and Analyzing Linear Equations
Chapter 4- Solving and Graphing Inequalities
Chapter 5- Systems of Linear Equations and Inequalities
Chapter 6- Polynomials
Chapter 7- Rational Expressions
Chapter 8- Radical Expressions and Geometry
Chapter 9- Quadratic Equations
Chapter 10- Algebra Word Problems
So far we have had no issues. My son is working on one lesson per day. The videos run smoothly on our computer and he said he enjoys Mr. Fisher's teaching methods. I am enjoying the fact that I am not hearing my son complain. He is actually happy to do his math classes now.
There are also some more reviews for you to read over on the Homeschool Review Crew website. Before buying a new course of study for my children, we always like to see what others are saying. It's important to research before jumping into anything. That is why I love reading through the honest reviews on the Homeschoool Review Crew site. I hope they will be a help to you as you decide if this resource would be a good fit for your family | 677.169 | 1 |
Mini maths courses bridge the divide
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A GROUND-BREAKING maths qualification for sixth-formers and adults has been praised by government advisers for encouraging students with poor GCSE grades to improve their skills.
The bite-size courses were piloted last year as an alternative to GCSEs and A-levels, offering "bolt-on" maths to non-mathematicians.
Since then the number of candidates taking the 60-hour maths programmes has increased nearly tenfold. Popular courses include "managing money" and "making sense of data".
Now in the second year of the pilot, the so-called "free-standing maths units" are the first qualifications to bridge the divide between academic and vocational courses.
They were devised after concerns that maths skills in Britain were inferior to those in other countries and that too few 16-year-olds chose to continue with the subject.
The courses have also been unexpectedly popular with A-level maths students. They complement A-level studies by teaching sixth-formers how to apply their maths skills, an analysis by the Government's exam quango, the Qualifications and Curriculum Authority, found.
Half of a course is assessed by a portfolio of coursework focusing on maths applications and half by a written exam testing knowledge of principles.
This year the courses are being taken by more than 3,700 candidates in 58 colleges and 15 schools, compared to 430 students in 21 centres in 1998.
The courses aim to meet the mathematical demands of different subjects or occupations. Ranging from foundation (equivalent to D to G grades at GCSE) to advanced level (nationally equivalent to an AS or A-level module), they focus on a narrow range of maths skills relevant to other subjects - calculus for scientists, for example, or three-dimensional problem-solving for art and design students.
The QCA evaluation of the first phase of the pilot found it was well-received by students and teachers alike.
The report said: "The units are encouraging post-16 students to study mathematics. At best the intermediate level students in the pilot study would have repeated the GCSE in an attempt to improve grades. Many of those at foundation level would have studied no further mathematics."
Teachers reported that the courses boosted students' interest in maths and gave teenagers who achieved only low GCSE grades an opportunity to apply their maths. Students who took the foundation courses had got at best an E at maths GCSE while intermediate units attracted pupils who had B to G grades.
The most popular foundation courses were managing money and making sense of data, while calculus and understanding mathematical thinking attracted the most candidates at advanced level.
Teachers were generally pleased with the content of the courses and considered they were pitched at the right level.
The QCA evaluation of the first phase of the pilot is | 677.169 | 1 |
All parts of the theory whicJi are beyond the comprehension
of
the student or wliicli are logically
unsound are
omitted.
manufactured for this purpose. etc. not only taxes a student's memory unduly but in variably leads to mechanical modes of study.
Such a large number of methods. are omitted."
this book.
All
practical
teachers
know how few
students understand and
appreciate the more difficult parts of the theory. and conse-
. Until recently the tendency was to multiply as far as possible.
Elementary Algebra.
"
While
in
many
respects
similar to the author's
to its peculiar aim.
however. chief
:
among
These
which are the following
1.
specially
2.
omissions serve not only practical but distinctly pedagogic " cases " ends. The entire study of algebra becomes a mechanical application of memorized
rules.. in order to make every example a
social
case of a memorized method. short-cuts that solve only examples
real value. but "cases" that are taught only on account of tradition.
giving to the student complete familiarity
with
all
the essentials of the subject.
and ingenuity
while the cultivation of the student's reasoning power is neglected.PREFACE
IN
this
book the attempt
while
still
is
made
to shorten the usual course
in algebra.
All unnecessary methods
and "cases" are
omitted. Typical in this respect is the
treatment of factoring in
many
text-books
In this book
all
methods which are of
and which are applied in advanced work are given.
owing
has certain distinctive features.
and it is hoped that this treatment will materially diminish the difficulty of this topic for young students. " The book is designed to meet the requirements for admis-
sion to our best universities
and
colleges. a great deal of the theory offered in the avertext-book is logically unsound . are placed early in the course.
differ
With very few
from those
exceptions
all
the exer
cises in this
book
in the
"Elementary Alge-
bra". The presenwill be found to be tation of problems as given in Chapter
V
quite a departure from the customary way of treating the subject. For the more ambitious student.g.
two negative numbers. the following
may
be quoted from the author's "Elementary Algebra":
which
"Particular care has been bestowed upon those chapters in the customary courses offer the greatest difficulties to
the beginner. in particular the
requirements of the College Entrance Examination Board.
Topics of practical importance.
In regard
to
some other features of the book.
The best way to introduce a beginner to a new topic is to offer Lim a large number of simple exercises.vi
PREFACE
quently hardly ever emphasize the theoretical aspect of alge bra. all proofs for the sign age
of the product of
of the binomial
3. however.
TJie exercises are slightly simpler than in the larger look.
enable students
who can devote only a minimum
This arrangement will of time to
algebra to study those subjects which are of such importance for further work. especially problems and factoring. hence either book
4. all elementary proofs theorem for fractional exponents. This made it necessary to introduce the theory of proportions
. there has been placed at the end of the book a collection of exercises which contains an abundance
of
more
difficult
work. e.
may
be used to supplement the other. etc. Moreover. as quadratic equations and
graphs.
but the true study of algebra has not been sacrificed in order to make an impressive display of sham
life
applications.
is
based principally upon the alge-
.
to solve a
It is
undoubtedly more interesting for a student
problem that results in the height of Mt. such examples. The entire work in graphical methods has been so arranged that teachers who wish
a shorter course
may omit
these chapters. in
"
geometry
.
viz. and hence the student is more easily led to do the work by rote
than when the arrangement
braic aspect of the problem. an innovation which seems to mark a distinct gain from the pedagogical point
of view.'
This topic has been preit is
sented in a simple. and commercial are numerous.
and they usually involve difficult numerical calculations.
By studying proportions during the first year's work. Moreover. and
of the
hoped that some
modes of representation given
will be considered im-
provements upon the prevailing methods.PREFACE
vii
and graphical methods into the first year's work."
Applications taken from geometry.
McKinley
than one that gives him the number of Henry's marbles. But on the other hand very few of such applied examples are
genuine applications of algebra.
while in the usual course proportions
are studied a long time after their principal application. based upon statistical abstracts. are
frequently arranged in sets that are algebraically uniform. physics.
nobody would find the length Etna by such a method. the student will be able to utilize this knowledge where it is most
needed. elementary way.
but they unquestionably furnish a very good antidote against 'the tendency of school algebra to degenerate into a mechanical application of
memorized
rules. " Graphical methods have not only a great practical value.
of the Mississippi or the height of Mt.
is such problems involves as a rule the teaching of physics by the
teacher of algebra.
.
William P.
NEW
YORK. desires to acknowledge his indebtedness to Mr. however.
genuine applications of elementary algebra work seems to have certain limi-
but within these limits the author has attempted to
give as
many
The author
for
simple applied examples as possible.
pupil's knowlso small that an extensive use of
The average
Hence the
field of
suitable for secondary school
tations.
April.viii
PREFACE
problems relating to physics often
offer
It is true that
a field
for genuine applications of algebra.
edge of physics. 1910. Manguse for the careful reading of the proofs and
many
valuable suggestions.
ARTHUR SCHULTZE.
Six
2
. a = 4.
. 38. a =4. of this exercise?
What kind of expressions are Exs.
26.
27. 6 = 1. if
:
a = 2.
34.
a =3. 6 = 2.
6.c) (a .6 .
Express in algebraic symbols 31.
23. physics. 6 = 7. 6.
The quantity a
6
2 by the quantity a
minus
36. 6 = 5. and the area of the
is
triangle
S
square feet (or squares of other units selected).
29. a = 3.6 -f c) (6
a
+ c).
Twice a3 diminished by 5 times the square root of the quantity a minus 6 square.
35. 25.
a.
Read the expressions
of Exs.
30. a
a=3.
37. 30.
and
If the three sides of a triangle contain respectively c feet (or other units of length).
:
6. a = 4. 2-6 of the exercise. 33.
28. = 3.
a = 3. geometry. 6 = 3.
w
cube plus three times the quantity a minus
plus 6 multiplied
6. and other sciences.
12 cr6
-f-
6 a6 2
6s. 10-14
The
representation of numbers by letters makes it posvery briefly and accurately some of the principles of arithmetic.
sible to state
Ex. 6 = 6.
24. 6 = 5.
6 = 4.
then
8 = \ V(a + 6 + c) (a 4. Six times a plus 4 times
32. a = 2.
22. 6 = 6.
6=2.
Six times the square of a minus three times the cube of Eight x cube minus four x square plus y square.12
17
&
*
ELEMENTS OF ALGEBRA
18
'
8
Find the numerical value of 8 a3
21.
4.e.
(c) 4.INTRODUCTION
E. b.
=
(a)
How
far does
a body fall from a state of rest in 2
seconds ?
(b)
*
stone dropped from the top of a tree reached the ground in 2-J.
9
distance s passed over by a body moving with the uniform velocity v in the time t is represented by the formula
The
Find the distance passed over by A snail in 100 seconds.seconds. and 5 feet. and 13 inches.
84 square
EXERCISE
1. b 14. 14. A carrier pigeon in 10 minutes.
.
S = | V(13-hl4-fl5)(13H-14-15)(T3-14-i-15)(14-13-f-15)
= V42-12-14. An electric car in 40 seconds. if v = 30 miles per hour. and 15 feet. the three sides of a triangle are respectively 13. if v
:
a. then a 13. c.
the area of the triangle equals
feet.g.) Assuming g
.
(b) 5. Find the height of the tree. count the resistance of the atmosphere. if v = 50 meters per second 5000 feet per minute. 15 therefore feet. 13.16 centimeters per second.
i. if v . A body falling from a state of rest passes in t seconds 2 over a space S (This formula does not take into ac^gt 32 feet. A train in 4 hours.
2. d.
By
using the formula
find the area of a triangle
whose
sides are respectively
(a) 3. and c
13
and
15
=
=
=
. 12.16
1
= 84. How far does a body fall from a state of rest in T ^7 of a (c)
A
second ?
3.
fo
If
i
represents the simple interest of
i
p
dollars at r
in
n
years.
is
H
2
units of length (inches.
ELEMENTS OF ALGEBRA
If
the radius of a circle
etc. and the value given above is only an
surface
$=
2
approximation.
.
diameter of a sphere equals d
feet.
2 inches.
the area
etc.14 is frequently denoted by the Greek letter TT.14d (square units).
of this formula
:
The The
interest on interest
$800
for 4 years at
ty%.
~
7n
cubic feet.
$ = 3.
If the
(b) 1 inch. This number cannot be expressed exactly.
to Centigrade readings:
(b)
Change the following readings
(a)
122 F.
on $ 500 for 2 years at 4 %.
denotes the number of degrees of temperature indi8.
(c)
5 miles.).
5.
If the diameter of a sphere equals d units of length.14
square meters. (The number 3.
(c)
8000 miles.
(c)
10
feet.
(c)
5
F.
:
8000 miles.).
32 F. the equivalent reading C on the Centigrade scale may be found by the formula
F
C
y
= f(F-32). the
3. If cated on the Fahrenheit scale. Find the area of a circle whose radius is
It
(b)
(a) 10 meters.
6
Find the volume of a sphere whose diameter equals:
(b)
3
feet.) Find the surface of a sphere whose diameter equals
(a)
7.
square units (square inches.
meters.14
4. then
=p
n
*
r
%>
or
Find by means
(a)
(b)
6.
then the
volume
V=
(a) 10 feet.
the fact that a loss of
loss of
+ $2.CHAPTER
II
ADDITION. in algebra this word includes also the results obtained by adding negative. Or in the symbols of algebra
$4) =
Similarly. AND PARENTHESES
ADDITION OF MONOMIALS
31. SUBTRACTION. however.
of
$6 and a gain
$4
equals a
$2 may be
represented thus
In a corresponding manner we have for a loss of $6 and a
of
loss
$4
(. but we cannot add a gain of $0 and a loss of $4. we call the aggregate value of a gain of 6 and a loss of 4 the sum of the two.$6) + (-
$4) = (-
$10). we define the sum of two numbers in such a way that these results become general.
. In algebra.
Since similar operations with different units always produce analogous results. or that
and
(+6) + (+4) = +
16
10.
While
in arithmetic the
word sum
refers only to
the
result obtained
by adding positive numbers. In arithmetic we add a gain of $ 6 and a gain of $ 4. Thus a gain of $ 2 is considered the sum of a gain of $ 6 and a loss of $ 4. or positive and negative numbers.
'. 23-26.
+ (-9).
5.
22. 12. find the numerical values of a + b
-f c-j-c?.
. the average of 4 and 8
The average The average
of 2. d = 0.
6
6
= 3. the one third their sum.
(_
In Exs. subtract their absolute values and
.
(-17)
15
+ (-14). 24. 5. c =
4.
19.
33.16
32.
(always) prefix the sign of the greater.
ELEMENTS OF ALGEBRA
These considerations lead to the following principle
:
If two numbers have the same sign. add their absolute values if they have opposite signs. if :
a
a
= 2. 10.
21. and the sum of the numbers divided by n.
23. 4.
l-f(-2).3.
d = 5.
of 2. = 5.
The average
of two
numbers
is
average of three numbers average of n numbers is the
is one half their sum.
-
0.
is 0.
of:
20. is 2.
EXERCISE
Find the sum
of:
10
Find the values
17.
Thus. c = = 5.
4
is
3 J. + -12.
18.
6..ADDITION. Find the average gain per year of a merchant.
33.
. \\ Add 2 a.
'
Find the average of the following
34.
which are not
similar.
AND PARENTHESES
d = l. -4. 12. $500 loss.
42. Find the average temperature of New York by taking the average of the following monthly averages 30. 7 yards. 7 a. 1.. -11 (Centigrade).
30. 37. 32. 74.
d=
3. 3.
:
Find the average temperature of Irkutsk by taking the average of the following monthly temperatures 12. 10. and $4500 gain. 10.
4
F.4.5.
&
28. 60.
29. $3000 gain. 6.
-'
1?
a
26.
:
48.
3 and 25.
27. . 5 and
12. 38. 32. and 4. 43.
and
-8
F.
31. 41. 09.
What number must be added to 9 to give 12? What number must be added to 12 to give 9 ? What number must be added to 3 to give 6 ? C* What number must be added to 3 to give 6? **j Add 2 yards. or
and
.. 7 a. . 10.
2.
^
'
37.
:
and
1. and 3 a.
.5. 55.
25. and 3 a.13.7. 34.
.
6. }/ Add 2 a.
36.
:
34.
Find the average of the following temperatures 27 F. $1000 loss. & = 15. affected
by the same exponents. .
40. if his yearly gain or loss during 6 years was $ 5000 gain.
= 22.7.
35. 72.
5 a2 &
6 ax^y and
7 ax'2 y. and 3 F. 0.
Similar or like terms are terms which have the same
literal factors. c=14.
Dissimilar or unlike terms are terms
4 a2 6c and o
4 a2 6c2 are dissimilar terms.
. SUBTRACTION. 66.
sets of
numbers:
13. and 3 yards.
.
= -23.
or 16
Va + b
and
2Vo"+~&.3. = -13. 39. $7000 gain. c = 0.
are similar terms.
2.
13.
5 a2
.
The sum The sum
of a of a
Dissimilar terms cannot be united into a single term.
-f
4 a2.
EXERCISE
Add:
1.
+ 6 af
.
Vm
-f.
12(a-f b)
12. or
a
6.
9(a-f-6).
ab
7
c
2
dn
6.
2
.
2 a&.
.
In algebra the word sum is used in a 36.
sum of two such terms can only be them with the -f.
12
13
b sx
xY xY 7 #y
7.
14
. Algebraic sum.
The sum
x 2 and
f
x2 .
2(a-f &).
b
a
-f (
6). The indicated by connecting and a 2 and
a
is
is
-f-
a2
. b wider sense than in arithmetic.
10.
1
\
-f-
7 a 2 frc
Find the sum of
9.18
35. While in arithmetic a denotes a difference only. and 4 ac2
is
a
2 a&
-|-
4 ac2.
-3a
.
:
2 a2.
13.ii.
11.
5Vm + w.
5l
3(a-f-6).13 rap
25 rap 2.
12
2 wp2 . 7 rap2.
11
-2 a +3a -4o
2. in algebra it may be considered b.sign.
12Vm-f-n. either the difference of a and b or the sum of a and
The sum
of
a.
ELEMENTS OF ALGEBRA
The sum
of 3
of
two similar terms
x2
is
is
another similar term.
the other number is required.
Ex.
may
be stated in a
:
5 take form e. from What 3. In addition.
3 gives
3)
The number which added
Hence. Subtraction is the inverse of addition.
AND PARENTHESES
23
subtraction of a negative
positive number.
7. SUBTRACTION.3.
(-
6)
-(-
= .
From
5 subtract
to
. The student should perform mentally the operation of chang8 2 6 from 6 a 2 fc. change the sign
of the subtrahend and
add. In subtraction.
1. the given number the subtrahend. ing the sign of the subtrahend thus to subtract 6 a 2 6 and 8 a 2 6 and find the sum of change mentally the sign of
.
State the other practical examples which show that the number is equal to the addition of a
40. Or in symbols. a-b =
x.
5
is
2.ADDITION.
41.
2. and the
required number the difference. two numbers are given.
NOTE.
6
-(-3) = 8. the algebraic sum and one of the two numbers is
The algebraic sum is given.2.
To
subtract. Therefore any example in subtraction
different
. called the minvend.
The
results of the preceding examples could be obtained
by the following
Principle.
+b
3. may be stated number added to 3 will give 5? To subtract from a the number b means to find the number which added to b gives a.
3 gives 5
is
evidently 8.g.
Ex. and their algebraic sum is required.
a.
3.
.
From
5 subtract
to
The number which added
Hence.
if
x
Ex.
From
5 subtract
+ 3.
This gives by the same method.
2 b . I.
6
o+(
a
+ c) = a =a 6 c) ( 4-. one occurring within the other.c.
a+(b-c) = a +b .
(b
c)
a
=a
6 4-
c.
tractions
By using the signs of aggregation.
II.
changed.6 b -f (. Ex. we may begin either at the innermost or outermost. A sign of aggregation preceded by the sign -f may be removed or inserted without changing the sign of any term.g.c.
66
2&-a + 6
4a
Answer. SUBTRACTION.
4a-{(7a + 6&)-[-6&-f(-2&.a~^~6)]} = 4 a -{7 a 6 b -[.& c
additions
and sub-
+ d) = a + b
c
+ d.
46. may be written as follows:
a
-f ( 4. If we wish to remove several signs of aggregation. Simplify 4 a f
+ 5&)-[-6& +(-25.a
-f-
= 4a
sss
7a
12
06
6.
45. the sign
is
understood. The beginner will find it most convenient
at every step to
remove only those parentheses which contain
(7 a
no others.ADDITION.
AND PARENTHESES
27
SIGNS OF AGGREGATION
43.
A
moved
w
may be resign of aggregation preceded by the sign inserted provided the sign of evei'y term inclosed is
E.
.
& -f
c.a^6)]
-
}
.b c = a a
&
-f-
-f.
If there is no sign before the first term within a paren*
-f-
thesis.
Hence
the
it is
sign
may
obvious that parentheses preceded by the -f or be removed or inserted according to the fol:
lowing principles
44.
7. The product of the sum and the difference
of
m and n.
6 diminished
.
2m-n + 2q-3t.
The product The product
m
and
n.
8.
m
x
2
4.
The
difference of a
and
6.
5.
z
+ d.
terms
5.
The sum
of the fourth powers of a of
and
6. )X
6.
p + q + r-s.
The sum
of tKe squares of a
and
b.
a-\-l>
>
c
+
d.
y
-f-
8
.ADDITION. SUBTRACTION.
II.
The The
difference of the cubes of
m
and
n.
5 a2
2. 9.
In each of the following expressions inclose the last three in a parenthesis preceded by the minus sign
:
-27i2 -3^ 2 + 4r/.
The square of the difference of a and b.
of the cubes of
m and
n.
4.
Three times the product of the squares of The cube of the product of m and n. The minuend is always the of the two numbers mentioned.
3.
2.
EXERCISES
IN"
ALGEBRAIC EXPRESSION
17
:
EXERCISE
Write the following expressions
I.7-fa.
12.
m and n.2 tf .
difference of the cubes of n and m.
4 xy
7 x* 4-9 x + 2.
5^2
_ r .
first.
and the subtrahend the second.
6.
Nine times the square of the sum of a and by the product of a and b.
7. .
10.4 y* .
3.
13.
EXERCISE
AND PARENTHESES
16
29
In each of the following expressions inclose the last three terms in a parenthesis
:
1.
'
NOTE.1.
The sum^)f
m
and
n.
d.30
14. The difference of the squares of two numbers divided by the difference of the numbers is equal to the sum of the two numbers. (Let a and b represent the numbers.
6 is equal to the square of
b.
and
c
divided by the
ference of a and
Write algebraically the following statements:
V 17.
difference of the cubes of a
and
b divided
by the
difference of
a and
6.)
. 6.
a plus the prod-
uct of a and
s
plus the square of
-19.
b.
18.
ELEMENTS OF ALGEBRA
The sum
x.
dif-
of the squares of
a and
b increased
by the
square root of
15.
x cube minus quantity 2 x2 minus 6 x plus
The sum
of the cubes of a. 16.
The sum
The
of a
and
b multiplied
b is equal to the difference of
by the difference of a and a 2 and b 2
.
2.
5. applied at let us indicate a downward pull at by a positive sign. is 5 x ( 3) ?
7. what force
31
is
produced by tak(
ing away 5 weights from
B ? What therefore is
5)
x(
3) ?
.
If the
two loads balance.CHAPTER
III
MULTIPLICATION
MULTIPLICATION OF ALGEBRAIC NUMBERS
EXERCISE
18
In the annexed diagram of a balance. weights at A ? Express this as a multibalance. If the two loads balance.
3. therefore. is
by
taking away 5 weights from
A?
5
X 3?
6.
two loads balance. weight at A ? What is the sign of a 3 Ib. what force is produced by the addition of 5 weights at B ? What.
A
A
A
1.
If the two loads
what
What. what force is produced by the Ib. and forces produced at by 3 Ib. weight at B ?
If the
addition of five 3
plication example.
4. let us consider the and JB.
force is produced
therefore. weights.
By what sign is an upward pull at A represented ? What is the sign of a 3 Ib.
5x(-4).
Thus.
9
9.
.
x
11. This definition has the additional advantage of leading to algenumbers which are identical with those for positive numbers.
examples were generally
method of the preceding what would be the values of
(
5x4. or
4x3 =
=
(_4) X
The preceding
3=(-4)+(-4)+(-4)=-12. make venient to accept the following definition
:
con-
49.4)-(-4) = +
12.
however. or plied by 3.
(-
9)
x (-
11) ?
State a rule by which the sign of the product of
two
fac-
tors can be obtained.9) x
11. 4 multi44-44-4 12.
Multiplication
by a negative
integer is a repeated
sub-
traction.
In multiplying integers we have therefore four cases
trated
illus-
by the following examples
:
4x3 = 4-12.4) x
braic laws for negative
~ 3> = -(. To take a number 7 times. 4 multiplied by 3.
the multiplier is a negative number.
(. a result that would not be obtained by other assumptions.32
8. thus. Multiplication by a positive integer is a repeated addition.
4x(-3)=-12.
NOTE. such as given in the preceding exercise. 9 x
(-
11).
(
(.
48.
ELEMENTS OF ALGEBRA
If
the signs obtained by the
true. times is just as meaningless as to fire a gun
tion
7
Consequently we have to define the meaning of a multiplicaif the multiplier is negative. Practical examples^
it
however. and we may choose any definition that does not lead to contradictions.
4
x(-8) = ~(4)-(4)-(4)=:-12.4)-(. becomes meaningless
if
definition. (-5)X4.
Multiply 2
+ a -a.1.
2. this method
tests only the values of the coefficients
and not the values of
the exponents. however.4.
a2
+ a8 + 3 .3 a 2 + a8
a
a = =-
I
1
=2
-f
2
a
4. To multiply two polynomials.
.M UL TIP LICA TION
37
58.
59.3 ab
2
2 a2
10 ab
-
13 ab
+ 15 6 2 + 15 6 2
Product. as illustrated in the following example
:
Ex.3 a 2 + a8 . Multiply 2 a .
Ex.3 a
3
2
by 2 a
:
a2 + l.2 a2 6 a8
2 a*
*
-
2"
a2
-7
60. 1 being the most convenient value to be substituted for all letters.a6 4 a 8 + 5 a* . the work becomes simpler and more symmetrical by arranging these expressions according to either ascending or descending powers. If the polynomials to be multiplied contain several powers of the same letter.
Check.
If
Arranging according to ascending powers
2
a
. Since errors. multiply each term of one by each term of the other and add the partial products thus formed. the student should
apply this test to every example.
The most convenient way of adding the partial products is to place similar terms in columns.a
.3
b
by a
5
b.a6
=2
by numerical
Examples
in multiplication can be checked
substitution.
2a-3b a-66 2 a . Since all powers of 1 are 1. are far more likely to occur in the coefficients than anywhere else.
sum of the
cross products. (100 + 3)(100 + 4).
6. ) (2
of a polynomial.
7.
3. 9. plus the product of the
EXERCISE
Multiply by inspection
1. (4s + y)(3-2y).
.
2
2
+ 2) (10 4-3).
65.
or
The student should note
minus
signs. the product of two binomials whose corresponding terms are similar is equal to the product of the first two
terms.
11.
7%e square of a polynomial is equal to the sum of the squares of each term increased by twice the product of each term with each that follows it. (3m + 2)(m-l).42
ELEMENTS OF ALGEBRA
of the result is obtained
product of 5 x
follows:
by adding the These products are frequently called the cross products.
(2a-3)(a + 2). 14. plus the
last terms.
2
10.
The square
2
(a 4.
8.
(5a-4)(4a-l).
(x
i-
5
2
ft
x 2 -3 6 s).
2
(2x y
(6
2
2
+ z )(ary + 2z ).
:
25
2.
5.
2
(2m-3)(3m + 2).
(5a6-4)(5a&-3).
that the square of each term is while the product of the terms may have plus always positive. 13.&
+ c) = a + tf + c
.
((5a?
(10
12.-f
2 a&
-f
2 ac
+ 2 &c.
4. and are represented as
2 y and 4y 3 x.
2 2 2 2 (2a 6 -7)(a & +
5).
The middle term
or
Wxy-12xy
Hence in general.
is the process of finding one of two factors and the other factor are given. The dividend is the product of the two factors, the divisor the given factor, and the quotient is the required factor.
67.
Division
if
their product
is
Thus
by
-f
to divide
12.
12
by
+
3,
we must find
is
the
;
number which
3 gives
But
this
number
4
hence
_
multiplied
12 r +3
=4.
68.
Since
-f
a
-
-f b
-fa
_a
and
it
-f-
a
= -f ab = ab b = ab b = ab,
b
-f-
follows that
4-a
=+b
ab
a
ab
a
69.
Hence the law
:
of signs
is
the same in division as in
multiplication
70.
Like signs produce plus, unlike signs minus.
Law
of
,
a8 -5- a5
=a
3
for a 3
It follows from the definition that Exponents. X a5 a8
=
.
Or
in general, if
greater than
m n, a
-f-
and n are positive integers, and m ~ n an = a m a" = a'"-", for a
<
m
m
is
45
46
ELEMENTS OF ALGEBRA
71. TJie exponent of a quotient of two powers with equal bases equals the exponent of the dividend diminished by the exponent
of the divisor.
DIVISION OF MONOMIALS
7 3 72. To divide 10x y z by number which multiplied by number is evidently
2x y
6
2
,
we have
z
to
find
the
2x*y
gives 10 x^ifz.
This
Therefore,
the quotient
*
,
= - 5 a*yz.
is
Hence,
sign,
of two monomials of their
part
coefficients,
is the
a monomial whose
coefficient is the quotient
preceded by the proper
literal
and whose
literal
found
in accordance with the
quotient of their law of exponents.
parts
73. In dividing a product of several factors by a number, only one of these factors is divided by that number. Thus (8 12 20)-?-4 equals 2 12 20, or 8 3 20 or 8 12 5.
-
-
.
-
.
-
.
EXERCISE
Perform the divisions indicated
'
:
28
'
2
.
76-H-15.
-39-*- 3.
2
15
3"
7
7'
3.
-4*
'
4.
5.
-j-2
12
.
4
2
9
5 11
68
3 19 -j-3
5
10.
(3
38
-
-2 4 )^(3 4 .2 2).
56
'
11.
3
(2
.3*.5 7 )-f-(
2
'
12
'
2V
14
36 a
'
13
''
y-ffl-g
35
-5.25
-12 a
2abc
15
-42^
'
-56aW
'
UafiV
DIVISION
lg
47
-^1^. 16 w
7
20>
7i
9
_Z^L4L.
22.
10 iy.
132 a V* 14 1
*
01
-240m
120m-
40
6c
fl
/5i.
3J)
c
23.
2 (15- 25. a ) -=- 5.
25. 26.
(18
(
.
5
.
2a )-f-9a.
2
24.
(7- 26 a
2
)
-f-
13.
DIVISION OF POLYNOMIALS BY MONOMIALS
To divide ax-}- fr.e-f ex by x we must find an expression which multiplied by x gives the product ax + bx -J- ex.
74.
But
TT
x(a
aa?
Hence
+ b e) ax + bx + ex. + bx -f ex = a 4- b +
-\.
,
.
c.
a?
To divide a polynomial by a monomial, cfc'wde each term of the dividend by the monomial and add the partial quotients thus
formed.
3 xyz
EXERCISE
Perform the operations indicated
1.
:
29
2.
5.
fl
o.
(5*
_5* + 52)
-5.
52
.
3.
97
.
(2
(G^-G^-G^-i-G
(11- 2
4.
(8- 3
+
11 -3
+ 11
-5)-*- 11.
18 aft- 27 oc
Q y.
9a
4
-25 -2 )^-2
<?
2
.
+8- 5 + 8-
7) -*-8.
5a5 +4as -2a
2
-a
-14gV+21gy
Itf
15 a*b
-
12
aW + 9 a
2
2
3a
48
,
ELEMENTS OF ALGEBRA
22
4,
m n - 33 m n
4
s
2
-f
55
mV
- 39 afyV + 26 arVz 3
- 49 aW + 28 a -W - 14 g 6 c
4 4
15. 16.
2 (115 afy -f 161 afy
- 69
4
2
a;
4
?/
3
- 23 ofy
3
4
)
-5-
23 x2y.
(52
afyV - 39
4
?/
oryz
- 65 zyz - 26 tf#z)
-5-
13 xyz.
-f-
,
17.
(85 tf
- 68 x + 51 afy - 34 xy* -f 1 7
a;/)
- 17
as.
DIVISION OF A POLYNOMIAL BY A POLYNOMIAL
75.
Let
it
be required to divide 25 a
- 12 -f 6 a - 20 a
3
2
by
2 a 2 -f 3 a, divide
4
a, or, arranging according to
2
descending powers of
6a3 -20a
-f
25a-12
2 by 2a -
The term containing the highest power of a in the dividend (i.e. a 8 ) is evidently the product of the terms containing respectively the highest power of a in the divisor and in the quotient.
Hence the term containing the highest power
of a in the quotient is
If
the product of 3 a and 2
2
4 a
+
3, i.e.
6 a3
12 a 2
-f
9 a, be sub-
8 a 2 -f 16 a tracted from the dividend, the remainder is 12. This remainder obviously must be the product of the divisor and the rest of the quotient. To obtain the other terms of the quotient we have
therefore to divide the remainder,
8 a2
-f-
16 a
12,
2 by 2 a
4 a
+
3.
consequently repeat the process. By dividing the highest term in the new dividend 8 a 2 by the highest term in the divisor 2 a 2 we obtain
,
We
4,
the next highest term in the quotient. 4 by the divisor 2 a2 4 a Multiplying
-I-
+ 3, we
obtain the product
8 a2
16 a
12,
which subtracted from the preceding dividend leaves
the required quotient.
no remainder. Hence 3 a
4
is
DIVISION
The work
is
49
:
usually arranged as follows
- 20 * 2 + 3 0a-- 12 a 2 +
a3
25 a
{)
-
12
I
2 a2 8 a
-
4 a 4
a
_
12
+3
I
-
8 a? 4- 16
a-
76. The method which was applied in the preceding example may be stated as follows 1. Arrange dividend and divisor according to ascending or
:
descending powers of a common letter. 2. Divide the first term of the dividend by the first term of the divisor, and write the result for the first term of the quotient.
3.
Multiply this term of the quotient by the whole divisor, and
subtract the result
4.
from
it
the dividend.
the same order as the given new dividend, and proceed as before.
Arrange
the
remainder in
as a
expression, consider
5.
until the highest poiver
Continue the process until a remainder zero is obtained, or of the letter according to which the dividend
is less
was arranged
the divisor.
than the highest poiver of the same
letter in
77.
Checks.
Numerical substitution constitutes a very con-
venient, but not absolutely reliable check. An absolute check consists in multiplying quotient and divisor. The result must equal the dividend if the division
was
exact, or the dividend diminished by the remainder division was not exact.
which
is
true for all values
a2 6 2 no matter what values we assign to a Thus.
81.
.
in
Thus x
12 satisfies the equation x
+
1
13.
the
80.
Thus.
hence
it
is
an equation
of
condition. y y or z) from its relation to
63
An
known numbers. (a + ft) (a b) and b.
The
first
member
or left side of an equation
is
that part
The secof the equation which precedes the sign of equality.
A set of numbers which when substituted for the letters an equation produce equal values of the two members. in the equation 2 x 0.
the
first
member
is
2 x
+
4.
ber
equation is employed to discover an unknown num(frequently denoted by x. An identity is an equation of the letters involved. second member is x
+
4
x
9.
y
=
7 satisfy
the equation x
y
=
13.
=11.
x
20. An equation of condition is usually called an equation.
82.
.r
-f9
= 20
is
true only
when
a. is said to satisfy an equation. ond member or right side is that part which follows the sign of
equality. An equation of condition is an equation which is true only for certain values of the letters involved.
83.CHAPTER V
LINEAR EQUATIONS AND PROBLEMS
79.
.
(rt+6)(a-ft)
=
2
-
b'
2
. The sign of identity sometimes used is = thus we may write
.
2.
Like powers or
like roots
of equals are equal. called axioms
1. the quotients are equal.
Axiom
4
is
not true
if
0x4
= 0x5. the products are equal.
89. A
2
a.
3.
90.
one member to another by changing
x + a=.
If equals be added
to equals.
NOTE.
fol-
A
linear equation is also called a simple equation.
2
= 6#-f7.
E.g.
5.
expressed in arithmetical numbers
literal
is
as (7
equation
is
one in which at least one of the
known
quantities as x -f a letters
88. 87.
86.
a.
.
A
term may be transposed from
its sign.
ELEMENTS OF ALGEBRA
If
value of the
an equation contains only one unknown quantity.
. the
sums are
equal.b.
A numerical
equation is one in
which
all
.
To
solve
an equation
to find its roots.
4.
Transposition of terms.
9
is
a root of the equation 2 y
+2=
is
20.
(Axiom
2)
the term a has been transposed from the left to thQ
right
member by changing
its
sign. Consider the equation b Subtracting a from both members.2.
= bx
expressed by a letter or a combination of
c. x
I.
85.
but 4 does not equal
5.
A
linear equation or
which when reduced
first
to its simplest
an equation of the first degree is one form contains only the
as
9ie
power of the unknown quantity.
the
known quan
x) (x -f 4)
tities are
=
.
If equals be multiplied by equals. the remainders are
equal.e.
If equals be subtracted from equals.
the divisor equals zero.
The process
of solving equations depends upon the
:
lowing principles.54
84.
If equals be divided by equals. an^ unknown quantity which satisfies the equation is
a root of the equation.
33
2.
a.
$> 100 yards cost one hundred dollars.
1.
Ex. and the smaller one
parts.
14.
find the cost of one yard.
one part equals
is 10. so that one part
The
difference between
is s. 11.
EXERCISE
1. Divide 100 into two
12. is d.
13.58
Ex.
10.
3. greater one is g. or 12 7. one yard will cost
100
-dollars.
If 7
2.
4.
6. so that one part Divide a into two parts.
What number divided by 3 will give the quotient a? ? What is the dividend if the divisor is 7 and the quotient
?
.
is
a?
2
is
c?.
9.
Divide a into two parts.
smaller one
16. Hence 6 a must be added
to a to give
5.
Find the greater one. 15.
By how much does a exceed 10 ? By how much does 9 exceed x ? What number exceeds a by 4 ? What number exceeds m by n ? What is the 5th part of n ? What is the nth part of x ? By how much does 10 exceed the third part of a? By how much does the fourth part of x exceed b ? By how much does the double of b exceed one half Two numbers differ by 7. The difference between two numbers Find the smaller one.
5. one yard will cost -
Hence
if
x
-f
y yards cost $ 100.
7.
x
-f-
y yards cost $ 100
.
two numbers
and the and the
2
Find the greater one. so that
of c ?
is
p.
17.
ELEMENTS OF ALGEBRA
What must
be added to a to produce a sum b ?
:
Consider the arithmetical question duce the sum of 12 ?
What must
be added to 7 to pro-
The answer is 5. 6. is b.
feet wider than the one
mentioned in Ex.
How many
cents
has he ?
27. 28.
How many
cents had he left ?
28.
What What What What
is
the cost of 10 apples at x cents each ?
is
is is
x apples cost 20 cents ? the price of 12 apples if x apples cost 20 cents ? the price of 3 apples if x apples cost n cents ?
the cost of 1 apple
if
. find the
has ra dollars.
numbers
is x. Find the sum of their ages
5 years ago.
26. find the of their ages 6 years hence.
32.
59
What must
The
be subtracted from 2 b to give a?
is a. and B's age is y years. The greatest of three consecutive the other two.
20.
A man
had a
dollars.
?/
31.
24. amount each will then have. 28.
and spent
5 cents.
How many
cents are in d dollars ? in x dimes ?
A has
a
dollars. A room is x feet long and y feet wide.
If
B
gave
A
6
25. 33. rectangular field is x feet long and the length of a fence surrounding the field. b dimes. and B has n dollars.
smallest of three consecutive numbers
Find
the other two.
Find
35. square feet are there in the area of the floor ?
How many
2 feet longer
29.
Find
21.
and
c cents.
and 4
floor of a room that is 3 feet shorter wider than the one mentioned in Ex.
Find the area of the Find the area of the
feet
floor of
a room that
is
and 3
30.
y years
How
old was he 5 years ago ?
How
old will he be 10 years hence ?
23.
is
A A
is
# years
old.
and
B
is
y years old.
A
feet wide.
22.
19.
How many years
A
older than
is
B?
old.
sum
If A's age is x years.
A
dollars.LINEAR EQUATIONS AND PROBLEMS
18.
34.
" we have to consider that in this by statement "exceeds" means minus ( ). If a man walks n miles in 4 hours. find the fraction.
miles does
will
If a man walks r miles per hour.
m is the
denominator.
If a
man walks
?
r miles per hour. he walk each hour ?
39.
How many
x years ago
miles does a train
move
in
t
hours at the
rate of x miles per hour ?
41.
a.
per
Find 5 Find 6
45.
of m.60
ELEMENTS OF ALGEBRA
wil\
36.
A
cistern
is
filled
43.
Find x
% %
of 1000. If a man walks 3 miles per hour.
% % %
of 100
of
x. in how many hours he walk n miles ?
40.
.50.
Find
a.
Find the
number.
A
was 20 years
old.
The two
digits of a
number
are x and
y.
of 4.
-46. and the second pipe alone fills it in
filled
y minutes. Find a
47. and "by as much as" Hence we have means equals (=)
95.
c
a
b
=
-
9. The first pipe x minutes.
48.
49. What fraction of the cistern will be filled by one pipe in one minute ?
42.
b
To express in algebraic symbols the sentence: " a exceeds much as b exceeds 9.
as
a exceeds
b
by as much
as c exceeds 9. how many miles he walk in n hours ?
37. What fraction of the cistern will be second by the two pipes together ?
44.
The numerator
If
of a fraction exceeds the denominator
by
3.
-.
How
old
is
he
now ?
by a pipe in x minutes.
how many
how many
miles will
he walk in n hours
38.
A
cistern can be filled
in
alone
fills it
by two pipes.
5.
9.
c.
8
-b ) + 80 = a
.
The double
as
7.
The product
of the
is
diminished by 90 b divided by 7. Four times the difference of a and b exceeds c by as
d exceeds
9.
In
many
word
There are usually several different ways of expressing a symbolical statement in words.
The
excess of a over b
is c.
c.
of x increased by 10 equals
x.
equal to the
sum and the difference of a and b sum of the squares of a and
gives the
Twenty subtracted from 2 a
a.
=
2
2
a3
(a
-
80.
EXERCISE
The The double The sum
One
34
:
Express the following sentences as equations
1.
same
result as 7
subtracted from
.
double of a
is
10. 2.
by one third of b equals 100. etc.
a exceeds b by c.
80.
4. a is greater than b by b is smaller than a by
c.LINEAR EQUATIONS AND PROBLEMS
Similarly.
cases it is possible to translate a sentence word by in algebraic symbols in other cases the sentence has to be changed to obtain the symbols.
of a and 10 equals 2
c.
of a increased
much
8.
3.
6.
-80.
third of x equals
difference of x
The
and y increased by 7 equals
a. thus:
a
b
= c may
be expressed as follows
difference between a
:
The
and
b is
c. the difference of the squares of a
61
and
b increased
-}-
a2
i<5
-
b'
2
'
by 80 equals the excess of a over
80
Or.
m is x %
of n.
first
00
x % of the
equals one tenth of the third sum. B's. (d) In 10 years A will be n years old. they have equal
of A's. pays to C $100. 14.
a.
->. of 30 dollars.
amounts. the
sum
and C's
money
(d)
(e)
will be $ 12.
12. a third sum of 2 x + 1 dollars.*(/)
(g)
(Ji)
Three years ago the sum of A's and B's ages was 50..
5x
A sum of money consists of x dollars.
x
4-
If A. A is 4 years older than
Five years ago A was x years old. and
(a)
(6)
A
If
has $ 5 more than B. the first sum exceeds b % of the second sum by
first
(e)
%
of the first plus 5
%
of the second plus 6
%
of the
third
sum
equals $8000. B.
(a)
(b)
(c)
A is twice as old as B. 3 1200 dollars.
A
gains
$20 and B
loses
$40. and C's ages will be 100. B's. In 10 years the sum of A's.
11.000.
and C's age
4
a. sum equals $20. B's age
20.62
10.
symbols
B. express in algebraic
3x
:
10. the first sum equals 6 % of the third sura.
is
If A's age is 2 x.
A
If
and
B
B together have $ 200 less than C.
18. and C have respectively 2 a.
as 17
is
is
above
a. (e) In 3 years A will be as old as B is now. In 3 years A will be twice as old as B.
. they have equal amounts.
x
is
100
x%
is
of 700.
(c)
If each
man
gains $500. express in algebraic symbols
:
-700.
6
%
of m.
a. Express as
:
equations of the (a) 5
(b)
(c)
% a%
of the second
(d)
x c of / a % of
4
sum equals $ 90.
17. 16.
ELEMENTS OF ALGEBRA
Nine
is
as
much below a
13.
#is5%of450. a second sum.
50
is
x % of
15.
denote the unknown
96.
Transposing. x + 15 = 3 x
3x 16
15. etc. x = 20.
Simplifying. 2. = x x
3x
-40
3x
40-
Or.
A
will
Check.
x+16 = 3(3-5). 15.
Let x
The
(2)
= A's present age.
but
30
=3
x
years. the
. In order to solve them.
Dividing.
Let x = the number.
Three times a certain number exceeds 40 by as Find the number. exceeds 40 by as much as 40 exceeds the no. The solution of the equation (jives the value of the unknown number.
3
x
+
16
=
x
x
(x
-
p)
Or. x= 15. 4 x = 80.
Uniting.
by 20
40 exceeds 20 by 20.
much
as 40 exceeds the number.LINEAR EQUATIONS AND PROBLEMS
63
PROBLEMS LEADING TO SIMPLE EQUATIONS
The simplest kind of problems contain only one unknown number.
Ex. be three times as old as he was 5 years ago. In 15 years A will be three times as old as he was 5 years ago.
Uniting.
Transposing.
6 years ago he was 10
.
-23 =-30.
1. Three times a certain no.
NOTE.
3z-40:r:40-z.
be 30
. verbal statement (1)
(1) In 15 years
A
will
may be expressed in symbols (2). The equation can frequently be written by translating the sentence word by word into algebraic symbols in fact.
Ex. Write the sentence in algebraic symbols. The student should note that x stands for the number of and similarly in other examples for number of dollars.
. the required
.
number. Find A's present age.
In 15 years
10.
Check.
number by x (or another letter) and express the yiven sentence as an equation.
equation is the sentence written in alyebraic shorthand. number of
yards.
3 x or 60 exceeds 40
+ x = 40 + 40.
47 diminished by three times a certain number equals 2. How long is the Suez
Canal?
10. Find the number. 14.
4.
300
56. then the
problem expressed in symbols
W
or.
What number
7
%
of
350?
Ten times the width of the Brooklyn Bridge exceeds 800 ft. % of
120.64
Ex.
Forty years hence
his present age.
.
Hence
40
= 46f. 11.
35
What number added
to twice itself gives a
sum
of
39?
44. How many miles per hour does it run ?
.
Find the width of the Brooklyn Bridge.
13.
Uldbe
66
| x x
5(5 is
=
-*-.
How
old
is
man will be he now ?
twice as old as he was
9.
A will
be three times as old as to-da3r
. Four times the length of the Suez Canal exceeds 180 miles by twice the length of the canal.
120.
Find the number whose double exceeds 30 by as much
as 24 exceeds the number.
Find the number.2.
Find the number whose double increased by 14 equals Find the number whose double exceeds 40 by 10.
twice the number plus
7. A train moving at uniform rate runs in 5 hours 90 miles more than in 2 hours. exceeds the width of the bridge.
5. by as much as 135 ft.
to
42 gives a
sum
equal to 7 times the
original
6.
ELEMENTS OF ALGEBRA
56
is
what per cent
of 120 ?
= number
of per cent.
Let x
3.
Dividing.
3.
A
number added
number.
14 50
is
is
4
what per cent of 500 ? % of what number?
is
12.
Find
8.
EXERCISE
1.
Six years hence a
12 years ago.
while in the more complex probWe denote one of the unknown
x. and
as
15. written in algebraic
symbols.
A
and
B
have equal amounts of money. Vermont's population increased by 180. and
B
has $00.
Ex. Ill the simpler examples these two
lems they are only implied.
97.
how many
acres did he wish to
buy
?
19.
How many dollars
must
?
B
give to
18.
14.000.
make A's money
equal to 4 times B's
money
wishes to purchase a farm containing a certain He found one farm which contained 30 acres too many. If the first farm contained twice as many acres as
A man
number
of acres. During the following 90 years. If A gains A have three times as much
16.
the second one.
.
is
the equation. The other verbal statement.
F
8.
65
A
and
B
$200. In 1800 the population of Maine equaled that of Vermont.
numbers (usually the smaller one) by
and use one of the
given verbal statements to express the other unknown number in terms of x. Find
the population of Maine in 1800.
How many
dol-
A has
A
to
$40. One number exceeds the other one by II.
The problem consists of two statements I. Maine's population increased by 510. and another which lacked 25 acres of the required number.
statements are given directly. two verbal statements must be given.
x. then dollars has each ? many
have equal amounts of money. B will have lars has A now?
17.
One number exceeds another by
:
and their sum
is
Find the numbers.
five
If
A
gives
B
$200. The sum of the two numbers is 14.
times as
much
as A. which gives the value of
8. and Maine had then twice as many inhabitants as Vermont.
B
How
will loses $100.
1.LINEAR EQUATIONS AND PROBLEMS
15. If a problem contains two unknown quantities.000.
. B will have twice as many as A. 8 = 11.
If
A gives
are
:
A
If
II. I. = 14.
in algebraic
-i
symbols produces
#4a.
Dividing. unknown quantity
in
Then.66
ELEMENTS OF ALGEBRA
Either statement may be used to express one unknown number in terms of the other.
has three times as many marbles as B. 25 marbles to B.
o\
(o?-f 8)
Simplifying.
x
= 8. the greater number. 26 = A's number of marbles after the exchange. 2. the smaller number.
26
= B's number of marbles after the exchange.
Uniting. consider that by the
exchange
Hence. 8 the greater number. the sum of the two numbers is 14.
x
x =14
8. = A's number of marbles.
Statement
x
in
=
the larger number. = 3.
+
a-
-f
-f
8
= 14.
Another method for solving this problem is to express one unknown quantity in terms of the other by means of statement II viz. A gives B 25 marbles.
A will lose.
<
Transposing. although in general the simpler one should be selected. to
Use the simpler statement. B will have twice as
viz.
If
we
select the first one. A has three times as many marbles as B.
Let
x
14
I
the smaller number.
.
The two statements
I. Let
x
3x
express one
many as A.
which leads
ot
Ex.
Then.= The second statement written
the equation ^
smaller number.
and
Let x
= the
Then x -+.
/
.
.
To
express statement II in algebraic symbols.
expressed symbols is (14 x) course to the same answer as the first method.
2x
a?
x
-j-
= 6.
x
3x
4-
and
B
will gain.
= B's number of marbles.
terms of the other.
A's number of marbles.
1.
greater
is
.
x
from
I. have a value of $3. x = 15. Eleven coins.10. 6 half dollars = 260 cents.
*
'
. consisting of half dollars and dimes.
Uniting. 11 x = 5.240. of dollars to the number of cents.
Simplifying..
Let
11
= the number of dimes. but 40 = 2 x 20. 15 + 25 = 40.. B's number of marbles.
we
express the statement II in algebraic symbols.
The numbers which appear
in the equation should
always
be expressed in the same denomination.$3.75.
50 x
Transposing. their sum
+ +
10 x 10 x
is
EXERCISE
36
is five
v
v. the number of half dollars.5 x . .
The sum of two numbers is 42. then. How many are there of each ?
The two statements are I. the number of dimes.
60. and the Find the numbers. etc.
by 44. 3 x = 45. x = 6.10. 3..
Find
the numbers.
.25 = 20.
is 70.10.
w'3.
67
x
-f
25 25
Transposing.
(Statement II)
Qx
.
Two numbers
the smaller.
6 times the smaller.
2.
Dividing.
Selecting the cent as the denomination (in order to avoid fractions).
dollars
and dimes
is
$3. The value of the half
:
is 11. Dividing.
Never add the number number of yards to their
Ex.
50.
Uniting.
differ
differ
and the greater and their sum
times
Two numbers
by
60.
6 dimes
= 60
= 310.
Simplifying. The number of coins II.
Check. the
price.
50(11 660 50 x
-)+ 10 x = 310. Find the numbers. Check. x = the number of half dollars.550 -f 310.
*
98. cents.
x x
+
= 2(3 x = 6x
25
25).LINEAR EQUATIONS AND PROBLEMS
Therefore. 40 x . 45 .
one of which increased by
9.
A's age is four times B's.
it
If the smaller
one contained 11 pints more. Mount Everest is 9000 feet higher than Mt. the larger part exceeds five times the smaller part by 15 inches.
Two numbers
The number
differ
by
39. and the
greater increased by five times the smaller equals 22.
tnree times the smaller by 65.
How many
14 years older than B. Twice 14. McKinley.000
feet.
as the larger one.
5.
United
States.
?
Two
vessels contain together 9 pints.
would contain three times as
pints does each contain ?
much
13.
Find
Find two consecutive numbers whose sum equals 157. McKinley exceeds the altitude of
Mt. the night in
Copenhagen
lasts 10 hours
longer than the day.
and twice the greater exceeds Find the numbers.
cubic foot of iron weighs three times as much as a If 4 cubic feet of aluminum and
Ibs.
2 cubic feet of iron weigh 1600 foot of each substance.
7.
3 shall be equal to the other increased by
10.68
4.
and in Mexico
?
A
cubic foot of aluminum. How many inches are in each part ?
15. the number.
of volcanoes in
Mexico exceeds the number
of volcanoes in the United States by 2. Everest by 11. and in 5 years A's age will be three times B's.
6. and four times the former equals five times the latter. Find their ages. and twice the altitude of Mt.
ELEMENTS OF ALGEBRA
One number
is
six
times another number.
How many
hours does the day last ?
.
On December
21. How many volcanoes are
in the
8. and B's age is as below 30 as A's age is above 40.
9.
What
is
the altitude of each
mountain
12.
11.
find the
weight of a cubic
Divide 20 into two parts. What are their ages ?
is
A A
much
line 60 inches long is divided into two parts..
then three times the sum of A's and B's money would exceed C's money by as much as A had originally.
5
5
Expressing in symbols Three times the sum of A's and B's money exceeds C's money by A's 3 x ( x _5 + 3z-5) (90-4z) = x. = number of dollars B had after giving $5. or 66 exceeds 58 by 8. 4 x = number of dollars C had after receiving $10. and 68. let us consider the words ** if A and B each gave $ 5 to C.
sum of A's and B's money would exceed much as A had originally. three One of the unknown num-
two are expressed in terms by means of two of the verbal statements. B has three times as much as A. number of dollars A had.
the
the
number
of dollars of dollars of dollars
A
B
C
has. and C together have $80. III.
If
4x
= 24.
A
and B each gave $ 5
respectively."
To
x
8x
90
= number of dollars A had after giving $5.
1. = 48. II. they would have 3. B. try to obtain
it
by a
series of successive steps. and the other
of x
problem contains three unknown quantities.
Let
x
II. x = 8.
has.
.
times as
much
as
A.
I. and B has three as A. B. Tf it should be difficult to express the selected verbal state-
ment
directly in algebraical symbols. The solution gives
:
3x
80
Check.LINEAR EQUATIONS AND PROBLEMS
99.
number
of dollars of dollars
B
C
had.
number
had.
are
:
C's
The three statements
A.
69
If a
verbal statements must be given.
first
According to
3 x
number
number
and according
to
80
4
x
=
the
express statement III by algebraical symbols.
8(8
+ 19)
to C. If A and B each gave $5 to C. bers is denoted by x. 19. If A and B each gave $5 to C. and C together have $80. The third
verbal statement produces the equation.
Ex. then
three times the
money by
I.
has. original amount.
x 35
-f
+
=
+
EXERCISE
1. The number of sheep is equal to twice tho number of horses and
x 4
the
cows together. x = 5.
85 (x 15 (4 x
I
+ 4)
+
8)
= the number of sheep. 90
may
be written. and 28 sheep would cost 6 x 90 -f 9 + 316 420 = 1185. and each sheep $ 15. + 35 x 4. number of cows. 90 x -f 35 x + GO x = 140 20 + 1185. each horse costing $ 90.
1 1
Check. 28 2 (9 5).
first. number of sheep. 2. the third five times the first. 2 (2 x -f 4) or 4 x
Therefore.
37
Find three numbers such that the second is twice the first.
III. = the number of dollars spent for sheep
Hence statement
90 x
Simplifying.
Uniting. Let
then. 9 cows. The number of cows exceeds the number of horses by 4.
+ 35 (x +-4) -f 15(4z-f 8) = 1185. cows. and the sum of the
.
The
I.
and. sheep.
28 x 15 or 450
5 horses.140 + (50 x x 120 = 185.
first
the third exceeds the second by and third is 20.
x
-j-
= the
number of horses.
The total cost equals $1185. number of horses. 9 -5 = 4 . according to II.
three statements are
:
IT. number of cows. and Ex. = the number of dollars spent for horses. according to III.
x
Transposing.
+
8
90 x
and.70
ELEMENTS OF ALGEBRA
man spent $1185 in buying horses. x -f 4 = 9.
Dividing. The number of cows exceeded the number of horses by
4. 4 x -f 8 = 28.
Find three numbers such that the second is twice the 2. 185 a = 925. = the number of dollars spent for cows. each cow $ 35.
A
and the number of sheep was twice as large as the number How many animals of each kind did he buy ?
of horses and cows together. and the difference between the third and the second is 15
2.
If the second angle of a triangle is 20 larger than the and the third is 20 more than the sum of the second and
first.
v
-
Divide 25 into three parts such that the second part first.
A
is
Five years ago the What are their ages ?
C. the copper.000. In a room there were three times as many children as If the number of women.
twice as old as B.
Find three consecutive numbers whose sum equals
63. equals 49 inches. If twice
The sum
the third side.
is five
numbers such that the sum of the first two times the first.
The
three angles of any triangle are together equal to
180. and is 5 years younger than sum of B's and C's ages was 25 years. The gold.
A
12.
9.
the
first
Find three consecutive numbers such that the sum of and twice the last equals 22.000 more inhabitants than Philaand Berlin has 1. and the third exceeds the
is
second by
5.
-
4.
what are the three angles ?
10. and the third part exceeds the second by 10.LINEAR EQUATIONS AND PROBLEMS
3.
13. the second one is one inch longer than the first.000 more than Philadelphia (Census 1905). increased by three times the second side.
twice the
6. and 2 more men than women. and children together was 37. women. and the sum of the first and third is 36.
New York
delphia. and the pig iron produced in one year (1906) in the United States represented together a value
. what is the population of each city ?
8.
the third
2.
7. how many children
were present ?
x
11.
71
the
Find three numbers such that the second is 4 less than the third is three times the second. men.000. what is the length of each?
has 3. If the population of New York is twice that of Berlin.
first.
v
.
first.
"Find three
is 4.
and
of the three sides of a triangle is 28 inches.
e. of 3 or 4 different kinds. After how many hours will they meet and how
E.
The copper had twice
the value of the gold.
and Massachusetts has one more than California and Colorado If the three states together have 31 electoral votes. and 4 (x But the 2) for the last column. and distance. but stops 2 hours on the way. then x 2 = number of hours B walks.
Find the value of each.72
of
ELEMENTS OF ALGEBRA
$ 750.
of
arid the value of the iron
was $300. 8 x = 15. we obtain 3 a.000 more than that
the copper.
has each state
?
If the example contains Arrangement of Problems. width. First fill in all the numbers given directly.
A and B
apart.
how many
100.
14. speed. 3 and 4. or time. and quantities
area.g. number of hours.
California has twice as
many
electoral votes as Colorado.
Hence
Simplifying.
.
3z + 4a:-8 = 27. and A walks at the rate of 3 miles per hour without stopping.000.
start at the same hour from two towns 27 miles walks at the rate of 4 miles per hour. it is frequently advantageous to arrange the quantities in a systematic manner. statement "A and B walk from two towns 27 miles apart until they meet " means the sum of the distances walked by A and B equals 27 miles. i.
B
many
miles does
A
walk
?
Explanation.000.
= 35. = 5. such as length. number of miles A
x
x
walks.
3x
+
4 (x
2)
=
27.
Dividing. together. Since in uniform motion the distance is always the product of
rate
and time.
7
Uniting.000. Let x = number of hours A walks.
but four men failed to pay their shares. How many pounds of each kind did he buy ?
8. twice as large.
Twenty men subscribed equal amounts
of
to raise a certain
money. The second is 5 yards longer than the first. were increased by 3 yards.
1.
as a
4. and follows on horseback traveling at the rate of 5 miles per hour. but as two of them were unable to pay their share.
sum $ 50
larger invested at 4
brings the same interest Find the first sum.74
ELEMENTS OF ALGEBRA
EXERCISE
38
rectangular field is 10 yards and another 12 yards wide.
A
If its length
rectangular field is 2 yards longer than it is wide. and the cost
of silk
of the auto-
and 30 yards of cloth cost together much per yard as the cloth.
A
of each. and a second sum.55.
3. How much did each man subscribe ?
sum
walking at the rate of 3 miles per hour. of coffee for $ 1. and the sum Find the length of their areas is equal to 390 square yards. how much did each cost per yard ?
6.
A sum
?
invested at 4 %.
sions of the field. and how far will each then have traveled ?
9.
invested at 5 %.
2. and in order to raise the required sum each of the remaining men had to pay one dollar more. together bring $ 78 interest. each of the others had to pay
$ 100 more.
Ten yards
$
42. paid 24 ^ per pound and for the rest he paid 35 ^ per pound. After how many hours will B overtake A.
Find the share of each.
Six persons bought an automobile. A man bought 6 Ibs.
Find the dimen-
A
certain
sum
invested at 5
%
%.
mobile. What are the
two sums
5. the area would remain the same. and its width decreased
by 2 yards.
If the silk cost three times as
For a part he 7.
A
sets out later
two hours
B
.
and another train starts at the same time from New York traveling at the rate of 41 miles an hour.LINEAR EQUATIONS AND PROBLEMS
v
75
10.
The
distance from
If a train starts at
. but
A has
a start of 2 miles. After how many hours.
walking at the same time in the same If A walks at the rate
of 2
far
miles per hour. and from the same point.will they be 36 miles apart ?
11.
A
and
B
set out
direction. traveling by coach in the opposite direction at the rate of 6 miles per hour. how many miles from New York will they meet?
X
12.
A
sets out
two hours
later
B
starts
New York to Albany is 142 miles. and B at the rate of 3 miles per hour. how must B walk before he overtakes A ?
walking at the rate of 3 miles per hour. Albany and travels toward New York at the rate of 30 miles per hour without stopping.
An
expression
is integral
and rational with respect
and rational. a. if it is integral to all letters contained in it.
it is
composite.
\-
V&
is
a
rational with respect to
and
irrational with respect
102.
this letter.
-f-
db
6
to b.
expression is rational with respect to a letter. if it does contain
some indicated root of
.
a factor of a 2
A
factor is said to be prime.
we
shall not.
76
. at this
6
2
.
An
after simplifying. An expression is integral with respect to a letter. as.
a-
+
2 ab
+ 4 c2
. but fractional with respect
103.CHAPTER
VI
FACTORING
101.
a2
to 6.
The
factors of
an algebraic expression are the quantities
will give the expression. if this letter does not occur in any denominator.
104.
+ 62
is
integral with respect to a.
irrational.
The prime
factors of 10 a*b are 2.
J Although Va'
In the present chapter only integral and rational expressions
b~
X
V
<2
Ir
a2
b'
2
2
?>
.
vV
. 6. it contains no indicated root of this letter
. 5.
a. consider
105.
which multiplied together
are considered factors. if.
stage of the work. if it contains
no other
factors (except itself
and unity)
otherwise
.
. 2.62 + &)(a 2
.
Ex.
Divide
6
a% .
factors of 12
&V
is
are 3.
The factors
of a
monomial can be obtained by inspection
2
The prime
108.
It (a.
2. 8) (s-1).
?/.)
Ex.62
can be
&).
x.3 sy + 4 y8).
77
Factoring
is
into its factors.
107.
in the
form
4)
+3.
An
the process of separating an expression expression is factored if written in the
form of a product.9 x2 y 8 + 12
3 xy
-f
by
3
xy\
and the quotient
But.
Hence
6 aty 2
= divisor x quotient.
y.9 x2^ + 12 sy* = 3 Z2/2 (2 #2 . or that a
=
6)
(a
= a .
.
Factor
14 a*
W-
21 a 2 6 4 c2
+ 7 a2 6
2
c2
7
a2 6 2 c 2 (2 a 2
.
109. for this result is a sum.
POLYNOMIALS ALL OF WHOSE TERMS CONTAIN A COMMON FACTOR
(
mx + my+ mz~m(x+y + z). since (a + 6) (a 2 IP factored.
Since factoring
the inverse of multiplication.
it
follows
that a 2
.
E.FACTORING
106.
or
Factoring examples may be checked by multiplication by numerical substitution.
it fol-
lows that every method of multiplication will produce a method
of factoring.
110. dividend
is
2 x2
4
2
1/
.g.
2 4 x + 3) is factored if written (x' would not be factored if written x(x and not a product.
TYPE
I. x.
01.3 6a + 1).
55.9 x if + 12 xy\
2
The
greatest factor
common
2
to all terms
flcy*
is
8
2
xy'
. 2.
1.
Factor G ofy 2
.
however.6 = 20. can be factored. Hence fc -f 10 ax
is
10 a are 11 a
-
12 /.5) (a 6). it is advisable to consider the factors of q first.4 .G) = .
We may consider
1. If q is negative.
.
m -5m + 6.4 x . + 30 = 20. the two numbers
have opposite
signs.
tfa2 -
3.
+
112. If q is positive.
3.1 afy 8 The two numbers whose product is equal to 12 yp and whose sum equals 3 8 7 y are -4 y* and -3 y*. of this type.
is
The two numbers whose product and -6. the student should first all terms contain a common monomial factor.
4. or 11 and 7 have a sum equal to 4. Factor x? .5) (a .
5. the two numbers have both the same sign as p. and (a .
79
Factor a2
-4 x .
and the greater one has the same sign
Not every trinomial
Ex.
Ex.
Factor a2
.
11 a2 and whose sum The numbers whose product is and a.
Therefore
Check.11 a
2
.
or
77
l.77 =
(a.
Factor
+ 10 ax .
as p.
determine whether
In solving any factoring example.
Ex. or 7 11.a).
If
30 and whose
sum
is
11 are
5
a2
11
a = 1.30 = (a .
77 as the product of 1 77. but only in a limited number
of ways as a product of two numbers.
but of these only
a:
Hence
2
. a 2 .1 1 a
tf
a 4. 2 11 a?=(x + 11 a) (a.11 a + 30.
.11.
EXERCISE
Besolve into prime factors :
40
4.FACTORING
Ex.
11
7.
Since a number can be represented in an infinite number of ways as the sum of two numbers.11) (a
+
7)..
.
2.
2
6. Hence z6 -? oty+12 if= (x -3 y)(x*-4 y ).
or
G
114. Since the first term of the first factor (3 x) contains a 3.FACTORING
If
81
we consider that the
factors of -f 5
as
must have
is
:
like signs. the signs of the second terms are minus.
sible
13 x
negative. 9 x 6.
The
and
factors of the first term consist of one pair only. 27 x 2.
11 x
2x. Hence only 1 x 54 and 2 x 27 need
be considered.17 x
2o?-l
V A
5
-
13 a
combination
the correct one.5) (2 x . X x 18. 2. 64 may be considered the
:
product of the following combinations of numbers 1 x 54. If py? -\-qx-\-r does not contain any monomial factor. 6 x 9.
the
If p and r are positive.83 x
-f-
54.
.
all
it is not always necessary to write down combinations. exchange the
signs of the second terms of the factors. which has the same absolute value as the term qx. we have to reject every combination of factors of 54 whose first factor contains a 3.
Ex. 3 x and x. viz. 2 x 27.31 x
Evidently the
last
2
V A
6. the second terms of the factors have same sign as q. If p is poxiliw.1). all pos-
combinations are contained in the following
6x-l
x-5 .e-5
V A
x-1
3xl \/ /\
is
3
a.13 x + 5 = (3 x .
and that they must be negative.
3. 54 x 1.
If
the factors
a combination should give a sum of cross products. but the opposite sign. then the second terms of
have opposite signs. and after a little practice the student possible should be able to find the proper factors of simple trinomials
In actual work
at the first trial. none of the binomial factors can contain a monomial factor.
.
The work may be shortened by the
:
follow-
ing considerations
1. 18 x 3.5 .
Factor 3 x 2
.
a. and r is negative.
C. C. F. F.
expressions which have no are prime to one another.
15
aW. C.
25
W.
2. The H. The H. C. F.
13 aty
39 afyV. aW.
II
2
. Thus the H. and GO aty 8 is 6 aty. F.
the algebraic factor of highest degree common expressions to these expressions thus a 6 is the II.
Two
common
factor except unity
The H. of (a
and (a
+
fc)
(a
4
is
(a
+ 6)
2
. F. 12 tfifz. of a 7 and a e b 7
.
EXERCISE
Find the H.
C. F.
are prime can be found by inspection.
122. of a 4 and a 2 b is a2
The H. of 6 sfyz.
C.
3.
6.
+
8
ft)
and
cfiW is
2
a 2 /) 2
ft)
. If the expressions have numerical coefficients.
8
.
33
2
7
3
22 3 2
. of
two or more monomials whose factors
.
5.
The
highest
is
common
factor (IT.
-
23 3
. of the algebraic expressions.
5
2
3
. C.
5
s
7
2
5.
5
7
34 2s
. of
:
48
4.
F.
. F. find by arithmetic the greatest common factor of the coefficients.
2
2
.
54
-
32
.
3
. of
aW. is the lowest
that the power of each factor in the power in which that factor occurs in any
of the given expressions.
24
s
.CHAPTER
VII
HIGHEST COMMON FACTOR AND LOWEST COMMON MULTIPLE
HIGHEST COMMON FACTOR
120.
89
.
121. F.
C.) of
two or more
. and prefix it as a coefficient to H.
The student should note
H. F. C. C.
6
c6 is
C a*b*c*.
126.
NOTE. C.
C. C.
Ex. M. C. L. of the
general.
M of the algebraic expressions. ory is the L.
etc. of several expressions which are not completely factored. C.
of 12(a
+
ft)
and (a
+ &)*( -
is
12(a
+ &)( .
of 3
aW. To find the L. If the expressions have a numerical coefficient.
M.M. which
also
signs.
M. resolve each expression into prime factors and apply the method for monomials.
of 4 a 2 6 2 and 4 a 4
-4 a 68
2
.
M.
&)
2
M.C. M.
Find the L.
=4 a2 62 (a2 .
128. two lowest common multiples.M.
= (a -f
last
2
&)'
is
(a
-
6) . L.(a + &) 2 (a
have the same absolute value.
300 z 2 y. of as -&2 a2 + 2a&-f b\ and 6-a.
. C.
Hence the L. M. 60
x^y'
2
.
.6 3 ).) of
two or more
expressions is the common multiple of lowest degree.
2. find by arithmetic their least common multiple and prefix it as a coefficient to the L. C.
2 multiples of 3 x
and 6 y are 30 xz y.
4 a 2 &2
_
Hence. thus.
Find the L. M. but opposite
.
The
lowest
common
multiple (L.
A
common
remainder.
Ex. C. M.
The
L.6)2. is equal to the highest power in which it occurs in any of the
given expressions. C.
127. C. each set of expressions has
In example ft).C.
2
The The
L.LOWEST COMMON MULTIPLE
91
LOWEST COMMON MULTIPLE
multiple of two or more expressions is an which can be divided by each of them without a expression
124. of tfy and xy*.
a^c8
3
.
Common
125. Obviously the power of each factor in the L.
1.
All operations with fractions in algebra are identical
with the corresponding operations in arithmetic.
Reduce
~-
to its lowest terms.
Remove
tor. If both terms of a fraction are multiplied or divided by the same number) the value of the fraction is not altered. only positive integral numerators shall assume that the
all
arithmetic principles are generally true for
algebraic numbers.
an indicated quotient.
Ex.ry ^
by
their H.
a?. F.
The dividend a is called the numerator and the The numerator and the denominator
are the terms of the fraction.
130.
rni
Thus
132.
A
-f-
fraction is
b.
common
6
2
divisors of
numerator and denomina-
and z 8
(or divide the terms
. as 8.
successively all
2
j/' . however. but we
In arithmetic.
131.CHAPTER
VIII
FRACTIONS
REDUCTION OF FRACTIONS
129.
and
i
x mx = my y
terms
A
1. C.
fraction
is
in its lowest
when
its
numerator
and
its
denominator have no
common
factors.
the product of two fractions is the product of their numerators divided by the product of their denominators. thus -
is identical
with a
divisor b the denominator.
a b
= ma
mb
.
and denominators are considered. the value of a fraction is not altered by multiplying or dividing both its numerator and its denominator by the same number. Thus. etc.
TT
Hence
24
2 z = --
3x
.
C. take the L.
and
Tb reduce fractions to their lowest common denominator.
TheL.
we may extend this method
to integral expressions.
1. multiplying the terms of
22
.
Multiplying these quotients by the corresponding numerators and writing the results over the common denominator.
.
M.3) (-!)'
=
. we have
(a
+ 3) (a -8) (-!)'
NOTE.D. of the denominators for the common denominator. =(z
(x
+ 3)(z. C.
-
of
//-*
2
.C.
3 a\ and 4
aW
is
12 afo 2 x2 . C.M.
-
by 4
6' .~16
(a
+ 3) (x.r
2
2
.
Ex
-
Reduce
to their lowest
common
denominator.
.
we have
the quotients (x
1). by the denominator of each fraction.
+
3).
1).
.
Since a
(z
-6 + 3)(s-3)O-l)'
6a.
ELEMENTS OF 'ALGEBRA
Reduction of fractions to equal fractions of lowest common Since the terms of a fraction may be multiplied
denominator. we may use the same process as in arithmetic for reducing fractions to the lowest
common
denominator.
we have
-M^.
^
to their lowest
com-
The
L. Divide the L.-1^22
' .
mon
T denominator.
Reduce -^-.
To reduce
to a fraction with the
denominator 12 a3 6 2 x2 numerator
^lA^L O r 2 a 3
'
and denominator must be multiplied by
Similarly.
Ex.
by any quantity without altering the value of the fraction.
multiply each quotient by the corresponding numerator.3)O -
Dividing this by each denominator.
2>
. and 6rar 3 a? kalr
.M. and
135.
and
(a-
8).by 3 ^
A
2
' .96
134.
and the terms of
***.
Fractions are multiplied by taking the product of tht numerators for the numerator.102
ELEMENTS OF ALGEBRA
MULTIPLICATION OF FRACTIONS
140. Common factors in the numerators and the denominators should be canceled before performing the multiplication.
-x
b
c
=
numerator by
To multiply a fraction by an
that integer.
expressed in symbols:
c
a
_ac b'd~bd'
principle proved for
b
141. and the product of the denominators for the
denominator.
Simplify 1 J
The
expreeaion
=8
6
. each
numerator and denomi-
nator has to be factored.)
Ex.
2
a
Ex. multiply the
142.
we may extend any
e.
!.
2.g. (In
order to cancel
common
factors.
Since -
= a.
fractions to integral numbers.
F J Simplify
.
or.
integer.
The
reciprocal of a
number
is
the quotient obtained by
dividing 1
by that number. To divide an expression by a fraction.y3
+
xy*
x*y~ -f y
8
y
-f
3
2/
x3
EXERCISE 56*
Simplify the following expressions
2
x*
'""*'-*'
:
om
2 a2 6 2
r -
3
i_L#_-i-17
ar
J
13 a& 2
5
ft2
'
u2
+a
.104
ELEMENTS OF ALGEBRA
DIVISION OF FRACTIONS
143. expression by the reciprocal of the fraction. To divide an expression by a fraction.
144. invert the divisor and multiply it by the dividend. and the principle of division follows
may
be expressed as
145. Integral or mixed divisors should be expressed in fractional form before dividing.
.
The reciprocal of ?
Hence the
:
+*
x
is
1
+ + * = _*_.
Divide X-n?/
. x a + b
obtained by inverting
reciprocal of a fraction
is
the fraction. :
a 4-1
a-b
* See page 272.
* x* -f xy 2
by
x*y
+y
x'
2
3
s^jf\ =
x'
2
x*
.
The The
reciprocal of a
is
a
1
-f-
reciprocal of J
is
|
|.
1.
8
multiply
the
Ex.
1. 2.180.
.
PROBLEMS LEADING TO FRACTIONAL AND LITERAL EQUATIONS 152. 2 3
.
is
36. When between 3 and 4 o'clock are the hands of
a clock together
?
is
At
3 o'clock the hour hand
15 minute spaces ahead of the minute
:
hand.
ELEMENTS OF ALGEBRA
(a) Find a formula expressing degrees of Fahrenheit terms of degrees of centigrade (<7) by solving the equation
(F)
in
(ft)
Express in degrees Fahrenheit 40
If
C.
then
= 2 TT#.
Multiplying by
Dividing. = the number of minute spaces the minute hand moves
over.
12.
days by x and the piece of work while in x days they would do
respectively
ff
~ and and hence the sentence written in algebraic symbols ^.
~^ = 15
11 x
'
!i^=15.
Ex..minutes after x=
^
of
3 o'clock.
and
12
= the number
over. Ex.
A would do
each day ^ and
B
j. = 16^.20
C. A can do a piece of work in 3 days and B in 2 days. hence the question would be formulated After how many minutes has the minute hand moved 15 spaces more than the hour hand ?
Let then
x x
= the required number of minutes after 3 o'clock.114
35.
.
of minute spaces the
hour hand moves
Therefore x
~ = the number of minute spaces the minute hand moves
more than the hour hand.
Find
R in terms of C and
TT.
C
is
the circumference of a circle whose radius
R. In how many days can both do it working together ?
If
we denote
then
/-
the required
number
by
1.
100
C.
x
Or
Uniting..
180
Transposing. hours more than the express train to travel 180 miles.
Clearing.FRACTIONAL AND LITERAL EQUATIONS
A
in symbols the following sentence
115
more symmetrical but very similar equation is obtained by writing ** The work done by A in one day plus the work done by B in one day equals the work done by both in one day.
or 1J.
32
x
= |."
:
Let
x -
= the
required
number
of days. u The accommodation train needs 4 hours more than the express train. The speed of an express train is $ of the speed of an If the accommodation train needs 4 accommodation train.
in
Then
Therefore.
But
in
uniform motion Time
=
Distance
.
fx
xx*
=
152
+4
(1)
Hence
=
36
= rate
of express train.
Ex. what is
the rate of the express train
?
180
Therefore.
the rate of the express train.
= the
x
part of the
work both do
one day. 4x = 80. then
Ox
j
5
a
Rate Hence the rates can be expressed.
= 100 + 4 x. and the statement.
Solving. 3. the required
number
of days." gives the equation /I).
Explanation
:
If
x
is
the rate of the accommodation train.
its
Find the number whose fourth part exceeds part by 3.
by 3.
of his present age. a man had How much money had he
at first?
. one half of What is the length
of the post ?
10
ter. How
did the
much money
man
leave ?
11.
9
its
A
post
is
a fifth of
its
length in water.
is
equal
7. and J of the greater Find the numbers. which was $4000.
by 6.
Twenty years ago A's age was |
age. to his daughand the remainder. How much money had he at first?
12
left
After spending ^ of his
^ of his money and $15. -|
Find their present ages.
Two numbers
differ
l to s of the smaller.
fifth
Two numbers
differ
2.
Find two consecutive numbers such that
9.
ex-
What
5. and one half the greater Find the numbers.
are the
The sum of two numbers numbers ?
and one
is
^ of the other.
The sum
10 years hence the son's age will be
of the ages of a father and his son is 50. A man lost f of his fortune and $500.
money and $10.
J-
of the greater
increased by ^ of the smaller equals
6.
ceeds the smaller by
4. and found that he had \ of his original fortune left.
Find a number whose third and fourth parts added
together
2.
A man left ^ of his property to his wife.
is oO.
make
21.
3.
and 9
feet above water.116
ELEMENTS OF ALGEBRA
EXERCISE
60
1. to his son.
Find A's
8.
length in the ground. and of the father's age.
)
22. 1.
117
The speed
of an accommodation train
is
f of the speed
of an express train. ^ at 5%.
investments. and B In how many days can both do it working together
in
?
12 days. and after traveling 150 miles overtakes the accommodation train.
152.
At what time between 7 and
8 o'clock are the hands of
?
a clock in a straight line and opposite
18. If the rate of the express train is -f of the rate of the accommodation train.
How much money
$500?
4%.
?
In
how many days can both do
working together
23.
.
A can
A
can do a piece of work in 2 days.
A man
has invested
J-
of his
money
at
the remainder at
6%. and an ounce of silver -fa of an ounce.
air. A can do a piece of work in 4 clays.FRACTIONAL AND LITERAL EQUATIONS
13. what is the
14. Ex.
at 4J % and P> has invested $ 5000 They both derive the same income from their How much money has each invested ?
20. Ex.)
At what time between 7 and 8
o'clock are the
hands of
a clock together ?
17. and
it
B in 6 days.
after
rate of the latter ?
15. An ounce of gold when weighed in water loses -fa of an How many ounce.
and losing
1-*-
ounces when weighed in water?
do a piece of work in 3 days.
A has invested capital
at
more
4%. If the accommodation train needs 1 hour more than the express train to travel 120 miles. ounces of gold and silver are there in a mixed mass weighing
20 ounces in
21. and B in 4 days.
At what time between 4 and
(
5 o'clock are the hands of
a clock together?
16. Ex. 2. and has he invested if
his animal interest therefrom is
19.) (
An express train starts from a certain station two hours an accommodation train. what is the rate of the express train? 152. 3. In how many days can both do it working together ? ( 152.
6
I
3
Solve the following problems
24.
26.
they can both do
in 2 days. and n = 3.g. A in 4. therefore.
ELEMENTS OF ALGEBRA
The
last three questions
and their solutions differ only two given numbers.009 918. is 57.
. Find three consecutive numbers whose sum equals m.
we
obtain the equation
m m
-. 3. A in 6.414. n x
Solving. Answers to numerical questions of this kind may then be found by numerical substitution. B in 16.
:
In
how many days
if
can
A
and
it
B
working together do a
piece of
work
each alone can do
(a)
(6)
(c)
in the following
number
ofdavs:
(d)
A in 5.
.=
m
-f-
n
it
Therefore both working together can do
in
mn
-f-
n
days. m and n. A in 6. Find the numbers if m = 24 30. B in 12. it is possible to solve all examples of this type by one example.
is 42.
To
and
find the numerical answer. Ex. Then
ft
i. e. if
B
in 3 days. 2.
Find three consecutive numbers whose sum
Find three consecutive numbers whose sum
last
:
The
two examples are
special cases of the following
problem 27. by taking for these numerical values two general algebraic numbers. In how
in the numerical values of the
:
many days
If
can both do
we
let
x
= the
it working together ? required number of days. is A can do a piece of work in m days and B in n days. Hence.
B in 5. B in 30. . The problem to be solved.= -.
25.
make
it
m
6
A can do this work in 6 days Q = 2.e. and apply the
method of
170.118
153.
. (a) 20 and 5 minutes. (b) 8 and 56 minutes.
A cistern can
be
filled
(c)
6 and 3 hours.
34. and how many miles does each travel ?
32.
squares
29. the rate of the
first. After how many hours do they rate of n miles per hour.
is ?n
. d miles the first traveling at the rate of m.
33. (c) 16. respectively. Find the side of the square. the area would be increased by 19 square feet. and how many miles does each travel ? Solve the problem if the distance.
:
(c)
64 miles. solve the following ones Find two consecutive numbers the difference of whose squares
:
find the smaller number. 2 miles per hour.
Find two consecutive numbers -the difference of whose
is 21. 88 one traveling 3 miles per hour. 2 miles per hour. (d) 1.
meet.
Two men
start at the
first
miles
apart. 5 miles per hour. (b) 149.
is (a)
51. 3J miles per hour. two pipes together ? Find the numerical answer.
last three
examples are special cases of the following
The
difference of the squares of
two consecutive numbers
By using the result of this problem.
The
one:
31.
119
Find two consecutive numbers the difference of whose
is 11.
by two pipes in m and n minutes In how many minutes can it be filled by the respectively.
same hour from two towns.001. respectively (a) 60 miles. the
Two men start at the same time from two towns. If each side of a square were increased by 1 foot. After how many hours do they meet. 3 miles per hour.
squares
30.721.FRACTIONAL AND LITERAL EQUATIONS
28. and the second 5 miles per hour. the second at the apart. (b) 35 miles.000.
and
the rate of the second are. if m and n are.
4J-
miles per hour.
.CHAPTER X
RATIO AND PROPORTION
11ATTO
154.
b
is
a
Since a ratio
a
fraction.
The
first
156.5. the denominator
The
the
157. a ratio
is
not changed
etc.
term of a ratio
a
the
is
is
the antecedent.
A
ratio
is
used to compare the magnitude of two
is
numbers.
the symbol
being a sign of division."
we may
write
a
:
b
= 6.
antecedent.
" a Thus.
:
:
155.
terms are multiplied or divided by the same number.or a *
b
The
ratio is also frequently
(In most European countries this symbol is employed as the usual sign of division.
1.
In the ratio a
:
ft. 6 12 = .g.
b. instead of writing
6 times as large as
?>. 158. all principles
relating
to
fractions
if its
may
be af)plied to ratios.
Ex.
The
ratio of
first
dividing the
two numbers number by the
and
:
is
the quotient obtained by
second.
:
A somewhat shorter way
would be to multiply each term by
120
6. the second
term the consequent. the antecedent.) The ratio of 12 3 equals 4.
Thus the
written a
:
ratio of a
b
is
.
The
ratio -
is
the inverse of the ratio -.
is
numerator of any fraction
consequent. b is the consequent.
E. etc.
Simplify the ratio 21 3|.
b.
In the proportion a b
:
=
b
:
c. and c.
10.
4|-:5f
:
5.
3:4.
Transform the following
unity
15.
8^-
hours.
AND PROPORTION
ratio 5
5
:
121
first
Transform the
3J so that the
term will
33
:
*~5
~
3
'4*
5
EXERCISE
Find the value of the following
1.
term
is
the fourth proportional to the
:
In the proportion a b = c c?.
3.
16.
J:l.
11. b and c the means.
3
8.
159.
16 x*y
64 x*y
:
24 48
xif.
3:1}.
two
|
ratios.
:
is
If the means of a proportion are equal.
5 f hours
:
2.
9.
extremes. b.
61
:
ratios
72:18.
4.
proportional between a
and
c.
18.
17.
62:16.
12.
Simplify the following ratios
7.RATIO
Ex. and the last term the third proportional to the first and second
161. a and d are the extremes.
and
c
is
the third proportional to a and
.
:
ratios so that the antecedents equal
16:64.
:
a-y
.
terms.
7f:6J. the second
and fourth terms of a proportion are the and third terms are the means.
A
proportion
is
a statement expressing the equality of
proportions.
16a2 :24a&.
6.
1.
equal
2. The last term d is the fourth proportional to a. The last
first three.
:
1.
$24: $8.
27 06: 18 a6.
= |or:6=c:(Z are
The
first
160.
b is the
mean
b. either mean the mean proportional between the first and the last terms.
7|:4 T T
4
.
If the product of two numbers is equal to the product of two other numbers^ either pair may be made the means. Instead of u
If 4
or 4 ccm.e.)
mn = pq.
:
c. 3 4. if the ratio of any two of the first kind.
ad =
be.
briefly. then 8 men can do it in 3 days. 6 ccm. and the time necessary to do it. Hence the number of men required to do some work.__(163.
ccm. Hence the weight of a mass of iron is proportional to its volume.122
162.
The mean proportional
of their product.
t/ie
product of the means
b
is
equal
to the
Let
a
:
=c
:
d.30 grams.'*
Quantities of one kind are said to be inversely proportional to quantities of another kind. if the ratio of any two of the first kind is equal \o the inverse ratio of the corresponding two of
the other kind.
163.
2
165. " we " NOTE. and we
divide both
members by
we have
?^~ E. is equal to the ratio of the corresponding two
of the other kind. = 30 grams 45 grams. of a proportion. or 8 equals the inverse ratio of 4 3.
163.
If 6 men can do a piece of work in 4 days. In any proportion product of the extremes. of iron weigh .
!-.
164. then G ccm. a b
:
bettveen two
numbers
is
equal to
the square root
Let the proportion be
Then Hence
6
=b = ac.
:
:
directly proportional
may say. are
: : :
inversely proportional.
ELEMENTS OF ALGEBRA
Quantities of one kind are said to be directly proper
tional to quantities of another kind.
If
(Converse of
nq.
Clearing of fractions. of iron weigh 45 grams.
q~~ n
. and the
other pair the extremes. i.) b = Vac.
pro-
portional.
and the
:
total cost. and the area
of the rectangle.inches long represents
map corresponds to how many miles ?
The
their radii. (e) The distance traveled by a train moving at a uniform rate.
the squares of their radii
(e)
55.
areas of circles are proportional to the squares of If the radii of two circles are to each other as
circle is
4
:
7. The number of men (m) is inversely proportional to the number of days (d) required to do a certain piece of work.
(d)
The sum
of
money producing $60
interest at
5%.
1
(6) The circumferences (C and C ) of two other as their radii (R and A").
56.
57.126
54. under a pressure of 15 pounds per square inch has a volume of
gas
is
A
16 cubic
feet.
(d)
The
areas
(A and
A') of two circles are to each other as
(R and R').
and the area of the smaller
is
8 square inches. othei
(a) Triangles
as their basis (b
and
b'). and
the time necessary for it. and the time. State whether the quantities mentioned below are directly or inversely proportional (a) The number of yards of a certain kind of silk.
(c)
of a rectangle of constant width.
(b)
The time a
The length
train needs to travel 10 miles.
A
line 11 inches long
on a certain
22 miles.
ELEMENTS OF ALGEBEA
State the following propositions as proportions : T (7 and T) of equal altitudes are to each.
the area of the larger? the same. the volume of a
The temperature remaining
body of gas inversely proportional to the pressure.
A
line 7^. (c) The volume of a body of gas (V) is
circles are to
each
inversely propor-
tional to the pressure (P).
what
58.
and the speed
of the train.
What
will be the
volume
if
the pressure
is
12 pounds per square inch ?
.
7 x = 42 is the second number.RATIO AND PROPORTION
69.
Therefore
7
=
14
= AC. AB = 2 x. 4 inches long.
Hence
or
Therefore
Hence
and
= the first number. produced to a point C. 11 x = 66 is the first number.
Let
A
B
AC=1x.
Divide 108 into two parts which are to each other
7.
is
A line AB. so that
Find^K7and BO.
Then
Hence
BG = 5 x.
11
x
x
7
Ex. What is the greatest distance a person can see from an elevation of 5 miles ? From h miles
the
Metropolitan
Tower (700
feet high) ?
feet
high) ?
From Mount
McKinley (20.
x=2. x = 6.
:
Ex.
as 11
Let
then
:
1.
4
'
r
i
1
(AC): (BO) =7: 5. = the second number. 18 x = 108.000
168. 11 x -f 7 x = 108.
. it is advisable to represent these unknown numbers by mx and nx.
127
The number
is
of miles one can see from an elevation of
very nearly the mean proportional between h and the diameter of the earth (8000 miles). When a problem requires the finding of two numbers which are to each other as m n. 2.
2 x
Or
=
4.
7.
What
are the parts ?
5.
13.
How many
grams of hydrogen are contained in 100
:
grams
10.
of water?
Divide 10 in the ratio a
b.
12. and c inches.
14.000. and 15 inches.
3.
The
total area of land is to the total area of
is
water as
7 18.
A line 24 inches
long
is
divided in the ratio 3
5.)
.
:
Divide 39 in the ratio 1
:
5.
m
in the ratio x:
y
%
three sides of a triangle are 11. Water consists of one part of hydrogen and 8 parts of
If the total surface of the earth
oxygen. How
The
long are the parts ? 15.
consists of 9 parts of copper and one part of ounces of each are there in 22 ounces of gun-
metal ?
Air is a mixture composed mainly of oxygen and nitrowhose volumes are to each other as 21 79. 2. The three sides of a triangle are respectively a.
:
Divide a in the ratio 3
Divide
:
7. How many ounces of copper and zinc are in 10 ounces of brass ?
6.000 square miles.
:
197. find the number of square miles of land and of water. How many gen.
Divide 44 in the ratio 2
Divide 45 in the ratio 3
:
9.
11.
Gunmetal
tin. 9. Brass is an alloy consisting of two parts of copper and one part of zinc.
How many
7. and the longest is divided in the ratio of the other two. 6. cubic feet of oxygen are there in a room whose volume is 4500
:
cubic feet?
8. If c is divided in the ratio of the other two. what are
its
parts ?
(For additional examples see page 279.
Divide 20 in the ratio 1 m.128
ELEMENTS OF ALGEBRA
EXERCISE
63
1. 12.
:
4.
=.
However. there is only one solution. such as
+ = 10.-. x = 1. y =
5
/0 \ (2)
of values.
the equations have the two values of
y must be equal.y=--|.
2 y = .e.
Hence.
a?
(1)
then
I.
y
(3)
these
unknown numbers can be found.
Hence
2s -5
o
= 10 _ ^
(4)
= 3. expressing a y.
The
root of (4)
if
K
129
.
If
satisfied
degree containing two or more by any number of values of
2oj-3y =
6. etc. the equation is satisfied by an infinite number of sets Such an equation is called indeterminate. y = 1.CHAPTER XI
SIMULTANEOUS LINEAR EQUATIONS
169.
An
equation of the
first
unknown numbers can be the unknown quantities. which substituted in (2) gives y both equations are to be satisfied by the same Therefore. is x = 7.-L
x
If
If
= 0. values of x and y.
if
there
is
different relation
between x and
*
given another equation. if
. From (3) it follows y 10 x and since
by the same values of x and
to be satisfied
y.
cannot be reduced to the same form.24. Any set of values satisfying 5 x + 6 y = 60 will also satisfy the equation 3 x -f. 3.
~ 50. 26 y = 60.130
170. 30 can be reduced to the same form -f 5 y Hence they are not independent. the last set inconsistent.
x
-H
2y
satisfied
6 and 7 x 3y = by the values x = I. for they are 2 y = 6 are But 2 x 2.
Substitution.
unknown
quantity.
(3)
(4)
Multiply (2) by
-
Subtract (4) from (3).
4y
.
By By
Addition or Subtraction. y
I
171. for they express the x -f y 10.
ELIMINATION BY ADDITION OR SUBTRACTION
175.
Therefore. for they cannot be satisfied by any value of x and y. The first set of equations is also called consistent.
A
system of two simultaneous equations containing two
quantities is solved by combining them so as to obtain
unknown
one equation containing only one
173.
21 y
.
172.
174. and 3 x + 3 y =. y = 2.X.
The process of combining several equations so as make one unknown quantity disappear is called elimination.
to
The two methods
I.
of elimination
most frequently used
II.
6x
.
are simultaneous equations.26.
viz. 6 and 4 x y not simultaneous. same relation.3 y = 80.
ELEMENTS OF ALGEBRA
A
system
of simultaneous equations is
tions that can be satisfied
a group of equa by the same values of the unknown
numbers.
Solve
-y=6x
6x
-f
Multiply (1) by
2.
E. Independent equations are equations representing different relations between the unknown quantities such equations
.
= .
=
l.
Let
x
y z
= the
the digit in the hundreds' place. 1. however.
.
y
*
z
30.y
125
(3)
The solution of these equations gives x Hence the required number is 125.
2
= 6.
unknown quantity by
every verbal statement as an equation.
and Then
100
+
10 y
+z-
the digit in the units' place. Problems involving several unknown quantities must contain. the number.
Obviously
of the other
.
The
three statements of the problem can
now be
readily expressed in
.
1
=
2.SIMULTANEOUS LINEAR EQUATIONS
143
x
29.
(1)
100s
+ lOy + z + 396 = 100* + 10y + x. + z = 2p. The digit in the tens' place is | of the sum of the other two digits.
x
:
z
=1
:
2.
(
99. and if 396 be added to the number. Simple examples of this
kind can usually be solved by equations involving only one
unknown
every
quantity.
Check. z + x = 2 n.
symbols:
x
+
y
+z-
8. y
31.
# 4. either directly or implied.
M=i. and to express
In complex examples.
Find the number.
to express
it is difficult
two of the required
digits in
terms
hence we employ 3
letters for the three
unknown
quantities.)
it is advisable to represent a different letter.
Ex. 1 digit in the tens place.2/
2/
PROBLEMS LEADING TO SIMULTANEOUS EQUATIONS
183.
+2+
6
= 8.
. the first and the last digits
will be interchanged.
= 2 m. The sum of three digits of a number is 8.
2
= 1(1+6). as many verbal statements as there are unknown quantities.
+
396
= 521.
x 3x-4y = 12.
+
I
2
(1)
and
These equations give x
Check. 5_
_4_
A. and C travel from the same place in the same B starts 2 hours after A and travels one mile per hour faster than A.
x 3
= 24. the fraction
Let and
then y
is
reduced to
nurn orator.
Or
(4)-2x(3).
ELEMENTS OF ALGE13KA
If both numerator and denominator of a fraction be
.
2.
3.
8
= xy + x xy = xy -f 3 x 2 y = 2.
4
x
= 24.
3
xand y
I
1
(2)
5. the fraction is reduced to | and if both numerator and denominator of the reciprocal of the fraction be dimin-
ished by one.
By
expressing the two statements in symbols.
2. = 8.
. B.
Ex.
Since the three
men
traveled the
same
distance.
=
Hence the
fraction
is
f.
x
y
= the = the
x
denominator
.
(1) (2)
12.
3+1 5+1
4_2. C.144
Ex.
we
obtain. who travels 2 miles an hour faster than B.
= the
fraction.
Find the
fraction.
6
x 4
= 24.
increased by one. From (3)
Hence xy
Check.
xy
a:
2y 4y
2.
(3) C4)
=
24 miles. starts 2 hours after B and overtakes A at the same How many miles has A then traveled? instant as B.
direction. the distance traveled by A. y = 3.
The sum
18
is is
and
if
added
of the digits of a number of two figures is 6.
the
number
(See Ex. the fraction is reduced
fraction.
6.
If the
numerator of a fraction be trebled. and the second increased by 2 equals three times the first.
If 4 be
Tf 3 be
is J. if its numerator and its denominator are increased by 1.
?
What
9.
5. If
9 be added to the number.)
added to a number of two digits.
183.
number by
the
first
3. Find the number. and the numerator increased by 4.
Half the sum of two numbers equals 4.
1. Four times a certain number increased by three times another number equals 33. the fraction equals . and the two digits exceeds the third digit by 3. Find the numbers. and the second one increased by 5 equals twice
number. Find the numbers. to the number the digits will be interchanged.
7.
.
Five times a certain number exceeds three times another 11. and four times the first digit exceeds the second digit by 3.
2. Find the number.
to
L
<>
Find the
If the
numerator and the denominator of a fraction be If 1 be subtracted from increased by 3. If the denominator be doubled. and
its
denomi-
nator diminished by one. fraction is reduced to \-.
If 27 is
10. its value added to the denominator. A fraction is reduced to J. both terms. and twice the numerator What is the fracincreased by the denominator equals 15. the digits will be interchanged.
part of their difference equals
4. the last two digits are interchanged.
The sum
of the first
sum
of the three digits of a number is 9.SIMULTANEOUS LINEAR EQUATIONS
EXERCISE
70
145
1. Find the fraction. the Find the fraction.}.
added to the numerator of a fraction. and the fourth 3.
Find the numbers.
tion ?
8. the value of the fraction is fa. it is reduced to J.
What was the amount of each investment ?
15. Two cubic centimeters of gold and three cubic centimeters of silver weigh
together 69 J. now.146
ELEMENTS OF ALGEBRA
11.grams. 12.
. and B's age is \ the sum of A's and C's ages.
Ten years ago A was B was as
as old as
B
is
old as
will be 5 years hence .
14. Twice A's age exceeds the sum of B's and C's ages by 30.
What was
the
sum and
rates
est
The sums of $1500 and $2000 are invested at different and their annual interest is $ 190. How 6 %.
A
sum
of $10.
and
money and
17. If the rates of interwere exchanged.
and 5 years ago
their ages is 55.000
is
partly invested at
6%.
in 8 years to $8500. and partly at 4 %. bringing a total yearly interest of $530. Ten years ago the sum of their ages was 90.
much money
is
invested at
A sum
of
money
at simple interest
amounted
in 6 years
to $8000. partly at 5% and partly at 4%. a part at 6 and the remainder bringing a total yearly interest of $260. and 4 %. and the 5% investment brings $15 more interest than the 4 % investment. respectively ?
16. and The 6 investment brings $ 70 more interest than the 5
%
%
4%
investments together. If the sum of
how
old
is
each
now ?
at
invested $ 5000.
partly at
5 %. the rate of interest?
18.
19. and in 5 years to $1125. What was the amount of each investment ?
A man
%
5%. Find the weight of one cubic centimeter of gold and one cubic centimeter of silver. the annual interest would be $ 195. A man invested $750. the rate of interest ?
What was
the
sum
of
A sum
of
money
at simple interest
amounted
in 2 years
to $090. Find
the rates of interest. Three cubic centimeters of gold and two cubic centimeters of silver weigh together 78 grains. Find their present ages.
13. 5 %.
SIMULTANEOUS LINEAR EQUATIONS
147
20. A farmer sold a number of horses. and their difference by GO .
Find the parts of the
ABC touching the three sides if AB = 9. How many did he sell
of each if the total
number
of animals
was 24?
21.
Find their
rates of walking. B find angles a.
.
BC=7. If one angle exceeds the sum of the other two by 20.
the three sides of a triangle E. then AD = AF.
E. and $15 for each sheep. ED = BE.
1
NOTE. for $ 740. and F. and F '(see diagram). BD = HE. what is
that
=
OF.
23.
A
r
^
A
circle is inscribed in triangle
sides in D. angle c = angle d. If angle ABC = GO angle BAG = 50. andCL4 = 8.
An C touch ing the sides in D. what are the angles of the triangle ?
22. three
AD = AF. he would walk it in two hours less than
than
to travel
B
B.
the length of
NOTE. and e. and sheep. and F. and CF?
is
a circle
inscribed in the
7<7.
.
24.
It takes
A two hours
longer
24 miles. c.
25.
is
the center of the circum-
scribed circle. but if A would double his pace.
In the annexed diagram angle a = angle b. receiving $ 100 for each horse. cows. and CE If AB = G inches.
points. and GE = CF.
On
/). and angle BCA = 70. are taken so
ABC. The number of sheep was twice the number of horses and cows together.
respectively. BE. and AC = 5 inches. The sum of the 3 angles of a triangle is 180. $ 50 for each cow. BC = 7 inches. and angle e angle/.
triangle
Tf
AD.
-3).
* This chapter
may
be omitted on a
148
reading.
It'
Location of a point. YY' they-axis.
or its equal
OM. B.
2). and ordinates abore the x-axis are considered positive . and point the origin. and r or its equal OA is
.
The
abscissa
is
usually denoted by
line XX' is called the jr-axis.
is the abscissa.
PN are given. then
the position of point is determined if the lengths
of
P
P3f and
185. PM.
is
The point whose abscissa is a.
first
3).
(3.
the ordinate of point P. and PN _L YY'. and
respectively represented
Dare
and
by
(3 7 4). jr.
?/. (7. Thus the points A. and PJ/_L XX'.CHAPTER
XII*
GRAPHIC REPRESENTATION OF FUNCTIONS AND
EQUATIONS
184.. Abscissas measured to the riyht of the origin.
(2.
lines
PM
the
and P^V are
coordinates
called
point P.
186.
two fixed straight lines XX' and YY' meet in at right angles. the ordinate by ?/.
(2.
The
of
Coordinates.
.
PN. hence
The
coordinates lying in opposite directions are negative. and whose ordinate is usually denoted by (X ?/).
GRAPHIC REPRESENTATION OF FUNCTIONS
The
is
149
process of locating a point called plotting the point.(!.
(-5.
Where do Where do
Where do
all
points
lie
whose ordinates
tfqual
4?
9.
.
What
Draw
is
the distance of the point
(3. (0.
Draw
the triangle whose vertices are respectively
(-l. i.
Plot the points: (-4. 0). (4.)
EXERCISE
1.
all all
points
points
lie
lie
whose abscissas equal zero ?
whose ordinates equal zero?
y) if y
10.
and measure
their
distance. paper ruled with two sets of equidistant and parallel linos intersecting at right angles. 11.
What
are the coordinates of the origin ? If
187.and(l.3). (See diagram on page 151.
=3?
is
If a point lies in the avaxis. (-4.
4).
Plot the points
:
(0.
two variable quantities are so related that
changes of the one bring about definite changes of the other. which of its coordinates
known ?
13.1). 0).
4.
6. (-4.
whose coordinates are given
NOTE.
the quadrilateral whose vertices are respectively
(4.
0).
2. (0. (4.e. (4. 3).
-2). (-2.
2J-). -2).2). -4). -3).
1).
8. Graphic constructions are greatly facilitated by the use of cross-section paper.
12.
(4. 0).
Graphs. 3).
3. 1). the mutual dependence of the two quantities may be represented
either by a table or
by a diagram.
(-1.
.
What
is
the locus of
(a?.
Plot the points:
(4.
Plot the points
(6.
(-3.
71
2).
4)
and
(4. -!).
6.
4)
from the
origin ?
7.4).
and the amount of gas subjected to pressures from
pound
The same data.
ically
each representing a temperature at a certain date.
A graphic
and
it
impresses upon the eye
all
the peculiarities of
the changes better and quicker than any numerical compilations. however.
from January 1 to December 1.
Thus the average temperature on May
on April 20. D. Thus the first table produces 12 points. may be represented graphby making each number in one column the abscissa.
By representing
of points.. but it indicates in a given space a great many more
facts than a table. 10
.
we meas1
.
1. C.
ure the ordinate of F. and the corresponding number in the adjacent column the ordinate of a point. we obtain an uninterrupted sequence
etc.
may be found
on Jan. B.150
ELEMENTS OF ALGEBRA
tables represent the average temperature
Thus the following
of
New
volumes
1
Y'ork City of a certain
to 8 pounds. A.
15. in like manner the average temperatures for every value of the time. or the curved line the temperature.
.
188.
representation does not allow the same accuracy of results as a numerical table.
ABCN
y
the so-called graph of
To
15
find
from the diagram the temperature on June
to be 15
.
(b) July 15. The engineer. the rise and fall of wages. the merchant. the
matics. (d) November 20.
:
72
find approximate answers to the following
Determine the average temperature of New York City on (a) May 1. Whenever a clear. uses them. (c) January 15. the graph
is
applied. as the prices and production of commodities.
EXERCISE
From the diagram
questions
1.
physician.GRAPHIC REPRESENTATION OF FUNCTIONS
151
i55$5St5SS 3{utt|s33<0za3
Graphs are possibly the most widely used devices of applied matheThe scientist uses them to compile the data found from experiments.
. Daily papers represent ecpnoniical facts graphically. and to deduce general laws therefrom. concise representation of a
number
of numerical data
is
required. etc.
During what months
above 18 C.
is
ture
we would denote the time during which the temperaabove the yearly average of 11 as the warm season.152
2. (1)
10
C.
During what month does the temperature decrease most
rapidly ?
13. (freezing
point) ?
7. At what date is the average temperature highest the highest average temperature?
?
What What
is
4.
June
July
During what month does the temperature increase most
?
rapidly
12. (c)
the average temperature oi 1 C. ?
-
3. from what date to what date would it extend ?
If
.
During what month does the temperature change least?
14.. 1 to Oct.
When
the average temperature below
C.
1 ?
does
the
temperature
increase from
11.
is
10. 1?
11
0. At what date is the average temperature lowest? the lowest average temperature ?
5.
15.
ELEMENTS OF ALGEKRA
At what date
(a) G
or dates
is
New York
is
C.
on
1 to
the
average.
How
much.
Which month
is
is
the coldest of the year?
Which month
the hottest of the year?
16.. ?
9.
When
What
is
the temperature equal to the yearly average of
the average temperature from Sept. (d) 9 0.?
is
is
the average temperature of
New York
6.
From what
date to what date does the temperature
increase (on the average)?
8..
How much warmer
1 ?
on the average
is it
on July 1 than
on
May
17.
in a similar manner as the temperature graph was applied in examples 1-18. One meter equals 1.
153
1?
When is the average temperature the same as on April
Use the graphs of the following examples for the solution of concrete numerical examples.
Hour
Temperature
.09 yards.GRAPHIC REPRESENTATION OF FUNCTIONS
18.
19.
a temperature chart of a patient.
NOTE.
Draw
.
Represent graphically the populations
:
(in
hundred thou-
sands) of the following states
22. From the table on page 150 draw a graph representing the volumes of a certain body of gas under varying pressures.
20.
Draw
a
graph for the
23. transformation of meters into yards. Construct a diagram containing the graphs of the mean temperatures of the following three cities (in degrees Fahren-
heit)
:
21.
.
+7
If
will
respec-
assume the values 7.
then
C
irJl.
A
10 wheels a day.
. represent his daily gain (or loss).
2
is
called
x
2 xy
+ 7 is a function of x.. 2 8 y' + 3 y is a function of x and
y.. An expression involving one or several letters a function of these letters. Represent graphically the cost of butter from 5 pounds if 1 pound cost $.
29.
function
If the value of a quantity changes.
ELEMENTS OF ALGEBRA
If
C
2
is
the circumference of a circle whose radius
is J2.50 per copy
(Let 100 copies = about \. if x assumes
successively the
tively
values
1.
3.
28. Represent graphically the distances traveled by a train in 3 hours at a rate of 20 miles per hour.
26.
The
initial cost of
cost of manufacturing a certain book consists of the $800 for making the plates. if he sells 0. the daily average expenses for rent.
If
dealer in bicycles gains $2 on every wheel he sells. 3.
to 20 Represent graphically the weight of iron from cubic centimeters. if 1 cubic centimeter of iron weighs 7. e. and $. the value of a of this quantity will change.5
grams.50.
Show
graphically the cost of the
REPRESENTATION OF FUNCTIONS OF ONE VARIABLE
189.g.inch.
2. 1 to 1200 copies. x
7 to 9. binding.
(Assume ir~
all circles
>2
2
. 9. 2 x -f 7 gradually from 1 to 2.
books from
for printing. if each copy sells for $1.154
24.
190.50.
x increases will change gradually from
13. 2 .)
T
circumferences of
25. gas.
x*
x
19.
4.
to
27. amount to $8. etc.
from
R
Represent graphically the = to R = 8 inches.) On the same diagram represent the selling price of the books. etc.
however.
1
the points (-3. to con struct the graph x of x 2 construct a series of -3 points whose abscissas rep2 resent X) and whose ordi1
tions
. for x=l.
3
(0. hence
various values of x
The values of a function for the be given in the form of a numerical table.
(1^.
values of x2
nates are the corresponding i.2 x
may
4 from x
=
4.
may
. (1. to
x = 4.
9).
2
(-1. Thus the table on page 1G4 gives the values of the functions x 2 x3 and Vsr. (2.
may.
etc. If a more exact diagram
is
required.
Draw the graph of x2 -f.
To
obtain the values of the functions for the various values of
the
following arrangement
be found convenient
:
. 3 50.
Graph
of a function.1).
(-
2. E. The values of func192. be also represented by a graph.g.
it is
In the example of the preceding article. 2.0). x a variable.
a*. plot points which
lie
between those constructed above.GRAPHIC REPRESENTATION OF FUNCTIONS
191.
Ex. 4).
Q-.
-J).
and join
the
points in order.e. construct
'.
155
-A
variable is a quantity
whose value changes in the
same
discussion.
is
A
constant
a quantity whose value does not change in the
same discussion.
.
is
supposed to change.1). while 7 is a constant. as
1. and (3.
2).
9). 4).
(-3.
If
If
Locating
ing
by a
3) and (4.156
ELEMENTS OF ALGEBRA
Locating the points(
4.. hence two points are sufficient for the construction
of these graphs.20).
194.4).
2.
It can be
proved that the
graph is a straight
of a function of the first degree
line. and joining in order produces
the graph
ABC. Thus 4x + 7. Thus in the above example. (-2. (4..
A
Y'
function of the
first
degree is an integral
rational function
involving only
the
power of the variable.. = 4. j/=-3. 2 4 and if y = x -f. etc. 5). and join(0. as y.2 x
. or ax + b -f c are funclirst
tions of the first degree.)
For brevity.-. 4). straight line produces the required graph.
. y = 6.
if
/*
4
>
1i >
>
?/
=
193..
(To avoid
very large ordinatcs.
Ex.
r
*/
+*
01
.
7
. the function
is
frequently represented
by a single letter.
4J.
rf
71
. -1).
Draw
y
z x
the graph of
= 2x-3.
= 0. the scale unit of the ordinatcs is taken smaller than that of the x.
y=
formula graphically.
1
C.
.
that
graph with the o>axis.
that the graph of two variables that are directly proportional is a straight line passing through the origin (assume
for c
27. 14 F.where x
c is
a constant.) scale are
expressed in
degrees of the Centigrade (C.
25. we have to measure the abscissas of the intersection of the
195.24.. what values of x make the function x2 + 2x 4 = (see 192). Therefore x = 1.24 or x =
P and
Q. it is evidently possible Thus to find to find graphically the real roots of an equation.
C.
A body
moving with a uniform
t
velocity of 3 yards per
second moves in
this
seconds a distance d
=3
1.. then
cXj
where
c is a constant. 9 F.. i.
if c
Draw
the locus of this equation
= 12..) scale by the formula
(a)
Draw
the graph of
C = f (F-32)
from
to
(b)
4 F F=l. 32 F.
Show
any convenient number).158
24. the abscissas of 3.
ELEMENTS OF ALGEBRA
Degrees of the Fahrenheit
(F.
From
grade equal to
(c)
the diagram find the number of degrees of centi-1 F..
to Fahrenheit readings
:
Change
10
C.
GRAPHIC SOLUTION OF EQUATIONS INVOLVING ONE
UNKNOWN QUANTITY
Since we can graphically determine the values of x make a function of x equal to zero. Represent 26. then
y = .e. If two variables x and y are directly proportional.
If
two variables x and y are inversely proportional.
If
x
=
0.
Ex.
Ex.
i.
(f
.
y=
A
and construct
x
(
-
graphically.
unknown
quantities. locate points
(0. that can be reduced
Thus
to represent
x
-
-
-L^-
\
x
=2
-
graphically.
Thus
If
in
points without solving the equation for the preceding example:
3x
s
.
X'-2
Locating the points
(2. Represent graphically
Solving for
y ='-"JJ y.
3x
_
4
. solve for
?/.e.
1)
and
0).
produces the
7*
required locus.
4)
and
(2.
if
y
=
is
0. Equations of the first degree are called linear equations. Draw the locus of 4 x + 3 y = 12. because their graphs are straight lines.
y y
2.2 y ~ 2.2. 2).
.
Hence
if
if
x
x
-
2. 0).
?/.
== 2.
and joining by a straight
line. = 0. and join the required graph.
NOTE.
199.
T
.
?/
=4
AB.
If the given equation is of the we can usually locate two
y.
fc
= 3.
represent graphically equations of the form y function of x ( 1D2).
Hence.
Graph
of
equations involving two
unknown
quantities.
first
degree.1. we can construct the graph or locus of any
Since
we can
=
equation involving two
to the above form.
Hence we may
join (0. y = -l.160
ELEMENTS OF ALGEBRA
GRAPHIC SOLUTION OF EQUATIONS INVOLVING TWO UNKNOWN QUANTITIES
198.
4) and
them by
straight line
AB
(3.
201. The coordinates of every point of the graph satisfy the given equation.
Solve graphically the equations
:
(1)
\x-y-\.
equation
x=
By measuring
3.
AB
y
= . the point of intersection of the coordinate of P.
parallel have only one point of intersection.
AB
but only one point
in
AB
also satisfies
(2). viz.
Graphical solution of a linear system.
By
the
method
of
the preceding article construct the graphs
AB
and
and
CD
of
(1)
(2) respectively.
3.
Since two straight lines which are not coincident nor simultaneous
Ex.
202.GRAPHIC REPRESENTATION OF FUNCTIONS
161
200.
and CD. The roots of two simultaneous equations are represented by the coordinates of the point (or points) at which their
graphs intersect.
To
find the roots of
the system.
(2)
.
203. and every set of real values of x and y satisfying the given equation is represented by a point in
the locus.
The
every
coordinates
of
point in satisfy the equation
(1). we obtain the roots.57. linear equations have only one pair of roots.15. P.1=0.
(-2.
4. the point
we
obtain
Ex.
Solve graphically the
:
fol-
lowing system
= =
25. and
. and joining by a
straight line.
3.162
ELEMENTS OF ALGEBRA
graph. 2. (4. if x equals
respectively
0.
Inconsistent equations.
The equations
2
4
= 0.g. i..
Locating the
points
(5.
4.
Solving (1) for y.
obtain the graph (a circle)
AB C
joining.5.0).
5.
1.
There can be no point of
and hence no
roots. 4.
(1)
(2)
cannot be satisfied by the same values of x and y.
. we of the
+
y*
= 25. 1.
y equals
3.
4.e. there are two pairs of By measuring the coordinates of
:
P and Q we find
204. 0.
4.
4.
In general. they are inconsistent. the graph
of
points
roots.9.
(1)
(2)
-C. e. 0.
Locating two points of equation (2).
V25
5.
3x
2 y = -6.5.
x2
. etc.
-
4. (-4. parallel graphs indicate inconsistent equations.
2.
3.
Measuring the coordinates
of P.
intersection. 3). This is clearly shown by the graphs of (1) arid (2). construct CD the locus of (2)
of intersection. 5. = 0. which consist of a
pair of parallel lines.
Using the method of the preceding para.
2 equation x
3). Since the two
-
we obtain DE.
4.y~ Therefore.
4. 0) and (0.
P
graphs meet in two and $. 3.0. AB the locus of (1).
and
+ 3).
215.
Evolution
it is
is the operation of finding a root of a quan the inverse of involution.
2.
numbers. quantity
may
the
be either 2wsitive
or negative.
27
=y
means
r'
=
27. for (+ a) = a \/32 = 2.
V9 = +
3. or y
~
3.
V
\/P
214.
Thus
V^I is an imaginary number.CHAPTER XIV
EVOLUTION
213. Since even powers can never be negative. for distinction. and ( v/o* = a.
1.
tity
.
or
-3
for
(usually written
3)
.
\/a
=
x means x n
=
y
?>
a.
for (-f 3) 2
(
3)
equal
0.
109
.
or x
&4 . and
all
other numbers are.
It follows
from the law of signs
in evolution that
:
Any
even root of a positive.
Every odd root of a quantity has
same sign as
and
2
the
quantity. etc.
\/"^27=-3.
4
4
. called real
numbers. it is evidently impossible to express an even root of a negative quantity by Such roots are called imaginary the usual system of numbers. (_3) = -27.
a)
4
= a4
.
= x means
= 6-. which can be simplified no further.
8
.
#2
a2
-
16.
+ 6 + 4a&.
a2
+ & + c + 2 a& .
The
term
a'
first
2
. the that 2 ab -f b 2
=
we have then to consider sum of trial divisor 2 a.
10. 11.
term a of the root
is
the square root of the
first
The second term
of the root can be obtained
a.172
7.
15.
The work may be arranged
2
:
a 2 + 2 ab
+ W \a + b
.
the given expression is a perfect square.>
13. let us consider the relation of a -f.
ELEMENTS OF ALGEBEA
4a2 -44a?> + 121V2 4a
s
.
mV-14m??2)-f 49. however.
2
.2 ac .
multiplied by b must give the last two terms of the
as follows
square. it is not known whether the given
expression is a perfect square.e. In order to find a general method for extracting the square root of a polynomial.
and
b.b 2 2 to its square.72 aW + 81 &
4
.
2
49a 8 16 a 4
9.2 ab + b
. i.
12.
. and b (2 a -f b).2 &c.
second term 2ab by the double of
by dividing the the so-called trial divisor.
2 2
218.
a-\-b
is
the root
if
In most cases.
2ab
. a -f.
14.
The process of the preceding article can be extended to polynomials of more than three terms.
and so
forth. 8 /-.EVOLUTION
Ex. 24# 2 y 3 by the trial divisor Dividing the first term of the remainder.
173
x*
Extract the square root of 1G
16x4
10 x*
__
. the first term of the answer.
4 x2
3
?/
8 is
the required square foot. 8 a 2 .
by division we
term of the
root.
is
As
there
is
no remainder. 8 a 2 Second complete divisor.
*/''
.
-
24 a
3
+
25 a 2
-
12 a
+4
Square of 4 a First remainder.24 a + 4 -12 a + 25 a8
s
. 8 a 2
2.
Arranging according to descending powers of
10 a
4
a. the required root
(4
a'2
8a
+
2}. As there is no remainder. Second trial divisor. By doubling 4x'2 we obtain 8x2 the trial divisor.
. 6 a.
10 a 4
8
a.
The square
.
.
double of this term
find the next
is
the
new
trial divisor.
\
24 a 3
4-f
a2
10 a 2
Second remainder. We find the first two terms of the root by the method used in Ex.
1.
Explanation.
First complete divisor. we obtain the next term of the root 3 y 3 which has to be added to 2 the trial divisor.
.
Arrange the expression according to descending powers root of 10 x 4 is 4 # 2 the lirst term of the root.
Extract the square root of
16 a 4
. 8 a 2
-
12 a
+4
a
-f 2.
.
219. Multiply the complete divisor Sx' 3y 3 by Sy 8 and subtract the product from the remainder.
First trial divisor. and consider Hence the their sum one term.24 afy* -f 9 tf.
Ex.
of x.
2. 1. 2 Subtracting the square of 4x' from the trinomial gives the remainder '24 x'2 + y.
.
= 80. the first of which is 8. Hence if we divide the digits of the number into groups.
the
consists of
group is the first digit in the root.
Ex. the first of which is 4.
From
A
will
show the
comparison of the algebraical and arithmetical method given below identity of the methods.
Ex. which may contain one or two). the square root of 7744 equals 88. the first of which is 9 the square root of 21'06'81 has three digits.
a
f>2'41 '70
6
c
[700
+ 20 + 4 = 724
2 a
a2 = +6=
41)
00 00
1400
+ 20 = 1420
4
341 76
28400
=
1444
57 76
6776
. and the square root of the greatest square in
units.EVOLUTION
220.000.
first
.
The
is
trial divisor
=
160.
square root of arithmetical numbers can be found to the one used for algebraic
Since the square root of 100 is 10.
the preceding explanation it follows that the root has two digits.1344.
a 2 = 6400. etc.000 is 1000.
1. then the number of groups is equal to the number of digits in the square root.
175
The
by a method very similar
expressions.
Find the square root of 7744.
As
8
x 168
=
1344..000 is 100. of a number between 100 and 10. beginning at the
and each group contains two digits (except the last. and the first remainder is. Thus the square root of 96'04' two digits. and we may apply the method used in algebraic process.000.176. of 1. two figures. of 10.
7744 80 6400
1
+8
160
+ 8 = 168
1344
1344
Since a
2 a
Explanation.
2. the integral part of the square root of a number less than 100 has one figure. etc. Therefore 6 = 8. Hence the root is 80 plus an unknown number. and the complete divisor
168.
Find the square root of 524.
Find the square root of
6/.70
6.
in .
EXERCISE
Extract the square roots of
:
82
.688
4
45 2 70
2 25
508
4064
6168 41)600
41344
2256
222.1 are
Ex.
The groups
of 16724.GO'61.7 to three decimal places.
ELEMENTS OF ALGEKRA
In marking
off groups in a number which has decimal begin at the decimal point.
3. annex a cipher. or by transforming the common fraction into a decimal.
Roots of common fractions are extracted either by divid-
ing the root of the numerator by the root of the denominator.0961
are
'.1T6
221.
we must
Thus the groups
1'67'24.10.
places. and if the righthand group contains only one digit.
12.
If the hypotenuse
whose angles
a
units of length.
27.)
of their squares
5.
and they con-
tain together 30G square feet.
Find
is
the number.
If 2
-f 2 b*
= 4w
2
-f c
sol ve for
m. The sides of two square fields are as 7 2.
2
.
2a
-f-
1
23.
.
:
6.
28.
find a in terms of 6
.
4.
2
:
3.
Find the side
of each field.
solve for d.
84
is
Find a positive number which
equal to
its
reciprocal
(
144). Find the side of each field. and the first exceeds the second by 405 square yards.
2
.
solve for
r.
may
be considered one half of a
rec-
square units.
29.
'
4.
The
two numbers
(See
is
2
:
3.
Find the numbers.b 2 If s
If
=c
. 24. then
Since such a triangle
tangle.
22
a.
If s
= 4 Trr
'
2
.
r.
If
G=m m
g
.
is
one of
_____
b
The side right angle.
228. 25.
A
number multiplied by
ratio of
its fifth
part equals 45.
26.180
on
__!_:L
ELEMENTS OF ALGEBRA
a.
If 22
= ~^-.
.
solve for v. its area contains
=a
2
-f-
b2
.
Three numbers are to each other as 1 Find the numbers.
9
&
-{-
c#
a
x
+a
and
c. and the two other sides respectively
c
2
contains
c
a and b units.
A
right triangle is a triangle. is 5(5.
and the sum
The
sides of
two square
fields are as
3
:
5. opposite the right angle is called the hypotenuse (c in the diagram).
EXERCISE
1.
=
a
2
2
(' 2
solve for solve for
= Trr
.
2.
108.
If a 2 4.
3.
and their product
:
150.
)
13. the radius of a sphere whose surface equals
If the radius of a sphere is r. and the third side is 15 inches.
The hypotenuse
of a right triangle is 2. in how many seconds will a body fall (a) G4 feet.
and the two smaller
11.
2m.
The area $
/S
of a circle
2
. let us compare x 2
The
left
the perfect square x2
2
mx -f m
to
2
. (b) 100 feet?
=
.
member can be made a complete square by adding 7 x with another term.7 x -f 10 = 0.
The following
ex-
ample
illustrates the
method
or
of solving a complete quadratic
equation by completing the square.
sides. The hypotenuse of a right triangle is to one side as 13:12.)
COMPLETE QUADRATIC EQUATIONS
229.
Method
of completing the
square.2
7
. (b) 44 square feet.
Find the
radii.
radii are as 3
14.
Solve
Transposing. A body falling from a state of rest.
4. Find the unknown sides and the area.
.
7r
(Assume
and their
=
2 7
2
.
make x2
Evidently 7 takes the place 7x a complete square
to
to
which corresponds
m
2
.
9.
8 = 4 wr2 Find 440 square yards. and the
other two sides are as 3
4.
8.
.
of a right triangle Find these sides.
The area
:
sides are as 3
4. we have
of
or
m = |. To find this term.
Find the
sides.
add
(|)
Hence
2
.
181
The hypotenuse
of a right triangle
:
is
35 inches. its surface
(Assume
ir
=
2 .
.
the formula
= Trr
whose radius equals r is found by Find the radius of circle whose area S
equals (a) 154 square inches.
-J-
=
12.
Two
circles together contain
:
3850 square
feet.
Find these
10.
24.
is
and the other
two
sides are equal. x* 7 x=
10. passes in t seconds 2 over a space s yt Assuming g 32 feet.QUADRATIC EQUATIONS
7.
and c in the general answer.
2
Every quadratic equation can be
reduced to the general form.
x
la
48.
=8
r/io?.
Solution
by formula.
= 12.
2x
3
4.
any quadratic equation may be obtained by 6.
=0.
article.184
ELEMENTS OF ALGEBRA
45
46.
231.
.c
= 0.
ao.
o^
or
-}-
3 ax == 4 a9
7 wr
.
Solving this equation we obtain
by the method of the preceding
2a
The
roots of
substituting the values of a.
49. -\-bx-\.
The
difference of
|.
area
A
a perimeter of 380
rectangular field has an area of 8400 square feet and Find the dimensions of the field.
7.
189
the equations whose roots are
53.
.
6.
G.
:
3.
two numbers is 4. -5.
2. Find the number.
1.
-2.3.0.
5. The
11.
Problems involving quadratics have
lems of this type have only one solution.
2.QUADRATIC EQUATIONS
Form
51.
-2.
PROBLEMS INVOLVING QUADRATICS
in general two answers. and equals 190 square inches.
What
are the
numbers
of
?
is
The product
two consecutive numbers
210.
-2.
and whose
product
9. and consequently many prob-
235.0.
88
its reciprocal
A
number increased by three times
equals
6J.1.
3.
The sum
of the squares of
two consecutive numbers
85.
its
sides of a rectangle differ by 9 inches.
8.
54.
58.
52.
57.
Find the number.
1.
3.2.
EXERCISE
1.
56. but frequently the conditions of the problem exclude negative or fractional answers. and the difference Find the numbers. feet.
Find
the numbers.
is
Find two numbers whose product
288.3.
Find a number which exceeds
its
square by
is
-|.
of their reciprocals is
4.3.
and whose sum
is
is 36.
Find two numbers whose difference
is 40.
Divide CO into two parts whose product
is 875.
Twenty-nine times a number exceeds the square of the 190.
0.
55.
number by 10. Find the sides. -2.9.
-4.
and lost as many per cent Find the cost of the watch.10.
ELEMENTS OF ALGEBRA
The length
1
B
AB of a rectangle. a distance One steamer travels half a mile faster than the two hours less on the journey. A man bought a certain number of apples for $ 2.
The diagonal
:
tangle as 5 4. exceeds its widtK AD by 119 feet. ABCD.
and gained as many per Find the cost of the horse.
ply between the same two ports. What did he pay for each
apple ?
A man bought a certain number of horses for $1200.
.
15. If a train had traveled 10 miles an hour faster.
If he
each horse ?
. and Find the sides of the rectangle.
19. and lost as many per cent Find the cost of the watch. it would have needed two hours less to travel 120 miles.
of a rectangle is to the length of the recthe area of the figure is 96 square inches.
other.
A man
A man
sold a
as the watch cost dollars. What did he pay for
21. sold a horse for $144.
watch for $ 24.
17.
14. dollars.
as the
16.190
12.
13.
c equals 221
Find
AB and AD.
Two steamers
and
is
of 420 miles. start together on voyages of 1152 and 720 miles respectively. he had paid 2 ^ more for each apple. Two vessels.
A man
cent as the horse cost dollars. At what rates do
the steamers travel ?
18. he would have received 12 apples less for the same money. and the slower reaches its destination one day
before the other. and the line BD joining
two opposite
vertices (called "diagonal")
feet.
watch cost
sold a watch for $ 21. he would have received two horses more for the same money.
vessel sail ?
How many
miles per hour did the faster
If 20. one of which sails two miles per hour faster than the other. Find the rate
of the train. had paid $ 20 less for each horse.
Find the side of an equilateral triangle whose altitude
equals 3 inches.) 25. is surrounded by a walk of uniform width.
B
AB
AB
-2
191
grass plot.
Find
TT r (Area of a circle . so that the rectangle. 30 feet long and 20 feet wide. If the area of the walk is equal to the area of the plot.
=9
Therefore
x
=
\/8
= 2. The number of eggs which can be bought for $ 1 is equal to the number of cents which 4 eggs cost.
EQUATIONS IN THE QUADRATIC FORM An equation is said to be in the quadratic form
if it
contains only two unknown terms. constructed with and CB as sides. a point taken.
Solve
^-9^ + 8 =
**
0. In how many days can B do the work ?
=
26. Find and CB.
By formula. and working together.
and the area of the path
the radius of the basin.
237.
. and the unknown factor of one of these terms is the square of the unknown factor of the
other. contains B 78 square inches.
Ex.
is
On the prolongation of a line AC.
1.
27.
(tf. How many eggs can be bought for $ 1 ?
236.QUADRATIC EQUATIONS
22. as
0. Equations in the quadratic form can be solved by the methods used for quadratics. how wide is the walk ?
23. the two men can do it in 3 days.
or x
= \/l = 1. 23 inches long.I) -4(aj*-l)
2
= 9.
24.
^-3^ = 7.
of the area of the basin.
A rectangular
A
circular basin is surrounded
is
-
by a path 5
feet wide. A needs 8 days more than B to do a certain piece of work.
It is. we let these quantities be what they must be if the exponent law of multiplication is generally true.
must be
*The symbol
smaller than.CHAPTER XVI
THE THEORY OF EXPONENTS
242.
>
m therefore. that a
an
= a m+n
.
(ab)
. (a m ) w
. instead of giving a formal definition of fractional and negative exponents.
The
first
of these laws
is
nition of power.
for all values
1
of
m and n. however.
no
Fractional and negative exponents. hence.
provided
w > n. very important that all exponents should be governed by the same laws. such as 2*.
Then the law
of involution. The following four fundamental laws for positive integral exponents have been developed in preceding chapters
:
I.*
III.
the direct consequence of the defiand third are consequences
FRACTIONAL AND NEGATIVE EXPONENTS
243.
we may choose
for such
symbols any definition that
is
con-
venient for other work."
means "is greater than"
195
similarly
means "is
.
We assume.
m
IV.
= a""
<
.
II.
a m a" = a m+t1 . (a ) s=a m = aw bm
a
.a" = a m n
mn .
244. while the second of the first. 4~ 3 have meaning according to the original definition of power. ~ a m -f. and
.
ml. as.196
ELEMENTS OF ALGEBRA
true for positive integral values of n.
we
try to discover the
let the
meaning of
In every case we
unknown quantity
and apply to both members of the equation that operation which makes the negative.
To
find the
meaning
of
a fractional exponent.
^=(a^)
3*
3
. 3*.
Hence
Or
Therefore
Similarly.
'&M
A
27.
23.
a*. a . fractional.
Assuming these two
8*.
-
we
find
a?
Hence we
define a* to be the qth root of of.
m$.
29. or zero exponent
equal
x.
= a.
a\
26. since the raising to a positive integral power is only a repeated multiplication.
28.
31.
disappear.
e.
. 4~ .
0?=-^. 25.
(bed)*.
30.
a?*.
Write the following expressions as radicals :
22.
n 2 a.
laws. at.g.
245. (xy$.
Let
x
is
The operation which makes the fractional exponent disappear evidently the raising of both members to the third power.
24.
etc.
Multiplying both members by
a".
each
is
The
fact that a
if
=
we
It loses its singularity
1 sometimes appears peculiar to beginners.
248.
ELEMENTS OF ALGEBRA
To
find the
meaning
of a negative exponent.
cr n. in which
obtained from the preceding one by dividing both
members by
a. consider the following equations.198
247.
vice versa.
a
a
a
= =
a a a
a1
1
a.g.
Let
x=
or". Factors
may
be transferred
from
the
numerator
to
the
denominator of a fraction.2
=
a2
.
Or
a"#
= l. e.
etc.
an x = a.
.
a8 a
2
=
1
1
.
by changing the sign of
NOTE. or
the exponent.
1 Multiply 3 or
+x
5 by 2 x
x. the term which does not contain x may be considered as a term containing #.2 d
.
lix
=
2x-l
=+1
Ex.
40.
Divide
by
^
2a
3 qfo
4.
we wish to arrange terms according to descending we have to remember that.202
ELEMENTS OF ALGEBRA
32.
If
powers of
a?.
1.
powers of x arranged are
:
Ex.
34.
6
35.
V ra
4/
3
-\/m
33. The
252.
2.
Arrange in descending powers of
Check.
1.
60.
46.y.
-v/a
-
DIVISION OF RADICALS
267.
E.
47.
44.
(2
45.
Va
-v/a.
it
more convenient to multiply dividend and divisor by a factor which makes the divisor rational.
(3V5-2V3)(2V3-V3).
(3V3-2Vo)(2V3+V5).
Ex.
a fraction.
(V50-f 3Vl2)-4-V2==
however.
43.
53.
.
(5V7-2V2)(2VT-7V2).V5) ( V3 + 2 VS).
V3 .
52.
48.
is
1
2.
268. Monomial surdn of the same order may be divided by multiplying the quotient of the coefficients by the quotient of the
surd factors. the quotient of the surds
is
If.
49.
Ex.
ELEMENTS OF ALGEHRA
(3V5-5V3)
S
.
(5V2+V10)(2V5-1).
all
monomial surds may be divided by
method.214
42.
51. a
VS
-f-
a?Vy
= -\/ -
x*y
this
Since surds of different orders can be reduced to surds of
the same order.
however.
by V7. To show that expressions with rational denominators are simpler than those with irrational denominators. e.
1. the rationalizing factor
x
' g
\/2.
3.57735.RADICALS
This method.
Hence
in arithmetical
work
it
is
always best to
rationalize the denominators before dividing. is illustrated
by
Ex.73205
we
simplify
JL-V^l
V3
*>
^>
division
Either quotient equals .
is
Since \/8
12 Vil
=
2 V*2.
Divide
VII by v7.
Divide 4 v^a by
is
rationalizing factor
evidently \/Tb
hence. Evidently.by the usual arithmetical method.g. called rationalizing the
the following examples
:
215
divisor.
.
The
2. the by 3 is much easier to perform than the division by
1.
/~
}
Ex.. arithTo find.
VTL_Vll '
~~"
\/7_V77
.
. metical problems afford the best illustrations. we have
V3
But
if
1.
4\/3~a'
36
Ex.73205.
+ 4\/5 _ 12v 3 + 4\/5 V8 V8
V2 V2
269.
we have
to multiply
In order to make the divisor (V?) rational.
Divide 12 V5
+ 4V5 by V.
.
Factor 27 a* -f
27 a 6
8.
Ex.
and have
for
any positive integral value of
If n
is
odd.
xn -f.
By
we obtain the other
factors.
2. if n For ( y) n -f y n = 0.
Two
special cases of the preceding propositions are of
viz.
:
importance.
Factor
consider
m
m
6
n9
.
It
y is
not divisible by
287.
The
difference of
two even powers should always be
considered as a difference of two squares.
We may
6
n 6 either a difference of two squares or a
dif-
* The symbol
means " and so forth to.
is
odd. For substituting y for x.
2.
it
follows from the Factoi
xn y n is always divisible by x y.
1.xy +/).
2 8 (3 a )
+8=
+
288. if n is even.
actual division
n.
-
y
5
=
(x
-
can readily be seen that #n -f either x + y or x y.y n is divisible by x -f ?/.
x* -f-/
= (x +/)O . if w is odd.
286.
2
Ex. If n is a Theorem that
1.
ar
+p=
z6
e.230
285.g. xn y n y n y n = 0.
ELEMENTS OF ALGEBRA
positive integer."
.
By making x
any * assigned
zero.
(1).
Let
2.
of the second exceeds the product of the first
Find three consecutive numbers such that the square and third by 1.
(1)
is
an
identity.g.increases
if
x
de-
x
creases.x'2 2 x =
1.
1.
I.i
solving
a problem
the result
or oo indicates that the
all
problem has no solution.
be the numbers.
. it
is
an
Ex.
Interpretation of
QO
The
fraction
if
x
x
inis
infinitely large.
i. the answer is indeterminate.
= 10.e.
(a:
Then
Simplifying.
great.
creases.
.
The
~~f
fraction .
is satisfied
by any number.
Or.
1.
ELEMENTS OF ALGEBRA
Interpretation of ?
e. cancel.
TO^UU"
sufficiently small.can be
If
It is
made
larger than
number.242
303. or that x may equal any finite number.
oo is
= QQ.
ToU"
^-100 a.
i.000
a. without exception.decreases
X
if
called infinity.
306. while the
remaining terms do not
cancelj the root is infinity.
Hence such an equation
identity.
customary to represent this result
by the equation ~
The symbol
304.
the
If in an equation
terms containing
unknown quantity
cancel.
and
. or infinitesimal) This result is usually written
:
305.e.
as
+ l.
+
I)
2
x2
'
-f
2x
+
1
-x(x + 2)= .
Hence any number will satisfy equation the given problem is indeterminate.
x
-f 2.
and becomes infinitely small. equation.
The
solution
x
=-
indicates that the problem
is indeter-
If all terms of an minate.
(1)
= 0. however
x approaches the value
be-
comes
infinitely large.
the area becomes
-f%
of
the original area.
rectangle is 360 square Find the lengths of the sides.
the
The mean proportional between two numbers sum of their squares is 328.
and the sum of
(
228.
ELEMENTS OF ALGEBRA
The
difference between
is
of their squares
325.
Find the
sides of the rectangle.
is
is
17 and the
sum
4.
9. and the side of one increased by the side of the other e.
The area of a
nal 41 feet.
8. and the edge of one exceeds the edge of the other by 2 centimeters.
255 and the sum of
5. Find the side of each square. Find the edge of each cube. The volumes of two cubes differ by 98 cubic centimeters. 148 feet of fence are required. 190.
103.
14.
and the diago(Ex.)
The area
of a right triangle is 210 square feet. equals 4 inches.
.
146 yards.
of a right triangle is 73. 12.
of a rectangular field
feet.
13.)
53 yards. Find the numbers. Find the dimensions of the
field. To inclose a rectangular field 1225 square feet in area.
and
is
The area of a rectangle remains unaltered if its length increased by 20 inches while its breadth is diminished by 10 inches. Find the edges. Find two numbers whose product whose squares is 514. p.
is
the breadth
diminished by 20 inches.
10.244
3. Two cubes together contain 30| cubic inches. Find these sides.quals 20 feet. The sum of the areas of two squares is 208 square feet.
The hypotenuse
is
the other two sides
7. But if the length is increased by 10 inches and
12.
Find the other two
sides. and the edge of one.
two numbers Find the numbers. and
its
The diagonal
is
is
perimeter
11.
and the
hypotenuse
is 37.
is 6.
6.
Find the
sides. increased by the edge of the other.
SIMULTANEOUS QUADRATIC EQUATIONS
15.
differ by 8 inches.
the quotient
is 2. (Surface of sphere
If a
number
of
two
digits be divided
its digits. Find the radii. Find the number.
and
if
the digits will be interchanged.) (Area of circle
and
=
1
16.
. irR *. their areas are together equal to the area of a circle whose radius is 37 inches.
by the product of 27 be added to the number. and the equal to the surface of a sphere Find the radii.
The
radii of
two spheres
is
difference of their surfaces
whose radius = 47T#2.
is
20 inches.)
17.
245
The sum of the radii of two circles is equal to 47 inches.
added to each term to obtain the next one.
..
11.
.
.
-f
.
The common differences are respectively 4. the second a descending. The first is an ascending. 12.
:
7..
-4. and d.
of the following series is
3. 3 d must be added to a.. P.
of a series are its successive numbers.
Since d
is
a
-f
3
d.
a
+
d.) is a series. progression.
309. a + 2 d.. . (n 1) d must be added to a. to produce the nth term.. 19.. P. to produce the 4th term.
a. 16.
Hence
/
= a + (n . is derived from the preceding by the addition of a constant number.11 246
(I)
Thus the 12th term of the
3
or 42.CHAPTER XX
PROGRESSIONS
307..
The terms
ARITHMETIC PROGRESSION
308. The progression is a. each term of which. P.
10. to produce the 3d term. An arithmetic progression (A.1) d.
To
find the
nth term
/
of an A. 2 d must be added to a.. 15 is 9 -f.7. a -f d. the first
term a and
the
common difference d being given.
The common
Thus each
difference is the
number which added
an A.
to each
term produces the next term.
+
2 d. except the first.
17. 3.
series 9. a
3d.. a
11..
to
A series
is
a succession of numbers formed according
some
fixed law.
Find the 7th term of the Find the 21st term
series
. . 7. the
term
a.
1.
247
first
To
find the
sum s
19
of the first
n terms of an A. 2..
6.
Find the 101th term of the
series 1. d
.
.
2
EXERCISE
1. -7.16..3 a = -l.8..
. 5..
= a + (a
Reversing the order.
7.4. 8.
= 99.
Adding. ?
(a) 1.
8. -4^.. 5. P.
Which
(6)
(c)
of the following series are in A. 2J.
series 2. 19..
9.
.PROGRESSIONS
310.
1-J. 99) = 2600.
5.
5. 3.
Find the 12th term of the
-4.. 2
sum
of the first 60
I
(II)
to find the
' '
odd numbers.. 6.
if
a = 5. 21. -24.. P. P. 8. a = 2.
Find the nth term of the
series 2. d = 3.
Find the 5th term of the
4.
2*=(a + Z) + (a + l) + (a + l)
2s = n
*
.-. 3.
of the series 10. 6.
= -2.
1..
the last term
and the common difference d being given. 5.
Or
Hence
Thus
from
(I)
= (+/)..-..
first
2
Write down the
(a)
(6)
(c)
6 terms of an A.
= I + 49 = *({ +
... -10.-
(a
+ + (a +
l)
l).
3.
6
we have
Hence
.
9. 3.
..
2.
-|. 4.
..
series
.
-3.' cZ == .
Find the 10th term of the
series 17.
115..
(d) 1J.
-.
16.7 -f
to 12 terms.
20.
1. 11.
15.
13.
ELEMENTS OF ALGEBRA
last
term and the sum of the following series :
.
33. and for each than for the preceding one. hence if any three of them are given.
3.
to 8 terms. 1|.
18.
$1
For boring a well 60 yards deep a contractor receives yard thereafter 10^ more How much does he receive all
together ?
^S5 A bookkeeper accepts a position at a yearly salary of $ 1000.
to 7 terms. and a yearly increase of $ 120.
7.
23.
.
(i)
(ii)
.
19.
2.
.
4.
1. strike
for the first yard.
+ 2-f-3 + 4 H
hlOO. In most problems relating to A.
.
to 20 terms.
to 20 terms.
>
2-f
2. 2J. 11. How much does he receive (a) in the 21st year (6) during the first 21 years ?
j
311. 7.
6. the other two may be found by the solution of the simultaneous equations
. P.248
Find the
10. 12. 7.
1+2+3+4H
Find the sum of the
first
n odd numbers.
(x +"l) 4.
to 20 terms.
\-n. 31.
to 15 terms..5
H + i-f
-f-
to 10 terms.
21.
Q^) How many times
in 12 hours ?
(&fi)
does a clock.
.
rf. striking hours only.
11. 29.
to 10 terms.
1J.
:
3.
15.
8.
. 11.
.
12.
Sum
the following series
14.1 -f 3.
.
+ 3.
22. 15.
17. 16.
to 16 terms. Jive quantities are involved.
'.(#
1
2) -f (x -f 3) H
to
a terms.
Find d. n = 16. Given a = |.
6?
9. 78. Find?.
f
J 1 1
/
.
has the series 82. P.
13.
4.
a+
and
b
a
b
5. = 17.
A
$300
is
divided
among 6 persons
in such a
way
that each
person receives $ 10 did each receive ?
more than the preceding
one.
I
Find
I
in terms of a. n = 17. Given a = 4. of 5 terms
6.
17.
15. = 83.
12. Find d. = 45. = 1870.
7.
produced.250
ELEMENTS OF ALGEBRA
EXERCISE
116
:
Find the arithmetic means between
1. Given a = .
f?
. 74.
11.
14.
Between 10 and 6
insert 7 arithmetic
means
. Find d and Given a = 1700. n.
.
How much
.
and
s.
8. = 52.
m
and
n
2.
Between 4 and 8
insert 3 terms (arithmetic
is
means)
so
that an A. n = 20.
10. and all his savings in 5 years amounted to $ 6540.3.
How many terms How many terms
Given d = 3. Find w.
y and #-f-5y. ceding one. n
has the series
^
j
. d = 5. = ^ 3 = 1. Find a Given a = 7.
= 16.
16.
man saved each month $2 more than in the pre 18.
I. s == 440. n = 4. s = 70. n = 13.
a x
-f-
b
and a
b. Find n. Find a and Given s = 44. Given a = 1.
T?
^. How much did he save the first month?
19.
3.
If
n
is less
:
than unity.. P. 24.)
is
a series each term of
which. -I.
is
16(f)
4
..
4. rs =
s
2
-..
. 24.
the following form 8
nf +
q(l-r")
1
r
.
A geometric progression
first. 36.arn ~ l .. the first term a and
the ratios r being given.g. 108.
Therefore
Thus the sum
= ^ZlD.
NOTE. P.
ratios are respectively 3.. 12. To find the sum s of the first n terms term a and the ratio r being given.
4. +1.. ar. or.
g==
it is
convenient to write formula' (II) in
*.
The progression is a.
4-
(1)
.PROGRESSIONS
251
GEOMETRIC PROGRESSION
313.
.
r
n~ l
.
(II)
of the
8 =s
first
6 terms of the series 16..
s(r
1)
8
= ar"
7*
JL
a.
fl
lg[(i)
-l]
==
32(W -
1)
= 332 J.
.
The
314.
(I)
of the series 16. the first
= a + ar -for ar -f ar Multiplying by r.
Hence
Thus the 6th term
l
= ar
n~l
. 36. a?*2 To obtain the nth term a must evidently be multiplied by
.
-2..
E.
|..
2 a.
. P.
of a G. 36.
and
To
find the
nth term
/ of
a G.
or 81
315. except the
multiplying
derived from the preceding one by by a constant number.
.
2
arn
(2)
Subtracting (1) from
(2). called the ratio.
ar8
r. is
it
(G.. <zr .
9.
series 5. 1..
first
5.
f..
144. f.
l. Jive quantities are in.
0.
Find the 7th term of the Find the 6th term of the
Find the 9th term of the
^.
.
.
a
=
I.18..72.
7.
(d) 5..
i 288..
9.. whose and whose second term is 8. 72.l.
series 6.
And the
required
means are
18. P.
series
Find the llth term of the Find the 7th term of the
ratio is
^.-.
6.
.288.6. In most problems relating to G.18.
36.
Evidently the total
number
of terms is 5
+ 2.
117
Which
(a)
of the following series are in G. 3.
is 16.
Write down the first 6 terms of a G.*.
To
insert 5 geometric
means between 9 and 576.
first
term
4.
-fa.
.
Find the 5th term of a G. I
= 670.
series
.
.
(b) 1. 25. 36. hence.
8.
is 3.
.
576. + 5. 144. 18. P.
-fa.
or
7.
.252
ELEMENTS OF ALGEBRA
316. 9.
volved .
\
t
series
.
2
term
3.5.
..
10.
Hence the
or
series is
0.
(it.
676
t
Substituting in
= r6 = 64.. 4.
..4.
72. 144. ?
(c)
2. P.
first
term
is
125 and
whose common
..
Write down the first 5 terms of a G. P.
.
4. 36._!=!>.
Ex.
EXERCISE
1..
. P. 676. 80.
Hence n
=
7. if any three of them are given.
.
.
r^2.54.
Find the 6th term of the
series J. whose and whose common ratio is 4. the other two be found by the solution of the simultaneous equations :
may
(I)
/=<!/-'.5.
+-f%9 %
...
|. 20.
288. whose
.
A
the
boy
is
as old as his father
and
3 years
sum
of the ages of the three is 57 years.
is
What are their ages ? Two engines are together
more than the
of 80 horse
16 horse power
other.
187. 6 in each row the lowest row has 2 panes of glass in each window more than the middle row. + 11 ~ 6.
ELEMENTS OF ALGEBRA
A A
number increased by
3.-36. How many are there in each window ?
.
father.
.
younger than his Find the age of
the father.
The age
of the elder of
it
three years ago of each.266
173.
A
boy
is
father.
.
-ll?/-102.
176.
power one of the two Find the power of each.
A
each
177. and the father's present age is twice what the son will be 8 years
hence.
sister
.
188. 10x 2 192. and 5 h. Four years ago a father was three times as old as his son is now. Find the dimensions of the floor.
178.
3 gives the
174. and the middle row has 4 panes in each window more than the upper row there are in all 168 panes of glass. 12 m. the ana of the floor will be increased 48 square feet.
.
180.
number divided by
3.
and | as old as his Find the age of the
Resolve into prime factors
:
184.
The length
is
of a floor exceeds its width
by 2
feet.
179.
3 gives the
same
result as the
numbet
multiplied by
Find the number.56. z 2
-92. was three times that of the younger. Find the age
5 years older than his sister
183.
2
2
+
a
_ no.
An
The two
express train runs 7 miles an hour faster than an ordinary trains run a certain distance in 4 h.
186. aW + llab-2&.
train.
respectively.
190. x*
185.
two boys is twice that of the younger.
Find the number. 189. z 2 + x .
181.
7/
191.
dimension
182. A house has 3 rows of windows.
What
is
the distance?
if
square grass plot would contain 73 square feet more Find the side of the plot.
if
each
increased 2 feet. 15 m. 4 a 2
y-y
-42. side were one foot longer.
+
a.
+x-
2.
same
result as the
number
diminished by
175.
13 a + 3. the
sum
of the ages of all three is 51.
Tn 6 hours
.a)(x b
b)
(x
b
~
)
412. far did he walk all together ?
A
. x
1
a
x
x1
ab
1
1
a
x
a
c
+
b
c
x
a
b
b
~
c
x
b
416
417. 411.
A man
drives to a certain place at the rate of 8 miles an
Returning by a road 3 miles longer at the rate of 9 miles an hour.(5 I2x
~r
l
a)
.
A
in 9 hours
B walks
11 miles
number of two digits the first digit is twice the second.
(x
-f
ELEMENTS OF ALGEBRA
a)(z
-
b)
=
a
2 alb
=
a
(x
-f
b)(x
2
.
-f
a
x
-f
x
-f c
1
1
a-b
b
x
415.
a
x
a
x
b
b
x
c
b
_a
b
-f
x
414.(c rt
a)(x
-
b)
=
0.
-
a)
-2
6 2a.
hour. 18 be subtracted from the number.
mx ~
nx
(a
~
mx
nx
c
d
d
c)(:r
lfi:r
a
b)(x
. How long is each road ?
423.
a
x
)
~
a
2 b
2
ar
a
IJ a.
2 a
x
c
x
6
-f c
a
+
a
+
a
+
6
-f
walks 2 miles more than B walks in 7 hours more than A walks in 5 hours. he takes 7 minutes longer than in going.
down again
How
person walks up a hill at the rate of 2 miles an hour.
4x
a
a
2 c
6
Qx
3 x
c
419. and at the rate of 3^ miles an hour. Find the number of miles an hour that A and B each walk.c) .
418 ~j-o.278
410.
420. and was out 5 hours.
(x
.
421. In a
if
and
422. the order of the digits will be inverted. Find the number.
if
the
sum of
the digits be multiplied by
the digits will be inverted. Find the sum and the rate of
interest. Of the ages of two brothers one exceeds half the other by 4 is equal to an eighth of 482.
486.
to
. and in 20 months to $275. had each at first?
B
B
then has
J
as
much
spends } of his money and as A. What is that fraction which becomes f when its numerator is doubled and its denominator is increased by 1.
481.
479. the Find their ages. A sum of money at simple interest amounted in 10 months to $2100. Find the numbers. and a fifth part of one brother's age that of the other. Find their ages.
least
The sum
of three
numbers
is
is
21.
A
spends \ of his. Find the principal and the rate of
interest. In a certain proper fraction the difference between the nu merator and the denominator is 12. and 5 times the less exceeds the greater by 3.
477.
by 4.
half the
The greatest exceeds the sum of the greatest and
480. and becomes when its denominator is doubled and its numerator increased by 4 ?
j|
478. Find the
fraction.
485.
whose difference
is
4.
Find
the number. A sum of money at simple interest amounts in 8 months to $260.
years. Find two numbers such that twice the greater exceeds the by 30.
.
487. and if each be increased by 5 the Find the fraction.
A
number
consists of
two
digits
4. also a third of the greater exceeds half the less by 2. Find the numbers. There are two numbers the half of the greater of which exceeds the less by 2. fraction becomes equal to |. and in 18 months to $2180. age.
If 1
be added to the numerator of a fraction
it
if 1
be added to the denominator
it becomes equal becomes equal to ^. and the other number least.
latter
would then be twice the
son's
A
and B together have $6000. How much money
less
484.
thrice that of his son
and added to the father's.282
ELEMENTS OF ALGEBRA
476.
483. If 31 years were added to the age of a father it would be also if one year were taken from the son's age
.
they would have met in 2 hours. N. 37 pounds of tin lose 5 pounds.
532. Tf and run together. in 28 minutes. and one overtakes the other in 6 hours. his father is half as old again as his mother was c years ago. If they had walked toward each other.
touches
and
F respectively.
and B together can do a piece of work in 2 days.
it
separately
?
531.REVIEW EXERCISE
285
525. it is filled in 35 minutes. A vessel can be filled by three pipes. and 23 pounds of lead lose 2 pounds. (a) How many pounds of tin and lead are in a mixture weighing 120 pounds in air.
. 90. and losing 14 pounds when weighed in water? (b) How many pounds of tin and lead are in an alloy weighing 220 pounds in air and 201 pounds in water ?
in 3 days. B and C and C and A in 4 days. In
circle
A ABC. Find the present ages of his father and mother.
527.
530. How long will B and C take to do
. Two persons start to travel from two stations 24 miles apart. AB=6. What are their rates of
travel?
. if and L. CD.
. M. L. A boy is a years old his mother was I years old when he was born. the first and second digits will change places. Find the numbers. Tu what time will it be filled if all run
M
N
N
t
together?
529.
and CA=7. A can do a piece of work in 12 days B and C together can do the same piece of work in 4 days A and C can do it in half the time in which B alone can do it.
and third equals \\ the sum third equals \.
sum of the reciprocals of of the reciprocals of the first of the reciprocals of the second and
the
sum
528. and BE. A number of three digits whose first and last digits are the same has 7 for the sum of its digits.
AC
in /). Throe numbers are such that the
A
the
first
and second equals
. When weighed in water.
E
533. BC = 5. An (escribed) and the prolongations of BA and BC in Find AD. if L and Af in 20 minutes. if the number be increased by Find the number. In how many days can each alone do the same work?
526.
2 541. The value of x that produces the greatest value of y.
545.
e. FRANCE. GERMANY. formation of dollars into marks. x 8
549.
d.
546.286
ELEMENTS OF ALGEBRA
:
534.10 marks.
542.
The values of y.
from x
=
2 to x
= 4.
-
3 x.
of
Draw
a graph for the trans-
The number
in
of
workmen Draw
required to finish a certain piece
the graph
work
D
days
it
is
from
D
1 to
D=
12. the function. z 2
-
x x
-
5.
2. x 2 544. x
2
+
x.
.
x*. 3 x
539. 2 x
+
5.
-
3 x. if x = f 1.
b.
How
is
t /
long will
I
take 11
men
2
t' . x *-x
+
x
+
1. AND BRITISH ISLES
535.
543. 2|.
c. One dollar equals 4. The values of x if y = 2.
540.
550.
547.
to
do the work? pendulum.
-
7. x*
-
2
x.3
Draw
down
the time of swing for a
pendulum
of length
8 feet. then / = 3 and write
=
3. The greatest value of the function.
+
3.
Draw
the graphs of the following functions
:
538. 536. The roots of the equation 2 + 2 x x z = 1.
548.
a. the time of whose swing a graph for the formula from / =0
537.e. i.
. If
to
feet is the length of a
seconds. Draw the graph of y 2 and from the diagram determine
:
+
2 x
x*. 2
-
x
-
x2
. Represent the following table graphically
TABLE OF POPULATION (IN MILLIONS) OF UNITED STATES.
725.
. Find the altitude of an equilateral triangle whose side equals a.
What two numbers
are those whose
sum
is
47 and product
A man
bought a certain number of pounds of tea and
10 pounds more of coffee. Find two consecutive numbers whose product equals 600.
716.l
+
8
-8
+
ft)'
(J)-*
(3|)*
+
(a
+
64-
+ i.
723.
722.
in value. 16 x* . needs 15 days longer to build a wall than B. How shares did he buy ?
if
726. paying $ 12 for the tea and $9 for the coffee. What number exceeds its reciprocal by {$.44#2 + 121 = 0.
___ _ 2* -5 3*2-7
715.
of a rectangle is 221 square feet and its perimeter Find the dimensions of the rectangle.
sum is a and whose product equals J.
The
difference of the cubes of
two consecutive numbers
is
find them. **-13a: 2
710. 721.
217
.292
709. if 1 more for 30/ would diminish
720. 724.
ELEMENTS OF ALGEBRA
+36 = 0. A man bought a certain number of shares in a company for
$375. he
many
312?
he had waited a few days until each share had fallen $6. Find two numbers whose 719. If a pound of tea cost 30 J* more than a pound of coffee. Find four consecutive integers whose product is 7920.
729.
A
equals CO feet. 2n n 2 2 -f-2aar + a -5 = 0. 3or
i
-16 .
The area
the price of 100 apples by $1. what is the
price of the coffee per
pound ?
:
Find the numerical value of
728.
a:
713.
714
2
*2
'
+
25
4
16
|
25 a2
711. In how many days can A build the wall?
718.40 a 2* 2 + 9 a 4 = 0. and working together they can build it in 18 days. 12
-4*+
-
8. 727. Find the price of an apple.25 might have bought five more for the same money. 717.
The
sum
of the circumferences of
44 inches. A and B run a race round a two-mile course. Tf there had been 20 less rows.
two squares equals 140 feet.
The sum
of the perimeters of
sum
of the areas of the squares is 16^f feet.square inches. and also contains 300 square feet. there would have been 25 more trees in a row. In the second
heat
A
.
2240.300
930. 152. The difference of two numbers cubes is 513. If each side was increased by 2 feet.
A
is
938.
The sum of two numbers Find the numbers. and B diminishes his as arrives at the winning post 2 minutes before B.
+ z)=18.000 trees.102.
is
3
.
and the sum of
their areas 78$.
34
939. Find the side of each two
circles is
IT
square. Find the numbers.
Assuming
= -y. In the first heat B reaches the winning post 2 minutes before A.
944.
is 3.
ELEMENTS OF ALGEBRA
(*+s)(* + y)=10. How many rows are there?
941.
diagonal
940.
find
the radii of the two circles.
much and A then
Find at what
increases his speed 2 miles per hour.
. = ar(a? -f y + 2) + a)(* + y
933.
942.
rate each
man
ran in the
first
heat. a second rec8 feet shorter. and 10 feet broader. A plantation in rows consists of 10. z(* + y + 2) = 76.
943. (y + *) = . (3 + *)(ar + y + z) = 96. y(x + y + 2) = 133. and the Find the sides of the and
its
is
squares.
931.
937.
two squares is 23 feet.
and the
difference of
936. *(* + #) =24. The diagonal of a rectangle equals 17 feet.
and the sum of their cubes
is
tangle
certain rectangle contains 300 square feet.
two numbers Find the numbers. the area of the new rectangle would equal 170 square feet.
the
The sum
of the perimeters of
sum
of their areas equals 617 square feet.
the difference of their
The
is
difference of
their cubes
270. Find the length and breadth of the first rectangle. (y
(* + y)(y +*)= 50. The perimeter of a rectangle is 92 Find the area of the rectangle.
935. Find the sides of the rectangle. feet. + z) =108. s(y
932.
feet.
is 20.
y(
934.
the area lengths of the sides of the rectangle.
is
407 cubic feet.
953.
.
overtook
miles. Find the width of the path if its area is 216 square yards. The square described on the hypotenuse of a right triangle is 180 square inches. The area of a certain rectangle is 2400 square feet. triangle is 6. The diagonal of a rectangular is 476 yards.
unaltered.
.
952.
sum
Find an edge of
954. set out from two places. the difference in the lengths of the legs of the Find the legs of the triangle. was 9 hours' journey distant from P. The sum of the contents of two cubic blocks
the
of the heights of the blocks is 11 feet.
P and
Q. at Find the his rate of traveling.
Find the
eter
947.
950. and the other 9 days longer to perform the work than if both worked together. Two men can perform a piece of work in a certain time one takes 4 days longer. A certain number exceeds the product of its two digits by 52 and exceeds twice the sum of its digits by 53.
whose
946. and if 594 be added to the number.REVIEW EXERCISE
301
945. What is its area?
field is
182 yards. and travels in the same direction as A. at
the same time
A
it
starts
and
B
from
Q
with the design to pass through Q.
Two
starts
travelers. A rectangular lawn whose length is 30 yards and breadth 20 yards is surrounded by a path of uniform width.
. A number consists of three digits whose sum is 14. Find the number.
A and
B. and
its
perim-
948. If the breadth of the rectangle be decreased by 1 inch and its
is
length increased by 2 inches. the square of the middle digit is equal to the product of the extreme digits.
951. distance between P and Q. its area will be increased 100 square feet. When
from
P
A
was found that they had together traveled 80 had passed through Q 4 hours before. Find in what time both will do it. The area of a certain rectangle is equal to the area of a square side is 3 inches longer than one of the sides of the rectangle. if its length is decreased 10 feet and its breadth increased 10 feet.
Find the number. and that B. Find two numbers each of which
is
the square of the other. Find its length and breadth.
949. each block. the digits are reversed. that
B
A
955.
all
A
perfect number
is
a number which equals the sum
divisible.04
+
.
to
n terms. Find four numbers in A.. then this
sum multiplied by
(Euclid.-.
989. P.)
the last term
the series
a perfect number.
to oo. and the sum of the first nine terms is equal to the square of the sum of the first two. such that the product of the and fourth may be 55. The 21st term of an A. 2 grains on the 2d. Insert 8 arithmetic means between
1
and
-. and of the second and third 03.
of n terms of 7
+
9
+ 11+
is
is
40.
What
2 a
value must a have so that the
sum
of
+
av/2
+
a
+
V2
+
.REVIEW EXERCISE
978. The
term. 5
11.
first
984.
to infinity
may
be 8?
.
992. and the
common
difference.
to 105?
981.
1.2
.. is 225.
303
979.1
+
2. The sum
982.
Find four perfect
numbers.
990. Find the value of the infinite product 4
v'i
v7-!
v^5
.
987. Insert 22 arithmetic means between 8 and 54.-.
How many
sum
terms of 18
+
17
+
10
+
amount
. who rewarded the inventor by promising to place 1 grain of wheat on
Sessa for the
the 1st square of a chess-board..001
4.3 '
Find the 8th
983.
v/2
1
+ +
+
1
4
+
+
3>/2
to oo
+
+
.+ lY L V.
Find the number of grains which Sessa should have received.
of n terms of an A.
Find
n... Find the
first
term. Find the sum of the series
988. and so on.
986.
980. P. The Arabian Araphad reports that chess was invented by amusement of an Indian rajah. doubling the number for each successive square on the board. P.01
3.
If
of
2
of
integers + 2 1 + 2'2
by which
is
it
is
the
sum
of
the series
2 n is prime.
.
"(.001
+
.---
:
+
9
-
-
V2
+
. 4 grains on the 3d..
0.
985. named Sheran.
998.
1001. Find (a) the sum of all circumferences. (a) after 5 strokes.
ABC
A A
n
same
sides. P. 1003.
and the
fifth
term
is
8 times
the second . and so forth to infinity. P. areas of all triangles. In an equilateral triangle second circle touches the first circle and the sides AB and AC. are
45 and 765
find the
numbers.
994. are 28 and find the numbers. The
fifth
term of a G. (6) the sum of the infinity. prove that they cannot be in A. Insert 4 geometric means between 243 and 32. 995. and G.
997.
AB =
1004.
third circle touches the second circle and the
to infinity. P.
and
if
so forth
What
is
the
sum
of the areas of all circles.
999. In a circle whose radius is 1 a square is inscribed. the sides
of a third triangle equal the altitudes of the second. If a. are unequal. P. in this circle a square. P.
pump removes
J
of the
of air
is
fractions of the original amount contained in the receiver. find the series. Each stroke of the piston of an air
air contained in the receiver. at the same time.
. The sides of a second equilateral triangle equal the altitudes of the first.
of squares of four
numbers
in
G.
The sum and sum
. (I) the sum of the perimeters of
all
squares.304
ELEMENTS OF ALGEBRA
993. and so forth to Find (a) the sum of all perimeters. The sum and product of three numbers in G.
512
996. ft. c. Insert 3 geometric means between 2 and 162. The side of an equilateral triangle equals 2. The other travels 8 miles the first day and After how increases this pace by \ mile a day each succeeding day. inches. Two travelers start on the same road. One of them travels uniformly 10 miles a day.
1000.
is 4. (6) after n
What
strokes?
many
1002. after how strokes would the density of the air be xJn ^ ^ ne original
density ?
a circle is inscribed. in this square a circle. Under the conditions of the preceding example. many days will the latter overtake the former?
.
so that the Logarithms.
xi 4-
373 pages. The author
has emphasized Graphical Methods more than is usual in text-books of this grade. 64-66 FIFTH AVBNTC.
which have been omitted from the body of the work Indeterminate Equahave been relegated to the Appendix. great many
work. which has been retained to serve as a basis for higher work.
i2mo. save Inequalities. The more important subjects
tions. and commercial life. but these few are treated so thoroughly and are illustrated by so many varied examples that the student will be much better prepared for further
The Exercises are superficial study of a great many cases.
not
The Advanced Algebra is an amplification of the Elementary.
$1. but the work in the latter subject
has been so arranged that teachers
who
wish a shorter course
may omit
it
ADVANCED ALGEBRA
By ARTHUR SCHULTZE. Ph.
THE MACMILLAN COMPANY
PUBLISHERS. etc. comparatively few methods are heretofore. Particular care has been bestowed upon those chapters which in the customary courses offer the greatest difficulties to the beginner. proportions and graphical methods are introduced into the first year's course.25
lamo.
xiv+563
pages. than by the
. without the sacrifice of scientific accuracy and thoroughness. book is a thoroughly practical and comprehensive text-book. given. To meet the requirements of the College Entrance Examination Board. physics. but none of the introduced illustrations is so complex as to require the expenditure of
time for the teaching of physics or geometry.
HEW TOSS
. especially
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Work
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Problems and Factoring. and the Summation of Series is here presented in a novel form.
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examples are taken from geometry.
$1.ELEMENTARY ALGEBRA
By ARTHUR SCHULTZE.D. very numerous and well graded there is a sufficient number of easy examples of each kind to enable the weakest students to do some work. All subjects now required for admission by the College Entrance Examination Board
have been omitted from the present volume.
Half
leather.
Half leather. The introsimpler and more natural than the
methods given
In Factoring.10
The treatment of elementary algebra here is simple and practical.
12010. than by the superficial study of a great many cases. save Inequalities.D.
$1. Logarithms.
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Problems and Factoring.25
i2mo.
In Factoring. The Exercises are very numerous and well graded.10
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of elementary algebra here
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has emphasized Graphical Methods more than is usual in text-books of this and the Summation of Series is here presented in a novel form. and commercial life. there is a sufficient number of easy examples of each kind to enable the weakest students to do some work. All subjects now required for admission by the College Entrance Examination Board
have been omitted from the present volume. To meet the requirements of the College Entrance Examination Board.
HEW YOKE
. great many
A
examples are taken from geometry. comparatively few methods are
given.
THE MACMILLAN COMPANY
PUBLISHBSS. but the work in the latter subject
has been so arranged that teachers
who
wish a shorter course
may omit
it
ADVANCED ALGEBRA
By ARTHUR SCHULTZE. The introsimpler and more natural than the
methods given heretofore. physics. but these few are treated so thoroughly and are illustrated by so many varied examples that the student will be much better prepared for further
work. bestowed upon those chapters which in the customary courses offer the greatest difficulties to the beginner. but none of the introduced illustrations is so complex as to require the expenditure of
time for the teaching of physics or geometry. which has been retained to serve as a basis for higher work.
Half leather.
HatF leather. so that the tions. 64-66
7HTH
AVENUE. proportions and graphical methods are introduced into the first year's course.ELEMENTARY ALGEBRA
By ARTHUR Sen ULTZE.
$1.
xi
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373 pages. The author
grade. book is a thoroughly practical and comprehensive text-book. without
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The Advanced Algebra is an amplification of the Elementary. Ph. etc.
Many proofs are presented in a simpler and manner than in most text-books in Geometry 8. 9. Proofs that are special cases of general principles obtained from the Exercises are not given in detail. SCHULTZE.
SEVENOAK. Ph. at the
It
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provides a course which stimulates him to do original time.10
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This Geometry introduces the student systematically to the solution of geometrical exercises. 6.
Half
leather. The Schultze and Sevenoak Geometry is in use in a large number of the leading schools of the country. The Analysis of Problems and of Theorems is more concrete and practical than in any other
distinct pedagogical value.
ments from which General Principles may be obtained are inserted in the " Exercises.
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PLANE GEOMETRY
Separate.10
By ARTHUR
This key will be helpful to teachers who cannot give sufficient time to the Most solutions are merely outsolution of the exercises in the text-book.
PLANE AND SOLID GEOMETRY
F.
By ARTHUR SCHULTZE and
370 pages.
. Difficult Propare made somewhat? easier by applying simple Notation . under the heading Remarks". more than 1200 in number in 2. aoo pages.
wor. Pains have been taken to give Excellent Figures
throughout the book.
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.
NEW YORK
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Algebraic Solution of Geometrical Exercises is treated in the Appendix to the Plane Geometry .
KEY TO THE EXERCISES
in
Schultze and Sevenoak's Plane and Solid Geometry.
THE MACMILLAN COMPANY
PUBLISHERS. These are introduced from the beginning 3. Preliminary Propositions are presented in a simple manner .D. Cloth.
text-book in Geometry
more
direct
ositions
7. The numerous and well-graded Exercises the complete book.
xii
+ 233 pages. 64-66 FIFTH AVENUE.
and no attempt has been made
to present these solutions in such form
that they can be used as models for class-room work.
$1.
lines.
i2mo.
Cloth.
25
The
author's long
and successful experience as a teacher
of mathematics in secondary schools and his careful study of the subject from the pedagogical point of view. and Assistant Professor of Mathematics in New York University
of
Cloth. 370 pages. New York City.
THE MACMILLAN COMPANY
64-66 Fifth Avenue.
New York
DALLAS
CHICAGO
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" The chief object of the speak with unusual authority. Most teachers admit that mathematical instruction derives its importance from the mental training that it But in affords.
causes of the inefficiency of mathematical teaching.
methods of teaching mathematics the first propositions in geometry the original exercise parallel lines methods of the circle attacking problems impossible constructions applied problems typical parts of algebra.
.
.The Teaching
of
Mathematics
in
Secondary Schools
ARTHUR SCHULTZE
Formerly Head of the Department of Mathematics in the High School Commerce.
. a great deal of mathematical spite
teaching
is
still
informational.
. " is to contribute towards book/ he says in the preface.
.
. $1. of these theoretical views. and not from the information that it imparts. Typical topics the value and the aims of mathematical teach-
ing
.
.
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to
still
learn
demon-
strations instead of learning
how
demonstrate. making mathematical teaching less informational and more disciplinary."
The treatment
treated are
:
is concrete and practical.
12mo.
is
an excellent example of the newer type of
school histories.
New York
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This book
pupil's mind the general movements in American history and their relative value in the development of our nation.
Cloth.
diagrams. This book is up-to-date not only in its matter and method. Maps. diagrams.
$1.
but in being fully illustrated with
many excellent maps.
i2mo. supply the student with plenty of historical
narrative on which to base the general statements and other classifications made in the text.AMERICAN HISTORY
For Use
fa
Secondary Schools
By ROSCOE LEWIS ASHLEY
Illustrated.40
is distinguished from a large number of American text-books in that its main theme is the development of history the nation. which put the main stress upon national development rather than upon military campaigns.
An exhaustive system of marginal references. Studies and Questions at the end of each chapter take the place of the individual teacher's lesson plans.
Topics.
THE MACMILLAN COMPANY
64-66 Fifth Avenue. All
smaller movements and single events are clearly grouped under these general movements. and a full index are provided.
photographs. " This volume
etc. which have been selected with great care and can be found in the average high school library. The book deserves the attention
of history teachers/'
Journal of Pedagogy. | 677.169 | 1 |
It is easy to confuse the processes for solving for the rate versus the number of years in the compound interest formula. The two MathFAQs compare the process of solving for the rate (using roots or powers) with solving for years (using logarithms)
Although a relative extrema may seem to be very similar to an absolute extrema, they are actually quite different. The term "relative" means compared to numbers nearby…so a relative extrema is either a bump or a dip on the function.
The term "absolute" means the most extreme on the entire function. An absolute extrema is the very highest or lowest point on the function. This may occur at a bump or a dip. They may also occur at the ends of the function if it is defined on a closed interval.
The basic algorithm for solving a standard minimization problem is covered in Section 4.3. This process, called the Simplex Method, uses matrices and row operations to gauge whether an objective function is maximized at corner points.
In the example below, I write out a standard maximization problem from an application and then solve it with the Simplex Method.
In your classes, you might hear about instructors who grade on "a curve". There is an idea that this might somehow benefit you when it comes to grading. Let's take a look how that might work if the curve is a normal curve. | 677.169 | 1 |
1356626
ISBN: 0521356628
Publication Date: 1988
Publisher: Cambridge University Press
AUTHOR
Reid, Miles, Series, C. M., Bruce, J. W.
SUMMARY
Algebraic geometry is, essentially, the study of the solution of equations and occupies a central position in pure mathematics. This short and readable introduction to algebraic geometry will be ideal for all undergraduate mathematicians coming to the subject for the first time. With the minimum of prerequisites, Dr Reid introduces the reader to the basic concepts of algebraic geometry including: plane conics, cubics and the group law, affine and projective varieties, and non-singularity and dimension. He is at pains to stress the connections the subject has with commutative algebra as well as its relation to topology, differential geometry, and number theory. The book arises from an undergraduate course given at the University of Warwick and contains numerous examples and exercises illustrating the theory.Reid, Miles is the author of 'Undergraduate Algebraic Geometry', published 1988 under ISBN 9780521356626 and ISBN 05213566 | 677.169 | 1 |
IB Math Studies Advice
Showing 1 to 3 of 7
This course teaches math in a way that is easy to understand. However, the course is still challenging enough to be interesting. Additionally, the math taught is useful for real life. The course teaches financial math and statistics.
Course highlights:
Through this course I learned how statistics works. This is useful when I read studies; I know understand what terms like 'statistically significant' mean. I also learned financial math which is useful for knowing how to manage finances later in life.
Hours per week:
3-5 hours
Advice for students:
Pay attention in class since the explanations of the math are very helpful and thorough. Make sure to use your graphing calculator as much as possible since it will help you get precise answers. Lastly, don't be scared to ask questions.
Course Term:Fall 2017
Professor:ANNA ZUNIGA
Course Required?Yes
Course Tags:Math-heavyGreat Intro to the SubjectMany Small Assignments
Jul 26, 2017
| Would highly recommend.
Not too easy. Not too difficult.
Course Overview:
IB Math Studies is a class for many students who aren't that strong in math and do not think they should go on to Pre-Calculus.
Course highlights:
Something to look forward to in this class is the chance to relearn many subjects already taught to you in your previous years (ex:algebra). You get to go over algebra 1 & 2, trigonometry, statistics, and much more.
Hours per week:
6-8 hours
Advice for students:
Students who are interested in taking this course should know that like every other classes you must study and work hard to see an actual improvement. One other thing they should take into consideration is to never give up and always keep trying especially since its math.
Course Term:Fall 2016
Professor:Ms. Zuniga
Course Required?Yes
Course Tags:Math-heavyGreat Intro to the SubjectMany Small Assignments
Jun 16, 2017
| Would recommend.
Not too easy. Not too difficult.
Course Overview:
If you're not the best at math, this course is fairly doable, and Ms.Zuniga will ensure you understand the topics as long as you put in some effort, and you show up to class.
Course highlights:
Trigonometry, graphs, statistics, systems of equations, how to use a graphing calculator. And IMPORTANT KNOWLEDGE about finances, credit card, investments, etc.
Hours per week:
3-5 hours
Advice for students:
Be in class, do your homework. Zuniga will be more than happy to help you, so don't be afraid to ask questions | 677.169 | 1 |
Having trouble understanding algebra? Do algebraic concepts, equations, and logic just make your head spin? We have great news: Head First Algebra is designed for you. Full of engaging stories and practical, real-world explanations, this book will help you learn everything from natural numbers and exponents to solving systems of equations and graphing polynomials.
Along the way, you'll go beyond solving hundreds of repetitive problems, and actually use what you learn to make real-life decisions. Does it make sense to buy two years of insurance on a car that depreciates as soon as you drive it off the lot? Can you really afford an XBox 360 and a new iPhone? Learn how to put algebra to work for you, and nail your class exams along the way.
Your time is way too valuable to waste struggling with new concepts. Using the latest research in cognitive science and learning theory to craft a multi-sensory learning experience, Head First Algebra uses a visually rich format specifically designed to take advantage of the way your brain really works.
Table of Contents
Chapter 1
What is Algebra?: Solving for unknowns...
It all started with a big gaming sale
What does a system really
cost?
Algebra is about solving for
unknowns
Jo's got more unknowns
X marks the spot unknown
Equations are math
sentences
Now SOLVE for the
unknown
So which operation do you use when?
Jo is ready to accessorize!
Equation training
Jo has an awesome setup!
Math Toolbox
Chapter 2
(More) Complicated Equations: Taking Algebra on the road
Paul loves "Pajama Death"
Always start with what you know
There's a COST for each
guy
Replace your words with numbers
Now solve for g... one step at a time
... but you have to keep the equation equal!
If you follow the rules, you'll ALWAYS get the right answer
Whole numbers are usually easier to work with
A variable can appear in an equation MORE THAN ONE TIME
Checking your work proves your answer
What's a road trip without some girls?
We need another variable
A term is a chunk of an algebraic equation
Tools for your Algebra Toolbox
Chapter 3
Rules for Numeric Operations: Follow the rules
Math or No Math
There's an order for working
expressions
You can re-group your equations
It's an important round...
Distributing a value over a grouping doesn't change a problem's
value
A constant stands in for a
number
Roll the credits...
Tools for your Algebra Toolbox
Chapter 4
Exponent Operations: Podcasts that spread like the plague (that's a
good thing...)
Addie's got a podcast
Let's mobilize Addie's listeners
Can Addie and Alex get enough hits?
Alex is flaking out on his sister
There's always a villain...
The order of operations says exponents FIRST
A root is the INVERSE of an
exponent
Tools for your Algebra Toolbox
Chapter 5
Graphing: A picture's worth 1,000 words
Edward's Lawn Mowing needs help...
Why don't you just SHOW me the
money?
Now we can LOOK at Ed's cash
pattern
Graphs show an ENTIRE
relationship
Let's graph Ed's equation on the Cartesian Plane
Ed's figuring out the SLOPE of
lawns
Linear equations in point-slope form
How does a point and a slope get you a line?
Let's use the point-slope form
Equations also have a standard form
The slope-intercept form is EASY to graph
Tools for your Algebra Toolbox
Chapter 6
Inequalities: Can't quite get enough?
Kathleen really loves football
The cost of all players can't be more than $1,000,000
Inequalities are COMPARISONS
Inequalities involving some negative number operations need
special treatment
Negative inequalities work BACKWARD
FLIP the inequality sign with
negative multiplication and division
When you're working with an inequality and negative
multiplication or division... | 677.169 | 1 |
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this file type before downloading and/or purchasing.
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Product Description
In this small set there are three worksheets and a handout which gives students plenty of practice translating English and Algebraic Expressions, to distinguish between expressions and equations, and to identify various types of linear equations. These worksheets can be used for additional classwork, homework, quizzes or assessments, or Do Nows (morning work).
Included Worksheets are:
Translating English and Algebraic Expressions
Expressions versus Equations, Types of Linear Equations, and Linear Equations (three part worksheet; one part is read only)
Conditional Equations, Identities, and Contradictions
There is also a handout to assist students with Steps for Solving Equations in One Variable.
In addition, there are answer keys for each worksheet. | 677.169 | 1 |
Similar
Written primarily for students who have completed the standard first courses in calculus and linear algebra, Elementary Differential Geometry, Revised 2nd Edition, provides an introduction to the geometry of curves and surfaces.
The Second Edition maintained the accessibility of the first, while providing an introduction to the use of computers and expanding discussion on certain topics. Further emphasis was placed on topological properties, properties of geodesics, singularities of vector fields, and the theorems of Bonnet and Hadamard.
This revision of the Second Edition provides a thorough update of commands for the symbolic computation programs Mathematica or Maple, as well as additional computer exercises. As with the Second Edition, this material supplements the content but no computer skill is necessary to take full advantage of this comprehensive text.
Over 36,000 copies sold worldwideAccessible, practical yet rigorous approach to a complex topic--also suitable for self-studyExtensive update of appendices on Mathematica and Maple software packagesThorough streamlining of second edition's numbering systemFuller information on solutions to odd-numbered problemsAdditional exercises and hints guide students in using the latest computer modeling tools
One of the most widely used texts in its field, this volume introduces the differential geometry of curves and surfaces in both local and global aspects. The presentation departs from the traditional approach with its more extensive use of elementary linear algebra and its emphasis on basic geometrical facts rather than machinery or random details. Many examples and exercises enhance the clear, well-written exposition, along with hints and answers to some of the problems. The treatment begins with a chapter on curves, followed by explorations of regular surfaces, the geometry of the Gauss map, the intrinsic geometry of surfaces, and global differential geometry. Suitable for advanced undergraduates and graduate students of mathematics, this text's prerequisites include an undergraduate course in linear algebra and some familiarity with the calculus of several variables. For this second edition, the author has corrected, revised, and updated the entire volume.
This book is about the light like (degenerate) geometry of submanifolds needed to fill a gap in the general theory of submanifolds. The growing importance of light like hypersurfaces in mathematical physics, in particular their extensive use in relativity, and very limited information available on the general theory of lightlike submanifolds, motivated the present authors, in 1990, to do collaborative research on the subject matter of this book. Based on a series of author's papers (Bejancu [3], Bejancu-Duggal [1,3], Dug gal [13], Duggal-Bejancu [1,2,3]) and several other researchers, this volume was conceived and developed during the Fall '91 and Fall '94 visits of Bejancu to the University of Windsor, Canada. The primary difference between the lightlike submanifold and that of its non degenerate counterpart arises due to the fact that in the first case, the normal vector bundle intersects with the tangent bundle of the submanifold. Thus, one fails to use, in the usual way, the theory of non-degenerate submanifolds (cf. Chen [1]) to define the induced geometric objects (such as linear connection, second fundamental form, Gauss and Weingarten equations) on the light like submanifold. Some work is known on null hypersurfaces and degenerate submanifolds (see an up-to-date list of references on pages 138 and 140 respectively). Our approach, in this book, has the following outstanding features: (a) It is the first-ever attempt of an up-to-date information on null curves, lightlike hypersur faces and submanifolds, consistent with the theory of non-degenerate submanifoldsThis unique monograph by a noted UCLA professor examines in detail the mathematics of Kerr black holes, which possess the properties of mass and angular momentum but carry no electrical charge. Suitable for advanced undergraduates and graduate students of mathematics, physics, and astronomy as well as professional physicists, the self-contained treatment constitutes an introduction to modern techniques in differential geometry. The text begins with a substantial chapter offering background on the mathematics needed for the rest of the book. Subsequent chapters emphasize physical interpretations of geometric properties such as curvature, geodesics, isometries, totally geodesic submanifolds, and topological structure. Further investigations cover relativistic concepts such as causality, Petrov type, optical scalars, and the Goldberg-Sachs theorem. Four helpful appendixes supplement the text.
Clarity, readability and rigor combine in the second edition of this widely-used textbook to provide the first step into general relativity for undergraduate students with a minimal background in mathematics. Topics within relativity that fascinate astrophysical researchers and students alike are covered with Schutz's characteristic ease and authority - from black holes to gravitational lenses, from pulsars to the study of the Universe as a whole. This edition now contains discoveries by astronomers that require general relativity for their explanation; a revised chapter on relativistic stars, including new information on pulsars; an entirely rewritten chapter on cosmology; and an extended, comprehensive treatment of modern detectors and expected sources. Over 300 exercises, many new to this edition, give students the confidence to work with general relativity and the necessary mathematics, whilst the informal writing style makes the subject matter easily accessible. Password protected solutions for instructors are available at
Riemannian geometry is characterized, and research is oriented towards and shaped by concepts (geodesics, connections, curvature,...) andobjectives,inparticularto understand certain classes of (compact) Riemannian manifolds de?ned by curvature conditions (constant or positive or negative curvature,...). Bywayofcontrast,g- metric analysis is a perhaps somewhat less systematic collection of techniques, for solving extremal problems naturally arising in geometry and for investigating and characterizing their solutions. It turns out that the two ?elds complement each other very well; geometric analysis o?ers tools for solving di?cult problems in geometry, and Riemannian geometry stimulates progress in geometric analysis by setting am- tious goals. It is the aim of this book to be a systematic and comprehensive introduction to Riemannian geometry and a representative introduction to the methods of geometric analysis. It attempts a synthesis of geometric and analytic methods in the study of Riemannian manifolds. The present work is the ?fth edition of my textbook on Riemannian geometry and geometric analysis. It has developed on the basis of several graduate courses I taught at the Ruhr-University Bochum and the University of Leipzig. The main new features of the present edition are the systematic inclusion of ?ow equations and a mathematical treatment of the nonlinear sigma model of quantum ?eld theory. These new topics also led to a systematic reorganization of the other material. Naturally, I have also included several smaller additions and minor corrections (for which I am grateful to several readers).
This book presents research on the latest developments in differential geometry of lightlike (degenerate) subspaces. The main focus is on hypersurfaces and a variety of submanifolds of indefinite Kählerian, Sasakian and quaternion Kähler manifolds.
Einstein's General Theory of Relativity leads to two remarkable predictions: first, that the ultimate destiny of many massive stars is to undergo gravitational collapse and to disappear from view, leaving behind a 'black hole' in space; and secondly, that there will exist singularities in space-time itself. These singularities are places where space-time begins or ends, and the presently known laws of physics break down. They will occur inside black holes, and in the past are what might be construed as the beginning of the universe. To show how these predictions arise, the authors discuss the General Theory of Relativity in the large. Starting with a precise formulation of the theory and an account of the necessary background of differential geometry, the significance of space-time curvature is discussed and the global properties of a number of exact solutions of Einstein's field equations are examined. The theory of the causal structure of a general space-time is developed, and is used to study black holes and to prove a number of theorems establishing the inevitability of singualarities under certain conditions. A discussion of the Cauchy problem for General Relativity is also included in this 1973 book.
This book provides an introduction to Riemannian geometry, the geometry of curved spaces, for use in a graduate course. Requiring only an understanding of differentiable manifolds, the author covers the introductory ideas of Riemannian geometry followed by a selection of more specialized topics. Also featured are Notes and Exercises for each chapter, to develop and enrich the reader's appreciation of the subject. This second edition, first published in 2006, has a clearer treatment of many topics than the first edition, with new proofs of some theorems and a new chapter on the Riemannian geometry of surfaces. The main themes here are the effect of the curvature on the usual notions of classical Euclidean geometry, and the new notions and ideas motivated by curvature itself. Completely new themes created by curvature include the classical Rauch comparison theorem and its consequences in geometry and topology, and the interaction of microscopic behavior of the geometry with the macroscopic structure of the space.
Manifolds, the higher-dimensional analogs of smooth curves and surfaces, are fundamental objects in modern mathematics. Combining aspects of algebra, topology, and analysis, manifolds have also been applied to classical mechanics, general relativity, and quantum field theory. In this streamlined introduction to the subject, the theory of manifolds is presented with the aim of helping the reader achieve a rapid mastery of the essential topics. By the end of the book the reader should be able to compute, at least for simple spaces, one of the most basic topological invariants of a manifold, its de Rham cohomology. Along the way, the reader acquires the knowledge and skills necessary for further study of geometry and topology. The requisite point-set topology is included in an appendix of twenty pages; other appendices review facts from real analysis and linear algebra. Hints and solutions are provided to many of the exercises and problems. This work may be used as the text for a one-semester graduate or advanced undergraduate course, as well as by students engaged in self-study. Requiring only minimal undergraduate prerequisites, 'Introduction to Manifolds' is also an excellent foundation for Springer's GTM 82, 'Differential Forms in Algebraic Topology'.
This book is an exposition of "Singular Semi-Riemannian Geometry"- the study of a smooth manifold furnished with a degenerate (singular) metric tensor of arbitrary signature. The main topic of interest is those cases where the metric tensor is assumed to be nondegenerate. In the literature, manifolds with degenerate metric tensors have been studied extrinsically as degenerate submanifolds of semi Riemannian manifolds. One major aspect of this book is first to study the intrinsic structure of a manifold with a degenerate metric tensor and then to study it extrinsically by considering it as a degenerate submanifold of a semi-Riemannian manifold. This book is divided into three parts. Part I deals with singular semi Riemannian manifolds in four chapters. In Chapter I, the linear algebra of indefinite real inner product spaces is reviewed. In general, properties of certain geometric tensor fields are obtained purely from the algebraic point of view without referring to their geometric origin. Chapter II is devoted to a review of covariant derivative operators in real vector bundles. Chapter III is the main part of this book where, intrinsically, the Koszul connection is introduced and its curvature identities are obtained. In Chapter IV, an application of Chapter III is made to degenerate submanifolds of semi-Riemannian manifolds and Gauss, Codazzi and Ricci equations are obtained. Part II deals with singular Kahler manifolds in four chapters parallel to Part I.
Albert Einstein is the unquestioned founder of modern physics. His theory of relativity is the most important scientific idea of the modern era. In this book Einstein explains, using the minimum of mathematical terms, the basic ideas and principles of the theory which has shaped the world we live in today. Unsurpassed by any subsequent books on relativity, this remains the most popular and useful exposition of Einstein's immense contribution to human knowledge.
In this work Einstein intended, as far as possible, to give an exact insight into the theory of relativity to those readers who, from a general and scientific philosophical point of view, are interested in the theory, but who are not conversant with the mathematical apparatus of theoretical physics. The theory of relativity enriched physics and astronomy during the 20th century.
(Relativity: The Special and the General Theory by Albert Einstein, 9789380914220)Einstein authors irrelevancies and going straight to the heart of the problem.
Ever since Albert Einstein's general theory of relativity burst upon the world in 1915 some of the most brilliant minds of our century have sought to decipher the mysteries bequeathed by that theory, a legacy so unthinkable in some respects that even Einstein himself rejected them.
Which of these bizarre phenomena, if any, can really exist in our universe? Black holes, down which anything can fall but from which nothing can return; wormholes, short spacewarps connecting regions of the cosmos; singularities, where space and time are so violently warped that time ceases to exist and space becomes a kind of foam; gravitational waves, which carry symphonic accounts of collisions of black holes billions of years ago; and time machines, for traveling backward and forward in time.
Kip Thorne, along with fellow theorists Stephen Hawking and Roger Penrose, a cadre of Russians, and earlier scientists such as Oppenheimer, Wheeler and Chandrasekhar, has been in the thick of the quest to secure answers. In this masterfully written and brilliantly informed work of scientific history and explanation, Dr. Thorne, a Nobel Prize-winning physicist and the Feynman Professor of Theoretical Physics Emeritus at Caltech, leads his readers through an elegant, always human, tapestry of interlocking themes, coming finally to a uniquely informed answer to the great question: what principles control our universe and why do physicists think they know the things they think they know? Stephen Hawking's A Brief History of Time has been one of the greatest best-sellers in publishing history. Anyone who struggled with that book will find here a more slowly paced but equally mind-stretching experience, with the added fascination of a rich historical and human component.
This is a book about physics, written for mathematicians. The readers we have in mind can be roughly described as those who: I. are mathematics graduate students with some knowledge of global differential geometry 2. have had the equivalent of freshman physics, and find popular accounts of astrophysics and cosmology interesting 3. appreciate mathematical elarity, but are willing to accept physical motiva tions for the mathematics in place of mathematical ones 4. are willing to spend time and effort mastering certain technical details, such as those in Section 1. 1. Each book disappoints so me readers. This one will disappoint: 1. physicists who want to use this book as a first course on differential geometry 2. mathematicians who think Lorentzian manifolds are wholly similar to Riemannian ones, or that, given a sufficiently good mathematical back ground, the essentials of a subject !ike cosmology can be learned without so me hard work on boring detaiis 3. those who believe vague philosophical arguments have more than historical and heuristic significance, that general relativity should somehow be "proved," or that axiomatization of this subject is useful 4. those who want an encyclopedic treatment (the books by Hawking-Ellis [1], Penrose [1], Weinberg [1], and Misner-Thorne-Wheeler [I] go further into the subject than we do; see also the survey article, Sachs-Wu [1]). 5. mathematicians who want to learn quantum physics or unified fieId theory (unfortunateIy, quantum physics texts all seem either to be for physicists, or merely concerned with formaI mathematics).
The most accessible, entertaining, and enlightening explanation of the best-known physics equation in the world, as rendered by two of today's leading scientists.
Professor Brian Cox and Professor Jeff Forshaw go on a journey to the frontier of 21st century science to consider the real meaning behind the iconic sequence of symbols that make up Einstein's most famous equation, E=mc2. Breaking down the symbols themselves, they pose a series of questions: What is energy? What is mass? What has the speed of light got to do with energy and mass? In answering these questions, they take us to the site of one of the largest scientific experiments ever conducted. Lying beneath the city of Geneva, straddling the Franco-Swiss boarder, is a 27 km particle accelerator, known as the Large Hadron Collider. Using this gigantic machine—which can recreate conditions in the early Universe fractions of a second after the Big Bang—Cox and Forshaw will describe the current theory behind the origin of mass.
Alongside questions of energy and mass, they will consider the third, and perhaps, most intriguing element of the equation: 'c' - or the speed of light. Why is it that the speed of light is the exchange rate? Answering this question is at the heart of the investigation as the authors demonstrate how, in order to truly understand why E=mc2, we first must understand why we must move forward in time and not backwards and how objects in our 3-dimensional world actually move in 4-dimensional space-time. In other words, how the very fabric of our world is constructed. A collaboration between two of the youngest professors in the UK, Why Does E=mc2? promises to be one of the most exciting and accessible explanations of the theory of relativity in recent years.
E=mc2 is the world's most famous equation. Discover the thought process and physics behind general relativity and Einstein's contribution to science, in this authorized edition.
In this collection of his seven most important essays on physics, Einstein guides his reader
This remarkable collection allows the general reader to understand not only the significance of Einstein's masterpiece, but also the brilliant mind behind it.
This authorized ebook features a new introduction by Neil Berger and an illustrated biography of Albert Einstein, which includes rare photos and never-before-seen documents from the Albert Einstein Archives at the Hebrew University of Jerusalem.
Since the publication of Einstein's Special Theory of Relativity in 1905, the discovery of such astronomical phenomena as quasars, pulsars, and black holes — all intimately connected to relativity — has provoked a tremendous upsurge of interest in the subject.This volume, a revised version of Martin Gardner's earlier Relativity for the Million, brings this fascinating topic up to date. Witty, perceptive, and easily accessible to the general reader, it is one of the clearest and most entertaining introductions to relativity ever written. Mr. Gardner offers lucid explanations of not only the special and general theories of relativity, but of the Michelson-Morley experiment, gravity and spacetime, Mach's principle, the twin paradox, models of the universe, and other topics. A new Postscript, examining the latest developments in the field, and specially written for this edition, is also included. The clarity of the text is especially enhanced by the brilliant graphics of Anthony Ravielli, making this "by far the best layman's account of this difficult subject." — Christian Science Monitor.
Foundations of Differentiable Manifolds and Lie Groups gives a clear, detailed, and careful development of the basic facts on manifold theory and Lie Groups. It includes differentiable manifolds, tensors and differentiable forms. Lie groups and homogenous spaces, integration on manifolds, and in addition provides a proof of the de Rham theorem via sheaf cohomology theory, and develops the local theory of elliptic operators culminating in a proof of the Hodge theorem. Those interested in any of the diverse areas of mathematics requiring the notion of a differentiable manifold will find this beginning graduate-level text extremely usefulThis practical, friendly guide focuses on critical concepts taught in a typical geometry course, from the properties of triangles, parallelograms, circles, and cylinders, to the skills and strategies you need to write geometry proofs. Geometry Essentials For Dummies is perfect for cramming or doing homework, or as a reference for parents helping kids study for exams.
Get down to the basics — get a handle on the basics of geometry, from lines, segments, and angles, to vertices, altitudes, and diagonals
"It would be hard to imagine a better guide to this difficult subject."--Scientific American In Three Roads to Quantum Gravity, Lee Smolin provides an accessible overview of the attempts to build a final "theory of everything." He explains in simple terms what scientists are talking about when they say the world is made from exotic entities such as loops, strings, and black holes and tells the fascinating stories behind these discoveries: the rivalries, epiphanies, and intrigues he witnessed firsthand.
Lockhart's Mathematician's Lament outlined how we introduce math to students in the wrong way. Measurement explains how math should be done. With plain English and pictures, he makes complex ideas about shape and motion intuitive and graspable, and offers a solution to math phobia by introducing us to math as an artful way of thinking and living.
There are two scientific theories that, taken together, explain the entire universe. The first, which describes the force of gravity, is widely known: Einstein's General Theory of Relativity. But the theory that explains everything else—the Standard Model of Elementary Particles—is virtually unknown among the general public.
In The Theory of Almost Everything, Robert Oerter shows how what were once thought to be separate forces of nature were combined into a single theory by some of the most brilliant minds of the twentieth century. Rich with accessible analogies and lucid prose, The Theory of Almost Everything celebrates a heretofore unsung achievement in human knowledge—and reveals the sublime structure that underlies the world as we know it.
With this reader-friendly book, it doesn't take an Einstein to understand the theory of relativity and its remarkable consequences. In clear, understandable terms, physicist Richard Wolfson explores the ideas at the heart of relativity and shows how they lead to such seeming absurdities as time travel, curved space, black holes, and new meaning for the idea of past and future. Drawing from years of teaching modern physics to nonscientists, Wolfson explains in a lively, conversational style the simple principles underlying Einstein's theory.
Relativity, Wolfson shows, gave us a new view of space and time, opening the door to questions about their flexible nature: Is the universe finite or infinite? Will it expand forever or eventually collapse in a "big crunch"? Is time travel possible? What goes on inside a black hole? How does gravity really work? These questions at the forefront of twenty-first-century physics are all rooted in the profound and sweeping vision of Albert Einstein's early twentieth-century theory. Wolfson leads his readers on an intellectual journey that culminates in a universe made almost unimaginably rich by the principles that Einstein first discovered.
Physicists have been exploring, debating, and questioning the general theory of relativity ever since Albert Einstein first presented it in 1915. Their work has uncovered a number of the universe's more surprising secrets, and many believe further wonders remain hidden within the theory's tangle of equations, waiting to be exposed. In this sweeping narrative of science and culture, astrophysicist Pedro Ferreira brings general relativity to life through the story of the brilliant physicists, mathematicians, and astronomers who have taken up its challenge. For these scientists, the theory has been both a treasure trove and an enigma, fueling a century of intellectual struggle and triumph..
Einstein's theory, which explains the relationships among gravity, space, and time, is possibly the most perfect intellectual achievement of modern physics, yet studying it has always been a controversial endeavor. Relativists were the target of persecution in Hitler's Germany, hounded in Stalin's Russia, and disdained in 1950s America. Even today, PhD students are warned that specializing in general relativity will make them unemployable.
Despite these pitfalls, general relativity has flourished, delivering key insights into our understanding of the origin of time and the evolution of all the stars and galaxies in the cosmos. Its adherents have revealed what lies at the farthest reaches of the universe, shed light on the smallest scales of existence, and explained how the fabric of reality emerges. Dark matter, dark energy, black holes, and string theory are all progeny of Einstein's theory.
We are in the midst of a momentous transformation in modern physics. As scientists look farther and more clearly into space than ever before, The Perfect Theory reveals the greater relevance of general relativity, showing us where it started, where it has led, and where it can still take us——when it succeeds as well as when it fails."—Neil deGrasse Tyson
"Science writing at its best. This book is not just learned, passionate, and witty—it is profoundly wise."—Junot Díaz
A paperback edition of a classic text, this book gives a unique survey of the known solutions of Einstein's field equations for vacuum, Einstein-Maxwell, pure radiation and perfect fluid sources. It introduces the foundations of differential geometry and Riemannian geometry and the methods used to characterize, find or construct solutions. The solutions are then considered, ordered by their symmetry group, their algebraic structure (Petrov type) or other invariant properties such as special subspaces or tensor fields and embedding properties. Includes all the developments in the field since the first edition and contains six completely new chapters, covering topics including generation methods and their application, colliding waves, classification of metrics by invariants and treatments of homothetic motions. This book is an important resource for graduates and researchers in relativity, theoretical physics, astrophysics and mathematics. It can also be used as an introductory text on some mathematical aspects of general relativity.
After completing the final version of his general theory of relativity in November 1915, Albert Einstein wrote a book about relativity for a popular audience. His intention was "to give an exact insight into the theory of relativity to those readers who, from a general scientific and philosophical point of view, are interested in the theory, but who are not conversant with the mathematical apparatus of theoretical physics." The book remains one of the most lucid explanations of the special and general theories ever written. In the early 1920s alone, it was translated into ten languages, and fifteen editions in the original German appeared over the course of Einstein's lifetime.
This new edition of Einstein's celebrated book features an authoritative English translation of the text along with an introduction and a reading companion by Hanoch Gutfreund and Jürgen Renn that examines the evolution of Einstein's thinking and casts his ideas in a broader present-day context. A special chapter explores the history of and the stories behind the early foreign-language editions in light of the reception of relativity in different countries. This edition also includes a survey of the introductions from those editions, covers from selected early editions, a letter from Walther Rathenau to Einstein discussing the book, and a revealing sample from Einstein's handwritten manuscript.
Published on the hundredth anniversary of general relativity, this handsome edition of Einstein's famous book places the work in historical and intellectual context while providing invaluable insight into one of the greatest scientific minds of all time.
Through Euclid's Window Leonard Mlodinow brilliantly and delightfully leads us on a journey through five revolutions in geometry, from the Greek concept of parallel lines to the latest notions of hyperspace. Here is an altogether new, refreshing, alternative history of math revealing how simple questions anyone might ask about space -- in the living room or in some other galaxy -- have been the hidden engine of the highest achievements in science and technology.
Based on Mlodinow's extensive historical research; his studies alongside colleagues such as Richard Feynman and Kip Thorne; and interviews with leading physicists and mathematicians such as Murray Gell-Mann, Edward Witten, and Brian Greene, Euclid's Window is an extraordinary blend of rigorous, authoritative investigation and accessible, good-humored storytelling that makes a stunningly original argument asserting the primacy of geometry. For those who have looked through Euclid's Window, no space, no thing, and no time will ever be quite the sameThis book is the American Mathematical Society printing of this title, which was first published in 1907 by W. A. Benjamin and whose second edition was published by Benjamin Cummings in 1978. The book was also distributed by Perseus Press for the last decade. It is the updated 1985 (fifth) printing that is reproduced here. It includes most of the basic results in manifold theory, as well as some key facts from point set topology and Lie group theory. Introductory chapters offer background in differential theory and calculus on manifolds. Later chapters are organized in sections on analytical dynamics, qualitative dynamics, and celestial mechanics. Chapter exercises are included. The book can be used as a textbook and as a basic reference for the foundations of differentiable and Hamiltonian dynamics. Readership includes mathematicians, physicists, and engineers interested in geometrical methods in mechanics, assuming a background in calculus, linear algebra, some classical analysis, and point set topology. Author information is not given.
Each page in the Common Core Math Workouts for grade 6 TheyThis is a highly readable, popular exposition of the fourth dimension and the structure of the universe. A remarkable pictorial discussion of the curved space-time we call home, it achieves even greater impact through the use of 141 excellent illustrations. This is the first sustained visual account of many important topics in relativity theory that up till now have only been treated separately. Finding a perfect analogy in the situation of the geometrical characters in Flatland, Professor Rucker continues the adventures of the two-dimensional world visited by a three-dimensional being to explain our three-dimensional world in terms of the fourth dimension. Following this adventure into the fourth dimension, the author discusses non-Euclidean geometry, curved space, time as a higher dimension, special relativity, time travel, and the shape of space-time. The mathematics is sound throughout, but the casual reader may skip those few sections that seem too purely mathematical and still follow the line of argument. Readable and interesting in itself, the annotated bibliography is a valuable guide to further study. Professor Rucker teaches mathematics at the State University of New York in Geneseo. Students and laymen will find his discussion to be unusually stimulating. Experienced mathematicians and physicists will find a great deal of original material here and many unexpected novelties. Annotated bibliography. 44 problems.
". . . Nothing less than a major contribution to the scientific culture of this world." — The New York Times Book Review This major survey of mathematics, featuring the work of 18 outstanding Russian mathematicians and including material on both elementary and advanced levels, encompasses 20 prime subject areas in mathematics in terms of their simple origins and their subsequent sophisticated developement. As Professor Morris Kline of New York University noted, "This unique work presents the amazing panorama of mathematics proper. It is the best answer in print to what mathematics contains both on the elementary and advanced levels." Beginning with an overview and analysis of mathematics, the first of three major divisions of the book progresses to an exploration of analytic geometry, algebra, and ordinary differential equations. The second part introduces partial differential equations, along with theories of curves and surfaces, the calculus of variations, and functions of a complex variable. It furthur examines prime numbers, the theory of probability, approximations, and the role of computers in mathematics. The theory of functions of a real variable opens the final section, followed by discussions of linear algebra and nonEuclidian geometry, topology, functional analysis, and groups and other algebraic systems. Thorough, coherent explanations of each topic are further augumented by numerous illustrative figures, and every chapter concludes with a suggested reading list. Formerly issued as a three-volume set, this mathematical masterpiece is now available in a convenient and modestly priced one-volume edition, perfect for study or reference. "This is a masterful English translation of a stupendous and formidable mathematical masterpiece . . ." — Social Science
The Banff NATO Summer School was held August 14-25, 1989 at the Banff Cen tre, Banff, Albert, Canada. It was a combination of two venues: a summer school in the annual series of Summer School in Theoretical Physics spon sored by the Theoretical Physics Division, Canadian Association of Physi cists, and a NATO Advanced Study Institute. The Organizing Committee for the present school was composed of G. Kunstatter (University of Winnipeg), H.C. Lee (Chalk River Laboratories and University of Western Ontario), R. Kobes (University of Winnipeg), D.l. Toms (University of Newcastle Upon Tyne) and Y.S. Wu (University of Utah). Thanks to the group of lecturers (see Contents) and the timeliness of the courses given, the school, entitled PHYSICS, GEOMETRY AND TOPOLOGY, was popular from the very outset. The number of applications outstripped the 90 places of accommodation reserved at the Banff Centre soon after the school was announced. As the eventual total number of participants was increased to 170, it was still necessary to tum away many deserving applicants. In accordance with the spirit of the school, the geometrical and topologi cal properties in each of the wide ranging topics covered by the lectures were emphasized. A recurring theme in a number of the lectures is the Yang-Baxter relation which characterizes a very large class of integrable systems including: many state models, two-dimensional conformal field theory, quantum field theory and quantum gravity in 2 + I dimensions.
A no-nonsense practical guide to geometry, providing concise summaries, clear model examples, and plenty of practice, making this workbook the ideal complement to class study or self-study, preparation for exams or a brush-up on rusty skills.
About the Book Established as a successful practical workbook series with more than 20 titles in the language learning category, Practice Makes Perfect now provides the same clear, concise approach and extensive exercises to key fields within mathematics. The key to the Practice Makes Perfect series is the extensive exercises that provide learners with all the practice they need for mastery.
Not focused on any particular test or exam, but complementary to most geometry curricula Deliberately all-encompassing approach: international perspective and balance between traditional and newer approaches. Large trim allows clear presentation of worked problems, exercises, and explained answers. Features No-nonsense approach: provides clear presentation of content. Over 500 exercises and answers covering all aspects of geometry Successful series: "Practice Makes Perfect" has sales of 1,000,000 copies in the language category – now applied to mathematics Workbook is not exam specific, yet it provides thorough coverage of the geometry skills required in most math tests.
Getting Inside the Mind of Nature: Discover the simple but powerful mathematics of the underlying geometric figures that shape our world
Can geometry be exciting?
It can for those who can appreciate the beauty of numbers and their relationships.
This book contains a meticulous geometric investigation of the 5 Platonic Solids and 5 other important polyhedra, as well as reference charts for each solid.
Poly (many) hedron (face) means "many faces." Polyhedra are 3 dimensional figures with 4 or more faces, or sides. These polyhedra are reflections of Nature herself, and a study of them provides insight into the way the world is structured. Nature is not only beautiful, but highly intelligent. As you explore the polyhedra in this book, this will become apparent over and over again.
The book contains a geometric explanation of the Phi Ratio and Fibonacci series, and a detailed analysis of the pentagon, which forms the basis for many of these solids. The pentagon is composed entirely of Phi relationships, and is integral to a proper understanding of sacred geometry.
With over 140 full-color illustrations, this book is perfect for teachers and students of geometry alike. It is a must for those who are serious about sacred geometry.
This book is dedicated to those who can appreciate the logic of numbers and the beauty of nature, for they are both aspects of the same unifying principle.
Prerequisites: knowledge of simple algebra and elementary trigonometry. No brainiac math skills required! Only the ability to appreciate nature's own logic.
This concise book introduces nonphysicists to the core philosophical issues surrounding the nature and structure of space and time, and is also an ideal resource for physicists interested in the conceptual foundations of space-time theory. Tim Maudlin's broad historical overview examines Aristotelian and Newtonian accounts of space and time, and traces how Galileo's conceptions of relativity and space-time led to Einstein's special and general theories of relativity. Maudlin explains special relativity with enough detail to solve concrete physical problems while presenting general relativity in more qualitative terms. Additional topics include the Twins Paradox, the physical aspects of the Lorentz-FitzGerald contraction, the constancy of the speed of light, time travel, the direction of time, and more. Introduces nonphysicists to the philosophical foundations of space-time theory Provides a broad historical overview, from Aristotle to Einstein Explains special relativity geometrically, emphasizing the intrinsic structure of space-time Covers the Twins Paradox, Galilean relativity, time travel, and more Requires only basic algebra and no formal knowledge of physics
First published in 1973, Gravitation is a landmark graduate-level textbook that presents Einstein's general theory of relativity and offers a rigorous, full-year course on the physics of gravitation. Upon publication, Science called it "a pedagogic masterpiece," and it has since become a classic, considered essential reading for every serious student and researcher in the field of relativity. This authoritative text has shaped the research of generations of physicists and astronomers, and the book continues to influence the way experts think about the subject.
With an emphasis on geometric interpretation, this masterful and comprehensive book introduces the theory of relativity; describes physical applications, from stars to black holes and gravitational waves; and portrays the field's frontiers. The book also offers a unique, alternating, two-track pathway through the subject. Material focusing on basic physical ideas is designated as Track 1 and formulates an appropriate one-semester graduate-level course. The remaining Track 2 material provides a wealth of advanced topics instructors can draw on for a two-semester course, with Track 1 sections serving as prerequisites.
This must-have reference for students and scholars of relativity includes a new preface by David Kaiser, reflecting on the history of the book's publication and reception, and a new introduction by Charles Misner and Kip Thorne, discussing exciting developments in the field since the book's original publication.
The book teaches students to:Grasp the laws of physics in flat and curved spacetimePredict orders of magnitudeCalculate using the principal tools of modern geometryUnderstand Einstein's geometric framework for physicsExplore applications, including neutron stars, Schwarzschild and Kerr black holes, gravitational collapse, gravitational waves, cosmology, and so much more
The fundamental outlines of the physical world, from its tiniest particles to massive galaxy clusters, have been apparent for decades. Does this mean physicists are about to tie it all up into a neat package? Not at all. Just when you think you're figuring it out, the universe begins to look its strangest. This eBook, "Ultimate Physics: From Quarks to the Cosmos," illustrates clearly how answers often lead to more questions and open up new paths to insight. We open with "The Higgs at Last," which looks behind the scenes of one of the most anticipated discoveries in physics and examines how this "Higgs-like" particle both confirmed and confounded expectations. In "The Inner Life of Quarks," author Don Lincoln discusses evidence that quarks and leptons may not be the smallest building blocks of matter. Section Two switches from the smallest to the largest of scales, and in "Origin of the Universe," Michael Turner analyzes a number of speculative scenarios about how it all began. Another two articles examine the mystery of dark energy and some doubts as to whether it exists at all. In the last section, we look at one of the most compelling problems in physics: how to tie together the very small and the very large – quantum mechanics and general relativity. In one article, Stephen Hawking and Leonard Mlodinow argue that a so-called "theory of everything" may be out of reach, and in another, David Deutsch and Artur Ekert question the view that quantum mechanics imposes limits on knowledge, arguing instead that the theory has an intricacy that allows for new, practical technologies, including powerful computers that can reach their true potential.
If you've ever thought that mathematics and art don't mix, this stunning visual history of geometry will change your mind. As much a work of art as a book about mathematics, Beautiful Geometry presents more than sixty exquisite color plates illustrating a wide range of geometric patterns and theorems, accompanied by brief accounts of the fascinating history and people behind each. With artwork by Swiss artist Eugen Jost and text by math historian Eli Maor, this unique celebration of geometry covers numerous subjects, from straightedge-and-compass constructions to intriguing configurations involving infinity. The result is a delightful and informative illustrated tour through the 2,500-year-old history of one of the most important branches of mathematicsWelcome, intrepid temporal explorers, to the world's first and only field manual/survival guide to time travel!DON'T LEAVE THIS TIME PERIOD WITHOUT IT! Humans from H. G. Wells to Albert Einstein to Bill & Ted have been fascinated by time travel-some say drawn to it like moths to a flame. But in order to travel safely and effectively, newbie travelers need to know the dos and don'ts. Think of this handy little book as the only thing standing between you and an unimaginably horrible death-or being trapped forever in another time or alternate reality. You get: Essential time travel knowledge: Choosing the right time machine, from DeLoreans to hot tubs to phone booths-and beyond What to say-and what NOT to say-to your doppelganger Understanding black holes and Stephen Hawking's term "spaghettification" (no, it's not a method of food preperation; yes, it is a horrifically painful way to meet your end) The connection between Einstein's General Theory of Relativity, traversing wormholes and the 88 mph speed requirement The possible consequences of creating a time paradox-including, but not limited to, the implosion of the universe Survival tips for nearly any sticky time travel situation: How to befriend a dinosaur and subsequently fight other dinosaurs with that dinosaur Instructions to build your very own Rube Goldberg Time Machine Crusading-for fun and profit Tips on battling cowboys, pirates, ninjas, samurai, Nazis, Vikings, robots and space marines How to operate a microwave oven Enjoying the servitude of robots and tips for living underground when they inevitably rise up against us
A Princeton astrophysicist explores whether journeying to the past or future is scientifically possible in this "intriguing" volume (Neil deGrasse Tyson).
It was H. G. Wells who coined the term "time machine"—but the concept of time travel, both forward and backward, has always provoked fascination and yearning. It has mostly been dismissed as an impossibility in the world of physics; yet theories posited by Einstein, and advanced by scientists including Stephen Hawking and Kip Thorne, suggest that the phenomenon could actually occur.
Building on these ideas, J. Richard Gott, a professor who has written on the subject for Scientific American, Time, and other publications, describes how travel to the future is not only possible but has already happened—and contemplates whether travel to the past is also conceivable. This look at the surprising facts behind the science fiction of time travel "deserves the attention of anyone wanting wider intellectual horizons" (Booklist).
"Impressively clear language. Practical tips for chrononauts on their options for travel and the contingencies to prepare for make everything sound bizarrely plausible. Gott clearly enjoys his subject and his excitement and humor are contagious; this book is a delight to read." —Publishers Weekly
Based on the ideas of Einstein and Minkowski, this concise treatment is derived from the author's many years of teaching the mathematics of relativity at the University of Michigan. Geared toward advanced undergraduates and graduate students of physics, the text covers old physics, new geometry, special relativity, curved space, and general relativity. Beginning with a discussion of the inverse square law in terms of simple calculus, the treatment gradually introduces increasingly complicated situations and more sophisticated mathematical tools. Changes in fundamental concepts, which characterize relativity theory, and the refinements of mathematical technique are incorporated as necessary. The presentation thus offers an easier approach without sacrifice of rigor.
Redesigned inside and out to have a fresh, appealing look, this new edition of a classic Crown Trade Paperback is a collection of Einstein's own popular writings on his work and describes the meaning of his main theories in a way virtually everyone can understand.
From acclaimed science author Jim Baggot, a pointed critique of modern theoretical physics In this stunning new volume, Jim Baggott argues that there is no observational or experimental evidence for many of the ideas of modern theoretical physics: super-symmetric particles, super strings, the multiverse, the holographic principle, or the anthropic cosmological principle. These theories are not only untrue; they are not even science. They are fairy-tale physics: fantastical, bizarre and often outrageous, perhaps even confidence-trickery. This book provides a much-needed antidote. Informed, comprehensive, and balanced, it offers lay readers the latest ideas about the nature of physical reality while clearly distinguishing between fact and fantasy. With its engaging portraits of many central figures of modern physics, including Paul Davies, John Barrow, Brian Greene, Stephen Hawking, and Leonard Susskind, it promises to be essential reading for all readers interested in what we know and don't know about the nature of the universe and reality itself.
This outstanding textbook by a distinguished mathematical scholar introduces the differential geometry of curves and surfaces in three-dimensional Euclidean space. The subject is presented in its simplest, most essential form, but with many explanatory details, figures and examples, and in a manner that conveys the geometric significance and theoretical and practical importance of the different concepts, methods and results involved. The first chapters of the book focus on the basic concepts and facts of analytic geometry, the theory of space curves, and the foundations of the theory of surfaces, including problems closely related to the first and second fundamental forms. The treatment of the theory of surfaces makes full use of the tensor calculus. The later chapters address geodesics, mappings of surfaces, special surfaces, and the absolute differential calculus and the displacement of Levi-Cività. Problems at the end of each section (with solutions at the end of the book) will help students meaningfully review the material presented, and familiarize themselves with the manner of reasoning in differential geometry.
"A lucid and masterly survey." — Mathematics GazetteProfessor Pedoe is widely known as a fine teacher and a fine geometer. His abilities in both areas are clearly evident in this self-contained, well-written, and lucid introduction to the scope and methods of elementary geometry. It covers the geometry usually included in undergraduate courses in mathematics, except for the theory of convex sets. Based on a course given by the author for several years at the University of Minnesota, the main purpose of the book is to increase geometrical, and therefore mathematical, understanding and to help students enjoy geometry.Among the topics discussed: the use of vectors and their products in work on Desargues' and Pappus' theorem and the nine-point circle; circles and coaxal systems; the representation of circles by points in three dimensions; mappings of the Euclidean plane, similitudes, isometries, mappings of the inversive plane, and Moebius transformations; projective geometry of the plane, space, and n dimensions; the projective generation of conics and quadrics; Moebius tetrahedra; the tetrahedral complex; the twisted cubic curve; the cubic surface; oriented circles; and introduction to algebraic geometry.In addition, three appendices deal with Euclidean definitions, postulates, and propositions; the Grassmann-Pluecker coordinates of lines in S3, and the group of circular transformations. Among the outstanding features of this book are its many worked examples and over 500 exercises to test geometrical understanding.
Each page in the Common Core Math Workouts for grade 7 | 677.169 | 1 |
Calculus. There, I said it. If your heart skipped a beat, you might be one of the roughly 1 million students–or the parent of one of these brave souls–that will take the class this coming school year. Math is already tough, you might have been told, and calculus is supposed to be the "make or break" math class that may determine whether you have a future in STEM (science, technology, engineering, or mathematics); no pressure huh?
But you've got a little under two months to go. That's plenty of time to brush up on your precalculus, learn a bit of calculus, and walk in on day one well prepared–assuming you know where to start.
That's where this article comes in. As a math professor myself I use several free to low-cost resources that help my students prepare for calculus. I've grouped these resources below into two categories: Learning Calculus and Interacting with Calculus.
This online site from Paul Dawkins, math professor at Lamar University, is arguably the best (free) online site for learning calculus. In a nutshell, it's an interactive textbook. There are tons of examples, each followed by a complete solution, and various links that take you to different parts of the course as needed (i.e., instead of saying, for example, "recall in Section 2.1…" the links take you right back to the relevant section). I consider Prof. Dawkins' site to be just as good, if not better, at teaching calculus than many actual calculus textbooks (and it's free!). I should also mention that Prof. Dawkins' site also includes fairly comprehensive precalculus and algebra sections.
A non-profit run by educator Salman Khan, the Khan academy is a popular online site featuring over 6,000 (according to Wikipedia) video mini-lectures–typically lasting about 15 minutes–on everything from art history to mathematics. The link I've included here is to the differential calculus set of videos. You can change subjects to integral calculus, or to trigonometry or algebra once you jump onto the site.
One of the earliest institutions to do so, MIT records actual courses and puts up the lecture videos and, in some cases, homeworks, class notes, and exams on its Open Courseware site. The link above is to the math section. There you'll find several calculus courses, in addition to more advanced math courses. Clicking on the videos may take you to iTunes U, Apple's online library of video lectures. Once there you can also search for "calculus" and you'll find other universities that have followed in MIT's footsteps and put their recorded lectures online.
If you're looking for something in print, this book is a great resource. The book will teach you calculus, probably have you laughing throughout due to the authors' good sense of humor, and also includes content not found in other calculus books, like tips for taking calculus exams and interacting with your instructor. You can read the first few pages on the book's site.
There are many sites that include java-based demonstrations that will help you visualize math. Two good ones I've come across are David Little's site and the University of Notre Dame's site. By dragging a point or function, or changing specific parameters, these applets make important concepts in calculus come alive; they also make it far easier to understand certain things. For example, take this statement: "as the number of sides of a regular polygon inscribed in a circle increases, the area of that polygon better approximates the area of the circle." Even if you followed that, text is no comparison to this interactive animation.
One technological note: Because these are java applets, some of you will likely run into technology issues (especially if you're on a Mac). For example, your computer may block these applets because it thinks that they are malicious. Here is a workaround from Java themselves that may help you in these cases.
Self-promotion aside, calculus teachers often sell students (and parents) on the need to study calculus by telling them about how applicable the subject is. The problem is that the vast majority of the applications usually discussed are to things that many of us will likely never experience, like space shuttle launches and the optimization of company profits. The result: math becomes seen as an abstract subject that, although has applications, only become "real" if you become a scientist or engineer.
In Everyday Calculus I flip this script and start with ordinary experiences, like taking a shower and driving to work, and showcase the hidden calculus behind these everyday events and things. For example, there's some neat trigonometry that helps explain why we sometimes wake up feeling groggy, and thinking more carefully about how coffee cools reveals derivatives at work. This sort of approach makes it possible to use the book as an experiential learning tool to discover the calculus hidden all around you.
With so many good resources it's hard to know where to start and how to use them all effectively. Let me suggest one approach that uses the resources above synergistically.
For starters, the link to Paul's site takes you to the table of contents of his site. The topic ordering there is roughly the same as what you'd find in a calculus textbook. So, you'd probably want to start with his review of functions. From there, the next steps depend on the sort of learning experience you want.
1. If you're comfortable learning from Paul's site you can just stay there, using the other resources to complement your learning along the way.
2. If you learn better from lectures, then use Paul's topics list and jump on the Khan Academy site and/or the MIT and iTunes U sites to find video lectures on the corresponding topics.
3. If you're more of a print person, then How to Ace Calculus would be a great way to start. That book's topics ordering is pretty much the same as Paul's, so there'd be no need to go back and forth.
Whatever method you decided on, I still recommend that you use Paul's site, the interactive java applets, and Everyday Calculus. These three resources, used together, will allow you to completely interact with the calculus you'll be learning. From working through examples and checking your answer (on Paul's site), to interacting directly with functions, derivatives, and integrals (on the java applet sites), to exploring and experiencing the calculus all around you (Everyday Calculus), you'll gain an appreciation and understanding of calculus that will no doubt put you miles ahead of your classmates come September.
March 20th. Don't recognize that date? You should, it's the official start of spring! I won't blame you for not knowing, because after the unusually cold winter we've had it's easy to forget that higher temperatures are coming. But why March 20th, and not the 21st or the 19th? And while we're at it, why are there even seasons at all? Read on to find out the answers.
The answer has to do with 2 numbers. Don't worry, they're simple numbers (not like pi [1]). Stick around and I'll show you some neat graphs to help you understand where they come from, and hopefully entertain you in the process too.
The first star of this show is the number 92 million. No, it's not the current Powerball jackpot; it's also not the number of times a teenager texts per day. To appreciate its significance, have a look at our first chart:
Figure 1: The average surface temperature (on the vertical axis) of the planets in our solar system sorted by their distance from our sun (the horizontal axis). From left to right: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune, and Pluto (technically, Pluto lost its planet status in 2006).
That first planet on the left is Mercury. It's about 36 million miles away from the sun and has an average surface temperature of 333o. (Bring LOTS of sunscreen.) Fourth down the line is the red planet, Mars. At a distance of about 141 million miles from the sun, Mars' average temperature is -85o. (Bring LOTS of hot chocolate.) We could keep going, but the general trend is clear: planets farther away from the sun have lower average temperatures [2].
If neither 333o nor -85o sound inviting, I've got just the place for you: Earth! At a cool 59o this planet is … drumroll please … 92 million miles from the sun.
We actually got lucky here. You see, it turns out that a planet's temperature T is related to its distance r from the sun by the formula , where k is a number that depends on certain properties of the planet. I've graphed this curve in Figure 1. Notice that all the planets (except for the pesky Venus) closely follow the curve. But there's more here than meets the eye. Specifically, the formula predicts that a 1% change in distance will result in a 0.5% change in temperature [3]. For example, were Earth just 3% closer to the sun—about 89 million miles away instead of 92 million—the average temperature would be about 1.5o higher. To put that in perspective, note that at the end of the last ice age average temperatures were only 5o to 9o cooler than today [4].
So our distance from the sun gets us more reasonable temperatures than Mercury and Mars have, but where do the seasons come from? That's where our second number comes in: 23.4.
Imagine yourself in a park sitting in front of a bonfire. You're standing close enough to feel the heat but not close enough to feel the burn. Now lean in. Your head is now hotter than your toes; this tilthas produced a temperature difference between your "northern hemisphere" and your "southern hemisphere." This "tilt effect" is exactly what happens as Earth orbits the sun. More specifically, our planet is tilted about 23.4o from its vertical axis (Figure 2).
Figure 2. Earth is tilted about 23.4o from the plane of orbit with the sun (called the ecliptic plane).
Because of its tilt, as the Earth orbits the sun sometimes the Northern Hemisphere tilts toward the sun—roughly March-September—and other times it tilts away from the sun—roughly September-March (Figure 3) [5].
Figure 3. Earth's tilt points toward the sun between mid-March and mid-September, and points away from it the remaining months of the year. The four marked dates describe how this "tilt effect" changes the number of daylight hours throughout the year. Assuming you live in the Northern Hemisphere, days are longest during the summer solstice (shorter nights) and shortest during the winter solstice (longer nights). During the equinoxes, daytime and nighttime are about the same length.
Now that you know how two numbers—92 million and 23.4—explain the seasons, let's get back to spring in particular. As Figure 3 shows, there are two days each year when Earth's tilt neither points toward nor away from the sun. Those two days, called the equinoxes, divide the warmer months from the colder ones. And that's exactly what happened on March 20th: we passed the spring equinox.
Before you go, I have a little confession to make. It's not entirely true that just two numbers explain the seasons. Distance to the sun and Earth's tilt are arguably the most important factors, but other factors—like our atmosphere—are also important. But that would've made the title a lot longer. And anyway, I would've ended up explaining those factors using more numbers. The takeaway: math is powerful, and the more you learn the better you'll understand just about anything [6].
[1] The ratio of a circle's circumference to its diameter, pi is a never-ending, never repeating number. It is approximately 3.14.
[2] Venus is the exception. Its thick atmosphere prevents the planet from cooling.
[3] Here's the explanation for the mathematically inclined. In calculus, changes in a function are described by the function's derivative; the derivative of T is . This tells us that for a small change dr in r the temperature change dT is . Relative changes are ratios of small changes in a quantity to its original value. Thus, the relative change in temperature, dT/T, is
which is minus 0.5 times the relative change in distance, dr/r. The minus sign says that the temperature decreases as r increases, confirming the results of Figure 1. | 677.169 | 1 |
This book is a completely revised and updated version of this invaluable text which allows science students to extend necessary skills and techniques, with the topics being developed through examples in science which are easily understood by students from a range of disciplines. The introductory approach eases students into the subject, progressing to cover topics relevant to first and second year study and support data analysis for final year projects. The revision of the material in the book has been matched, on the accompanying website, with the extensive use of video, providing worked answers to over 200 questions in the book plus additional tutorial support. The second edition has also improved the learning approach for key topic areas to make it even more accessible and user-friendly, making it a perfect resource for students of all abilities. The expanding website provides a wide range of support material, providing a study environment within which students can develop their independent learning skills, in addition to providing resources that can be used by tutors for integration into other science-based programmes. Hallmark Features: Applied approach providing mathematics and statistics from the first to final years of undergraduate science courses.Second edition substantially revised to improve the learning approach to key topics and the organisation of resources for ease of use in teaching.Companion website at providing: Over 200 videos showing step-by-step workings of problems in the book.Additional materials including related topic areas, applications, and tutorials on Excel and Minitab.Interactive multiple-choice questions for self-testing, with step-by-step video feedback for any wrong answers.A developing resource of study plans for useful topics and applications.Figures from the book for downloading.
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We use a variety of resources on this site. The
lesson pages link to the resources as needed, but on this page
we list them by category. Click on a title to open a resource
document. The documents open in separate windows.
Presentations
The lessons use a number of presentations. These are
slides prepared by the instructor, many with naration.
They are published as quicktime movies and we use that
icon for access. Click on the icon to see a list of presentations.
The presentations can be opened from the list.
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Interactive movies have been prepared in Flash. Click
the icon to the left of a title to open a separate window.
If the movie doesn't work, you probably need a more recent
Flash Player.
Calculator
Factor Calculator
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Money Around
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materials.
These are the problems listed below except they are presented
in Frames rather than Flash. Click the icon at the left to
open the index page.
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the lessons. They are presented in Flash documents. Neglect
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Linear Programming and Game Theory
Mathematical methods for application in management science. Setting up optimization problems for management applications, techniques of linear programming including simplex method, sensitivity analysis, and introduction to game theory. | 677.169 | 1 |
Real Estate Math
Description
In this course the real estate professional will have the opportunity to review some basic mathematical fundamentals such as fractions, decimals, and percentages, as well as review their respective conversions.
Other topics include:
Prorations, including those found in real estate taxation.
How to determine the rates of return on investments.
Rectangles, triangles, and other shapes.
Using to loan-to-value ratios to calculate loan points and origination fees. | 677.169 | 1 |
Optimisation problems are concerned with optimising an objective function subject to a set of constraints. When optimisation problems are translated in algebraic form, we refer to them as mathematical programs. Mathematical programming, as an area within Operational Research / Management Science (OR/MS), is concerned with strategies and methods for solving mathematical programs. In this course, we address model building and validation in OR/MS, present a variety of typical OR/MS problems and their formulations, provide general tips on how to model certain managerial situations, and discuss solution strategies and present solution methods for linear programs, non-linear programs and integer programs. Last, but not least, students are encouraged to use computer software for solving mathematical programs and to interpret computer output.
Note: For Economics with Management Science, and Mathematics and Business Studies programmes EITHER Mathematical Programming OR Decision Making Under Uncertainty (BUST10013) is a mandatory course in Year 4.
On completion of the course students should:
(i) be able to assess critically the utility of a number of mathematical programming techniques
(ii) be able to describe mathematical programming solution techniques
(iii) be able to use mathematical programming methods to model management decision problems.
Cognitive Skills
On completion of the course students should:
(i) demonstrate ability in deciding whether a problem is amenable to solution by mathematical programming techniques
(ii) demonstrate ability in using mathematical programming solution techniques
(iii) demonstrate ability in explaining the solution to mathematical programming models.
Key Skills
On completion of the course students should:
(i) be able to formulate problems in mathematical programming terms
(ii) be able to solve mathematical programming problems using commercial software.
(iii) be able to communicate mathematical programming solutions to non-specialists.
Subject Specific Skills
On completion of the course students should:
(i) have extended their model building skills
(ii) have increased their model solution skills. | 677.169 | 1 |
Numerical Modeling. Issue Study. Discovering Verification Support. Computation Issues. Create MY Papers Personally. There's not really a shadow of suspect that it's a very difficult career to clarify what's happening about us working with the assistance of mathematics.<p>Nevertheless, statistical modeling is considered the most common tasks when choosing your self in higher education. The key notion is you need to get involved in a unique problem in the world close to us... | 677.169 | 1 |
Pre-Calculus B
How do I use math to explain my world?
Why Take This Course?
Calculus is the mathematical reasoning that we use to describe change. It is the reason that we were able to fly to the moon and back. It is used in business to make economic predictions. It is used to maximize the effectiveness of various systems and it has many other applications. In PreCalculus B, we will study two classical texts that defined the history of mathematics, specifically in the field of calculus. We will also study matrices and statistics during the course of this semester.
Course Framework
Unit 1
Unit 2
Unit 3
Unit 4
How can I use the definitions of conic sections to derive their standard equations?
How can I discover the foundational ideas of calculus by reading classical texts?
How are matrices used to design and animate cartoon characters?
How can I use statistics to describe and understand my world?
How can I derive the equation of a parabola?
How can I use convergence to prove that curves are similar?
How can I add and subtract matrices?
How can I calculate the expected value?
How can I write the standard equation of circle?
How are spaces covered and the times during which they are covered related in orbital paths? | 677.169 | 1 |
for courses that require the use of a graphing calculator, ALGEBRA AND TRIGONOMETRY: REAL MATHEMATICS, REAL PEOPLE, 6th Edition, features quality exercises, interesting applications, and innovative resources to help you succeed. Retaining the book's emphasis on student support, selected examples include notations directing students to previous sections where they can review concepts and skills needed to master the material at hand.
The book also achieves accessibility through careful writing and design--including examples with detailed solutions that begin and end on the same page, which maximizes readability. Similarly, side-by-side solutions show algebraic, graphical, and numerical representations of the mathematics and support a variety of learning styles. Reflecting its new subtitle, this significant revision focuses more than ever on showing readers the relevance of mathematics in their lives and future careers. | 677.169 | 1 |
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$51.19 solid introduction to the entire modeling process, A FIRST COURSE IN MATHEMATICAL MODELING, 4th Edition delivers an excellent balance of theory and practice, and gives you relevant, hands-on experience developing and sharpening your modeling skills. Throughout, the book emphasizes key facets of modeling, including creative and empirical model construction, model analysis, and model research, and provides myriad opportunities for practice. The authors apply a proven six-step problem-solving process to enhance your problem-solving capabilities -- whatever your level. In addition, rather than simply emphasizing the calculation step, the authors first help you learn how to identify problems, construct or select models, and figure out what data needs to be collected. By involving you in the mathematical process as early as possible -- beginning with short projects -- this text facilitates your progressive development and confidence in mathematics and modeling. | 677.169 | 1 |
Interactive Math Book
By G95 Associates
Description
The Interactive Math Book is for high school pre-algebra and algebra. Step by step instructions are given on solving problems with examples, problems and solutions. Interactive calculators and interactive graphs show concepts in simple and intuitive ways.
The text is split up into small sections, which are connected by hyperlinks. The hyperlink colors indicate more advanced or more basic material. The text is searchable, with forward and backwards navigation. | 677.169 | 1 |
The content of the course is to expand students' knowledge gained in basic mathematical objects of a number of application examples and historical notes, and in German language. Specifically, the following areas of mathematics: linear algebra, Bool algebra, graphs theory, linear optimization, differential equations, numeric methods.
The students are expected outcomes matematily range of high school and first semester of college and an active knowledge of German language, ie the ability to communicate in that language, and knowledge of basic mathematical terminology in the German language equivalent of the course Mathematik (KMA / MATD).
learning outcomes
Ability to communicate technical issues in a foreign language - German. Ability to develop in a foreign language, brief scholarly text. After completing the course, students will be able to work with specialized texts in German and prepare technical contribution in the German language as a seminar or conference. | 677.169 | 1 |
Search La Trobe
DISCRETE MATHEMATICS FOR COMPUTER SCIENCE
CSE2DMX
2018
Credit points: 15
Subject outline
This subject is an introduction to discrete mathematics. Concepts in discrete mathematics are useful in several branches of computer science as computers operate in discrete binary states. Among the topics covered are: sets, functions, relations, counting, sequences, modular arithmetic and Big-O notation. All these topics are central to the application of mathematics in electronics and computer science.
SchoolSchool Engineering&Mathematical Sciences
Credit points15
Subject Co-ordinatorFei Liu
Available to Study Abroad StudentsNo
Subject year levelYear Level 2 - UG
Exchange StudentsNo
Subject particulars
Subject rules
Prerequisites Must be admitted in SBAIO and have passed CSE1ITXDiscrete Mathematics for computer science
Prescribed
Computer Power Institute, 2018
Didasko digital
Readings
Discrete Mathematics with Applications (4th Ed)
Recommended
Susanna S. Epp, 2011
Cengage learning
Graduate capabilities & intended learning outcomes activities.Specific algorithms in computer science which use modular arithmetic will be used to illustrate its purpose.
Online chapters from the prescribed textbook will form the core learning.
In addition, webinars will provide students with opportunity to perform guided problem solving actives.
04. Compare algorithms arising in computer science for time complexity
Activities: actives. Common sorting, searching algorithms will be used to illustrate concepts.
Online, 2018, Study Block 1, Online
Overview
Class requirements
Unscheduled Online ClassWeek:
02
-
13
One 3.0 hours unscheduled online class per week
on any day including weekend
during the day
from week 02 to week 13 206
-
17
One 3.0 hours unscheduled online class per week
on any day including weekend
during the day
from week 06 to week 17 310
-
21
One 3.0 hours unscheduled online class per week
on any day including weekend
during the day
from week 10 to week 2115
-
26
One 3.0 hours unscheduled online class per week
on any day including weekend
during the day
from week 15 to week 26 519
-
30
One 3.0 hours unscheduled online class per week
on any day including weekend
during the day
from week 19 to week 30 623
-
34
One 3.0 hours unscheduled online class per week
on any day including weekend
during the day
from week 23 to week 34 728
-
39
One 3.0 hours unscheduled online class per week
on any day including weekend
during the day
from week 28 to week 39 832
-
43
One 3.0 hours unscheduled online class per week
on any day including weekend
during the day
from week 32 to week 4336
-
47
One 3.0 hours unscheduled online class per week
on any day including weekend
during the day
from week 36 to week 47 1041
-
52
One 3.0 hours unscheduled online class per week
on any day including weekend
during the day
from week 41 to week 52 1145
One 3.0 hours unscheduled online class per week
on any day including weekend
during the day
in week 45 1249
One 3.0 hours unscheduled online class per week
on any day including weekend
during the day
in week 49 | 677.169 | 1 |
Geometry for Teaching
From MathWiki
This page is just barely under construction!
Clearly Geometry for Teaching will include some profound understanding of big ideas, or bridges to big ideas, in geometry. It will include a profound sense of the connections between dimensions, and some grasp of the child's capacities and understandings.
One recommended text for exploring this, with university students would be David Henderson and Daina Taimina: Experiencing Geometry: Euclidean, Non-Euclidean with Strands of History (Prentice Hall), 2004. It has an appropriate emphasis on hands on explorations, reflection on activities, and reasoning, as well as some nice historical links and thoughts. | 677.169 | 1 |
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Overview
Master Math: Pre-Calculus: Pre-Calculus / Edition 2Product Details
Read an ExcerptFirst ChapterMost Helpful Customer Reviews
This is the best book on learning basic algebra. It is thorough yet concise. The information is presented very clearly. The author has obviously tried to explain the concepts so that they `make sense¿ to students - and their parents. I use the book to explain algebra to my students. Like the other Master Math books by Ross, the topics flow logically and build in difficulty. What a breath of fresh air after the often confusing text books students are given in school. This book is helpful for students struggling with algebra and the parents who are tutoring them. This book is also extremely useful for older students who did not adequately learn algebra, yet find they need to know it later. Topics can easily be looked up and reviewed or learned. I highy recommend this book!
Log-IC
More than 1 year ago
Splendid clarity and progression.
A cherry on top of this fine sundae,
maybe a single page of exponential,
radical,and complex properties for
a quick reference,( but that's nitpicking).
Guest
More than 1 year ago
This is the best book out there on learning trigonometry. I especially appreciate the visually-oriented focus. Each concept is described in all its forms, such as sine. Do you know each of the different ways sine can be described? Like the other Master Math books by Ross, the topics flow logically and are in context with what precedes and follows. It is thorough yet concise, and packed full of everything you, as tutor, or your kids need to know. The real world and fun applications are wonderful! The information is explained clearly and in a way that makes sense, so that a given concept is explained in such a way you understand what is being discussed rather than just memorizing formulas. What a breath of fresh air after the often confusing text books I was and my children are given in school. I really feel I can explain trigonometry to young people using this book! if I were going back to school, and taking math or science, this book would be in my backpack.
Anonymous
More than 1 year ago
Anonymous
More than 1 year ago
i am not sure if i am missing something but on page 14 the book says x times $1.00 per glass will equal 20. then shows this equation
X + ($1.00 per glass)=$20.00 terrible. There are many books that are better. I found websites more helpful.
This revised second edition provides an introductory guide through the maze of interdisciplinary themes that
comprise 'biodiversity.' It combines biological sciences with insights into the origins, variety and distribution of biodiversity, analysis of the social and political context, and the ...
Appropriate for business schools and community colleges, this textbook reviews basic math functions, fractions, and
percents, and applies them to the calculation of payroll, taxes, product pricing, interest, mortgages, and depreciation. Each section concludes with step-by-step instructions for performing the ...
This book is for instructors who think that most calculus textbooks are too long. In
writing the book, James Stewart asked himself: What is essential for a three-semester calculus course for scientists and engineers? ESSENTIAL CALCULUS: EARLY TRANSCENDENTALS, Second Edition, ...
The major concepts and discoveries of science are comprehensible to everyone with keen interest and
patience. But to really participate in fundamental science, particularly in physics, one must master a lot of math.Math is what prevents most science enthusiasts from ...
Differential geometry and topology have become essential tools for many theoretical physicists. In particular, they
are indispensable in theoretical studies of condensed matter physics, gravity, and particle physics. Geometry, Topology and Physics, Second Edition introduces the ideas and techniques of ...
INSIDE RHINOCEROS 5, is the ideal introduction to using the latest version of Rhino. This
well-designed book bridges the gap between theoretical and software-oriented approaches to computer modeling by providing a balanced presentation of theory, concepts, and hands-on tutorials. It ...
Incorporating new problems and examples, the second edition of Linear Systems and Signals features MATLAB®
material in each chapter and at the back of the book. It gives clear descriptions of linear systems and uses mathematics not only to prove ... | 677.169 | 1 |
About this product
Description
The main intended audience for this book is undergraduate students in pure and applied sciences, especially those in engineering. Chapters 2 to 4 cover the probability theory they generally need in their training. Although the treatment of the subject is surely su?cient for n-mathematicians, I intentionally avoided getting too much into detail. For instance, topics such as mixed type random variables and the Dirac delta function are only brie?y mentioned. Courses on probability theory are often considered di?cult. However, after having taught this subject for many years, I have come to the conclusion that one of the biggest problems that the students face when they try to learn probability theory, particularly wadays, is their de?ciencies in basic di?erential and integral calculus. Integration by parts, for example, is often already forgotten by the students when they take a course on probability. For this reason, I have decided to write a chapter reviewing the basic elements of di?erential calculus. Even though this chapter might t be covered in class, the students can refer to it when needed. In this chapter, an e?ort was made to give the readers a good idea of the use in probability theory of the concepts they should already kw. Chapter 2 presents the main results of what is kwn as elementary probability, including Bayes' rule and elements of combinatorial analysis. | 677.169 | 1 |
FUNDAMENTALS OF ALGEBRAIC MODELING strives to show the student connections between math and their daily lives. Algebraic modeling concepts and solutions are presented in non-threatening, easy-to-understand language with numerous step-by-step examples to illustrate ideas. Whether they are going on to study early childhood education, graphic arts, automotive technologies, criminal justice, or something else, students will discover that the practical applications of mathematical modeling will continue to be useful well after they have finished this course.
Available with InfoTrac® Student Collections
Sonya McCook,
Alamance Community College
What's New
A new four-color design helps further distinguish the features of the text.
Examples and exercises have been updated brand new Chapter R A Review of Algebra Fundamentals has been added and it gives students an opportunity to review the algebra skills needed to be successful in a modeling course.
Reviews
"I give this text an A. I have looked at numerous other texts for the MAT 115 course, but none have been better suited than this book, in my opinion, to the Mathematical Modeling course."
"I have also asked my students for their opinions on the textbook, and I heard the following things: They really like that the definitions and formulas they will need are highlighted in blue. They also said that mathematics textbooks either make the sections impossible to follow on your own, or so easy to follow on your own that you feel you do not need to attend class, so they stop going. They felt this textbook was right in the middle. They said they need to come to my lecture because it helps them to understand exactly what the examples are doing in the book, but they wouldn't be able to follow completely on their own. They like the way the book works directly with lecture and vise versa. "
FOR INSTRUCTORS
Instructors Resource Manual
Student Solutions Manual
ISBN: 9781285420424
FOR STUDENTS
Student Solutions Manual
ISBN: 9781285420424
Prepare for exams and succeed in your mathematics course with this comprehensive solutions manual! Featuring worked out-solutions to the problems in FUNDAMENTALS OF ALGEBRAIC MODELING, 6th Edition, this manual shows you how to approach and solve problems using the same step-by-step explanations found in your textbook examples. | 677.169 | 1 |
Homework: Weekly problem
sets, to be handed in at
lab.
(Bring two copies to lab: one to hand in at the beginning, and the
other
to consult when presenting solutions at the board. Students are
encouraged
to talk about the homework with others in their study group,
but should
write up and hand in their homeworks
individually.)
Description of course:
This is the second
semester of a year-long undergraduate course in algebra. This semester
will emphasize topics concerning groups, rings, and fields. The basics of
linear algebra (as covered in Math 370) will be assumed, and will be used
where appropriate in the course.
The course will be varied,
involving theory, computations, and
examples. It is open to
undergraduate students, both to math majors and
others. Math 503 is a
more advanced and more theoretical course in
algebra than Math 371, and
that course is open both to undergraduate and
graduate students.
Students are expected to have taken Math 370 or the equivalent, and to be
familiar with vector spaces and matrices. Students who would like a
reference on linear algebra may wish to look at the book "Linear Algebra"
by K. Hoffman and R. Kunze, or the book of the same title by Lipschutz and
Lipson in the Schaum Outline series. Some prior familiarity with the
concepts of groups, rings, algebras, and fields would be helpful but not
essential. Students who would like another reference to the topics of
Math 371 may wish to consult "A First Course in Abstract Algebra" by John
Fraleigh, or "Schaum's Outline of Abstract Algebra" by Lloyd Jaisingh. | 677.169 | 1 |
Machine Design Research
Using Design Research and History to Tackle a Fundamental Problem with School Algebra By Sinan Kanbir English | PDF,EPUB | 2017 (2018 Edition) | 339 Pages | ISBN : 3319592033 | 16.05 MB
In this well-illustrated book the authors, Sinan Kanbir, Ken Clements, and NeridaEllerton, tackle a persistent, and universal, problem in school mathematics—why doso many middle-school and secondary-school students find it difficult to learn algebrawell? What makes the book important are the unique features which comprise thedesign-research approach that the authors adopted in seeking a solution to the problem.
Showcasing exemplars of how various aspects of design research were successfully transitioned into and influenced, design practice, this book features chapters written by eminent international researchers and practitioners from industry on the Impact of Design Research on Industrial Practice.
Machine Design continues 80 years of engineering leadership by serving the design engineering function in the original equipment market and key processing industries. It is a respected source for design and engineering products, technology and reference information, for use in the field of design engineering. | 677.169 | 1 |
precalculo 1 Advice
Showing 1 to 3 of 3
Although it might look like a simple course precalculus is the basics to calculus. You get to start to see a profound beauty for numbers and that can make you decide if mathematics is the major youre looking for.
Course highlights:
You basically learn simple things like: unit circle, sums, basic ideas for limits which is really important in calculus. You will start learning sets and you'll gather all of what you know to make something quite interesting.
Hours per week:
12+ hours
Advice for students:
Just don't get nervous about it, it's not hard. Pretty calculus can't be different but it's an interesting course and try to enjoy it. Because the things you learn there are going to be used in discrete mathematics and calculus. But it is really important that you study, it might be an easy course but learning the ideas clearly is important.
The professor knows how to explain math in a way everyone can understand. She is truly great. Everyone that knows her says how lucky I am to take class with her. You have study daily and really commit to study for the test during the semester.
Course highlights:
I love math, therefore I enjoyed the class very much. The highlight of the class was learning so well how to do graphs. We learned all the basics we need to know to move up to Pre cal 2. The course itself can be a little tiring during the semester but extremely worth it.
Hours per week:
9-11 hours
Advice for students:
The way to succeed this class is to practice everyday and go along with the syllabus. Math is an area where you have to study daily or else you forget quickly what you learned. Study intensely for the test with weeks of anticipation; makes reviews, practice problems from the book, and print past exams provided on the university's website to practice as well.
Course Term:Spring 2017
Professor:Ivelisee Rubio
Course Required?Yes
Course Tags:Math-heavyBackground Knowledge Expected
Feb 04, 2017
| Would highly recommend.
Not too easy. Not too difficult.
Course Overview:
I recommend this course to students that would like to pursue a science bachelor. This professor is the best explaining math. I took the course with him, and other students from different professors go to him to understand the material. When I took the continuation course MATE 3024 - Precálculo 2, I went to him to understand correctly because my professor did not actually explain in class. The course is really good, especially for students in Physics, Chemistry, and Biology.
Course highlights:
I learn lots of things. Especially, factorizing, analyzing verbal problems, logic, functions, rate and average, and lots of other things. Is a great class to have in your curriculum for future references.
Hours per week:
6-8 hours
Advice for students:
You should be revised your notes every week. Maybe do from three to five problems every time you took something new. It helps to remember for the next class. Is good to do at least 15 problems in the weekend, this helps to practice and if you have a doubt about a problem you know what to ask the professor. If you all this, one week before the exam you should be ready to take it. That week look for old exams in the course page and do them by yourself and then ask a tutor or the professor if you are right.
Course Term:Fall 2017
Professor:Aponte, José
Course Tags:Great Intro to the SubjectMany Small AssignmentsGreat Discussions | 677.169 | 1 |
An accessible treatment of the modeling and solution of integer programming problems, featuring modern applications and software
In order to fully comprehend the algorithms associated with integer programming, it is important to understand not only how algorithms work, but also why they work. Applied Integer Programming features a unique emphasis on this point, focusing on problem modeling and solution using commercial software. Taking an application-oriented approach, this book addresses the art and science of mathematical modeling related to the mixed integer programming (MIP) framework and discusses the algorithms and associated practices that enable those models to be solved most efficiently.
The book begins with coverage of successful applications, systematic modeling procedures, typical model types, transformation of non-MIP models, combinatorial optimization problem models, and automatic preprocessing to obtain a better formulation. Subsequent chapters present algebraic and geometric basic concepts of linear programming theory and network flows needed for understanding integer programming. Finally, the book concludes with classical and modern solution approaches as well as the key components for building an integrated software system capable of solving large-scale integer programming and combinatorial optimization problems.
Throughout the book, the authors demonstrate essential concepts through numerous examples and figures. Each new concept or algorithm is accompanied by a numerical example, and, where applicable, graphics are used to draw together diverse problems or approaches into a unified whole. In addition, features of solution approaches found in today's commercial software are identified throughout the book.
Thoroughly classroom-tested, Applied Integer Programming is an excellent book for integer programming courses at the upper-undergraduate and graduate levels. It also serves as a well-organized reference for professionals, software developers, and analysts who work in the fields of applied mathematics, computer science, operations research, management science, and engineering and use integer-programming techniques to model and solve real-world optimization problemsThoroughly classroom-tested, Applied integer programming is an excellent book for integer programming courses at the upper-undergraduate and graduate levels." (Mathematical Reviews, 2011)
"The book is intended as a textbook for an application oriented course for senior undergraduate or postgraduate students, mainly with an engineering, business school, or applied mathematics background. Each chapter comes with several exercises, solutions of which are provided in an appendix. Many figures illustrate the flow of algorithms and other concepts." (Zentralblatt MATH, 2010)
About the Author
Der-San Chen, PhD, is Professor Emeritus in the Department of Industrial Engineering at The University of Alabama. He has over thirty years of academic and consulting experience on the applications of linear programming, integer programming, optimization, and decision support systems. Dr. Chen currently focuses his research on modeling optimization problems arising in production, transportation, distribution, supply chain management, and the application of optimization and statistical software for problem solving.
Robert G. Batson, PhD, PE, is Professor of Construction Engineering at The University of Alabama, where he is also Director of Industrial Engineering Programs. A Fellow of the American Society for Quality Control, Dr. Batson has written numerous journal articles in his areas of research interest, which include operations research, applied statistics, and supply chain management.
Yu Dang, PhD, is Qualitative Manufacturing Analyst at Quickparts.com, a manufacturing services company that provides customers with an online e-commerce system to procure custom manufactured parts. She received her PhD in operations management from The University of Alabama in 2004.
Top customer reviews
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A timely, comprehensive, easy-to-read, and self-contained application textbook for integer programming - the first readable text this 30 year veteran has seen in a decade - a must-have for every practitioner. The book is formatted as a traditional textbook, with problems at the end of each chapter, and solutions in the back of the book for many of the more difficult problems. This text is a natural extension of the well-known introductory texts: e.g. Winston. An extensive reference list bridges the practical solutions to the underlying theory. The references are linked from the historical notes at the end of each chapter. The text covers integer programming in 3 major sections: modeling, linear programming theory, and classical and modern solutions.
The modeling section covers all the classical problems: knapsack, production planning, and scheduling - followed by the network models: assignment, transshipment, maxflow, and shortest path.
Since linear programming based branch and bound solutions are state-of-the-art, the second section reviews linear programming fundamentals as both a traditional linear algebra formulation and, in preparation for branch and bound cuts, a geometrical formulation where the columns are the basis vectors spanning the feasible solution space. Figures are extensively used to crystallize the geometric concepts.
In the final, integer programming methods are covered in general: branch and bound, cutting plane, and group theoretic - focusing on using the methods in combinations, especially, branch and bound with cutting plane. Detailed, but tractable, examples with figures are included every step of the way emphasizing how and why the algorithms work. Rarely introduced in a text are 3 modeling languages that can actually be used in commercial applications.
This is a good book if you combine it with other resources. The book does cover a lot of topics with up-to-date information. However 450 pages is not enough for in-depth discussion of many topics such as the geometric concepts of simplex or heuristics. I totally agree with another reviewer about mistakes in book and they are very, very annoying. Too bad there is no errata available on publisher's website.
I have ready many books in mathematics in my day, but I have never seen one with as many errors as this book. Apart from being littered with spelling and grammar errors, there are a number of concerning mathematical mistakes as well.
The authors spend an inordinate amount of paragraphs explaining relatively simple concepts like introduction of slack variables, but avoid elaborating on more complex topics such as decomposition and transformations.
That said, the book is a good general introduction to the field as long as you read it carefully and notice their mistakes. | 677.169 | 1 |
2 Contents PARTI Fundamentals of Discrete Mathematics 1 1 Fundamental Principles of Counting The Rules of Sum and Product Permutations Combinations: The Binomial Theorem Combinations with Repetition: Distributions An Application in the Physical Sciences (Optional) Summary and Historical Review 44 2 Fundamentals of Logic Basic Connectives and Truth Tables Logical Equivalence: The Laws of Logic Logical Implication: Rules of Inference The Use of Quantifiers Quantifiers, Definitions, and the Proofs of Theorems Summary and Historical Review Set Theory Sets and Subsets Set Operations and the Laws of Set Theory Counting and Venn Diagrams A Word of Probability Summary and Historical Review Properties of the Integers: Mathematical Induction The Well-Ordering Principle: Mathematical Induction Recursive Definitions The Division Algorithm: Prime Numbers The Greatest Common Divisor: The Euclidean Algorithm 4.5 The Fundamental Theorem of Arithmetic Summary and Historical Review 238
3 XIV Contents 5 Relations and Functions Cartesian Products and Relations Functions: Piain and One-To-One Onto Functions: Stirling Numbers of the Second Kind Special Functions The Pigeonhole Principle Function Composition and Inverse Functions Computational Complexity Analysis of Algorithms Summary and Historical Review Languages: Finite State Machines Language: The Set Theory of Strings Finite State Machines: A First Encounter Finite State Machines: A Second Encounter Summary and Historical Review Relations: The Second Time Around Relations Revisited: Properties of Relations Computer Recognition: Zero-One Matrices and Directed Graphs Partial Orders: Hasse Diagrams Equivalence Relations and Partitions Finite State Machines: The Minimization Process Summary and Historical Review 394 PART 2 Further Topics in Enumeration The Principle of Inclusion and Exclusion The Principle of Inclusion and Exclusion Generalizations of the Principle Derangements: Nothing Is in Its Right Place Rook Polynomials Arrangements with Forbidden Positions Summary and Historical Review Generating Functions Introductory Examples Definition and Examples: Calculational Techniques Partitions of Integers The Exponential Generating Function The Summation Operator Summary and Historical Review 456
Discrete Mathematics Grades 9-12 Prerequisite: Precalculus G/T 1 credit This course is an introduction to the study of Discrete Mathematics, a branch of contemporary mathematics that develops reasoning
I The Real and Complex Number Systems 1. Identify subsets of complex numbers, and compare their structural characteristics. 2. Compare and contrast the properties of real numbers with the properties of
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MAT 051 Pre-Algebra Mathematics (MAT) MAT 051 is designed as a review of the basic operations of arithmetic and an introduction to algebra. The student must earn a grade of C or in order to enroll in MAT
MATHEMATICS (MATH) This is a list of the Mathematics (MATH) courses available at KPU. For information about transfer of credit amongst institutions in B.C. and to see how individual courses transfer, goInstructions. Answer each of the questions on your own paper, and be sure to show your work so that partial credit can be adequately assessed. Credit will not be given for answers (even correct ones) withoutModern Algebra, Review for Final Exam As with the first two one-hour exams, the final exam be divided into three sections: a section on definitions and statements of important results, a section on examplesThe primary objective of the Ph.D. Mathematics program is to provide training for the development of research capabilities in Mathematics and its possible applications. Complementary to this main objective
FIT 302: Mathematics for ICT INTRODUCTION This is one of the three courses designed for the Foundation in Information Technology (FIT) programme. This course consists of two parts which play an important
Lecture 5. Introduction to Set Theory and the Pigeonhole Principle A set is an arbitrary collection (group) of the objects (usually similar in nature). These objects are called the elements or the membersMath 113, Summer 2015 Prof. Haiman Review guide and exercises 1. Outline of topics Questions on the final exam will cover some subset of the topics listed below. Approximately one half of the exam will
Math 213 F1 Final Exam Solutions Page 1 Prof. A.J. Hildebrand 1. For the questions below, just provide the requested answer no explanation or justification required. As always, answers should be left in
Math Department Student Learning Objectives Updated April, 2014 Institutional Level Outcomes: Victor Valley College has adopted the following institutional outcomes to define the learning that all students
Lecture 3 Mathematical Induction Induction is a fundamental reasoning process in which general conclusion is based on particular cases It contrasts with deduction, the reasoning process in which conclusion482 Common Mistakes in Discrete Mathematics Common Mistakes in Discrete Mathematics In this section of the Guide we list many common mistakes that people studying discrete mathematics sometimes make. The
Chapter 0 Prerequisites All topics listed in this chapter are covered in A Primer of Abstract Mathematics by Robert B. Ash, MAA 1998. 0.1 Elementary Number Theory The greatest common divisor of two integers
Advanced Higher Mathematics Course Assessment Specification (C747 77) Valid from August 2015 This edition: April 2016, version 2.4 This specification may be reproduced in whole or in part for educational
Standard 1: Relations and Functions Students use polynomial, rational, and algebraic functions to write functions and draw graphs to solve word problems, to find composite and inverse functions, and to
MATH 30 Final Exam, May 3, 03, WTT Student Name and ID Number Note. The questions posed here appeared on the Spring 03 Final Exam, although in some instances, the format of the question has been altered.Discrete Mathematics Lecture 2 Logic of Quantified Statements, Methods of Proof, Set Theory, Number Theory Introduction and General Good Times Harper Langston New York University Predicates A predicate
Mathematics (offered by the Department of Mathematical Sciences) Telephone number 012 429 6202 1 Introduction All attempts to give a definition of Mathematics have suffered from one shortcoming or another, 1, a 0 F. The a i, 0 i n are the | 677.169 | 1 |
Math 412: Abstract Algebra
Fall 2017 Section 3
Math 412 is an introduction to abstract algebra,
required for all math majors but possibly of interest also to physicists, computer scientists,
and lovers of mathematics. We
will begin with ring theory: our first goal is to prove the Fundamental Theorem of Algebra,
about the ring you've been studying since elementary school, the integers. In the second half,
we will study group theory. In addition to developing many examples,
students will prove nearly all statements in this course.
Warning: we differ from the book by including in our definition of ring that every ring contains 1.
Sections:
Course Assistant:
All sections will use the same syllabus, do the same classwork, take the same exams, and do the same homework,
regardless of instructor.
Prerequisites:
Math 217. Students are expected to know linear algebra and to have had a course in which they have been trained in rigorous proof techniques (induction, proof by contradiction, etc).
Course Description:
This class is an introduction to the basic concepts of algebra.
The topics covered are approximately Chapters 1-9 and 13 in the textbook. The class
is roughly structured as follows: we begin with a rigorous study of arithmetic of the integers
(division algorithm, primes, and unique factorization, congruences, modular arithmetic) culminating with the proof of the
Fundamental Theorem of Arithmetic. Part II is about basic properties of rings and
ring homomorphism (ideals, quotient rings, fields). Here, another
important example, which shares many properties of Z, the integers, is the ring of polynomials over a field.
Finally, in part III we study the basics of group theory
(groups, group homomorphisms, symmetry groups, the symmetric group,
normal subgroups, quotient groups, and group actions on sets).
The parts are not evenly spaced: I is shorter than II, and II is shorter than III.
Office Hours:
My office hours are Mondays 1-3, Wednesdays 12-1, in my office 4827 East Hall
I am available also by appointment if you need me and can't make regular office hours.
Karen's office hours: Mondays 11-12, Wednesdays 1-3, in her office 3074 East Hall
Yifan's office hours: Thursdays 5-6 in the Math Atrium (East Hall second floor)
Daily Update:
Wednesday 11/29: Started with a quiz, then did A-C(2) of worksheet #22.
Will finish that, then go on to simple groups and structure of finite abelian groups next time.
Monday 11/27: We finished
finding the normal subgroups and the quotient groups of the symmetric groups on 4 and 5 elements, front side of worksheet #21.
Next time, we will start with the quiz, then go
on to the First Isomorphism Theorem for groups, and simple groups.
Please read the Diffie-Hellman part of the RSA worksheet / Thanksgiving leftovers worksheet before next week.
Wednesday 11/22: We discussed RSA today,
following the RSA worksheet #20 below. We didn't get to Diffie-Hellman. Will resume with normal subgroups after the break.
Wednesday 11/14, Friday 11/16: Discussed normal subgroups and quotient groups, worksheet #19.
Monday 11/12: Started with a quiz.
Discussed the bonus problem on testing whether an element is a cyclic generator for the multiplicative group of Z/p.
Started discussing normal subgroups, following the normal subgroups worksheet.
Review sections 8.2 and 8.3, as well as problem D from HW#7 for next time. A summary of what we covered on group actions is posted below.
Friday 11/9: Finished the worksheet on group actions. Discussed the Orbit-Stabilizer theorem and examples. Read 8.3 for Monday.
Wednesday 11/7: Discussed the fact that even though we know multiplicative groups of finite fields are cyclic, what generates them is subtle.
Started on the group actions worksheet. Review 8.1 and 8.2 for next time.
Monday 11/5: We reviewed the proof of the theorem that the multiplicative group of any finite field is cyclic. We then defined a group action on a set, and disucssed the example of the integers acting on the real number line. Review the definitions on today's worksheet for next time.
Friday 11/3: Worked on group homomorphisms and Lagrange's theorem. Read 8.2 for next time. Know the main definitions for the quiz.
Wednesday 11/1: Finished classification of small groups. No reading assigned.
Monday 10/30: Started classification of small groups. Read 8.1.
Friday 10/27: We focused on permutation groups today. Did worksheet #15 on
permutation groups. If you didn't get through B, look at that, and see if you can do C. Read 7.4 next time. Quiz on Monday.
Wednesday 10/25: Took a quiz, discussed cyclic subroups. Did worksheet #14
on subgroups and generators. Do up through C if you didn't get through it
in class. Read 4.5 for next time.
Monday 10/23: Collected some
examples of groups:
additive groups of rings, multiplicative groups of units, and
groups of symmetries/permutations. Did worksheet #13 on Groups 2.
Altogether, we found evidence for the conjecture that the order of any
element divides the order of the group. Read 7.3 for next time.
There is a quiz on Wedensday.
Friday 10/20: Did worksheet #12 on Groups. Read 7.1 and 7.2 for next time. No quiz on Monday.
Wednesday 10/18: Reviewed what elements are units in quotients of polynomial rings, and how to solve linear equations there. Showed that the quotient of a polynomial ring over a field by an irreducible polynomial is a field, and used the quotient ring construction to construct fields where polynomials have roots.
Friday 10/13: Discussed the First Isomorphism Theorem.
Wednesday 10/11: Continued with quotient rings, and introduced the First Isomorphism Theorem.
Monday 10/9: Started to discuss quotient rings. Showed that addition and multiplication are well-defined. Read 6.2 for next time.
Friday 10/6: Division algoithm in general. Ideals in polynomial ring, both over fields and other rings.
Wednesday 10/4: We reviewed some of the peculiarities about the defintions of ideals/subrings.
We talked a bit about polynomial rings and the division algorithm for polynomial rings over fields.
Got through C on worksheet #11. Will continue with polynomial rings and ideals on Friday. Read 4.3.
Monday 10/2: Started talking about ideals, and generators of ideals on worksheet #10. We showed that every ideal in the ring of integers is generated by one element. More ideals Wednesday. Read 4.1 and 4.2.
Friday 9/29: Discussed ring homomorphisms and did A, B from worksheet #9. Discussed C and F briefly. Defined ideal. Note that an ideal is NOT a subring in terms of our definition of ring, because 1 is usually not in an ideal.
Wednesday 9/27: Did worksheet #8 on Rings 2. On Friday, we discuss homomorphisms.
Monday 9/25: Did worksheet #7 on Rings. Will continue talking about rings on Wednesday. Read section 3.2.
Friday 9/22: Went through worksheet #5 on Systems of Congruences, and then did
worksheet #6 on operations. Reread Section 3.1 for Monday.
Wednesday 9/20: Did worksheet #4 on solving the equation [a]x=[b] in Z_N.
We all discussed through Problem B, and Problem C1, which said solutions may
not exist and may not be unique if N is not prime. Read 14.1 and 3.1 for Friday.
Monday 9/18: Did Quiz #2, and and mostly finished the Congruence worksheet.
Read Section 2.3 and start 14.1 for Wednesday.
Friday 9/15: Did worksheet #3 on Congruence up through part D. Will
continue with E and F on Monday. Finish reading section 2.2 for Monday.
Wednesday 9/13: Started with quiz #1, then discussed equivalence relations
and equivalence classes. Read 2.1 and start 2.2 for Friday.
Monday 9/11: Did worksheet #2 on the Fundamental Theorem of Algebra. We
recapped through part D. You may want to go over part E in the solutions.
Read Appendix D and start 2.1 for Wednesday.
Friday 9/8: Did worksheet #1 on the Euclidean algorithm. We discussed how
to find the gcd of two numbers as a linear combination, why the
Euclidean algorithm gives a proof of this fact, and the difference between
a constructive proof and a nonconstructive proof. Dicussed Theorem 1.4 of
Section 1.2. Review 1.2 and read 1.3 for next time.
Wednesday 9/6: We discussed syllabus matters, loosely discussed the notion
of a ring, and covered Section 1.1, inculding the proof of the division
algorithm in detail. Read Sections 1.1 and 1.2 for next time. Know
definitions of divides, GCD, and go over proof of Theorem 1.2.
Quizzes:
Course Expectations:
Math 412 students are responsible for learning the material on their own through individual reading of
the textbook before coming to class. Like in Math 217, you will often work together on
more theoretical concepts in small groups using worksheets in class.
You will be expected to work out more computational exercises on your own,
which will be supplemented with some webwork when possible.
You will also have a graded, written problem set (think Math 217 Part B) due Fridays. Attendance is required.
There will be two exams (one midterm and a final).
There will be many quizzes, some on the reading.
Review Session:
Grades:
Testing and Disability:
If you think you need an accommodation for a disability, please let me know as soon as
possible. In particular, a Verified Individualized Services and Accommodations (VISA)
form must be provided to me at least two weeks
prior to the need for a test/quiz accommodation.
The Services for Students with Disabilities (SSD) Office (G664 Haven
Hall) issues VISA forms. | 677.169 | 1 |
Main menu
5 Takeaways That I Learned About Resources
Education is the key to success in life. Mathematics is vital in most people's careers. Growing individuals are aware of mathematics that awaits them. They start counting any number they hear. Mathematics is taught to persons in colleges, universities, and also primary school children. Studying mathematics is vital to the person who tends to run their businesses. Mathematics syllabus has several units, one of them being algebra. It helps person to deal with any calculation in their lives. It will be fast and calm for most individuals to calculate their money in their business. The formula that you use in primary school will be the formula you will use in college. Below are the advantages of studying algebra in school.
Algebra will control your money
Businesses are sources of additional money that business persons have. Characters work extra hard to be able to take of their bills. The amount of money you earn needs to be controlled so that it will be enough to buy the most vital needs. Having studied algebra, it will be easy to plan your money. Algebra calculation will enable you budget your money without any challenge. Algebra will ensure that you do not spend your money in a wasteful ways. Algebra will enable you know the amount of money you spend on different occasions. Valuable Lessons I've Learned About Resources
Algebra helps you deal with most difficult situation that comes your way. Most individuals do not have ways to deal with life difficulties. Algebra will give you the needed solution to these issues. The community can use the persons with algebra knowledge to solve the community problems. Algebra ensures that persons think more than it is expected of them. Persons with skills will guide individuals in the correct direction. They will be used to set the goals and objectives by business firms. Studying algebra will ensure that you have a good paying job.
Algebra creates a foundation of other subjects
It is easy to handle any question with algebra knowledge. It is assumed that, it is necessary to attend all algebra lessons. Physics and chemistry will not be a challenge to you when you have enough skills on algebra. With algebra knowledge will make your school life easy.
Algebra enables decision making
Good choices are made with the help of Algebra diagrams. Algebra involves drawing of diagrams and graphs. It is vital to compare things before you make decision on which is the best.Best choices are made after a long time of thinking. Algebra enables you deal with most persons and situations. | 677.169 | 1 |
Every mathematician and student of mathematics needs a familiarity with mathematical induction. This volume provides advanced undergraduates and graduate students with an introduction and a thorough exposure to these proof techniques. 2017 edition.
Classic by prominent mathematician offers a concise introduction to set theory using language and notation of informal mathematics. Topics include the basic concepts of set theory, cardinal numbers, transfinite methods, more. 1960 edition. | 677.169 | 1 |
This book has been designed to meet the latest syllabus prescribed by Anna University for the students of First Year Programme of B.E./B.Tech. A large number of worked out examples, clear and inter woven throughout the text to reinforce the concepts. The steps and procedures are very friendly so that students have a plenty of exercises with short and long answers for every topics in consecutive solved problems. All the topics are presented in an elaborate manner with numerous solved problems. This book will surely enable the students of successfully diagnose, plan, implement and evaluate their learning experiences. Sample lessons and learning activities provide teachers with practical and simple ideas to support students' development. This book remains a hallmark with the features and eludication of mathematical concepts making it a tool for effective practising. | 677.169 | 1 |
Student Audience:
Prerequisite:
The prerequisite is successful completion of Math 121 with a "C" or better or permission of the Associate Dean of the Industrial Technology and Mathematics Division.
Course Description:
Mathematics 122, Calculus and Analytic Geometry II, includes instruction in Calculus topics common to the standard college second semester Calculus course. General objectives in the course are to increase the student's mastery of the deductive nature of reasoning, to understand the nature of critical thinking, to increase the student's ability in problem solving, and to increase the student's ability to work with others towards a common goal.
Type of Instruction:
Lecture, discussion, problem solving, and group work will be used. Students should come to class with a prepared list of questions.
Attendance Policy:
Regular attendance is essential for satisfactory completion of this course. If you have excessive absences, you cannot develop to your fullest potential in the course. Students who, because of excessive absences, cannot complete the course successfully, will be administratively dropped from the class.
The student is responsible for all assignments, changes in assignments, or other verbal information given in the class, whether you are in attendance or not.
Do NOT assume all that will be done on review days is review for the exam. New material may be covered the day before the exam.
Test Policy:
If a student must miss class, a call to the instructor (RCC's phone system has an answering system) is to be made. If an exam is to be missed, a phone call is to be made and a written notice given. If the instructor is not contacted, the grade will be zero. If a student misses an exam, and gives written notice, the percent score of the final exam will be used in its place. The student should be careful in exercising this policy, as it is very rare when a student gets a noticeably higher grade on the final exam. This substitution of the final exam percent will be done once, and only once. Any other examination missed will receive a grade of 0. If a student does not give written notice of missing the exam, the option of using the final exam score as a substitute grade will not be done, and the exam grade will be zero.
Grading Policy:
There will be several one hour examinations and a comprehensive final examination. Announced and unannounced quizzes may be given. Laboratory and homework exercises (to be announced) may be used in grading. Collected assignments will lose 10% of the grade for each class period late. A grade will be taken on your notebook. Note: Homework is essential to the study of mathematics.
Letter grades will be assigned to final adjusted scores as follows:
A = 90 - 100%
B = 80 - 89%
C = 70 - 79%
D = 60 - 69%
F = 0 - 59%
Consideration will be given to such qualities as attendance, class participation, attentiveness, attitude comments that are appropriate. In general, it should contain everything written on the chalkboard. Be sure to bring your notebook if you come to the instructor or a tutor for help. A grade will be taken on the notebook. Things that will be considered for this grade are: 1) Organization, 2) Completeness of notes, and 3) Amount of homework attempted. I strongly urge you to get a three-ring binder to keep your papers in.
Calculators:
Calculators may be used to do homework. Calculators may be used on exams and/or quizzes in class unless otherwise announced. The calculator should be a scientific calculator capable of doing trigonometric work. A graphing calculator, such as the TI-82 or TI-85, is also a useful tool and highly recommended, but not required.
Additional Supplies:
The student should have a red pen, ruler, graph paper, stapler, and paper punch. The student is expected to bring calculators and supplies as needed to class.
Additional Help:
Office hours will be announced. Anytime I am in my office, feel free to stop and get help. The student is encouraged to seek additional help when the material is not comprehended. Mathematics is a cumulative subject; therefore, getting behind is a very difficult situation for the student.
If your class(es) leave you puzzled, the Study Assistance Center is a service that Richland Community College offers you.
Calculus is a very time intensive subject. An average student will need to spend two hours outside of class for every one hour in class. Working a full-time job and taking Calculus is very difficult and stressful. There is very little time spent reviewing the material previous to calculus. If you find yourself in trouble, see the instructor, the Study Assistance Center, or other class members immediately. Do not wait until you are too far behind.
Technology Laboratory Guide
Assigned Problems
The assigned problems from the Technology guide will be due on the day of the exam for the corresponding chapter. Deviations from this schedule will be announced in class. Be sure to pay attention. Each chapter will be worth 20 points, for a total of 100 points. | 677.169 | 1 |
Course Overview
Algebra is an area of mathematics that permits us to describe and model the real world. You will explore the big ideas of algebraic thinking that are currently taught in elementary and middle schools in the United States. You will complete activities that build conceptual understanding between and among these big ideas. Topics covered include properties of real numbers, mental math, equal vs equivalent, growing patterns, functions, linear and non-linear models, recursive functions, linear inequalities and systems, and simplifying, combining, evaluating and factoring quadratic expressions and equations.
Technology Requirements
Participation in this course will require the basic technology for all online classes at Columbia College:
Course Objectives
To progress from a procedural/computational understanding of mathematics to a broad understanding encompassing logical reasoning, generalization, abstraction, and formal proof.
To use technology (calculator and computer) as a learning and teaching tool for mathematics.
To learn the algorithmic approach to problem solving.
To display an understanding of the nature of rigorous proof. To write elementary proofs, especially proofs by induction and basic number theory proof.
Measurable Learning Outcomes
Know the basic properties of the real numbers including commutativity, associativity, identity, distributivity.
Use the basic properties of the real numbers to determine equivalent algebraic equations and solve algebraic equations.
Use equations to model problem solving situations
Understand and use a variable to generalize a pattern, to represent a fixed but unknown number, to represent a quantity varies in relation to another quantity and that a variable can be a discrete or continuous quantity.
Use quantitative reasoning to generalize relationships
Use functional thinking to generalize relationships between covarying quantities and to express those relationships in words, symbols, tables, or graphs and reason with those relationships to analyze function behavior.
Use functional thinking to generalize relationships between covarying quantities and to express those relationships in words, symbols, tables, or graphs and reason with those relationships to analyze function behavior.
Compare and contrast the concepts of equality or equivalence.
Compare and contrast the concepts of equality or equivalence.
Understand that an inequality can describe a relationship between equalities and solve these inequalities.
Understand and describe recursive relationships
Classify functions based on the rate at which the variables change and the situations that they model
Solve equations using symbolic, graphical and numerical methods.
UG
Grading
Grading Scale
Grade
Points
Percent
A
720-800
90-100%
B
640-719
80-89%
C
560-639
70-79%
D
480-559
60-69%
F
0-479
0-59%
Grade Weights
Assignment Category
Points
Percent
Discussion (9)
210
26%
Homework (8)
200
25%
Quizzes (6)
90
11%
Midterm Exam (1)
150
19%
Final Exam (1)
150
19%
Total
800
100%
Schedule of Due Dates
Week 1
Assignment
Points
Due
Introduction Discussion
10
Thursday/Saturday
Discussion 1
25
Homework 1
25
Friday
Quiz 1
15
Sunday
Week 2
Assignment
Points
Due
Discussion 2
25
Wednesday/Saturday
Homework 2
25
Friday
Quiz 2
15
Sunday
Proctor Information
N/A
Week 3
Assignment
Points
Due
Discussion 3
25
Wednesday/Saturday
Homework 3
25
Friday
Quiz 3
15
Sunday
Week 4
Assignment
Points
Due
Discussion 4
25
Wednesday/Saturday
Homework 4
25
Friday
Midterm Exam
150
Sunday
Week 5
Assignment
Points
Due
Discussion 5
25
Wednesday/Sunday
Homework 5
25
Friday
Quiz 4
15
Sunday
Week 6
Assignment
Points
Due
Discussion 6
25
Wednesday/Saturday
Homework 6
25
Friday
Quiz 5
15
Sunday
Week 7
Assignment
Points
Due
Discussion 7
25
Wednesday/Saturday
Homework 7
25
Friday
Quiz 6
15
Sunday
Week 8
Assignment
Points
Due
Discussion 8
25
Wednesday/Saturday
Homework 8
25
Friday
Final Exam
150
Saturday
Total Points
800
Assignment Overview
Discussions
Each week you will participate in an online discussion worth 25 points each. In Week 1, you will have have an additional 10 point introductory discussion. You should respond fully to the initial discussion question and post at least two responses. Initial posts are due by 11:59 pm Wednesday, except for Week 1 when they are due on Thursday. Responses are due by 11:59 pm on Saturdays.
Your posts should be supported with examples with evidence of synthesis of readings and/or outside sources. Formal paragraph structure with college-level grammar and spelling is expected. Additional grading criteria will be available in the course.
Homework
Weekly Homework is required to allow you to practice mathematical skills. Each homework assignment is worth 25 points and due by 11:59 PM on Fridays. The Homework worksheets will be available in the Dropbox, and should be completed, saved, and resubmitted to their respective Dropbox.
Quizzes
There are six quizzes containing 5 short, objective-style questions. You will have 30 minutes to complete the quizzes and up to two attempts. The highest score will be recorded. Quizzes open on Monday and are due by 11:59 pm Sundays. They are worth 15 points.
Exams
The course has a Midterm and a Final Exam. Both must be proctored. They will contain M/C, T/F, or short answer questions. You will have two hours to complete the exams and just one attempt. They are available beginning on Monday of their respective weeks. The Midterm is due by 11:59 pm Sunday of week 4, the Final is due by 11:59 Saturday of week 8. Each exam is worth 150 points.
The Midterm covers material from weeks 1-4.
The Final covers material from weeks 5-8
Course Outline
Click on each week to view details about the activities scheduled for that week.
Introduction Discussion Tell us a little about yourself. Items you might share are your interest in math, your career ambitions or family. Let's get to know each other! (Reminder: Two responses are expected in this and all discussions.)
Discussion 1 The concepts of even and odd numbers seem to be simple to us, but for children this may not always be the case.
Watch the video clip at thisUniversity of Michigan siteand respond. You can use these questions to help, but feel free to respond in other ways.
1. Were you surprised by the students' various interpretations of odd and even numbers?
2. What do you think is the importance of allowing the students to share and defend their thinking?
3. What role is the teacher playing in the classroom?
4. Is this how you remember learning about even and odd numbers and or what you have seen in the classrooms?
5. Why do you think the teacher is letting the kids develop the definition? (i.e. They have a working definition of even number)
Discussion 2 View the algebraic puzzles (Word download) that were posted on my Facebook page, prompting quite a discussion between me and my friends. (Note: puzzles will be available in the course.) Many answers were found and many paths were taken. People got excited about it even if they were not "math" people.
Share these images with at least 10 people and then report your findings in the discussion for this week. What were their thought processes and reactions? Were they able to solve them?
Are you able to solve the puzzles? Discuss what the equal sign is representing in these puzzles.
Proctor Information Submit your proctor form to the appropriate Dropbox folder by the end of the week. Remember to "Save" the form before placing it in Dropbox. See the Content area for more information.
In the article found on the Ontario Ministry of Education site,Paying Attention to Algebra, on pages 19-21, an activity and student work is shown. Consider the use of representations, tables, and graphs in helping students understand the roles of variables and to eventually write, use, and solve equations. As you do you may wish to respond to all of these prompts:
Do you believe these types of experiences are important in helping children develop their algebraic reasoning skills?
Did you have experiences such as this when developing your algebraic reasoning?
Discussion 4 The 5 Big Ideas are summarized on pages 12- 13 of your text (Developing Essential Understanding of Algebraic Thinking 4 Homework 4 covers body ratios.
Midterm Exam The Midterm Exam is proctored and covers material learned in weeks 1-4. See the course for additional information.
Discussion 6 This week we have been modeling with recursive functions. In past weeks we have modeled with both linear and non-linear functions that are called explicit functions.
Describe the differences between these two types of functions (explicit and recursive).
Sometimes functions are more easily described using an explicit function; give an example of such a function.
In other cases, functions can be more are more easily described using recursive notation. Give an example of such a function.
Homework 6 Complete Homework 6.
Quiz 5 Quiz 5 covers week 6 readings 8 The 5 Big Ideas are summarized on pages 7-11 of your text (Developing Essential Understanding of Expressions, Equations & Functions 8 Complete Homework 8.
Final Exam The Final Exam must be proctored and covers material from weeks 5-8. See the course for additional information However late initial posts may receive partial credit provided they are posted before the end of the week deadlines (Saturdays).
Late exams are not accepted without prior approval. Approval is only given under significant extenuating circumstances and must be requested before the due date.
I do not accept late weekly quizzes.
I do accept late dropbox assignments, but will deduct 5% for each day they are late | 677.169 | 1 |
Houghton Mifflin Harcourt's Go Math series is the textbook that will be used for 7th grade math. Every student will be responsible for bringing this book to class everyday.
Online access at my.hrw.com
Required Supplies
Pencils
Notebook Paper
3-Ring Binder
Composition Notebook
Math Workbook/Textbook
1 pack of 5 tab dividers
Course Overview
In Grade 7, instructional time will focus on four critical areas: (1) developing understanding of and applying proportional relationships; (2) developing understanding of operations with rational numbers and working with expressions and linear equations; (3) solving problems involving scale drawings and informal geometric constructions, and working with two- and three-dimensional shapes to solve problems involving area, surface area, and volume; and (4) drawing inferences about populations based on samples. The standards are fully explained TN.gov website.
Classwork: Participation in class activities and class assignments are important to informally assess student learning and understanding. All students are expected to participate in class. Participating in note-taking is a part of the class experience. Students are expected to take notes in their composition notebooks/3-ring binder.
Homework: Students will receive homework regularly to reinforce key math concepts. Students are expected to complete and turn in all homework in a timely manner. Homework is usually due the next day, unless otherwise specified by the teacher.
Tests: A test will be given at the conclusion of every Module and Unit.
Tests should be completed independently. Using notes, talking, or cheating during a test is NOT allowed. If a student is found doing such things, he/she will receive a ZERO on the test and cannot make up the test.
Late/Makeup work: Ten points will be deducted per day for all late work. A student will have one day for everyday they are absent to complete absent work.
Grading: Grades will be updated on Gradebook Wizard once per week. All students and parents have a username and password to use to check grades and communicate with teachers. | 677.169 | 1 |
This beautifully written text starts with proofs and sets in the first
40 pages and continues in the rest of Parts I and II to maintain an
ongoing emphasis on the construction of proofs, demonstrating proper
skills through detailed examples using the "forward-backward"
method.
one of the text's greatest strengths are the problem sets,
which are many and varied. In addition to a large number of more
traditional problems, students are asked to complete partial proofs,
find flaws in incorrect proofs, and modify proofs in the light of
new information.
offers a wide range of problem material—many of which
follow a "stream of consciousness" format—guiding readers
through large projects and allowing them to explore and develop interest
in near research-level topics.
presents the most abstract subject matter in terms that
relate to students' experience in calculus, rather than ignoring or
downplaying the value of this experience.
depicts the structure of the real number system as a collection
of closely interrelated properties, rather than simply a list of theorems.
covers a wide variety of topics in Part IV, each explored
using a "discovery" process.
conveys concepts in an interesting, conversational tone,
presenting the subject as one that is open and appealing to everyone. | 677.169 | 1 |
Based on extensive research, SmartGraphs: Algebra is designed for students studying Algebra 1 or Algebra 2. Activities are aligned with the Common Core. Teacher resources are available online.
KEY FEATURES
- SmartGraphs develops understanding of graphs
- Hints guide your work
- Includes dozens of algebraic problems and hours of engagement
- No third-party advertising
- No in-app purchases
For additional science, math and engineering activities, visit us at like us on Facebook: or follow us on Twitter: | 677.169 | 1 |
Transformational Mathematics
Instructor: Tom McGarrity
Mathematics underpins much more of our reality than most of us realize. It contains
many of life's design secrets… how all the different elements of nature connect and
grow, what proportions contain greatest power and vitality, and how our digital and
logical thinking works. The magic of math is often lost in classroom tedium. This
course attempts to bring the magic back to life. This course will be different from any
math course that most have ever taken. Together, we will walk through many of the basic languages and principles of mathematics and end up not only more
comfortable with mathematical concepts, but hopefully, enjoying and finding
amazement in these most ancient forms of human inquiry.
Course Learning Outcomes
By the end of the course students should be able to:
Demonstrate comfort and competence in working with variables, exponents,
fractions, decimals and percents en route to generating appropriate equations
leading to the solving of complex word problems.
Appreciate and speak the foundational language of mathematics as heard
through numbers theory and calculus. And comfortably begin to explore and
examine the qualitative and quantitative, sacredly geometric patterns of this
spectacular living universe. | 677.169 | 1 |
£22 introduction to mathematical analysis and linear algebra for economists. This work presents a balance between mathematics and economic examples. It includes topics ranging from elementary algebra to more advanced material, whilst focusing on the core topics usually taught in undergraduate courses on mathematics for economists. | 677.169 | 1 |
Mathematics
Peter O'Shaughnessy Head of Faculty, Mathematics
Mathematics is a unique and powerful way of viewing the world to investigate patterns, order, generality and uncertainty. Mathematics assists individuals to make meaning of their world. The use of mathematics allows individuals to analyse events using universally true abstractions and, at the same time, to apply these abstract ideas to interpret new situations in the real world.
All students study Mathematics in Years 7, 8, 9 and 10 under the Australian Curriculum guidelines. This program provides students with essential mathematical skills and knowledge in Number and Algebra, Measurement and Geometry and Statistics and Probability. Classes in the Middle School are organised according to individual student progress. The curriculum caters for students of all ability levels and aims to prepare them for the skills required for basic numeracy as well as the fundamentals required for senior and tertiary study. In Year 10, students study a core course for the first semester, and then specialise into one of three electives in the second semester according to their skills and possible options in the senior school.
The senior school offers Authority subjects Maths A, B and C as well as the Authority Registered Prevocational Mathematics. This range of subjects is aimed at providing students with the necessary skills outside of school life.
Students can participate in the Annual Maths Teams Challenge involving Year 8, 9 and 10 teams from schools throughout the region, and a team of Year 8 students participates in the annual Year 8 QAMT team competition. After school maths tutorials are held each week on both campuses. | 677.169 | 1 |
Professor of Mathematics
Interdisciplinary Center for Applied Mathematics @ Virginia Tech
Math 3054
Programming and Mathematical Problem Solving
Course Description
This course serves as an introductory programming course for
Mathematics majors. This course introduces elementary programming
skills that are applied to solve mathematical problems. The
course covers a basic understanding of how computers store and
programs manipulate data, procedural and object-oriented
programming techniques, parallel programming, and as well as
graphics and visualization. The main programming language for
the course is Matlab.
This course may be used to satisfy the Mathematics undergraduate
programming requirement.
We will cover programming solutions to mathematical problems,
including those that emphasize
This course will be managed using the Virginia Tech
Canvas site. Example
programs are posted under the Files menu.
Students are required to have access to Matlab in order to
perform computations, edit, and run programs. Matlab is available
for $25 to all Virginia Tech mathematics and engineering students
through the Virginia Tech Computing Center. You may can find
purchasing details
at the Information Technology Acquisitions site.
You may also find one of the many MATLAB tutorials and Learning
Resources useful: link
Course Contract
Math 3054 - Programming and Mathematical Problem Solving - Fall 2016
Professor
Jeff Borggaard, regular office hours will be held in McBryde 528. Students
are invited to e-mail me
at (jborggaard@vt.edu) to arrange an
appointment.
Evaluation and Grading
The grade will be determined by homework and projects as well as a
midterm exam. Assignments during the semester will make up 70% of the
grade. A midterm exam will be worth 15% of the grade and the remaining
15% will be determined through the final project that takes the place
of the final exam. Homework will be assigned weekly to bi-weekly.
Assignments will be managed during class or through Cody coursework.
All grades will be posted on the course Scholar site. To encourage
everyone to keep up with the material and lectures, late homework
will not be accepted unless by prior agreement.
Academic Integrity and Honor Code
Students enrolled in this course are responsible for abiding by the
Undergraduate Honor Code pledge that each member of the university
community agrees to abide by and which states:
As a Hokie, I will conduct myself with honor and integrity at all times.
I will not lie, cheat, or steal, nor will I accept the actions of those who do.
You are encouraged to discuss assignments with other members of the
class. However, the objective of this course is to teach everyone how to
program and use computer programs to solve mathematical problems, thus
any submitted write-up and code to be graded (including Cody coursework)
should be your work alone. Direct copying of solutions or parts of
solutions from any source is a violation of the honor code, as is
sharing your solutions with others.
All exams and the final project are to be done without any assistance
from anyone else. A student who has doubts about how the Honor Code
applies to any assignment is responsible for obtaining specific
guidance from the course instructor before submitting the assignment
for evaluation. Ignorance of the rules does not exclude any member
of the University community from the requirements and expectations
of the Honor Code. | 677.169 | 1 |
Unit 8 : Electrochemistry
Electrolytic and metallic conduction, conductance in electrolytic solutions, specific and molar conductivities and their variation with concentration : Kohlrausch's law and its applications. Electrochemical cells- Electrolytic and Galvanic cells, different types of electrodes, electrode potentials including standard electrode potential, half-cell and cell reactions, emf of a galvanic cell and its measurement; Nernst equation and its applications; dry cell and lead accumulator; fuel cells; corrosion and its prevention.
Unit 13 : Organic Compounds Containing Nitrogen
General methods of preparation, properties, reactions and uses. Amines : Nomenclature, classification, structure, basic character and identification of primary, secondary and tertiary amines and their basic character.
Unit 14 : Polymers
General introduction and classi_cation of polymers, general methods of polymerization-addition and condensation, copolymerization; natural and synthetic rubber and vulcanization; monomers and uses - polythene, nylon, polyester and bakelite.
Unit 15 : Chemistry in Everyday Life
Mathematics Syllabus
Unit 1 : Sets, Relations and Functions
Sets and their representations, union, intersection and complements o f sets and their algebraic properties.
Unit 2 : Complex Numbers
Complex numbers in the form a+ib and their representation in a plane. Argand diagram. Algebra of complex numbers, modulus and argument ( or amplitude ) of a complex number, square root of a complex number. Cube roots of Unity, triangle inequality.
Unit 3 : Matrices and Determinants
Determinants and matrices of order two and three, properties of determinants, evaluation of determinants. Addition and multiplication of matrices, adjoint and inverse of matrix.
Unit 4 : Applications of Matrices and Determinants
Computing the rank of a matrix – test of consistency and solution of simultaneous linear equations using determinants and matrices.
Unit 5 : Quadratic Equations
Quadratic equations in real and complex number system and their solutions. Relation between roots and coefficients, nature of roots, formation of quadratic equations with given roots; symmetric functions of roots, equations reducible to quadratic equations.
Unit 6 : Permutations and Combinations
Fundamental principle of counting: permutation as an arrangement and combination as selection, meaning of P(n,r) and C(n,r). Simple applications.
Unit 7 : Mathematical Induction and its Applications
Stating and interpreting the principle of mathematical induction. Using it to prove formula and facts.
Unit 9 : Sequences and Series
Arithmetic, geometric and harmonic progressions. Insertion of arithmetic, geometric and harmonic means between two given numbers. Relation between A.M., G.M. and H.M. arithmetic, geometric series, exponential and logarithmic series.
Unit 10 : Differential Calculus
Polynomials, rational, trigonometric, logarithmic and exponential functions. Inverse functions. Graphs of simple functions. Limits, continuity, differentiation of the sum, difference, product and quotient of two functions, differentiation of trigonometric, inverse trigonometric, logarithmic, exponential, composite and implicit functions, derivatives of order up to two.
Unit 11 : Applications of Differential Calculus
Rate of change of quantities, monotonic- increasing and decreasing functions, maxima and minima of functions of one variable, tangents and normals, Rolle's theorem and Lagrange's theorem mean value theorem.
Unit 12 : Integral Calculus
Fundamental integrals involving algebraic, trigonometric, exponential and logarithmic functions. Integration by substitution, by parts and by partial fractions. Integration using trigonometric identities. Integral as limit of a sum. Properties of definite integrals. Evaluation of definite integrals; Determining areas of the regions bounded by simple curves.
Unit 13 : Differential Equations
Ordinary differential equations, their order and degree. Formation of differential equations. Solution of differential equations by the method of separation of variables. Solution of homogeneous and linear differential equations and those of the type d2y/dx2= f(x).
Unit 14 : Straight Lines in Two Dimensions
Equation of family of lines passing through the point of intersection of two lines, homogeneous equation of second degree in x and y, angle between pair of lines through the origin, combined equation of the bisectors of the angles between a pair of lines, condition for the general second degree equation to represent a pair of lines, point of intersection and angle between two lines.
Unit 15 : Circles in Two Dimensions
Standard form of equation of a circle, general form of the equation of a circle, its radius and centre, equation of a circle in the parametric form, length of the tangent, equation of the tangent, equation of a family of circles through the intersection of two circles, condition for two intersecting circles to be orthogonal.
Unit 16 : Conic Sections in Two Dimensions
Sections of cones, equations of conic sections ( parabola, ellipse and hyperbola ) in standard form, condition for y = mx+ c to be a tangent and point(s) of tangency.
Unit 17 : Vector Algebra
Vectors and scalars, addition of vectors, components of a vector in two dimensions and three dimensional space, scalar and vector products, scalar and vector triple product. Application of vectors to plane geometry.
Unit 18 : Measures of Central Tendency and Dispersion
Calculation of mean, median and mode of grouped and ungrouped data. Calculation of standard deviation, variance and mean deviation for grouped and ungrouped data.
Unit 19 : Probability
Probability of an event, addition and multiplication theorems of probability and their applications; Conditional probability; Baye's theorem, probability distribution of a random variate; binomial and Poisson distributions and their properties. | 677.169 | 1 |
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