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Mathematics
Mathematics has many aspects. It is the language and tool of the sciences, a key part of cultural development since ancient times, and a model of abstract reasoning. The course offerings and the major in Mathematics reflect these multiple facets. The Mathematics program provides a broad education in various areas of mathematics and is flexible enough to accommodate many interests. Mathematics majors have numerous options after graduation, including graduate study in mathematics or in various fields of application, government or corporate laboratory work, consulting, finance and banking, and teaching. Students considering a major in Mathematics are encouraged to consult with the director of undergraduate studies (DUS) during their freshman year.
A variety of resources is available to students who want additional help in any mathematics course:
Each course instructor holds office hours for students multiple times a week.
Courses
A full description of all mathematics courses can be found in Yale College Program of Study. The department offers several courses that satisfy the QR distributional requirement and do not assume knowledge of mathematics beyond the basic high-school level. They include:
MATH 107, Mathematics in the Real World
MATH 108, Estimation and Error
The courses below include the ones most frequently taken by freshmen. Enrollment requires appropriate placement based on the online placement exam.
MATH 112, Calculus of Functions of One Variable I, is the introductory course for students with a strong foundation in high school mathematics. It focuses primarily on differentiation and assumes no previous exposure to calculus.
MATH 115, Calculus of Functions of One Variable II, builds on MATH 112 and focuses primarily on integration and infinite series. Knowledge of topics covered in MATH 112 is assumed.
MATH 120, Calculus of Functions of Several Variables, combines the ideas and techniques of one-variable calculus with vector geometry and algebra to deal with geometrical situations in planes and in three-dimensional space. Knowledge of topics covered in MATH 112 and MATH 115 is assumed.
The courses below serve as options after MATH 112 for calculus students who have a specific focus. Students interested in the biological sciences should consider MATH 116, Mathematical Models in the Biosciences I: Calculus Techniques; those interested in economics should consider MATH 118, Introduction to Functions of Several Variables.
MATH 116 is designed for bioscience and premedical students. It combines differential equations with geometrical modeling and applications in biology. Knowledge of topics covered in MATH 112 is assumed.
MATH 118 is designed for students interested in economics and social sciences who do not intend to take additional mathematics courses. It covers basic ideas in linear algebra as well as differential calculus of several variables. Knowledge of topics covered in MATH 112 is assumed.
The higher-level courses below assume at least a year of calculus and are available to unusually well prepared freshmen.
MATH 222, Linear Algebra with Applications, and MATH 225, Linear Algebra and Matrix Theory, deal with linear algebra, the common language for a wide variety of applications involving many variables. MATH 222 emphasizes computations and applications of linear algebra, while MATH 225 focuses on geometric and conceptual issues and the logical structure of the subject. Students normally take linear algebra after completing MATH 120, Calculus of Functions of Several Variables; however, students who are sufficiently prepared and motivated may take MATH 222 or MATH 225 concurrently with MATH 120.
MATH 230, Vector Calculus and Linear Algebra I, and MATH 231, Vector Calculus and Linear Algebra II, is a demanding, two-term course sequence on calculus of many variables, designed for students with a firm grasp of one-variable calculus and a strong interest in mathematics. It emphasizes conceptual and logical structure and pays considerable attention to proofs and challenging problems. This course covers with greater rigor the material studied in MATH 120, MATH 225, and MATH 250, Vector Analysis, but assumes no prior knowledge of those topics.
MATH 250, Vector Analysis, is a rigorous introduction to calculus of many variables, based on linear algebra. It focuses on proofs of calculus results and their generalizations to higher dimensions; knowledge of topics covered in MATH 120, and MATH 222 or MATH 225 is assumed. The sequence of MATH 120, MATH 222 or MATH 225, and MATH 250 serves as an alternative to MATH 230 and MATH 231 for the requirements of the Mathematics major.
Additional higher-level courses may be open to exceptionally well prepared freshmen. Interested students should seek advice from the director of undergraduate studies (DUS) or another knowledgeable faculty member. | 677.169 | 1 |
Number Theory
Customize your search:
The PACE Mathematical Foundations MOOC is an online program designed to enhance your mathematics skills in the areas of Number Theory, Algebra, Geometry, Probability and Statistics. This MOOC is designed to help develop the skills needed to be successful in college-level mathematics | 677.169 | 1 |
Underground Mathematics
Welcome to Underground Mathematics! This site brings you free resources that will help students explore the connections that underpin different math concepts.
This site is designed to help students learn the math concepts they need to pass their A levels. An A level is short for Advanced level and is taken by students primarily in the UK around age 18. I think this site would have been very helpful to me when I took math in my undergraduate studies.
Before we dive into navigation, I want to rave about the infographic on the main page. It is color coded by math subject (Number, Geometry, Algebra, Functions, and Calculus) and shows how they are all connected.
If you scroll down the main page, you can learn more about the site including how to use it. Now back to the map, it also serves as the menu to help you navigate the site. You can click any of the hyperlinked concepts (those without a hammer icon or circle icon) and be whisked away to a full range of content for that subject. You'll notice the sections with hammer icons are also linked, but they might be under construction or not have a full range of content yet.
At the top of the section, you'll find a list of Key Questions and a list of nearby stations (related concepts). As you scroll down you'll find introductory content, developing content, and then review questions.
One thing I'd like to point out is the navigation strip along the top of the page. There you'll find a search field that you can use to locate specific concepts that you might be working on. You'll also find the user section where you can create a free account which allows you to comment on resources and receive the newsletter. And the More button that has a variety of links, but most importantly a link to the How to Guide.
I think this is a really awesome site to help review and connect math concepts! Go check it out for yourself today!
This entry was posted
on Thursday, August 11th, 2016 at 10:29 AM and is filed under Cool Sites.
You can follow any responses to this entry through the RSS 2.0 feed.
You can leave a response, or trackback from your own site.
87770 | 677.169 | 1 |
Schaum's Outline of Mathematical Methods for Business and Economics reviews the mathematical tools, topics, and techniques essential for success in business and economics today. The theory and solved problem format of each chapter provides concise explanations illustrated by examples, plus numerous problems with fully worked-out solutions. And you don't have to know advanced math beyond what you learned high school. The pedagogy enables you to progress at your own pace and adapt the book to your own needs | 677.169 | 1 |
books.google.com - This... Derivatives Modeling
Financial Derivatives Modeling
This reader has a solid mathematical background and is a graduate/final-year undergraduate student specializing in Mathematical Finance, or works at a financial institution such as an investment bank or a hedge fund. | 677.169 | 1 |
Summary and Info
Students must prove all of the theorems in this undergraduate-level text, which features extensive outlines to assist in study and comprehension. Thorough and well-written, the treatment provides sufficient material for a one-year undergraduate course. The logical presentation anticipates students' questions, and complete definitions and expositions of topics relate new concepts to previously discussed subjects.Most of the material focuses on point-set topology with the exception of the last chapter. Topics include sets and functions, infinite sets and transfinite numbers, topological spaces and basic concepts, product spaces, connectivity, and compactness. Additional subjects include separation axioms, complete spaces, and homotopy and the fundamental group. Numerous hints and figures illuminate the text. | 677.169 | 1 |
This is a digital textbook for a first course (sequence of courses) in Abstract Algebra covering the essentials of groups, rings and fields. The book is not an electronic version of a traditional print textbook but rather makes use of the digital environment to enhance student learning. One way this is achieved is by spiraling through the material, periodically returning to previous concepts to reinforce students' understanding.
Matrices, Vectors, and 3D Math:
A Game Programming Approach with MATLAB®
Scott Stevens
Written for undergraduate students, Matrices, Vectors, and 3D Math: A Game Programming Approach with MATLAB provides a resource to learn standard topics in linear algebra and vector calculus in a single course within the context of game programming applications and projects. Topics presented in this text are constructed to enable students to succeed after completing a single college-level calculus course.
This is an e-textbook for a first course in linear algebra. The topics covered include: Linear Systems, The Vector Space R^n, Matrix Algebra, Determinants, Abstract Vector Spaces, Linear Transformations, Eigenvalues and Eigenvectors, and Orthogonality in R^n. It introduces linear transformations in R^n quite early and uses them to motivate the addition and multiplication of matrices.
There is no shortage of books on Commutative Algebra, but the present book is different. Most books are monographs, with extensive coverage. There is one notable exception: Atiyah and Macdonald's 1969 classic. It is a clear, concise, and efficient textbook, aimed at beginners, with a good selection of topics. So it has remained popular. However, its age and flaws do show. So there is need for an updated and improved version, which the present book aims to be. | 677.169 | 1 |
Task #1
Determining Income, Savings & Investment DO THE MATH!
Using the following steps, calculate your disposable income.
1. Draw a gross annual salary from the salary basket and write it on the line below.
_52,500_
2. Divide your gross annual salary by
Showing 1 to 3 of 4
I would recommend this course because Calculus is easy if you have good algebra skills and having a teacher as thorough as Mrs. Holmes makes it even easier.
Course highlights:
Although I have only taken the first semester of this course, I learned a lot about derivatives and limits that I will probably never forget. This class was really fun in the sense that when I walked in at 8:00 AM every morning, it didn't feel like a burden. We didn't get a lot of homework but we were given enough to help us understand the topic.
Hours per week:
3-5 hours
Advice for students:
To succeed, do your homework. You cannot go into this class expecting an A or a B without doing homework, it counts! Also, make sure you study for the tests, they are worth 85% of your grade. | 677.169 | 1 |
Product Description:
This book offers a very interesting panorama of the development of mathematics from the ancient Babylonians and Greeks to the present. It is written in a lucid style with very readable mathematical content. Understanding the material requires some broad mathematical education, but not a lot of specialized knowledge. One of the strongest sections deals with the accomplishments of the Greeks. The author clearly explains the problems tackled in ancient Greece, places them in context, outlines the accomplishments (some with concise proofs), and compares these with our present understanding of the subject. He also places the mathematical achievements of ancient Greece in the context of early Ionian Philosophy, Platonism, Aristotelism, or in the mindset of the Alexandrians. The chapters on the seventeenth and eighteenth centuries are presented clearly with emphasis on the great figures of these two centuries. Mathematics of the nineteenth and twentieth centuries are presented more thematically than chronologically. Analysis, in particular functional analysis, receives a very good overview. An appendix contains a transcript of the talk by Laurent Schwartz on the historical roots and basic
REVIEWS for History of | 677.169 | 1 |
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Introduction to School Algebra [DRAFT
]
H. Wu
September 4, 2005.
Department of Mathematics, #3840
University of California
Berkeley, CA 94720-3840
wu@math.berkeley.edu
c _Hung-Hsi Wu, 2005
1
Contents
1 Symbolic Expressions 5
2 Transcription of Verbal Information into Symbolic Language 22
3 Linear Equations in One Variable 28
4 Linear Equations in Two Variables and Their Graphs 34
5 Some Word Problems 55
6 Simultaneous Linear Equations 65
7 Linear Inequalities and Their Graphs 82
8 Exponents and Absolute Value 110
9 Functions and Their Graphs 128
10 Proportional Reasoning (outlined) 146
11 Quadratic Functions and Their Graphs 149
12 The Quadratic Formula and Applications (outlined) 162
2
Preface
These are the lecture notes of a three-week summer institute on beginning algebra
that I gave for middle school teachers. There is nothing fancy about their content:
with minor exceptions, it covers only the standard topics of Algebra I. These notes may
therefore be called Algebra I from a somewhat advanced point of view. If there is
any merit to be claimed for them, it may be the sequencing of the topics and the logical
coherence of the presentation. The exposition is formally self-contained, in the sense
that the reader is not assumed to know anything about algebra. In practice, though, the
reader is likely to have taught, or will be teaching Algebra I so that he or she is already
familiar with the more routine aspect of the subject. For this reason, while I have made
an effort to comment on the standard computations whenever it is appropriate, I have
on the whole slighted the drills that unavoidably accompany any such presentation.
Many of our algebra teachers are experiencing real difficulty in carrying out their
duties, mainly because they have been told to emphasize certain ideas that they (like
most mathematicians) cannot relate to, such as that of a quantity that "changes" or
"varies" or that the equal sign is a "method that expresses equivalence". At the same
time, they are denied the explanations of key facts which form the backbone of intro-
ductory algebra, such as why the graph of a linear equation is a straight line, or why the
solution of a system of linear equations is the point of intersection of the lines defined by
the equations. The fact that many teachers do not even recognize these as key facts in
algebra speaks volume about the present state of pre-service professional development
in mathematics. The main impetus behind the writing of these notes is to propose a
remedy. They give an exposition of algebra ab initio, assuming only a knowledge of the
rational number system. It is an exposition that is consistent with the basic require-
ments of any exposition in mathematics. The reader will therefore find in these notes
the unfolding of ideas based not on abstruse philosophical discussions but on clear and
precise definitions and logical reasoning.
An integral part of the learning of algebra is learning how to use symbols precisely
and fluently. I believe if there is any meaning at all to the phrase "algebraic thinking"
in school mathematics, this would be it. In this regard, there is a need to single out
the first two sections of these notes for some special comments. A principal object of
study in introductory algebra is polynomials or, in techincal language, elements of the
polynomial ring R[x]. Every exposition of school algebra must come to grips with the
problem of how to properly introduce this concept to beginners. The mathematical
decision I made (which is of course not mentioned in the notes) is to exploit the theorem
3
that R[x] is isomorphic to the ring of real-valued polynomial functions, so that in the
context of introductory algebra, the x in a polynomial a
n
x
n
+ + a
1
x + a
0
may be
simply taken to be a number. The purpose of Section 1 is to demonstrate how one
can do algebra by taking x to be just a number, and algebra then becomes generalized
arithmetic, literally. Formal algebra in the sense of R[x] can be left to a later date, e.g.,
in Algebra II.
Section 1 is therefore entirely elementary: it is nothing more than a direct extension
of arithmetic, and the exposition intentionally emphasizes its similarity to arithmetic.
There is a danger, however, that precisely because of its elementary character, the main
thrust of this section would be lost on some readers, namely, the fact that what is in
Section 1 is genuine algebra, and that the simplicity of its content demonstrates how
algebra can be taught without any fanfare. At a time when Algebra for All is the clarion
call of the day in mathematics education, this section may be considered a contribution
to this national goal.
Section 2 addresses one of the main obstacles in the teaching of algebra: students'
seeming inability to solve word problems. It seems to me sensible to separate this
difficulty into two stages: first teach how to translate verbal information into symbolic
expressions, and then teach the necessary symbolic manipulations to extract the solution
from the symbolic expressions. The need of such a separation does not seem to be
recognized at present in mathematics education, and the main purpose of Section 2 is
to promote this recognition.
Elsewhere, I have called attention to the need in school mathematics to emphasize
precise definitions and to respect the WYSIWYG characteristic of mathematics.
1
My
personal experience is that this need is real and urgent. The exposition in these notes
fully reflects this sense of urgency.
1
What is so difficult about the preparation of mathematics teachers?
4
1 Symbolic Expressions
Why symbols? When we try to assert that something is valid for a collection of numbers
(e.g., for all positive integers, or for all rational numbers) instead of just for a few specific
numbers, we have to resort to the use of symbols to express this assertion correctly and
succinctly. For example, we know that 2 3 = 3 2, 3 4 = 4 3,
6
17
4
9
=
4
9
6
17
,
(−
8
3
) 82 = 82 (−
8
3
), etc. In fact, what we wish to assert is the commutative law
of multiplication, namely,
for any two numbers, if we multiply them one way, and switching the order
and multiplying them again, then we would get the same number.
The above collection of examples, 2 3 = 3 2, etc., give an indication that this law
is true, but fail to convey, completely and unambiguously, what this law says because
a finite number of examples (even with "etc." thrown in) can only suggest but do not
convey clearly the fact that the law is valid for all numbers. It remains therefore to use
symbols to state the commutative law of multiplication succinctly as:
ab = ba for all numbers a and b
Compared with the preceding indented verbal statement, the brevity resulting from the
use of symbols should be obvious.
It is time to recall that in arithmetic, there are many occasions when the use of
symbols achieves both clarity and brevity of the mathematical message. In addition
to the commutative law of multiplication, the statements of the same law for addition,
the associative laws for addition and multiplication, and also the distributive law
require a similar use of symbols. In addition, the formulas for the addition, subtraction,
multiplication, and division of fractions likewise cannot be stated without the use of
symbols. We repeat these formulas here to emphasize this point: let
k
m
n
=
kn
m
5
We emphasize that in each of these formulas, we don't need to know what exact value
each of k, , m, n has, but so long as they are whole numbers, they will have to satisfy
k
±
m
n
=
kn±m
n
, etc. This is no different from what takes place in a detective novel:
A murder has been committed and the detective is charged with finding the murderer.
Usually the detective has no idea who the murderer is, but knowing that the murderer
is a person who does what ordinary human beings do (such as having to eat and sleep
and needing a car or a plane to go from city to city, and being susceptible to the usual
temptations) is enough to track down the murderer in most cases. In the same way,
without knowing what the integers k, , m, n are, these formulas are established by
making use of only the properties that can be logically deduced from the definition
of a rational number and the associative, commutative and distributive laws. Once
established in such generality, these formulas can be used for any specific values of
k
5
23
=
11 5
−7 23
= −
55
161
11
−7
5
23
=
11 23
5 (−7)
= −
253
35
As a natural extension of these ideas, we now give some well-known algebraic identi-
ties. The term identity is used in mathematics to indicate, informally, that an equality is
valid for "most" numbers of interest, and what "most" means will be clearly indicated
in each case and, in addition, is usually clear from context. Thus "most" could mean
all whole numbers, all the primes, all positive real numbers, all numbers except inte-
ger multiples of π/2, or all numbers. Identity is definitely not a technical mathematical
concept that requires a 100% precise definition. However since the meaning of this term
seems at present to be endlessly (and one may say, unnecessarily) debated, we will now
try to clarify its meaning as best we can. By a number expression or more simply
an expression in a given collection of numbers x, y, . . . w, we mean a number obtained
from these x, y, . . . w by performing a specific combination of arithmetic operations (i.e.,
addition, subtraction, multiplication, and division). For example, if x, y, z are numbers,
6
then
xy
xyz + 2
+ x
3
(16z −y
2
) −z
21
,
x −y
3
8 +
5
(yz)
2
, x
4
+ y
4
+ z
4
−xyz
are examples of expressions in the numbers x, y, z (we have to assume xyz ,= −2 in the
first expression and y ,= 0 and z ,= 0 in the second expression to avoid dividing by 0).
Later on in Section 8, we shall expand the meaning of "expression" after we have defined
"n-th root". You may have noticed that the above expressions would be ambiguous unless
a notational convention concerning the arithmetic operations among the symbols
is understood. With the help of parentheses, the correct order in carrying out the
arithmetic operations in, for example,
xy
xyz + 2
+ x
3
(16z −y
2
) −z
21
will always be understood to be
¦xy (xyz + 2)
−1
¦ +¦(x
3
)[(16z) −(y
2
)]¦ −¦z
21
¦
The ungainly sight of the last expression should be reason enough for the adoption of this
notational convention. Postponing the exact description of this notational convention to
a later passage so as not to disrupt the flow of the exposition, we may roughly describe
this convention as follows: do the multiplication indicated by the exponents first, then
the multiplications (including the division in
xy
xyz+2
, as this is nothing other than the
multiplication xy (xyz + 2)
−1
), and finally the additions.
2
Now we give "an approximate definition" of an algebraic identity, or more simply
an identity, as a statement that two given number expressions are equal for a clearly
defined collection of numbers (such as all whole numbers, all positive numbers, all num-
bers, etc.) allowing for a small set of exceptions. Recall again that an identity is not a
precise concept within mathematics, but is no more than a terminology used loosely for
convenience. In specific situations, there will be plenty of opportunities to discern what
the "small set of exceptions" means. The assertion that ab = ba is true for all numbers
a and b is an example of an identity, and so is
k
±
m
n
=
kn±m
n
for all integers k, , m, n
provided ,= 0 and n ,= 0. Right here, we see that the latter identity is one which make
allowance for the exceptions of = 0 and n = 0. More is true. We have just stated
this equality
k
±
m
n
=
kn±m
n
for integers k, , m, n, but we know from considerations of
2
Recall in this connection that subtraction is nothing but addition in disguise: a −b = a +(−b), by
definition, for any two rational numbers a, b.
7
complex fractions
3
that this equality remains true even if k, , m, n are arbitrary rational
numbers. Therefore, in this form, this identity is valid for all rational numbers k, , m,
n provided ,= 0 and n ,= 0. (The fact that the identity remains valid for all irrational
numbers as well requires more advanced considerations, though this is a fact routinely
used but rarely explained in school mathematics. But even here, the exceptions to this
general identity are still ,= 0 and n ,= 0.)
In case it helps to further illustrate the cavalier manner in which the terminology
of identity is used, we give two advanced examples without attempting to define the
relevant concepts. The equality log xy = log x + log y is an identity for all positive
numbers x and y. The equality 1 + cot
2
x = csc
2
x is an identity for all numbers x
except for all integral multiples of π.
We want to get more interesting identities. Consider the computation of the square,
104
2
, for example. One can compute it directly, of course. But one can also proceed by
appealing to the distributive law, as follows:
104
2
= (100 + 4)
2
= (100 + 4)(100 + 4)
= ¦(100 + 4) 100¦ +¦(100 + 4) 4¦ (dist. law)
= ¦100
2
+ (4 100)¦ +¦(100 4) + 4
2
¦ (dist. law again)
= 100
2
+ 2 (100 4) + 4
2
At this point, it should be possible to mentally finish the computation as 10000 +800 +
16 = 10816. More than a trick, this idea of computing the square of a sum using the
distributive law turns out to be almost omnipresent in algebraic manipulations of all
kinds. It is a good idea to formalize it once and for all. We therefore have, in an
identical fashion:
(a + b)
2
= a
2
+ 2ab + b
2
for all numbers a and b.
This is our first identity of note.
A similar consideration, but worth pointing out in any case, is the computation of
the square of 497, for example. We recognize it as (500 −3)
2
, so that
497
2
= (500 −3)
2
= (500 −3)(500 −3)
= ¦(500 −3) 500 −(500 −3) 3¦ (dist. law)
= ¦500
2
−(3 500) −¦(500 3) −3
2
¦ (dist. law again)
= 500
2
−2 (500 3) + 3
2
3
Recall that these are quotients
A
B
where A and B are rational numbers.
8
(Note that the preceding computation furnishes a good review of the basic arithmetic
of rational numbers: the distributive law for a difference, a(b − c) = ab = ac for all
numbers a, b, c, and the removal of parentheses by −(a − b) = −a + b for all a, b.)
Again, we stop the calculation at this point because it can now be finished in one's
head: 250000 −3000 + 9 = 247009. The same computation also leads to:
(a −b)
2
= a
2
−2ab + b
2
for all numbers a and b.
It is a good illustration of the use of symbols, and the attendant generality the
symbolic method brings, that the identity for (a −b)
2
can be obtained directly from the
identity for (a + b)
2
. Indeed, since the identity (a + b)
2
= a
2
+ 2ab + b
2
is valid for all
numbers, we may replace b by an arbitrary number −c to get
(a + (−c))
2
= a
2
+ 2a(−c) + (−c)
2
= a
2
−2ac + c
2
Since a + (−c) = a − c by definition, we get (a − c)
2
= a
2
− 2ac + c
2
, and since c is
arbitrary anyway, we may replace c by b to obtain (a − b)
2
= a
2
− 2ab + b
2
for any
number b. Thus we retrieve the second identity by way of the first.
A third common identity can be introduce by a computation of another kind: 409
391 =? We recognize that 409 391 = (400 + 9)(400 −9), so that
409 391 = ¦(400 + 9) 400¦ −¦(400 + 9) 9¦ (the dist. law)
= 400
2
+ (9 400) −(400 9) −9
2
(the dist. law again)
= 400
2
−9
2
It follows that 409 391 = 160000 − 81 = 159919. The same reasoning carries over to
any two numbers a and b, so that
(a + b)(a −b) = (a + b)a −(a + b)b
= a
2
+ ba −ab −b
2
= a
2
−b
2
When the symbolic computation is given in such detail, we see that in the second line,
the commutative law for multiplication was used. We have obtained our third identity:
(a + b)(a −b) = a
2
−b
2
for all numbers a and b.
This particular product (a+b)(a−b) for any two numbers a and b, turns out to be very
common, and the usefulness of this identity far transcends the preceding computation
9
of 409 391 and others like it. For example, if we read this identity backwards, i.e.,
a
2
−b
2
= (a + b)(a −b) for all numbers a and b
then we obtain what is known as a factorization or factoring of a
2
−b
2
, which merely
means expressing a
2
−b
2
as a product, in the same sense that 24 = 38 is a factorization
of 24. Knowing such a factorization for any two numbers a and b can be very useful.
Thus, if a + b ,= 0, we can simplify the division
a
2
−b
2
a+b
to
a
2
−b
2
a + b
= a −b
because a
2
−b
2
= (a+b)(a−b), so that we can cancel the number a+b in the numerator
and the denominator. We explicitly point out that, insofar as a and b can be rational
numbers (say,
17
5
and
2
7
), we are using the cancellation law for complex fractions here.
4
One cannot over-emphasize the importance of the role played by complex fractions in
school mathematics.
There is another identity that is equally elementary and has very far reaching appli-
cations in mathematics. This time, we start with a symbolic calculation: if a, b are any
two numbers, then
(a
3
+ a
2
b + ab
2
+ b
3
)(a −b) = (a
3
+ a
2
b + ab
2
+ b
3
)a −(a
3
+ a
2
b + ab
2
+ b
3
)b
= (= a
4
−b
4
Notice two features in the preceding calculation. First, if we call any of the products
separated by two consecutive +'s,
5
e.g., a
3
, a
2
b, ab
2
, . . . a
4
, b
4
, a term of the number
expression, then the way to remember the expressions a
3
+ a
2
b + ab
2
+ b
3
, a
4
+ a
3
b +
a
2
b
2
+ab
3
, and a
3
b +a
2
b
2
+ab
3
+b
4
is to observe that the power of a decreases by 1 and
the power of b increases by 1 as we go through the terms from left to right. Second, the
cancellation in the second line
(is due to the matching of each term in the first pair of parentheses with a term in the
second pair of parentheses, except for the first term a
4
and the last term b
4
, and this is
4
See, for example, Section 9 of Chapter 2: Fractions (Draft),
5
Recall that since a subtraction is an addition in disguise, this reference to + includes automatically
all the −'s.
10
why the only survivors at the end are the two terms a
4
−b
4
. The same pattern repeats
itself if we multiply
(a
4
+ a
3
b + a
2
b
2
+ ab
3
+ b
4
)(a −b),
or
(a
5
+ a
4
b + a
3
b
2
+ a
2
b
3
+ ab
4
+ b
5
)(a −b),
or for that matter, for any positive integer n,
(a
n
+ a
n−1
b + a
n−2
b
2
+ a
n−3
b
3
+ + ab
n−1
+ b
n
)(a −b).
The results would be a
5
−b
5
, a
6
−b
6
, and a
n+1
−b
n+1
, respectively. Let us write these
down. For any two numbers a and b, and for any positive integer n, we have:
(a −b)(a
4
+ a
3
b + a
2
b
2
+ ab
3
+ b
4
) = a
5
−b
5
(a −b)(a
5
+ a
4
b + a
3
b
2
+ a
2
b
3
+ ab
4
+ b
5
) = a
6
−b
6
( = a
n+1
−b
n+1
Let us rewrite the general case a
n+1
−b
n+1
given by the last line as:
a
n+1
−b
n+1
= (
for any two numbers a and b, and any positive integer n
This gives a very useful factorization of the difference of two numbers raised to the same
power. If n = 1, it gives back a
2
− b
2
= (a − b)(a + b). Therefore the general identity
for an arbitrary positive integer n is nothing but a generalization of the earlier simple
identity a
2
− b
2
= (a − b)(a + b). The case n = 2 of this identity also comes up often
and we should state it separately:
a
3
−b
3
= (a −b)(a
2
+ ab + b
2
) for any numbers a, b.
For us, the special case of b = 1 is extremely interesting. It reads:
(a
n+1
−1) = (a −1)(a
n
+ a
n−1
+ a
n−2
+ + a
2
+ a + 1) for any number a.
This factorization says that if you raise a whole number a which is bigger than 2 (e.g.,
a = 215, 608) to any power (e.g., n = 87) and diminish it by 1, then the resulting number
a
n+1
−1 is the product of two whole numbers both of which are bigger than 1 (this is
where we need a > 2 as otherwise a−1 would be either 0 or 1). Therefore a
n+1
−1 is
11
never a prime number. We will discuss the case a = 2 presently. But with a = 215, 608
and n = 87, we see that a −1 = 215, 607, so that
215, 608
87
−1 = (215, 607)(215, 608
86
+ + 215608 + 1)
This shows that 215, 608
87
−1 is not a prime, because it is the product of 215, 607 with
another number. Thus, even if we have no idea of what kind of a number 215, 608
87
−1
is except that it has 426 digits(!), we know at least that it is not a prime. We can
illustrate this factorization another way: if we know that 1, 453, 933, 567 = 68
5
−1, then
the number 1, 453, 933, 567 is not a prime because it is equal to 68
5
−1 = (68 −1)(68
4
+
68
3
+ 68
2
+ 68 + 1) and is therfore divisible by 67.
If a = 2, then the preceding factorization only says
2
n+1
−1 = 2
n
+ 2
n−1
+ + 2
2
+ 2 + 1
and there is no factorization. In fact, for some n, 2
n+1
− 1 is a prime. For example,
2
2
− 1 = 3 (n = 1) and 2
3
− 1 = 7 (n = 2), and both 3 and 7 are primes, although
2
4
− 1 = 15 and 15 is certainly not a prime. Those numbers 2
p
− 1 which are primes
for a positive integers p are called Mersenne primes. We do not know if there are an
infinite number of Mersenne primes, but the largest Mersenne prime that is known at
present has 7, 816, 230 digits, corresponding to p = 25, 964, 951 (discovered on February
18, 2005).
But we are far from exhausting the charm of this identity; we can read it backwards
to obtain
(1 + a + a
2
+ + a
n−1
+ a
n
) =
a
n+1
−1
a −1
for any number a ,= 1.
This now becomes a summation formula for the first n + 1 powers of a number a,
starting with a
0
= 1 up to a
n
. For example, if a = 5 and n = 11, then
1 + 5 + 5
2
+ 5
3
+ + 5
10
+ 5
11
=
5
12
−1
5 −1
=
5
12
−1
4
.
Since 5
12
−1 = 244, 140, 624, we have
1 + 5 + 5
2
+ 5
3
+ + 5
10
+ 5
11
= 61, 035, 156.
If a = −3 and n = 15, then
1 −3 + 3
2
−3
3
+ 3
4
− + 3
14
−3
15
=
3
16
−1
−3 −1
= −
43, 046, 720
4
= −10, 761, 480.
12
And finally, if a =
3
4
and n = 10, we have
1 +
3
4
+ (
3
4
)
2
+ (
3
4
)
3
+ + (
3
4
)
10
= ¦(
3
4
)
11
−1¦/(
3
4
−1)
which is equal to
16, 068, 628
4, 194, 304
.
It is roughly 3.8.
A sum of the form 1 +a +a
2
+ +a
n−1
+a
n
is called a finite geometric series
of n terms in a. We have just learned how to sum a finite geometric series. Geometric
series appear everywhere in both science and mathematics.
We next introduce polynomials. Underlying the whole discussion of polynomials
will be a simple observation based on the distributive law, and we deal with this first.
Suppose we have a sum
(18 5
3
) + (5
3
23) + (69 5
3
)
One can compute this sum by multiplying out each product 185
3
, 5
3
23, and 695
3
,
(also called a term) and then add the resulting numbers to get
(18 5
3
) + (5
3
23) + (69 5
3
) = 2250 + 2875 + 8625 = 13750
Now if we reflect for a moment, we would realize that we wasted precious time doing three
multiplications before adding. If we apply the distributive law, then the computation
becomes easier:
(18 5
3
) + (5
3
23) + (69 5
3
) = (18 + 23 + 69) 5
3
= 110 125 = 13750
(Notice that we have made use of the commutative law of multiplication to change 5
3
23
to 235
3
in the process.) You may think that with the advent of high speed computers,
it doesn't make any difference whether we get the answer by multiplying three times
and then add once, or (as in the second case) add three times and multiply once. This
is true, but the difference in conceptual clarity between
(18 5
3
) + (5
3
23) + (69 5
3
)
and
(18 + 23 + 69) 5
3
13
is enormous. This is because multiplication is a far more complicated concept than
addition: 234 + 677 merely means lumping 234 and 677 together, but 234 677
means adding 677 to itself 234 times. It is therefore conceptually simpler to add three
times and multiply once than to multiply three times and add once. Because conceptual
clarity is very important in learning and doing mathematics, whenever we see terms
involving the same numbers raised to a fixed power in the future (such as 5
3
in (18
5
3
) + (5
3
23) + (69 5
3
)), we shall always collect them together by the use of the
distributive law. For example, we shall always rewrite
(181 2
5
) + (67 2
5
) + (2
5
96) −(257 2
5
)
as
87 2
5
(= (181 + 67 + 96 −257) 2
5
),
Similarly, we write
[24 59
14
] −[(
3
5
)
8
89] + [59
14
73] + [59
14
66] + [25 (
3
5
)
8
] + [(
3
5
)
8
11]
as
[163 59
14
] −[53 (
3
5
)
8
],
where 163 = 24 + 73 + 66 and −53 = −89 + 25 + 11. Recall yet again that we talk
about both of the above expressions as a "sum" even in the presence of the terms
−(2572
5
) and −[(
3
5
)
8
89] because −(2572
5
) = +¦−(2572
5
)¦ and −[(
3
5
)
8
89] =
+¦−[(
3
5
)
8
89]¦.
In an entirely similar manner, suppose we are given a sum of multiples of nonnegative
integer powers of a fixed number x, where multiple here means simply multiplication by
any number and not necessarily by a whole number, and "nonnegative integers" refers
to whole numbers 0, 1, 2, . . . . Then we would automatically collect together the terms
involving the same power of x as before. For example, we would rewrite
1
2
x
3
+ 16 −8x
2
+
1
3
x
3
−x
5
−6x
2
+ 75x + 2x
3
as
−x
5
+
17
6
x
3
−14x
2
+ 75x + 16.
Observe that we have followed three conventions in writing the latter sum involving the
powers of a fixed number x: (i) the earlier convention that parentheses are suppressed
14
with the understanding that exponents be computed first, multiplications second, and
additions third, (ii) the power of x is placed last in each term (so that instead of −x
2
14,
we write −14x
2
), and (iii) the terms are written in decreasing powers of the number x
in question. (The term 16 is the term 16x
0
, and incidentally, this is where we need
the zeroth power of x.) The number in front of a power of x is called the coefficient
of that particular power of x, and a sum of multiples of nonegative integer powers of
x is called a polynomial in x. A multiple of a single nonnegative power of x, such
as 58x
12
is called a monomial. Thus, a monomial is a polynomial with only one
term. The highest power of x with a nonzero coefficient in a polynomial is called the
degree of the polynomial. The terminology about "nonzero coefficient" refers to the
fact that the preceding polynomial −x
5
+
17
6
x
3
−14x
2
+ 75x + 16 could be written as
0 x
37
− x
5
+
17
6
x
3
− 14x
2
+ 75x + 16, but the 37-th power of x clearly doesn't count.
This polynomial has degree 5, and not 37 (and not any whole number different from 5,
for that matter.) Moreover, −1 is the coefficient of x
5
, 0 is the coefficient of x
4
, and −14
is the coefficient of x
2
, because, strictly as a sum of the powers of x, this polynomial is
in reality
(−1)x
5
+ 0x
4
+
17
6
x
3
+ (−14)x
2
+ 75x + 16x
0
.
Similarly, 16 is the coefficient of x
0
.
As is well-known, a polynomial of degree 1 is called a linear polynomial, and one
of degree 2 is called a quadratic polynomial. Because a general quadratic polynomial
has only three terms, ax
2
+bx +c, it is sometimes called a trinomial in school math-
ematics. We will discuss quadratic polynomials in some detail in the last two sections.
A polynomial of degree 3 is called a cubic polynomial.
There is no reason why we must restrict ourselves to polynomials in one variable. If x,
y, z, etc., are numbers, then sums of multiples of the products of nonnegative powers of x,
y, z, etc., are called polynomials in x, y, z, etc. For example, 19x
3
y
21
−8y
9
z
5
−xyz+31
is such a polynomial.
[Here we should address the issue of order of operations, a topic as wrongly
over-emphasized in school mathematics as the insistence on having all frac-
tions in lowest terms.
6
Just learn it, and go on to more important topics,
such as those on which we will be spending a lot of time in the days ahead.
6
See "Order of operations" and other oddities in school mathematics,
15
Also need to bring out the use of in place of , the fact that we usually write
42x
2
instead of 42 x
2
unless we wish to achieve an extra degree of clarity,
the fact that we write 1x simply as x, and the omission of all terms of the
form 0x
m
. Also explain why, for example, −14x
2
is equal to +(−14)x
2
.]
You have seen polynomials before. The so-called expanded form of the five-digit
whole number 75018, for example, is
(7 10
4
) + (5 10
3
) + (0 10
2
) + (1 10
1
) + (8 10
0
)
which is a fourth-degree polynomial in the number 10. Of course the expanded form of
any k-digit whole number is a polynomial of degree (k − 1) in 10. On the other hand,
the so-called complete expanded form of a decimal such as 32.58,
(3 10
1
) + (2 10
0
) + (5 10
−1
) + (8 10
−2
),
is not a polynomial in 10, for the reason that it contains negative powers of 10.
It should be pointed out that, as a polynomial in 10, the expanded form of a whole
number is very special: each coefficient is a single digit whole number. Thus none of the
following polynomials in 10 is the expanded form of a whole number:
(35 10
2
) + (2 10
1
) + (7 10
0
)
(3 10
3
) −(1 10
2
) + (7 10
1
) + (4 10
0
)
(5 10
2
) + (
2
3
10
1
) + (7 10
0
)
The first is not the expanded form of a whole number because 35 is not a single digit,
the second because the coefficient of 10
2
is −1, which is not a whole number, and the
third because
2
3
is not a whole number. However if we choose to rewrite the first of these
three polynomials in 10 as
(3 10
3
) + (5 10
2
) + (2 10
1
) + (7 10
0
),
then it is the expanded form of 3527.
Because polynomials are just numbers, we can add, subtract, multiply, and divide
them as usual. With the exception of division, the other three arithmetic operations
produce another polynomial in a routine manner. (Division of polynomials does not
generally produce a polynomial and will be looked at separately.) Consider the product
16
of two first degree polynomials, for example. If a, b, c, d are the coefficients of two linear
polynomials (i.e., polynomials of degree 1) in x, then
(ax + b)(cx + d) = (ax + b)(cx) + (ax + b)d (dist. law)
= acx
2
+ bcx + adx + bd (dist. law)
= acx
2
+ (ad + bc)x + bd (dist. law)
Of course we had to collect terms of the same degree using the distributive law and
rearranging the terms so that they are in descending powers of x at the end. By the
definition of a polynomial, this has to be done automatically anyway. The main point
is to emphasize the role played by the distributive law and to showcase the fact that
multiplying polynomials is no different from the usual operations with numbers. If the
arithmetic of numbers (whole numbers and fractions) is taught correctly, such operations
with polynomials are just more of the same and would not be a problem. In particular,
the uncivilized mnemonic device called FOIL is to be studiously avoided.
We have mentioned more than once the need to read an equality backwards, and
now we have to repeat this message. What we obtained above,
(ax + b)(cx + d) = acx
2
+ (ad + bc)x + bd
is nothing but routine applications of the distributive law. However, when this equality
is read backwards, it becomes
acx
2
+ (ad + bc)x + bd = (ax + b)(cx + d)
In general, we say that for polynomials p(x), q(x), r(x) in x, p(x) = q(x)r(x) is a
factorization of p(x) if the degrees of both q(x) and r(x) are positive. (Thus
5
3
x
3
−
2x
2
+
2
3
= (
1
3
)(5x
3
− 6x
2
+ 2) is not a factorization of
5
3
x
3
− 2x
2
+
2
3
.) We have just
factored the trinomial acx
2
+ (ad + bc)x + bd (i.e., a polynomial with three terms)
as a product (ax + b)(cx + d). For example, we get
1
2
x
2
+
5
4
x −3 = (2x −3)(
1
4
x + 1)
by letting a = 2, b = −3, c =
1
4
, and d = 1, but of course with some practice the
factorization could be done directly. For example, since it is much easier to deal with
integers rather than rational numbers, we rewrite
1
2
x
2
+
5
4
x −3 =
1
4
(2x
2
+ 5x −12)
17
Then we recognize that
(2x
2
+ 5x −12) = (2x −3)(x + 4)
because the zero-degree term (i.e., 12) of 2x
2
+ 5x − 12 has to be the product of the
zero-degree terms of 2x −3 and x + 4, and the coefficient 2 of 2x
2
+ 5x −12 has to be
the product of the coefficients 2 and 1 of 2x −3 and x + 4, respectively. So a few trials
and erors would get it done.
At present, the teaching of factoring trinomials with integer coefficients figures promi-
nently, not to say obsessively, in an algebra course. For this reason, some perspective on
this subject is called for. One should keep in mind that all it is is learning to decompose
two whole number A and C into a product so that a given trinomial Ax
2
+ Bx + C
can be written as acx
2
+ (ad + bc)x + bd (which equals (ax + b)(cx + d)). There is
no denying that beginning students ought to acquire some facility with decomposing
numbers into products. It is also important that they can effortlessly rewrite a simple
trinomials such as x
2
+ 2x − 35 as (x + 7)(x − 5). But it sometimes happens that if
a little bit of something is good, a lot of it can actually be bad for you. This seems to
be the case here, when the teaching of a small skill gets blown up to be a major topic,
with the consequence that other topics that are more central and more substantial (such
as learning about the graphs of linear equations or solving rate problems correctly) get
slighted. The teaching of algebra should avoid this pitfall. Please also keep in mind the
fact that once the quadratic formula becomes available (see Section 12), there will be
a two-step algorithm to acomplish this factorization no matter what the coefficients of
the trinomial may be.
We give one more illustration of the multiplication of polynomials; each step except
the last makes use of the distributive law.
(5x
3
−
1
2
x)(x
2
+ 2x −4) = (5x
3
−
1
2
x)x
2
+ (5x
3
−
1
2
x)2x −(5x
3
−
1
2
x) 4
= (5x
5
−
1
2
x
3
) + (10x
4
−x
2
) −(20x
3
−2x)
= 5x
5
+ 10x
4
−
41
2
x
3
−x
2
+ 2x
By the way, reading this equality backwards gives a factorization that is (for a change)
not so trivial:
5x
5
+ 10x
4
−
41
2
x
3
−x
2
+ 2x = (5x
3
−
1
2
x)(x
2
+ 2x −4)
18
Note the fact that if p(x) and q(x) are polynomials of degree m and n, respectively,
then the degree of the product p(x)q(x) is (m + n). In other words, the degree of
a product is the sum of the degrees of the individual polynomials. For example, the
preceding calculation which multiplies a degree 3 polynomial with a degree 2 polynomial
yields a polynomial of degree 5 (= 3 + 2).
Is the sum of two n-th degree polynomials always an n-th degree polynomial?
A quotient (i.e., division) of two polynomials in a number x is called a rational
expression in x. Here is an example:
3x
5
+ 16x
4
−25x
2
−7
x
2
−1
We note that in the case of rational expressions, we need to exercise some care in not
allowing division by 0 to take place. For example, in the preceding rational expression, x
can be any number except ±1 because if x = ±1, then x
2
−1 = 0 and the denominator
would be 0.
In writing rational expressions, it is usually understood that only those values for which
the denominator is nonzero are considered.
In middle school, we are mainly interested in rational numbers only and, as a conse-
quence, all computations with numbers tacitly assume that the numbers involved are rational
numbers. With this mind, since x is a (rational) number, a rational expression is just
a complex fraction and can therefore be added, subtracted, multiplied, and divided
like any other complex fraction. For example, in case x =
1
2
in the foregoing rational
expression, we would be looking at the complex fraction
3(
1
32
) + 16(
1
16
) −25(
1
4
) −7
(
1
4
) −1
,
which is equal to 16
5
24
. In general, no matter what x is, we can compute with rational
expressions in x in the usual way:
5x
3
+ 1
x
8
+ x −2
+
2x
7
x
3
+ 4
=
(5x
3
+ 1)(x
3
+ 4) + (2x
7
)(x
8
+ x −2)
(x
8
+ x −2)(x
3
+ 4)
and
x
2
+ 1
x
2
+ 4x −7
6
3x
4
−5
=
(x
2
+ 1)(6)
(x
2
+ 4x −7)(3x
4
−5)
and
19
2x+1
x
2
−3
4x
3
−x+11
2x
=
(2x + 1)(2x)
(x
2
−3)(x
3
−x + 11)
These are the same as any computation with complex fractions. It is important to realize
that these computations ar exactly the same as those in complex fractions and not just
"analogous to" them. There is so much in algebra that is just a revisit of arithmetic.
Because the cancellation law is valid for complex fractions (i.e.,
AB
AC
=
B
C
for all
rational numbers A, B, C, with A ,= 0, C ,= 0),
7
some rational expressions can be
simplified. Sometimes the cancellation presents itself, as in
(5x
4
−x
3
+ 2)(2x −15)
(14x
2
+ 3x −28)(5x
4
−x
3
+ 2)
Here, the number (5x
4
−x
3
+2) in both the numerator and denominator can be cancelled,
resulting in
(5x
4
−x
3
+ 2)(2x −15)
(14x
2
+ 3x −28)(5x
4
−x
3
+ 2)
=
2x −15
14x
2
+ 3x −28
Sometimes, the cancellation can be less obvious. For example, the rational expression
x
3
−8
x
2
+ 2x + 4
can be simplified to x −2 because, by an earlier identity,
x
3
−8 = x
3
−2
3
= (x −2)(x
2
+ 2x + 4)
and we can cancel the number (x
2
+2x +4) from the numerator and denominator. (As
you will learn when we come to quadratic equations, it turns out that in the case of this
particular rational expression in x, x can be any number because x
2
+ 2x + 4 is never
equal to 0. Therefore, we actually have an identity
x
3
−8
x
2
+2x+4
= x −2 for all x.)
In beginning algebra, often there is too much emphasis on simplifying rational ex-
pressions. This is a left-over from the questionable practice of teaching fractions by
insisting on the reduction of all fractions to lowest terms at all costs.
It remains to round off this discussion by mentioning that, just as one can easily de-
fine polynomials in x, y, z, etc., one can likewise define rational expressions in x, y, z, etc.
EXERCISES
7
Again, compare Section 9 of Chapter 2: Fractions (Draft),
20
1. If a, b are two numbers, what are (a + b)
3
and (a − b)
3
? (These are useful
identities to bear in mind.)
2. Is the following true or false for any numbers s and t ?
(s
2
−t
2
)
2
+ (2st)
2
= (s
2
+ t
2
)
2
Do you see why such an identity could be of interest?
3. Sum 5
6
+ 5
7
+ 5
8
+ + 5
27
.
4. If y is a nonzero number, what is 1 +
1
y
+
1
y
2
+
1
y
3
+ +
1
y
19
?
5. If y is a nonzero number and n is a positive integer, what is
1
y
3
+
1
y
4
+ +
1
y
n
?
6. Sum
1
4
3
−
1
4
5
+
1
4
7
−
1
4
9
+ −
1
4
33
.
7. If x is a nonzero number and n is a positive integer, what is
−1 +
1
x
3
−
1
x
6
+
1
x
9
− −
1
x
2n·3
?
8. If a is a number, what is (
2
3
a
2
−
3
4
a − 1)(
1
2
a
2
+
1
3
a −
4
9
) ? What is (a
2
−
5
3
a −
2
3
)(a
2
+
5
3
a −
2
3
) ?
9. If x and y are numbers and x ,= y,
1
x
2
−y
2
−
1
x
2
+ y
2
= ?
10. If x is a number different from 2, −3, and −1, what is
2
x −2
+
3
x + 3
−
1
x + 1
= ?
11. If x is a number that makes all the denominators nonzero in the following,
simplify:
2x
3
−9x
2
−5x
(x−2)
2
x
2
−3x−10
x
4
−16
12. Factor 4x
2
−12x + 9, and 25x
2
+ 10x + 1 for any x.
13. Factor a
3
+ b
3
for any numbers a and b. Factor a
2n+1
+ b
2n+1
for any positive
integer n. Show that 5
89
+ 6
89
is never a prime.
14. Factor s
4
+s
2
t
2
+t
4
for any numbers s and t. Factor s
4k
+s
2k
t
2k
+t
4k
for any
positive integer k.
15. Simplify (assuming x and y are not both 0):
(i)
4x
4
−9y
4
4x
4
+ 12x
2
y
2
+ 9y
4
, (ii)
15x
3
y
4
−60x
2
y
7
21
2 Transcription of Verbal Information into Symbolic
Language
Word problems are the bugbears of many students and teachers alike, and most of
the difficulty stems from the inability to correctly transcribe verbal information into
equations or inequalities. We pause to formally define an equation in a number x to be
a statement that two given number expressions involving x are equal. This equality may
be valid for all x, for some x, or for no x. For example, the equality x
3
−1 = −
5
2
x
2
−
1
2
x
for a number x is an equation which happens to be valid for exactly x =
1
2
, −1, −2.
On the other hand, the equation x
2
+ 1 = x
2
is valid for no x. After a while, the fact
that an equation is about some number x yet to be specified will be understood and
reference to this fact will be omitted. Because only one number x is involved, these
equations are sometimes called equations of one variable out of respect for tradition.
In mathematics, the word "variable" is loosely understood to be an element in the so-
called domain of definition of a function, and there is no precise definition of the term.
Indeed, none is called for. It is retained in the mathematical literature only because, as
we said, it was used in the past and, like "identity", it is convenient to have around.
Yet in school mathematics, it is an extreme irony that this term is taken seriously while
others that should be (such as fractions or decimals or the graph of a function) are not.
The psychological damage that the terminology of a "variable" can do to beginners,
conjuring as it does the mental image of a number that "varies", is something you as a
teacher have to be careful about. A number is a number. It never varies.
Equations in a collection of (yet-to-be-determined) numbers are similarly defined.
We will deal with equations in two variables in ¸4.
In a similar vein, an inequality in a number x is defined as a statement that one
number expression in x is bigger than (or, bigger than or equal to) the other. Again,
the inequality may be valid for all x, for some x, or for no x. The inequality x
2
+1 < 0,
for example, is satisfied by no x. Again, as in the case of equations, the fact that an
inequality in x is a statement about the validity of one number expression bigger than
another one for a certain collection of numbers x will sometimes be omitted in the future.
There is no hope of getting a correct solution to a word problem if we work with the
wrong equations or inequalities. Therefore the focus of this section is on this critical
link in the solution of word problems, namely, the process of transcribing verbal infor-
mation into equations or inequalities. The importance of a correct transcription of the
verbal information into symbolic expressions in the context of problem-solving seems to
22
be not fully recognized by teachers and educators alike. Many teacher jump directly
from reading the verbal information in a problem to attempting to get a solution. The
fact that there is an intermediate step between the "reading" and the "solving" is one
that deserves emphasis. Our purpose here is to isolate this intermediate step and call
attention to the need of addressing this critical issue. Without worrying about how to
solve equations or inequalities (i.e., the exact determination of all the numbers x that
make the equality valid), our task is to make sure that students learn to set up problems
correctly. To this end, you as a teacher must take the lead. You will be asked to do
plenty of such transcriptions, with the hope that the extra practice would enable you to
gain the needed facility and confidence to help the students.
We will give a few illustrative examples. In these examples, notice that the starting
point is always a systematic, sentence by sentence transcription of the verbal data into
the symbolic language. Then all the information is pulled together at the end to arrive
at the correct equation(s). Let us begin with a simple one.
Let
a
b
be a fraction. If
a
b
of 57 is taken away from 57, what remains exceeds
2
3
of 57 by 4. Express this information as an equation in
a
b
.
Solution We know from the meaning of the multiplication of fractions that "
a
b
of 57"
is just
a
b
57. (We put in the symbol here for clarity, as
a
b
57 would look somewhat odd,
while writing it as 57
a
b
might confuse it with a mixed number. The main purpose of the
symbolic language is to add clarity and brevity to the verbal expression. Consequently,
whatever convention there is concerning the use of symbols should be ignored when
clarity or brevity is jeopardized.) Thus the statement, "If a fraction
a
b
of 57 is taken
away from 57", becomes
57 −(
a
b
57)
According to the given information, this number is 4 bigger than
2
3
of 57, i.e., 4 bigger
than
2
3
57. So we transcribe this information directly:
¦57 −(
a
b
57)¦ −(
2
3
57) = 4
This is one answer. Obviously, one could equally well express this equation as
57 −(
a
b
57) = (
2
3
57) + 4
The following example is a bit more complicated.
23
Johnny has three siblings, two brothers and a sister. His sister is half the
age of his older brother, and three fourths the age of his younger brother.
Johnny's older brother is four years older than Johnny, and his younger
brother is two years younger than Johnny. Let J be the age of Johnny, A the
age of Johnny's older brother, and B the age of his younger brother. Express
the above information in terms of J, A, and B.
Solution The first thing to take note is that the given data of the problem involves
Johnny's sister, but we are asked to "express the above information in terms of J, A,
and B", i.e., the sister is left out. There are many ways to deal with this situation, and
one of them is to transcribe all the information by bringing in the sister, and then try
at the end to omit any reference to her and still faithfully transcribe the given verbal
data.
So let S be the age of the sister. "His sister is half the age of his older brother" then
becomes S =
1
2
A, and "His sister is . . . three fourths the age of his younger brother"
becomes S =
3
4
B. "Johnny's older brother is four years older than Johnny" becomes
A = J + 4, while "his younger brother is two years younger than Johnny" translates to
B = J −2.
At this point, the two equations, A = J + 4 and B = J −2, would appear to be the
answer because they are the only equations directly involving A, B, and J. But these
two equations fail to capture the part of the given information about how the brothers
are related to the sister, which gives information on how the brothers are related to each
other. So we go back to look at S =
1
2
A and S =
3
4
B. They show that both
1
2
A and
3
4
B
are equal to S, and therefore equal to each other. Thus we also have
1
2
A =
3
4
B. Now we
have collected all the given information concerning J, A and B:
A = J + 4, B = J −2,
1
2
A =
3
4
B.
We next give an example requiring the use of inequalities:
Erin has 10 dollars and she wants to buy as many of her two favorite pastries
as possible. She finds that she can buy either 10 of one and 9 of the other,
or 13 of one and 6 of the other, and in both cases she will not have enough
money left over to buy more of either pastry. If the prices of the pastries are
x dollars and y dollars, respectively, write down the inequalities satisfied by
x and y.
24
Solution With x and y understood, Erin spends a total of $(10x + 9y) in the first
option, and then $(13x +6y) in the second option. The key point is that in either case,
"she will not have enough money left over to buy more of either pastry". Consider then
the first option: the total number of dollar left over is 10 − (10x + 9y). If this amount
exceeds or equals x, then Erin would be able to purchase one more of this pastry. Such
not being the case, we have 10−(10x+9y) < x. Moreover, she cannot spend more than
$10, and so 10 − (10x + 9y) ≥ 0. We combine these two inequalities into the following
double ineqiality:
0 ≤ 10 −(10x + 9y) < x
Note that it is strict inequality and not 10−(10x+9y) ≤ x, as the possibility of equality
(implied by ≤ ) would mean that Erin could buy one more of the x-pastry. Similarly,
0 ≤ 10 −(10x + 9y) < y
The same consideration applies to the second case. The answer is then the collection of
four double inequalities:
0 ≤ 10 −(10x + 9y) < x, 0 ≤ 10 −(10x + 9y) < y,
0 ≤ 10 −(13x + 6y) < x, 0 ≤ 10 −(13x + 6y) < y
As a final illustration, we do a problem that is a trifle more sophisticated than the
previous three. It is very instructive.
Two women started at sunrise and each walked at constant speed. One went
from City A to City B while the other went from B to A. They met at noon
and, continuing with no stop, arrived respectively at B at 4 pm and at A at
9 pm.
If the sunrise was x hours before noon, and if L is the speed of the woman
going from A to B and R is the speed of the woman going from B to A,
transcribe the information above into equations using the symbols L, R and
x.
Solution For ease of discussion, we will refer to the woman going from City A to
City B as the First Woman, and the other as the Second Woman. Before looking at a
correct solution, we first look at one that may be what most people would write down,
25
and explain why it is not good enough. The reasoning goes as follows. The distance
between City A and City B is fixed, and both the First Woman and Second Woman
walked this distance in the time given: the First Woman walked x hours before noon,
and then another 4 hours (from noon till 4 pm), while the Second Woman also walked
x hours before noon but continued for another 9 hours (from noon till 9 pm). So the
First Woman walked a total of x + 4 hours while the Second Woman walked a total of
x + 9 hours. Given that the former walked with speed L (let us say, miles per hour),
the total distance she walked in x + 4 hours is of course L(x + 4) miles. Similarly the
total distence the Second Woman walked in x+9 hours is R(x+9) miles. By a previous
remark, both distances are the same as both women walked between cities A and B.
Therefore we get
L(x + 4) = R(x + 9)
This is supposed to be the answer to the problem. But is it?
What this equation fails to capture is the information that the two women met at
noon after walking x hours in opposite directions from City A and City B. This means
that the total distance they covered after walking the first x hours (which is Lx + Rx
miles) is the distance between the cities, which is L(x+4) or R(x+9) miles, as we have
seen. Therefore the additional piece of information that must be incorporated into the
symbolic transcription is Lx + Rx = L(x + 4) or Lx + Rx = R(x + 9). Either one
would do. Therefore, a solution to the problem is the following set of equations:
L(x + 4) = R(x + 9), R(x + 9) = Lx + Rx
Now one may reason slightly differently. Let the meeting point of the two women be
C:
A C B
¸ ¡
Consider the distance between A and C. The First Woman covered it in x hours
(before noon) while the Second Woman covered it in 9 hours (after noon). But in x
hours the First Woman would walk Lx miles, while in 9 hours the Second Woman would
walk 9R miles. Therefore Lx = 9R. Similarly, if we consider the distance between C
and B, we get in exactly the same fashion that 4L = Rx. A correct solution is then:
Lx = 9R, 4L = Rx
We will leave as an exercise that these two solution are "the same", in a precise sense.
26
EXERCISES
1. A train travels s km in t hours. At the same constant speed, how far does it travel
in 5 hours? In T hours? How long does it take the train to travel 278 km? x km? If the
speed of the train is tripled (3 times as fast), how long will it take to travel s km? And
if the speed is n times faster?
2. The sum of the squares of three consecutive integers exceeds three times the
square of the middle integer by 2. If the middle integer is x, express this fact in terms
of x. If the smallest of the three integers is y, express the same fact in terms of y.
3. Paulo read a number of pages of a book with N pages, then he read 43 pages
more and finished three-fifths of the book. If p is the number of pages Paulo read the
first time, write an equation using p and N to express the above information.
4. A whole number has the property that when the square of half this number is
subtracted from 5 times this number, we get back the number itself. If y is this number,
write down an equation for y.
5. Helena bought two books. The total cost is 49 dollars, and the difference of
the squares of the prices is 735. If the prices are x and y dollars, express the above
information in terms of x and y.
6. I have two numbers x and y. Take 20% of x from x, then what remains would be
4 less than y. If however I enlarge y by 20%, then it would exceed x by 5. Express this
information in equations in terms of x and y.
7. I have $4.60 worth of nickels, dimes and quarters. There are 40 coins in all, and
the number of nickels and dimes is three times the number of quarters. If N, D, and Q
denote the number of nickels, dimes, and quarters, respectively, write equations in terms
of these symbols to capture the given information.
8. There are two whole numbers. When the large number is divided by the smaller
number, the quotient is 9 and the remainder is 15. Also, the larger number is 97.5% of
ten times the smaller number. If x is the larger number and y is the smaller number,
express the given information in equations in terms of x and y.
9. We look for two whole numbers so that the larger exceeds the the smaller by at
least 10, but that the cube of the smaller exceeds the square of the larger number by at
least 500. If the larger number is x and the smaller number is y, transcribe the above
information in terms of x and y.
10. If the digits of a three-digit number are reversed, the sum of the new number
and the original number is 1615. If 99 is added to the original number, the digits are
27
reversed. Let the hundreds, tens, and ones digits of the original numbers be a, b, and c,
respectively. Write equations in a, b and c to express the given information.
11. Two marathon runners run at constant speeds. The first runner runs D kilome-
ters in A hours, and the second runner in B hours. If they start running at the same
time from separate cities, D kilometers apart, towards each other, they are 11 kilome-
ters apart after 1 hour. If the first runner runs twice as fast and, again, they do the
same, then they would be 5 kilometers apart after one hour. Express this information
in symbolic language in terms of A, B, and D.
12. A sum of money is to be divided equally among x people, each receiving y dollars.
If there are 3 more people, each person would receive 1 dollar less, and if there are 6
fewer people, each would receive 5 dollars more. Write equations in x and y to express
this information.
13. A man walked from one place to another in 5
1
2
hours. If he had walked
1
4
of a
mile an hour faster, the walk would have taken 36
2
3
fewer minutes. If the distance he
walked is x miles and his speed was v miles an hour, express the above information in
terms of x and v.
14. The denominator of a fraction exceeds twice the numerator by 2, and the dif-
ference between the fraction and its reciprocal is
55
24
. If the numerator is x and the
denominator y, write equations in terms of x and y to express the above information.
15. A video game manufacturer sells out every game he brings to a game show. He
has two games, an A Game and a B Game. He can bring 50 of A Games and B Games
in total to the show. Each A Game costs $75 to manufacture and will bring in a profit
of $ 185. Each B Game costs $165 to manufacture and will bring in a profit of $185.
However, he only has $6, 000 to spend on manufacturing. If he brings x A Games and
y B Games, describe in terms of x and y how he can maximize his profit.
3 Linear Equations in One Variable
In the preceding section, we have seen how equations involving a number x arise natu-
rally. Recall that we call such an equation an equation in one variable, which is merely a
statement that two number expressions in a number x (as yet undetermined) are equal.
The statement may be true for all values of x, some values of x, or no value of x. In this
section, we make a first attempt at solving an equation, i.e., determining all possible
values of x that make the equation valid. The numbers that make the equation valid are
28
called the solution set of the equation, and they are also said to satisfy the equation.
Sometimes one refers to the solution set more simply as the solution or the solutions,
as the case may be. If it turns out that any number would make the equation valid,
i.e., if the solution set comprise all the numbers, then the equation would be what we
have called earlier an identity. For example, x
2
− 1 = (x − 1)(x + 1) is an example of
an equation in x that is in fact an identity; it is one that we have encountered before.
At the opposite extreme, there are equations, such as x
5
= x
5
−3, that would be valid
for no number x. In this case, we say the equation has no solution. The interesting
equations fall somewhere in between: they are valid for a finite collection of numbers.
For example, consider all the numbers whose squares are equal to 4. If x denotes such
a number, then we are given that x
2
= 4. It is intuitively clear that there are only two
possibilities: x = 2 and x = −2. The solutions of x
2
= 4 are then x = ±2. (Later on,
we shall rigorously prove that ±2 are indeed the solution set of x
2
= 4.)
In this section, we deal with the simplest kind of equations which turn out to be
basic, namely, equations which state that two polynomials of degree at most 1 in the
same number x are equal. These are called linear equations of one variable. Examples
are: 12x −7 = 5x +13, −
5
6
x +1 = 23x −4, and 9 = 27x −4. Some general principles
which lead us to the solutions of linear equations are however valid for all equations,
and we will point these out in due course.
Suppose a linear equation is in the following form:
3x = 15 for a number x.
Then we can read off what x must be, namely 5 because 5 is the only number whose
product with 3 is 15. For a reason that will be obvious, we want to produce this obvious
solution more systematically. We multiply both sides of this equation by
1
3
, and the
left side becomes
1
3
(3x) = (
1
3
3)x = (1)x = x, while the right side becomes
1
3
5 = 5.
Because multiplying two equal numbers by the same number yields two equal numbers,
the resulting two numbers, x and 5, must be equal. Thus we get x = 5, and this is the
solution.
You will be bored if I tell you there are important lessons to be learned here, but
this happens to be the truth. To begin with, the systematical procedure of solving the
preceding problem points to the way of solving a class of linear equations in general,
namely, equations in x of the form ax = b, where a, b are coefficients of the polynomials
with a ,= 0; a and b are traditionally referred to as the constants of the equation. (By
letting a = 3 and b = 15, we get the preceding example again.) We now show that the
29
only number x that satisfies this equation is
b
a
, because multiplying both sides of ax = b
by
1
a
, we get
1
a
(ax) =
1
a
b. By the associative law, the left side is
1
a
(ax) = (
1
a
a)x = 1 x = x
while the right side is
b
a
. Therefore x =
b
a
.
If we are careful, we would notice that what the above shows is that if x is any
number that satisfies ax = b, then it has to be
b
a
. We must also ensure that if x =
b
a
,
then indeed ax = b. But this is an easy direct verification.
Intuitively, what we did above was to "free up" the x on the left side of ax = b by
getting rid of the coefficient a of ax; this we accomplished by multiplying both sides of
ax = b by
1
a
. This is an important technique. What interests us right now are the two
principle behind this technique. One is the critical role played by the associative law for
multiplication, in its full strength, in a sense we now explain. We applied the associative
law in the step
1
a
(ax) = (
1
a
a)x
It must be remembered that a is a number that may be very small or very big, so the
eventual solution x of the equation can also be very small or very big. We don't know
a priori what it is. But the associative law, being true for all numbers, is nevertheless
applicable here to validate the preceding step. In subsequent discussions about solution
of equations, the associative, commutative, and distributive laws will be applied in an
entirely analogous manner to unknown numbers, and we will almost never take the
trouble to highlight such applications of these laws again. But on this occasion, we want
to emphasize why the fact that these laws are valid for all numbers is important, as
otherwise the solving of equations would be impossible.
The other general principle behind the above method of solution is this:
Two numbers that are equal remain equal when multiplied by equal numbers.
In symbols: if a = b and c = d for numbers a, b, c, d, then ac = bd. In our definition
of multiplication as area of rectangles, this is the assertion that two rectangles with the
length of sides being pair-wise equal have equal area. This is clear. We will enunciate
presently the analogous principle for addition because the latter also plays a key role in
the solution of equations.
We are now in a position to solve any linear equation of one variable. It is simpler
than you think. We have solved all linear equations of the form ax = b. We now show
30
how to bring any linear equation to this form, thereby solving the linear equation. First
let us look at the issue of solving linear equations from an experimental point of view.
Consider the equation 12x − 5 = 6x. Recall what this means: we have numbers x
for which the two expression 12x − 5 and 6x are equal, and we have to determine
all possible values of such x. Is 2 one of them, or as we say, is 2 is a solution of
12x −5 = 6x ? In other words, if x = 2 in 12x −5 = 6x, are the two sides equal? The
answer is no, because 19 (= 12 2 −5) is not equal to 12 (6 2). What about
1
3
? No
again, because −1 (= 12
1
3
− 5) is not equal to 2 (= 6
1
3
). We can save ourselves
some work, instead of guessing, by isolating x on one side so that the equation will
look like ax = b, which can then be solved. This can be done by appealing to another
basic principle:
Two numbers which are equal remain equal when added to equal numbers.
In symbols: if a, b, c, d are numbers with a = b and c = d, then a + c = b + d. Using
the vector model for the addition of rational numbers, we see that this must be true.
To isolate x in 12x − 5 = 6x, we "move" or "transpose" 12x from the left side
to the right side by adding the number −12x to both sides. The left side becomes,
by an application of the associative law of addition, (−12x) + (12x − 5) = (−12x) +
[12x +(−5)] = (−12x +12x) +(−5) = −5, while the right side is (−12x) +6x. By the
preceding basic principle, −5 must equal (−12x) + 6x, i.e.,
−5 = (−12x) + 6x
The change from 12x−5 = 6x to −5 = (−12x) +6x is what accounts for the common
terminology of transposing 12x from left to right, but we know that it is really adding
−12x to both sides. Since (−12x) + 6x = −6x, we get −5 = −6x, or −6x = −5. This
is in the form of ax = b with a = −6 and b = −5. We therefore know right away that
x =
−5
−6
=
5
6
.
To recapitulate, starting with 12x − 5 = 6x, we add −12x to both sides to ob-
tain −6x = −5, from which we conclude x =
5
6
. In short, if a number x satisfies
12x − 5 = 6x, then necessarily x =
5
6
. To round off the picture, it remains to show
that if x =
5
6
, then the equality 12x − 5 = 6x is valid, but again this is simple to
check as both sides are equal to 5. This then shows that x =
5
6
is the only solution to
12x −5 = 6x.
Before proceeding with the discussion of solving linear equations, it is a good idea
to pause and reflect on the above procedure of transposing a term from one side of the
31
equation to another. The preceding consideration shows that if A, B, C are numbers,
then when we transpose the term B of A+B = C to the right side, we get −B. More
precisely,
(T1) A + B = C implies A = C −B
Now consider, for example, 6x = 7 −2x. What happens when we try to transpose the
term −2x to the left side of the equal sign? Let us make sure how to do this correctly,
and why. We first recall the meaning of subtraction among rational numbers: 7 −2x =
7 + (−2x), by definition. Therefore, when we add 2x to both sides of 6x = 7 − 2x,
we may apply the associative law of addition to transform 6x + 2x = (7 −2x) + 2x to
6x + 2x = ¦7 + (−2x)¦ + 2x = 7 +¦(−2x) + 2x¦ = 7 + 0 = 7, i.e.,
6x = 7 −2x implies 6x + 2x = 7
In general, the same reasoning shows that for numbers A, B, C,
(T2) A −B = C implies A = C + B
The two facts (T1) and (T2) together are usually summarized by saying that when we
transpose, we change the sign. What needs greater emphasis in classroom instructions
is the underlying reasoning of why this is true.
We can now suppress some of the explanations and solve another linear equation to
give another demonstration of the general procedure. Consider 5x + 1 = 2x − 11. We
first transpose 2x to the left side: (5x + 1) −2x = −11, so that 3x + 1 = −11. Now we
have to transpose +1 to the right side: 3x = −11 −1, and 3x = −12. Thus the solution
is
(−12)
3
, which is −4.
Here is another example. To solve −
2
3
x+4 = −
1
5
x+5
1
3
, we collect all the x's to the
left: (−
2
3
x+4) −(−
1
5
x) = 5
1
3
. On the left side we have −
2
3
x+4+
1
5
x = −
2
3
x+
1
5
x+4 =
−7
15
x + 4. Thus the equation becomes
−7
15
x + 4 = 5
1
3
. We next transpose 4 to the right
side:
−7
15
x = 5
1
3
−4, and so
−7
15
x =
4
3
. Thus the solution is
4
3
/
−7
15
= −
20
7
.
We conclude with an exposition of the solution of any linear equation. To solve
ax + b = cx + d, where a, b, c, d are constants and a ,= c, we first transpose cx
to the left: (ax + b) − cx = d. The associative and commutative laws of addition
imply that the left side is equal to (ax −cx) +b, and the distributive law implies that
(ax − cx) + b = (a − c)x + b. Therefore we have (a − c)x + b = d. Transposing b to
the right yields (a − c)x = d − b. From this, we conclude immediately that x =
d−b
a−c
.
32
What we have done is to show that any number x satisfying ax + b = cx + d is equal
to
d−b
a−c
. We still need to check that
d−b
a−c
is a solution, i.e., if x =
d−b
a−c
, then it is true
that ax + b = cx + d. This can be done directly:
ax + b = a
_
d −b
a −c
_
+ b =
ad −ab
a −c
+ b
=
ad −ab
a −c
+
ab −bc
a −c
=
ad −bc
a −c
while
cx + d = c
_
d −b
a −c
_
+ d =
cd −bc
a −c
+ d
=
cd −bc
a −c
+
ad −cd
a −c
=
ad −bc
a −c
So ax +b = cx +d when x =
d−b
a−c
. This then completes the reasoning that
d−b
a−c
is the
only solution of ax + b = cx + d.
To summarize, solving a linear equation in a number x depends on two simple ideas:
by transposing terms, we isolate x on one side of the equation, and then we solve an
equation of the type ax = b. From this point of view, the common practice of classifying
linear equations into one-step equations, two-step equations, three-step equations and
fourth-step equations, and then teach the solving of linear equations according to this
classification simply does not make sense. You should avoid using this classification
when you teach this topic.
It remains to point out that sometimes a linear equation is disguised as one involving
rational expressions. For example, consider a number x that satisfies
2
3x −1
=
4
x +
1
3
by the cross-multiplication algorithm (which is valid also for complex fractions), this
equation is equivalent to
2(x +
1
3
) = 4(3x −1)
33
We can either use the distributive law on both sides directly, or we can avoid computa-
tions with fractions at this early stage and simply multiply both sides by 3 to get
2(3x + 1) = 4(9x −3)
Then 6x + 2 = 36x −12 and 14 = 30x. In other words, x =
7
15
.
EXERCISES
1. Solve: (i) 2x −8 = 15 +
4
3
x. (ii)
7
3
x + 2 =
3
2
−
2
5
x. (iii)
11
9
−
5
3
x = −6x +
1
18
.
(iv) ax +6 = 8 −7ax, where a is a nonzero number. (v) 4bx +13 = 2x +26b, where
b is a number not equal to
1
2
. (vi)
1
2
−
8
3
x =
5
6
x +
2
3
. (vii)
2
5
ax −17 =
1
3
ax −
15
2
.
4 Linear Equations in Two Variables and Their Graphs
An equation such as x −2y = −2 is an example of a linear equation in the two numbers
(usually called variables) x and y. A solution of this equation is an ordered pair of
numbers (x
0
, y
0
) so that x
0
and y
0
satisfy the equation x−2y = −2, in the sense that
x
0
− 2y
0
= −2. We observe that in this situation, it is easy to find all the solutions
with a prescribed first number x
0
or a prescribed second number y
0
. For example, with
the first number prescribed as 3, then we solve the linear equation in y, 3 −2y = −2,
to get y =
5
2
. Therefore (3,
5
2
) is the sought-for solution. Or, if the second number
is prescribed to be −1, then we solve the linear equation in x, x − 2(−1) = −2, to
get x = −4. The solution is now (−4, −1). Relative to a pair of coordinate axes
in the plane, the collection of all the points (x
0
, y
0
) in the coordinate plane so that
each pair (x
0
, y
0
) is a solution of the equation x − 2y = −2 is called the graph of
x − 2y = −2 in the plane. Using the above method of getting all the solutions of the
equation x − 2y = −2, we can plot as many points of the graph as we please to get a
good idea of the graph. For example, the following picture contains the following points
(given by the dots) on the graph, going from left to right:
(−5, −1.5), (−4, −1), (−2, 0), (0, 1), (2, 2),
(2.5, 2.25), (4, 3), (6, 4), (7, 4.5).
These points strongly suggest that the graph of x − 2y = −2 is a (straight) line,
and we will presently prove that such is the case.
34
O
Y
X
2 4 6 −2
−4
2
−2
4
s
s
s
s
s
s
s
s
s
Consider next the graph of the linear equation in two variables y = 3 which, as an
equation in two variables, is in reality the abbreviated form of the equation 0x+1y = 3.
The collection of all solutions of y = 3 is then exactly all the pairs (s, 3), where s
is an arbitrary number, for the following reason. Every one of these (s, 3)'s is clearly a
solution because 0 s + 1 3 = 3. Are there perhaps other pairs of numbers which are
also solutions? For example, (s, 3.1)? But 0 s + 1 3.1 = 3.1 ,= 3, so (s, 3.1) is not a
solution of y = 3. Similarly, if a number t is not equal to 3, then (s, t) is not a solution
of y = 3 no matter what t may be. This shows that the preceding assertion about the
pairs (s, 3) is true. In terms of the graph, the points with coordinates (s, 3) always
lie on the horizontal line (i.e., parallel to the x-axis) passing through the point (0, 3)
on the y-axis, and since s is arbitrary, these points (s, 3) then comprise the complete
horizontal line passing through (0, 3). In short, the graph of the equation y = 3 in the
plane is exactly the horizontal line passing through the point (0, 3) on the y-axis.
X
Y
3
O
35
Similarly, the graph of the equation x = −2 is the vertical line (i.e., parallel to
the y-axis) passing through the point (−2, 0) on the x-axis. In a similar manner,
the graph of x = c for any number c is the vertical line passing through the point
(c, 0) on the x-axis, and the graph of y = b for any number b is the horizontal
line passing through the point (0, b) on the y-axis.
Since there is only one horizontal (respectively, vertical) line passing through a given
point of the plane (do you know why?), it follows that
every vertical line is the graph of the equation x = c, where (c, 0) is the point of
intersection of the line and the x-axis, and every horizontal line is the graph of
an equation y = b, where (0, b) is the point of intersection of the line with the
y-axis.
Both of these simple facts are probably known to you, but the precise reasoning may
have been missing. We have supplied so much detail to explain them because your stu-
dents should understand that these facts are consequences of careful reasoning and the
precise definition of a graph.
We next treat the general case. A linear equation in two variables x and y is
an equation in the numbers x and y which is either of the form ax + by = c where
a, b, c are given numbers, commonly referred to as constants and at least one of a
and b is nonzero, or can be rewritten in this form after transposing and using the four
arithmetic operations. Thus −2x =
2
5
y + 7 and 6 +
3
8
y = 179 − 5x are examples
of linear equations of two variables. A solution of this equation is an ordered pair of
numbers x
0
and y
0
, written in the expected fashion as (x
0
, y
0
), so that they satisfy the
equation ax + by = c, in the sense that ax
0
+ by
0
= c. The graph of ax + by = c is
the collection of all the points in the plane with coordinates (x
0
, y
0
) (relative to a given
pair of coordinate axes), so that each ordered pair of numbers x
0
and y
0
is a solution
of ax + by = c. As we have seen, and as we shall continue to bear witness, the study
of linear equations of two variables is grounded in the study of linear equation of one
variable.
Armed with these precise definitions, we will now explain why the following theorem
is true.
36
Theorem The graph of a linear equation is a straight line. Conversely, every
straight line is the graph of a linear equation.
This theorem establishes a correspondence between lines and the graphs of linear
equations in two variables: the graph of a linear equation ax + by = c is a line L, and
every line L is the graph of some linear equation ax + by = c. It is customary to call
ax + by = c the equation of the line L if L is the graph of ax + by = c, and say
that L is defined by ax + by = c. Incidentally, this theorem explains why equations
of the form ax + by = c are called linear equations (because their graphs are lines).
Moreover, the reasoning in the proof of this theorem is very important. This reasoning
is the key to the understanding of almost everything about linear equations in introductory
algebra.
We want to make a minor, but important simplification in the subsequent discusion
of this Theorem. Suppose we start with a linear equation ax+by = c. If b = 0, then by
the definition of a linear equation in two variables, a ,= 0. The equation may therefore
be rewritten as x = c
, where c
is the constant c
=
c
a
. In this case, we have seen that
the graph is a vertical line. The first part of the Theorem is therefore true in this case.
We may therefore assume from now on that b ,= 0 in a given equation ax + by = c.
Such being the case, we may rewrite the equation as by = −ax + c, and therefore
y = mx + k, where m = −
a
b
and k =
c
b
. On the other hand, we have seen that a
vertical line is the graph of x = c
(i.e., x + 0 y = c
), where (c
, 0) is the point at
which intersects the x-axis. In other words, the second part of the Theorem is true for
vertical lines. We may therefore assume from now on that a given line is not a vertical
line.
To summarize, what we have shown is that
in the subsequent discussion of this Theorem, we may assume that a given linear
equation is of the form y = mx + k, where m, k are constants, and that a
given straight line is non-vertical.
Let us first begin the consideration of the Theorem by looking at a concrete case
such as y =
2
3
x + 2. Why is the graph of this equation a straight line? The reasoning
in this case will shed light on the general case. So let G be the graph of y =
2
3
x + 2.
37
Notice that the point (0, 2) on the y-axis and the point (−3, 0) on the x-axis are on
G, because 2 =
2
3
0 + 2 and 0 =
2
3
(−3) + 2. Let L be the (straight) line joining
(0, 2) and (−3, 0). We are going to prove that G is the line L.
X
Y
2
−3
r
r
L
O
How to show that the two sets G and L are the same? Obviously, we first have to
show that every point on the graph G lies on the line L. But this is not enough because
G could just be part of L and not all of L. So we must also show that every point of L
is a point of G. Therefore, we must show two things:
(α) Every point on the graph G is a point on the line L.
(β) Every point on the line L is a point on the graph G.
Before proving either of these assertions, we review some facts concerning similar
triangles and a basic property of straight lines (concerning slope) which is almost uni-
versally mishandled in introductory algebra textbooks. First recall the definition of two
geometric figures being similar: one is mapped onto the other by a rigid motion followed
by a dilation. One then proves that two triangles are similar if and only if the corre-
sponding angles are equal and the ratios of (the lengths of) the corresponding pairs of
sides are equal. The proof of this fact follows quite easily from the definition of a dilation
and the so-called fundamental theorem of similarity. However the following is less
obvious:
The AA criterion of similarity: If two triangles have two pairs of equal
angles, they are similar.
In order to use this criterion effectively, one needs to know when two angles are equal.
In this context, we recall also the following fact:
38
If a line L meets two other lines
1
and
2
, then the following three conditions
are equivalent:
(1)
1
|
2
.
(2) A pair of alternate interior angles of the line L are equal.
(3) A pair of corresponding angles of the line L are equal.
We apply these facts to the study of straight lines by proving that, once a pair of
coordinate axes have been set up, associated to each non-vertical straight line L is a
number called slope which measures the amount of "slant" of the line. Thus coordinate
axes as given, let a pair of distinct points P, Q on L be chosen.
X
Y
r
r
Q
P
L
O
X
Y
`
`
`
`
`
`
`
`
`
`
`
``
r
r
Q
P
L
O
Let the coordinates of P, Q be (p
1
, p
2
) and (q
1
, q
2
), respectively. Consider the
quotient
p
2
−q
2
p
1
−q
1
_
=
q
2
−p
2
q
1
−p
1
_
Observe that the denominator is never 0 because if q
1
− p
1
= 0, then p
1
= q
1
and the
fact that P, Q are distinct would mean that the line L is a vertical line. Because we
have excluded vertical lines, this is impossible. Thus the denominator of this quotient is
never zero, and the quotient makes sense. Also observe that this quotient is positive for
the line on the left, which is slanted this way /, and negative for the line on the right,
which is slanted this way ¸, because, for the line on the left, p
2
> q
2
and p
1
> q
1
while
for the line on the right, p
2
> q
2
but p
1
< q
1
. This observation concerning the sign
can be seen perhaps more transparently in terms of the geometry. Form a right triangle
´PQR so that the lines PR and QR are parallel to the coordinate axes, as shown:
39
X
Y
r
r
Q
P
R
L
O
X
Y
`
`
`
`
`
`
`
`
`
`
`
``
r
r
Q
P
R
L
O
If we denote the length of the segment PR as usual by [PR[, etc., then
p
2
−q
2
p
1
−q
1
=
[PR[
[QR[
for the line on the left, and
p
2
−q
2
p
1
−q
1
= −
[PR[
[QR[
for the line on the right. Thus the quotient
p
2
−q
2
p
1
−q
1
has been directly expressed as a
positive or negative quotient of the lengths of the legs of a right triangle, positive for
lines slanting like / and negative for lines slanting like ¸.
We claim that this quotient
p
2
−q
2
p
1
−q
1
in terms of the coordinates of P and Q does not
depend on the position of the points P and Q on the line L. Let us prove this claim
for the line on the left; the proof for the line on the right is entirely similar. Let points
P
and Q
be chosen on the line L as shown and let their coordinates be (p
1
, p
2
) and
(q
1
, q
2
)s, and let right triangle ´P
Q
R
be formed so that the lines P
R
and Q
R
are
parallel to the coordinate axes.
X
Y
r
r
Q
P
R
r
r
P
Q
R
L
O
40
Our goal is to show that
p
2
−q
2
p
1
−q
1
=
p
2
−q
2
p
1
−q
1
By rewriting these quotients in terms of the lengths of the legs of ´PQR and ´P
Q
R
,
this equality becomes
[P
R
[
[Q
R
[
=
[PR[
[QR[
This is the same (using the cross-multiplication algorithm) as
[P
R
[
[PR[
=
[Q
R
[
[QR[
Now if we can show that the triangles ´PQR and ´P
R
Q
are similar, then this equality
would be valid because it merely states that the ratios of corresponding sides of these
similar triangles are equal. So why are these triangles similar? By the AA criterion
for similarity, we only have to show that two pairs of corresponding angles are equal.
´PQR and ´P
R
Q
being right triangles, there is already one pair of equal angles. We
get a second pair by observing that the line L meets both line PR and line P
R
and
PR | P
R
because they are by definition parallel to the y-axis. Thus the corresponding
angles ∠QPR and ∠Q
P
R
are equal, and the AA criterion implies the similarity of
´PQR and ´P
R
Q
. The desired equality of the quotients is now completely proved.
It should be remarked that in the picture, we have made P higher than Q, and also
P
higher than Q
to make the argument more intuitive. But the validity of the argument
by no means depends on having P
higher than Q
. Why this is so is left as a classroom
activity.
The important conclusion to draw from the preceding discussion is this: given any
straight line which is not vertical, choose any two points P and Q on the line with
coordinates (p
1
, p
2
) and (q
1
, q
2
), respectively, and form the quotient
p
2
−q
2
p
1
−q
1
,
then we can be confident that the resulting number will always be the same. In other
words, for a given line L, this number does not depend on the choice of the P and Q on
L and is therefore associated with the line L itself. We call this number the slope of the
line L. By a previous observation, the slope of a line that slants this way / is positive,
and the slope of a line that slants this way ¸ is negative. A little experimentation
41
would reveal that if the slope of a line is close to 0 (be it positive or negative), then the
line is close to being horizontal, and if it is large be it positive or negative (in the sense
of being far away from 0 in either direction on the number line), then the line would be
close to being vertical.
Activity Suppose a line L passes through (2, −3) and (−4, 1). If P is a point on L
with x-coordinate
2
3
, what is the y coordinate of P?
Textbooks usually define the slope of a line by picking two points on the line and then
declaring the quotient formed from the coordinates of these two points in the manner
shown above to be the slope of the line. A priori, the quotient resulting from a different
choice of points on the line could be a different number so that a line could have many
slopes. This would render any discussion of "the slope" of the line nonsensical. For
example, suppose instead of a straight line we have the circle of radius 5, denoted by (.
If we take the two points (−5, 0) and (3, 4) on (, and form the usual quotient, we get
4−0
3−(−5)
= −
1
2
. On the other hand, taking another pair of points (5, 0) and (0, 5) on (
leads to the quotient of
5−0
0−5
= −1. For the curve (, the quotients formed from different
pairs of points on it are therefore not always the same. Yet for a line, these quotients
are always be the same. The question is Why?. We have answered this question by the
use of similar triangles. Be sure your students know the answer too, because
the exploitation of the fact that one can compute the slope of a line by using
any two points to one's liking is a powerful tool in dealing with all kinds of
questions related to linear equations.
The following discussion will amply bear out this assertion.
We next consider another seemingly obvious question: if two lines have the same
slope and pass through the same point, are they identical?
If two straight lines have the same slope and pass through the same point, then
they are the same line.
Let the lines be L and L
, let their slopes be equal, and let them both pass through the
point P. A priori, they are distinct lines and will therefore be schematically represented
as such. Our task is to show that they are in fact the same. In the following, we only
take up the case of positive slope. The case of negative slope can be handled the same
way.
42
X
Y
O
r
r
r
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
L
L
P
Q
Q
R
Take an arbitrary point Q
on L
, and form right triangle ´PQ
R
so that the line
PR
is parallel to the x-axis, and the line Q
R
is parallel to the y-axis. Let the line Q
R
intersect L at a point Q. Recall that we are trying to show the lines L and L
coincide,
and this would be true as soon as we can show that Q and Q
coincide because there is
only one line passing through two given points. We shall make use of the fact that the
slope of a line can be computed using any two points on the line. The equality of the
slopes of L
and L is now expressed as:
[Q
R
[
[PR
[
=
[QR
[
[PR
[
Since the denominators of these two quotients are equal, the numerators must be equal
as well, i.e., [Q
R
[ = [QR
[. Therefore Q = Q
and L and L
coincide, as claimed.
We are now in a position to resume our task of showing that the first part of the
Theorem is true, i.e., the graph of a linear equation y = mx + k is a straight line.
As previously suggested, we first look at the special case of m =
2
3
and k = 2, i.e., the
equation y =
2
3
x + 2. Denote as before the line joining (−3, 0) and (0, 2) by L, and
denote the graph of y =
2
3
x + 2 by G. Recall that we will prove that G coincides with
L by proving:
(α) Every point on the graph G is a point on the line L.
(β) Every point on the line L is a point on the graph G.
43
Let the point (0, 2) on the y-axis be denoted by P. Take an arbitrary point Q
on
the graph G distinct from P, and let L
be the line joining P to Q
. If we can show that
L and L
coincide, then this arbitrary point Q
on the graph G would belong to L, and
therefore step (α) would be proved. So why do L and L
coincide? There is no a priori
reason that they do, because if the graph G is a curve, then the picture could look like
this:
G
O
L
L
r
r
r
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
P
Q
In this case, L and L
would be distinct. What we are going to show is that, because
G is the graph of a linear equation y =
2
3
x + 2, L and L
would have the same slope.
Since they both pass through P, they coincide as we have just observed. To explain the
latter, we refer to the following picture:
X
Y
2
−3
r
r
r
>
>
>
>
>
>
>
>
>
>
>
>
>
>
L
P
Q
L
O
The slope of L can be computed from any pair of points on L, in particular, from (0, 2)
(i.e., P) and (−3, 0): it is
0−2
−3−0
=
2
3
. What about the slope of L
being distinct from (0, 2) (= P) implies that x
0
,= 0.)
So both lines have the same slope
2
3
and pass through the same point P. They must
coincide and we have proved step (α).
To prove step (β) for the equation y =
2
3
x +2, we must show that if a point Q lies
on the line L joining P (= (0, 2)) and (−3, 0), then Q also lies on G. This means, if the
coordinates of Q are (x
0
, y
0
), then we must show y
0
=
2
3
x
0
+ 2.
X
Y
2
−3
r
r
r
L
P
Q = (x
0
, y
0
)
O
Now we already know from the preceding paragraph that the slope of L is
2
3
. But
we may also compute the slope of L using instead the points Q and P, obtaining:
y
0
−2
x
0
−0
=
y
0
−2
x
0
Since
y
0
−2
x
0
is the slope of L, we get
y
0
−2
x
0
=
3
2
. This implies y
0
− 2 =
2
3
x
0
, so that
y
0
=
2
3
x
0
+ 2, as desired. This completes the proof of step (β), and therewith also the
proof that the graph of y =
2
3
x + 2 is the line L joining (0, 2) and (−3, 0).
The proof for the graph of the general equation y = mx+k is no different. Let any
two points P and Q be chosen on the graph G of y = mx + k, and let L be the line
joining P and Q. For simplicity, we simply take P to be the point (0, k) on the y-axis
(check that (0, k) is on the graph of y = mx + k !). We will use the same method to
show that L and G are the same, i.e., we go through the same two steps:
(α) Every point on the graph G is a point on the line L.
(β) Every point on the line L is a point on the graph G.
We begin with step (α). Take a random point Q
by showing
that they have the same slope. Since they also pass through the same point P, it follows
that L = L
. We could compute the two slopes as in the special case of y =
2
3
x + 2 to
arrive at our conclusion, but the virtue of having worked through a special case is that
we can now afford to think more abstractly. We claim:
For any two distinct points on the graph of a linear equation y = mx +k, the
slope of the line joining them is always equal to m.
(Caution: it is tempting to assert instead that "the slope of the graph of y = mx+k is
m", but at this particular juncture, we do not as yet know that the graph of y = mx+k
is a line, so we cannot talk about the slope of the graph of y = mx+k.) Indeed let the
two points on the graph of y = mx + k be (p
1
, p
2
) and (q
1
, q
2
). The slope of the line
joining them is then (q
2
− p
2
)/(q
1
− p
1
).
8
Being on the graph, the coordinates of these
points satisfy, by definition, the equations
p
2
= mp
1
+ k and q
2
= mq
1
+ k
Therefore, the slope is
q
2
−p
2
q
1
−p
1
=
(mq
1
+ k) −(mp
1
+ k)
q
1
−p
1
)
=
m(q
1
−p
1
)
q
1
−p
1
= m
This proves our claim.
It immediately follows from this claim that since L and L
, as desired.
The proof of step (α) is complete.
8
Note that this quotient always makes sense because q
1
− p
1
is never 0. The reason for the latter
is that, if it were, we would have p
1
= q
1
. But (p
1
, p
2
) and (q
1
, q
2
) being points on the graph of
y = mx + k, we have p
2
= mp
1
+ k and q
2
= mq
1
+ k. Thus also p
2
= q
2
and the two points (p
1
, p
2
)
and (q
1
, q
2
) would not be distinct. Hence q
1
−p
1
is never 0.
46
Now, step (β): why every point of L lies on the graph G. The reason is very simple
now. Take any point R on L distinct from P and let the coordinates of P and R be
(p
1
, p
2
) and (r
1
, r
2
), respectively. We have just seen that the slope of L is m. Thus if we
compute the slope of L using P and R, it is still equal to m. Thus
r
2
−p
2
r
1
−p
1
= m
This implies r
2
−p
2
= m(r
1
−p
1
). But P being on G means p
2
= mp
1
+k. So we obtain
r
2
− (mp
1
+ k) = m(r
1
− p
1
), or r
2
− k = mr
1
, which is the same as r
2
= mr
1
+ k.
But the last equality is exactly the statement that the point (r
1
, r
2
) is a point on the
graph G of y = mx + k. Thus R lies on G, and we have completely proved the first
part of the Theorem.
We proceed to finish the proof of the Theorem by showing that every straight line
is the graph of a linear equation. Let us begin as usual with a special case: we look at
our standby, the line L that joins the points (−3, 0) and (0, 2). What equation is it the
graph of? Recall that the slope of L (obtain by computing with (−3, 0) and (0, 2)) is
2
3
.
X
Y
2
−3
r
r
r
L
(x, y)
O
Therefore, for any point (x, y) on L, but not equal to (−3, 0), we can compute the
slope of L by
y−0
x−(−3)
=
2
3
. This leads us to consider the linear equation in two variables
y =
2
3
(x −(−3))
Observe first of all that this is a linear equation in two variables because, after an
application of the distributive law and after transposing, we can rewrite it as
2
3
x −y =
−2. Let be the graph of y =
2
3
(x − (−3)). By the first part of the Theorem, is
a line. Since both (−3, 0) and (0, 2) are easily seen to be solutions of y =
2
3
(x −(−3)),
the line passes through both of these points (−3, 0) and (0, 2). Thus both L and
47
are lines which pass through the same two points (−3, 0) and (0, 2); they have to be the
same line. It follows that L is the graph of y =
2
3
(x −(−3)).
It remains to tackle the general case. The idea of the proof is essentially the same as
in the special case. Given a (non-vertical) straight line L, we must find a linear equation
whose graph is exactly L. Since L and the y-axis are not parallel, they must meet at
some point. Let L intersect the y-axis at (0, k). Let the slope of L be m. We are going
to show that L is the graph G of the equation y = mx +k. For the sake of variety, let
us assume m is negative so that we have the following picture:
X
Y
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
k
r
L
O
Let the graph of y = mx +k be . By the first part of the Theorem, is a line. By
an observation that was made in the proof of the first part of the Theorem, the slope
of is m. Since (0, k) is obviosuly a solution of y = mx + k, also passes through
the point (0, k). Therefore and L are two lines which have the same slope m and pass
through the same point (0, k). By an earlier observation, and L are the same line. It
follows that the given line L is the graph of y = mx + k. This completes the proof of
the Theorem.
The preceding proof of the Theorem may seem long, but every piece of reasoning in
the proof will turn out to be important in subsequent considerations of linear equations
and straight lines. We proceed to justify this statement by extracting four useful facts
from the proof.
Recall that we are only dealing with equations of the form y = mx + k and lines
which are not vertical. Now that we know the graph of a linear equation y = mx + k
is a non-vertical straight line, we see from the proof of step (α) that the following holds:
(i) The graph of y = mx +k is obtained as follows: take any two solutions
(p
1
, p
2
) and (q
1
, q
2
) of the equation, then the line joining the points (p
1
, p
2
)
48
and (q
1
, q
2
) is the graph. Furthermore, the slope of the graph is m and the
graph intersects the y-axis at the point (0, k).
A second useful fact is taken from the proof of the second part of the Theorem. For its
statement, we introduce two standard concepts. If a line intersects the y-axis at (0, k),
then the number k is called the y-intercept of the line . Similarly, if intersects the
x-axis at (c, 0), then c is called its x-intercept. Observe that if a non-vertical line has
0 y-intercept, then it also has 0 x-intercept. Now the fact in question is:
(ii) The equation of a non-vertical line is y = mx + k, where m is the
slope of and k is y-intercept of .
The third fact is a direct consequence of the first two, and is usually glossed over in
the standard texts at all levels.
(iii) The lines defined by the two equations ax +by = c and a
x +b
y = c
,
are the same if and only if there is a nonzero number λ so that a
= λa,
b
= λb, and c
= λc.
We prove (iii) as follows. First of all, if there is a nonzero number λ so that a
= λa,
b
= λb, and c
= λc, then the solutions of ax + by = c and a
x + b
y = c
are
clearly identical because a
x +b
y = c
may be rewritten as λ(ax +by) = λc. Since the
line defined by a linear equation is just the set of all its solutions, the lines defined by
ax +by = c and a
x +b
y = c
have to be the same. Conversely, suppose the two lines
defined by ax +by = c and a
x +b
y = c
coincide. Let us call it . Let the y-intercept
of be µ. Rewrite ax + by = c as y = −
a
b
x +
c
b
. Then since (0, µ) is on , µ =
c
b
.
Furthermore, by (i) above, the slope of is −
a
b
. Reasoning likewise with the equation
a
x + b
y = c
, we get that µ =
c
b
and the slope of is −
a
b
. Hence we get
c
b
=
c
b
and
a
b
=
a
b
. This is the same as saying
b
b
=
c
c
and
b
b
=
a
a
A straightforward computation then shows that letting λ =
a
a
would get the job done.
In the situation of (iii), we can retrieve the equation ax + by = c of from the
equation a
x + b
y = c
by multiplying both sides by
1
λ
. For this reason, one normally
49
regards any two equations defining a line as "the same", and speaks of the defining
equation of a line.
The final and fourth fact is a consequence of (i) and (iii):
(iv) A line defined by ax + by = c with a ,= 0 and b ,= 0 has slope −
a
b
and x-intercept
c
a
.
By (iii), is also defined by y = (−
a
b
)x +
c
b
because this equation is obtained by mul-
tiplying both sides of ax + by = c with
1
b
. By (i), has slope −
a
b
. It is also obvious
that (
c
a
, 0) is a solution of y = (−
a
b
)x +
c
b
, so has x-intercept
c
a
.
We give some examples of how to write down the equation of a line.
Example 1 What is the equation of the line passing through the point (2, −1) with
slope
2
3
?
According to (ii), the equation has the form y =
2
3
x + k, but we need to find out
what k is. It is not necessary to directly compute the y-intercept. Since the line contains
(2, −1) (and since the line is the graph of y =
2
3
x + k), we know −1 =
2
3
2 + k, from
which k = −
7
3
. Thus the equation is y =
2
3
x −
7
3
.
This is the proper place to comment on a common misconception. Sometimes it is
taught in classrooms, and written up in textbooks, that the equation of this line is
y −(−1)
x −2
=
2
3
This is not correct because the point (2, −1) would not satisfy this equation as we would
have (2 −2) in the denominator on the left side. Therefore the graph of
y−(−1)
x−2
=
2
3
con-
tains every point of the line passing through the point (2, −1) with slope
2
3
except the
point (2, −1) itself. However, if we re-express this equation as y−(−1) =
2
3
(x−2), then
certainly this is an equation whose graph is the desired line. That said, we should add
that the equation
y−(−1)
x−2
=
2
3
contains the correct geometric conception of the desired
line, because what it says is that this line consists of all the points (x, y) so that the
slope of the line containing (x, y) and (2, −1) is
2
3
. The point of this comment is there-
fore that even correct thinking needs to be complemented by correct technical execution.
50
Example 2 What is the equation of the line passing through (−1, 3) and (
1
2
, 4) ?
Call this line . The slope of is
4 −3
1
2
−(−1)
=
2
3
so the equation of has the form y =
2
3
x +k, where the constant k is determined by
observing that since (−1, 3) lies on , we have 3 =
2
3
(−1) + k. Thus k =
11
3
and the
equation of is y =
2
3
x +
11
3
.
The preceding solution needs to be complemented by two remarks. The first is that,
if we use the point (
1
2
, 4) instead of the point (−1, 3) to determine k, we would get the
same equation (of course). Indeed, since (
1
2
, 4) lies on the graph of y =
2
3
x+k, we have
4 =
2
3
1
2
+ k, so that
k = 4 −
1
3
=
11
3
as before. A second remark is that, since we are given two points on to begin with,
we can use a slightly different method to obtain the equation of if we follow the
reasoning in the proof of the second part of the Theorem. We begin by computing the
slope of the line as before, obtaining
2
3
. Now we claim that the equation of , if we use
(−1, 3) as the point of reference, is
y −3 =
2
3
(x −(−1))
It is to be observed that this is indeed a linear equation in two variables because, applying
the distributive law and transposing, we may rewrite it as
2
3
x − y = −
11
3
. This said,
let L be the line defined by y −3 =
2
3
(x−(−1)). Then one verifies routinely that both
(−1, 3) and (
1
2
, 4) are solutions of y −3 =
2
3
(x −(−1)) and therefore L passes through
both of the points (−1, 3) and (
1
2
, 4). Since there is only one line passing through two
given points, is equal to L, and therefore y −3 =
2
3
(x −(−1)) is in fact the equation
of .
If we use the point (
1
2
, 4) as the point of reference, then we would write the equation
of as
y −4 =
2
3
(x −
1
2
)
It is simple to see, after multiplying both sides by (x −
1
2
) and simplifying, that this
equation is again y =
2
3
x +
11
3
.
51
In general, the equation of the line joining two given points (p
1
, p
2
) and (q
1
, q
2
) is
y −p
2
= m(x −p
1
),
where m is the slope of the line, m =
p
2
−q
2
p
1
−q
1
. Equivalently, the equation is
y −q
2
= m(x −q
1
).
The reasoning is the same as before.
It must be emphasize that none of these equations should be memorized by brute
force beyond the fact that the equation of a non-vertical line is of the form y = mx +k
for some constants m and k, where m is the slope of the line. Instead, one should get
to know the reasoning underlying these procedures and do a simple computation each
time to get at the equation. One additional comment is that, if two points are given,
the first solution using y = mx +k to get the equation of the line is the more basic of
the two methods.
Example 3 What is the x-intercept of the line joining the points (−4, 6) and (2, 1) ?
The slope of the line is
6−1
−4−2
= −
5
6
, so the equation of the line is y = −
5
6
x +k, for
some constant k. Since it contains the point (2, 1), we have 1 = −
5
6
2 + k, and so
k = 1 +
5
3
=
8
3
. The equation of the line is therefore y = −
5
6
x+
8
3
. The point where this
line intersects the x-axis has y-coordinate equal to 0; let it be (c, 0). Since (c, 0) lies on
the line, we also get 0 = −
5
6
c +
8
3
, which is the same as
5
6
c =
8
3
. Multiplying through
by
6
5
, we get c =
16
5
. So the x-intercept is
16
5
.
We conclude this section with a comment on the concept of dilation in the plane.
Let D be a dilation of the coordinate plane with center at the origin O and with scale
factor r (r > 0). If (A, B) is a point and if D maps (A, B) to another point (A
, B
),
i.e., if D(A, B) = (A
, B
), we want to show that
A
= rA and B
= rB
If we define the multiplication of a point (x, y) by a number c as
c(x, y) = (cx, cy),
52
then we can rewrite the preceding result as
D(A, B) = r (A, B)
for a dilation D with scale factor r and centered at O. The reason is as follows. Denote
the points (A, B) and (A
, B
) by P and P
, respectively. First recall the definition of
P
: on the ray
L from O to P, P
is the point so that the distance [OP
[ from O to P
is r times the distance [OP[ from O to P, i.e.,
[OP
[ = r [OP[
For simplicity, first assume both A and B are positive. Then we drop perpendiculars
PQ and P
Q
from P and P
to the x-axis, as shown.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
L
P
P
Q Q
O
Notice that the coordinates of Q are (A, 0) and those of Q
are (A
, 0). Because PQ |
P
Q
, the theorem on corresponding angles of parallel lines implies that the triangles
OPQ and OP
Q
are similar because the corresponding angles are equal. Therefore
[OQ[
[OQ
[
=
[OP[
[OP
[
But we have seen that
|OP|
|OP
|
= r, so we get
[OQ[
[OQ
[
= r,
or what is the same things, [OQ[ = r [OQ
[. But [OQ[ = A and [OQ
[ = A
, therefore
A
= rA. But using the sides PQ and P
Q
in place of OQ and OQ
, we get B
= rB
in a similar manner. This shows (A
, B
) = (rA, rB) = r (A, B), as claimed.
In case one or both of A and B is negative, consider the case A < 0 for definiteness.
Then the preceding picture beomes
53
L
P
P
Q Q
O
In this case, the only difference is that [OQ[ = −A and [OQ
[ = −A
, so that from
[OQ[ = r [OQ
[, we conclude −A = r(−A
). Then we get A = rA
as before. The proof
is complete.
EXERCISES
1. Write down three linear equations of two variables whose graphs all pass through
(−2, 1).
2. Write down three linear equations of two variables whose graphs all have slope
3
2
,
and compute the y-intercept in each case.
3. (i) Find the equation of the line joining (2, −1) and (−3, −11). What is its x-
intercept? (ii) Find the equation of the line joining (−
1
4
,
2
3
) and (5,
3
2
). What is its
y-intercept?
4. (i) What is the equation of the line with x-intercept equal to −2 and slope −
1
3
?
What is its y-intercept? (ii) What is the equation of the line with x-intercept equal to
2
5
and y-intercept equal to −
4
3
?
5.(i) Find the equation of the line passing through (1, −1) with slope −1. What is
its y-intercept? (ii) Find the equation of the line passing through (−
2
5
, 3) with slope
1
7
.
What is its x-intercept?
6. (i) What is the y-intercept of the graph of x = −5y +7? What is its slope? Does
the point (352
2
5
, −70
1
5
) lie on the graph? (ii) What is the x-intercept of the line passing
through (5, −
2
3
) and (−
4
3
,
1
2
)? What is its slope?
7. Do the graphs of 6x − 2y = 7 and
1
5
x −
1
15
y = 329 intersect? Explain why or
why not using what we have done so far.
8. (i) Let L be the line passing through (1, −2) with slope m. For which value of
m would L pass through (20, 72)? (ii) Let be the line with slope m passing through
(
1
2
,
3
4
). For which value of m would pass through (
5
3
,
1
3
)?
9. (i) Let L be the line joining (1, 2) to (p, −4), where p is some number. For what
value of p would L pass through (10, 25)? (ii) Let be the line joining (−
3
2
, 4) and (
4
5
, q),
54
where q is some number. For what value of q would pass through (2, −3)?
10. Does the line joining (3, −2) and (6, 2) contain the point (9, 6)? Explain it two
different ways.
11. Practice explaining to an eighth grader why the line joining the origin and the
point (−1, −1) is the graph of a linear equation of two variables. Do it once assuming
that the student knows the graph of a linear equation of two variables is a line, and do
it also without making that assumption. In any case, be clear about what you assume
the student knows, and make it as simple as possible.
12. Practice explaining to an eighth grader why the graph of y = x − 1 is a line.
Obviously the explanation would vary depending on how much s/he knows about similar
triangles, or whether s/he knows that isosceles right triangles have only 45 degree angles.
In any case, be clear about what you assume the student knows, and make it as simple
as possible.
5 Some Word Problems
Here are some examples of word problems involving the solution of linear equations in
one variable.
Example 1 There are 39 coins made up of quarters and pennies, and they are
worth $4.47. How many quarters are there?
We follow the practice of ¸2 and simply transcribe the information faithfully into
symbolic language before doing anything. So if there are Q quarters, then there are
39 −Q pennies. In terms of cents, we have 447 cents, of which Q 25 cents come from
the quarters and 39 −Q cents coming from the pennies. Obviously,
25Q + (39 −Q) = 447
This is a linear equation in one variable, so the technique of ¸3 allows us to solve this
easily. Transposing, we have 25Q−Q = 447 −39, therefore 24Q = 408, and therefore
Q = 17. There are 17 quarter.
One should always check: 17 quarters lead to 17 25 = 425 cents. Added to
39 −17 = 22 cents (from the pennies) does give 447 cents.
55
Example 2 Find four consecutive odd integers so that the product of the second
and fourth integers exceeds the product of the first and third integers by 64.
Let the smallest of the four odd integers be x. At the moment we do not worry about
whether x is even or odd; we just transcribe the given information and wait to see what
happens. No reason to do more than you have to! Thus the next three integers are x+2,
x +4, and x +6. The given data is that (x +2)(x +6) is bigger than x(x +4) by 64. So
(x + 2)(x + 6) −x(x + 4) = 64
The solution of this equation, which is not linear, begins with a simplification of the left
side by the use of the distributive law. We get x
2
+8x +12 −x
2
−4x = 4x +12. Thus
the equation becomes 4x+12 = 64, which is a linear equation in one variable after all.
From 4x = 52, we obtain x = 13. Thus the four integers are 13, 15, 17, 19. We check
that (15 19) −(13 17) = 285 −221 = 64.
Example 3 Break 48 into two parts so that the smaller part is
2
3
of the greater part.
Let s be the smaller part; then the greater part is 48−s. It is given that s =
2
3
(48−s).
Thus
3
2
s = 48 − s, and
5
2
s = 48. It follows that s =
96
5
= 19
1
5
. We check that the
greater part is 48 −
96
5
=
144
5
, and
96
5
=
2
3
144
5
.
The next few problems are about so-called constant rates: the constant rate of walk-
ing (which we call constant speed), the constant rate of water pouring into a tub, the
constant rate of work, such as the number of square feet a lawn is mowed, etc. Unfor-
tunately, it is not only that the concept of rate is mangled in the standard materials,
but the concept of constancy, which is central to the solution of this class of problems,
is hardly ever clearly defined and, worse, it is an assumption that is often omitted in
the formulation of such problems. Our discussion must therefore begin with precise
definitions of these concepts.
We will concentrate on speed ; it will be seen that the extrapolation of the speed
discussion to other kinds of rates is not difficult. Suppose Lisa takes a walk, and she
walks at a constant rate of 3
1
2
miles per hour. This is a common enough expression,
but what does it mean? It means at least that, the distance she walks from t = 0 hour to
t = 1 hour is 3
1
2
miles, the distance from t = 1 hour to t = 2 hours is another 3
1
2
miles,
the distance from t = 2 hour to t = 3 hours is yet another 3
1
2
miles, etc. Therefore, in t
56
hours, where t is a whole number, it is clear that she walks t 3
1
2
miles if we use the fact
that, for a whole number t, repeated addition of the number 3
1
2
t times is just t 3
1
2
.
So at least for a whole number t, if y is the number of miles Lisa walks in t hours, then
y = 3
1
2
t miles.
What happens if t is not a whole number? Let us say t =
4
3
hours. Intuitively, in
4
3
hours, Lisa should walk
4
3
of 3
1
2
miles if her speed is a constant 3
1
2
miles per
hour. From what we know about the multiplication of fractions, "
4
3
of 3
1
2
miles" is
expressed in precise mathematical terms as
4
3
3
1
2
miles. In like manner, in
m
n
hours,
where m, n are whole numbers, we expect Lisa to walk
m
n
3
1
2
miles. Such is our
intuitive understanding of the concept of constant speed.
We now formalize the intuitive expectation by defining a constant speed of 3
1
2
miles
an hour to mean: starting with time t = 0, the number of miles y traveled in t hours
is y = 3
1
2
t miles, for any positive number t. Please take note that this is a definition,
i.e., we are giving meaning to the concept of constant speed, and you should also teach
your students this piece of vital information. More generally, we say a person walks at
a constant speed of v miles per hour (mph), for a number v, if the number of miles
y that she walks after t hours is given by y = vt miles for any positive number t. An
important point here is that v is a fixed number, i.e., a constant.
At least two comments have to be made about the defining equation of y = vt of
constant speed. First, consider y = vt as a linear equation in the two numbers t and
y; its graph is a straight line passing through the origin of the TY -plane.
T
Y
y = vt
O
Observe that we do not extend the graph to the left of y-axis, because we start with
t = 0. The fact that the graph is a straight line with slope v (see preceding section)
gives an interpretation of the speed v (in the case of constant speed) in terms of the
slope of the graph of y = vt. Moroever, because the slope of this line can be computed
by using any two points on the line means that if we take any two points (s, vs) and
57
(t, vt) on the graph, where s < t, then
vt −vs
t −s
= v
T
Y
(s, vs)
(t, vt)
r
r
y = vt
O
But vt is the distance traveled from time 0 to time t, and vs is the distance traveled
from time 0 to time s. Therefore the distance traveled from time s to time t is the
numerator in the preceding equality, vt − vs. Obviously t − s is the length of the time
interval from time s to time t. Hence this equality, which is a geometric statement about
the slope of a straight line, now translates into the well-known relationship between the
distance traveled and the time duration of the travel when the speed is constant:
the distance traveled in the time interval from s to t divided by the length of
the time interval t −s for any numbers s and t, is always the speed v.
In the usual terminology of school texts, we have
speed =
distance traveled in a time interval of length T
T
,
which is valid for any time interval of length T, where T is any number. This leads to
the usual formulas: "speed is distance divided by time", "distance is speed multiplied
by time", etc. The difference between the standard treatment of these formulas and
what we have done is that we laid bare the fact that the truth of these formulas depends
on the constancy of the speed, whereas the standard treatment simply dictates a for-
mula without any explanation. Our explanation of these formulas would not have been
possible if we had not given a clear definition of "constant speed".
Naturally, the truth of
vt−vs
t−s
= v also follows from a simple algebraic fact:
vt −vs
t −s
=
v(t −s)
(t −s)
= v
58
But it is important to also understand it in terms of the geometry of the graph of y = vt.
A second comment is about the terminology of rate. The precise concept is rate of
change, and it is impossible to give a precise meaning to it without the use of calculus.
Very roughly, it means "the amount of change (whatever it is) over a time interval
divided by the length of the time interval". For example, the quotient above
vt−vs
t−s
is
an example of the "amount of change of the position of the moving object (or person)
over the time interval from s to t divided by the length of the interval t −s". Speed is
therefore the rate of change of the position of the moving object. It is tempting to say
that if this is so simple to state, what is the subtlety? It is this: "speed" as a rate of
change in the above naive sense is meaningful only because we luck out when the speed
is constant: the quotient
vt−vs
t−s
is always the same so that s and t become irrelevant.
In general, the "amount of change of the position of the moving object over the time
interval from s to t divided by the length of the interval t − s changes as either s or
t changes, so that all we get would be just what is usually called "the average rate of
change from s to t", and this would be a cumbersome concept dependent on both s and
t. Again, the constancy of the speed makes it unnecessary to make any reference to s
and t, thereby simplifying life for us.
The above complicated explanation should make it very clear that the term "rate"
as used in school mathematics is in fact never explained precisely. In mathematics,
whatever is not precisely explained should be ignored. So just ignore this word, but
understand that on the intuitive level, rate has something to do with amount-of-change-
divided-by-time, or more simply, rate is just a division of one quantity by (usually)
time length. Such an abbreviated understanding is good enough, because the real effort
should be spent on understanding the equation y = vt instead.
Example 4 Two cars A and B move at constant speed. A starts from P to Q, 150
mile apart, at the same time that B starts from Q to P. They meet at the end of 1
1
2
hours. If A moves 10 mile per hour faster than B, what are their speeds?
P Q
¸ ¡
A B
Let the speed of A be v mph. Then the speed of B is v − 10 mph. In 1
1
2
hours, A
covers 1
1
2
v miles while B covers 1
1
2
(v −10) miles. If they meet after 1
1
2
hours, then the
distances they have covered, combined, would be the same as the distance between P
59
and Q, which is 150 miles. Therefore
1
1
2
v + 1
1
2
(v −10) = 150
There are many ways to solve this equation, one of them is to first multiply both sides
by
2
3
(noting that 1
1
2
=
3
2
) to get v + (v − 10) = 100. Thus 2v = 110, and
v = 55. Therefore the speed of A is 55 mph, and the speed of B is 45 mph. We check:
3
2
55 +
3
2
45 =
165
2
+
135
2
= 150.
The discussion about speed can be carried over, verbatim, to the discussion of the
rate of water flow or the rate of getting work done (painting a house, mowing a lawn,
etc.), keeping in mind that we always idealize that these processes are evolving at a
constant rate, in the sense that water flows out of the faucet at a constant rate, houses
are painted at a constant rate, lawns are mowed at a constant rate, etc. In school math-
ematics, without the tool of calculus at our disposal, this is the only kind of problems
we can handle. With this in mind, the following are some related examples.
Example 5 Water flows out of two faucets A and B at constant rate. Suppose the
water flow from faucet A is 10 gallons per minute more than that from faucet B, and
suppose a container has a capacity of 150 gallons. If both faucets are turn on at the
same time and the container is filled in 1
1
2
minutes, what are the rates of the water
flows in both faucets?
In accordance with the preceding discussion of constant speed, the concept of a
constant-rate water flow (from a faucet) is that, for some fixed constant r, the amount
of water coming out of the faucet after t minutes is rt gallons. The number r is the so-
called rate of the water flow and it is in terms of gallone per minutes. See the preceding
interpretation of speed as a division of distance by time. Now let the rate from faucet
A be x gallons per minute. The the rate from faucet B is x − 10 gallons per minute.
After 1
1
2
minutes, the amount of water coming out of faucet A is 1
1
2
x gallons, and
that of faucet B is 1
1
2
(x −10) gallons. Since after 1
1
2
minutes, the container is filled,
the total amount of water from both faucets is equal to 150 gallons. Therefore
1
1
2
x + 1
1
2
(x −10) = 150
This is the same equation as before, and the answer is: the rate of faucet A is 55 gallons
of water per minute and that of faucet B is 45 gallons of water per minute.
60
Example 6 Karen and Lisa paint houses at a constant rate. Suppose Karen paints
10 square meters more per hour than Lisa, and suppose a wall has an area of 150 sqare
meters. If both Karen and Lisa paint this wall at the same time and they finish it in
1
1
2
hours, what are the rates at which each paints?
Let Karen paint x square meters per hour. Then Lisa paints x − 10 square meters
per hour. After 1
1
2
hours, Karen has painted 1
1
2
x square meters and Lisa, 1
1
2
(x−10)
square meters. So in 1
1
2
hours they have painted a combined area of 1
1
2
x +1
1
2
(x −10)
square meters. Since the area of the wall is 150 square meters, we have
1
1
2
x + 1
1
2
(x −10) = 150
Same equation. So Karen paints 55 square meters per hour and Lisa 45 square meters
per hour.
Example 7 Two trains A and B run at constant speed. A goes from P to Q in 2
hours while B goes from P to Q in 3 hours. If A leaves P for Q at the same time that
B leaves Q for P (on a separate but identical rail!), after how many hours will they meet?
This example is very similar to Example 4, although for this problem we do not know
the distance between P and Q. Just call it D in order to finish the transcription of the
given data into symbolic language.
D
P Q
¸ ¡
A B
As we saw in the solution of Example 4, we need the speed of both A and B to
compute the distances they have traveled before they meet. By the definition of constant
speed, the speed of A is
D
2
mph and that of B is
D
3
mph, because "speed is distance
divided by time". Let the two trains meet after x hours. Then in x hours, the distance
traveled by A is
D
2
x and that by B is
D
3
x. The total distance must be equal to the
distance between P and Q, which is D. So
D
2
x +
D
3
x = D
61
This seems to be an equation in both D and x until we realize that if we multiply both
sides by
1
D
, we'd be left with
1
2
x +
1
3
x = 1
Thus (
1
2
+
1
3
)x = 1 , and x =
6
5
hours.
Example 8 Paul and Genevi`eve walk at a constant rate. Paul walks from their
house to the train station in 30 minutes while Genevi`eve needs only 24 minutes to do
the same. Genevi`eve gives Paul a head start of 4 minutes and then she starts off. Does
she catch up with Paul, and if so, after how many minutes?
Since it takes Genevi`eve only 24 minutes to get to the station, it takes only 4+24 = 28
minutes after Paul took off before she gets to the station. But 28 minutes after Paul
took after, he is still on his way to the station. Therefore Genevi`eve overtakes him at
some point on her way to the station. We have to find out exactly when.
We can do this problem at least two ways. First, suppose Genevi`eve catches up with
Paul t minutes after her departure from the house. How far has she walked? We have
only the time (t minutes) but not her speed; we do know it takes her 24 minutes to get
to the station. It is always safe to try doing the most obvious thing, namely, if we let the
distance from the house to the station be D miles, then we get immediately her speed:
D
24
miles per minute (mpm). So in t minutes she has walked
D
24
t miles. Paul's speed is
D
30
mpm, and by the time Genevi`eve has walked t minutes, he would have walked t + 4
minutes. Therefore when Genevi`eve catches up with him, he would have walked
D
30
(t+4)
miles. Since the two will have walked the same distance at that point,
D
24
t =
D
30
(t + 4)
This again looks like an equation in the two numbers D and t, but once more the D
goes away as soon as we multiply both sides by
1
D
. So
1
24
t =
1
30
(t + 4) and therefore
30
24
t = t + 4. This leads to
1
4
t = 4 and t = 16. So 16 minutes after Genevi`eve leaves
the house, she catches up with Paul.
We can also do the same problem from the other end, i.e., the station instead of the
house. Again, suppose Genevi`eve catches up with Paul t minutes after her departure
from the house. Then in (24 −t) more minutes she would be at the station. Therefore
the distance from where she catches with Paul to the station is
D
24
(24−t) miles. For Paul,
he will have walked 4 + t minutes when Genevi`eve catches up with him, and therefore
62
he has 30 −(4 +t) = 26 −t minutes to go before he reaches the station. Since his speed
is
D
30
mpm, the remaining distance to the station can also be computed by
D
30
(26 − t).
Thus
D
24
(24 −t) =
D
30
(26 −t)
As before, we get
1
24
(24 −t) =
1
30
(26 −t), which implies (upon multiplying both sides
by 24 30) 30(24 −t) = 24(26 −t). Hence 6t = 24(30 −26) and t = 16 once again.
It is instructive to look at the graphs of the equations describing the motion of Paul
and Genevi`eve. We will use the first set of equations for illustration. For the sake of
clarity , we need to give the distance D between the house and the station a definite
value, say D = 6. If y denotes the distance from the house t minutes after Genevi`eve
leaves the house, then for her, the equation is y =
6
24
t =
1
4
t. For Paul, the equation is
y =
6
30
(t + 4) =
1
5
t +
4
5
. We now graph both linear equations in two variables,
y =
1
4
t and y =
1
5
t +
4
5
,
on the same set of coordinate axes:
T
Y
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
r
(16, 4)
1
2
3
4
5
6
0
1 12 16 24 26
Genevi`eve
Paul
The y-intercept of the graph of Paul (which is
4
5
, as we saw above) now has a graphic
interpretation: it gives his distance from the house when Genevi`eve leaves the house.
The point of intersection of the two graphs, which is (16, 4), also has an interpretation:
the x-coordinate tells the time when Genevi`eve catches up with Paul because at that
instant, both are exactly the same distance (4 miles, the y-coordinate of the point) from
the house.
This is the first time we discuss two graphs on the same set of axes simultaneously.
We will pursue this discussion vigorously in the next section.
63
EXERCISES
1. A man has six hours at his disposal. How far can he ride in a car going at a
constant speed of 25 mph if he has to ride a bicycle back at the rate of 6 mph?
2. A train loses
1
6
of its passengers at the first stop, 25 at the second, 20% of the
remainder at the third, three quarters of the remainder at the fourth; 25 remain, What
was the original number?
3. Water flows out of two faucets A and B at constant rate. Faucet A fills a given
container in 5 minutes, while faucet B fills it in 6 minutes. How long would it take to
fill the container if both faucets are turned on at the same time?
4. The numerator of a fraction is 7 less than the denominator. If 4 is subtracted
from the numerator and 1 added to the denominator, the resulting fraction equals
1
3
.
What is the fraction?
5. A had twice as much money as B, but after giving B $28, he has
2
3
as much as
B. How much did each have at first?
6. Find two numbers whose sum is
7
6
and difference
1
6
.
7. Lisa and Karen mow lawns at a constant rate. Lisa finishes mowing a certain
lawn in 4
1
2
hours, and Karen does it in 3 hours. If they mow the same lawn together,
how long will it take them to finish it?
8. There are two heaps of coins, one containing nickels and the other dimes. The
second heap is worth 20 cents more than the first, and has 8 fewer coins. Find the
number in each heap.
9. If A has $566 and B $370, how much money must A give B in order that B may
have
4
5
as much as A?
10. A tank with a capacity of 150 gallons can be filled by one pipe in 15 minutes,
and emptied by another in 25 minutes. After the first pipe has been open a certain
number of minutes, it is closed, and the second pipe opened. The tank is then found to
be empty 24 minutes after the first pipe was open. Assuming that water flow is always
at a constant rate, how many minutes is each pipe open?
11. A train running at 30 mph requires 21 minutes longer to go a certain distance
than does a train running at 36 mph. How great is the distance?
12. A woman drives a car for 3
1
2
hours and she finds that she has covered a distance
of 130 miles. If she drives at a constant speed of 45 mph in the country and 20 mph
within city limits, how many miles of her trip is in the country?
13. Paul can mow a certain lawn by all himself in 11 hours. After working for 2
1
2
64
hours, however, Paul is joined by Henry and the two together finish mowing the lawn in
another 5 hours. Assume as always that both mow the lawn at constant rate, how long
would it take Henry to mow the lawn alone? Explain clearly how you get the solution.
14. Aaron and Bill together can mow a lawn in 36 minutes. If Aaron mows it alone,
it takes 58 minutes. Suppose the lawn is 280 square yards, and suppose both Bill and
Aaron mow the lawn at a constant rate, how long would it take Bill to mow it alone?
15. Driving from A to B at a constant speed of 45 mph is 25 minutes faster than
doing it at 39 mph. How great is the distance from A to B?
16. A solution consisting of water and alcohol has 70% alcohol. If 25 cc of water is
added to the solution, how much alcohol must be added in order for the solution to still
contain 70% alcohol?
17. Aaron, Bill, and Carl all mow a given lawn at a constant rate. If it is mowed
individually, they can finish it in A, B, C hours, respectively. If Aaron and Bill mow
it together, they finish mowing it in 2
2
9
hours, and if Aaron and Carl mow it together,
they finish mowing it in 2
2
3
hours. However, if all three mow it together, they finish it in
1
17
23
hours. Write equations in terms of A, B, C to capture the above information, and
explain how you arrive at these equations.
18. Fifteen minutes after Colin leaves for school, his mother discovers that he forgot
to take his homework. She drives at a constant rate and it takes her 6 minutes to get
to school. Colin walks to school at a constant rate and it takes him 24 minutes to get
there. Use mental math to decides if Colin's mother can catch up with him, and if she
does, compute how soon this happens after Colin leaves.
6 Simultaneous Linear Equations
In Example 8 of the preceding section, we found ourselves looking at the graphs of two
linear equations in two variables. We saw that the point of intersection of the graphs
corresponds to the solution of the problem. In this section, we inspect more closely the
situation of two linear equations.
Recall that a solution to a linear equation such as 4x+5y = −3 is a pair of numbers
(x
0
, y
0
) so that 4x
0
+5y
0
= −3. For example, (0, −
3
5
) is a solution. There are an infinite
number of solutions to a linear equation of two variables and their totality constitutes
the graph of the equation, which is a straight line. Still with 4x+5y = −3, suppose we
also consider another equation −2x+y = 5 and ask if there could be a pair of numbers
65
(x
0
, y
0
) so that it is a solution of both 4x + 5y = −3 and −2x + y = 5. Indeed there
is, for example, (−2, 1), as it is easy to check. We say the pair of linear equations
_
4x + 5y = −3
−2x + y = 5
is a system of linear equations, or sometimes, simultaneous (linear) equations in
the numbers x and y. To be precise, one would have to refer to such a pair of equations
as a linear system of two equations in two unknowns or two variables. As
in the case of a single equation in one variable, implicit in the writing down of such
a system is the claim that the given number expressions involving x and y are equal,
and we have to check whether the equality holds for all x and y, some x and y, or no
x and y. This implicit statement will be taken for granted and will not be repeated in
subsequent discussions. To solve the system is to find all the ordered pair of numbers
(x
0
, y
0
) which are solution of both equations. Such an (x
0
, y
0
) is called a solution of
the system. Sometimes we also call the collection of all these (x
0
, y
0
)'s the solution
of the system. Thus (−2, 1) is a solution of the above system. A priori, there may be
others, but it will turn out that the solution of the system consists only of (−2, 1). At
present we are only concerned with systems consisting of two equations in two variables,
but note that there will be occasion to consider systems of equations consisting of many
equations in any number of unknowns.
Postponing for the moment the discussion of how to get a solution such as (−2, 1)
to the above system, let us first give a geometric interpretation of this (−2, 1). Now
(−2, 1), being a solution of the first equation 4x +5y = −3, lies on the line defined by
4x + 5y = −3. Similarly, (−2, 1) also lies on the line defined by the second equation
−2x + y = 5. This means the solution (−2, 1) is the point of intersection of the two
lines defined by the equations in the linear system, as shown:
X
Y
`
`
`
`
`
`
`
r
O
−2
1
−2x + y = 5
4x + 5y = −3
66
Conversely, suppose (A, B) is not at the intersection of these two lines, then it cannot
be a solution of the linear system because if (for example) the point (A, B) is not on the
line defined by −2x + y = 5, then −2A + B ,= 5, and (A, B) is not a solution of both
equations. Therefore, the point of intersection of the lines defined by the equations of
the linear system is the exact solution of the linear system.
This reasoning is perfectly general. Suppose we are given a linear system Suppose (x
0
, y
0
) is a solution of the system. We wish
to interpret this solution geometrically. Since (x
0
, y
0
) is a solution of ax + by = e, the
point (x
0
, y
0
) lies on
1
. For the same reason, (x
0
, y
0
) lies on
2
as well. Therefore (x
0
, y
0
)
is the point of intersesction of
1
and
2
. If a point (A, B) is not at the intersection of
these two lines, then let us say (A, B) does not lie on
1
. By definition of
1
, aA+bB ,= e
and the ordered pair of numbers (A, B) is not a solution of the equation ax + by = e
and therefore not a solution of the linear system either. Thus, if a linear system has a
solution, it is the point of intersection of the lines defined by the equations of the system,
and conversely, a point in the plane that is not on the intersection of these lines cannot
be a solution of the linear system. In particular, inasmuch as two lines intersect at at
most one point, a linear system of two equations in two unknowns has at most one
solution.
We have just given the precise reasoning why there is a one-one correspondence be-
tween the solution of a linear system of two linear equations in two unknowns and the
point of intersection of the two lines defined by the linear equations of the system. Such a
correspondence is usually decreed by fiat in standard texts, probably because the precise
definition of the graph of an equation is rarely given or, if given, is not put to use. It is
very important that you learn to make use of definitions in your teaching. In particular,
please do not forget to explain why the solution of a linear system can be obtained from
the intersection of the lines.
Next, how to get the solution (−2, 1)? We adopt the time-honored strategy of first
assuming that there is a solution to the given linear system and use this information to
find out what it has to be. Then we turn around to verify that the presumptive solution
is indeed a solution of the linear system. The reasoning given below is perfectly general,
67
but we choose to explain it in terms of the concrete linear system above, namely,
_
4x + 5y = −3
−2x + y = 5
(1)
We want to shown that if there is an ordered pair of numbers (x, y) satisfying the
system (1), then (x, y) = (−2, 1) in the sense that x = −2 and y = 1. We begin by
multiplying both sides of the second equation in (1) by −2 with a view to making the
coefficient of x in this equation equal to the coefficient of x in the first equation, i.e., 4.
We get:
_
4x + 5y = −3
4x − 2y = −10
Now rewrite the system as
_
4x = −5y − 3
4x = 2y − 10
(2)
We emphasize that systems (1) and (2) are equivalent in the sense that every solution
of one of them is a solution of the other; the validity of this assertion should be clear
after a moment's reflection. Now, since both number expressions −5y −3 and 2y −10
are equal to the number 4x, we have
−5y −3 = 2y −10 (3)
We can solve this linear equation in y, obtaining easily y = 1. This leads to the
important consequence that when y = 1, (3) is true and therefore both equations in (2)
become identical linear equations in x, or what is the same, when y = 1, both equations
of the system in (1) are identical linear equations in x, so that a solution of one equation
(in the number x) in (1) is automatically a solution of the other equation (in x). We
can be more explicit in the present case. When y = 1, the system (1) becomes:
_
4x = −8
−2x = 4
(4)
and both equations are of course the same. So we get y = 1 and x = −2, as claimed.
We should emphasize that thus far, what we have shown is merely that if a solution
exists for the system in (1), it would have to be (−2, 1). This is not the same as
saying that (−2, 1) is a solution of (1). Now we proceed to fill this gap by showing that
(−2, 1) is a solution. In one sense, this is trivial: letting x = −2 and y = 1 in the
equations in (1), we get −3 = −3 and 5 = 5, so that is that. But this is unsatisfactory
because this simple numerical verification does not reveal why (−2, 1) is a solution and,
68
more importantly, it does nothing to give us confidence that this method of obtaining
a solution (via equations (1) through (4)) also works in other situations. So we do it
differently. We begin by analyzing the concept of a solution to a system. The meaning
of (−2, 1) being a solution of the system (1) is that when y = 1 in (1), then the two
solutions of the two linear equations in x turn out to be the same (in fact, equal to −2,
as we know). To drive home this point, if y is equal to 2 instead of 1, then the system
(1) becomes
_
4x + 10 = −3
−2x + 2 = 5
The first equation yields x = −
13
4
and the second equation yields x = −
3
2
. So the
ordered pair (−
13
4
, 2) solves the first equation of (1) but not the second, and the ordered
pair (−
3
2
, 2) solves the second equation but not the first. In short, there is no solution
of (1) which has y = 2.
It remains to explain why, if y = 1, both equations in x of the system in (1) yield
the same solution. This is because y = 1 is the solution of equation (3), and therefore
both equations in (2) are identical when y = 1. Since systems (1) and (2) are equivalent,
we see that both equations in (1) become identical linear equations in x when y = 1.
Of course that solution is x = −2, as claimed.
We have written out the method of getting (−2, 1) via the steps associated with
(2)–(4) to facilitate the explanation of why (−2, 1) is a solution. In practice, however,
one achieves some simplificaton by skirting (2), as follows. Working directly with
_
4x + 5y = −3
4x − 2y = −10
we subtract both sides of the second equation from the corresponding sides of the first
(or, to conform with the basic principle enunciated in ¸3, we add the negatives of both
sides of the second equation to the corresponding sides of the first), obtaining
5y −(−2y) = (−3) −(−10) (5)
This leads to y = 1 as before. More important is the fact that equation (5), if we
transpose 5y and −10 to the opposite sides, is the same equation as (3). Therefore,
this way of "bringing the coefficient of the number x in both equations to be the same
and then eliminate x by subtraction" achieves the same result as the method of solution
in (2)–(4). This "subtraction method" is what is usually done to solve simultaneous
69
equations.
We bring closure to this discussion by revisiting Example 8 of the last section. In
retrospect, we had arrived at a linear system in the numbers t and y in that Example:
_
y =
1
4
t
y =
1
5
t +
4
5
(6)
This is a linear system the minute it is rewritten as
_
−
1
4
t + y = 0
−
1
5
t + y =
4
5
When the system (6) is given as is, there is no question as to the fastest method to solve
it: since both
1
4
t and
1
5
t +
4
5
are equal to the number y, we get
1
4
t =
1
5
t +
4
5
,
so that
1
20
=
4
5
, and t = 16. The first equation of (6) then gives y = 4. Thus (16, 4)
is the solution to (6), which was already obtained earlier by another method having
nothing to do with linear systems. Moreover, the fact that the point (16, 4) is the point
of intersection of the two lines defined by the equations in (6) was already pointed out
in Example 8.
It is time to point out that not every linear system of two equations in two unknowns
has a solution. For example, obviously the system
_
x + 0 y = 1
x + 0 y = 2
(7)
has no solution. To achieve a better understanding of this phenomenon, we have to
introduced a basic property of a pair of lines in the plane.
Theorem 1 Two distinct, non-vertical lines in the plane are parallel if and only if
they have the same slope.
Proof Let the lines be
1
and
2
. We first assume that they are parallel and prove
that they have the same slope. If either of
1
and
2
is horizontal, then since
1
|
2
, the
other is also horizontal and both would have 0 slope. There would be nothing to prove
70
in this case. So we may assume both
1
and
2
are not horizontal. Take a point P on
1
and let a vertical line through P intersect
2
at Q. (This vertical line must intersect
2
because the latter is not vertical.) Since the lines are distinct, P ,= Q. Go along this
vertical line from P to Q and stop at a point R so that [PQ[ = [QR[. From Q and R,
draw horizontal lines which meet
1
and
2
at S and T, respectively.
Y
X
OP
Q
R T
S
2
1
Because ´PQS and ´QRT are right triangles with legs parallel to the coordinate
axes, the slopes of
1
and
2
are
[PQ[
[SQ[
and
[QR[
[TR[
respectively. We have to show that these two numbers are equal. We do so by showing
that ´PQS and ´QRT are congruent. Once this is done, then [SQ[ = [TR[ because
they are corresponding sides of congruent triangles. Since [PQ[ = [QR[ by construction,
we get
[PQ[
[QR[
=
[SQ[
[TR[
because both are equal to 1. By the cross-multiplication algorithm, this implies
[PQ[
[SQ[
=
[QR[
[TR[
which is what we are after. So it remains to prove that the triangles are congruent.
One way is to translate the whole plane in the direction from P to Q. Call this
translation τ. So by definition, τ(P) = Q. Now translation moves a line (not parallel
71
to the direction of the translation) to a line parallel to itself. Thus τ(
1
) is a line passing
through Q and parallel to
1
. Since
2
is already such a line, the Parallel Postulate says
τ(
1
) =
2
. Now consider the line SQ. By construction, SQ | TR. Since τ(Q) = R
because by construction [PQ[ = [QR[, the same reasoning therefore yields the fact that
the line τ(SQ) coincides with the line TR. But S is the intersection of the lines
1
and
SQ, while T is the intersection of the lines
2
and TR. Hence the fact that τ(
1
) =
2
and τ(SQ) = TR means that τ(S) = T. This says the translation τ maps P to Q, Q
to R, and S to T. By definition, ´PQS
∼
= ´QRT.
A more traditional argument is as follows. Observe that ∠SPQ and ∠TQR are equal
because they are corresponding angles of the parallel lines
1
and
2
with respect to the
transversal PR. Of course ∠PQS and ∠QRT are equal because they are both right
angles. The equality [PQ[ = [QR[ (which results from the construction OF R) then
shows that ´PQS
∼
= ´QRT because of the ASA criterion of congruence. In any case,
this shows that nonvertical parallel lines have the same slope.
Conversely, suppose two distinct, non-vertical lines
1
and
2
have the same slope,
and we have to show that they are parallel. We give two proofs: the first one being
a direct continuation of the preceding line of geometric reasoning, and the second an
algebraic one. First, if they have slope 0, then they are horizontal and are therefore
parallel. We may therefore assume that they have nonzero slope so that they are both
non-horizontal. We now perform the same construction as before to get triangles ´PQS
and ´QRT. The fact that
1
and
2
have the same slope then implies that
[PQ[
[SQ[
=
[QR[
[TR[
These quotients have equal numerators because we constructed the point R so that
[PQ[ = [QR[. The denominators are consequently equal as well, i.e., [SQ[ = [TR[.
Again, we let τ be the translation from P to Q, then τ(P) = Q by the definition
of τ, and τ(Q) = R because [PQ[ = [QR[. As above, the fact that SQ | TR
and the Parallel Postulate imply that τ maps the line SQ to the line TR. Since also
[SQ[ = [TR[, we see that, necessarily, τ(S) = T. Recall that already τ(P) = Q. This
means τ maps the line
1
to the line
2
. Because translation maps a line to another line
parallel to itself, we conclude that
1
|
2
.
A more traditional proof would run as follows. Since ∠PQS and ∠QRT are equal
because they are both right angles, the triangles ´PQS and ´QRT are congruent by
the SAS criterion of congruence. Their corresponding angles ∠SPQ and ∠TQR are
therefore equal. This implies
1
|
2
because their corresponding angles relative to the
72
transversal PR are equal. The proof of Theorem 1 is complete.
Here is a second algebraic proof. Since
1
and
2
are both nonvertical, say they have
slope m. Then let the equations defining them be y = mx + k and y = mx + k
,
respectively, where k ,= k
because by assumption the lines are distinct. Suppose they
intersect at a point (x
0
, y
0
), then y
0
= mx
0
+k and y
0
= mx
0
+k
, which then imply
that mx
0
+ k = mx
0
+ k
, which in turn implies that k = k
. This is a contradiction
to the earlier conclusion that k ,= k
. Thus the proof is concluded again.
Now we can analyze when a linear system of two equations in two unknowns has a
solution. Let the system be There are several cases to consider.
Case 1. b = d = 0. Then by the definition of a linear equation of two variables,
a ,= 0 and c ,= 0, and the system may be rewritten as
_
x =
e
a
x =
f
c
In this case, the system has an infinite number of solutions if
e
a
=
f
c
, and the solutions
are all ordered pairs of the form (
e
a
, y) for any number y. The system has no solution
if
e
a
,=
f
c
. These assertions should be understood in terms of the graphs
1
and
2
.
These are two vertical lines passing through (
e
a
, 0) and (
f
c
, 0). If
e
a
=
f
c
, these lines
are identical, and therefore any point on the line is a solution of both equations, and
therefore of the system. If however
e
a
,=
f
c
, then
1
and
2
are distinct vertical lines; they
are parallel and therefore have no intersection. By an earlier remark in this section, the
linear system has no solution.
Case 2. Not both b = d = 0. Let us say, b ,= 0. The consideration then breaks up
into two sub-cases.
Case 2a. d = 0. Then the graph
2
is a vertical line. Since b ,= 0,
1
is not
vertical and so
1
and
2
will intersect at one point. The system therefore has exactly
one solution corresponding to the point of intersection.
Case 2b. d ,= 0. Then the system may be rewritten as
_
y = m
1
x + k
1
y = m
2
x + k
2
73
where m
1
= −
a
b
, k
1
=
e
b
, m
2
= −
c
d
, and finally, k
2
=
f
d
. Now both
1
and
2
are
non-vertical line. If
a
b
,=
c
d
, the lines
1
and
2
are not parallel according to Theorem 1,
and they will intersect so that the system will have exactly one solution. If
a
b
=
c
d
, then
the lines
1
and
2
are parallel or indentical, according to whether
e
b
,=
f
d
or
e
b
=
f
d
.
If the former,
1
and
2
do not intersect and the system has no solution. If the latter,
then the lines
1
and
2
coincide and the system will have an infinite number of solutions
corresponding to each point on the line.
We can summarize all this into one theorem.
Theorem 2 Given a linear system
_
ax + by = e
cx + dy = f
where a, b, . . . f are constants. Then:
(i) If b = d = 0, the system has an infinite number of solutions if
e
a
=
f
c
, and has no
solution if
e
a
,=
f
c
.
(ii) If b ,= 0 but d = 0, the system has exactly one solution.
(iii) If d ,= 0 but b = 0, the system has exactly one solution.
(iv) If b ,= 0 and d ,= 0, then:
the system has exactly one solution if
a
b
,=
c
d
,
the system has no solution if
a
b
=
c
d
, but
e
b
,=
f
d
, and
the system has an infinite number of solutions if both
a
b
=
c
d
and
e
b
=
f
d
.
We now give two applications of simultaneous equations. The first one is to express
certain rational expressions in a number x as a sum of simpler rational expressions also
in x. Consider the simple sum:
5
x −2
+
4
x + 3
=
5(x + 3) + 4(x −2)
(x −2)(x + 3)
After simplifying the numerator of the right side and multiplying out (x − 2)(x + 3) =
x
2
+ x −6, we get the identity
5
x −2
+
4
x + 3
=
9x + 7
x
2
+ x −6
(8)
This is straightforward. Things get interesting, however, if you happen not to know
identity (8) ahead of time but ask whether
9x+7
x
2
+x−6
can be expressed as a sum of
74
(constant) multiples of the simple rational expressions
1
x−2
and
1
x+3
. In general terms
this question may be understood as part of our overall desire to express complicated
objects in terms of simpler ones (think of the prime decomposition of a whole number,
for example). On a more concrete level, this question arises naturally in calculus, and is
a special case of the partial fraction decomposition of a rational expression. The answer
to this question is by no means obvious, for two reasons. One is that even knowing
x
2
+x−6 = (x−2)(x+3) ahead of time, one would be inclined to believe that
9x+7
x
2
+x−6
is a sum of more complicated rational expressions such as
ax+b
x−2
and
cx+d
x+3
, for some
appropriately chosen constants a, b, c, d, rather than just a sum of
5
x−2
and
4
x+3
without any x's in the numerators. The other reason is that even if you believe that
such an expression is possible, there remains the question of how to get the precise values
of the numerators, i.e., 4 and 5.
In order to answer this question, we have to first quote two facts without proof; the
proofs are not difficult but they do take up time and space that we cannot afford at this
point.
(A) Let (a
1
x +b
1
), (a
2
x +b
2
), . . . (a
n
x +b
n
) be n linear polynomials in x
(n is a positive integer) so that none is a constant multiple of another. Let
p(x) be a polynomial in x of degree less than n. Then there are constants
c
1
, c
2
, . . . c
n
so that
p(x)
(a
1
x + b
1
)(a
2
x + b
2
) (a
n
x + b
n
)
=
c
1
a
1
x + b
1
+
c
2
a
2
x + b
2
+ +
c
n
a
n
x + b
n
(B) Suppose the following two n-th degree polynomials in x (n is a postive
integer) are equal for all x:
a
n
x
n
+ a
n−1
x
n−1
+ + a
1
x + a
0
= b
n
x
n
+ b
n−1
x
n−1
+ + b
1
x + b
0
Then the coefficients of the polynomials are pairwise equal: a
n
= b
n
, a
n−1
=
b
n−1
, . . . a
0
= b
0
.
Using Fact (A), we see that there must be constants a and b so that
9x + 7
(x −2)(x + 3)
=
a
x −2
+
b
x + 3
which is valid no matter what x may be. We now use Fact (B) to recover equation
(8), i.e., to obtain the values of the constants a and b as 5 and 4, respectively. By the
75
addition of rational expressions,
a
x −2
+
b
x + 3
=
a(x + 3) + b(x −2)
(x −2)(x + 3)
=
(a + b)x + (3a −2b)
(x −2)(x + 3)
Combining the two equalities, we get
9x + 7
(x −2)(x + 3)
=
(a + b)x + (3a −2b)
(x −2)(x + 3)
and therefore the numerators must be equal. Hence,
9x + 7 = (a + b)x + (3a −2b)
for all x. From Fact (B), we know that the coefficients a + b and 3a − 2b must be
equal to 9 and 7, respectively. In other words, we have the following simultaneous linear
equations in a and b:
_
a + b = 9
3a − 2b = 7
We can solve this system by simply multiplying the first equation by 2 and then adding
it to the second equation. This yields 5a = 25 and therefore a = 5. The first equation
now gives b = 4, as claimed.
The method is clearly sufficient to deal with the general situation.
We now give the second application of simultaneous linear equations. We say three
positive integers a, b, c form a Pythagorean triple ¦a, b, c¦ if a
2
+ b
2
= c
2
.
In other words, a, b and c are the lengths of three sides of a right triangle. Note that
the third member of a Pythagorean triple is, by definition, the length of the hypotenuse
of the right triangle. It goes without saying that the key point of the definition of a
Pythagorean triple is that all three numbers are positive integers. Everybody knows
that 3, 4, 5 form a Pythagorean triple; some may even know that ¦5, 12, 13¦ is
another Pythagorean triple, or even that ¦8, 15, 17¦ is yet another example. But are
there others?
Our purpose is to produce Pythagorean triples at will by solving an extremely sim-
ple linear system of equations. It will be obvious that we will get an infinite number
of Pythagorean triples by this method. It is even true that the method produces all
the Pythagorean triples, though we will not prove this fact here. One would like to
76
say that this method is due to the Babylonians some thirty-eight centuries ago (circa
1800 B.C.), but a more accurate statement would be that it is the algebraic rendition of
the method one infers from a close reading of the celebrated cuneiform tablet, Plimp-
ton 322, which lists fifteen Pythagorean triples.
9
See Eleanor Robson, Neither Sherlock
Holmes nor Babylon: A reassessment of Plimpton 322, Historia Mathematica, 28 (2001),
167-206.
Let us first perform a conceptual simplification. Take ¦3, 4, 5¦, for example. Once
we are in possession of this triple, we will in fact be in possession of an infinite number
of Pythagorean triples, namely, ¦6, 8, 10¦, ¦9, 12, 15¦, ¦12, 16, 20¦, and in general,
¦3n, 4n, 5n¦ for any positive integer n. Clearly, if you already have the Pythagorean
triple ¦3, 4, 5¦, there is not much glory in claiming that you also have another
Pythagorean triple, namely, ¦6, 8, 10¦. Accordingly, we define a Pythagorean triple
¦a, b, c¦ to be primitive if the integers a, b, and c have no common divisor other than
1 (i.e., if k is a positive integer that divides all three a, b and c, then k = 1), and will
henceforth concentrate on getting primitive Pythagorean triples. We say a Pythagorean
triple ¦a, b, c¦ is a multiple of another Pythagorean triple ¦a
, b
, c
¦ if there is
a positive integer n so that a = na
, b = nb
, and c = nc
. In this terminology, a
given Pythagorean triple is either primitive, or is a multiple of a primitive Pythagorean
triple. Therefore whenever a Pythagorean triple is given, we lose nothing by replacing
it with the primitive Pythagorean triple of which the first Pythagorean triple is a mul-
tiple. Therefore, instead of dealing with ¦15, 36, 39¦, we will replace it by ¦5, 12, 13¦.
We will give a proof of the following theorem. Observe that its statement makes
use of the fact that any two fractions can be written as two fractions with the same
denominator (fundamental fact of fraction pairs).
Theorem 3 Let (u, v) be the solution of the linear system
_
¸
¸
_
¸
¸
_
u + v =
t
s
u − v =
s
t
where s, t are positive integers with s < t. If we write u and v as two fractions with the
9
Lest you entertain for even a split second the idle thought that people couldn't have known such
advanced mathematics thirty-eight centuries ago and that these triples were probably hit upon by trials
and errors, let it be noted that the largest triple is ¦12709, 13500, 18541¦.
77
same denominator, u =
c
b
and v =
a
b
, then ¦a, b, c¦ form a Pythagorean triple.
There is extra incentive in providing a proof of this theorem, not only because the
proof is very simple, but also because it actually tells us why the solution (u, v) of the
linear system furnishes a Pythagorean triples. So with (u, v) as the solution of the linear
system, we multiply the corresponding sides of the two equations in the theorem to get
(u + v)(u −v) =
t
s
s
t
, or u
2
−v
2
= 1. So with u =
c
b
and v =
a
b
, we have
_
c
b
_
2
−
_
a
b
_
2
= 1
Multiplying through both sides of this equality by b
2
gives c
2
−a
2
= b
2
and therefore,
a
2
+ b
2
= c
2
We have our Pythagorean triple and the proof of Theorem 3 is complete.
It is easy to explain how Theorem 3 came about, but before doing that, let us put it
to use to produce some new Pythagorean triples.
Example 1 Consider
_
u + v = 2
u − v =
1
2
Adding the equations gives 2u =
5
2
so that u =
5
4
. From the second equation, we get
v = u −
1
2
=
5
4
−
1
2
=
3
4
. Thus we have retrieved the grandfather of all Pythagorean
triples, ¦3, 4, 5¦.
Example 2 Consider
_
u + v =
3
2
u − v =
2
3
Adding the equations gives 2u =
13
6
so that u =
13
12
. From the second equation, we get
v = u −
2
3
=
13
12
−
2
3
=
5
12
. By Theorem 3, ¦5, 12, 13¦ is a Pythagorean triple, which
of course we already know.
Example 3 Consider
_
u + v =
4
3
u − v =
3
4
Adding the equations gives 2u =
25
12
so that u =
25
24
and the second equation gives
v =
25
24
−
3
4
=
7
24
. By Theorem 3, ¦7, 24, 25¦ is a Pythagorean triple. Since this is new
78
to most people, one should check directly that 7
2
+ 24
2
= 25
2
.
Example 4 Consider
_
u + v =
69
2
u − v =
2
69
Adding the two equations gives 2u =
4765
138
so that u =
4765
276
. From the second equa-
tion, we get v =
4765
276
−
2
69
=
4757
276
. This time we get an unfamiliar Pythagorean triple
¦276, 4757, 4765¦. Although Theorem 2 guarantees that this is indeed a Pythagorean
triple, it would be good for your soul to directly check that 276
2
+4757
2
= 4765
2
is in
fact true.
Observe that thus far, every single Pythagorean triple has been primitive. Now
consider:
Example 5 Consider
_
u + v = 5
u − v =
1
5
Adding the equations, we obtain 2u =
26
5
and multiplying both sides by
1
2
gives u =
26
10
.
From the second equation of the system, we get v =
26
10
−
1
5
=
24
10
. By Theorem 3,
¦10, 24, 26¦ is a Pythagorean triple. This is not a primitive triple because it is a mul-
tiple of ¦5, 12, 13¦, which is clearly primitive.
Observe, however, that if we had taken the trouble to reduce u =
26
10
to its lowest
terms, then we would obtain u =
13
5
, and then we would get v =
13
5
−
1
5
=
12
5
, and the
primitive triple ¦5, 12, 13¦ would be the result. Thus we see that different values of
s and t do not always lead to distinct primitive Pythagorean triples.
We now explain the genesis of Theorem 3. We will follow the time-honored method
of assuming that we already have a Pythagorean triple ¦a, b, c¦, and proceed to find
out what it must be. By assumption, a
2
+ b
2
= c
2
, and by dividing through by b
2
, we
get (a/b)
2
+ 1 = (c/b)
2
. We now adopt the convention of letting
u =
c
b
and v =
a
b
(9)
Thus u and v are both fractions (i.e., positive rational numbers) and, by our convention
of letting c be the length of the hypotenuse of the right triangle, we always have u > v.
Now v
2
+ 1 = u
2
and therefore,
u
2
−v
2
= 1
79
Since u
2
−v
2
= (u +v)(u −v), we get (u +v)(u −v) = 1. But we are assuming a, b, c
to be known quantities, so both u and v are also known quantities and therefore so are
u + v and u −v. Let s, t be positive integers with s < t so that u + v =
t
s
. (We are
letting u + v be a fraction bigger than 1 because (u + v)(u −v) = 1 and therefore one
of u +v and u −v is greater than 1 and the other is less than 1. Clearly, u +v > u −v,
so u + v > 1.) Because (u + v)(u −v) = 1, necessarily, u −v =
s
t
. Therefore we have:
_
¸
¸
_
¸
¸
_
u + v =
t
s
u − v =
s
t
(10)
where, we recall, the s and t are positive integers with s < t. We may regard this
system as a system of linear equations in the variables u and v, and it is exactly the
system in the statement of Theorem 3. From this point of view, Theorem 3 becomes
inevitable.
We give a refinement of Theorem 3 by directly solving system (10). Adding the two
equations, we get 2u =
t
s
+
s
t
=
t
2
+s
2
st
, and therefore
u =
t
2
+ s
2
2st
From the second equation of (10), we then obtain
v = u −
s
t
=
s
2
+ t
2
2st
−
s
t
=
s
2
+ t
2
2st
−
2s
2
2st
=
t
2
−s
2
2st
so that
v =
t
2
−s
2
2st
Since u
2
−v
2
= 1, we have
_
t
2
+ s
2
2st
_
2
−
_
t
2
−s
2
2st
_
2
= 1
Multiplying both sides by (2st)
2
, we get
(t
2
+ s
2
)
2
−(t
2
−s
2
)
2
= (2st)
2
or,
(s
2
−t
2
)
2
+ (2st)
2
= (s
2
+ t
2
)
2
(Compare item 2 in the Exercises of Section 1.) This shows that if s, t are positive integers
and t > s, then
¦2st, t
2
−s
2
, t
2
+ s
2
¦ is a Pythagorean triple
80
We have therefore presented two ways of obtaining Pythagorean triples: by giving values
of s and t into the preceding formula, or by using Theorem 2 and solving the linear system
there. Of course the former is a consequence of the latter, but for school mathematics,
the latter is more instructive.
With a little more work, one can prove that if s and t are relatively prime (i.e., no
common divisor other than ±1), and if one of them is even and the other odd, then the
triple ¦2st, t
2
−s
2
, t
2
+s
2
¦ is primitive. With more work still, it can be shown that ev-
ery primitive Pythagorean triple is represented in terms of suitable s and t in this manner.
EXERCISES
1. Solve:
(a)
_
7x −3y = 10
3x −5y = −5
(b)
_
7x −9y = 15
8y −5x = −17
(c)
_
2
5
x −
5
6
y = −
1
2
1
6
x +
5
9
y =
5
2
(d)
_
12x + 11y = 172
28x −17y = 60
(e)
_
0.08x + 0.9y = 0.46
0.1x −0.04y = 0.16
(f)
_
5x −
3
4
y = 2
x + 2y =
11
6
2. Solve:
(a)
_
_
_
6
x
+
12
y
= −1
8
x
−
9
y
= 7
(b)
_
_
_
9
x
−3y = 4
3
x
+ 2y =
10
3
3. Alan's age is
6
5
of Bill's, and 15 years ago, his age was
13
10
of Bill's. Find their ages.
4. If 3 is added to the numerator of a fraction and 7 subtracted from the denominator,
its value is
6
7
. But if 1 is subtracted from the numerator and 7 added to the denominator,
its value is
2
5
. Find the fraction.
5. The contents of one barrel is
5
6
wine, and of another
8
9
wine. How many gallons
must be taken from each to fill another barrel whose capacity is 24 gallons, so that the
mixture may be
7
8
wine?
6. Express
8x+2
x
2
−1
as a sum of multiples of
1
x+1
and
1
x−1
.
7. Express
−5(x+1)
3(x
2
+x−12)
as a sum of multiples of
1
x+2
and
1
x−3
.
8. In each of the following, you are asked to solve the linear system in Theorem 2
with the given values of s and t to obtain Pythagorean triples. You may use a scientific
calculator, especially for (i) and (j) below.
(a) s = 2, t = 5.
(b) s = 4, t = 5.
81
(c) s = 1, t = 4.
(d) s = 1, t = 3.
(e) s = 3, t = 13.
(f) s = 1, t = 12.
(g) s = 3, t = 13.
(h) s = 1, t = 6.
(i) s = 2, t = 69.
(j) s = 54, t = 125.
(k) s = 8, t = 9907.
9. In (k) of the the last problem, the largest number in the Pythagorean triple
has 8 digits. Suppose you have a calculator with only a 12-digit display on the screen.
Explain how you can use such a calculator to directly verify that the triple of numbers
so obtained is a Pythagorean triple.
10. The second digit of a two-digit number is
1
3
of the first digit. If the number is
divided by the difference of the digits, the quotient is 15 and the remainder is3. Find
the number.
Appendix
Theorem 4 Two lines are perpendicular if and only if the product of their slopes is −1.
The proof will be written up at a later date. The idea is to use Theorem 1 of this
section to move the lines to the origin by parallelism, then we only need to prove the
theorem for lines passing through the origin. By dropping perpendiculars from the lines
to the x-axis, we can explicitly compute the slopes of the lines by the distances between
the origin and the feet of the perpendiculars. By suitably adjusting these distances and
making use of congruent triangles, we arrive at the desired conclusion.
7 Linear Inequalities and Their Graphs
So far we have only discussed equations and indeed, algebra, in the sense it is usually
understood in mathematics, is primarily about equations. However, school algebra in its
current usage includes everything related to number computations. As such, inequalities
occupy a prominent position in school algebra because in real life two numbers (e.g., two
82
people's weights, two people's incomes, etc.) are more likely to be different than the
same. One will generally be bigger than the other, and that is where inequalities come
in.
As an example, consider the following Manufacturing Problem:
A video game manufacturer is invited to a game show, and is told that she
can bring up to 50 games. She has two games, A and B, and has up to $6000
to spend on manufacturing costs. Game A costs $75 to manufacture and will
bring in a profit of ¸125, while Game B costs $165 to manufacture but will
bring in a profit of $185. Assuming that she sells every game she brings, how
many games of each kind should she manufacture if she wants to maximize
her profit?
It is clear that in this case there is no equation to solve, because the answer to the
problem is a pair of number, a certain number of A games and a certain number of
B games, so that this combination brings in a profit bigger than any other possible
combination. The emphasis here is on the words bigger than, i.e., inequality. One way
to understand what is involved in a problem of this nature is to approach it in a naive
way and see why it doesn't work. For example, a casual glance at the data would suggest
that it is more profitable to sell Game A than Game B, in the following sense. Suppose
you have $165. Then if you manufacture one B game, you only make $185, but if you
use the same amount to manufacture two A games (each costing $75), you'd not only
make $250 (= 2 1215) but would have $15 to spare.
A precise way to think about this is to notice that each Game A brings in
a profit that is 1
2
3
of its manufacturing cost (because
125
75
= 1
2
3
), but each
Game B only brings in a profit that is only about 1
1
8
of its manufacturing
cost (because
185
165
= 1
4
33
, which is about 1
4
32
= 1
1
8
).
One's first impulse is therefore to say that the manufacturer should bring only A games
to the show. But, $6000 is good for manufacturing 80 A games as it takes only $75
to manufacture one A game, whereas she is allowed to bring only 50 games. So if she
brings 50 A games, she would be using only 50 75 = 3750 dollars for manufacturing,
thereby wasting $2250 (= 6000 − 3750) of her $6000 budget. We see right away that
this may be a bad strategy in terms of maximizing profit, and we can easily confirm
it, as follows. With 50 A games, she would make 50 125 = 6250 dollars, but with
48 A games and 2 B games instead (which still add up to 50 games), she would make
(48 125) + (2 185) = 6370 dollars, and 6370 > 6250. It may be pointed out that
83
the cost of manufacturing 48 A games and 2 B games is
(48 75) + (2 165) = 3930
dollars, which is well within her $6000 budget.
She can also approach this problem from the opposite end, namely, knowing that
each Game B brings a profit of $185 whereas each Game A brings a mere $125, she
could rightly decide to concentrate on Game B and forget about Game A. The problem
now is that she cannot bring 50 B games to the show because her budget of $6000
wouldn't allow it: the cost of manufacturing 50 B games is 50 165 = 8250 dollars,
which is more than $6000. Again, this would suggest that bringing all B games to the
show is a poor strategy for maximizing profit. For confirmation, notice that a budget of
$6000 can produce at most 36 B games because
6000
165
= 36
4
11
, and her profit from 36 B
games would be 36 = 6660 dollars, whereas the same budget can equally well produce
34 B games and 5 A games (because (34 165) + (5 75) = 5985 < 6000) and these
39 games now bring a profit of
(34 185) + (5 125) = 6915
dollars, which is $255 more than $6660.
It is now clear that there is an inherent push-pull in this problem: bringing too many
A games would under-utilize the $6000 manufacturing budget because of the 50-game
quota, and bringing too many B games would under-utilize the 50-game quota because
of the $6000 manufacturing budget. Neither of these options would bring in the max-
imum profit. Intuitively, the combination of A games and B games that brings in the
maximum profit must be a kind of "equilibrium" between the number of A games and
the number of B games. What we need to understand in mathematical terms is how to
negotiate this push-pull to arrive at this "equilibrium". The main theme of this chapter
is about this mathematical understanding which, as adumbrated above, will require a
good grounding on inequalities.
The need for a better understanding of inequalities would be more in evidence if we
begin with a transcription of the given data of the Manufacturing Problem into symbolic
language. (Review the Section 2 at this point if necessary.)
Suppose the manufacturer produces x A games and y B games. We have to find
values of x and y so that the profit P,
P = 125x + 185y,
84
is a maximum. Now x and y are not arbitrary but are under constraints (a technical
terms for "restrictions") that come with the problem. Because she can bring at most 50
games, x and y are constrained by the inequality
x + y ≤ 50
Her manufacturing budget imposes another constraint in that she has at most $6000 to
spend on the production:
75x + 165y ≤ 6000
There are also two other obvious but indispensable constraints: x ≥ 0 and y ≥ 0. In
summary then,
we want to maximize P = 125x + 185y among all x's and y's which satisfy:
_
¸
¸
_
¸
¸
_
x ≥ 0
y ≥ 0
x + y ≤ 50
75x + 165y ≤ 6000
Even in this symbolic language, we still need a framework to understand the prob-
lem. What we propose to do is to interpret these four inequalities as constraints on
a point (x, y) in the xy-plane. The collection of all the points (x, y) satisfying these
four constraints is a certain region 1 in the plane. In a language that will be formally
introduced presently, 1 is called the graph of these inequalities (in the plane). The
profit P = 125x + 185y can now be thought of as an assignment to (or at) each point
(x
, y
) in 1 a number 125x
+185y
. As such, we shall refer to P as the profit function
on this 1. The Manufacturing Problem now become a purely mathematical problem
independent of context:
Among the points in the graph 1 of the inequalities
_
¸
¸
_
¸
¸
_
x ≥ 0
y ≥ 0
x + y ≤ 50
75x + 165y ≤ 6000
at which point does the profit function P = 125x+185y assign the maximum
value?
The virtue of this reformulation of the Manufacturing Problem is that it points to
clearly defined mathematical tasks:
85
(i ) What does the graph 1 of a collection of inequalities look like?
(ii ) Can we achieve enough of an understanding of the profit function P =
125x+185y to predict where it might assignment its maximum value in 1?
We now do the spade work necessary for analyzing graphs of inequalities. Our first
concern is with the behavior of inequalities with respect to arithmetic operations. Let
a, b, c, d be arbitrary numbers in the following discussion. Recall that the inequality
a < b (or written differently as b > a) means, by definition, that a is to the left of b on
the number line. There is also a weaker notion of inequality in the form of a ≤ b, which
means a is less than or equal to b, or in symbols, a < b or a = b. For emphasis, we may
sometimes explicitly refer to "≤" as weak inequality. Observe the following simple
consequences of the definitions which will be used in subsequent discussions without
comment (a, b, c are numbers):
a < b, b < c =⇒a < c
a ≤ b, b ≤ a =⇒a = b
In the following five assertions about inequalities, (α) – (), we state everything
in terms of weak inequality, but they will all remain valid if "≤" is replace by "<"
throughout, for the same reasons. Moreover, while each of these five inequalities is valid
as stated for arbitrary numbers a, b, c, d, for school mathematics it suffices most of
the time to know the validity of these inequalites for rational numbers. Accordingly, we
tacitly assume that all the numbers are rational numbers in the proofs.
(α) b ≥ a is equivalent to b −a ≥ 0.
If both a and b are positive, then (α) is quite obvious. We can also verified directly
that (α) must be true when a and b are explicitly known. For example, −2 > −5 is the
same as −2 − (−5) > 0, i.e., 3 > 0. Similarly, 1 > −7 is the same as 1 − (−7) > 0,
i.e., 8 > 0. However, when negative quanities are involved, such an assertion requires a
little thought. The proof given below can be skipped on first reading.
For the explanation, the case of equality in (α) is easy: it is obvious that b = a
is equivalent to b−a = 0. Therefore it suffices to deal with the case of >. We
have to show two things: b > a implies b−a > 0, and also b−a > 0 implies
b > a. First, we prove: b > a implies b −a > 0. If b > a, let c = b −a and
we must show c > 0. By adding a to both sides of c = b −a, we get a+c = b.
Using the vector interpretation of addition for rational numbers, the following
pictures, corresponding to the three cases of b > a > 0, b > 0 > a, and
0 > b > a, respectively, show why c > 0.
86
0
0
0
a
a
a
b
b
b
¸
¸
¸
c
c
c
(We briefly review the vector representation of a+c. Let [c[ denote the distance
of c from 0. To locate the point a+c on the number line, go from 0 to a, then
go from a a distance of [c[ to left if c is negative, to right if c is positive. The
point of final destination is then a +c. Now if b = a +c, and b is to the right
of a, then this means in going from a to b we must go right a distance of [c[.
Therefore c has to be positive).
Conversely, suppose b −a > 0, then we have to show that b > a, i.e., that
b is to the right of a. Again let c = b − a, then b = a + c. Since c > 0 by
hypothesis, the vector interpretaion of a + c means that from a we go right a
distance of [c[ to arrive at the point a + c. But a + c = b, so b is to the right
of a. The proof of (α) is complete.
(β) If a ≤ b and c ≤ d, then a + c ≤ b + d.
By (α), if suffices to show that (b + d) −(a + c) ≥ 0. But
(b + d) −(a + c) = (b + d) −a −c = b + d + (−a) + (−c) = (b + (−a)) + (d + (−c))
= (b −a) + (d −c)
By hypothesis, a ≤ b, and therefore b −a ≥ 0 (using (α) again). Similarly, d −c ≥ 0.
Therefore (b −a) + (d −c) ≥ 0. This implies (b + d) −(a + c) ≥ 0, as desired.
(γ) a + b ≤ c is equivalent to a ≤ c −b.
If a+b ≤ c, (β) implies that (a+b)+(−b) ≤ c+(−b). Thus a ≤ c−b. Conversely,
if a ≤ c −b, then (β) implies that a + b ≤ (c −b) + b, which implies a + b ≤ c.
It is common to paraphrase (γ) by saying that one can transpose a term to the
other side of an inequality.
87
(δ) If a > 0 and b ≤ c, then ab ≤ ac.
This is obvious if both b and c are positive, but less so if they are both negative.
The way to explain (δ) in general is to appeal to (α). Proving ab ≤ ac then becomes
proving ac −ab ≥ 0. But ac −ab = a(c −b) and since a > 0 and c −b ≥ 0 (because by
hypothesis b ≤ c), we see that a(c −b) ≥ 0 by the definition of fraction multiplication.
Hence ac −ab ≥ 0 and therefore ab ≤ ac.
() If a < 0 and b ≤ c, then ab ≥ ac.
This is the most mysterious among the basic facts concerning inequalities. Let us
illustrate by letting a = −3, b = −7, and c = −4. Since (−3)(−7) = 21 and
(−3)(−4) = 12, we have 21 > 12. Therefore (−3)(−7) > (−3)(−4).
Let us prove directly why, although 2 < 5, we have (−3)2 > (−3)5. From 2 < 5,
we get 3 2 < 3 5, by (γ). Now −(3 2) is the mirror reflection of 3 2 with respect to
0, while −(3 5) is the mirror reflection of 3 5 with respect to 0. So −(3 5) is further
away from 0 than −(3 2). However, −(3 2) = (−3)2 and −(3 5) = (−3)5. Therefore
the inequality (−3)2 > (−3)5 is clear from the following picture:
−(3 5) −(3 2) (3 5) (3 2) 0
The proof of (δ) can be obtained in a similar manner if 0 < b ≤ c. For the proof
of the general case, this line of reasoning would require the separate considerations of
three cases: (i) 0 < b ≤ c, (ii) b ≤ 0 ≤ c, and (iii) b ≤ c ≤ 0. It can get tedious.
We will instead rely on a skillful use of (α) and prove () in one stroke. Thus assume
b ≤ c and a < 0 are given, we let a = −a
)(c −b) = a(c −b) = ac −ab,
therefore ac−ab ≤ 0. Using (γ), we obtain ac ≤ ab, which is what we set out to prove.
We will make frequent use of (α) – () in the succeeding discussion. Here is one
simple application.
Example 1 Exhibit all the numbers x on the number line which satisfy (5−2x) +
12 > 4 −(3x −5).
88
The set of all these x's is called the graph of (5 −2x) +12 > 4 −(3x −5) on the
number line, and Example 1 is usually expressed as: graph (5−2x)+12 > 4−(3x−5)
on the number line. As in the case of solving linear equations, one simply isolates the
variable x in the inequality, in the sense of transposing all the x's to one side of the
inequality by making repeated use of (γ). Thus (5−2x)+12 > 4−(3x−5) is equivalent to
(5−2x) > 4−(3x−5)−12. Now 4−(3x−5)−12 = −8−(3x−5) = −8−3x+5 = −3−3x.
So (5 − 2x) + 12 > 4 − (3x − 5) is equivalent to 5 − 2x > −3 − 3x, which, by (γ)
again, is equivalent to 5−2x+3x > −3, and is equivalent to −2x+3x > −3 −5, i.e.,
x > −8. Thus we see that (5 −2x) +12 > 4 −(3x −5) is equivalent to the inequality
x > −8. This means all the x's which satisfy (5 −2x) +12 > 4 −(3x −5) are exactly
the same as those which satisfy x > −8, which therefore can be represented by the
thickened semi-infinite line segment below.
−8 0
Now we are finally in a position to deal with task (i) above, i.e., what does the graph
of a collection of inequalities look like in the plane? Formally, the graph of a linear
inequality of two variables ax+by ≥ c (where a, b, c are given constants) is the set
of all the points (x
, y
) in the plane whose coordinates x
and y
satisfy this inequality.
i.e., ax
+by
≥ c. The graph of ax +by > c is defined in like manner. Similarly, one
defines the graphs of ax +by ≤ c and ax +by < c. It is customary in mathematics to
denote the graph of an inequality such as ax+by ≥ c by the notation {ax+by ≥ c},
and we will use this notation below.
The graph of a collection of inequalities is by definition the set of all the points
which satisfy each of the inequalities in the collection. It follows that the graph of a
collection of inequalities is the intersection of all the graphs of the individual inequalities.
Note that the concept of the graph of an inequality or of a collection of inequalities
is usually used without an explicit definition in textbooks and professional development
materials. When this happens, no mathematical reasoning is possible in any discussion
concerning graphs of linear inequalities. Please be sure to teach your students this defi-
nition.
89
To describe the graph of an inequality, we have to discuss the separation property of
lines in the plane. We are going to show that every line L in the plane separates the
plane into two parts, called half-planes, and they are defined as follows. First assume
L is not horizontal. Then L meets the x-axis at a point (x
L
, 0). Strictly speaking, the
number x
L
is not the point (x
L
, 0), but by common practice, we identify x
L
with
(x
L
, 0). This point separates the x-axis into two parts: those points to the left of x
L
,
to be denoted by X
−
, and those to the right of x
L
, to be denoted by X
+
. Note that
the x-axis is the union of X
−
, the set consisting of the single point x
L
, and X
+
. If x
L
is the origin O, then X
−
and X
+
are of course the usual negative and positive x-axes,
respectively. The half-plane L
+
of L is by definition all the points P so that the line
passing through P and parallel to L meets the x-axis at X
+
. Similarly, the half-plane
L
−
of L is by definition all the points P so that the line passing through P and parallel
to L meets the x-axis at X
−
.
r r r
r
r
x
L
P
P
X
−
X
+
L
L
+
L
−
Recall that every non-horizontal line has an x-intercept, which is the x-coordinate of
the point of intersection of the line with the x-axis. Therefore, we may rephrase the
definitions of L
+
and L
−
in the following equivalent way:
L
+
(resp. L
−
) is the set of all the points P in the plane so that the x-
intercepts of the line passing through P and parallel to L is bigger than x
L
(resp. smaller than x
L
).
It is clear that the three subsets of the plane, L
−
, L, and L
+
are disjoint and that
their union is the whole plane. A little thought will also reveal that, implicitly, we have
invoked the Parallel Postulate throughout the foregoing discussion.
If L is vertical, then L
+
(resp. L
−
) is nothing other than all the points (a, b) so that
the x-coodinate a is positive (resp. negative). Pictorially, L
+
(resp. L
−
) in this case is
90
literally the part of the plane to the right (resp. left) of the L. For this reason, we call
this L
+
(resp. L
−
) the right half-plane (resp. left half-plane) of the vertical line L.
L
x
L
L
−
L
+
x
On the other hand, if L is horizontal, then of course none of these considerations apply.
In that case, however, we simply replace the x-axis by the y-axis. Then the point x
L
would be replaced by y
L
, the point of intersection of L with the y-axis. The right and
left half-planes in the case of a vertical line are now replaced by the upper half-plane
L
+
and the lower half-plane L
−
, as shown.
y
L
y
L
L
−
L
+
The occasion will arise when more precision regarding half-planes is needed, for the
following reason. The half-planes L
+
and L
−
do not include L, but we shall see presently
that there is a need to consider half-planes together with L itself. We will refer to L
+
∪L
and L
−
∪L as the two closed half-planes of L. The closed half-planes are not disjoint
as they have L in common. If there is any fear of confusion, we will refer to L
+
and
L
−
as the two open half-planes of L for emphasis. (The terminology of "open" and
"closed" is standard in mathematics.)
The reason for our interest in half-planes is that they allow us to identify the graphs
¦ax + by > c¦ and ¦ax + by < c¦ in terms of the half-planes of the line L defined by
ax+by = c (a, b, c are given constants). We first dispose of two extreme cases. If a = 0,
then the linear equation becomes y =
c
b
and L is horizontal. The consideration of the
91
linear inequalities themselves separates into two cases, b > 0 and b < 0, on account of
the difference between assertions (δ) and () above.
So assuming a = 0, if b > 0, the inequality ax + by > c is equivalent to y >
c
b
(by
(δ)), and the graph ¦ax + by > c¦ is just ¦y >
c
b
¦, which is the upper half-plane of L.
Similarly, the graph ¦ax + by < c¦ in this case is just the lower half-plane.
On the other hand, still with a = 0, if b < 0, then the inequality ax + by > c is now
equivalent to y <
c
b
by virtue of (), so that the graph ¦ax + by > c¦ is now ¦y <
c
b
¦,
which is the lower half-plane of L. Similarly, the graph ¦ax +by < c¦ in this case is the
upper half-plane.
Now if b = 0, then the equation ax + by = c becomes x =
c
a
and its graph L
is therefore vertical. An entirely analogous discussion then shows that in this case, for
a > 0, the graph ¦ax + by > c¦ is the right half-plane and the graph ¦ax + by < c¦ is
the left half-plane, while for a < 0, the graph ¦ax + by > c¦ is the left half-plane and
the graph ¦ax + by < c¦ is the right half-plane.
What we have seen are special cases of the following general theorem.
Theorem 1 Given constants a, b, and c, the graph of ax + by > c (respectively,
ax+by < c) is one of the two half-planes L
+
and L
−
of the line defined by ax+by = c.
The proof of this theorem contains all the key ingredients needed for the solution of the
Manufacturing Problem.
The statement of this theorem gives the impression of sloppiness: if we focus on, for
example, ¦ax + by > c¦, then the theorem does not seem to specify which of the two
half-planes of the line L defined by ax+by = c is equal to ¦ax+by > c¦. Such explicit
information will in fact come with the proof, — in the form of Theorem 1bis below, —
but in practice, there is no need for such a general statement. There are, after all, only
two half-planes of L, and if it is a matter of deciding which of the two is ¦ax +by > c¦,
it can be done very simply, as follows. Take a point (x
, y
) of L
+
and we check whether
the inequality ax
+by
> c is true or not. If it is, then (x
, y
) belongs to ¦ax+by > c¦
by definition, and we have ¦ax+by > c¦ = L
+
. If it is not, then (x
, y
) does not belong
to ax
+by
> c so that ¦ax+by > c¦ cannot be L
+
. But since Theorem 1 guarantees
that ¦ax + by > c¦ must be a half-plane, we conclude that ¦ax + by > c¦ = L
−
.
One more general comment on Theorem 1 before beginning the proof. This the-
orem shows why the concept of a closed half-plane is relevant. Indeed, suppose we
want to know the graph of the weak inequality ax + by ≤ c. By Theorem 1, we know
92
¦ax+by < c¦ is one of L
+
and L
−
. Let us say for definiteness that ¦ax+by < c¦ = L
+
.
It follows that the graph of ax+by ≤ c is just L
+
together with the graph of ax+by = c,
which is L. Therefore the graph of ax +by ≤ c is simply the closed half-plane L
+
∪L.
Now, let us give the proof of Theorem 1.
Having disposed of the special cases of Theorem 1 when either a = 0 or b = 0,
we may henceforth assume that a ,= 0 and b ,= 0. Without loss of generality, we may
further assume that a > 0, for the following reason. Suppose a < 0 in the inequality
ax + by > c. Multiplying both sides of the inequality by (−1) and applying (δ), we
obtain (−a)x −by < −c, so that
¦ax + by > c¦ = ¦ (−a)x −by < −c¦
Similarly, under the same assumptions,
¦ax + by < c¦ = ¦ (−a)x −by > −c¦
Thus instead of considering the graphs ¦ax + by > c¦ and ¦ax + by < c¦, we may
equivalently consider the graphs ¦ (−a)x − by < −c¦ and ¦ (−a)x − by > −c¦ instead.
But of course the coefficient (−a) of x is now positive and the considerations in the case
of a > 0 now apply. Therefore we shall henceforth assume that a > 0.
To prove Theorem 1, it therefore suffices to prove the following assertion, which for
convenience will be referred to as Theorem 1bis:
Let L be the line defined by ax +by = c, where a, b, c are given constants,
and a > 0, b ,= 0. Then the graph ¦ax + by > c¦ is the half-plane L
+
, and
the graph ¦ax + by < c¦ is the half-plane L
−
.
As mentioned above, Theorem 1 in the form of Theorem 1bis carries the explicit
information about which half-plane of L is the graph of (for example) ¦ax+by > c¦. Let
us give an illustration of how to put Theorem 1bis to use in a concrete case. Suppose
we are trying to determine the graphs of −3x + y < 9 and −3x + y > 9. Let L be the
line defined by −3x + y = 9. Because −3x + y = 9 is equivalent to 3x −y = −9, L
is also the line defined by 3x −y = −9. Then we apply Theorem 1bis to obtain:
¦3x −y > −9¦ = L
+
¦3x −y < −9¦ = L
−
as shown:
93
/
/
/
/
/
/
/
/
/
/
/
//
x
y
L
−3
9
O
L
−
L
+
In terms of the original question of what the graphs of −3x+y < 9 and −3x+y > 9 are,
we use (). So 3x −y > −9 is equivalent to −3x +y < 9 and 3x −y < −9 is equivalent
to −3x + y > 9. Therefore, we get
¦−3x + y < 9¦ = L
+
¦−3x + y > 9¦ = L
−
We now return to the proof of Theorem 1bis. The following fact, proved in Section
4 (Fact (iv) near the end), will prove to be useful:
(∗) A line L defined by ax + by = c with a ,= 0 and b ,= 0 has slope −
a
b
and x-intercept
c
a
.
We will refer to it as (∗).
So given a linear equation ax+by = c with graph L, where a > 0, we have to prove
that ¦ax+by > c¦ is equal to L
+
. (The proof of ¦ax+by < c¦ = L
−
is entirely similar
and will be left as an exercise.) Recall that L
+
is by definition all the points (x
, y
) so
that the line parallel to L and passing through (x
, y
) has x-intercept bigger than µ, the
x-intercept of L. We note that, by (∗),
µ =
c
a
To prove ¦ax + by > c¦ = L
+
, we have to prove
(A) every point of ¦ax + by > c¦ is a point of L
+
, and
(B) every point of L
+
is a point of ¦ax + by > c¦.
94
We first prove (A). Let (x
, y
) be a point of ¦ax +by > c¦, and let be the line passing
through (x
, y
) and parallel to L. If µ
is the x-intercept of , we have to show µ
> µ.
We begin by determining the equation of . The slope of L is −
a
b
, by (∗). Let
c
= ax
+ by
, and let
be the line defined by ax + by = c
. The slope of
is also −
a
b
(by (∗) again), so that
is parallel to L, by Theorem 1 of Section 6. Moreover, (x
, y
) is
obviously a solution of ax + by = c
so that
also passes through (x
, y
). We conclude
(by the Parallel Postulate) that
is just , i.e., is defined by ax + by = c
. Using (∗)
once more, we see that the x-intercept, µ
, of is
µ
=
c
a
=
ax
+ by
a
Now (x
, y
) belongs to ¦ax+by > c¦, so by definition, ax
+by
> c. But a > 0 implies
1
a
> 0, so that by (δ),
_
1
a
_
(ax
+ by
) >
_
1
a
_
c
and this means
ax
+ by
a
>
c
a
Since µ =
c
a
, we have µ
> µ, as desired.
Next we prove (B). Suppose (x
, y
) is in L
+
, then we must prove that it is also a
point of ¦ax + by > c¦, i.e., we must show that ax
+ by
> c. Let be the line
passing through (x
, y
) and parallel to L, and let have an x-intercept of µ
. Then by
the definition of L
+
, µ
> µ, where as before, µ is the x-intercept of L. Now as in the
proof of part (A), we can prove µ
=
ax
+by
a
; we have already seen that µ =
c
a
. Therefore
ax
+by
a
>
c
a
. Multiplying both sides of the last inequality by a and making use of (δ),
we get ax
+ by
> c.
We have completed the proof of Theorem 1bis, and therewith, also Theorem 1.
In the proof of (A), we tacitly assumed that, because a > 0,
1
a
> 0. A precise
proof of this fact goes as follows. If
1
a
is not positive, then it is either 0 or
negative. It suffices to show that neither possibility is admissible.
1
a
cannot
be 0 because if it were, then
1
a
a = 0 a = 0, whereas we know
1
a
a = 1, and
1 ,= 0. It cannot be negative either because if it were, then the product
1
a
a
would be the product of a negative number and a positive number (using the
assumption a > 0) and is therefore negative. But again, we know
1
a
a = 1,
and 1 is positive. We have now showed that neither possibility can hold, so
1
a
has to be positive after all.
The following examples illustrate how to make use of Theorem 1.
95
Example 2 Graph 3x −2y < −5 in the plane.
The line L defined by 3x −2y = −5 is shown below.
´
´
´
´
´
´
´
´
´
´
´
x
y
L
−
5
3
5
2
O
L
−
L
+
Since L does not pass through the origin O, then either O is in ¦3x − 2y < −5¦ or it
is not. Since it is not true that 3 0 − 2 0 < −5, O is not in ¦3x − 2y < −5¦. Thus
by Theorem 1, ¦3x −2y < −5¦ is a half-plane of L that does not contain O. From the
picture, ¦3x −2y < −5¦ has to be L
−
.
Of course there is no harm if you remember exactly what Theorem 1bis says and
use that to decide which half-plane of L is ¦3x −2y < −5¦. Because the coefficient 3 of
x in 3x −2y = −5 is positive, Theorem 1bis is directly applicable. So we know right
away that ¦3x −2y < −5¦ = L
−
, as before.
If we ask for the graph of 3x − 2y ≤ −5, then we would get the closed half-plane
L
−
∪ L.
Example 3 Graph −x −2y < 4 in the plane.
The graph of the line defined by −x −2y = 4 is the following:
O
+
−
−4
−2
96
We first check to see if the origin O belongs to ¦−x−2y < 4¦: the inequality −0−20 < 4
being true, O is in ¦−x − 2y < 4¦. Therefore by Theorem 1, ¦−x − 2y < 4¦ is a
half-plane of that contains O. Looking at the picture, we see that O lies in
+
. So
¦−x −2y < 4¦ has to be
+
. we conclude that ¦−x −2y < 4¦ =
+
.
Again, we can also directly appeal to Theorem 1bis. The equation −x−2y = 4 is the
same as x+2y = −4, so that Theorem 1bis implies that ¦x+2y > −4¦ =
+
. Now by
(), the inequality x+2y > −4 is equivalent to −x−2y < 4. Thus ¦−x−2y < 4¦ =
+
as before.
If we ask for the graph of −x−2y ≤ 4, then we would get the closed half-plane
+
∪.
Example 4 Find the graph of the pair of inequalities −x − 2y < 4 and
−2x + 3y > 0.
This example asks for the intersection of the graphs of the individual inequalities.
We already know the graph of −x −2y < 4 from Example 3. It remains to determine
the graph of −2x + 3y > 0. Let L be the line defined by −2x + 3y = 0. Then the
picture is the following:
LThe graph in Example 4 is an "infinite region" in a sense that is self-explanatory
(although "infinite" in this context can be precisely described in advanced mathematics).
In applications such as the Manufacturing Problem of this section, however, the graph
would tend to be a polygon with the edges included. We can illustrate this phenomenon
by elaborating on Example 4. The graph of the two weak inequalities −x−2y ≤ 4 and
−2x +3y ≥ 0 is now the intersection of the closed half-planes L
−
∪L and
+
∪, and is
the same dotted region as above plus the two semi-infinite boundary line segments. Call
this region o. The graph of the three weak inequalities
−x −2y ≤ 4, −2x + 3y ≥ 0, and y ≤ 0
is the intersection of o with the closed lower half-plane of the x-axis, and is therefore
the following dotted triangular region together with the three edges.
O
L
q
q
q q
q
q
q
q
q
At this point, we have all the needed information to tackle task (ii) above, i.e., to
understand the behavior of the profit function P = 125x + 185y of the Manufacturing
Problem. Recall that we have introduced the region 1 as the graph of the following four
98
inequalities:
_
¸
¸
_
¸
¸
_
x ≥ 0
y ≥ 0
x + y ≤ 50
75x + 165y ≤ 6000
Sometimes we say 1 is a region defined by these inequalities. This graph is the
dotted region below:
x
y
O
1
x + y = 50
75x + 165y = 6000
q
q
q
q
q
q
q
q
q
q
Our reformulation of the Manufacturing Problem now asks at which point (x
0
, y
0
) of 1
the profit function P = 125x + 185y attains (or achieves) its maximum in 1, in
the sense that if (x
, y
) is any other point of 1, then the profit P at (x
0
, y
0
) (which is
125x
0
+185y
0
) is at least as big as the profit at (x
, y
) (which is 125x
+185y
). In other
words,
125x
0
+ 185y
0
≥ 125x
+ 185y
The point (x
0
, y
0
) is called a maximum point of the profit function. We pause to
point out that in the context of the Manufacturing Problem, it makes sense to talk
about the profit at a point (x, y) (i.e., 125x + 185y) only if x and y are both whole
numbers. After all, there is no such thing as the profit derived from 1.27 A games and
13
7
B games. However, as a mathematical problem, it makes perfect sense to see, for
any two numbers x and y regardless of whether or not they are whole numbers, what
the value of 125x + 185y is so long as at the end, we come up with a solution (x
0
, y
0
)
of the Manufacturing Problem where the coordinates x
0
and y
0
are whole numbers. So
by freeing x and y from the constraint of being only whole numbers, we are able to
consider the profit function P = 125x + 185y for any (x, y) in the region 1. Part
99
of learning algebra includes learning when to take an abstract approach to a problem,
and the extension of the concept of "profit" to include 125x +185y for any numbers x
and y is a good example of the needed abstraction for the solution of many problems in
algebra, including the Manufacturing Problem.
To continue the discussion, we will see that there is no advantage to gain by studying
the specific profit function P = 1215x + 185y instead of something more general.
Accordingly, we let a, b, e be three fixed numbers, with a ,= 0 or b ,= 0, and consider
the general expression P = ax + by + e, which assigns to any point (x
, y
) in the
plane the value ax
+ by
+ e. We call such a P a linear function of two variables.
Then the concepts of a linear function attaining (or achieving) a maximum in 1, and
a point in 1 being a maximum point of P in 1 are defined similarly. For example, we
say P = ax +by +e achieves a maximum at a point (x
0
, y
0
) in 1 if for any point
(x
, y
) of 1, we always have ax
0
+ by
0
+ e ≥ ax
+ by
+ e. And of course, (x
0
, y
0
) is a
maximum point of P in 1. Naturally, the corresponding concepts for a minimum
in place of a maximum is similarly defined.
With such a linear function P = ax +by +e understood, take a point (p, q) and let
P assign to (p, q) the value c
, i.e., ap +bq +e = c
. Our first observation is that the line
L defined by ax +by = c
−e, which is ax +by +e = c
in disguise, consists of all the
points (x
, y
) to which P assigns the same value c
. To see this, we show first of all that
P assigns to each point of L the value c
. This is because if (x
, y
) is a point on the line
L, then by the very definition of L, ax
+by
= c
−e, so that ax
+by
+e = c
, which
is the statement that P assigns to (x
, y
) the value c
. Next we show that if (x
, y
) is a
point to which P assigns the value c
, then (x
, y
) is already in L. This is so because if
P assigns c
to (x
, y
), then ax
+by
+e = c
, so that ax
+by
= c
−e. Thus (x
, y
)
satisfies ax + by = c
−e and by the definition of L, (x
, y
) is a point on L.
If we let c = c
− e in the preceding paragraph, then we may rephrase the above
information in terms of our more familiar notation as follows: given a line ax + by = c
and a linear function P = ax + by + e, where the coefficients a and b are the same in
both L and P, then the line L consists of all the points to which P assigns the value
c + e.
Up to this point, we have only looked at a line L as the graph of a linear equation,
such as ax +by = c. Now we get a different perspective: we may also look at L as the
totality of all the points to which the linear function P = ax +by +e assigns the value
where c + e. Given a fixed number c
to a point (x
0
, y
0
),
then the level set of P containing (x
0
, y
0
) is the graph of ax+by = c, where
c = c
−e.
We can now translate Theorem 1bis into the language of the level set of a linear
function P = ax + by + e.
Theorem 2 Given P = ax + by + e with a > 0. Let (x
0
, y
0
) be an arbitrary point
in the plane and let P assign to it the value c
. Let L be the line which is the level set
of P containing (x
0
, y
0
). Then the half-plane L
+
consists of all the points to which P
assigns values > c
, and the half-plane L
−
consists of all the points to which P assigns
values < c
.
The proof of Theorem 2 is very simple indeed. We will concentrate on the state-
ment relating to L
+
, and leave that relating to L
−
to an exercise. First we show that
P assigns to every point of L
+
a value > c
. Let (x
, y
) be a point of L
+
. Since L is
the graph of ax +by = c, where c = c
−e, Theorem 1bis implies that (x
, y
) belongs
to ¦ax + by > c¦, which is ¦ax + by > c
−e¦. Therefore, ax
+ by
> c
−e, and by
item (γ) on inequalities, we have ax
+ by
+ e > c
, which is then the statement that
P assigns to (x
, y
) a value > c
. We finish the proof by showing that if there is any
point (X, Y ) to which P assigns a value > c
, then (X, Y ) must belong to L
+
. Thus we
are given that aX + bY + e > c
. Again by (γ), we get aX + bY > c
− e = c. By
Theorem 1bis, we see that (X, Y ) belongs to L
+
. This then completes the proof.
Theorem 2 immediately leads to our first conclusion concerning where a linear func-
tion P = ax + by + e can achieve its maximum in a region 1.
With a > 0, if a linear function P = ax +by +e achieve its maximum in 1
at a point (p, q) of 1, then 1 lies completely in the closed half-plane L
−
∪L
of the level set L of P containing (p, q). If on the other hand P achieves
its minimum in 1 at (p, q), then 1 lies completely in the closed half-plane
L
+
∪ L.
101
This is because, by Theorem 2, if 1 contains a point (s, t) of L
+
, then P assigns to
(s, t) a value exceeding the value it assigns to (p, q), and therefore P does not achieve
its maximum at (p, q), as shown:
x
r (p, q)
r
(s, t)r r
L
L
+
1
The statement about minimum is proved in the same way.
In a similar manner, we have:
If a < 0, if a linear function P = ax +by +e achieve its maximum in 1 at
a point (p, q) of 1, then 1 lies completely in the closed half-plane L
+
∪ L
of the level set L of P containing (p, q). If on the other hand P achieves its
minimum at (p, q), then 1 lies completely in the closed half-plane L
−
∪ L.
It is instructive to give a proof of this assertion which makes use of the preceding
one. Because the leading coefficient a is now negative rather than positive, we consider
instead the linear function Q = −(ax +by +e). Then Q = (−a)x +(−b)y −e, and the
coefficient of x in Q is now positive. We can therefore apply to Q what we have learned
thus far. We know P achieves its maximum in 1 at (p, q), but this means Q achieves
its minimum in 1 at (p, q) because: ap + bq + e ≥ ax
+ by
+ e for all (x
, y
) in 1
means −(ap +bq +e) ≤ −(ax
+by
+e) for all (x
, y
) in 1, by item () on inequalities,
which is then precisely the statement that Q achieves its minimum in 1 at (p, q). Now
suppose P achieves its maximum in 1 at (p, q), and yet 1 contains a point (s
¦ of Q.
Therefore, we know that L
−
consists of all the points to which Q assigns values < −c
.
10
In particular then, Q assigns to (s
, t
) a value smaller than −c
, i.e., −(as
+bt
+e) < −c
.
By item () on inequalities again, we have as
+ bt
+ e > c
, and this contradicts the
given data that P achieves its maximum in 1 at (p, q). Hence no such (s
, t
) exists.
The statement about minimum is proved in the same way.
We can summarize the foregoing discussion as follows. Let us call a point (p, q) an
interior point of 1 if (p, q) is in 1 and is disjoint from the boundary of 1. Then:
Theorem 3 If a linear function P = ax +by +e achieves a maximum or a min-
imum at a point (p, q) of 1, then 1 lies completely in one of the two closed half-planes
of L, where L is the line which is the level set of P passing through (p, q). In particular,
a linear function P = ax + by + e cannot achieve a maximum or a minimum at an
interior point of a region 1.
In view of what we have already discussed, only two minor comments on this theo-
rem are called for. First, although we have always assumed in the above arguments that
a > 0 or a < 0, the case of a = 0 in P = ax+by +e (which is not excluded in Theorem
3) can be easily taken care of by noting that resulting level set L passing through a max-
imum point (p, q) would be horizontal. Then we simply replace the erstwhile L
+
and
L
−
by the upper and lower half-planes of L, respectively, in all the arguments. Second,
10
It may help for the understanding of this statement to recall that the definition of L
−
depends only
on the fact that L is a line and has nothing to do amy linear functions P or Q.
103
the reason for the last statement of Theorem 3 is that if (p, q) is an interior point of a
region 1, then any line L passing through (p, q) would have the property that both of
its open half-planes contain points of 1.
For regions which are intersections of closed half-planes such as the one arising from
the Manufacturing Problem, Theorem 3 has an important refinement. What is important
for our purpose about intersections of closed half-planes is that their boundary consists
of straight line segments and that the boundary is part of the region.
11
The points of
intersection of these boundary line segments are called corners. The corners of the
region arising from the Manufacturing Problem are indicated with a dot in the picture
below.
x
y
O
t t
t
t
x + y = 50
75x + 165y = 6000
1
q
q
q
q
q
q
q
q
q
q
The refinement in question is:
Theorem 4 Let a region 1 be an intersection of closed half-planes. If a linear
function P = ax +by +e achieves its maximum or minimum in 1, then it does so at
a corner.
Please note what the theorem does not say. It does not say that P only achieves
its maximum (resp., minimum) at a corner; it merely asserts that if there is a maxi-
11
If the region is the intersection of open half-planes, the boundary of the region would not be part
of the region. For example, consider ¦x > 0¦ ∩ ¦y > 0¦. This region is the first quadrant but without
the two (positive) coordinate axes, so that this region does not contain its boundary.
104
mum point (resp., minimum point) in 1 for P, then a corner would already serve as a
maximum point (resp., minimum point).
Proof We will deal with the case of a maximum; the case of a minimum is entirely
similar. Consider a typical such region:
`
`
`
`
`
``
x
We will prove the theorem for the case a > 0. The case of a < 0 is similar and will
be left as an exercise (but see the proof Theorem 3). The case of a = 0 is simple and
can be left to the reader.
Suppose P achieves its maximum at a point (p, q) of 1. Let L be the level set
¦P = c
¦ passing through (p, q). By Theorem 3, 1 must lie in the closed half-plane
L
−
∪ L. Because the boundary of 1 consists of line segments, there can only be three
posibilities for this to happen: either (p, q) is already a corner and L contains no point
of 1 except the corner, or (p, q) is a corner but L also contains an edge of the boundary,
or (p, q) is not a corner but lies inside an edge of the boundary and then L contains this
edge, as shown.
t
t
t
(p, q)
(s, t)
(u, v)
x
If (p, q) is already a corner, as in the case on the left and the case in the middle, there
would be nothing to prove. If (p, q) is not a corner, as in the case on the right, then the
fact that L = ¦P = c
, and therefore P also achieves
its maximum in 1at the corner (s, t) (or for that matter, (u, v)). This proves Theorem 4.
The case on the right points to the subtlety of the mathematical language. In the
context of ordinary communication, Theorem 4 would seem to imply that P achieves
a maximum in 1 at a corner but nowhere else, but in fact (as we have already pointed
out) what it says is merely that there is a corner at which P achieves a maximum.
Nothing in the statement of the theorem precludes the possibility that P achieves a
maximum somewhere else too so that, as we know from the WYSIWYG characteristic
of mathematics, the case on the right in no way contradicts Theorem 4.
A further comment is that one has to pay attention to the careful wording of Theorem
4: it says that if P achieves a maximum in 1, then it does so at a corner, but it does
not say in any 1 that is the intersection of closed half-planes, a linear function will
always achieve a maximum in 1 at a corner. Indeed, this assertion is not even true.
Consider for example the linear function P = x+y in the region 1
0
which is the graph
of the pair of inequalities x ≥ 0 and y ≥ 0, i.e., the first quadrant including the
positive coordinate axes, as shown:
O
x
y
k
k
¦P = k¦
(x
, y
)
q
Clearly the value P assigns to a point (x
, y
) increases without bound as (x
, y
)
moves up in the direction of the upper right corner, and so P cannot achieve a max-
iumum in 1
0
. However, if 1 is a finite region (in the sense of being contained inside
some circle) which is the intersection of closed half-planes, then any linear function will
always achieve a maximum in 1. While this fact is fairly believable, its proof requires
some advanced mathematics.
The solution of the Manufacturing Problem is now relatively simple.
106
We are looking at the profit function P = 125x + 185y on the graph 1 of the
following weak inequalities:
_
¸
¸
_
¸
¸
_
x ≥ 0
y ≥ 0
x + y ≤ 50
75x + 165y ≤ 6000
We have already seen a picture of 1:
x
y
O
t t
t
t
(x
0
, y
0
)
50
80
50
36
4
11
1
2
1
q
q
q
q
q
q
q
q
q
q
In the context of the Manufacturing Problem, this 1 is called the feasibility region
of the problem. In this case, 1 is a quadrilateral together with its interior bounded by
the positive x- and y-axes, the line
1
defined by x +y = 50 and the line
2
defined by
75x + 165y = 6000. The four corners of 1 are
(0, 0), (50, 0), (x
0
, y
0
), (0, 36
4
11
)
where every point is there for the obvious reason except for (x
0
, y
0
), which is the point of
intersection of
1
and
2
and is therefore the solution (by Section 6) of the simultaneous
system:
_
x + y = 50
75x + 165y = 6000
Solving this system in the standard way (but note that a simplification can be achieved
by reducing the second equation to x + 2.2 y = 80, we get x
0
= y
0
= 25. So (x
0
, y
0
) =
(25, 25).
Now the profit in Manufacturing Problem must have a maximum; for example, try
out all possible profits with n A games and 50 − n B games for n = 0, 1, 2, . . . , 50
107
and the maximum is among these. Theorem 4 therefore tells us where to look for the
maximum point: check the profit at each of the four corners above. The profits at
(0, 0), (50, 0), (25, 25), (0, 36
4
11
) are, respectively, 0, 6250, 7750, and 6725
3
11
. Thus the
maximum point of the profit function is (25, 25), and the solution is: manufacture ex-
actly 25 A games and 25 B games to make the maximum profit of $7750.
It remains to round off the preceding discussion by tying up a loose end. It is
obvious that we could have left out any consideration of the corner (0, 0) in search of
the maximum possible profit. But the corner (0, 36
4
11
) had to be accounted for even if
it makes no sense to talk about 36
4
11
B games. This is because there is always the
possibility that if (0, 36
4
11
) is a maximum point, then one of the nearby points (inside
the feasibility region) with whole number coordinates, i.e., (0, 36) and (1, 35), may be
a maximum point among those points with whole number coordinates. Thus, it could
be that 36 B games or 1 A game and 35 B games produces the maximum profit. In
our case, this did not happen because even the profit at (0, 36
4
11
) is not large enough.
In other situations, one would need to check such a possibility. (See Problem 10 in the
following Exercises.)
To give some intuitive content to this way of approaching the problem, consider the
solution of 25 each of A and B games. Why did we not check to see if the $6000 man-
ufacturing budget is enough to cover the production of 25 A games and 25 B games?
This is because (25, 25) is in the feasibility region of the problem, so that in particular
it is in the half-plane of 75x + 165y ≤ 6000. Therefore, with x = 25 and y = 25,
(7525) +(16525) ≤ 6000, which is precisely the statement about the manufacturing
cost of these 50 games being at most $6000.
EXERCISES
1. (a) Graph the inequality
2
3
x −(2 +7x) ≥ (6 +x) −(1 −
1
2
x) on the number line.
(b) Graph the inequality
2
5
−
1
2
x ≥
1
5
x +
1
6
on the number line.
2. Graph the following inequalities in the plane:
_
_
_
5
2
x +
3
4
y ≤ 2
−2x + 3y ≤ 12
1
3
x −y ≤ 5
3. Give a proof of the other half of Theorem 1, i.e., the fact that ¦ax+by < c¦ = L
−
.
4. Give a proof of the half of Theorem 2 concerning L
−
.
108
5. Give a proof of the case of a < 0 in Theorem 4.
6. Given a linear function P = ax +by +e with a > 0 and b ,= 0. Suppose we have
two lines both with slope −
a
b
as shown:
x
yr r
r
r
r
(p
).
7. Let o be the region defined by the inequalities
_
3x −3y ≤ 5
2x + y ≤ −4
Does the linear function P =
2
3
x + y achieve a maximum in o? Does it achieve a
minimum in o? Explain.
8. Let 1 be the graph of the following inequalities:
_
_
_
1
2
x +
4
5
y ≥ 4
−2x + 5y ≤ 10
1
3
x −
1
4
y ≤ 2
Where would the linear function 2x +
1
3
y −4 attain its maximum in 1? Its minimum?
What are the maximum and minumum values of the function?
9. Find the point (x
0
, y
0
) at which the linear function 10x +3y achieves a maximum
in the graph of the inequalities:
_
¸
¸
_
¸
¸
_
x ≥ 0
x ≤ 10
y ≥ 5
0.5x + y ≤ 20
Where does 10x + 3y achieve a minimum in this graph?
10. A farmer sells 100 bushels of Crop A for $450, and the same amount of Crop B
for $650. The available resources and restrictions are as follows:
109
(a) He has 150 acres of land. It takes 1 acre to produce 100 bushels of Crop
A but 2 acres to produce the same amount of Crop B.
(b) He has a capital of $9500. It takes $125 to produce 100 bushels of Crop
A and $95 to produce the same amount of Crop B.
(c) He has to complete the harvesting of the crops in 200 hours. Each 100
bushels of Crop A require 4 hours of work, and the same amount of Crop B
requires only 1 hour.
Suppose business transactions are done only in terms of 100 bushels. How many 100
bushels of Crop A and Crop B should he produce in order to maximize his revenue
(i,e., total dollar intake) assuming he can sell everything he produces? Use a scientific
calculator.
11. The nutritional values of a basic unit of two food items are tabulated below:
calorie vitamin C (i.u.) iron (mg) protein (mg)
A 156 50 4.1 13
B 112 200 1.2 15
A mountain climber wants to bring enough of both items for her trip so that she
would get at least 2700 calories, 1500 i.u. of vitamin C, 50 mgs of iron, and 300 mgs of
protein.
Suppose each unit of Item A costs $4.20 and each unit of Item B costs $2.80. How
many units of each should she buy so that the total cost is minimum and her nutritional
requirements are met? Use a scientific calculator.
8 Exponents and Absolute Value
So far we have dealt with linear equations, linear inequalities, and linear functions of
two variables. Linear objects are important because they are the basic building blocks of
mathematics, but life is often not linear. A good example is Kepler's famous Third Law
governing the motion of an object around the sun: the square of the period
12
divided
by the cube of the so-called semi-major axis of the elliptic orbit
13
is a fixed constant no
12
The time it takes the object to complete a revolution around the sun.
13
The maximum distance of the object from the sun. In some school mathematics and physics
textbooks, this law is stated using "mean distance" in place of "major axis", and that is an error.
110
matter what the object may be (e.g., any planet, any meteor, any asteroid). In symbols,
this means there is a number C so that, if T is the period and D is the semi-major axis
of an object revolving around the sun,
T
2
D
3
= C
Thus if the object is far from the sun compared with the earth (e.g., Pluto), then it
would take much more than a year for tht object to complete a revolution around the
sun. By multiplying both sides with D
3
, we can rewrite this equality as
T
2
−CD
3
= 0
You can see that this is not a linear equation in T and D. What this means is that, to
progress further into mathematics, we would have to deal with powers of numbers, such
as T
3
and even x
−3/5
, as well as the absolute value of a number, i.e., [x[. These are the
most basic nonlinear quantities.
Let us start from the beginning. Unless stated to the contrary, let us agree that for
the remainder of this section, α, β, γ will always stand for positive numbers, and m, n,
k, will stand for positive integers. Recall that, by definition, α
2
= αα, α
3
= ααα,
α
4
= αααα, . . . and in general,
α
n
= αα α (n times)
The positive integer n is called the exponent or power of α
n
. One also speaks of α
n
as raising α to the n-th power. Here are the most basic facts concerning exponents:
(8.1) α
m
α
n
= α
m+n
(8.2) (α
m
)
n
= α
mn
(8.3) (αβ)
m
= α
m
β
m
where m, n are positive integers. These three facts are, simultaneously, trivial to prove
and "fun" to use due to their simplicity. For example, (8.1) says that, in some vague
sense, exponents are additive under multiplication. As to the triviality of their proofs,
there is no doubt of that. For example,
(α
3
)
5
= (ααα)
5
= (ααα)(ααα)(ααα)(ααα)(ααα)
= α
5×3
= α
3×5
111
The general proof of (8.2) is almost identical, and the same is true of the proofs of the
other two identities.
Having established the desirability of (8.1)–(8.3), we may ask why something so good
should be restricted to positive integer values of m and n. What should α
0
mean? What
about fractions in the exponents? In fact, what about rational numbers? We now ap-
proach these three questions systematically.
First, what could 5
0
mean? While we do not know what meaning to give 5
0
, we know
what we want out of it: (8.0)–(8.3) should still hold even when m, or n is allowed to
be 0. Therefore we shall perform a bold experiment by throwing out all the usual rules
in mathematics. As we have emphasized, mathematics only deals with concepts with
precise definitions. Right now we do not know what 5
0
means but for once will pretend
that we do. We further assume that this object 5
0
that we know nothing about satisfies
(8.1) even when n is 0. The boldness of this experiment therefore lies in the fact that we
are assuming that we know the very thing we are trying to find out. This doesn't matter
to us because we are merely trying to make the right guess about what 5
0
ought to be,
and we are not claiming to be proving anything. So with this understood, the version
of (8.1) with n = 0 gives:
5
1
5
0
= 5
1+0
In other words, we should have 55
0
= 5. Multiplying both sides by
1
5
gives
1
5
55
0
=
1
5
5,
so that 5
0
= 1.
To recapitulate: if (8.1)–(8.3) are to remain meaningful and continue to be valid
even when one or more of the exponents is allowed to be 0, then it is necessary to define
5
0
= 1. This heuristic argument is the same if 5 is replaced by any positive number α: if
the equality α
1
α
0
= α
1+0
makes sense and is correct, then αα
0
= α, so that necessarily
α
0
= 1.
On the basis of this heuristic argument, we now define the 0-th power of a positive
numbex α to be
α
0
= 1
We pause to make a point. The presentation of the definition of α
0
in some books
gives the impression that one can prove α
0
= 1. The "proof", such as it is, is exactly
the argument we gave above. It is therefore necessary to underscore once more the fact
that the preceding heuristic argument is not a proof of why α
0
= 1. What it does is
to give us confidence that we have probably made the correct definition to ensure the
112
validity of (8.1)–(8.3) even when one or both of m, n is 0.
Fractional exponent is next. We start with the simplest case: what could α
1
2
mean?
We will be guided by (8.1)–(8.3) again, and (8.2) suggests that
(α
1
2
)
2
= α
1
2
2
= α
If we write γ for α
1
2
, this says γ should be a number so that γ
2
= 2. You recognize this
as saying γ is the square root of 2.
A good mathematics education sometimes has the beneficial effect of making you
stop and think about things that you may have taken for granted all along, and gain
new understanding in the process. A case in point for most of us is our first serious
encounter with the number π; we learn what it actually is and how to directly estimate
it. The "square root of 2", something most of us have been familiar with since perhaps
primary school, may yet be another such case. How do we know that there is a number
whose square is exactly 2? One can rattle off 1.4142135 . . . as that number, but one may
also be aware that the decimal expansion of the square root of 2 is non-repeating, so that
no matter how many decimal digits one writes down, it will just be an approximation.
For example, 1.142135
2
= 1.99999091405925. So what gives us the confidence that there
is a "square root of 2"?
This is where mathematical knowledge is helpful by providing the answer we need.
There is a theorem, proved in advanced courses, that not only square roots, but any
so-called n-th roots exist and are unique. Precisely, let n be a positive integer. Then
given a positive number α, a positive number γ is said to be a positive n-root of α if
γ
n
= α. (Recall that γ
n
= γγ γ (n times), by definition.)
Note the emphasis throughout on the positivity of α and γ. This is because
if α = −2, then there is no number on the number line whose square is a
negative number. (Do you know why?) Moreover, in case α > 0, e.g., α = 4,
there will at least two numbers whose square is 2, namely, 2 and −2. This is
why we have to specify the positivity of γ in the preceding paragraph.
Then the theorem that resolves all doubts in this context is the following.
Theorem Given a positive number α and a positive integer n, there is one and
only one positive number γ so that γ
n
= α.
113
It is to be remarked that the uniqueness part of the theorem, which says that there
is at most one such γ, is actually not too difficult to prove, but we will postpone this
proof so as not to interrupt our discussion. Henceforth, we shall refer to the γ in the
theorem as the positive n-th root of α and, if there is no fear of confusion, more simply
as the n-root of α. The standard notation for the positive n-th root of α is
n
√
α.
Note that the case n = 2 is distinguished and the notation for the positive square root
is
√
α rather than the more elaborate
2
√
α. Please remember that
n
√
α is always
positive, by convention. Therefore
√
4 = 2, and never −2. In any case, one message
of the theorem is that there is such a thing as the positive square root of 2.
The third root of α is traditionally called its cube root.
We are now in a position to resume our discussion of rational exponents. We will
continue to use (8.1)–(8.3) as a guide for the correct definition of rational exponents.
Thus, if the analogue of (8.2) is valid in general, we must have (α
1
2
)
2
= α
1
2
·2
= α
1
= α
for any positive α. By Theorem 1, α
1
2
must be the positive square root of α. The same
heuristic argument would yield the fact that, for any positive integer n, (α
1
n
)
n
= α,
and therefore α
1
n
has to be the positive n-th root of α. Thus for any positive integer
n, we feel sufficiently confident to define the
1
n
-th power of a positive number α
to be
α
1
n
= the positive n-th root
n
√
α of α
The next question is: what should 2
3
2
mean? Again, assuming that the analogue
of (8.2) for rational exponents is valid, we must have
2
3
2
= 2
1
2
·3
= (2
1
2
)
3
= 2
1
2
2
1
2
2
1
2
i.e., the cube of 2
1
2
, which we have just defined. This suggests that, in general, for any
positive integers m and n, and for any positive number α, we define
α
m
n
= α
1
n
α
1
n
· · · α
1
n
( m times)
In other words,
α
m
n
= (α
1
n
)
m
For example, 4
5
3
means the cube root of 4 raised to the fifth power, i.e.,
4
5
3
= 4
1
3
4
1
3
4
1
3
4
1
3
4
1
3
114
We are almost through with the definitions! There is only one more to go: negative
exponents. What should 4
−
5
3
mean? We appeal to the analogue of (8.1) for rational
exponents to obtain:
4
−
5
3
4
5
3
= 4
−
5
3
+
5
3
= 4
0
= 1
Thus we get
4
−
5
3
4
5
3
= 1
Multiplying both sides by
1
4
5
3
we get
4
−
5
3
=
1
4
5
3
There is nothing special about either the number 4 or the exponent −
5
3
, so if we insist
that (8.1) be valid for rational exponents, we should define for any positive number α
and for any fraction (i.e., positive rational number) A,
α
−A
=
1
α
A
We have now finished defining α
r
for any positive number α and for any rational
number r. But our work has barely begun! We do not want to get a general definition
of α
r
just to satisfy our idle curiosity; our goal is rather to generalize (8.1)–(8.3) to
rational exponents. With this in mind, we now claim that for any positive number α
and β, and for any rational numbers r and s, the following are valid:
(8.4) α
r
α
s
= α
r+s
(8.5) (α
r
)
s
= α
rs
(8.6) (αβ)
r
= α
r
β
r
These are the laws of exponents. In advanced courses, it is shown how to define α
r
for any number r, rational or irrational, and prove these laws in one fell swoop for all
numbers r and s using sophisticated reasoning. Our more modest task here is to at least
get a partial understanding of these laws for rational exponents.
The complete proofs of (8.4)–(8.6) are, without a doubt, unsuitable for school mathe-
matics, especially grade eight. These proofs are as unpleasant as the proofs of (8.1)–(8.3)
are trivial. Nevertheless, students should at least have a glimpse of why (8.4)–(8.6) are
115
true, because they will be using these laws often and it is altogether a bad idea if they are
total strangers to any basic tools they use. On the other hand, all prospective teachers
should at least go through the complete proof of at least one of these laws, because if
they will be telling their students that these proof are too difficult, then they should at
least know whereof they speak.
Let us consider the problem of how to give students a glimpse of the reasoning behind
the laws of exponents. We suggest first of all to check a few simple cases. For example:
we check
64
1
2
64
2
3
= 64
7
6
Note that
7
6
=
1
2
+
2
3
. Now 64 = 8
2
, so 64
1
2
= 8. Also 64 = 4
3
, so 64
2
3
= (64
1
3
)
2
= 4
2
=
16. Thus
64
1
2
64
2
3
= 8 16 = 128
On the other hand, 64 = 2
6
, so also
64
7
6
= (64
1
6
)
7
= 2
7
= 128
and we are done.
We also suggest giving a proof of a special case of each of (8.4)–(8.6), such as the
following:
_
¸
¸
¸
¸
¸
_
¸
¸
¸
¸
¸
_
5
−
1
2
5
−
1
4
= 5
−
3
4
(5
2
)
1
3
= 5
2
3
(αβ)
1
n
= α
1
n
β
1
n
(11)
when the last equality is for all positive numbers α and β and for all positive integer n.
The proofs of all three items in (11) make use of the following useful lemma. (In
the tradition of Euclid, a lemma is a theorem that is deemed to be of a lower status,
either because it is not as central or not as comprehensive as others already designated
as theorems. It is sometimes a very subjective judgment whether something should be
a lemma or a theorem.)
Lemma 1 If two positive numbers α and β satisfy α < β, then for any positive
integer n, α
n
< β
n
.
Proof Suppose α < β, then (because α > 0), multiplying both sides by α gives
α
2
< αβ
116
(see item (δ) of Section 7). Now we multiply both sides of α < β by β to get
αβ < β
2
Combining the preceding two displayed inequalities, we get α
2
< β
2
. Next we will
prove that α
3
< β
3
. Multiplying both sides of α
2
< β
2
by α, we get
α
3
< αβ
2
Now we make use of the just-proven inequality αβ < β
2
: multiply both sides by β to
get
αβ
2
< β
3
Combining the preceding two displayed inequalities, we get α
3
< β
3
. Continuing one
more step, we want to show next that α
4
< β
4
. Multiplying both sides of α
3
< β
3
by
α, we get
α
4
< αβ
3
Now multiply both sides of the just-proven inequality αβ
2
< β
3
by β and we get
αβ
3
< β
4
Combining the last two displayed inequalities, we get α
4
> β
4
. In this way, we will
eventually get to α
n
< β
n
no matter what n is. This completes the proof.
The following two corollaries (immediately drawn conclusions) of Lemma 1 are of
independent interest.
Corollary 1 For two positive numbers α and β, if α
n
< β
n
for some positive
integer n, then α < β.
This is of course the converse of Lemma 1, so it is a curious fact that it is also a
corollary of Lemma 1. Before explaining why, let us note first of all what it does not
say: it does not say that if any two numbers α and β satisfy α
n
< β
n
for some positive
integer n, then α < β. For example, 3
2
< (−4)
2
, but 3 > (−4). So the truth of
Corollary 1 depends critically on the positivity of both α and β.
As to the deduction of Corollary 1 from Lemma 1, we make use of what is called
the trichotomy law among numbers, which states that for any two numbers a and
b, one and only one of the three possibilities holds: either a = b, or a < b, or a > b.
117
If you recall that numbers are points on the number line, then this assertion becomes
obvious. The point is however that such an obvious fact can also be an effective tool in
problem-solving. For the case at hand, we want to prove, under the assumptions α > 0,
β > 0, and α
n
< β
n
, that α < β. By the trichotomy law, it is suficient to show that
neither α = β nor α > β is a possibility. Let us first rule out α = β: in this case,
clearly α
n
= β
n
for any positive integer n, thereby contradicting the hypothesis that
α
n
< β
n
. If α > β, then Lemma 1 implies α
n
> β
n
for any positive integer n, again
contradicting the hypothesis. Therefore, only α < β is possible.
The next corollary is the statement that the positive n-th root of a positive number
is unique.
Corollary 2 If two positive numbers α and β satisfy α
n
= β
n
for some positive
integer n, then α = β.
We use the trichotomy law again to eliminate the possibility of either α < β or
α > β. Lemma 1 says that if either is true, then α
n
,= β
n
, which contradicts the
hypothesis. Thus α = β, as desired.
We are now ready for the proof of (11). To prove the first item of (11), it suffices to
prove (according to Corollary 2) that
(5
−
1
2
5
−
1
4
)
4
= (5
−
3
4
)
4
Now,
(5
−
1
2
5
−
1
4
)
4
=
_
1
5
1
2
=
m+nk
n
here, and the use of this
formula in this setting may be as good a reason as any as to why one should not follow the
common method of defining fraction addition in terms of the LCM of the denominators
n and .) By Corollary 2, it suffices to prove
(α
m
n
α
k
)
which is exactly (a). The proof of case (c) is simpler. By (12), we know
α
m
n
α
k
= α
m
n
+
k
By the cross-multiplication algorithm, this is equivalent to
1
α
m
n
α
k
=
1
α
m
n
+
k
which is of course the same as
α
−
m
n
α
−
k
= α
−
m
n
−
k
The proof of (8.4) is complete.
Remark We have chosen to give a complete proof of (8.4) because until one goes
through such a proof, one doesn't know what it means that the proofs of (8.4)–(8.6)
are unpleasant. Now in mathematics, when things get unpleasant, 90% of the time it is
because they are not done "the right way". In this case, the right way is to prove (8.4)–
(8.6) for all number r, s all at once without restricting oneself to only rational values of
r and s. Such a proof is achieved by appealing to basic properties of the exponential and
logarithmic functions and using differentiation. This will be done in advanced courses.
For our task at hand, it remains to emphasize that you should feel free to use (8.4)–(8.6)
for all values of r and s and worry about their proofs later.
122
We will put (8.4)–(8.6) to use in the next section.
Before we leave the discussion of exponents, we would like to expand on the meaning
of an expression as defined in Section 1. Recall that an expression (or number expression)
is simply a collection of numbers x, y, etc. connected by the four arithmetic operations.
Now that exponents are available, we can add to the meaning of an expression by
defining it to mean a collection of numbers x, y, etc. which are connected by the four
arithmetic operations and the use of rational exponents. In this context, an expression
is also called an algebraic expression. Thus, the following is an algebraic expression:
x
−3
+¦(yz)
2
+ 5¦
3
4
−(
xy
z
)
5
We now introduce the concept of absolute value of a number. For any number x, the
absolute value |x| of x is by definition:
[x[ = the distance of x from 0
Thus
[x[ =
_
x if x ≥ 0
−x if x < 0
For example, [ −
7
6
[ =
7
6
, [1728[ = 1728, and [ −62.9[ = 62.9.
Note in particular that
[x[ ≥ 0 for every number x, and [x[ = 0 is equivalent to x = 0.
The relevance of absolute value in the context of exponents lies in the following
alternate definition of the absolute value [x[ of a number x:
[x[ =
√
x
2
The fact that this is equivalent to the original definition is easy to see.
A question often asked is, "why bother with absolute value?" A superficial answer
is that very often one wishes to say that a number x is "close" to another number, and
the concept of absolute value allows us to express this "closeness" in a simple manner.
For example, a common expression is that "today's temperature is about 60 (degrees
Fahrenheit)". It means the temperature of today, say x degrees, is at most 5 degree
more than 60 or 5 degrees less than 60. Thus,
123
if x ≥ 60, then x −60 ≤ 5, and
if x < 60, then 60 −x ≤ 5.
We now show that one could express this much more simply as:
[x −60[ ≤ 5
For convenience of discussion, let us refer to these two ways of expressing "today's
temperature is about 60" as the First Expression and Second Expression, respectively.
We want to show that they are equivalent. So suppose x satisfies the First Expression,
then we will show it satisfies the Second Expression. On the number line, an x satisfying
the First Expression must stay within the thickened line segment below:
55 60 65
x
¡ ¸
Now consider [x−60[. If x ≥ 60, then [x−60[ = x−60, and we know x−60 ≤ 5 from
the First Expression. So [x −60[ ≤ 5 in this case. If x < 60, then [x −60[ = 60 −x
and again we know 60 −x ≤ 5 in this case. Thus also [x−60[ ≤ 5. Altogether, we have
[x −60[ ≤ 5 no matter what x is.
Conversely, suppose x satisfies the Second Expression, then [x −60[ ≤ 5. If x ≥ 60,
then [x −60[ = x −60 and therefore x −60 ≤ 5. If on the other hand, x < 60, then
[x − 60[ = 60 − x and the fact that [x − 60[ ≤ 5 now becomes the statement that
60 −x ≤ 5. Thus we see that x satisfies the First Expression.
The idea behind the preceding discussion has a wider application, and we isolate the
two key points. Define the distance between two numbers on the number line as the
length of the segment joining them. Then the first is that for any two numbers x and c,
[x −c[ is the distance between x and c on the number line.
This is easily seen, as in the preceding discussion, by first considering the case that
x ≥ c, and then the case x ≤ c, as shown:
c x c x
124
Thus to say "x is within units of c", we can simply say: [x −c[ ≤ .
A second key point is to point out that an inequality involving absolute value can
be expressed directly in terms of ordinary inequalities. We introduce a new notation for
this purpose: if three numbers a, b, c satisfy the following inequalities:
a ≤ b, and b ≤ c
then we abbreviate these inequalities into a double inequality:
a ≤ b ≤ c
We now prove:
Lemma 3 Let x, c be arbitrary numbers and let be a positive number. Then
[x −c[ ≤ is equivalent to the double inequality c − ≤ x ≤ c + .
The proof is best achieved through the use of a picture of the number line. We have
seen that the inequality [x −c[ ≤ is the statement that the distance of x from c is at
most . The latter means that x can move down the number line only as far as c − ,
and can move up only as far as c +. But then this is exactly the content of the double
inequality c − ≤ x ≤ c + .
c − c c +
x
¡ ¸
We now give a more satisfactory answer to the question of "why bother with absolute
value?" It is because in mathematics the distance of a number x (regardless of whether x
is positive or negative) from 0 is very often an issue of intense interest. Having a symbol
to denote this distance, i.e., [x[, then allows for precise computations with this symbol
to estimate this distance. We give two elementary examples to illustrate this kind of
computation. The first is the inequality
[x + y[ ≤ [x[ +[y[
for all number x and y. A second example is the inequality, also valid for all numbers x
and y:
[2xy[ ≤ x
2
+ y
2
125
Imagine how clumsy these inequalities would be if the concept of absolute value were
not available. Notice that the second inequality actually compares two absolute values,
i.e.,
[2xy[ ≤ [x
2
+ y
2
[
One gains an appreciation of the importance of absolute values if one experiments with
ways of formulating inequalities like these without having absolute values at one's dis-
posal. For example, compare an inequality such as
2xy ≤ x
2
+ y
2
with the preceding one. Which carries more information? The inequality 2xy ≤ x
2
+y
2
does not exclude the possibility that two numbers x and y satisfy x
2
+y
2
= 1 but their
product xy is −2, or −3, or even −150. However, the inequality [2xy[ ≤ [x
2
+ y
2
[
says this does not happen because the product 2[xy[ must satisfy [2xy[ ≤ 1 so that
the product xy can be as far from 0 as −
1
2
, but nothing further to the left on the
number line.
In advanced mathematics, most of the important inequalities are those involving ab-
solute values.
In Example 1 of Section 7, we graphed an inequality on the number line. Here is
more of the same, with the only difference that we now make use of absolute value.
Example Graph [6 + 2x[ > 1 on the number line.
The inequality is the same as
1
2
[6 + 2x[ >
1
2
1, which in turn is the same as
[3 + x[ >
1
2
(cf. Problem 6 (iv) of the following Exercises). Since [3 + x[ = [x −(−3)[,
the inequality is therefore the same as [x −(−3)[ >
1
2
. This means we have to find all
the points x so that its distance from −3 is bigger than
1
2
. From the picture,
r r
−3
1
2
−2
1
2
−3
we see that the graph is the union of two semi-infinite segments: the segment to the left
of −3
1
2
and including −3
1
2
, and the segment to the right of −2
1
2
and including −2
1
2
.
EXERCISES
126
1. Explain why there is no number on the number line whose square is a negative
number.
2. Prove that for all rational numbers r, s and t, (α
r
β
s
)
t
= α
rt
β
st
.
3. Verify the equality 729
1
2
729
1
3
= 729
5
6
by direct computation. Do the same to
117649
1
2
117649
1
6
= 117649
2
3
4. Prove using only the definition of α
m
n
, but without using (8.5), that (α
m
n
)
1
m
= α
1
n
.
5. Prove that for any rational number r, and for any positive number α,
α
−r
=
1
α
r
6. Verify each of the following about absolute values of arbitrary numbers x, y: (i)
[ −x[ = [x[, (ii) −[x[ ≤ x ≤ [x[, (iii) [x + y[ ≤ [x[ +[y[, (iv) [xy[ = [x[ [y[.
7. Prove: [2xy[ ≤ x
2
+ y
2
for all numbers x and y.
8. Given two similar triangles as shown:
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
a
a
If the ratio of the area of the smaller triangle to the area of the bigger triangle is s, what
is the ratio
a
a
in terms of s ?
9. Recall that an annual interest rate of x percent means that an account of P
dollars earns at the end of one full year an amount of
_
x
100
_
P
dollars. Derive a formula which gives the amount of money in an account at the end of
n years if the account has an initial deposit of P dollars and an annual interest rate of
x percent.
10. Graph on the number line each of the following: (i) [x[ − 14 > −8, (ii)
[x[ −4 <
1
3
, (iii) 9 −[3x −1[ < 4, (iv) [2x +
3
5
[ ≥
1
5
, (v) [6x + 1[ + 2
1
4
< 5.
127
9 Functions and Their Graphs
A function from a set A to a set B is a rule (i.e., a precise prescription) that assigns to
each element of A an element of B. If the function is denoted by f, then f : A → B
is the correct notation to capsulize this information. However, when A and B are
understood, the function f is often denoted generically by f(x). If f assigns the element
b of B to an element a of the set A, then we write
f(a) = b
For example, if F is the function
F : ¦all numbers¦ →¦all numbers¦
which assigns to each number its square, then F can be succinctly given as F(x) = x
2
for each number x. We note for emphasis that F(x) is always ≥ 0 for any x, so if we
write instead
F : ¦all numbers¦ →¦all numbers ≥ 0¦
then it would also be correct.
If G is the function
G : ¦a deck of cards¦ →¦club, diamond, heart, spade¦
which assigns to each card its suit, then what G does to each card would be difficult to
describe in symbols. One can illustrate by giving some examples, such as
G(King of diamonds) = diamond
G(Two of spades)= spade
G(Queen of hearts) = heart, etc.
In a course on introductory algebra, the kind of functions one encounters are usually
those from a set of numbers or a set of points in the plane to some other set of numbers.
For the moment, however, we want to give a general discussion and would not want to
limit ourselves, yet, to these common cases. First we want to say a few words about
why we study functions.
Functions are needed to describe the various processes and phenomena that occur in
the human and the natural worlds. Both worlds are in a state of change, and functions
are the tools to describe change. We will give three examples. Suppose you have just
128
brewed a cup of coffee and are waiting for it to cool down. So you watch out for the
temperature of the coffee. In this case, "the temperature" is a number that depends
on the time of the measurement; it is not a single number, but an infinite collection
of numbers, each corresponding to a particular time of measurement. Let us say it is
195
◦
(Fahrenheit) at the beginning. A minute later, the temperature drops to 180
◦
.
In another minute, it is 165
◦
; in succeeding minutes it is 153
◦
and 143
◦
, respectively.
Finally afer five minutes, it drops to 135
◦
and it becomes borderline drinkable.
Now this verbal description of the change in temperature is cumbersome and totally
unsuitable for scientific investigations. As we have emphasized all along, it is important
to render this information into symbolic form so that we can compute with it. Here then
is an alternative: denote by f(t) the temperature of the coffee (in degrees Fahrenheit)
t minutes after the coffee is brewed. The following table then summarizes every piece of
information in the last paragraph:
t f(t)
0 195
1 180
2 165
3 153
4 143
5 135
Of course, one can get much more information about the temperature of the coffee by
measuring it at shorter time intervals. For example:
t f(t)
1.5 172
2.5 159
3.5 148
3.8 145
4.5 139
4.8 137
Reading from the second table, we see that 4.8 minutes (i.e., 4 minutes and 48
seconds) after the coffee is brewed, the temperature is 137
◦
F. Thus f is a function which
assigns to every number t ≥ 0 another number, namely, the temperature of the coffee t
minutes after it is brewed, i.e.,
129
f : ¦all number ≥ 0¦ →¦all numbers¦
f(t) = the temperature of the coffee t minutes after it is brewed.
Consider a second example: a man drives to the airport which is 25 miles away. He
plans to leave his house two hours before departure time. If we want to see how far
he is from the airport, clearly one number won't get the job done because this distance
depends on the time when the distance is measured. Our experience with the coffee
problem suggests that we make use of a function F for this description, such that
F(t) = his distance (in miles) from the airport t minutes after he leaves his
house
Thus F(0) = 25. In general, F assigns to each number t ≥ 0 another number which
is his distance in miles from the airport at time t. Even a skeletal description of this
function in terms of a few values of t can tell a story, as for instance:
t f(t) t f(t)
0 25 30 13
5 24 35 19
10 22.5 40 24
15 21 43 25
20 16 44 25
25 10 45 24.5
26 9 55 13
27 10 60 7.5
28 11 67 0
We can see that he has to start his trip slowly probably because of city traffic, so
that after 10 minutes he only travels two and a half miles. Around the 26-th minute
after he leaves home, he turns around as the values of f(27) and f(28) and those of
subsequent minutes show that he is driving away from the airport. He forgets to bring
his photo-ID (a guess!). He manages to get home at 43 minutes after his departure and
it takes him only about a minute to get the necessary document. Then he speeds a bit
as he makes it to the airport in 23 minutes (67 − 44 = 23); not trivial considering the
traffic condition these days. He has a few minutes to spare.
As a final example, consider the problem of the temperature of the city of Berkeley
on a certain day. To say that Berkeley is 67
◦
(Fahrenheit) makes no sense, strictly
130
speaking. Is the temperature taken in the early dawn or in the afternoon? In Berkeley,
this could mean a 25
◦
difference. And where is the temperature measured: at the top of
the hill (about 1500 feet high), downtown, or by the Bay? The difference here could be
another 15
◦
. If we start measuring the time t in hours from midnight, then 0 ≤ t ≤ 24.
To specify the geographic location, we need two more numbers which may be thought
of as the idealized x and y coordinates. Berkeley being a small city, 5 miles from the
city center in any direction would include everything. Therefore, a scientifically usable
description of the temperature of Berkeley would make use of a function T, so that, if
o is the region in 3-space consisting of all ordered triples of numbers (x, y, t) so that x
and y satisfy [x[, [y[ ≤ 5 (miles), and 0 ≤ t ≤ 24 (hours), then
14
F : o → ¦all numbers¦
F(x, y, t) = the temperature of Berkeley, t hours past midnight at a spot
specified by the x and y coordinates
Incidentally, such a function F is said to be a function of three variables, because
three numbers x, y, and t are involved in its definition. It may not have escaped your
attention that we give F this name as an afterthought, without holding forth on the
philosophical implications of what a "variable" is. This is as it should be. (Compare
the discussion of the term "variable" given at the beginning of Section 2.)
Quite apart from the description of change, functions have already forced their way
into our work whether we know it or not. Transformations of the plane, including
rotations, reflections, and translations, are functions which assign to each point of the
plane another point of the plane, i.e., these are examples of function T , so that
T : ¦the plane¦ → ¦the plane¦
For example, the translation T which moves every point of the plane 2 units to the left
horizontally is precisely given by:
T(x, y) = (x −2, y)
These are among the simplest examples of how functions naturally arise. Of course
functions are everywhere as soon as you look around. For example, in the Manufacturing
Problem of Section 7, we had to use the profit function P to describe the possible
14
Notice also the use of absolute value to describe the physical extent of the city. It means of course
that −5 ≤ x ≤ 5 and −5 ≤ y ≤ 5. See Lemma 3 of Section 8.
131
profits of the game manufacturer in various scenarios. (Back then we slipped in the
nomenclature of a "function" without warning.) Now that we know the language and
notation of a function a bit better, we can improve on the notation as follows: if the
game manufacturer makes x A games and y B games, then
P(x, y) = the profit she makes by selling x A games and y B games.
Recall that P(x, y) = 125x + 185y, so that P(28, 22) = 125 28 + 185 22 = 7570.
Thus the profit of selling 28 A games and 22 B games is $7570.
We already saw in Section 8 that it was to our advantage to expand the role of
P by allowing it to assign to any point (x
, y
) in the feasibility region 1 the number
125x
+ 185y
. Then P becomes a function
P : 1 → ¦all numbers¦
P(x
, y
) = 125x
+ 185y
The purpose of this discussion is to make it plain that the concept of a function is not
something artifically concocted for the purpose of giving students a hard time. Rather,
it is a tool, created out of necessity, to effectively describe the phenomena around us be
they sociological or mathematical. It is indispensable.
The best way to get to know the concept of a function is to look at many examples
of functions and to examine their graphs. Precisely, let f be a function from a set of
numbers A to a set of numbers B, f : A → B. Then the graph of f is the set of all
the points (x, f(x)) in the plane, where x is an element of A. In general, of course,
the set A is infinite, so that the graph of f is an infinite set as well. In principle, it is
impossible to literally get hold of the whole graph of any function. However, it is usually
the case that plotting a finite number of well-chosen points in the graph is enough to
reveal the essential features of the graph, and therefore of the function itself. It should
be stated that plenty of practice in plotting points on a graph is an essential component in
the learning process of getting to know graphs and functions. So please remember: use the
graphing calculator only after you have acquired fluency in plotting points.
Before we give some examples, we introduce a few related concepts. Suppose we are
given a function H from the plane to the set of all numbers, i.e., for any point (x, y) in
the plane, H assigns to it the number H(x, y). Such a function H is called a function
of two variables. In Section 7, we already came across a special case of this concept,
132
namely, a linear function of two variables. The graph of the equation H = 0 is the
set of all the points (x
, y
) so that H(x
, y
) = 0. Where these concepts of graphs come
together is the following. Suppose f is a function from numbers to numbers,
f : ¦all numbers¦ → ¦all numbers¦
Define a function of two variables H
0
by H
0
(x, y) = y −f(x) for all x and y. Then
¦the graph of f¦ = ¦the graph of the equation H
0
= 0 ¦.
This is because (x, y) being in the graph of H
0
= 0 is equivalent to y − f(x) = 0,
which is equivalent to y = f(x), which is then equivalent to (x, y) = (x, f(x)). The
last is of course equivalent to (x, y) being in the graph of f itself.
This fact will be useful below.
Example 1 A linear function of one variable g is by definition a function
from numbers to numbers so that for some fixed constants a and b, g(x) = ax + b for
all numbers x. The graph of g is therefore the set of all points (x, ax + b). First look
at a special case, such as g
1
(x) = 2x + 3 (thus a = 2, b = 3). Imagine that we are
approaching the graph of g
1
for the first time, which points should we plot? We should
certainly find out where the graph crosses the coordinate axes. In other words, if (0, y
0
)
and (x
0
, 0) are on the graph, what are the values of x
0
and y
0
? Since all points on the
graph of g
1
are of the form (x, 2x + 3), we see that y
0
= 2 0 + 3 and 2x
0
+ 3 = 0.
Therefore y
0
= 3 and x
0
= −
3
2
, so that (−
3
2
, 0) and (0, 3) are on the graph. Once
these "obvious" points have been plotted, we pick some other points in random, e.g.,
letting x = −2, −1, 1, 2, 3, we get the following points on the graph of g
1
:
(−2, −1) (−1, 1) (1, 5) (2, 7) (3, 9)
133
q
q
q
q
q
q
−2 −1 O 1 2 3
1
3
5
7
8
q
If one has to guess on the basis of these seven points, one would say that the graph
of g
1
is a straight line. This turns out to be correct because the graph of g
1
is the same
as the graph of y = g
1
(x), i.e., y = ax+b, which is in turn the same as (−a)x+y = b.
Thus the graph of the linear function of one variable g
1
is the same as the graph of
the linear equation of two variables (−a)x + y = b (which is of course the same as the
equation (−a)x + y − b = 0). We already know from Section 4 that the latter is a
straight line. This is why the graph of g
1
is a straight line.
The general case is no different: Given f(x) = ax + b. Then the same reasoning
shows that the graph of the linear function of one variable f is the same as the graph
of the linear equation of two variables (−a)x + y = b, and is therefore a line. Thus the
graph of a linear function of one variable is always a line in the plane. By Section 4, we
know in addition that this line has slope a and y-intercept b.
Activity 1 Plot the given points on the graph of the function f in the coffee
problem above.
Activity 2 Plot the given points on the graph of the function F in the problem
of the man driving to the airport.
Example 2 Graph the function s(x) = x
2
.
134
The graph of s consists of all the points of the form (x, x
2
), where x is arbitrary.
Since (−x)
2
= x
2
, we see that the graph includes both (x, x
2
) and (−x, x
2
), no matter
what x may be. The point (0, 0) is the obvious point on the graph. We can put in values
of x = ±1, ±2, ±3, ±4 to get the points
(±1, 1) (±2, 4) (±3, 9) (±4, 16)
Let us also throw in
(±0.5, 0.125) (±1.5, 2.25) (±2.5, 6.25) (±3.5, 12.25)
for good measure, and we get the following sequence of points on the graph of s. Note
that in order to make the picture manageable, we have shrunk the scale of the y-axis by
half.
q q
q
q q
q q
q q
q q
q q
q q
q q
−4 −2 O 2 4
2
6
10
14
16
It is not difficult to extrapolate from these points to envision the graph as the following
curve:
135
This curve is an example of what is called a parabola. Parabolas are discussed more
fully in Sections 11 and 12.
We emphasize once more that it is the plotting of points on the graph, and the ability
to envision the whole graph on the basis of these points, that are our major concern here.
The beginning stage of learning about graph is best done being away from a graphing
calculator.
Example 3 Graph the function h given by h(x) =
1
x
.
We first address an issue concerning this h that we have not confronted thus far,
and it is this. Would it be correct to say that h is a function from all numbers to all
numbers? In other words, would it be correct to say that
h : ¦all numbers¦ → ¦all numbers¦
The answer is no, because h cannot assign any number to 0, as infinity is not a number.
So the correct statement is that
h : ¦all nonzero numbers¦ → ¦all numbers¦
That said, we start plotting points. Again, there are two obvious points: (1, 1) and
(−1, −1). Beyond that we take some random values of x and compute
1
x
, and we will
136
remark on the following choices in due course:
(0.1, 10) (0.2, 5) (0.4, 2.5) (0.5, 2) (2, 0.5) ((4, 0.25)
(5, 0.2) (8, 0.125) (10, 0.1) (−0.1, −10) (−0. −2, 5) (−0.4, −2.5)
(−0.5, −2) (−2, −0.5) (−4, −0.25) (−5, −0.2) (−8, −0.125) (−10, −0.1)
The corresponding picture is then:
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q 2 4 6 8 10
−10
10
8
6
4
2
−10
Notice that there are two separate curves here, and they are called the two branches of
a hyperbola. Regrettably, we will not pursue the study of hyperbolas here.
The plotted points above exhibit a pattern: if 0 < a < b or a < b < 0, then
1
a
>
1
b
.
(In an exercise, you are asked to prove this in general.) This pattern justifies that this
limited choice of the points on the graph is enough to reveal the general behavior of the
graph: it tells us that as the upper right curve extends to the right end of the positive
x-axis, all it does is to get closer and closer to the x-axis, and as it approaches 0 from
the positive x-direction, all it does is to get closer and closer to the positive y-axis. A
similar statement also applies to the lower left curve.
Example 4 Graph the function given by f(x) =
√
x.
We note, as in Example 3, that this is not a function from all numbers to all numbers,
but
f : ¦ all numbers ≥ 0¦ →¦ all numbers¦
137
The following sequence of points on the graph of f are self-explanatory:
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
O 1 4 9 16
1
2
3
4
This sequence of points exhibit a different pattern: when 0 < x < 1, x <
√
x, and
when 1 < x, x >
√
x. (This too will be an exercise.) Note also that if 0 < a < b,
then
√
a <
√
b. This fact then tells us that there is no need to see more points on the
graph because as we go towards the right end of the positive x-axis, the graph simply
rises in the same way as it does here for the values of 1 ≤ x ≤ 16.
Example 5 Graph the function G given by G(x) = x
3
−3x + 6.
This is a cubic polynomial and, at the same time, a function from all numbers to
all numbers. Since we have no idea what to expect, we try some obvious numbers, e.g.,
G(0), G(±1), G(±2), G(±3), G(±4), getting the following points on the graph of G:
(−4, −48) (−3, −12) (−2, 4) (−1, 8) (0, 6)
(1, 4) (2, 8) (3, 24) (4, 58)
By compressing the y-axis, we can exhibit these points as follows:
138
q
q
q
q
q
q
q
q
q
−4 −3 −2 −1 O 1 2 3 4
−10
−20
−30
−40
−50
−60
10
20
30
40
50
60
The graph seems to cross the x-axis between −3 and −2. Suppose it crosses the x-axis at
(x
0
, 0), then 0 = G(x
0
) by definition of the graph of G. This means x
3
0
−3x
0
+6 = 0.
Such an x
0
is called a root of the cubic polynomial equation x
3
− 3x + 6 = 0. The
roots of a polynomial equation are of great interest in mathematics (see the last two
sections on the roots of quadratic polynomials equations). For this reason, one may try
to get a better estimate of this x
0
. We have G(−2.5) = −2.125, G(−2.4) = −0.624,
G(−2.3) = 0.73, so it is intuitively clear that x
0
is between −2.4 and −2.3. By
experimenting with G(−2.31), etc., we can get even better estimates of this root.
Notice that the graph has a bump above (roughly) −1, and has a trough above
(roughly) 1. It is a known fact, proved in advanced courses, that the graph of a cubic
polynomial can have at most one bump and one trough. Therefore we are very fortunate
that with the choice of the nine obvious points on this graph, we already know that
there are no further troughs and bumps below x = −4 and above x = 4. So the graph
will continue to go down as we go to the left x-axis, and continue to go up as we go to
the right of the x-axis.
If we were less fortunate and the chosen points happen not to reveal the bump and
the trough of the graph, then we would have to plot more points, since these features
may not have appeared yet or may even not exist.
139
Example 6 Graph H(x) = 2
x
.
It is understood that we do not know as yet the meaning of 2
x
when x is an irrational
number, but as usual in school mathematics, we ignore this gap in our knowledge and
concentrate on the rational values of x. Observe that (0, 1) is on the graph of H. We
should obviously get the points of the graph above x = ±1, ±2, ±3, ±4. Note that
H(−1) =
1
2
= 0.5, H(−2) =
1
2
2
= 0.25,
H(−3) =
1
2
3
= 0.125, H(−4) =
1
2
4
= 0.0625.
We should get a few more points, e.g.,
H(
1
2
) =
√
2 ∼ 1.4
H(−
1
2
) =
1
√
2
∼ 0.7
H(
3
2
) = (
√
2)
3
∼ 2.8
H(−
3
2
) =
1
(
√
2)
3
∼ 0.35
H(
5
2
) = (
√
2)
5
∼ 5.7
H(−
5
2
) =
1
(
√
2)
5
∼ 0.18
H(
7
2
) = (
√
2)
7
∼ 11.3
H(−
7
2
) =
1
(
√
2)
7
∼ 0.09
where the symbol "∼" means "approximately equal to". Here is the picture of these
points:
140
q q q q q
q
q
q
q
q
q
q
q
q
q
q
q
−4 −3 −2 −1
O
4 3 2 1
2
4
6
8
10
12
14
16
Notice that the points rise steadily to the right, and this is because for all rational
numbers r and s so that r < s, we have 2
r
< 2
s
. The proof of a more general fact is
left as an exercise.
We do not have the time here to pursue the study of this function H, or the func-
tions α
x
for a positive α. It is worthwhile to mention nevertheless that these functions
are important in mathematics and the sciences. For our purpose, they give a hint of
the reason why we want to devote a whole section to the discussion of rational exponents,
With the concept of a function available, we are now in a position to revisit the
earlier discussion of rates and constant rates in Section 5, especially Examples 4–8 and
make better sense of that discussion. For definiteness, let us consider the case of motion.
An object in motion is described by a function f, where
f(t) = the distance traveled from time 0 to time t
Let us say t is measured in hours and f(t) in miles. We adopt the usual convention of
letting time 0 be the starting time and that f(0) = 0 by definition. Formally introduce
the concept of average speed during the time interval from time t
1
to time t
2
(0 ≤
t
1
< t
2
) as the distance traveled from time t
1
to time t
2
divided by the length of the
time interval (which is of course t
2
−t
1
). In terms of f, we therefore have:
average speed from t
1
to t
2
=
f(t
2
) −f(t
1
)
t
2
−t
1
Average speed therefore measures miles per hour (mph). Finally, we say the motion has
constant speed v mph if the average speed over any time interval is the fixed number
141
v. Here is the basic observation.
Theorem Notation as above, the fact that the motion has constant speed v mph is
equivalent to the fact that f is given by f(t) = vt miles for all t ≥ 0.
Proof First suppose the motion has constant speed v. Then by definition,
f(t
2
) −f(t
1
)
t
2
−t
1
= v
for any t
1
and t
2
so that 0 ≤ t
1
< t
2
. By letting t
1
= 0 and let t
2
be arbitrary, we get
f(t
2
)
t
2
= v
which is the same as
f(t
2
) = vt
2
Since t
2
is arbitrary, we have f(t) = vt for all t. Conversely, suppose f(t) = vt. By
definition,
average speed from t
1
to t
2
=
f(t
2
) −f(t
1
)
t
2
−t
1
=
vt
2
−vt
1
t
2
−t
1
=
v(t
2
−t
1
)
t
2
−t
1
= v
This then proves the theorem.
For your beginning students in algebra, it would be good to point out that the can-
cellation of (t
2
− t
1
) in the last calculation depends on the fact that it is a nonzero
number, which explains why we were so careful to specify that t
1
< t
2
.
Naturally, an entirely similar discussion can be given for water flow at a constant
rate, work done at a constant rate, etc. For example, in case of water flow from a faucet
(let us say), let F be the function so that F(t) is the amount of water (in gallons) coming
out of the faucet from time 0 to time t (in minutes). The rate of flow is constant, by
definition, if there is a fixed number r so that during any time interval from t
1
to t
2
(0 ≤ t
1
< t
2
) the total amount of water coming out of the faucet in this time interval
142
divided by the length of the time interval is always equal to r gallons per minute. In
other words,
F(t
2
) −F(t
1
)
t
2
−t
1
= r
for all t
1
and t
2
so that 0 ≤ t
1
< t
2
. One then proves in exactly the same way that
the rate of water flow being a constant r gallons per minute is equivalent to F(t) = rt
gallons for all t ≥ 0.
For work such as painting a house, let H be the function so that H(t) is the number of
square feet painted from time 0 to time t (in minutes). The rate of painting is constant,
by definition, if there is a fixed number c so that during any time interval from t
1
to t
2
(0 ≤ t
1
< t
2
) the total number of square feet painted in this time interval divided by the
length of the time interval is always equal to c square feet per minute. Thus,
H(t
2
) −H(t
1
)
t
2
−t
1
= c
for all t
1
and t
2
so that 0 ≤ t
1
< t
2
. One then proves as before that the rate of painting
the house being c square feet per minute is equivalent to H(t) = ct square feet.
We now do a prototypical "rate problem" from the perspective of linear functions.
This problem can be done without algebra, so it is the reasoning behind the solution
based on algebra that will be the main interest here. In the ensuing discussion, one
can also appreciate the importance of being able to transcribe verbal information into
symbolic expressions (Section 2).
Example 7 Joshua, Li and Manfred are going to paint a house together. It is
estimated that, individually, it will take them 18 hours, 15 hours, and 16 hours, respec-
tively, to paint the whole house. Assuming that each person paints at a constant rate,
how long will it take them to do it together?
By the constant rate assumption, we know that there are numbers j, , and m that
the number of square feet that each of Joshua, Li and Manfred paints in t hours are,
respectively,
J(t) = jt
L(t) = t
M(t) = mt
143
We can determine each of these constants j, and m. Let A be the the number of
square feet of the house that need painting. Since it take Joshua 18 hours to paint the
house, we see that J(18) = A. Thus 18j = A and j =
A
18
. In like manner, we get
=
A
15
and m =
A
16
. If all three paint together, then in t hours, each of Joshua, Li
and Manfred paints, respectively,
A
18
t,
A
15
t, and
A
16
t square feet. Therefore all three
together paint
A
18
t +
A
15
t +
A
16
t =
_
1
18
+
1
15
+
1
16
_
At
square feet in t hours. If they paint A square feet (i.e., the whole house) in t
0
hours,
then
_
1
18
+
1
15
+
1
16
_
At
0
= A
Multiplying both sides by
1
A
, we get
_
1
18
+
1
15
+
1
16
_
t
0
= 1
and therefore
t
0
=
1
1
18
+
1
15
+
1
16
= 4
32
37
hours.
Note two things. The preceding argument shows that, when the three work together,
they paint at a constant rate of
A
18
+
A
15
+
A
16
square feet per hour. A second thing of
note is that, if one is used to the use of symbols, then the preceding solution is entirely
straightforward and is devoid of subtlety. Compare with any solution that does not use
algebra.
Before we leave this section, we introduce a terminology. In a linear function
f(x) = ax+b, the number b is called the constant term of the function. If b = 0, then
we say the function f(x) = ax is without constant term. It remains to observe that
the linear functions that arise from rate problems are linear functions without constant
term. Such functions are the subject of the next section.
EXERCISES
1. Prove that if 0 < a < b or a < b < 0, then
1
a
>
1
b
.
2. Prove that when 0 < x < 1, x <
√
x, and when 1 < x, x >
√
x.
144
3. Plot enough points in the graph of each of the following functions to get an
accurate picture of the graph: (i) x
3
, (ii) x
4
, (iii) x
3
+5, (it iv) x
5
. (Use a scientific
calculator.)
4. Plot enough points in the graph of each of the following functions to get an accurate
picture of the graph: (i) x
3
−x, (ii) x
3
−4x
2
+x +6, (iii) x
3
−x
2
−4x +4, (it iv)
2x
3
+ 9x
2
−15x −22. (Use a scientific calculator.)
5. Plot enough points in the graph of 4
x
without the use of a calculator to get an
accurate picture of its graph.
6. Plot enough points in the graph of each of the following functions to get an accurate
picture of the graph: (i) (
1
3
)
x
, (ii) 3
x
, (iii) 5
x
. Compare the graphs of (i) – (iii) by
putting all three in the same coordinate grid if necessary: what are the similarities, and
what are the differences? (Use a scientific calculator.)
7. Prove that if 0 < a < b, then
n
√
a <
n
√
b, for any positive interger n. (The fact
that this is true for n = 2 is used in Example 6.)
8. Prove that if a > 1, then for all rational numbers r and s so that r < s, a
r
< a
s
,
while if 0 < b < 1, b
r
> b
s
.
9. Suppose Jessica can do a piece of work in 5 days and Jessica and Helena together
can do it in 3 days. Assuming that each works at a constant rate, in how many days
can Helena do the work alone?
10. A man walks from point A to point B at a constant rate. If he walks at the rate
of 1 yard per second, then it takes him 5
1
2
minutes more to get to point B than if he
walks at the rate of 4 yards per 3 seconds. How far is point A from point B?
11. A freight train runs 6 miles an hour less than a passenger train. It runs 80 miles
in the same time that the passenger trains runs 112 miles. Assuming that both trains
run at a constant rate, find the rate of each train.
12. A train leaves A for B, 112 miles apart, at 9 am, and one hour later a train
leaves B for A; they met at 12 noon. If the second train had started at 9 am and the
first at 9 : 50 am, they would also have met at noon. Assume that each train runs at a
fixed constant speed, find their speeds.
13. A can do a piece of work in
2
3
as many days as B, and B can do it in
4
5
as many
days as C. Together they can do it in 3
7
11
days. Assuming constant rate of work, in how
many days can each each do it alone?
145
10 Proportional Reasoning (outlined)
Consider the following problem (taken from NCTM Standards, p. 83):
A group of 8 people are going camping for three days and need to carry their
own water. They read in a guide book that 12.5 liters are needed for a party
of 5 persons for 1 day. How much water should they carry?
This problem is supposed to illustrate the importance of a concept called "propor-
tional reasoning" in middle school mathematics. Yet as it stands, the problem cannot
be done without the additional assumption that each person drinks the same amount of
water each day. What is noteworthy is that this assumption is missing from the problem.
The classical (traditional) approach to such a problem is to set up a proportion: if w
is the number of liters of water 8 people need per day, then 12.5 liters is to 5 people as
w is to 8 people, so
12.5
5
=
w
8
Use this equation to solve for w, and the answer to the problem is therefore 3w liters.
Unfortunately, there is no logical reasoning that can be used to explain this approach,
at least none that is based on what is given in the problem. Recall the WYSIWYG
characteristic of mathematics in this context!
A more recent approach is to isolate the concept of "proportional reasoning" and use
it as a way to understand problems of this type. It would seem that this concept does
not have a precise meaning but has something to do with thinking in terms of multiplica-
tion. One guess is that "proportional reasoning" means the ability to fluently apply the
concept of a linear function without constant term. Now for "proportional reasoning"
to be applicable in this case, one has to begin by making explicit the assumption that
if f(x) is the amount of water (in liters) x people need each day, then f(x) is a linear
function in x without constant term. (We proceed to allow x to be an arbitrary number
rather than just whole numbers; see the discussion of the profit function in Section 7.)
Therefore we may write
f(x) = cx
for some fixed number c. Since f(1) = c, the definition of f means that c is the amount
of water one person needs per day. What is given is that f(5) = 12.5. Thus 5c = 12.5,
and c = 2.5. It follows that f(8) = 8 2.5 = 20, and the answer is 3f(8) = 60 liters.
146
Incidentally, with the availability of the linear function f(x) = cx, we can at last
understand what "setting up a proportion" is all about. We know that
f(x)
x
= c no
matter what x may be, so for arbitrary x
1
and x
2
, we will always have
f(x
1
)
x
1
=
f(x
2
)
x
2
,
because both quotients are equal to c. The preceding proportion of
12.5
5
=
w
8
is now seen
to be nothing other than the statement that
f(5)
5
=
f(8)
8
Therefore one can begin to make sense of this venerable device of "setting up a propor-
tion" from the standpoint of a linear function without constant term.
Consider another problem (taken from Balanced Assessment for the Mathematics
Curriculum):
John's grandfather enjoys knitting. He can knit a scarf 30 inches in 10 hours.
He always knits for 2 hours each day.
1. How many inches can he knit in 1 hour?
2. How many days will it take Grandpa to knit a scarf 30 inches long?
3. How many inches long will the scarf be at the end of 2 days? Explain
how you figured it out.
4. How many hours will it take Grandpa to knit a scarf 27 inches long?
Explain your reasoning.
Think through and do this problem yourself before comparing your solution with the
suggested solution in Balanced Assessment, which is as follows:
1. 3 inches, by division: 30 ÷10.
2. 5 days, by division: 10 ÷2.
3. 12 inches. Give explanation such as: In one day he knits 3 2 = 6
inches. In 2 days he knits 2 6 inches.
4. 9 hours. Give explanation such as: To knit 27 inches takes 27÷3 hours.
147
As a problem in mathematics, with the exception of Part 2, this problem is not
doable as it stands because what is given cannot support any kind of logical reasoning
for its solution. The missing assumption is that grandfather knits at a constant rate.
Without this assumption, how can one begin to think about such a problem? Where to
begin? This problem is symptomatic of a generic failure of mathematics education at
the moment, namely, students are asked to develop conceptual understanding of a given
topic and yet the topic is not taught in a manner that meets the minimum requirements
of mathematics. For the case at hand, students are asked to develop an understanding
of a multiplicative process but they cannot do so because the given assumptions are not
sufficient to support the needed reasoning. Think of WYSIWYG again.
What must be made explicit is the fact that, if g(t) is the number of inches grandfa-
ther can knit in t hours, then g(t) is a linear function without constant term. The given
data is that g(10) = 30. Thus if we write
g(t) = t
for some constant , then from g(10) = 30, we get 10 = 30, and = 3. The answers
to the four parts are then, in succession: f(1) = 3,
10
2
= 5, f(4) = 12, and finally, the
value of t
0
so that f(t
0
) = 27, which results in t
0
= 9.
To drive home the point of the need to understand linear functions without constant
term in all such "proportional reasoning" problems, we make one more comment on the
proposed solution to Part 4 in Balanced Assessment. A little reflection would reveal that
this method of solution is merely a repackaging of the proportion
30
10
=
27
h
where h is the number of hours it takes grandfather to knit 27 inches. From our vantage
point, we know that
g(t)
t
is always equal to the constant no matter what t may be.
In particular, if h has the same meaning as before and we are given g(h) = 27, then we
have
g(10)
10
=
g(h)
h
,
and this is exactly the proportion implicitly used in Balanced Assessment to solve the
problem.
Note that there is a real difference between these two problems. The first problem is
a so-called discrete problem: there is a natural unit for people, namely, 1 person. There
148
is no such thing as 1.27 persons, for example. On the other hand, the knitting problem
is a standard rate problem in that there is no natural unit for time. Any time interval
can be broken down into smaller time intervals. The lack of a natural unit in this case
makes the knitting problem harder. This is the difference between the discrete and the
continuous.
In one way or another, students must be given the information that a certain function
is a linear function without constant term, because this is part of the basic assumption
of constant rate. See the discussion at the end of the preceding section. Mathematics
education should not engage in the practice of not telling students that there is such
a linear function without constant term, and then make judgement about their con-
ceptual understanding (or lack thereof) because they cannot guess that there is such a
linear function. Mathematics is not concerned with making wild guesses about a hidden
agenda.
It is not a matter of what kind of conceptual understanding students should have
about proportional reasoning, but rather a matter of being able to clearly and unam-
biguously articulate, in textbooks and in the classroom, and in a way that is grade
appropriate, what a linear function is and when a linear function is being assumed in
word problems. As we have seen in the preceding two examples, untill we make explicit
the fact that the functions f and g are linear functions without constant terms, students
have no way of doing the problems except by guessing. It makes no sense whatsoever
to demand that students have the conceptual understanding that those functions are
linear. Without an explicit assumption, there is no reason why they should be linear. We
must never forget the WYSIWYG characteristic of mathematics.
Further discussion of the role of proportional reasoning in the school mathematics
curriculum is given at the end of Section 4 in What is so difficult about the preparation
of mathematics teachers?,
11 Quadratic Functions and Their Graphs
If a function f from all numbers to all numbers is given by f(x) = ax
2
+bx+c for some
given numbers a, b, and c, we say f is a quadratic polynomial function, or more
commonly, a quadratic function. The graph of such a function is called a parabola.
149
The main theme of this section is quadratic functions and their graphs. We have come
across the simplest quadratic function f
0
(x) = x
2
in Example 2 of Section 9. At the
end of this section, we shall prove that all parabolas are similar (in the sense of the
composition of a congruence followed by a dilation) to the graph of f
0
.
We begin with two general observations. Given a quadratic function f(x) = ax
2
+
bx + c as above:
(A) [f(x)[ becomes arbitrarily large if [x[ gets sufficiently large.
(B) If a > 0, then f(x) > 0 for [x[ large. If a < 0, then f(x) < 0 for
[x[ large.
In a suggestive notation, we will paraphrase the combined statements of (A) and (B)
by writing:
lim
|x|→∞
f(x) = ∞ for a > 0
lim
|x|→∞
f(x) = −∞ for a < 0
Notice how we have to use absolute value to express both (A) and (B) suc-
cinctly. If the concept of absolute value were not at our disposal, we would
have to write something like the following, for example, for (A):
"f(x) > R for a given large positive number R, or f(x) < −R for a large
positive number R if the distance of x from 0 is sufficiently large"
The reason for (A) is as follows. Write
f(x) =
_
a +
b
x
+
c
x
2
_
x
2
Now if [x[ is sufficiently large, then
1
|x|
would be sufficiently small, and therefore so are
[
b
x
[ and [
c
x
2
[. Consequently, when [x[ is sufficiently large, (a+
b
x
+
c
x
2
) would be very close
to a. If we use ∼ to indicate "approximately equal to", then we have f(x) ∼ ax
2
when
[x[ is sufficiently large. Since x
2
is always positive, f(x) ∼ a[x
2
[ when [x[ is sufficiently
large. This proves (B), and (A) also follows from the same expression for f(x) when [x[
is sufficiently large.
It remains to find out how f(x) behaves when x is not very large. We will use a
symmetry property of the graph of f to cut down our effort by half. Recall that a set
S in the plane is said to be symmetric with respect to a line L if for every point Q
in S, the reflection R across L maps Q to a point R(Q) which also lies in S. If we use
150
the terminology of Section 7, S being symmetric with respect to L means the part of S
in the half-plane L
+
is congruent to the part of S in the other half-plane L
−
. If there is
such a symmetry, then the study of S itself reduces to a study of one of the two halves,
S ∩ L
+
or S ∩ L
−
. This explains the interest in such a symmetry. To describe this
symmetry more precisely, we introduce a concept already used informally in Section 7.
Given a function f from all numbers to all numbers (e.g., a quadratic function). We say
f achieves a minimum at a point β if f(x) ≥ f(β) for any number x. Similarly,
we say f achieves a maximum at a point β if f(x) ≤ f(β) for any number x. We
also say that x
0
is a zero of the function f is f(x
0
) = 0. The following is the main
theorem about quadratic functions.
Theorem 1 For any quadratic function f(x) = ax
2
+ bx + c, there is a unique
number p so that, if a > 0, then p is the number at which f achieves a minimum,
and if a < 0, then p is the number at which f achieves a maximum. Furthermore, f
can have at most two zeros, and the graph of f is symmetric with respect to the vertical
line defined by x = p.
We will break up the proof into two parts. The first part assumes that the quadratic
function can be written in the form f(x) = a(x − p)
2
+ q for some numbers p and q.
Let us examine what this means. If we write
a(x −p)
2
+ q = ax
2
−2apx + (ap
2
+ q),
then to say that f can be written as f(x) = a(x−p)
2
+q is to say that the coefficients
b and c in f(x) = ax
2
+ bx + c can be expressed in terms of two numbers p and q as
b = −2ap and c = ap
2
+ q
In the second part, we will show that in fact every quadratic function can be written in
the form f(x) = a(x −p)
2
+ q.
As usual, we give a detailed proof of this theorem because the ideas of the proof
contain important information about quadratic functions in general.
So suppose f(x) = a(x −p)
2
+ q. We observe that for any number k,
f(p + k) = f(p −k) = ak
2
+ q
This is easy:
f(p −k) = a((p −k) −p)
2
+ q = ak
2
+ q = a((p + k) −p)
2
+ q = f(p + k)
151
We now use the fact that f(p+k) = f(p−k) to show that the graph of f is symmetric
with respect to the vertical line L defined by x = p. Indeed, the reflection R with
respect to L maps a point (p +k, y
)
q
Now if a point (p − k, f(p − k)) is on the graph of f, then under the reflection R, this
point is moved to (p + k, f(p − k)), which is equal to (p + k, f(p + k)) because we
have seen that f(p −k) = f(p +k). Since (p +k, f(p +k)) is a point on the graph of f
(the set of all the points of the form (x, f(x)), by definition), we have proved our claim
of symmetry with respect to L.
To study the graph of f, this symmetry says it suffices to study (x, f(x)) where
x ≥ p; the rest (i.e., those (x, f(x)) where x ≤ p) can be obtained by reflection. Now if
x ≥ p, we can write x = p + k, where k ≥ 0, and from the above discussion, we have
f(p + k) = ak
2
+ q. Thus
if a > 0, then f(p + k) increases with k so that f(p + k) is smallest when
k = 0, that is, f attains its minimum only at x = p, and
if a < 0, then f(p + k) decreases with k so that f(p + k) is biggest when
k = 0, that is, f attains its maximum only at x = p.
The preceding assertions is about the behavior of f(x) for x ≥ p, but by the symmetry
of the graph of f with respect to L (i.e., the line x = p), they remain true for all x.
Moreover, for a > 0, if q > 0, then f(p + k) = ak
2
+ q > 0 and there would be no
zero for f when x ≥ p. By symmetry with respect to L, f has no zero in this case. If
q = 0, then f(p) = 0, and since p is the only minimum of f, there is exactly one zero
for f. Finally, if q < 0, then f(p) = q < 0. But from (A) and (B) above, we know
that f(p + k) becomes large positive when k is large. Because f(p + k) increases with
k, we see that f(p +k
0
) = 0 for some k
0
> 0. Thus f has one zero at p +k
0
among the
numbers p +k for k ≥ 0. By symmetry with respect to L, f(p −k
0
) is also zero, so that
152
f has another zero among the numbers p − k for k > 0. Altogether, f has exactly two
zeros in this case.
In a similar manner, we can show that if a < 0, then f has no zero, one zero, and
two zeros, according to whether q < 0, q = 0, and q > 0, respectively. So the proof of
Theorem 1 is complete in case f(x) = a(x −p)
2
+ q.
Activity Plot points on the graph of f(x) = 4(x −2)
2
+
1
3
which are symmetric
with respect to the line x = 2. Does the graph cross the x-axis?
Remark Before continuing with the proof of Theorem 1, we digress to interpret
the information from the preceding discussion in terms of the shape and the location of
the graph of f(x) = a(x−p)
2
+q relative to the coordinate axes. As mentioned above,
because every quadratic function will be shown presently to be expressible in the form
of a(x −p)
2
+ q, what we are going to say is valid for all quadratic functions.
With L as the line x = p and f(x) = a(x −p)
2
+ q, first assume a > 0. Then we
have seen that f(p+k) with k > 0 gets bigger as k gets bigger. In terms of the graph of
f, which consists of all the points (p+k, ak
2
+q) where k is arbitrary, the graph rises as
we go to the right. Because of the symmetry with repsect to L, this means the graph of
f also rises as we go to the left. Moreover, if q > 0, then f(p+k) = ak
2
+q > q > 0 for
every k. Thus the whole graph of f would be above the x-axis and it would not intersect
the x-axis. If on the other hand q < 0, then f(p) = q < 0 whereas lim
|x|→∞
f(x) = ∞.
Therefore the graph of f goes from below the x-axis at p to above the x-axis when [x[
is very large. Thus the graph of f crosses the x-axis at two points exactly, once in L
+
because the graph rises as we go to the right, and therefore also once in L
−
because of
the symmetry with respect to L. Finally, if q = 0, then f(p) = 0, and the graph would
touch the x-axis at exactly (p, 0).
Our interest in the intersection of the graph of f with the x-axis naturally coincides
with our interest in locating the zeros of the polynomial function f(x) = ax
2
+ bx + c.
(A zero of the polynomial ax
2
+bx +c is also called a root of the polynomial equation
ax
2
+ bx + c = 0.) If the graph of f intersects the x-axis at (x
0
, 0), then (x
0
, 0) being
on the graph means 0 = f(x
0
), so that x
0
is a zero of f. Conversely, if f(x
0
) = 0,
then the point (x
0
, 0) (which is of course (x
0
, f(x
0
))) is on both the graph of f and the
x-axis and is therefore a point of the intersection of the graph with the x-axis. Thus
locating the points of intersection of the graph of f with the x-axis is equiv-
alent to locating the zeros of f.
153
We will discuss the zeros of a quadratic function a bit more in the next section and
we will see that these zeros provide critical information in applications. What we have
established is that, in case a > 0 in f(x) = a(x −p)
2
+ q, f may have one zero, two
zeros, or no zero, depending on whether q = 0, q < 0, and q > 0, respectively.
Typically, these three cases lead to the following kinds of graphs, with the case of q < 0
on the left, q = 0 in the middle, and q > 0 on the right:
q
q
p
qq
p
q
q
p
In the case of a > 0, we call f(p) (which equals q) the minimum value of f,
and the point (p, f(p)) is then called the vertex of the parabola. The vertex is the
lowest point of the parabola in this case. Precisely, the x-coordinate of the vertex is the
point at which f achieves its minimum, and the y-coordinate is the minumum value of f.
Still with f(x) = a(x − p)
2
+ q, now assume a < 0. The same discussion can be
carried through, but now f achieves a maximum at p and, because f(p + k) = ak
2
+ q
and a < 0 so that f gets more and more negative as k goes to the right of the x-axis, and
the graph of f goes down as we go to the right of the x-axis. By symmetry with respect
to L, the graph of f also goes down as we go to the left of the x-axis. The point (p, f(p))
is still called the vertex of the parabola; it is now the highest point on the parabola
when a < 0. Furthermore, if q < 0, the whole graph would be below the x-axis and f
would have no zero. If q > 0, then because f(p) = q, the graph of f will now have
points both above and below the x-axis and will therefore intersect the axis at precisely
two points (the reason for "two" is the same as before). If q = 0, then the graph touches
the x-axis at exactly (p, 0). The three possibilites are given in the following pictures:
q
q
p
q
p
q
q
p
154
It is common to refer to the graph of a quadratic function in the case of a > 0 as a
up parabola, and in the case of a < 0 as a down parabola.
We can now return to the proof of the second part of Theorem 1.
15
Given f(x) = ax
2
+ bx + c, we want to find numbers p and q so that f(x) =
a(x − p)
2
+ q. We first do it algebraically. One should first work out some concrete
examples such as f(x) = 2x
2
+ 6x −7 or f(x) = 3x
2
−4x + 1 to get some feeling for
the problem before embarking on the general case. Be that as it may, here is the general
argument. Write
f(x) = a(x
2
+
b
a
x) + c
and compare it with what we want, which is
f(x) = a(x −p)
2
+ q = a(x
2
−2px + p
2
) + q
For these two expressions to be equal for all x, i.e.,
a(x
2
+
b
a
x) + c = a(x
2
−2px + p
2
) + q
for all x, we must have, at least heuristically, the equality of all the corresponding
coefficients.
16
In particular, the two coefficients of x must be equal. Thus
b
a
= −2p,
which would suggest that we let p = −
b
2a
. We try that and get:
a(x −p)
2
+ q = a(x +
b
2a
)
2
+ q = a(x
2
+
b
a
x +
b
2
4a
2
) + q = ax
2
+ bx + (
b
2
4a
+ q)
If we hope to make this last expression equal to = ax
2
+bx+c, we have to set
b
2
4a
+q = c,
or what is the same thing, set q = c −
b
2
4a
=
4ac−b
2
4a
. With this heuristic reasoning in the
background, we now get to work.
Formally, let
p = −
b
2a
and q =
4ac −b
2
4a
Then:
a(x −p)
2
+ q = a
_
x +
b
2a
_
2
+
4ac −b
2
4a
15
You will see that the identities (x +y)
2
= x
2
+2xy +y
2
and (x −y)
2
= x
2
−2xy +y
2
of Section
1 play a crucial role in the subsequent argument.
16
This fact can be proven in general, but for quadratic functions, the proof is not too difficult. See
the Exercises in the next section.
155
= a
_
x
2
+
b
a
x +
b
2
4a
2
_
+
4ac −b
2
4a
= ax
2
+ bx +
b
2
4a
+
4ac −b
2
4a
= ax
2
+ bx + c
It follows that with these values for p and q, we now know that for the given f(x) =
ax
2
+ bx + c,
f(x) = a(x −p)
2
+ q
Then the proof of the first part of Theorem 1 now takes over to completely prove the
theorem.
There are two comments on this proof that are of equal importance, and we go
through both carefully. With f(x) = ax
2
+ bx + c understood, the first one is that,
from the proof of the first part of Theorem 1, we know that p is the point at which f
achieves its maximum or minimum, depending on whether a < 0 or a > 0, and the
maximum or minimum value of f at p is q. But we have the explicit values of p and q
given above in terms of the coefficients a, b, and c of f. We can therefore summarize
these conclusions in the following important theorem about quadratic functions.
Theorem 2 The quadratic function f(x) = ax
2
+ bx + c achieves
its maximum at −
b
2a
if a < 0, and the maximum value f(−
b
2a
) is
4ac−b
2
4a
,
and achieves
its minimum at −
b
2a
if a > 0, and the minimum value f(−
b
2a
) is
4ac−b
2
4a
.
Furthermore, the quadratic function can be rewritten as
f(x) = a
_
x +
b
2a
_
2
+
4ac −b
2
4a
It goes without saying that the vertex of the graph of f has coordinates (−
b
2a
,
4ac−b
2
4a
).
A second comment is that, looking back, we can see that the most critical step in the
proof of the second part of Theorem 1, one that allows us to reduce the most general
156
quadratic function to the manageable special case of a(x −p)
2
+q, is the identity that
x
2
+
b
a
x =
_
x +
b
2a
_
2
−
_
b
2a
_
2
Indeed, if we knew this identity, then we would immediately recognize that ax
2
+bx+c
could be put in the form of a(x −p)
2
+q with p = −
b
2a
and q = c −
b
2
4a
, because it is
a simple computation to see that
ax
2
+ bx + c = a
_
x
2
+
b
a
x
_
+ c
= a
_
_
x +
b
2a
_
2
−
b
2
4a
2
_
+ c
= a
_
x +
b
2a
_
2
+
_
c −
b
2
4a
_
.
Such an identity has wide applicability in mathematics and should be stated more gen-
erally, as follows: for any number β, we have
x
2
+ βx =
_
x +
β
2
_
2
−
_
β
2
_
2
(13)
In this form, identity (13) is known under the name of completing the square. Its
validity is not in doubt if you start with the right side and simply expand it and simplify
to arrive at the left side. What is more interesting is how, starting with the left side
x
2
+ βx, one would be led into thinking that it can be expressed as the difference of
two squares, and what the name "completing the square" means. Both questions are
answered simultaneously if we retrace the steps of the Babylonians who made extensive
use of such an identity some 38 centuries ago (around 1800 B.C.).
17
Writing x
2
+ βx
as
x
2
+ 2
_
β
2
x
_
and remembering that β and x are numbers, we see that this expression is exactly the
area of the following figure consisting of a square with one side of length x and two
rectangles each with sides of length
β
2
and x.
17
No, the Babylonians did not state completing the square as identity (13), but if they had the
symbolic notation that we have today, they might have.
157
x
β
2
x
β
2
Looking at this picture, it would be natural to complete it to a bigger square with a side
of length x +
β
2
by adding the small dotted square at the lower right corner:
x
β
2
x
β
2 p
p
p
p
p
p
p
p
p
Now the dotted square has a side of length
β
2
, so its area is (
β
2
)
2
. With this dotted
square added to the original figure, the total area is now the area of the big square with
a side of length x +
β
2
. Thus:
(x
2
+ βx) +
_
β
2
_
2
=
_
x +
β
2
_
2
Transposing (
β
2
)
2
to the right, we get immediately identity (13).
We conclude this section by fulfilling the promise made at the beginning of the
section: we are going to prove that
the graph of any quadratic function is similar to the standard parabola G
1
,
which is by definition the graph of the quadratic function given by f
1
(x) = x
2
158
for all x.
A few comments before the proof. It is quite clear that this theorem is not part of
the standard algebra curriculum, whether it be Algebra I or Algebra II. So why spend
time on it here? There are three reasons. The first is that this is a surprising result,
and even mentioning something like this in an algebra class could be inspirational or
intriguing to students. A second reason is that the proof of the theorem shows why
it is essential that we know a precise definition of similarity. In this case, we have to
prove that two parabolas, which are not rectilinear figures, are similar. There are no
line segments to measure and no angles to compare; instead we are forced to use the
definition of similarity in terms of dilation. The final reason, and in fact the main one,
is that the proof deepens your understanding of the precise definition of the graph of a
function. You may have gotten the message by now that these notes consider learning
precise definitions to be a critical component of a mathematics education, and there is
no more convincing illustration of this message than this proof. Indeed, the proof hinges
on exactly what the graph of the function f
k
(x) = kx
2
is.
We break the proof up into two parts:
Part 1: Let G
k
be the graph of the function f
k
(x) = kx
2
where k > 0,
then the dilation T defined by T(x, y) = (kx, ky) maps G
k
to the standard
parabola G
0
.
Part 2: Let G be the graph of a quadratic function f(x) = ax
2
+bx +c. If
a > 0, then a translation would map G to a G
k
of Part 1; if a < 0, then a
reflection followed by a translation would map G to a G
k
of Part 1.
Because a similarity is by definition a rigid motion (a composition of translations,
rotations, or reflections) followed by a dilation, Part 1 and 2 together prove that the
graph of any quadratic function is similar to the standard parabola.
Part 1 is proved as follows. By definition, G
k
consists of all points of the form
(x, kx
2
), where x is arbitrary, while G
1
consists of all points of the form (x, x
2
) for all
x. For an arbitrary point (t, kt
2
) of G
k
, we have
T(t, kt
2
) = (kt, k
2
t
2
) = (kt, (kt)
2
)
which shows that T(G
k
) is part of G
1
. On the other hand, if (x, x
2
) is an arbitrary point
of G
1
, then T(
1
k
x,
1
k
x
2
) = (x, x
2
). Moreover, (
1
k
x,
1
k
x
2
) = ( ¦
1
k
x¦, k ¦
1
k
x¦
2
), which shows
159
that (
1
k
x,
1
k
x
2
) belongs to G
k
. Thus T(G
k
) is all of G
1
. This proves Part 1.
For Part 2, let the graph G of a quadratic function be given. If the vertex of G is
(p, q), then by Theorem 2, G is the graph of the quadratic function f(x) = a(x−p)
2
+q.
We are going to assume the worst case scenario so that both p and q are nonzero, and
that a < 0. In other words, G is a down parabola. We will prove that after a reflection
and a translation, G is a G
k
for some k.
q
(p, q)
Let R be the reflection with respect to the x-axis. Then R maps a point (x, y)
to (x, −y). Now a point on G, which is the graph of f = a(x − p)
2
+ q, is of the
form (x, f(x)) = (x, a(x − p)
2
+ q). Therefore each point (x, a(x − p)
2
+ q) of G
is mapped by R to (x, (−a)(x − p)
2
− q). Observe that the collection of all these
point ¦ (x, (−a)(x −p)
2
−q) ¦ is exactly the graph of the function g given by g(x) =
(−a)(x−p)
2
−q. Because a < 0, (−a) > 0. Write γ for (−a) for simplicity, and we
see that the reflected parabola R(G) is the graph of g given by g(x) = γ(x −p)
2
−q,
where γ > 0 so that R(G) is an up parabola, as shown:
q
(p, −q)
Notice that the vertex of R(G) is (p, −q).
Next we will use a translation to move (p, −q) to (0, 0). The translation T is
described algebraically by
T(x, y) = (x −p, y + q)
for all (x, y). Observe that T(p, −q) = (0, 0). The points (x, γ(x − p)
2
− q) of the
reflected parabola R(G) are now translated to (x−p, γ(x−p)
2
−q+q) = (x−p, γ(x−p)
2
).
160
So the translated parabola T(R(G)) consists of all points of the form (x−p, γ(x−p)
2
)
where x is arbitrary. This means that T(R(G)) is G
γ
, the graph of the function
f
γ
(x) = γx
2
.
qq
In case a > 0 in f(x) = a(x − p)
2
+ q, then in the preceding argument, we simply
dispense with the use of the reflection R and directly apply the translation T to this G.
Then T(G) is G
a
, the graph of the function f
a
(x) = ax
2
. This completes the proof.
EXERCISES
1. Determine whether any of the following quadratic functions has zeros by using
only what has been done in this section (in particular, don't use the quadratic formula!):
(i) f(x) = −3x
2
+ 4x −
8
3
. (ii) g(x) =
3
2
x
2
−x + 8. (iii) h(x) = −x
2
+
4
5
x −2. (iv)
f(x) = −
3
10
x
2
+ 7x −35.
2. Find the maximum or the minimum of each of the following quadratic functions:
(i) f(x) = 2x
2
+ 3x + 4. (ii) g(x) =
3
4
x
2
−2x +
8
5
. (iii) h(x) = −6x
2
+ x +
5
3
. (iv)
h(x) = 3x
2
−2x +
8
3
.
3. Write each of the following quadratic functions in the form a(x − p)
2
+ q for
suitably chosen numbers a, p and q: (i) f(x) = x
2
−8x+7. (ii) g(x) = −2x
2
+6x−21.
(iii) h(x) = 3x
2
+ 4x + 6. (iv) f(x) = −
2
3
x
2
+ x − 1. (v) g(x) = 5x
2
−
2
5
+ 2. (vi)
g(x) =
√
2x
2
−
√
6x + 5. (vii) h(x) = −3x
2
+
√
5x −1.
4. Graph each of the functions in the preceding problem; indicate whether it is an
up parabola or a down parabola, estimate the zeros of the function if there is any, and
locate the vertex.
5. Write down the explicit transformation of the plane that realizes the similarity of
the parabola f(x) = x
2
+ 2x with the standard parabola f
0
(x) = x
2
. Do the same
with the parabola g(x) = −4x
2
+ 12x −9.
6. In the text, it is asserted that the translation T which goes in the direc-
tion from a point (p, q) to the origin is the transformation of the plane given by
161
T(x, y) = (x −p, y −q). Explain why this is true.
12 The Quadratic Formula and Applications (out-
lined)
Consider the problem:
A rectangle has a perimeter 180 linear units and an area of 1800 area units.
What are its dimensions?
If the length of one side is x linear units, then we have an equation in x, namely,
x(90 − x) = 180. This is known as a quadratic equation in the variable x, and this
problem illustrates how such equations arise naturally without the intervention of a
quadratic function. We want to solve this equation, i.e., determine all the numbers
x
0
so that x
0
(90 − x
0
) = 180, or, −x
2
0
+ 90x
0
− 180 = 0. Such an x
0
is also called a
root of the polynomial equation −x
2
+90x −180 = 0. From the perspective of the last
section, solving this equation is the same as locating the zeros of the quadratic function
f(x) = −x
2
+90x−180. From Theorem 1 of Section 11, we know that there are at most
two zeros of f, and therefore the quadratic equation has at most two roots. The proof of
Theorem 1 in facts gives an algorithm (completing the square) for locating these zeros if
they exist. Thus an understanding of the quadratic function itself gives a comprehensive
understanding of the nature of its zeros. Our study of the quadratic function therefore
covers more ground than the solving of quadratic equations.
There is value, nevertheless, in studying a quadratic equation by itelf without ref-
erence to the associated quadratic function, because it gives a different perspective on
the zeros of the quadratic function. For this reason, we now approach the quadratic
polynomial ax
2
+ bx + c purely algebraically, without considering its graph, and show
again that it has at most two zeros. Thus, given
ax
2
+ bx + c = 0,
we start from the beginning and assume nothing about quadratic functions. Suppose
there is a root s. By completing the square, we get
_
s +
b
2a
_
2
=
b
2
−4ac
4a
2
162
Because the right side is equal to the square of the number s+
b
2a
, necessarily b
2
−4ac ≥ 0.
Hence we can take the square root of b
2
−4ac ≥ 0 to get
s =
−b ±
√
b
2
−4ac
2a
This shows that if there is a root s, then s must be one of the numbers on the right side.
Now we must actually verify directly that both
−b ±
√
b
2
−4ac
2a
are roots of ax
2
+bx +c = 0. We do so by a direct computation (and making use of the
identity (x −y)(x + y) = x
2
−y
2
in the process):
a
_
−b ±
√
b
2
−4ac
2a
_
2
+ b
_
−b ±
√
b
2
−4ac
2a
_
+ c =
_
−b ±
√
b
2
−4ac
2a
__
a
−b ±
√
b
2
−4ac
2a
+ b
_
+ c =
_
−b ±
√
b
2
−4ac
2a
__
+b ±
√
b
2
−4ac
2
_
+ c =
1
4a
(−b ±
√
b
2
−4ac)(+b ±
√
b
2
−4ac) =
1
4a
(−b
2
+ (b
2
−4ac)) + c = 0
We have therefore proved the quadratic formula (QF):
Theorem 1 If b
2
−4ac ≥ 0, then the equation ax
2
+ bx + c = 0 has the following
two roots (which coincide if b
2
−4ac = 0):
−b ±
√
b
2
−4ac
2a
Conversely, if ax
2
+ bx + c = 0 has a root, then b
2
−4ac ≥ 0.
Theorem 1 gives a direct proof that a quadratic equation has 0, 1, or 2 roots.
We continue with the study of the algebraic properties of the polynomial ax
2
+bx+c.
The following is the quadratic-polynomial analog of the division-with-remainder among
whole numbers.
163
Theorem 2 Given ax
2
+ bx + c and a number r, there is a linear polynomial
(ax + β) so that
ax
2
+ bx + c = (ax + β)(x −r) + γ
for some constant γ and for all x.
The linear polynomial is called the quotient of the division, and γ the remainder.
If this terminology seems strange, we review the division-with-remainder for whole num-
bers. When we say "59 divided by 7 has quotient 8 and remainder 3", we are making
a statement about multiplication, namely, among all whole number multiples of 7, the
one that is closest to 59 but not exceeding it is the 8th multiple (which is of course 56),
and it misses 59 by 3 (3 = 59 − (8 7)). The quotient is therefore this multiple, and
the remainder is determined by the quotient once the quotient is clearly defined. Anal-
ogously, we look for a "multiple" of (x − r) that is "closest" to ax
2
+ bx + c, but in
this case, "multiple" would have to mean "multiplication by another polynomial". Since
the degrees add when polynomials are multiplied, and since ax
2
+ bx + c has degree 2,
this "multiple" has degree either 1 or 0. But how close is "close"? We would obviously
welcome a multiple of (x−r) that matches ax
2
+bx+c coefficient-by-coefficient for as
many coeficients as possible. It turns out that if we let β = ar +b, then (ax+β)(x−r)
would match ax
2
+bx+c up to the coefficients of x
2
and x, i.e., a and b. So the remainder
ax
2
+ bx + c −(ax + β)(x −r)
in this case is just a number (or a constant, which is the common terminology). This
is what γ is all about.
Theorem 2 can appear to be deceptively simple (or perhaps plain opaque), but
through its many substantial consequences, such as those listed below, one gains a
better understanding of it.
(a) A number r is a root of ax
2
+ bx + c = 0 if and only if (x − r) divides
ax
2
+ bx + c. (We say (x − r) divides a polynomial p(x) if p(x) factors as
p(x) = q(x)(x − r) for some polynomial q(x). Thus to say (x − r) divides
ax
2
+ bx + c means precisely that γ = 0 in Theorem 2).
(b) If r
1
and r
2
are the roots of ax
2
+ bx + c = 0, then for all x,
ax
2
+ bx + c = a(x −r
1
)(x −r
2
)
(This identity may seem too obvious to you because some may even regard
this as the definition of the roots r
1
and r
2
. But it is not obvious. What it says
164
is that, if r
1
and r
2
are any two numbers which satisfy a(r
1
)
2
+br
1
+c = 0 and
a(r
2
)
2
+br
2
+c = 0, then the equality ax
2
+bx+c = a(x−r
1
)(x−r
2
) is valid
for all x. Such a striking statement undoubtedly demands an explanation.
One way is to get explicit expressions of r
1
and r
2
in terms of a, b, and c by
appealing to QF, and then explcitly compute a(x−r
1
)(x−r
2
) to show that
it is equal to a(r
1
)
2
+br
1
+c. It is also possible to obtain a more conceptual,
but also more sophisticated proof of this fact by a judicious use of (a).)
(c) Let r
1
and r
2
be the roots of x
2
+ bx + c = 0, then
b = −(r
1
+ r
2
) and c = r
1
r
2
(This needs (b) and Problem 2 in the following Exercises.)
(d) If a quadratic polynomial ax
2
+bx +c can be factored into a product of
linear polynomials, then the factorization is a(x −r
1
)(x −r
2
), where r
1
and
r
2
are the roots of ax
2
+bx+c = 0. (In short, the QF provides a factorization
for all trinomials.)
Activity Factor 10x
2
−13x −30.
Here is a typical example of how all this knowledge about quadratic functions and
quadratic equations is put to use in solving word problems.
Example If an object is thrown from a height of h meters from the ground with
an initial velocity of v
0
m/sec, then its distance f(t) above the ground t seconds after it
is thrown (in meters) is
f(t) = −4.9t
2
+ v
0
t + h
(This follows from Newton's second law.) Now if h = 20 meters and v
0
= 2 m/sec, what
is the highest point of the object above the ground, when does it get there, and when
does it hit the ground?
The highest point above the ground is the maximum of the quadratic function
f(t) = −4.9t
2
+ 2t + 20, which is 20
10
49
meters. The object hits the ground after
t
0
seconds if f(t
0
) = 0. Thus solving −4.9t
2
+ 2t + 20 = 0, we get t
0
= 2.2 seconds,
approximately.
165
EXERCISES
1. (Everybody must do this problem.) Starting with ax
2
+ bx + c = 0, give a self-
contained and coherent derivation of the quadratic formula.
2. Suppose we have two quadratic functions f(x) = ax
2
+ bx + c and g(x) =
a
x
2
+b
x+c
, and suppose f(x) = g(x) for all x. Prove that a = a
, b = b
, and c = c
.
3. Solve each of the following quadratic equations: (i) 6x
2
− 13x = 5. (ii)
s
2
= 6s + 3. (iii) −3x
2
+ 4
√
3 x − 4 = 0. (iv)
1
6
x
2
+ x −
1
3
= 0. (v) x
2
+
1
4
x =
1
4
.
(vi) −t
2
−
√
13t + 3 = 0. (vii) −x
2
−
√
13x = 3. (viii)
√
180x
2
+ 7x =
√
5. (ix)
x
2
−3
√
2x + 4 = 0. (x) 3s
2
−
4
√
3
s +
1
5
= 0.
4. Use mental math to decide whether each of the following quadratic functions
has two distinct zeros, only one zero, or no zero: (i) f(x) = 215x
2
− 87x + 21. (ii)
f(s) = 5s
2
+
22
3
s + 7. (iii) g(x) = −83x
2
+ 5.2x −
9
76
. (iv) h(s) =
1
2
s
2
−
_
15
7
s + 1.5.
(v) h(x) = 3.2x
2
−9.5x + 22.
5. In the quadratic function f(x) = 3x
2
−ux +2, u is a number. For what value of
u would f have two zeros? One zero? No zero?
6. In the quadratic function g(x) = 3x
2
+x + 2u, u is a number. For what value of
u would g have two zeros? One zero? No zero?
7. Factor these trinomials: (i) 30x
2
+ 13x − 36. (ii) 5x
2
− x − 7. (iii) 105x
2
+
766x + 72. (iv) 4x
2
−
11
6
x −3.
8. A hifi store sells only 35 CD players per month when the price is marked up to
make a pofit of $50 per player. Suppose for each $2 decrease in the price, the store can
sell 5 more players. What should the price be per player in order to maximize total
monthly profit? What will the total profit be per month?
9. Among all rectangles with a perimeter of s feet, which has the greatest area?
10. Find a quadratic polynomial with the indicated pair of zeros: (i) 2 and −5. (ii)
−
3
5
and 4. (iii)
3
4
and
7
3
. (iv) 2 +
√
5 and 2 −
√
5. (v)
√
6 and 5. (vi)
√
2 and
√
3. (vii)
2
3
+
√
5 and
2
3
−
√
5. (viii) 1 −
√
10
3
and 1 +
√
10
3
.
11. Find two numbers whose difference is 7, and the difference of their cubes 721.
12. A merchant has a cask full of wine. He draws out 6 gallons and fills the cask
with water. Again he draws out 6 gallons, and fills the cask with water. There are now
25 gallons of pure wine in the cask. How many gallons does the cask hold?
13. Two workmen can do a piece of work together in 6 days. In how many days can
each do it alone if it takes one of them 5 days longer than the other. (Assume both work
at a constant rate.)
166
14. A train makes a run of 120 miles. A second train starts one hour later and,
traveling at 6 mph faster, reaches the end of the same run 20 minutes later than the
first train. Find the time of the run of each train. (Assume that both trains make the
run at a constant rate.)
15. A tank can be filled by the larger of two faucets in 5 hours less time than by the
smaller one. It is filled by them both together in 6 hours. If the water flows from the
faucets at a constant rate, how many hours will it take to fill the tank by each faucet
separately?
167
Preface
These are the lecture notes of a three-week summer institute on beginning algebra that I gave for middle school teachers. There is nothing fancy about their content: with minor exceptions, it covers only the standard topics of Algebra I. These notes may therefore be called Algebra I from a somewhat advanced point of view. If there is any merit to be claimed for them, it may be the sequencing of the topics and the logical coherence of the presentation. The exposition is formally self-contained, in the sense that the reader is not assumed to know anything about algebra. In practice, though, the reader is likely to have taught, or will be teaching Algebra I so that he or she is already familiar with the more routine aspect of the subject. For this reason, while I have made an effort to comment on the standard computations whenever it is appropriate, I have on the whole slighted the drills that unavoidably accompany any such presentation. Many of our algebra teachers are experiencing real difficulty in carrying out their duties, mainly because they have been told to emphasize certain ideas that they (like most mathematicians) cannot relate to, such as that of a quantity that "changes" or "varies" or that the equal sign is a "method that expresses equivalence". At the same time, they are denied the explanations of key facts which form the backbone of introductory algebra, such as why the graph of a linear equation is a straight line, or why the solution of a system of linear equations is the point of intersection of the lines defined by the equations. The fact that many teachers do not even recognize these as key facts in algebra speaks volume about the present state of pre-service professional development in mathematics. The main impetus behind the writing of these notes is to propose a remedy. They give an exposition of algebra ab initio, assuming only a knowledge of the rational number system. It is an exposition that is consistent with the basic requirements of any exposition in mathematics. The reader will therefore find in these notes the unfolding of ideas based not on abstruse philosophical discussions but on clear and precise definitions and logical reasoning. An integral part of the learning of algebra is learning how to use symbols precisely and fluently. I believe if there is any meaning at all to the phrase "algebraic thinking" in school mathematics, this would be it. In this regard, there is a need to single out the first two sections of these notes for some special comments. A principal object of study in introductory algebra is polynomials or, in techincal language, elements of the polynomial ring R[x]. Every exposition of school algebra must come to grips with the problem of how to properly introduce this concept to beginners. The mathematical decision I made (which is of course not mentioned in the notes) is to exploit the theorem 3
e.
1
What is so difficult about the preparation of mathematics teachers? Elsewhere.berkeley. There is a danger. the main thrust of this section would be lost on some readers. The exposition in these notes fully reflects this sense of urgency. and that the simplicity of its content demonstrates how algebra can be taught without any fanfare. Formal algebra in the sense of R[x] can be left to a later date. in Algebra II. and the main purpose of Section 2 is to promote this recognition. namely.. and the exposition intentionally emphasizes its similarity to arithmetic. At a time when Algebra for All is the clarion call of the day in mathematics education. literally. Section 1 is therefore entirely elementary: it is nothing more than a direct extension of arithmetic. however. It seems to me sensible to separate this difficulty into two stages: first teach how to translate verbal information into symbolic expressions.edu/∼wu/
4
. Section 2 addresses one of the main obstacles in the teaching of algebra: students' seeming inability to solve word problems. and then teach the necessary symbolic manipulations to extract the solution from the symbolic expressions.g. I have called attention to the need in school mathematics to emphasize precise definitions and to respect the WYSIWYG characteristic of mathematics. that precisely because of its elementary character. the x in a polynomial an xn + · · · + a1 x + a0 may be simply taken to be a number. this section may be considered a contribution to this national goal. The purpose of Section 1 is to demonstrate how one can do algebra by taking x to be just a number. the fact that what is in Section 1 is genuine algebra. The need of such a separation does not seem to be recognized at present in mathematics education.that R[x] is isomorphic to the ring of real-valued polynomial functions.1 My personal experience is that this need is real and urgent. so that in the context of introductory algebra. and algebra then becomes generalized arithmetic.
. there are many occasions when the use of symbols achieves both clarity and brevity of the mathematical message. . we have to resort to the use of symbols to express this assertion correctly and 4 6 6 succinctly.1
Symbolic Expressions
Why symbols? When we try to assert that something is valid for a collection of numbers (e. In addition. multiplication." thrown in) can only suggest but do not convey clearly the fact that the law is valid for all numbers. for any two numbers. namely. the brevity resulting from the use of symbols should be obvious. 3 × 4 = 4 × 3. 2 × 3 = 3 × 2. what we wish to assert is the commutative law 3 3 of multiplication. We repeat these formulas here to emphasize this point: let k . subtraction. for all positive integers. k. m. etc. The above collection of examples. what this law says because a finite number of examples (even with "etc. In other words. and also the distributive law require a similar use of symbols. the statements of the same law for addition. In addition to the commutative law of multiplication. m be arbitrary n rational numbers.. if we multiply them one way. and division of fractions likewise cannot be stated without the use of symbols.g. the formulas for the addition. we know that 2 × 3 = 3 × 2. 9 (− 8 ) × 82 = 82 × (− 8 ). For example. In fact. but fail to convey. 17 × 4 = 9 × 17 . and switching the order and multiplying them again. the associative laws for addition and multiplication. n = 0. give an indication that this law is true. and = 0. or for all rational numbers) instead of just for a few specific numbers. It remains therefore to use symbols to state the commutative law of multiplication succinctly as: ab = ba for all numbers a and b
Compared with the preceding indented verbal statement. then we would get the same number. etc. Then: k ± m kn ± m = n n
k m km · = n n
k m n
=
kn m
5
. It is time to recall that in arithmetic. completely and unambiguously. n are integers.
but knowing that the murderer is a person who does what ordinary human beings do (such as having to eat and sleep and needing a car or a plane to go from city to city. In the same way.. Usually the detective has no idea who the murderer is. no matter what they are.e. is usually clear from context. in addition. = −7. all numbers except integer multiples of π/2. and division). we don't need to know what exact value each of k. they will have to satisfy k ± m = kn±m . we mean a number obtained from these x. we now give some well-known algebraic identities. w by performing a specific combination of arithmetic operations (i. all positive real numbers. m. and what "most" means will be clearly indicated in each case and. subtraction. n are. y. if x. Once established in such generality. unnecessarily) debated. etc. w. we will now try to clarify its meaning as best we can. addition. these formulas are established by making use of only the properties that can be logically deduced from the definition of a rational number and the associative. m. . these formulas can be used for any specific values of k . z are numbers. n For example. This is no different from what takes place in a detective novel: n n A murder has been committed and the detective is charged with finding the murderer. without knowing what the integers k. For example. or all numbers. but so long as they are whole numbers. . and being susceptible to the usual temptations) is enough to track down the murderer in most cases. However since the meaning of this term seems at present to be endlessly (and one may say. m = 5 and n = 23. 6
. n has. Identity is definitely not a technical mathematical concept that requires a 100% precise definition. .We emphasize that in each of these formulas. The term identity is used in mathematics to indicate. Thus "most" could mean all whole numbers. multiplication. . all the primes. . . informally. commutative and distributive laws. y. . m . . that an equality is valid for "most" numbers of interest. then the above formulas imply that 5 (11 × 23) ± (5 × (−7)) 218 288 11 ± = =− or − −7 23 (−7) × 23 161 161 11 5 11 × 5 55 · = =− −7 23 −7 × 23 161
11 −7 5 23
=
11 × 23 253 =− 5 × (−7) 35
As a natural extension of these ideas. By a number expression or more simply an expression in a given collection of numbers x. with k = 11. y.
or more simply an identity. but is no more than a terminology used loosely for convenience. Postponing the exact description of this notational convention to a later passage so as not to disrupt the flow of the exposition. 5 8 + (yz)2 x4 + y 4 + z 4 − xyz
are examples of expressions in the numbers x. In specific situations.) allowing for a small set of exceptions. etc.
2
7
. we may roughly describe this convention as follows: do the multiplication indicated by the exponents first. and finally the additions. z (we have to assume xyz = −2 in the first expression and y = 0 and z = 0 in the second expression to avoid dividing by 0). . for example. but we know from considerations of n n
Recall in this connection that subtraction is nothing but addition in disguise: a − b = a + (−b). xy + x3 (16z − y 2 ) − z 21 xyz + 2 will always be understood to be {xy · (xyz + 2)−1 } + {(x3 )[(16z) − (y 2 )]} − {z 21 } The ungainly sight of the last expression should be reason enough for the adoption of this notational convention. as a statement that two given number expressions are equal for a clearly defined collection of numbers (such as all whole numbers. You may have noticed that the above expressions would be ambiguous unless a notational convention concerning the arithmetic operations among the symbols is understood.2 Now we give "an approximate definition" of an algebraic identity. More is true. the correct order in carrying out the arithmetic operations in.then xy + x3 (16z − y 2 ) − z 21 . and so is k ± m = kn±m for all integers k. y. Later on in Section 8. m. we shall expand the meaning of "expression" after we have defined "n-th root". n. there will be plenty of opportunities to discern what the "small set of exceptions" means. m. we see that the latter identity is one which make allowance for the exceptions of = 0 and n = 0. as this is nothing other than the multiplication xy · (xyz + 2)−1 ). With the help of parentheses. We have just stated this equality k ± m = kn±m for integers k. for any two rational numbers a. by definition. . xyz + 2 x − y3 . The assertion that ab = ba is true for all numbers a and b is an example of an identity. b. n n n provided = 0 and n = 0. all numbers. Right here. then xy the multiplications (including the division in xyz+2 . all positive numbers. Recall again that an identity is not a precise concept within mathematics.
1042 . in this form. The equality log xy = log x + log y is an identity for all positive numbers x and y. It is a good idea to formalize it once and for all. this idea of computing the square of a sum using the distributive law turns out to be almost omnipresent in algebraic manipulations of all kinds. The equality 1 + cot2 x = csc2 x is an identity for all numbers x except for all integral multiples of π. but worth pointing out in any case. A similar consideration.complex fractions3 that this equality remains true even if k. in an identical fashion: (a + b)2 = a2 + 2ab + b2 for all numbers a and b. m. (dist. We therefore have. the exceptions to this general identity are still = 0 and n = 0. We want to get more interesting identities. though this is a fact routinely used but rarely explained in school mathematics. m. But even here. for example. we give two advanced examples without attempting to define the relevant concepts. law) (dist. n are arbitrary rational numbers. this identity is valid for all rational numbers k. But one can also proceed by appealing to the distributive law. Consider the computation of the square. law again)
Recall that these are quotients
A B
where A and B are rational numbers. of course. for example. n provided = 0 and n = 0. is the computation of the square of 497. it should be possible to mentally finish the computation as 10000 + 800 + 16 = 10816. One can compute it directly. . so that 4972 = (500 − 3)2 = (500 − 3)(500 − 3) = {(500 − 3) × 500 − (500 − 3) × 3} = {500 − (3 × 500) − {(500 × 3) − 3 } = 5002 − 2 × (500 × 3) + 32
3
(dist. law again)
This is our first identity of note. as follows: 1042 = (100 + 4)2 = (100 + 4)(100 + 4) = {(100 + 4) × 100} + {(100 + 4) × 4} = {1002 + (4 × 100)} + {(100 × 4) + 42 } = 1002 + 2 × (100 × 4) + 42 At this point.) In case it helps to further illustrate the cavalier manner in which the terminology of identity is used. More than a trick.
8
. We recognize it as (500 − 3)2 . (The fact that the identity remains valid for all irrational numbers as well requires more advanced considerations. . law)
2
2
(dist. Therefore.
since the identity (a + b)2 = a2 + 2ab + b2 is valid for all numbers.
It is a good illustration of the use of symbols.(Note that the preceding computation furnishes a good review of the basic arithmetic of rational numbers: the distributive law for a difference. The same reasoning carries over to any two numbers a and b. Indeed. a(b − c) = ab = ac for all numbers a. turns out to be very common. and the usefulness of this identity far transcends the preceding computation 9
. law again)
This particular product (a+b)(a−b) for any two numbers a and b. we stop the calculation at this point because it can now be finished in one's head: 250000 − 3000 + 9 = 247009. we may replace c by b to obtain (a − b)2 = a2 − 2ab + b2 for any number b. (the dist. we may replace b by an arbitrary number −c to get (a + (−c))2 = a2 + 2a(−c) + (−c)2 = a2 − 2ac + c2 Since a + (−c) = a − c by definition. we get (a − c)2 = a2 − 2ac + c2 . b. and since c is arbitrary anyway. so that 409 × 391 = {(400 + 9) × 400} − {(400 + 9) × 9} = 4002 + (9 × 400) − (400 × 9) − 92 = 4002 − 92 It follows that 409 × 391 = 160000 − 81 = 159919. the commutative law for multiplication was used. A third common identity can be introduce by a computation of another kind: 409 × 391 =? We recognize that 409 × 391 = (400 + 9)(400 − 9). law) (the dist. and the removal of parentheses by −(a − b) = −a + b for all a. so that (a + b)(a − b) = (a + b)a − (a + b)b = a2 + ba − ab − b2 = a2 − b2 When the symbolic computation is given in such detail. The same computation also leads to: (a − b)2 = a2 − 2ab + b2 for all numbers a and b.) Again. c. that the identity for (a − b)2 can be obtained directly from the identity for (a + b)2 . we see that in the second line. We have obtained our third identity: (a + b)(a − b) = a2 − b2 for all numbers a and b. and the attendant generality the symbolic method brings. Thus we retrieve the second identity by way of the first. b.
a2 − b2 = (a + b)(a − b) for all numbers a and b
then we obtain what is known as a factorization or factoring of a2 − b2 .edu/∼wu/. First. this reference to + includes automatically all the −'s. We explicitly point out that.
5 4
10
. and a3 b + a2 b2 + ab3 + b4 is to observe that the power of a decreases by 1 and the power of b increases by 1 as we go through the terms from left to right. a4 + a3 b + a2 b2 + ab3 . There is another identity that is equally elementary and has very far reaching applications in mathematics. if we read this identity backwards. 2 −b2 Thus.. which merely means expressing a2 −b2 as a product.g. Recall that since a subtraction is an addition in disguise. the cancellation in the second line (a4 + a3 b + a2 b2 + ab3 ) − (a3 b + a2 b2 + ab3 + b4 ) is due to the matching of each term in the first pair of parentheses with a term in the second pair of parentheses. a2 b. for example. .of 409 × 391 and others like it. if a + b = 0. if we call any of the products separated by two consecutive +'s. a term of the number expression. a4 . and this is
See. we are using the cancellation law for complex fractions here. then the way to remember the expressions a3 + a2 b + ab2 + b3 . we start with a symbolic calculation: if a. a3 . For example. . then (a3 + a2 b + ab2 + b3 )(a − b) = (a3 + a2 b + ab2 + b3 )a − (a3 + a2 b + ab2 + b3 )b = (a4 + a3 b + a2 b2 + ab3 ) − (a3 b + a2 b2 + ab3 + b4 ) = a4 − b 4 Notice two features in the preceding calculation. insofar as a and b can be rational numbers (say.5 e.. except for the first term a4 and the last term b4 . b are any two numbers. we can simplify the division aa+b to a2 − b2 a+b =a−b
because a2 − b2 = (a + b)(a − b).4 5 7 One cannot over-emphasize the importance of the role played by complex fractions in school mathematics. Knowing such a factorization for any two numbers a and b can be very useful. ab2 . b4 . Second. Section 9 of Chapter 2: Fractions (Draft). i. . 17 and 2 ).e. so that we can cancel the number a + b in the numerator and the denominator.berkeley. This time. in the same sense that 24 = 3×8 is a factorization of 24.
We can illustrate this factorization another way: if we know that 1. then the number 1. 60887 − 1 = (215. 816. we see that a − 1 = 215. 2005). 607)(215. We do not know if there are an infinite number of Mersenne primes. 60887 − 1 is except that it has 426 digits(!). 453. For example. 230 digits. we know at least that it is not a prime.never a prime number. so that 215. Those numbers 2p − 1 which are primes for a positive integers p are called Mersenne primes. 035. 5−1 4
12
. 607 with another number. we have 1 + 5 + 52 + 53 + · · · + 510 + 511 = 61. In fact. 046. 933. But with a = 215. 60887 − 1 is not a prime. 2n+1 − 1 is a prime. then 1 − 3 + 32 − 33 + 34 − · · · + 314 − 315 = 316 − 1 43. for some n. then the preceding factorization only says 2n+1 − 1 = 2n + 2n−1 + · · · + 22 + 2 + 1 and there is no factorization. 964. and both 3 and 7 are primes. If a = −3 and n = 15. 22 − 1 = 3 (n = 1) and 23 − 1 = 7 (n = 2). even if we have no idea of what kind of a number 215. 608 and n = 87.
This now becomes a summation formula for the first n + 1 powers of a number a. If a = 2. 951 (discovered on February 18. but the largest Mersenne prime that is known at present has 7. 720 =− = −10. 567 is not a prime because it is equal to 685 − 1 = (68 − 1)(684 + 683 + 682 + 68 + 1) and is therfore divisible by 67. then 1 + 5 + 52 + 53 + · · · + 510 + 511 = Since 512 − 1 = 244. corresponding to p = 25. For example. we can read it backwards to obtain (1 + a + a2 + · · · + an−1 + an ) = an+1 − 1 a−1 for any number a = 1. because it is the product of 215. 453. 933. if a = 5 and n = 11. −3 − 1 4 512 − 1 512 − 1 = . 140. 480. 607. 567 = 685 − 1. starting with a0 = 1 up to an . But we are far from exhausting the charm of this identity. We will discuss the case a = 2 presently. Thus. although 24 − 1 = 15 and 15 is certainly not a prime. 761. 156. 60886 + · · · + 215608 + 1) This shows that 215. 624.
and we deal with this first. 194. we have
3 3 3 3 1 + 4 + ( 3 )2 + ( 4 )3 + · · · + ( 4 )10 = {( 3 )11 − 1}/( 4 − 1) 4 4
which is equal to 16. (also called a term) and then add the resulting numbers to get (18 × 53 ) + (53 × 23) + (69 × 53 ) = 2250 + 2875 + 8625 = 13750 Now if we reflect for a moment. Underlying the whole discussion of polynomials will be a simple observation based on the distributive law. If we apply the distributive law. 53 × 23. or (as in the second case) add three times and multiply once.And finally. 304 It is roughly 3. we would realize that we wasted precious time doing three multiplications before adding. Suppose we have a sum (18 × 53 ) + (53 × 23) + (69 × 53 ) One can compute this sum by multiplying out each product 18 × 53 . then the computation becomes easier: (18 × 53 ) + (53 × 23) + (69 × 53 ) = (18 + 23 + 69) × 53 = 110 × 125 = 13750 (Notice that we have made use of the commutative law of multiplication to change 53 ×23 to 23 × 53 in the process. 068. 4. 628 . and 69 × 53 . if a =
3 4
and n = 10. A sum of the form 1 + a + a2 + · · · + an−1 + an is called a finite geometric series of n terms in a.8. but the difference in conceptual clarity between (18 × 53 ) + (53 × 23) + (69 × 53 ) and (18 + 23 + 69) × 53 13
. We next introduce polynomials. We have just learned how to sum a finite geometric series. Geometric series appear everywhere in both science and mathematics. it doesn't make any difference whether we get the answer by multiplying three times and then add once.) You may think that with the advent of high speed computers. This is true.
where 163 = 24 + 73 + 66 and −53 = −89 + 25 + 11. Recall yet again that we talk about both of the above expressions as a "sum" even in the presence of the terms −(257 × 25 ) and −[( 3 )8 × 89] because −(257 × 25 ) = +{−(257 × 25 )} and −[( 3 )8 × 89] = 5 5 3 8 +{−[( 5 ) × 89]}. In an entirely similar manner, suppose we are given a sum of multiples of nonnegative integer powers of a fixed number x, where multiple here means simply multiplication by any number and not necessarily by a whole number, and "nonnegative integers" refers to whole numbers 0, 1, 2, . . . . Then we would automatically collect together the terms involving the same power of x as before. For example, we would rewrite 1 1 3 x + 16 − 8x2 + x3 − x5 − 6x2 + 75x + 2x3 2 3 17 3 x − 14x2 + 75x + 16. 6 Observe that we have followed three conventions in writing the latter sum involving the powers of a fixed number x: (i) the earlier convention that parentheses are suppressed −x5 + 14 as
with the understanding that exponents be computed first, multiplications second, and additions third, (ii) the power of x is placed last in each term (so that instead of −x2 14, we write −14x2 ), and (iii) the terms are written in decreasing powers of the number x in question. (The term 16 is the term 16x0 , and incidentally, this is where we need the zeroth power of x.) The number in front of a power of x is called the coefficient of that particular power of x, and a sum of multiples of nonegative integer powers of x is called a polynomial in x. A multiple of a single nonnegative power of x, such as 58x12 is called a monomial. Thus, a monomial is a polynomial with only one term. The highest power of x with a nonzero coefficient in a polynomial is called the degree of the polynomial. The terminology about "nonzero coefficient" refers to the fact that the preceding polynomial −x5 + 17 x3 − 14x2 + 75x + 16 could be written as 6 17 3 37 5 2 0 · x − x + 6 x − 14x + 75x + 16, but the 37-th power of x clearly doesn't count. This polynomial has degree 5, and not 37 (and not any whole number different from 5, for that matter.) Moreover, −1 is the coefficient of x5 , 0 is the coefficient of x4 , and −14 is the coefficient of x2 , because, strictly as a sum of the powers of x, this polynomial is in reality 17 (−1)x5 + 0x4 + x3 + (−14)x2 + 75x + 16x0 . 6 0 Similarly, 16 is the coefficient of x . As is well-known, a polynomial of degree 1 is called a linear polynomial, and one of degree 2 is called a quadratic polynomial. Because a general quadratic polynomial has only three terms, ax2 + bx + c, it is sometimes called a trinomial in school mathematics. We will discuss quadratic polynomials in some detail in the last two sections. A polynomial of degree 3 is called a cubic polynomial. There is no reason why we must restrict ourselves to polynomials in one variable. If x, y, z, etc., are numbers, then sums of multiples of the products of nonnegative powers of x, y, z, etc., are called polynomials in x, y, z, etc. For example, 19x3 y 21 −8y 9 z 5 −xyz+31 is such a polynomial.
[Here we should address the issue of order of operations, a topic as wrongly over-emphasized in school mathematics as the insistence on having all fractions in lowest terms.6 Just learn it, and go on to more important topics, such as those on which we will be spending a lot of time in the days ahead.
6
See "Order of operations" and other oddities in school mathematics,
15
Also need to bring out the use of · in place of ×, the fact that we usually write 42x2 instead of 42 · x2 unless we wish to achieve an extra degree of clarity, the fact that we write 1x simply as x, and the omission of all terms of the form 0xm . Also explain why, for example, −14x2 is equal to +(−14)x2 .]
You have seen polynomials before. The so-called expanded form of the five-digit whole number 75018, for example, is (7 × 104 ) + (5 × 103 ) + (0 × 102 ) + (1 × 101 ) + (8 × 100 ) which is a fourth-degree polynomial in the number 10. Of course the expanded form of any k-digit whole number is a polynomial of degree (k − 1) in 10. On the other hand, the so-called complete expanded form of a decimal such as 32.58, (3 × 101 ) + (2 × 100 ) + (5 × 10−1 ) + (8 × 10−2 ), is not a polynomial in 10, for the reason that it contains negative powers of 10. It should be pointed out that, as a polynomial in 10, the expanded form of a whole number is very special: each coefficient is a single digit whole number. Thus none of the following polynomials in 10 is the expanded form of a whole number: (35 × 102 ) + (2 × 101 ) + (7 × 100 ) (3 × 10 ) − (1 × 102 ) + (7 × 101 ) + (4 × 100 ) (5 × 102 ) + ( 2 × 101 ) + (7 × 100 ) 3
3
The first is not the expanded form of a whole number because 35 is not a single digit, the second because the coefficient of 102 is −1, which is not a whole number, and the third because 2 is not a whole number. However if we choose to rewrite the first of these 3 three polynomials in 10 as (3 × 103 ) + (5 × 102 ) + (2 × 101 ) + (7 × 100 ), then it is the expanded form of 3527. Because polynomials are just numbers, we can add, subtract, multiply, and divide them as usual. With the exception of division, the other three arithmetic operations produce another polynomial in a routine manner. (Division of polynomials does not generally produce a polynomial and will be looked at separately.) Consider the product 16
e.) We have just 3 factored the trinomial acx2 + (ad + bc)x + bd (i. (ax + b)(cx + d) = acx2 + (ad + bc)x + bd is nothing but routine applications of the distributive law... but of course with some practice the 4 factorization could be done directly. By the definition of a polynomial. q(x). c = 1 . In particular. this has to be done automatically anyway. p(x) = q(x)r(x) is a factorization of p(x) if the degrees of both q(x) and r(x) are positive.of two first degree polynomials. b = −3. then (ax + b)(cx + d) = (ax + b)(cx) + (ax + b)d = acx + bcx + adx + bd = acx + (ad + bc)x + bd
2 2
(dist. r(x) in x. For example. However. What we obtained above. b. and now we have to repeat this message. For example. d are the coefficients of two linear polynomials (i. polynomials of degree 1) in x. we say that for polynomials p(x). c. If a. law) (dist. (Thus 5 x3 − 3 2 5 2 2x2 + 3 = ( 1 )(5x3 − 6x2 + 2) is not a factorization of 3 x3 − 2x2 + 3 . it becomes acx2 + (ad + bc)x + bd = (ax + b)(cx + d) In general. the uncivilized mnemonic device called FOIL is to be studiously avoided. when this equality is read backwards.e. The main point is to emphasize the role played by the distributive law and to showcase the fact that multiplying polynomials is no different from the usual operations with numbers. law) (dist. for example. we rewrite 1 2 5 1 x + x − 3 = (2x2 + 5x − 12) 2 4 4 17
. and d = 1. law)
Of course we had to collect terms of the same degree using the distributive law and rearranging the terms so that they are in descending powers of x at the end. we get 1 2 5 1 x + x − 3 = (2x − 3)( x + 1) 2 4 4 by letting a = 2. since it is much easier to deal with integers rather than rational numbers. We have mentioned more than once the need to read an equality backwards. such operations with polynomials are just more of the same and would not be a problem. If the arithmetic of numbers (whole numbers and fractions) is taught correctly. a polynomial with three terms) as a product (ax + b)(cx + d).
It is also important that they can effortlessly rewrite a simple trinomials such as x2 + 2x − 35 as (x + 7)(x − 5). and the coefficient 2 of 2x2 + 5x − 12 has to be the product of the coefficients 2 and 1 of 2x − 3 and x + 4. in an algebra course. For this reason. not to say obsessively. (5x3 − 1 1 1 1 x)(x2 + 2x − 4) = (5x3 − x)x2 + (5x3 − x)2x − (5x3 − x) 4 2 2 2 2 1 3 = (5x5 − x ) + (10x4 − x2 ) − (20x3 − 2x) 2 41 3 = 5x5 + 10x4 − x − x2 + 2x 2
By the way. At present. some perspective on this subject is called for. when the teaching of a small skill gets blown up to be a major topic.. One should keep in mind that all it is is learning to decompose two whole number A and C into a product so that a given trinomial Ax2 + Bx + C can be written as acx2 + (ad + bc)x + bd (which equals (ax + b)(cx + d)). But it sometimes happens that if a little bit of something is good. a lot of it can actually be bad for you. there will be a two-step algorithm to acomplish this factorization no matter what the coefficients of the trinomial may be. The teaching of algebra should avoid this pitfall. We give one more illustration of the multiplication of polynomials. 12) of 2x2 + 5x − 12 has to be the product of the zero-degree terms of 2x − 3 and x + 4. reading this equality backwards gives a factorization that is (for a change) not so trivial: 5x5 + 10x4 − 41 3 1 x − x2 + 2x = (5x3 − x)(x2 + 2x − 4) 2 2 18
. respectively.e. the teaching of factoring trinomials with integer coefficients figures prominently. Please also keep in mind the fact that once the quadratic formula becomes available (see Section 12). So a few trials and erors would get it done. This seems to be the case here. There is no denying that beginning students ought to acquire some facility with decomposing numbers into products. with the consequence that other topics that are more central and more substantial (such as learning about the graphs of linear equations or solving rate problems correctly) get slighted.Then we recognize that (2x2 + 5x − 12) = (2x − 3)(x + 4) because the zero-degree term (i. each step except the last makes use of the distributive law.
(1) − 1 4 5 which is equal to 16 24 . as a consequence. For example. then the degree of the product p(x)q(x) is (m + n). respectively.. a rational expression is just a complex fraction and can therefore be added. then x2 − 1 = 0 and the denominator would be 0. For example. in the preceding rational expression. and divided 1 like any other complex fraction. In middle school. For example. subtracted. division) of two polynomials in a number x is called a rational expression in x.Note the fact that if p(x) and q(x) are polynomials of degree m and n. Is the sum of two n-th degree polynomials always an n-th degree polynomial?
A quotient (i. the preceding calculation which multiplies a degree 3 polynomial with a degree 2 polynomial yields a polynomial of degree 5 (= 3 + 2). multiplied. since x is a (rational) number. In general. Here is an example: 3x5 + 16x4 − 25x2 − 7 x2 − 1 We note that in the case of rational expressions.e. we need to exercise some care in not allowing division by 0 to take place. we would be looking at the complex fraction
1 1 1 3( 32 ) + 16( 16 ) − 25( 4 ) − 7 . In other words. we can compute with rational expressions in x in the usual way:
2x7 (5x3 + 1)(x3 + 4) + (2x7 )(x8 + x − 2) 5x3 + 1 + 3 = x8 + x − 2 x +4 (x8 + x − 2)(x3 + 4) and x2 + 1 6 (x2 + 1)(6) · = x2 + 4x − 7 3x4 − 5 (x2 + 4x − 7)(3x4 − 5)
and 19
. all computations with numbers tacitly assume that the numbers involved are rational numbers. we are mainly interested in rational numbers only and. it is usually understood that only those values for which the denominator is nonzero are considered. With this mind. In writing rational expressions. no matter what x is. in case x = 2 in the foregoing rational expression. x can be any number except ±1 because if x = ±1. the degree of a product is the sum of the degrees of the individual polynomials.
the cancellation can be less obvious. y. EXERCISES
7
Again. C.e. the number (5x4 −x3 +2) in both the numerator and denominator can be cancelled. etc.2x+1 x2 −3 4x3 −x+11 2x
=
(x2
(2x + 1)(2x) − 3)(x3 − x + 11)
These are the same as any computation with complex fractions. just as one can easily define polynomials in x. one can likewise define rational expressions in x.7 some rational expressions can be simplified. with A = 0. as in (5x4 − x3 + 2)(2x − 15) (14x2 + 3x − 28)(5x4 − x3 + 2) Here. C = 0). AB = B for all AC C rational numbers A.. etc.berkeley.edu/∼wu/. For example. x3 − 8 = x3 − 23 = (x − 2)(x2 + 2x + 4) and we can cancel the number (x2 + 2x + 4) from the numerator and denominator.) +2x+4 In beginning algebra. This is a left-over from the questionable practice of teaching fractions by insisting on the reduction of all fractions to lowest terms at all costs.. There is so much in algebra that is just a revisit of arithmetic. by an earlier identity. z. Sometimes the cancellation presents itself.
20
. (As you will learn when we come to quadratic equations. we actually have an identity x2x −8 = x − 2 for all x. Because the cancellation law is valid for complex fractions (i. x can be any number because x2 + 2x + 4 is never 3 equal to 0. B. Therefore. it turns out that in the case of this particular rational expression in x. It remains to round off this discussion by mentioning that. y. compare Section 9 of Chapter 2: Fractions (Draft). resulting in (5x4 − x3 + 2)(2x − 15) 2x − 15 = 2 + 3x − 28)(5x4 − x3 + 2) 2 + 3x − 28 (14x 14x Sometimes. z. the rational expression x3 − 8 x2 + 2x + 4 can be simplified to x − 2 because. often there is too much emphasis on simplifying rational expressions. It is important to realize that these computations ar exactly the same as those in complex fractions and not just "analogous to" them.
2
Transcription of Verbal Information into Symbolic Language
Word problems are the bugbears of many students and teachers alike. After a while. A number is a number. conjuring as it does the mental image of a number that "varies". and most of the difficulty stems from the inability to correctly transcribe verbal information into equations or inequalities. the fact that an inequality in x is a statement about the validity of one number expression bigger than another one for a certain collection of numbers x will sometimes be omitted in the future. bigger than or equal to) the other. the word "variable" is loosely understood to be an element in the socalled domain of definition of a function. Again. −2. The inequality x2 + 1 < 0. In mathematics. is something you as a teacher have to be careful about. There is no hope of getting a correct solution to a word problem if we work with the wrong equations or inequalities. the equation x2 + 1 = x2 is valid for no x. The importance of a correct transcription of the verbal information into symbolic expressions in the context of problem-solving seems to
22
. for some x. it is convenient to have around. it was used in the past and. Therefore the focus of this section is on this critical link in the solution of word problems. for some x. the process of transcribing verbal information into equations or inequalities. is satisfied by no x. an inequality in a number x is defined as a statement that one number expression in x is bigger than (or. namely. We pause to formally define an equation in a number x to be a statement that two given number expressions involving x are equal. as in the case of equations. the equality x3 − 1 = − 5 x2 − 2 x 2 for a number x is an equation which happens to be valid for exactly x = 1 . The psychological damage that the terminology of a "variable" can do to beginners. Again. and there is no precise definition of the term. Indeed. these equations are sometimes called equations of one variable out of respect for tradition. the inequality may be valid for all x. for example. In a similar vein. Yet in school mathematics. 2 On the other hand. like "identity". the fact that an equation is about some number x yet to be specified will be understood and reference to this fact will be omitted. It never varies. it is an extreme irony that this term is taken seriously while others that should be (such as fractions or decimals or the graph of a function) are not. or for no x. Equations in a collection of (yet-to-be-determined) numbers are similarly defined. none is called for. or for no x. Because only one number x is involved. We will deal with equations in two variables in §4. For example. as we said. It is retained in the mathematical literature only because. This equality may 1 be valid for all x. −1.
Many teacher jump directly from reading the verbal information in a problem to attempting to get a solution. you as a teacher must take the lead. Express this information as an equation in a . (We put in the × symbol here for clarity.e. as a ·57 would look somewhat odd.. 3 b Solution We know from the meaning of the multiplication of fractions that " a of 57" b is just a ×57. the exact determination of all the numbers x that make the equality valid). We will give a few illustrative examples. Without worrying about how to solve equations or inequalities (i. Our purpose here is to isolate this intermediate step and call attention to the need of addressing this critical issue.e.be not fully recognized by teachers and educators alike. Let us begin with a simple one. In these examples. To this end. this number is 4 bigger than 2 of 57. whatever convention there is concerning the use of symbols should be ignored when clarity or brevity is jeopardized. with the hope that the extra practice would enable you to gain the needed facility and confidence to help the students. So we transcribe this information directly: 3 a 2 {57 − ( × 57)} − ( × 57) = 4 b 3 This is one answer. Obviously. You will be asked to do plenty of such transcriptions.
Let a be a fraction. "If a fraction a of 57 is taken b away from 57". Consequently. becomes a 57 − ( × 57) b According to the given information. our task is to make sure that students learn to set up problems correctly. The fact that there is an intermediate step between the "reading" and the "solving" is one that deserves emphasis. The main purpose of the b symbolic language is to add clarity and brevity to the verbal expression. Then all the information is pulled together at the end to arrive at the correct equation(s). b b while writing it as 57 · a might confuse it with a mixed number.) Thus the statement. 4 bigger 3 than 2 × 57. If a of 57 is taken away from 57. what remains exceeds b b 2 of 57 by 4. notice that the starting point is always a systematic. i. one could equally well express this equation as a 2 57 − ( × 57) = ( × 57) + 4 b 3 The following example is a bit more complicated..
23
. sentence by sentence transcription of the verbal data into the symbolic language.
and in both cases she will not have enough money left over to buy more of either pastry. and B". "His sister is half the age of his older brother" then 1 becomes S = 2 A. . and B the age of his younger brother. A the age of Johnny's older brother. If the prices of the pastries are x dollars and y dollars. They show that both 1 A and 4 B 4 2 1 are equal to S. A = J + 4 and B = J − 2. but we are asked to "express the above information in terms of J. three fourths the age of his younger brother" becomes S = 3 B. Johnny's older brother is four years older than Johnny. and J. A. She finds that she can buy either 10 of one and 9 of the other. and therefore equal to each other. So we go back to look at S = 2 A and S = 3 B. At this point. which gives information on how the brothers are related to each 3 1 other.. and one of them is to transcribe all the information by bringing in the sister. i. B. A and B: A = J + 4. 24
. B = J − 2. respectively. while "his younger brother is two years younger than Johnny" translates to B = J − 2. "Johnny's older brother is four years older than Johnny" becomes 4 A = J + 4. Express the above information in terms of J. A.Johnny has three siblings. two brothers and a sister. the two equations. Solution The first thing to take note is that the given data of the problem involves Johnny's sister. Let J be the age of Johnny. Thus we also have 2 A = 3 B. the sister is left out. and "His sister is . and his younger brother is two years younger than Johnny. 1 3 A = B. So let S be the age of the sister. or 13 of one and 6 of the other. and three fourths the age of his younger brother. There are many ways to deal with this situation. and then try at the end to omit any reference to her and still faithfully transcribe the given verbal data. and B. His sister is half the age of his older brother. write down the inequalities satisfied by x and y. Now we 4 have collected all the given information concerning J. would appear to be the answer because they are the only equations directly involving A. . But these two equations fail to capture the part of the given information about how the brothers are related to the sister. 2 4
We next give an example requiring the use of inequalities:
Erin has 10 dollars and she wants to buy as many of her two favorite pastries as possible.e.
"she will not have enough money left over to buy more of either pastry". If this amount exceeds or equals x. and then $(13x + 6y) in the second option. and so 10 − (10x + 9y) ≥ 0. R and x. and the other as the Second Woman. Before looking at a correct solution. 0 ≤ 10 − (13x + 6y) < x. Erin spends a total of $(10x + 9y) in the first option. If the sunrise was x hours before noon. One went from City A to City B while the other went from B to A. 25
. we do a problem that is a trifle more sophisticated than the previous three. The answer is then the collection of four double inequalities: 0 ≤ 10 − (10x + 9y) < x. Such not being the case. as the possibility of equality (implied by ≤ ) would mean that Erin could buy one more of the x-pastry. she cannot spend more than $10. Moreover. transcribe the information above into equations using the symbols L.Solution With x and y understood. Solution For ease of discussion. and if L is the speed of the woman going from A to B and R is the speed of the woman going from B to A.
Two women started at sunrise and each walked at constant speed. They met at noon and. Consider then the first option: the total number of dollar left over is 10 − (10x + 9y). continuing with no stop. 0 ≤ 10 − (10x + 9y) < y. We combine these two inequalities into the following double ineqiality: 0 ≤ 10 − (10x + 9y) < x Note that it is strict inequality and not 10 − (10x + 9y) ≤ x. we have 10 − (10x + 9y) < x. It is very instructive. Similarly. we will refer to the woman going from City A to City B as the First Woman. 0 ≤ 10 − (13x + 6y) < y
As a final illustration. 0 ≤ 10 − (10x + 9y) < y The same consideration applies to the second case. we first look at one that may be what most people would write down. arrived respectively at B at 4 pm and at A at 9 pm. then Erin would be able to purchase one more of this pastry. The key point is that in either case.
while in 9 hours the Second Woman would walk 9R miles. Given that the former walked with speed L (let us say. The First Woman covered it in x hours (before noon) while the Second Woman covered it in 9 hours (after noon). Let the meeting point of the two women be C: AE C
'
B
Consider the distance between A and C. as we have seen. This means that the total distance they covered after walking the first x hours (which is Lx + Rx miles) is the distance between the cities. and both the First Woman and Second Woman walked this distance in the time given: the First Woman walked x hours before noon. Therefore. Therefore the additional piece of information that must be incorporated into the symbolic transcription is Lx + Rx = L(x + 4) or Lx + Rx = R(x + 9). Similarly the total distence the Second Woman walked in x + 9 hours is R(x + 9) miles. But is it? What this equation fails to capture is the information that the two women met at noon after walking x hours in opposite directions from City A and City B. 4L = Rx
We will leave as an exercise that these two solution are "the same". both distances are the same as both women walked between cities A and B. Therefore we get L(x + 4) = R(x + 9) This is supposed to be the answer to the problem. Either one would do. in a precise sense. while the Second Woman also walked x hours before noon but continued for another 9 hours (from noon till 9 pm). The distance between City A and City B is fixed.
26
. But in x hours the First Woman would walk Lx miles. So the First Woman walked a total of x + 4 hours while the Second Woman walked a total of x + 9 hours. R(x + 9) = Lx + Rx
Now one may reason slightly differently. if we consider the distance between C and B. A correct solution is then: Lx = 9R.and explain why it is not good enough. By a previous remark. we get in exactly the same fashion that 4L = Rx. miles per hour). The reasoning goes as follows. and then another 4 hours (from noon till 4 pm). Similarly. which is L(x + 4) or R(x + 9) miles. the total distance she walked in x + 4 hours is of course L(x + 4) miles. Therefore Lx = 9R. a solution to the problem is the following set of equations: L(x + 4) = R(x + 9).
express the given information in equations in terms of x and y. and Q denote the number of nickels. At the same constant speed. 5. we get back the number itself. express the above information in terms of x and y. Also. If the middle integer is x.5% of ten times the smaller number. and quarters. 10. If N . how long will it take to travel s km? And if the speed is n times faster? 2. 8. the larger number is 97. the digits are
27
. write an equation using p and N to express the above information. If the digits of a three-digit number are reversed. write equations in terms of these symbols to capture the given information. write down an equation for y. If 99 is added to the original number.60 worth of nickels. Express this information in equations in terms of x and y. A train travels s km in t hours. There are 40 coins in all. Paulo read a number of pages of a book with N pages.EXERCISES 1. 9. and the difference of the squares of the prices is 735. the sum of the new number and the original number is 1615. 4. then he read 43 pages more and finished three-fifths of the book. There are two whole numbers. 3. then what remains would be 4 less than y. D. We look for two whole numbers so that the larger exceeds the the smaller by at least 10. If the prices are x and y dollars. dimes. 7. I have $4. how far does it travel in 5 hours? In T hours? How long does it take the train to travel 278 km? x km? If the speed of the train is tripled (3 times as fast). Helena bought two books. If however I enlarge y by 20%. If x is the larger number and y is the smaller number. If the smallest of the three integers is y. but that the cube of the smaller exceeds the square of the larger number by at least 500. A whole number has the property that when the square of half this number is subtracted from 5 times this number. Take 20% of x from x. I have two numbers x and y. the quotient is 9 and the remainder is 15. If y is this number. transcribe the above information in terms of x and y. then it would exceed x by 5. and the number of nickels and dimes is three times the number of quarters. express this fact in terms of x. If p is the number of pages Paulo read the first time. 6. If the larger number is x and the smaller number is y. dimes and quarters. When the large number is divided by the smaller number. express the same fact in terms of y. The total cost is 49 dollars. respectively. The sum of the squares of three consecutive integers exceeds three times the square of the middle integer by 2.
Recall that we call such an equation an equation in one variable. The denominator of a fraction exceeds twice the numerator by 2.reversed.e. D kilometers apart. B. we make a first attempt at solving an equation. each person would receive 1 dollar less. The first runner runs D kilometers in A hours. He has two games. and if there are 6 fewer people. If he brings x A Games and y B Games. describe in terms of x and y how he can maximize his profit. write equations in terms of x and y to express the above information. tens. b. A video game manufacturer sells out every game he brings to a game show. A man walked from one place to another in 5 1 hours. In this section. If the numerator is x and the 24 denominator y. they are 11 kilometers apart after 1 hour. determining all possible values of x that make the equation valid. However. Write equations in a. or no value of x. and c. and ones digits of the original numbers be a. A sum of money is to be divided equally among x people. each receiving y dollars. b and c to express the given information. and the second runner in B hours. they do the same. Write equations in x and y to express this information. 13. then they would be 5 kilometers apart after one hour. 12. The statement may be true for all values of x. He can bring 50 of A Games and B Games in total to the show. which is merely a statement that two number expressions in a number x (as yet undetermined) are equal. some values of x. Let the hundreds. 15. and D. The numbers that make the equation valid are 28
. Each B Game costs $165 to manufacture and will bring in a profit of $185. i. 14. he only has $6.. we have seen how equations involving a number x arise naturally. 11. Express this information in symbolic language in terms of A. towards each other.
3
Linear Equations in One Variable
In the preceding section. again. 000 to spend on manufacturing. If the distance he 3 walked is x miles and his speed was v miles an hour. Two marathon runners run at constant speeds. express the above information in terms of x and v. If the first runner runs twice as fast and. Each A Game costs $75 to manufacture and will bring in a profit of $ 185. an A Game and a B Game. If they start running at the same time from separate cities. If there are 3 more people. the walk would have taken 36 2 fewer minutes. each would receive 5 dollars more. If he had walked 1 of a 2 4 mile an hour faster. respectively. and the difference between the fraction and its reciprocal is 55 .
The solutions of x2 = 4 are then x = ±2. x and 5. The interesting equations fall somewhere in between: they are valid for a finite collection of numbers. and they are also said to satisfy the equation. and the 1 left side becomes 1 (3x) = ( 1 · 3)x = (1)x = x. equations which state that two polynomials of degree at most 1 in the same number x are equal. we shall rigorously prove that ±2 are indeed the solution set of x2 = 4. we deal with the simplest kind of equations which turn out to be basic. then we are given that x2 = 4. equations in x of the form ax = b. as the case may be. it is one that we have encountered before. It is intuitively clear that there are only two possibilities: x = 2 and x = −2. Some general principles 6 which lead us to the solutions of linear equations are however valid for all equations.e.) We now show that the 29
. such as x5 = x5 − 3. You will be bored if I tell you there are important lessons to be learned here. 3 3 Because multiplying two equal numbers by the same number yields two equal numbers. i. We multiply both sides of this equation by 3 . that would be valid for no number x. Sometimes one refers to the solution set more simply as the solution or the solutions. namely 5 because 5 is the only number whose product with 3 is 15. the systematical procedure of solving the preceding problem points to the way of solving a class of linear equations in general. (By letting a = 3 and b = 15. If x denotes such a number.. we say the equation has no solution. Thus we get x = 5.
Then we can read off what x must be. then the equation would be what we have called earlier an identity. while the right side becomes 3 · 5 = 5. For example. consider all the numbers whose squares are equal to 4. we want to produce this obvious 1 solution more systematically. x2 − 1 = (x − 1)(x + 1) is an example of an equation in x that is in fact an identity. if the solution set comprise all the numbers. For a reason that will be obvious.called the solution set of the equation. the resulting two numbers. and we will point these out in due course. In this case. (Later on. namely. To begin with. At the opposite extreme. If it turns out that any number would make the equation valid. must be equal. we get the preceding example again. namely.) In this section. Suppose a linear equation is in the following form: 3x = 15 for a number x. a and b are traditionally referred to as the constants of the equation. where a. These are called linear equations of one variable. − 5 x + 1 = 23x − 4. there are equations. but this happens to be the truth. and 9 = 27x − 4. Examples are: 12x − 7 = 5x + 13. b are coefficients of the polynomials with a = 0. For example. and this is the solution.
what we did above was to "free up" the x on the left side of ax = b by getting rid of the coefficient a of ax. in its full strength. It is simpler than you think. Intuitively. This is clear. the associative. this is the assertion that two rectangles with the length of sides being pair-wise equal have equal area. c. But the associative law.b only number x that satisfies this equation is a . In subsequent discussions about solution of equations. In our definition of multiplication as area of rectangles. the left side is
1 1 (ax) = ( · a)x = 1 · x = x a a
b b while the right side is a . In symbols: if a = b and c = d for numbers a. One is the critical role played by the associative law for multiplication. If we are careful. The other general principle behind the above method of solution is this:
Two numbers that are equal remain equal when multiplied by equal numbers. and distributive laws will be applied in an entirely analogous manner to unknown numbers. What interests us right now are the two principle behind this technique. then it has to be a . being true for all numbers. then indeed ax = b. and we will almost never take the trouble to highlight such applications of these laws again. We now show 30
. We are now in a position to solve any linear equation of one variable. we get a (ax) = a · b. is nevertheless applicable here to validate the preceding step. But this is an easy direct verification. By the associative law. But on this occasion. we want to emphasize why the fact that these laws are valid for all numbers is important. We will enunciate presently the analogous principle for addition because the latter also plays a key role in the solution of equations. b. d. as otherwise the solving of equations would be impossible. so the eventual solution x of the equation can also be very small or very big. this we accomplished by multiplying both sides of 1 ax = b by a . This is an important technique. because multiplying both sides of ax = b 1 1 1 by a . We have solved all linear equations of the form ax = b. then ac = bd. We don't know a priori what it is. in a sense we now explain. Therefore x = a . we would notice that what the above shows is that if x is any b b number that satisfies ax = b. commutative. We must also ensure that if x = a . We applied the associative law in the step 1 1 (ax) = ( · a)x a a It must be remembered that a is a number that may be very small or very big.
What about 1 ? No 3 1 again. Recall what this means: we have numbers x for which the two expression 12x − 5 and 6x are equal. by an application of the associative law of addition. Before proceeding with the discussion of solving linear equations.e. but we know that it is really adding −12x to both sides. or as we say. we "move" or "transpose" 12x from the left side to the right side by adding the number −12x to both sides. (−12x) + (12x − 5) = (−12x) + [12x + (−5)] = (−12x + 12x) + (−5) = −5. if a number x satisfies 12x − 5 = 6x. then necessarily x = 5 . which can then be solved. thereby solving the linear equation. This is in the form of ax = b with a = −6 and b = −5. while the right side is (−12x) + 6x. we see that this must be true. To round off the picture. b. c. We can save ourselves 3 some work. we add −12x to both sides to ob5 tain −6x = −5. by isolating x on one side so that the equation will look like ax = b. First let us look at the issue of solving linear equations from an experimental point of view. i. To isolate x in 12x − 5 = 6x. if x = 2 in 12x − 5 = 6x. then the equality 12x − 5 = 6x is valid. then a + c = b + d. because 19 (= 12 × 2 − 5) is not equal to 12 (6 × 2). In short. By the preceding basic principle. −5 = (−12x) + 6x The change from 12x − 5 = 6x to −5 = (−12x) + 6x is what accounts for the common terminology of transposing 12x from left to right. it is a good idea to pause and reflect on the above procedure of transposing a term from one side of the 31
. Is 2 one of them. it remains to show 6 5 that if x = 6 . d are numbers with a = b and c = d. In symbols: if a. is 2 is a solution of 12x − 5 = 6x ? In other words. from which we conclude x = 6 . We therefore know right away that −5 x = −6 = 5 . but again this is simple to 5 check as both sides are equal to 5. −5 must equal (−12x) + 6x. The left side becomes. or −6x = −5. Using the vector model for the addition of rational numbers.how to bring any linear equation to this form. we get −5 = −6x. and we have to determine all possible values of such x. starting with 12x − 5 = 6x. Since (−12x) + 6x = −6x. 6 To recapitulate. because −1 (= 12 × 3 − 5) is not equal to 2 (= 6 × 1 ). This then shows that x = 6 is the only solution to 12x − 5 = 6x. This can be done by appealing to another basic principle: Two numbers which are equal remain equal when added to equal numbers. instead of guessing.. are the two sides equal? The answer is no. Consider the equation 12x − 5 = 6x.
B. for example. we change the sign. To solve − 2 x + 4 = − 5 x + 5 1 . C are numbers. From this. which is −4. 32
. Thus the solution is 3 / −7 = − 20 . we collect all the x's to the 3 3 1 2 1 2 left: (− 3 x + 4) − (− 1 x) = 5 1 . and so 15 x = 3 . What happens when we try to transpose the term −2x to the left side of the equal sign? Let us make sure how to do this correctly. where a. i. Now we have to transpose +1 to the right side: 3x = −11 − 1. C. we may apply the associative law of addition to transform 6x + 2x = (7 − 2x) + 2x to 6x + 2x = {7 + (−2x)} + 2x = 7 + {(−2x) + 2x} = 7 + 0 = 7. by definition. so that 3x + 1 = −11. Transposing b to d−b the right yields (a − c)x = d − b. we first transpose cx to the left: (ax + b) − cx = d. We can now suppress some of the explanations and solve another linear equation to give another demonstration of the general procedure. and why. Therefore we have (a − c)x + b = d. We next transpose 4 to the right 15 15 3 −7 1 −7 4 4 side: 15 x = 5 3 − 4. On the left side we have − 2 x + 4 + 5 x = − 3 x + 5 x + 4 = 5 3 3 −7 x + 4. Consider 5x + 1 = 2x − 11. then when we transpose the term B of A + B = C to the right side. and 3x = −12. 6x = 7 − 2x. d are constants and a = c. when we add 2x to both sides of 6x = 7 − 2x. (T2) A−B =C implies A = C + B
The two facts (T1) and (T2) together are usually summarized by saying that when we transpose. b. we get −B. (T1) A + B = C implies A = C − B Now consider. We first transpose 2x to the left side: (5x + 1) − 2x = −11. More precisely.equation to another. and the distributive law implies that (ax − cx) + b = (a − c)x + b. the same reasoning shows that for numbers A. 6x = 7 − 2x implies 6x + 2x = 7
In general. To solve ax + b = cx + d. What needs greater emphasis in classroom instructions is the underlying reasoning of why this is true.e. 15 7 We conclude with an exposition of the solution of any linear equation. Thus the solution is (−12) . B. Therefore. we conclude immediately that x = a−c . Thus the equation becomes −7 x + 4 = 5 1 .. We first recall the meaning of subtraction among rational numbers: 7 − 2x = 7 + (−2x). The preceding consideration shows that if A. 3 1 Here is another example. c. The associative and commutative laws of addition imply that the left side is equal to (ax − cx) + b.
three-step equations and fourth-step equations. You should avoid using this classification when you teach this topic. From this point of view.e. only solution of ax + b = cx + d.What we have done is to show that any number x satisfying ax + b = cx + d is equal d−b d−b d−b to a−c . solving a linear equation in a number x depends on two simple ideas: by transposing terms. We still need to check that a−c is a solution. i. consider a number x that satisfies 2 4 = 3x − 1 x+
1 3
by the cross-multiplication algorithm (which is valid also for complex fractions). This can be done directly: ax + b = a d−b a−c +b = ad − ab +b a−c ad − ab ab − bc + = a−c a−c ad − bc = a−c
while cx + d = c d−b a−c +d = cd − bc +d a−c cd − bc ad − cd = + a−c a−c ad − bc = a−c
d−b a−c
d−b So ax + b = cx + d when x = a−c . For example. the common practice of classifying linear equations into one-step equations. and then we solve an equation of the type ax = b. then it is true that ax + b = cx + d.. if x = a−c . two-step equations. It remains to point out that sometimes a linear equation is disguised as one involving rational expressions. and then teach the solving of linear equations according to this classification simply does not make sense. we isolate x on one side of the equation. this equation is equivalent to 1 2(x + ) = 4(3x − 1) 3 33
.
This then completes the reasoning that
is the
To summarize.
We observe that in this situation. 5 ) is the sought-for solution. (−2. A solution of this equation is an ordered pair of numbers (x0 . (iii) 11 − 5 x = −6x + 18 . (vii) 2 ax − 17 = 3 ax − 15 . Using the above method of getting all the solutions of the equation x − 2y = −2. For example. (0. 2). 3 − 2y = −2. 5 to get y = 2 . (−4. 15
4
Linear Equations in Two Variables and Their Graphs
An equation such as x − 2y = −2 is an example of a linear equation in the two numbers (usually called variables) x and y. −1). 0). 3). (2. For example. 4. Therefore (3. x − 2(−1) = −2. 34
. where a is a nonzero number. Solve: (i) 2x − 8 = 15 + 4 x. with the first number prescribed as 3. 1). y0 ) is a solution of the equation x − 2y = −2 is called the graph of x − 2y = −2 in the plane. going from left to right: (−5.5). (ii) 3 x + 2 = 3 − 2 x. we can plot as many points of the graph as we please to get a good idea of the graph. −1. (7. y0 ) so that x0 and y0 satisfy the equation x − 2y = −2. where 1 8 1 b is a number not equal to 2 . 3 2 5 9 3 (iv) ax + 6 = 8 − 7ax.
These points strongly suggest that the graph of x − 2y = −2 is a (straight) line.We can either use the distributive law on both sides directly. The solution is now (−4. the following picture contains the following points (given by the dots) on the graph. to get x = −4. 2. (2. if the second number 2 is prescribed to be −1. then we solve the linear equation in x. (4. In other words. y0 ) in the coordinate plane so that each pair (x0 .5). (6. the collection of all the points (x0 . Relative to a pair of coordinate axes in the plane.5. (v) 4bx + 13 = 2x + 26b. and we will presently prove that such is the case. in the sense that x0 − 2y0 = −2. or we can avoid computations with fractions at this early stage and simply multiply both sides by 3 to get 2(3x + 1) = 4(9x − 3) Then 6x + 2 = 36x − 12 and 14 = 30x. Or. (vi) 1 − 3 x = 5 x + 2 .25). 2 6 3 5 2 7 . it is easy to find all the solutions with a prescribed first number x0 or a prescribed second number y0 . then we solve the linear equation in y. 4). x = EXERCISES
7 1 1. −1).
Y
s
4 s 2 s −4 s s −2 s −2 s s
s
O
2
4
6
X
Consider next the graph of the linear equation in two variables y = 3 which. 3). 3) on the y-axis. as an equation in two variables. for the following reason. Similarly. so (s. In short. Y
3
O
X
35
. In terms of the graph. 3) then comprise the complete horizontal line passing through (0. Are there perhaps other pairs of numbers which are also solutions? For example.1) is not a solution of y = 3. This shows that the preceding assertion about the pairs (s. the points with coordinates (s. 3). 3)'s is clearly a solution because 0 · s + 1 × 3 = 3.1 = 3.e. 3) is true. and since s is arbitrary. is in reality the abbreviated form of the equation 0·x+1·y = 3. parallel to the x-axis) passing through the point (0. these points (s. The collection of all solutions of y = 3 is then exactly all the pairs (s. 3) on the y-axis. the graph of the equation y = 3 in the plane is exactly the horizontal line passing through the point (0. t) is not a solution of y = 3 no matter what t may be. where s is an arbitrary number. if a number t is not equal to 3. then (s.1 = 3.1)? But 0 · s + 1 × 3.. 3) always lie on the horizontal line (i. Every one of these (s. (s. 3. 3.
Thus −2x = 5 y + 7 and 6 + 8 y = 179 − 5x are examples of linear equations of two variables. As we have seen. In a similar manner. and the graph of y = b for any number b is the horizontal line passing through the point (0.
36
. the graph of x = c for any number c is the vertical line passing through the point (c. b) is the point of intersection of the line with the y-axis. We have supplied so much detail to explain them because your students should understand that these facts are consequences of careful reasoning and the precise definition of a graph. the study of linear equations of two variables is grounded in the study of linear equation of one variable. where (0. it follows that every vertical line is the graph of the equation x = c. b. Both of these simple facts are probably known to you. and every horizontal line is the graph of an equation y = b. vertical) line passing through a given point of the plane (do you know why?). but the precise reasoning may have been missing. in the sense that ax0 + by0 = c. so that they satisfy the equation ax + by = c. A linear equation in two variables x and y is an equation in the numbers x and y which is either of the form ax + by = c where a.Similarly. A solution of this equation is an ordered pair of numbers x0 and y0 . where (c. Armed with these precise definitions. we will now explain why the following theorem is true. parallel to the y-axis) passing through the point (−2. 0) is the point of intersection of the line and the x-axis. The graph of ax + by = c is the collection of all the points in the plane with coordinates (x0 . so that each ordered pair of numbers x0 and y0 is a solution of ax + by = c. y0 ). 0) on the x-axis. Since there is only one horizontal (respectively. the graph of the equation x = −2 is the vertical line (i. commonly referred to as constants and at least one of a and b is nonzero. We next treat the general case. written in the expected fashion as (x0 . and as we shall continue to bear witness. 0) on the x-axis. y0 ) (relative to a given pair of coordinate axes).e. or can be rewritten in this form after transposing and using the four 2 3 arithmetic operations. b) on the y-axis. c are given numbers..
In this case. and say that L is defined by ax + by = c. This reasoning is the key to the understanding of almost everything about linear equations in introductory algebra. If b = 0. We want to make a minor. In other words. we may rewrite the equation as by = −ax + c. So let G be the graph of y = 3 x + 2. then by the definition of a linear equation in two variables.e. where m. We may therefore assume from now on that b = 0 in a given equation ax + by = c. we have seen that the graph is a vertical line. The first part of the Theorem is therefore true in this case. It is customary to call ax + by = c the equation of the line L if L is the graph of ax + by = c. but important simplification in the subsequent discusion of this Theorem. the second part of the Theorem is true for vertical lines. and therefore y = mx + k. and every line L is the graph of some linear equation ax + by = c.. Why is the graph of this equation a straight line? The reasoning 2 in this case will shed light on the general case. Such being the case. a = 0. what we have shown is that in the subsequent discussion of this Theorem. This theorem establishes a correspondence between lines and the graphs of linear equations in two variables: the graph of a linear equation ax + by = c is a line L.
Let us first begin the consideration of the Theorem by looking at a concrete case 2 such as y = 3 x + 2. every straight line is the graph of a linear equation. 37
. Moreover. Conversely. the reasoning in the proof of this theorem is very important. and that a given straight line is non-vertical. Suppose we start with a linear equation ax + by = c. Incidentally. We may therefore assume from now on that a given line is not a vertical line. 0) is the point at which intersects the x-axis. where c is the constant c = a . The equation may therefore c be rewritten as x = c . we have seen that a b b vertical line is the graph of x = c (i. where m = − a and k = c . k are constants.Theorem The graph of a linear equation is a straight line. where (c . x + 0 · y = c ). this theorem explains why equations of the form ax + by = c are called linear equations (because their graphs are lines). To summarize. On the other hand. we may assume that a given linear equation is of the form y = mx + k.
One then proves that two triangles are similar if and only if the corresponding angles are equal and the ratios of (the lengths of) the corresponding pairs of sides are equal. one needs to know when two angles are equal. The proof of this fact follows quite easily from the definition of a dilation and the so-called fundamental theorem of similarity. We are going to prove that G is the line L. we review some facts concerning similar triangles and a basic property of straight lines (concerning slope) which is almost universally mishandled in introductory algebra textbooks. 0) on the x-axis are on 2 G. Therefore. However the following is less obvious: The AA criterion of similarity: If two triangles have two pairs of equal angles. we first have to show that every point on the graph G lies on the line L. they are similar. Let L be the (straight) line joining 3 (0. So we must also show that every point of L is a point of G. Before proving either of these assertions. In order to use this criterion effectively. First recall the definition of two geometric figures being similar: one is mapped onto the other by a rigid motion followed by a dilation. 2) and (−3. (β) Every point on the line L is a point on the graph G. Y L
r 2 r −3
O
X
How to show that the two sets G and L are the same? Obviously.Notice that the point (0. because 2 = 3 × 0 + 2 and 0 = 2 × (−3) + 2. 0). we must show two things: (α) Every point on the graph G is a point on the line L. But this is not enough because G could just be part of L and not all of L. we recall also the following fact:
38
. In this context. 2) on the y-axis and the point (−3.
associated to each non-vertical straight line L is a number called slope which measures the amount of "slant" of the line. which is slanted this way /. Y
Pr Q r
L
Le Y
e e e P r e e e eQ r e e e e
O
X
O
X
Let the coordinates of P . q2 ). let a pair of distinct points P . Q on L be chosen. p2 > q2 and p1 > q1 while for the line on the right. Q are distinct would mean that the line L is a vertical line. as shown:
39
. Also observe that this quotient is positive for the line on the left. and the quotient makes sense. p2 ) and (q1 . Form a right triangle P QR so that the lines P R and QR are parallel to the coordinate axes. This observation concerning the sign can be seen perhaps more transparently in terms of the geometry. because. We apply these facts to the study of straight lines by proving that. Thus coordinate axes as given.
1
and
2. p2 > q2 but p1 < q1 . this is impossible. for the line on the left. respectively. and negative for the line on the right. (3) A pair of corresponding angles of the line L are equal. then p1 = q1 and the fact that P .If a line L meets two other lines are equivalent: (1)
1 2. Because we have excluded vertical lines. Consider the quotient q2 − p 2 p 2 − q2 = p 1 − q1 q1 − p 1 Observe that the denominator is never 0 because if q1 − p1 = 0. Q be (p1 . once a pair of coordinate axes have been set up.
then the following three conditions
(2) A pair of alternate interior angles of the line L are equal. Thus the denominator of this quotient is never zero. which is slanted this way \.
. and |P R| p 2 − q2 =− p 1 − q1 |QR|
p2 for the line on the right. p2 −q2 We claim that this quotient p1 −q1 in terms of the coordinates of P and Q does not depend on the position of the points P and Q on the line L.Y
Pr Q r R
L L
e
Y
e e eP r e e e R Q er e e e e
O
X
O
X
If we denote the length of the segment P R as usual by |P R|. and let right triangle P Q R be formed so that the lines P R and Q R are parallel to the coordinate axes. positive for lines slanting like / and negative for lines slanting like \. the proof for the line on the right is entirely similar.
Y
Pr Q r Pr Q r R R
L
O
X
40
. p2 ) and (q1 . q2 )s. Let us prove this claim for the line on the left. etc. Let points P and Q be chosen on the line L as shown and let their coordinates be (p1 . then p 2 − q2 |P R| = p 1 − q1 |QR| for the line on the left. Thus the quotient p1 −q2 has been directly expressed as a −q1 positive or negative quotient of the lengths of the legs of a right triangle.
and the slope of a line that slants this way \ is negative. p 1 − q1 then we can be confident that the resulting number will always be the same. The important conclusion to draw from the preceding discussion is this: given any straight line which is not vertical. we only have to show that two pairs of corresponding angles are equal. there is already one pair of equal angles.
. We call this number the slope of the line L. We get a second pair by observing that the line L meets both line P R and line P R and P R P R because they are by definition parallel to the y-axis. q2 ). The desired equality of the quotients is now completely proved. and also P higher than Q to make the argument more intuitive. p2 ) and (q1 . and the AA criterion implies the similarity of P QR and P R Q . respectively. It should be remarked that in the picture. and form the quotient p 2 − q2 . this number does not depend on the choice of the P and Q on L and is therefore associated with the line L itself. P QR and P R Q being right triangles. we have made P higher than Q. A little experimentation 41 P QR and P QR. Thus the corresponding angles ∠QP R and ∠Q P R are equal. for a given line L. Why this is so is left as a classroom activity.Our goal is to show that p 2 − q2 p 2 − q2 = p 1 − q1 p 1 − q1 By rewriting these quotients in terms of the lengths of the legs of this equality becomes |P R| |P R | = |Q R | |QR| This is the same (using the cross-multiplication algorithm) as |P R | |Q R | = |P R| |QR| Now if we can show that the triangles P QR and P R Q are similar. then this equality would be valid because it merely states that the ratios of corresponding sides of these similar triangles are equal. But the validity of the argument by no means depends on having P higher than Q . the slope of a line that slants this way / is positive. In other words. choose any two points P and Q on the line with coordinates (p1 . By a previous observation. So why are these triangles similar? By the AA criterion for similarity.
we get 1 4−0 = − 2 . and if it is large be it positive or negative (in the sense of being far away from 0 in either direction on the number line). 42
. are they identical? If two straight lines have the same slope and pass through the same point. A priori. −3) and (−4. 0) and (0. Yet for a line. This would render any discussion of "the slope" of the line nonsensical. If P is a point on L 2 with x-coordinate 3 . Activity Suppose a line L passes through (2. let their slopes be equal. what is the y coordinate of P ? Textbooks usually define the slope of a line by picking two points on the line and then declaring the quotient formed from the coordinates of these two points in the manner shown above to be the slope of the line. 0) and (3. then the line would be close to being vertical. We have answered this question by the use of similar triangles. 5) on C 3−(−5) leads to the quotient of 5−0 = −1. they are distinct lines and will therefore be schematically represented as such. the quotients formed from different 0−5 pairs of points on it are therefore not always the same. Let the lines be L and L . On the other hand. then they are the same line. suppose instead of a straight line we have the circle of radius 5. the quotient resulting from a different choice of points on the line could be a different number so that a line could have many slopes. A priori. we only take up the case of positive slope. taking another pair of points (5. The case of negative slope can be handled the same way. because the exploitation of the fact that one can compute the slope of a line by using any two points to one's liking is a powerful tool in dealing with all kinds of questions related to linear equations. In the following. Be sure your students know the answer too.would reveal that if the slope of a line is close to 0 (be it positive or negative). If we take the two points (−5. then the line is close to being horizontal. and form the usual quotient. For example. For the curve C. denoted by C. and let them both pass through the point P . Our task is to show that they are in fact the same. We next consider another seemingly obvious question: if two lines have the same slope and pass through the same point. The question is Why?. The following discussion will amply bear out this assertion. 1). 4) on C. these quotients are always be the same.
and form right triangle P Q R so that the line P R is parallel to the x-axis. and the line Q R is parallel to the y-axis. Recall that we are trying to show the lines L and L coincide. We are now in a position to resume our task of showing that the first part of the Theorem is true.e. (β) Every point on the line L is a point on the graph G. We shall make use of the fact that the slope of a line can be computed using any two points on the line. i. 2 As previously suggested. the graph of a linear equation y = mx + k is a straight line. 0) and (0.e. and 3 2 denote the graph of y = 3 x + 2 by G. i.. 2) by L. we first look at the special case of m = 3 and k = 2. Recall that we will prove that G coincides with L by proving: (α) Every point on the graph G is a point on the line L. 43
. the equation y = 2 x + 2.. |Q R | = |QR |. and this would be true as soon as we can show that Q and Q coincide because there is only one line passing through two given points. the numerators must be equal as well. Denote as before the line joining (−3.Y L
4 Qr 4 L 4 r 4 4 4 Q 4 4 4 4 4 4 4 4 P 4 r 4 R 4 4 4 4 4
O
X
Take an arbitrary point Q on L . as claimed. Let the line Q R intersect L at a point Q. Therefore Q = Q and L and L coincide. i.e.. The equality of the slopes of L and L is now expressed as: |Q R | |QR | = |P R | |P R | Since the denominators of these two quotients are equal.
3 Since they both pass through P . because if the graph G is a curve. P ) and (−3. Let the coordinates of Q be (x0 . from (0. But we know more: since Q is on the graph G of y = 2 x + 2. 2) 2 0−2 (i. 2) and Q . then this arbitrary point Q on the graph G would belong to L. x0 and y0 satisfy x0 −0 3 2 y0 = 3 x0 + 2.. L and L would have the same slope. What we are going to show is that. and let L be the line joining P to Q .e. If we can show that L and L coincide. Take an arbitrary point Q on the graph G distinct from P . by definition. because G is the graph of a linear equation y = 2 x + 2. It follows that the slope of L is
2 ( 2 x0 + 2) − 2 x0 y0 − 2 2 = 3 = 3 = x0 − 0 x0 x0 3
44
. L and L would be distinct. So why do L and L coincide? There is no a priori reason that they do. 0): it is −3−0 = 3 . and therefore step (α) would be proved. then the picture could look like this: L
r 4 5 r 4 5 4 Q 4 5 5 44 G 5 4 5 4 5 4 544 54 54 4 5 4 r 4 P5 4 5 4 45 5 O 5
L
4
In this case. y0 ). To explain the latter. What about the slope of L ? We compute it using the points (0. they coincide as we have just observed. we refer to the following picture: Y L
¨ ¨ L ¨ r¨ ¨ ¨ Q ¨¨ P r ¨¨ ¨ 2 ¨¨ ¨ r −3 O
X
The slope of L can be computed from any pair of points on L. 2) on the y-axis be denoted by P . Then the slope of L is y0 −2 .Let the point (0. in particular.
(We can cancel x0 because Q being distinct from (0. This means. Y L
Q r
= (x0 . k) is on the graph of y = mx + k !).) 2 So both lines have the same slope 3 and pass through the same point P . if the 2 coordinates of Q are (x0 .e. 2)) and (−3. we get y0 −2 = 3 .
45
. We will use the same method to show that L and G are the same. This implies y0 − 2 = 3 x0 .. then Q will be a point on L and step (α) would be proved. For simplicity. This completes the proof of step (β). and let L be the line joining P and Q. We begin with step (α). then we must show y0 = 3 x0 + 2. If we can prove that L and L coincide. Take a random point Q on the graph G distinct from P . y0 ). we go through the same two steps: (α) Every point on the graph G is a point on the line L. They must coincide and we have proved step (α). k) on the y-axis (check that (0. 0). we simply take P to be the point (0. so that x0 x0 2 y0 = 2 x0 + 2. 2) and (−3. obtaining: y0 − 2 y0 − 2 = x0 − 0 x0
2 Since y0 −2 is the slope of L. 2) (= P ) implies that x0 = 0. and therewith also the 3 proof that the graph of y = 2 x + 2 is the line L joining (0. To prove step (β) for the equation y = 2 x + 2. But 3 we may also compute the slope of L using instead the points Q and P . 0). i. as desired. then Q also lies on G. y0 )
P r 2
r −3
O
X
Now we already know from the preceding paragraph that the slope of L is 2 . and let L be the line joining P to Q . Let any two points P and Q be chosen on the graph G of y = mx + k. we must show that if a point Q lies 3 on the line L joining P (= (0. 3
The proof for the graph of the general equation y = mx + k is no different. (β) Every point on the line L is a point on the graph G.
(Caution: it is tempting to assert instead that "the slope of the graph of y = mx + k is m". But (p1 . the slope is q2 − p 2 (mq1 + k) − (mp1 + k) m(q1 − p1 ) = = =m q1 − p 1 q1 − p 1 ) q1 − p 1 This proves our claim. q2 ) being points on the graph of y = mx + k. The slope of the line joining them is then (q2 − p2 )/(q1 − p1 ). p2 ) and (q1 . Hence q1 − p1 is never 0. it follows that L = L . q2 ). p2 ) and (q1 . if it were. we will prove the coincidience L and L by showing 3 that they have the same slope.) Indeed let the two points on the graph of y = mx + k be (p1 . p2 ) and (q1 . we would have p1 = q1 . as desired. the coordinates of these points satisfy. but at this particular juncture. It immediately follows from this claim that since L and L are straight lines joining distinct points of the graph G of y = mx + k. The reason for the latter is that. The proof of step (α) is complete. the slope of the line joining them is always equal to m. We could compute the two slopes as in the special case of y = 2 x + 2 to 3 arrive at our conclusion. q2 ) would not be distinct.Y L
Q r ¨ ¨ L ¨ r¨ ¨ ¨ Q ¨ ¨ P¨ ¨r ¨ ¨ ¨ ¨ O
X
As in the case of y = 2 x + 2. Since they also pass through the same point P .8 Being on the graph. We claim: For any two distinct points on the graph of a linear equation y = mx + k.
Note that this quotient always makes sense because q1 − p1 is never 0. the equations p2 = mp1 + k Therefore. by definition.
8
and
q2 = mq1 + k
46
. So L = L . but the virtue of having worked through a special case is that we can now afford to think more abstractly. their slope is m. we have p2 = mp1 + k and q2 = mq1 + k. Thus also p2 = q2 and the two points (p1 . we do not as yet know that the graph of y = mx + k is a line. so we cannot talk about the slope of the graph of y = mx + k.
Thus R lies on G. The reason is very simple now. Thus r2 − p 2 =m r1 − p 1 This implies r2 − p2 = m(r1 − p1 ). But P being on G means p2 = mp1 + k. but not equal to (−3. y) on L. the line passes through both of these points (−3. So we obtain r2 − (mp1 + k) = m(r1 − p1 ). r2 ). 2)) is 3 . we can compute the y−0 2 slope of L by x−(−3) = 3 . Thus both L and 47
. We have just seen that the slope of L is m. Take any point R on L distinct from P and let the coordinates of P and R be (p1 . 0) and (0. the line L that joins the points (−3. This leads us to consider the linear equation in two variables y= 2 (x − (−3)) 3
Observe first of all that this is a linear equation in two variables because. and we have completely proved the first part of the Theorem. which is the same as r2 = mr1 + k. 2) are easily seen to be solutions of y = 3 (x − (−3)). 2). after an application of the distributive law and after transposing. respectively. 0) and (0. 0). for any point (x. What equation is it the 2 graph of? Recall that the slope of L (obtain by computing with (−3. Y L
y) (x. p2 ) and (r1 . By the first part of the Theorem. Since both (−3. step (β): why every point of L lies on the graph G. Let us begin as usual with a special case: we look at our standby. r2 ) is a point on the graph G of y = mx + k. we can rewrite it as 2 x − y = 3 2 −2. Let be the graph of y = 3 (x − (−3)). But the last equality is exactly the statement that the point (r1 . 0) and (0. Thus if we compute the slope of L using P and R. or r2 − k = mr1 . We proceed to finish the proof of the Theorem by showing that every straight line is the graph of a linear equation. 2). is 2 a line. it is still equal to m. r 2 r −3 r
O
X
Therefore.Now. 0) and (0.
Given a (non-vertical) straight line L. We proceed to justify this statement by extracting four useful facts from the proof. let us assume m is negative so that we have the following picture: Y L
rk
O
X
Let the graph of y = mx + k be . The idea of the proof is essentially the same as in the special case. We are going to show that L is the graph G of the equation y = mx + k. k) is obviosuly a solution of y = mx + k. 3 It remains to tackle the general case. 0) and (0. k). It follows that L is the graph of y = 2 (x − (−3)). we must find a linear equation whose graph is exactly L. they must meet at some point.are lines which pass through the same two points (−3. they have to be the same line. Since (0. and L are the same line. but every piece of reasoning in the proof will turn out to be important in subsequent considerations of linear equations and straight lines. Let the slope of L be m. The preceding proof of the Theorem may seem long. Therefore and L are two lines which have the same slope m and pass through the same point (0. Now that we know the graph of a linear equation y = mx + k is a non-vertical straight line. q2 ) of the equation. p2 ) 48
. k). Let L intersect the y-axis at (0. the slope of is m. Recall that we are only dealing with equations of the form y = mx + k and lines which are not vertical. By an earlier observation. k). For the sake of variety. Since L and the y-axis are not parallel. By an observation that was made in the proof of the first part of the Theorem. This completes the proof of the Theorem. is a line. we see from the proof of step (α) that the following holds: (i) The graph of y = mx + k is obtained as follows: take any two solutions (p1 . It follows that the given line L is the graph of y = mx + k. p2 ) and (q1 . By the first part of the Theorem. also passes through the point (0. then the line joining the points (p1 . 2).
we can retrieve the equation ax + by = c of from the 1 equation a x + b y = c by multiplying both sides by λ .
(iii) The lines defined by the two equations ax + by = c and a x + b y = c . the lines defined by ax + by = c and a x + b y = c have to be the same. First of all. we get that µ = c and the slope of is − a . suppose the two lines defined by ax + by = c and a x + b y = c coincide. then it also has 0 x-intercept. Hence we get c = c b b b b a a and b = b . then the number k is called the y-intercept of the line . are the same if and only if there is a nonzero number λ so that a = λa. µ) is on .and (q1 . if intersects the x-axis at (c. k). and c = λc. and c = λc. Now the fact in question is: (ii) The equation of a non-vertical line slope of and k is y-intercept of . the slope of the graph is m and the graph intersects the y-axis at the point (0. Let the y-intercept of be µ. b = λb. Since the line defined by a linear equation is just the set of all its solutions. the slope of is − a . we introduce two standard concepts. where m is the
The third fact is a direct consequence of the first two. q2 ) is the graph. is y = mx + k. A second useful fact is taken from the proof of the second part of the Theorem. b = λb. Conversely. Let us call it . Similarly. Furthermore. and is usually glossed over in the standard texts at all levels. For its statement. k). If a line intersects the y-axis at (0. For this reason. b b b Furthermore.
In the situation of (iii). Then since (0. Reasoning likewise with the equation b a x + b y = c . if there is a nonzero number λ so that a = λa. by (i) above. then c is called its x-intercept. one normally 49
. µ = c . 0). Rewrite ax + by = c as y = − a x + c .
We prove (iii) as follows. Observe that if a non-vertical line has 0 y-intercept. This is the same as saying b c = b c and b a = b a
a a
A straightforward computation then shows that letting λ =
would get the job done. then the solutions of ax + by = c and a x + b y = c are clearly identical because a x + b y = c may be rewritten as λ(ax + by) = λc.
It is not necessary to directly compute the y-intercept. if we re-express this equation as y −(−1) = 3 (x−2). The point of this comment is there3 fore that even correct thinking needs to be complemented by correct technical execution. Therefore the graph of y−(−1) = 2 conx−2 3 tains every point of the line passing through the point (2.regards any two equations defining a line as "the same". we should add that the equation y−(−1) = 2 contains the correct geometric conception of the desired x−2 3 line. so has x-intercept a . Example 1 What is the equation of the line passing through the point (2. then certainly this is an equation whose graph is the desired line. 0) is a solution of y = (− b )x + b . Since the line contains 2 2 (2.
50
. 3 3 3 This is the proper place to comment on a common misconception. that the equation of this line is
2 y − (−1) = x−2 3 This is not correct because the point (2. because what it says is that this line consists of all the points (x. but we need to find out what k is. We give some examples of how to write down the equation of a line. −1) with slope 2 except the 3 2 point (2. has slope − a . we know −1 = 3 · 2 + k. Thus the equation is y = 2 x − 7 . and written up in textbooks. However. By (i). −1) would not satisfy this equation as we would have (2 − 2) in the denominator on the left side. from which k = − 7 . and speaks of the defining equation of a line. The final and fourth fact is a consequence of (i) and (iii): (iv) A line defined by ax + by = c with a = 0 and b = 0 has slope − a b c and x-intercept a . −1) (and since the line is the graph of y = 3 x + k). Sometimes it is taught in classrooms. By (iii). y) so that the slope of the line containing (x. −1) with slope 2 ? 3
2 According to (ii). It is also obvious b b c a c c that ( a . −1) is 2 . y) and (2. the equation has the form y = 3 x + k. −1) itself. That said. is also defined by y = (− a )x + c because this equation is obtained by mulb b tiplying both sides of ax + by = c with 1 .
is equal to L. Then one verifies routinely that both 2 (−1. A second remark is that. 3) and ( 1 . Thus k = 11 and the 3 11 2 equation of is y = 3 x + 3 . Now we claim that the equation of . 4) as the point of reference. since we are given two points on to begin with. and therefore y − 3 = 3 (x − (−1)) is in fact the equation of . we may rewrite it as 3 x − y = − 11 .
51
. 4) are solutions of y − 3 = 3 (x − (−1)) and therefore L passes through 2 both of the points (−1. obtaining 2 . 3) to determine k. 3) lies on . is y−3= 2 (x − (−1)) 3
It is to be observed that this is indeed a linear equation in two variables because. 4) instead of the point (−1. where the constant k is determined by 3 2 observing that since (−1. after multiplying both sides by (x − 1 ) and simplifying. we would get the 2 1 same equation (of course). 4) lies on the graph of y = 2 x + k. The first is that. If we use the point ( 1 . 4). The preceding solution needs to be complemented by two remarks. 3 2 let L be the line defined by y − 3 = 3 (x − (−1)). The slope of
is
1 2
4−3 2 = 3 − (−1)
so the equation of has the form y = 2 x + k. Indeed. we can use a slightly different method to obtain the equation of if we follow the reasoning in the proof of the second part of the Theorem. that this 2 2 11 equation is again y = 3 x + 3 .1 Example 2 What is the equation of the line passing through (−1. 3) and ( 1 . Since there is only one line passing through two 2 2 given points. so that 3 1 11 k =4− = 3 3 as before. 3) and ( 2 . This said. 4) ?
Call this line
. we have 3 1 4 = 2 · 2 + k. if we use 3 (−1. we have 3 = 3 (−1) + k. applying 2 the distributive law and transposing. 3) as the point of reference. We begin by computing the slope of the line as before. if we use the point ( 1 . then we would write the equation 2 of as 1 2 y − 4 = (x − ) 3 2 It is simple to see. since ( 2 .
m =
p2 −q2 . 1). B) = (A . p1 −q1
Equivalently. which is the same as 5 c = 3 . we get c = 16 . Let D be a dilation of the coordinate plane with center at the origin O and with scale factor r (r > 0). 6) and (2. One additional comment is that.In general. p2 ) and (q1 . i. B ). Example 3 What is the x-intercept of the line joining the points (−4. the equation is
y − q2 = m(x − q1 ). 5 5
We conclude this section with a comment on the concept of dilation in the plane..e.
52
. we have 1 = − 6 · 2 + k. cy). the equation of the line joining two given points (p1 . q2 ) is y − p2 = m(x − p1 ). y) = (cx. where m is the slope of the line. If (A. It must be emphasize that none of these equations should be memorized by brute force beyond the fact that the equation of a non-vertical line is of the form y = mx + k for some constants m and k. let it be (c. the first solution using y = mx + k to get the equation of the line is the more basic of the two methods. Multiplying through 6 6 by 5 . where m is the slope of the line. one should get to know the reasoning underlying these procedures and do a simple computation each time to get at the equation. The reasoning is the same as before. y) by a number c as c(x. and so 5 8 k = 1 + 3 = 8 . Since (c. So the x-intercept is 16 . for 6 6 5 some constant k. B ). Instead. 0) lies on 8 8 5 the line. B) to another point (A . Since it contains the point (2. if two points are given. we want to show that A = rA and B = rB
If we define the multiplication of a point (x. 0). The point where this 3 6 line intersects the x-axis has y-coordinate equal to 0. The equation of the line is therefore y = − 5 x + 3 . so the equation of the line is y = − 5 x + k. B) is a point and if D maps (A. we also get 0 = − 6 c + 3 . if D(A. 1) ?
6−1 The slope of the line is −4−2 = − 5 .
as shown. |OQ| = r |OQ |. In case one or both of A and B is negative. First recall the definition of P : on the ray L from O to P . B ) by P and P . we get B = rB in a similar manner. But using the sides P Q and P Q in place of OQ and OQ . Therefore |OQ| |OP | = |OQ | |OP | But we have seen that
|OP | |OP |
= r. |OQ |
or what is the same things. |OP | = r |OP | For simplicity. so we get |OQ| = r. 0) and those of Q are (A . B) and (A . consider the case A < 0 for definiteness. first assume both A and B are positive. Because P Q P Q . This shows (A . B). therefore A = rA. But |OQ| = A and |OQ | = A . the theorem on corresponding angles of parallel lines implies that the triangles OP Q and OP Q are similar because the corresponding angles are equal. Denote the points (A. The reason is as follows. Then the preceding picture beomes 53
. rB) = r (A.then we can rewrite the preceding result as D(A. as claimed. respectively. L
4 4 4
P 4 4
4
4
P 4 4
4 4 4 4 4 4
4
O
Q
Q
Notice that the coordinates of Q are (A. P is the point so that the distance |OP | from O to P is r times the distance |OP | from O to P .e. B ) = (rA. i.. B) for a dilation D with scale factor r and centered at O. Then we drop perpendiculars P Q and P Q from P and P to the x-axis. 0). B) = r (A.
For which value of m would L pass through (20. What is 2 its y-intercept? (ii) Find the equation of the line passing through (− 5 . − 3 ) and (− 3 . the only difference is that |OQ| = −A and |OQ | = −A .(i) Find the equation of the line passing through (1. Write down three linear equations of two variables whose graphs all have slope 2 . 3) with slope 1 . Write down three linear equations of two variables whose graphs all pass through (−2. −2) with slope m. 2) to (p. 3 )? 4 3 9. we conclude −A = r(−A ). (i) Let L be the line joining (1. Do the graphs of 6x − 2y = 7 and 1 x − 15 y = 329 intersect? Explain why or 5 why not using what we have done so far. where p is some number. 2 ) and (5. (i) What is the y-intercept of the graph of x = −5y + 7? What is its slope? Does the point (352 2 . For what 3 value of p would L pass through (10. (i) What is the equation of the line with x-intercept equal to −2 and slope − 3 ? What is its y-intercept? (ii) What is the equation of the line with x-intercept equal to 2 and y-intercept equal to − 4 ? 5 3 5. 7 What is its x-intercept? 6. 3. −1) and (−3. 3 2. Then we get A = rA as before. 25)? (ii) Let be the line joining (− 2 .L ˜
˜ ˜ ˜P ˜ ˜ ˜ ˜P ˜ ˜ ˜ ˜
Q
Q
O
In this case. −11). q). (i) Let L be the line passing through (1. 3 ). −4). 4) and ( 4 . 3 ). 1 )? What is its slope? 2 1 7. and compute the y-intercept in each case. 5 54
. What is its 3 2 y-intercept? 1 4. What is its x1 intercept? (ii) Find the equation of the line joining (− 4 . (i) Find the equation of the line joining (2. 72)? (ii) Let be the line with slope m passing through 1 1 ( 2 . so that from |OQ| = r |OQ |. EXERCISES 1. 1). −1) with slope −1. The proof is complete. −70 1 ) lie on the graph? (ii) What is the x-intercept of the line passing 5 5 2 4 through (5. 8. For which value of m would pass through ( 5 .
and do it also without making that assumption.where q is some number. How many quarters are there? We follow the practice of §2 and simply transcribe the information faithfully into symbolic language before doing anything. So if there are Q quarters. and make it as simple as possible. There are 17 quarter. Does the line joining (3. therefore 24Q = 408. of which Q · 25 cents come from the quarters and 39 − Q cents coming from the pennies. and therefore Q = 17. Practice explaining to an eighth grader why the graph of y = x − 1 is a line. 2) contain the point (9. then there are 39 − Q pennies. Obviously the explanation would vary depending on how much s/he knows about similar triangles. we have 447 cents. One should always check: 17 quarters lead to 17 · 25 = 425 cents. For what value of q would pass through (2. be clear about what you assume the student knows. −1) is the graph of a linear equation of two variables. and they are worth $4. Transposing. Do it once assuming that the student knows the graph of a linear equation of two variables is a line.47. we have 25Q − Q = 447 − 39. In terms of cents.
55
. be clear about what you assume the student knows. Example 1 There are 39 coins made up of quarters and pennies. and make it as simple as possible. 12. −3)? 10. In any case. −2) and (6. so the technique of §3 allows us to solve this easily. 6)? Explain it two different ways. In any case. Obviously. 25Q + (39 − Q) = 447 This is a linear equation in one variable. 11.
5
Some Word Problems
Here are some examples of word problems involving the solution of linear equations in one variable. Added to 39 − 17 = 22 cents (from the pennies) does give 447 cents. Practice explaining to an eighth grader why the line joining the origin and the point (−1. or whether s/he knows that isosceles right triangles have only 45 degree angles.
So (x + 2)(x + 6) − x(x + 4) = 64 The solution of this equation. No reason to do more than you have to! Thus the next three integers are x + 2. which is central to the solution of this class of problems. such as the number of square feet a lawn is mowed. which is not linear. Thus the equation becomes 4x + 12 = 64. is hardly ever clearly defined and. 19. This is a common enough expression. and 96 = 2 · 144 . it is an assumption that is often omitted in the formulation of such problems. Thus the four integers are 13. but what does it mean? It means at least that. which is a linear equation in one variable after all. x + 4. the distance from t = 1 hour to t = 2 hours is another 3 2 miles. we just transcribe the given information and wait to see what happens. From 4x = 52. Unfortunately. We get x2 + 8x + 12 − x2 − 4x = 4x + 12. etc. and 2 s = 48. 17. We will concentrate on speed . then the greater part is 48−s. It is given that s = 2 (48−s). 3 5 96 1 3 Thus 2 s = 48 − s. in t 2 56
. 5 5 5 3 5 The next few problems are about so-called constant rates: the constant rate of walking (which we call constant speed). the constant rate of water pouring into a tub. Our discussion must therefore begin with precise definitions of these concepts. the constant rate of work. and she 1 walks at a constant rate of 3 2 miles per hour. Therefore.
Let s be the smaller part. etc. and x + 6. 15. begins with a simplification of the left side by the use of the distributive law.Example 2 Find four consecutive odd integers so that the product of the second and fourth integers exceeds the product of the first and third integers by 64. we obtain x = 13. the distance she walks from t = 0 hour to 1 1 t = 1 hour is 3 2 miles. We check that the greater part is 48 − 96 = 144 . it is not only that the concept of rate is mangled in the standard materials. it will be seen that the extrapolation of the speed discussion to other kinds of rates is not difficult. The given data is that (x + 2)(x + 6) is bigger than x(x + 4) by 64. worse. Example 3 Break 48 into two parts so that the smaller part is
2 3
of the greater part. but the concept of constancy. We check that (15 × 19) − (13 × 17) = 285 − 221 = 64. the distance from t = 2 hour to t = 3 hours is yet another 3 1 miles. Let the smallest of the four odd integers be x. Suppose Lisa takes a walk. It follows that s = 5 = 19 5 . At the moment we do not worry about whether x is even or odd.
. An important point here is that v is a fixed number. a constant.e. vs) and 57
. i. At least two comments have to be made about the defining equation of y = vt of constant speed. in 4 4 hours. we say a person walks at a constant speed of v miles per hour (mph). 3 2 n 1 m where m. Please take note that this is a definition. The fact that the graph is a straight line with slope v (see preceding section) gives an interpretation of the speed v (in the case of constant speed) in terms of the slope of the graph of y = vt. In like manner. then y = 3 1 t miles. for a number v. 1 We now formalize the intuitive expectation by defining a constant speed of 3 2 miles an hour to mean: starting with time t = 0. for a whole number t. where t is a whole number. and you should also teach your students this piece of vital information. Such is our intuitive understanding of the concept of constant speed. for any positive number t.
Y
y = vt
O
T
Observe that we do not extend the graph to the left of y-axis. " 3 of 3 1 miles" is 2 expressed in precise mathematical terms as 4 · 3 1 miles. its graph is a straight line passing through the origin of the T Y -plane. 2 i.e. Moroever. we expect Lisa to walk n · 3 2 miles. because the slope of this line can be computed by using any two points on the line means that if we take any two points (s. 2 2 So at least for a whole number t. repeated addition of the number 3 1 t times is just t · 3 1 . First. n are whole numbers. Lisa should walk 3 of 3 1 miles if her speed is a constant 3 1 miles per 3 2 2 4 hour. Intuitively..1 hours. we are giving meaning to the concept of constant speed. 2 4 What happens if t is not a whole number? Let us say t = 3 hours. if y is the number of miles Lisa walks in t hours. if the number of miles y that she walks after t hours is given by y = vt miles for any positive number t. the number of miles y traveled in t hours is y = 3 1 t miles. in m hours. consider y = vt as a linear equation in the two numbers t and y. it is clear that she walks t · 3 2 miles if we use the fact that. because we start with t = 0. More generally. From what we know about the multiplication of fractions.
In the usual terminology of school texts. etc. This leads to the usual formulas: "speed is distance divided by time". where T is any number. where s < t. Hence this equality. which is a geometric statement about the slope of a straight line. Naturally. vt) on the graph. Our explanation of these formulas would not have been possible if we had not given a clear definition of "constant speed". is always the speed v. and vs is the distance traveled from time 0 to time s. now translates into the well-known relationship between the distance traveled and the time duration of the travel when the speed is constant: the distance traveled in the time interval from s to t divided by the length of the time interval t − s for any numbers s and t. Obviously t − s is the length of the time interval from time s to time t. "distance is speed multiplied by time". we have speed = distance traveled in a time interval of length T . vs)
O
T
But vt is the distance traveled from time 0 to time t. Therefore the distance traveled from time s to time t is the numerator in the preceding equality. the truth of vt−vs = v also follows from a simple algebraic fact: t−s vt − vs v(t − s) = =v t−s (t − s) 58
. then vt − vs =v t−s Y y = vt
r (t. The difference between the standard treatment of these formulas and what we have done is that we laid bare the fact that the truth of these formulas depends on the constancy of the speed.(t. T
which is valid for any time interval of length T . vt) r (s. whereas the standard treatment simply dictates a formula without any explanation. vt − vs.
In mathematics. at the same time that B starts from Q to P . It is tempting to say that if this is so simple to state. and this would be a cumbersome concept dependent on both s and t. thereby simplifying life for us. For example. 150 mile apart. because the real effort should be spent on understanding the equation y = vt instead. then the 2 2 distances they have covered. rate has something to do with amount-of-changedivided-by-time. rate is just a division of one quantity by (usually) time length. the constancy of the speed makes it unnecessary to make any reference to s and t. In 1 2 hours. A starts from P to Q. or more simply. the "amount of change of the position of the moving object over the time interval from s to t divided by the length of the interval t − s changes as either s or t changes. If A moves 10 mile per hour faster than B. A second comment is about the terminology of rate. but understand that on the intuitive level. Example 4 Two cars A and B move at constant speed. combined. whatever is not precisely explained should be ignored. Very roughly. A 1 covers 1 2 v miles while B covers 1 1 (v − 10) miles. They meet at the end of 1 1 2 hours. If they meet after 1 1 hours. and it is impossible to give a precise meaning to it without the use of calculus. would be the same as the distance between P
59
. t−s In general. Speed is therefore the rate of change of the position of the moving object. So just ignore this word. so that all we get would be just what is usually called "the average rate of change from s to t". it means "the amount of change (whatever it is) over a time interval divided by the length of the time interval". what is the subtlety? It is this: "speed" as a rate of change in the above naive sense is meaningful only because we luck out when the speed is constant: the quotient vt−vs is always the same so that s and t become irrelevant. the quotient above vt−vs is t−s an example of the "amount of change of the position of the moving object (or person) over the time interval from s to t divided by the length of the interval t − s". The precise concept is rate of change. Then the speed of B is v − 10 mph. what are their speeds?
EA
B'
P
Q
1 Let the speed of A be v mph.But it is important to also understand it in terms of the geometry of the graph of y = vt. The above complicated explanation should make it very clear that the term "rate" as used in school mathematics is in fact never explained precisely. Again. Such an abbreviated understanding is good enough.
Therefore 1 1 1 v + 1 (v − 10) = 150 2 2 There are many ways to solve this equation. The number r is the socalled rate of the water flow and it is in terms of gallone per minutes. the concept of a constant-rate water flow (from a faucet) is that.). Suppose the water flow from faucet A is 10 gallons per minute more than that from faucet B. etc. what are the rates of the water flows in both faucets? In accordance with the preceding discussion of constant speed. See the preceding interpretation of speed as a division of distance by time. lawns are mowed at a constant rate. houses are painted at a constant rate. in the sense that water flows out of the faucet at a constant rate. this is the only kind of problems we can handle. one of them is to first multiply both sides 3 by 2 (noting that 1 1 = 2 ) to get v + (v − 10) = 100. which is 150 miles. 2 the total amount of water from both faucets is equal to 150 gallons. for some fixed constant r. mowing a lawn. Thus 2v = 110. Now let the rate from faucet A be x gallons per minute. etc. 2 2 2 The discussion about speed can be carried over. Since after 1 1 minutes. 1 After 1 1 minutes. the container is filled. verbatim. Therefore the speed of A is 55 mph. Example 5 Water flows out of two faucets A and B at constant rate. to the discussion of the rate of water flow or the rate of getting work done (painting a house. the amount of water coming out of the faucet after t minutes is rt gallons. Therefore 1 1 1 x + 1 (x − 10) = 150 2 2 This is the same equation as before. the amount of water coming out of faucet A is 1 2 x gallons.and Q.
60
. the following are some related examples. without the tool of calculus at our disposal. and 2 1 that of faucet B is 1 2 (x − 10) gallons. The the rate from faucet B is x − 10 gallons per minute. We check: 3 3 · 55 + 2 · 45 = 165 + 135 = 150. and the speed of B is 45 mph. and 3 2 v = 55. If both faucets are turn on at the 1 same time and the container is filled in 1 2 minutes. In school mathematics. and the answer is: the rate of faucet A is 55 gallons of water per minute and that of faucet B is 45 gallons of water per minute. With this in mind. and suppose a container has a capacity of 150 gallons. keeping in mind that we always idealize that these processes are evolving at a constant rate.
Then in x hours. Since the area of the wall is 150 square meters. The total distance must be equal to the 2 3 distance between P and Q. Example 7 Two trains A and B run at constant speed. If A leaves P for Q at the same time that B leaves Q for P (on a separate but identical rail!). Let the two trains meet after x hours. after how many hours will they meet? This example is very similar to Example 4. So Karen paints 55 square meters per hour and Lisa 45 square meters per hour. Suppose Karen paints 10 square meters more per hour than Lisa. and suppose a wall has an area of 150 sqare meters.Example 6 Karen and Lisa paint houses at a constant rate. we need the speed of both A and B to compute the distances they have traveled before they meet. So D D x+ x=D 2 3
61
. After 1 1 hours. because "speed is distance 2 3 divided by time". which is D. although for this problem we do not know the distance between P and Q. By the definition of constant speed. Karen has painted 1 2 x square meters and Lisa. the speed of A is D mph and that of B is D mph. A goes from P to Q in 2 hours while B goes from P to Q in 3 hours. we have 1 1 1 x + 1 (x − 10) = 150 2 2
Same equation. Just call it D in order to finish the transcription of the given data into symbolic language. 1 2 (x − 10) 2 1 1 1 square meters. So in 1 2 hours they have painted a combined area of 1 2 x + 1 2 (x − 10) square meters. what are the rates at which each paints? 2 Let Karen paint x square meters per hour.
EA B'
P
D
Q
As we saw in the solution of Example 4. Then Lisa paints x − 10 square meters 1 1 per hour. the distance traveled by A is D x and that by B is D x. If both Karen and Lisa paint this wall at the same time and they finish it in 1 1 hours.
e. we do know it takes her 24 minutes to get to the station. suppose Genevi`ve catches up with Paul t minutes after her departure e from the house. Paul walks from their e house to the train station in 30 minutes while Genevi`ve needs only 24 minutes to do e the same. suppose Genevi`ve catches up with e Paul t minutes after her departure from the house. we'd be left with 1 1 x+ x=1 2 3 1 6 1 Thus ( 2 + 3 )x = 1 . Genevi`ve gives Paul a head start of 4 minutes and then she starts off. Does e she catch up with Paul. First. and x = 5 hours.. Again. So in t minutes she has walked 24 t miles. and by the time Genevi`ve has walked t minutes. namely. Therefore Genevi`ve overtakes him at e some point on her way to the station. i. We can also do the same problem from the other end. and therefore e 62
. Example 8 Paul and Genevi`ve walk at a constant rate.This seems to be an equation in both D and x until we realize that if we multiply both 1 sides by D . But 28 minutes after Paul took after. then we get immediately her speed: D D miles per minute (mpm). So 24 t = 30 (t + 4) and therefore 30 t = t + 4. We can do this problem at least two ways. it takes only 4+24 = 28 e minutes after Paul took off before she gets to the station. D D t = (t + 4) 24 30 This again looks like an equation in the two numbers D and t. he will have walked 4 + t minutes when Genevi`ve catches up with him. he would have walked t + 4 e 30 D minutes. It is always safe to try doing the most obvious thing. How far has she walked? We have only the time (t minutes) but not her speed. if we let the distance from the house to the station be D miles. Paul's speed is 24 D mpm. after how many minutes? Since it takes Genevi`ve only 24 minutes to get to the station. the station instead of the house. she catches up with Paul. he would have walked 30 (t+4) e miles. and if so. Since the two will have walked the same distance at that point. Therefore when Genevi`ve catches up with him. So 16 minutes after Genevi`ve leaves e 24 4 the house. Therefore D the distance from where she catches with Paul to the station is 24 (24−t) miles. This leads to 1 t = 4 and t = 16. he is still on his way to the station. For Paul. but once more the D 1 1 1 goes away as soon as we multiply both sides by D . Then in (24 − t) more minutes she would be at the station. We have to find out exactly when.
he has 30 − (4 + t) = 26 − t minutes to go before he reaches the station. Thus D D (24 − t) = (26 − t) 24 30 1 1 As before. We now graph both linear equations in two variables. It is instructive to look at the graphs of the equations describing the motion of Paul and Genevi`ve. We will pursue this discussion vigorously in the next section. Since his speed D D is 30 mpm. 5 5 y= 1 t 4 and y= 1 4 t+ . For Paul.
63
. also has an interpretation: the x-coordinate tells the time when Genevi`ve catches up with Paul because at that e instant. the equation is y = 24 t = 4 t. the remaining distance to the station can also be computed by 30 (26 − t). 4). we need to give the distance D between the house and the station a definite value. we get 24 (24 − t) = 30 (26 − t). both are exactly the same distance (4 miles. This is the first time we discuss two graphs on the same set of axes simultaneously. which is (16. which implies (upon multiplying both sides by 24 · 30) 30(24 − t) = 24(26 − t). 4) 3¨ 33¨ 3¨¨ 33¨ 33¨¨ 33¨¨ 33¨¨ 3 ¨ ¨ ¨
1
12
16
24 26
T
4 The y-intercept of the graph of Paul (which is 5 . We will use the first set of equations for illustration. If y denotes the distance from the house t minutes after Genevi`ve e 1 6 leaves the house. then for her. as we saw above) now has a graphic interpretation: it gives his distance from the house when Genevi`ve leaves the house. the equation is 6 y = 30 (t + 4) = 1 t + 4 . e The point of intersection of the two graphs. say D = 6. Hence 6t = 24(30 − 26) and t = 16 once again. 5 5
on the same set of coordinate axes: Y 6 5 4 3 2 1 0
Genevi`ve e 3 ¨¨3 Paul 3 ¨3 ¨3 3 ¨¨ 33 ¨ 3 r 3 3¨ 3¨ 3¨ (16. For the sake of e clarity . the y-coordinate of the point) from the house.
Faucet A fills a given container in 5 minutes. and emptied by another in 25 minutes. he has 2 as much as 3 B. What was the original number? 3. 20% of the 6 remainder at the third. The tank is then found to be empty 24 minutes after the first pipe was open. the resulting fraction equals 1 . how many miles of her trip is in the country? 13. A woman drives a car for 3 1 hours and she finds that she has covered a distance 2 of 130 miles. A train running at 30 mph requires 21 minutes longer to go a certain distance than does a train running at 36 mph. 25 at the second. 9. and the second pipe opened. three quarters of the remainder at the fourth. After working for 2 1 2 64
. and Karen does it in 3 hours. Assuming that water flow is always at a constant rate. but after giving B $28. while faucet B fills it in 6 minutes. Lisa and Karen mow lawns at a constant rate. how many minutes is each pipe open? 11. After the first pipe has been open a certain number of minutes. Water flows out of two faucets A and B at constant rate. A had twice as much money as B. A tank with a capacity of 150 gallons can be filled by one pipe in 15 minutes. Find two numbers whose sum is 6 and difference 6 . 25 remain. The numerator of a fraction is 7 less than the denominator. How long would it take to fill the container if both faucets are turned on at the same time? 4. it is closed. How far can he ride in a car going at a constant speed of 25 mph if he has to ride a bicycle back at the rate of 6 mph? 2. one containing nickels and the other dimes.EXERCISES 1. There are two heaps of coins. If she drives at a constant speed of 45 mph in the country and 20 mph within city limits. and has 8 fewer coins. Paul can mow a certain lawn by all himself in 11 hours. Lisa finishes mowing a certain lawn in 4 1 hours. how much money must A give B in order that B may 4 have 5 as much as A? 10. If A has $566 and B $370. 3 What is the fraction? 5. If they mow the same lawn together. A man has six hours at his disposal. Find the number in each heap. A train loses 1 of its passengers at the first stop. The second heap is worth 20 cents more than the first. 7. 2 how long will it take them to finish it? 8. How much did each have at first? 1 7 6. If 4 is subtracted from the numerator and 1 added to the denominator. How great is the distance? 12.
his mother discovers that he forgot to take his homework. and suppose both Bill and Aaron mow the lawn at a constant rate. If Aaron mows it alone.hours. Use mental math to decides if Colin's mother can catch up with him. B. 14. B. and if Aaron and Carl mow it together. Recall that a solution to a linear equation such as 4x + 5y = −3 is a pair of numbers 3 (x0 .
6
Simultaneous Linear Equations
In Example 8 of the preceding section. Aaron and Bill together can mow a lawn in 36 minutes. In this section. If Aaron and Bill mow it together. they can finish it in A. However. suppose we also consider another equation −2x + y = 5 and ask if there could be a pair of numbers 65
. they finish it in 17 1 23 hours. which is a straight line. Aaron. if all three mow it together. and explain how you arrive at these equations. Paul is joined by Henry and the two together finish mowing the lawn in another 5 hours. how much alcohol must be added in order for the solution to still contain 70% alcohol? 17. however. A solution consisting of water and alcohol has 70% alcohol. Still with 4x + 5y = −3. respectively. − 5 ) is a solution. and Carl all mow a given lawn at a constant rate. We saw that the point of intersection of the graphs corresponds to the solution of the problem. Fifteen minutes after Colin leaves for school. C to capture the above information. compute how soon this happens after Colin leaves. Suppose the lawn is 280 square yards. it takes 58 minutes. how long would it take Bill to mow it alone? 15. Colin walks to school at a constant rate and it takes him 24 minutes to get there. Write equations in terms of A. we inspect more closely the situation of two linear equations. Assume as always that both mow the lawn at constant rate. and if she does. C hours. If 25 cc of water is added to the solution. how long would it take Henry to mow the lawn alone? Explain clearly how you get the solution. 18. 9 2 they finish mowing it in 2 3 hours. y0 ) so that 4x0 +5y0 = −3. Driving from A to B at a constant speed of 45 mph is 25 minutes faster than doing it at 39 mph. Bill. we found ourselves looking at the graphs of two linear equations in two variables. (0. She drives at a constant rate and it takes her 6 minutes to get to school. For example. How great is the distance from A to B? 16. If it is mowed individually. There are an infinite number of solutions to a linear equation of two variables and their totality constitutes the graph of the equation. they finish mowing it in 2 2 hours.
1). To be precise. As in the case of a single equation in one variable. Thus (−2. Now (−2.(x0 . Similarly. being a solution of the first equation 4x + 5y = −3. 1). y0 )'s the solution of the system. but note that there will be occasion to consider systems of equations consisting of many equations in any number of unknowns. or no x and y. To solve the system is to find all the ordered pair of numbers (x0 . 1) to the above system. y0 ) so that it is a solution of both 4x + 5y = −3 and −2x + y = 5. (−2. We say the pair of linear equations 4x + 5y = −3 −2x + y = 5 is a system of linear equations. 1). simultaneous (linear) equations in the numbers x and y. one would have to refer to such a pair of equations as a linear system of two equations in two unknowns or two variables. as shown: Y
¡−2x + y = 5 ¡ ¡ ¡ ¡ ¡ ¡ 4x + 5y = −3 ¡ ¡ ™ ™r ¡ 1 ¡™ ™ ¡ ™ ¡ −2 ™O ™ ¡
X
66
. 1) also lies on the line defined by the second equation −2x + y = 5. and we have to check whether the equality holds for all x and y. y0 ) is called a solution of the system. as it is easy to check. A priori. implicit in the writing down of such a system is the claim that the given number expressions involving x and y are equal. Indeed there is. lies on the line defined by 4x + 5y = −3. This implicit statement will be taken for granted and will not be repeated in subsequent discussions. for example. (−2. 1) is the point of intersection of the two lines defined by the equations in the linear system. some x and y. let us first give a geometric interpretation of this (−2. This means the solution (−2. Sometimes we also call the collection of all these (x0 . Such an (x0 . At present we are only concerned with systems consisting of two equations in two variables. 1). or sometimes. y0 ) which are solution of both equations. 1) is a solution of the above system. but it will turn out that the solution of the system consists only of (−2. Postponing for the moment the discussion of how to get a solution such as (−2. there may be others.
B) is not on the line defined by −2x + y = 5.
67
. Suppose (x0 . then −2A + B = 5. For the same reason. the point (x0 . This reasoning is perfectly general. We wish to interpret this solution geometrically. B) is not a solution of both equations. . 2 be the lines defined by the equations ax+by = e and cx + dy = f . how to get the solution (−2. if given. . y0 ) is a solution of ax + by = e. B) is not at the intersection of these two lines. probably because the precise definition of the graph of an equation is rarely given or. suppose (A. then it cannot be a solution of the linear system because if (for example) the point (A. and conversely. Suppose we are given a linear system ax + by = e cx + dy = f where a. y0 ) lies on 1 . a point in the plane that is not on the intersection of these lines cannot be a solution of the linear system. and (A. The reasoning given below is perfectly general. 1)? We adopt the time-honored strategy of first assuming that there is a solution to the given linear system and use this information to find out what it has to be. By definition of 1 . It is very important that you learn to make use of definitions in your teaching. B) is not a solution of the equation ax + by = e and therefore not a solution of the linear system either. Let 1 . (x0 . aA + bB = e and the ordered pair of numbers (A. In particular. if a linear system has a solution. y0 ) is a solution of the system. Thus. then let us say (A. is not put to use. Then we turn around to verify that the presumptive solution is indeed a solution of the linear system. the point of intersection of the lines defined by the equations of the linear system is the exact solution of the linear system. Therefore. Since (x0 . . Such a correspondence is usually decreed by fiat in standard texts. B) is not at the intersection of these two lines. If a point (A. respectively. y0 ) is the point of intersesction of 1 and 2 . In particular. inasmuch as two lines intersect at at most one point. We have just given the precise reasoning why there is a one-one correspondence between the solution of a linear system of two linear equations in two unknowns and the point of intersection of the two lines defined by the linear equations of the system. Therefore (x0 . a linear system of two equations in two unknowns has at most one solution. it is the point of intersection of the lines defined by the equations of the system. y0 ) lies on 2 as well. please do not forget to explain why the solution of a linear system can be obtained from the intersection of the lines. B) does not lie on 1 . Next. f are constants.Conversely. b.
But this is unsatisfactory because this simple numerical verification does not reveal why (−2. or what is the same. y) = (−2. both equations of the system in (1) are identical linear equations in x. since both number expressions −5y − 3 and 2y − 10 are equal to the number 4x. the validity of this assertion should be clear after a moment's reflection. obtaining easily y = 1. what we have shown is merely that if a solution exists for the system in (1). i. we have −5y − 3 = 2y − 10 (3)
We can solve this linear equation in y.e. as claimed. When y = 1. 1) is a solution of (1). so that a solution of one equation (in the number x) in (1) is automatically a solution of the other equation (in x). then (x. 4x + 5y = −3 −2x + y = 5 (1)
We want to shown that if there is an ordered pair of numbers (x. We should emphasize that thus far. namely. We begin by multiplying both sides of the second equation in (1) by −2 with a view to making the coefficient of x in this equation equal to the coefficient of x in the first equation. so that is that. 4.. we get −3 = −3 and 5 = 5. We get: 4x + 5y = −3 4x − 2y = −10 Now rewrite the system as 4x = −5y − 3 4x = 2y − 10 (2)
We emphasize that systems (1) and (2) are equivalent in the sense that every solution of one of them is a solution of the other. So we get y = 1 and x = −2. when y = 1. 68
. We can be more explicit in the present case. 1) in the sense that x = −2 and y = 1. Now we proceed to fill this gap by showing that (−2.but we choose to explain it in terms of the concrete linear system above. This is not the same as saying that (−2. This leads to the important consequence that when y = 1. In one sense. (3) is true and therefore both equations in (2) become identical linear equations in x. the system (1) becomes: 4x = −8 −2x = 4 (4)
and both equations are of course the same. it would have to be (−2. Now. 1) is a solution. 1) is a solution and. this is trivial: letting x = −2 and y = 1 in the equations in (1). 1). y) satisfying the system (1).
more importantly, it does nothing to give us confidence that this method of obtaining a solution (via equations (1) through (4)) also works in other situations. So we do it differently. We begin by analyzing the concept of a solution to a system. The meaning of (−2, 1) being a solution of the system (1) is that when y = 1 in (1), then the two solutions of the two linear equations in x turn out to be the same (in fact, equal to −2, as we know). To drive home this point, if y is equal to 2 instead of 1, then the system (1) becomes 4x + 10 = −3 −2x + 2 = 5 The first equation yields x = − 13 and the second equation yields x = − 3 . So the 4 2 ordered pair (− 13 , 2) solves the first equation of (1) but not the second, and the ordered 4 3 pair (− 2 , 2) solves the second equation but not the first. In short, there is no solution of (1) which has y = 2. It remains to explain why, if y = 1, both equations in x of the system in (1) yield the same solution. This is because y = 1 is the solution of equation (3), and therefore both equations in (2) are identical when y = 1. Since systems (1) and (2) are equivalent, we see that both equations in (1) become identical linear equations in x when y = 1. Of course that solution is x = −2, as claimed. We have written out the method of getting (−2, 1) via the steps associated with (2)–(4) to facilitate the explanation of why (−2, 1) is a solution. In practice, however, one achieves some simplificaton by skirting (2), as follows. Working directly with 4x + 5y = −3 4x − 2y = −10 we subtract both sides of the second equation from the corresponding sides of the first (or, to conform with the basic principle enunciated in §3, we add the negatives of both sides of the second equation to the corresponding sides of the first), obtaining 5y − (−2y) = (−3) − (−10) (5)
This leads to y = 1 as before. More important is the fact that equation (5), if we transpose 5y and −10 to the opposite sides, is the same equation as (3). Therefore, this way of "bringing the coefficient of the number x in both equations to be the same and then eliminate x by subtraction" achieves the same result as the method of solution in (2)–(4). This "subtraction method" is what is usually done to solve simultaneous 69
equations. We bring closure to this discussion by revisiting Example 8 of the last section. In retrospect, we had arrived at a linear system in the numbers t and y in that Example: y = y =
1 4 1 5
t t +
4 5
(6)
This is a linear system the minute it is rewritten as
1 −4 t + y = 0
−1 t + y = 5
4 5
When the system (6) is given as is, there is no question as to the fastest method to solve 1 it: since both 4 t and 1 t + 4 are equal to the number y, we get 5 5 1 1 4 t= t+ , 4 5 5
1 so that 20 = 4 , and t = 16. The first equation of (6) then gives y = 4. Thus (16, 4) 5 is the solution to (6), which was already obtained earlier by another method having nothing to do with linear systems. Moreover, the fact that the point (16, 4) is the point of intersection of the two lines defined by the equations in (6) was already pointed out in Example 8.
It is time to point out that not every linear system of two equations in two unknowns has a solution. For example, obviously the system x + 0·y = 1 x + 0·y = 2 (7)
has no solution. To achieve a better understanding of this phenomenon, we have to introduced a basic property of a pair of lines in the plane. Theorem 1 Two distinct, non-vertical lines in the plane are parallel if and only if they have the same slope. Proof Let the lines be 1 and 2 . We first assume that they are parallel and prove that they have the same slope. If either of 1 and 2 is horizontal, then since 1 2 , the other is also horizontal and both would have 0 slope. There would be nothing to prove 70
in this case. So we may assume both 1 and 2 are not horizontal. Take a point P on 1 and let a vertical line through P intersect 2 at Q. (This vertical line must intersect 2 because the latter is not vertical.) Since the lines are distinct, P = Q. Go along this vertical line from P to Q and stop at a point R so that |P Q| = |QR|. From Q and R, draw horizontal lines which meet 1 and 2 at S and T , respectively. Y
¨
P ¨ ¨
¨
¨ 1 ¨ ¨ ¨ 2
¨
¨
¨
¨
¨
¨
¨
¨
S ¨ ¨
¨
¨
¨ ¨
¨ ¨
¨ ¨¨
¨
¨ T ¨¨
¨
¨
¨
¨ ¨
¨ ¨
¨ ¨ ¨Q
¨
¨
¨
R X
O
Because P QS and QRT are right triangles with legs parallel to the coordinate axes, the slopes of 1 and 2 are |P Q| |SQ| and |QR| |T R|
respectively. We have to show that these two numbers are equal. We do so by showing that P QS and QRT are congruent. Once this is done, then |SQ| = |T R| because they are corresponding sides of congruent triangles. Since |P Q| = |QR| by construction, we get |P Q| |SQ| = |QR| |T R| because both are equal to 1. By the cross-multiplication algorithm, this implies |P Q| |QR| = |SQ| |T R| which is what we are after. So it remains to prove that the triangles are congruent. One way is to translate the whole plane in the direction from P to Q. Call this translation τ . So by definition, τ (P ) = Q. Now translation moves a line (not parallel 71
Conversely. Thus τ ( 1 ) is a line passing through Q and parallel to 1 . A more traditional proof would run as follows. Recall that already τ (P ) = Q. we let τ be the translation from P to Q. First. while T is the intersection of the lines 2 and T R. We may therefore assume that they have nonzero slope so that they are both non-horizontal. This says the translation τ maps P to Q. Their corresponding angles ∠SP Q and ∠T QR are therefore equal. Q to R. and we have to show that they are parallel. and τ (Q) = R because |P Q| = |QR|. By definition. and S to T . Since 2 is already such a line. the triangles P QS and QRT are congruent by the SAS criterion of congruence. The fact that 1 and 2 have the same slope then implies that |P Q| |QR| = |SQ| |T R| These quotients have equal numerators because we constructed the point R so that |P Q| = |QR|. the fact that SQ T R and the Parallel Postulate imply that τ maps the line SQ to the line T R. Since τ (Q) = R because by construction |P Q| = |QR|. The denominators are consequently equal as well. Since ∠P QS and ∠QRT are equal because they are both right angles. We give two proofs: the first one being a direct continuation of the preceding line of geometric reasoning. i. then they are horizontal and are therefore parallel. Again. we conclude that 1 2. Hence the fact that τ ( 1 ) = 2 and τ (SQ) = T R means that τ (S) = T . The equality |P Q| = |QR| (which results from the construction OF R) then shows that P QS ∼ QRT because of the ASA criterion of congruence. Of course ∠P QS and ∠QRT are equal because they are both right angles. = A more traditional argument is as follows. the Parallel Postulate says τ ( 1 ) = 2 .. This means τ maps the line 1 to the line 2 . |SQ| = |T R|. By construction. SQ T R. But S is the intersection of the lines 1 and SQ. We now perform the same construction as before to get triangles P QS and QRT . = this shows that nonvertical parallel lines have the same slope.e. P QS ∼ QRT . Now consider the line SQ. This implies 1 2 because their corresponding angles relative to the 72
. non-vertical lines 1 and 2 have the same slope. Observe that ∠SP Q and ∠T QR are equal because they are corresponding angles of the parallel lines 1 and 2 with respect to the transversal P R. suppose two distinct. τ (S) = T . Because translation maps a line to another line parallel to itself. necessarily.to the direction of the translation) to a line parallel to itself. we see that. if they have slope 0. then τ (P ) = Q by the definition of τ . and the second an algebraic one. Since also |SQ| = |T R|. the same reasoning therefore yields the fact that the line τ (SQ) coincides with the line T R. As above. In any case.
and e therefore of the system. . b = 0. 2 be the lines defined by the equations ax+by = e and cx + dy = f . a = 0 and c = 0. y) for any number y. Here is a second algebraic proof. and the solutions c e are all ordered pairs of the form ( a . This is a contradiction to the earlier conclusion that k = k . 0) and ( f . which then imply that mx0 + k = mx0 + k . These assertions should be understood in terms of the graphs 1 and 2 . these lines c c are identical. Case 2b. Not both b = d = 0. Case 1. respectively. b. they c are parallel and therefore have no intersection. Let 1 . There are several cases to consider.transversal P R are equal. . which in turn implies that k = k . respectively. the system has an infinite number of solutions if a = f . Let us say. Let the system be ax + by = e cx + dy = f where a. d = 0. 1 is not vertical and so 1 and 2 will intersect at one point. The consideration then breaks up into two sub-cases. If a = f . Case 2. then 1 and 2 are distinct vertical lines. Suppose they intersect at a point (x0 . By an earlier remark in this section. d = 0. say they have slope m. The proof of Theorem 1 is complete. b = d = 0. and therefore any point on the line is a solution of both equations. 0). . If however a = f . Then the graph 2 is a vertical line. Since 1 and 2 are both nonvertical. f are constants. Then let the equations defining them be y = mx + k and y = mx + k . Then the system may be rewritten as
y = m 1 x + k1 y = m 2 x + k2 73
. Since b = 0. The system has no solution e if a = f . c e e These are two vertical lines passing through ( a . Now we can analyze when a linear system of two equations in two unknowns has a solution. the linear system has no solution. then y0 = mx0 + k and y0 = mx0 + k . and the system may be rewritten as x = x =
e a f c
e In this case. Thus the proof is concluded again. Then by the definition of a linear equation of two variables. The system therefore has exactly one solution corresponding to the point of intersection. y0 ). where k = k because by assumption the lines are distinct. Case 2a.
and finally. the system has exactly one solution. the system has exactly one solution. b c and they will intersect so that the system will have exactly one solution. f are constants. but
= f . k1 = e . and has no c
c = d . d
We now give two applications of simultaneous equations. if you happen not to know 9x+7 identity (8) ahead of time but ask whether x2 +x−6 can be expressed as a sum of 74
. e b e a
= f . k2 = f .c where m1 = − a . c (ii) If b = 0 but d = 0. (iv) If b = 0 and d = 0. b. 1 and 2 do not intersect and the system has no solution. Things get interesting. (iii) If d = 0 but b = 0. we get the identity 5 4 9x + 7 + = 2 x−2 x+3 x +x−6 (8)
This is straightforward. b d b d If the former. The first one is to express certain rational expressions in a number x as a sum of simpler rational expressions also in x. Then: (i) If b = d = 0. and d
a b
the system has an infinite number of solutions if both
=
c d
and
e b
= f. m2 = − d . If the latter. If a = d . the lines 1 and 2 are not parallel according to Theorem 1. the system has an infinite number of solutions if e solution if a = f .
Theorem 2 Given a linear system ax + by = e cx + dy = f where a. then the lines 1 and 2 coincide and the system will have an infinite number of solutions corresponding to each point on the line. according to whether e = f or e = f . We can summarize all this into one theorem. then: the system has exactly one solution if the system has no solution if
a b a b c = d. then b the lines 1 and 2 are parallel or indentical. Now both 1 and 2 are b b d c non-vertical line. . If a = d . Consider the simple sum: 4 5(x + 3) + 4(x − 2) 5 + = x−2 x+3 (x − 2)(x + 3) After simplifying the numerator of the right side and multiplying out (x − 2)(x + 3) = x2 + x − 6. however. . .
(a2 x + b2 ).1 1 (constant) multiples of the simple rational expressions x−2 and x+3 . d. an−1 = bn−1 . In order to answer this question.e. (an x + bn ) be n linear polynomials in x (n is a positive integer) so that none is a constant multiple of another. We now use Fact (B) to recover equation (8). cn so that c1 c2 cn p(x) = + + ··· + (a1 x + b1 )(a2 x + b2 ) · · · (an x + bn ) a1 x + b1 a2 x + b2 an x + bn (B) Suppose the following two n-th degree polynomials in x (n is a postive integer) are equal for all x: an xn + an−1 xn−1 + · · · + a1 x + a0 = bn xn + bn−1 xn−1 + · · · + b1 x + b0 Then the coefficients of the polynomials are pairwise equal: an = bn . and is a special case of the partial fraction decomposition of a rational expression. a0 = b0 . In general terms this question may be understood as part of our overall desire to express complicated objects in terms of simpler ones (think of the prime decomposition of a whole number. 4 and 5. i. . One is that even knowing 9x+7 x2 + x − 6 = (x − 2)(x + 3) ahead of time. there remains the question of how to get the precise values of the numerators. c2 . . . The answer to this question is by no means obvious. .
(A) Let (a1 x + b1 ). for some x−2 x+3 4 5 appropriately chosen constants a. b. On a more concrete level. . for two reasons. we have to first quote two facts without proof. Then there are constants c1 . By the 75
. rather than just a sum of x−2 and x+3 without any x's in the numerators.. we see that there must be constants a and b so that a b 9x + 7 = + (x − 2)(x + 3) x−2 x+3 which is valid no matter what x may be. . this question arises naturally in calculus.e. Let p(x) be a polynomial in x of degree less than n. . . respectively. i. Using Fact (A). The other reason is that even if you believe that such an expression is possible. one would be inclined to believe that x2 +x−6 is a sum of more complicated rational expressions such as ax+b and cx+d . to obtain the values of the constants a and b as 5 and 4.. c. for example). the proofs are not difficult but they do take up time and space that we cannot afford at this point. .
17} is yet another example. It will be obvious that we will get an infinite number of Pythagorean triples by this method. In other words. or even that {8. From Fact (B).addition of rational expressions. Everybody knows that 3. b and c are the lengths of three sides of a right triangle. c form a Pythagorean triple {a. the length of the hypotenuse of the right triangle. We say three positive integers a. It is even true that the method produces all the Pythagorean triples. 15. though we will not prove this fact here. The method is clearly sufficient to deal with the general situation. One would like to 76 (a + b)x + (3a − 2b) (x − 2)(x + 3)
. In other words. by definition. 12. respectively. some may even know that {5. We now give the second application of simultaneous linear equations. 4. Hence. we know that the coefficients a + b and 3a − 2b must be equal to 9 and 7. 9x + 7 = (a + b)x + (3a − 2b) for all x. Note that the third member of a Pythagorean triple is. we have the following simultaneous linear equations in a and b: a + b = 9 3a − 2b = 7 We can solve this system by simply multiplying the first equation by 2 and then adding it to the second equation. 5 form a Pythagorean triple. we get 9x + 7 (a + b)x + (3a − 2b) = (x − 2)(x + 3) (x − 2)(x + 3) and therefore the numerators must be equal. The first equation now gives b = 4. a. This yields 5a = 25 and therefore a = 5. b. 13} is another Pythagorean triple. a b a(x + 3) + b(x − 2) + = x−2 x+3 (x − 2)(x + 3) = Combining the two equalities. c} if a2 + b2 = c2 . It goes without saying that the key point of the definition of a Pythagorean triple is that all three numbers are positive integers. as claimed. But are there others? Our purpose is to produce Pythagorean triples at will by solving an extremely simple linear system of equations. b.
In this terminology.9 See Eleanor Robson. b. namely.e. 5n} for any positive integer n. 28 (2001). or is a multiple of a primitive Pythagorean triple. Neither Sherlock Holmes nor Babylon: A reassessment of Plimpton 322. we will replace it by {5. 20}. 18541}. we will in fact be in possession of an infinite number of Pythagorean triples.say that this method is due to the Babylonians some thirty-eight centuries ago (circa 1800 B. 12. {6.C. 13500. Therefore. we lose nothing by replacing it with the primitive Pythagorean triple of which the first Pythagorean triple is a multiple.). and in general. 16. {9. We will give a proof of the following theorem. Theorem 3 Let (u. let it be noted that the largest triple is {12709. and will henceforth concentrate on getting primitive Pythagorean triples. Therefore whenever a Pythagorean triple is given. if k is a positive integer that divides all three a. but a more accurate statement would be that it is the algebraic rendition of the method one infers from a close reading of the celebrated cuneiform tablet. 10}. b. 15}. 167-206. {3n. c} to be primitive if the integers a. 10}. Once we are in possession of this triple. b . Historia Mathematica. 36. v) be the solution of the linear system u + v = t s u − v = s t where s. namely. 4. Accordingly. then k = 1). 5}. Let us first perform a conceptual simplification.
9
77
. and c = nc . there is not much glory in claiming that you also have another Pythagorean triple. 13}. 5}. we define a Pythagorean triple {a. If we write u and v as two fractions with the
Lest you entertain for even a split second the idle thought that people couldn't have known such advanced mathematics thirty-eight centuries ago and that these triples were probably hit upon by trials and errors. for example. c } if there is a positive integer n so that a = na . b. We say a Pythagorean triple {a. 12. instead of dealing with {15. 39}. t are positive integers with s < t. Plimpton 322. Clearly. Take {3. {6. a given Pythagorean triple is either primitive. 4. b and c. 8. and c have no common divisor other than 1 (i. which lists fifteen Pythagorean triples. b = nb .. 4n. {12. c} is a multiple of another Pythagorean triple {a . 8. if you already have the Pythagorean triple {3. Observe that its statement makes use of the fact that any two fractions can be written as two fractions with the same denominator (fundamental fact of fraction pairs).
b
There is extra incentive in providing a proof of this theorem. we get 6 12 2 13 5 v = u − 3 = 12 − 2 = 12 . 13} is a Pythagorean triple. not only because the proof is very simple. 12. 24. Thus we have retrieved the grandfather of all Pythagorean 4 4 triples. we have t b b c b
2
−
a b
2
=1
Multiplying through both sides of this equality by b2 gives c2 − a2 = b2 and therefore. 25} is a Pythagorean triple. {3. From the second equation. 5}. So with u = c and v = a . u =
c b
and v = a . or u2 − v 2 = 1. So with (u. Since this is new 24 4
78
. we get 1 1 v = u − 2 = 5 − 2 = 3 . Example 3 Consider
u + v = u − v =
4 3 3 4
25 Adding the equations gives 2u = 12 so that u = 25 and the second equation gives 24 7 v = 25 − 3 = 24 . b. a2 + b2 = c2 We have our Pythagorean triple and the proof of Theorem 3 is complete. let us put it to use to produce some new Pythagorean triples. {5. then {a. Example 1 Consider
u + v = 2 1 u − v = 2
5 5 Adding the equations gives 2u = 2 so that u = 4 . v) of the linear system furnishes a Pythagorean triples. From the second equation. {7. which 3 of course we already know. By Theorem 3. 4.
Example 2 Consider
u + v = u − v =
3 2 2 3
Adding the equations gives 2u = 13 so that u = 13 . c} form a Pythagorean triple. we multiply the corresponding sides of the two equations in the theorem to get t (u + v)(u − v) = s · s .same denominator. but also because it actually tells us why the solution (u. but before doing that. By Theorem 3. It is easy to explain how Theorem 3 came about. v) as the solution of the linear system.
e.. we get (a/b)2 + 1 = (c/b)2 . c}. by our convention of letting c be the length of the hypotenuse of the right triangle. Although Theorem 2 guarantees that this is indeed a Pythagorean triple. 12. that if we had taken the trouble to reduce u = 26 to its lowest 10 13 13 terms. we get v = 10 − 1 = 24 .
Observe. 5 10 26 From the second equation of the system. a2 + b2 = c2 . 5 10 {10. b. Thus we see that different values of s and t do not always lead to distinct primitive Pythagorean triples. Now v 2 + 1 = u2 and therefore.to most people. 24. 13}. We now explain the genesis of Theorem 3. we always have u > v. 26} is a Pythagorean triple. By assumption. it would be good for your soul to directly check that 2762 + 47572 = 47652 is in fact true. 4765}. 12. which is clearly primitive. We now adopt the convention of letting u= c b and v = a b (9)
Thus u and v are both fractions (i. From the second equa138 276 2 4757 4765 tion. 4757. we get v = 276 − 69 = 276 . This is not a primitive triple because it is a multiple of {5. however. positive rational numbers) and. and then we would get v = 5 − 1 = 12 . every single Pythagorean triple has been primitive. Observe that thus far. we obtain 2u = 26 and multiplying both sides by 2 gives u = 26 . By Theorem 3. and proceed to find out what it must be. one should check directly that 72 + 242 = 252 . We will follow the time-honored method of assuming that we already have a Pythagorean triple {a. then we would obtain u = 5 . This time we get an unfamiliar Pythagorean triple {276. u2 − v 2 = 1 79
. Example 4 Consider
u + v = u − v =
69 2 2 69
Adding the two equations gives 2u = 4765 so that u = 4765 . Now consider: Example 5 Consider u + v = 5 1 u − v = 5
1 Adding the equations. and by dividing through by b2 . 13} would be the result. and the 5 5 primitive triple {5.
Express 8x+2 as a sum of multiples of x+1 and x−1 . its value is 6 . Find their ages. and 15 years ago. so that the mixture may be 7 wine? 8 1 1 6. You may use a scientific calculator. 81
. especially for (i) and (j) below. (b) s = 4. In each of the following. If 3 is added to the numerator of a fraction and 7 subtracted from the denominator. (a) s = 2. and of another 8 wine. The contents of one barrel is 6 wine. t2 + s2 } is primitive. 7 its value is 2 .
EXERCISES 1. t = 5. Of course the former is a consequence of the latter.9y 0. Alan's age is 6 of Bill's. it can be shown that every primitive Pythagorean triple is represented in terms of suitable s and t in this manner. x−3
8.16
7x − 9y = 15 8y − 5x = −17 12x + 11y = 172 28x − 17y = 60 3 5x − 4 y = 2 x + 2y = 11 6
9 x 3 x
+ −
12 y 9 y
= −1 (b) = 7
− 3y = 4 + 2y =
10 3
13 3. With a little more work. But if 1 is subtracted from the numerator and 7 added to the denominator. Solve: (a) (c) (e) 2. but for school mathematics. t = 5.08x + 0.. one can prove that if s and t are relatively prime (i. the latter is more instructive. his age was 10 of Bill's. then the triple {2st.e. Solve: (a)
6 x 8 x
7x − 3y = 3x − 5y = 2 x − 5y = 5 6 5 1 x + 9y = 6 0. t2 − s2 .04y
10 −5 −1 2
5 2
(b) (d) (f)
= 0. 5 5 5.We have therefore presented two ways of obtaining Pythagorean triples: by giving values of s and t into the preceding formula. How many gallons 9 must be taken from each to fill another barrel whose capacity is 24 gallons. you are asked to solve the linear system in Theorem 2 with the given values of s and t to obtain Pythagorean triples. and if one of them is even and the other odd. With more work still.46 = 0. x2 −1
7. 5 4. or by using Theorem 2 and solving the linear system there. Find the fraction. Express
−5(x+1) 3(x2 +x−12)
as a sum of multiples of
1 x+2
and
1 . no common divisor other than ±1).1x − 0.
By dropping perpendiculars from the lines to the x-axis. The proof will be written up at a later date. The idea is to use Theorem 1 of this section to move the lines to the origin by parallelism. is primarily about equations. If the number is divided by the difference of the digits. t = 69. (e) s = 3. the largest number in the Pythagorean triple has 8 digits. (j) s = 54. t = 12. The second digit of a two-digit number is 3 of the first digit. Explain how you can use such a calculator to directly verify that the triple of numbers so obtained is a Pythagorean triple.(c) s = 1. (f) s = 1. In (k) of the the last problem. t = 13. we can explicitly compute the slopes of the lines by the distances between the origin and the feet of the perpendiculars. 9.. Find the number.
Appendix Theorem 4 Two lines are perpendicular if and only if the product of their slopes is −1. t = 125. (d) s = 1. t = 4. (h) s = 1. in the sense it is usually understood in mathematics. By suitably adjusting these distances and making use of congruent triangles. t = 9907. t = 13. As such. (g) s = 3. the quotient is 15 and the remainder is3. (k) s = 8.
7
Linear Inequalities and Their Graphs
So far we have only discussed equations and indeed. However. (i) s = 2. then we only need to prove the theorem for lines passing through the origin. school algebra in its current usage includes everything related to number computations. two 82
. 1 10. inequalities occupy a prominent position in school algebra because in real life two numbers (e.g. t = 3. we arrive at the desired conclusion. t = 6. Suppose you have a calculator with only a 12-digit display on the screen. algebra.
but with 48 A games and 2 B games instead (which still add up to 50 games). i. two people's incomes. A and B. It may be pointed out that 83
. inequality. a casual glance at the data would suggest that it is more profitable to sell Game A than Game B. and that is where inequalities come in. For example. With 50 A games. she would be using only 50 × 75 = 3750 dollars for manufacturing. while Game B costs $165 to manufacture but will bring in a profit of $185. As an example. she would make 50 × 125 = 6250 dollars. We see right away that this may be a bad strategy in terms of maximizing profit. $6000 is good for manufacturing 80 A games as it takes only $75 to manufacture one A game.people's weights. and we can easily confirm it. The emphasis here is on the words bigger than. in the following sense. as follows. Game A costs $75 to manufacture and will bring in a profit of §125. she would make (48 × 125) + (2 × 185) = 6370 dollars. but if you use the same amount to manufacture two A games (each costing $75). but each 3 75 3 1 Game B only brings in a profit that is only about 1 8 of its manufacturing 4 4 cost (because 185 = 1 33 .. Assuming that she sells every game she brings. whereas she is allowed to bring only 50 games. and has up to $6000 to spend on manufacturing costs. 165 8
One's first impulse is therefore to say that the manufacturer should bring only A games to the show.
A precise way to think about this is to notice that each Game A brings in a profit that is 1 2 of its manufacturing cost (because 125 = 1 2 ). One will generally be bigger than the other. Then if you manufacture one B game. thereby wasting $2250 (= 6000 − 3750) of her $6000 budget. She has two games. a certain number of A games and a certain number of B games.) are more likely to be different than the same. But. and 6370 > 6250. etc. how many games of each kind should she manufacture if she wants to maximize her profit? It is clear that in this case there is no equation to solve. One way to understand what is involved in a problem of this nature is to approach it in a naive way and see why it doesn't work.e. which is about 1 32 = 1 1 ). and is told that she can bring up to 50 games. because the answer to the problem is a pair of number. you only make $185. consider the following Manufacturing Problem: A video game manufacturer is invited to a game show. you'd not only make $250 (= 2 × 1215) but would have $15 to spare. Suppose you have $165. So if she brings 50 A games. so that this combination brings in a profit bigger than any other possible combination.
It is now clear that there is an inherent push-pull in this problem: bringing too many A games would under-utilize the $6000 manufacturing budget because of the 50-game quota. Intuitively. knowing that each Game B brings a profit of $185 whereas each Game A brings a mere $125. and bringing too many B games would under-utilize the 50-game quota because of the $6000 manufacturing budget. she could rightly decide to concentrate on Game B and forget about Game A. P = 125x + 185y. The need for a better understanding of inequalities would be more in evidence if we begin with a transcription of the given data of the Manufacturing Problem into symbolic language. (Review the Section 2 at this point if necessary. She can also approach this problem from the opposite end. as adumbrated above.) Suppose the manufacturer produces x A games and y B games. The main theme of this chapter is about this mathematical understanding which. whereas the same budget can equally well produce 34 B games and 5 A games (because (34 × 165) + (5 × 75) = 5985 < 6000) and these 39 games now bring a profit of (34 × 185) + (5 × 125) = 6915 dollars. What we need to understand in mathematical terms is how to negotiate this push-pull to arrive at this "equilibrium". Again. will require a good grounding on inequalities. which is more than $6000. the combination of A games and B games that brings in the maximum profit must be a kind of "equilibrium" between the number of A games and the number of B games. The problem now is that she cannot bring 50 B games to the show because her budget of $6000 wouldn't allow it: the cost of manufacturing 50 B games is 50 × 165 = 8250 dollars.the cost of manufacturing 48 A games and 2 B games is (48 × 75) + (2 × 165) = 3930 dollars. For confirmation. We have to find values of x and y so that the profit P . 84
. Neither of these options would bring in the maximum profit. which is well within her $6000 budget. which is $255 more than $6660. namely. and her profit from 36 B 165 games would be 36 = 6660 dollars. notice that a budget of 4 $6000 can produce at most 36 B games because 6000 = 36 11 . this would suggest that bringing all B games to the show is a poor strategy for maximizing profit.
Because she can bring at most 50 games. R is called the graph of these inequalities (in the plane). y) in the xy-plane. What we propose to do is to interpret these four inequalities as constraints on a point (x. we still need a framework to understand the problem. x and y are constrained by the inequality x + y ≤ 50 Her manufacturing budget imposes another constraint in that she has at most $6000 to spend on the production: 75x + 165y ≤ 6000 There are also two other obvious but indispensable constraints: x ≥ 0 and y ≥ 0. In summary then. Now x and y are not arbitrary but are under constraints (a technical terms for "restrictions") that come with the problem. The collection of all the points (x. The profit P = 125x + 185y can now be thought of as an assignment to (or at) each point (x .is a maximum. As such. In a language that will be formally introduced presently. we shall refer to P as the profit function on this R. y) satisfying these four constraints is a certain region R in the plane. The Manufacturing Problem now become a purely mathematical problem independent of context: Among the points in the graph R of the inequalities x≥0 y≥0 x + y ≤ 50 75x + 165y ≤ 6000 at which point does the profit function P = 125x+185y assign the maximum value? The virtue of this reformulation of the Manufacturing Problem is that it points to clearly defined mathematical tasks: 85
. we want to maximize P = 125x + 185y among all x's and y's which satisfy: x≥0 y≥0 x + y ≤ 50 75x + 165y ≤ 6000 Even in this symbolic language. y ) in R a number 125x +185y .
or in symbols. Accordingly. The proof given below can be skipped on first reading. d. Using the vector interpretation of addition for rational numbers. i. If b > a. Observe the following simple consequences of the definitions which will be used in subsequent discussions without comment (a. but they will all remain valid if "≤" is replace by "<" throughout. 1 > −7 is the same as 1 − (−7) > 0. the following pictures. b < c =⇒ a < c a ≤ b. b ≤ a =⇒ a = b In the following five assertions about inequalities. For example. 8 > 0.. There is also a weaker notion of inequality in the form of a ≤ b. b. c. Therefore it suffices to deal with the case of >. b. when negative quanities are involved. we state everything in terms of weak inequality. Similarly. we prove: b > a implies b − a > 0. respectively. We can also verified directly that (α) must be true when a and b are explicitly known. Moreover. Let a. First. while each of these five inequalities is valid as stated for arbitrary numbers a.(i ) What does the graph R of a collection of inequalities look like? (ii ) Can we achieve enough of an understanding of the profit function P = 125x + 185y to predict where it might assignment its maximum value in R? We now do the spade work necessary for analyzing graphs of inequalities. i. If both a and b are positive.. We have to show two things: b > a implies b−a > 0. for the same reasons. b. and also b−a > 0 implies b > a. By adding a to both sides of c = b − a.
For the explanation. d be arbitrary numbers in the following discussion. c are numbers): a < b. Our first concern is with the behavior of inequalities with respect to arithmetic operations. we may sometimes explicitly refer to "≤" as weak inequality. However. (α) b ≥ a is equivalent to b − a ≥ 0. b > 0 > a. then (α) is quite obvious.e. that a is to the left of b on the number line.e. For emphasis. we tacitly assume that all the numbers are rational numbers in the proofs. we get a + c = b. corresponding to the three cases of b > a > 0. let c = b − a and we must show c > 0. −2 > −5 is the same as −2 − (−5) > 0. which means a is less than or equal to b. by definition. such an assertion requires a little thought. and 0 > b > a. (α) – ( ). 3 > 0. show why c > 0. a < b or a = b.
86
. c. for school mathematics it suffices most of the time to know the validity of these inequalites for rational numbers. the case of equality in (α) is easy: it is obvious that b = a is equivalent to b − a = 0. Recall that the inequality a < b (or written differently as b > a) means.
e. Thus a ≤ c−b. and therefore b − a ≥ 0 (using (α) again). (β) implies that (a+b)+(−b) ≤ c+(−b). This implies (b + d) − (a + c) ≥ 0. Similarly. then a + c ≤ b + d. i. suppose b − a > 0. Therefore (b − a) + (d − c) ≥ 0.c 0 a E b
c a 0 E b
a
c E b
0
(We briefly review the vector representation of a+c. d − c ≥ 0. Conversely. If a+b ≤ c. Conversely. if a ≤ c − b. Let |c| denote the distance of c from 0. Since c > 0 by hypothesis. But (b + d) − (a + c) = (b + d) − a − c = b + d + (−a) + (−c) = (b + (−a)) + (d + (−c)) = (b − a) + (d − c) By hypothesis. and b is to the right of a. to right if c is positive. then b = a + c. as desired. then this means in going from a to b we must go right a distance of |c|. It is common to paraphrase (γ) by saying that one can transpose a term to the other side of an inequality. a ≤ b. The proof of (α) is complete. Therefore c has to be positive). which implies a + b ≤ c. that b is to the right of a. then go from a a distance of |c| to left if c is negative. so b is to the right of a. By (α). if suffices to show that (b + d) − (a + c) ≥ 0. the vector interpretaion of a + c means that from a we go right a distance of |c| to arrive at the point a + c. Now if b = a + c. (γ) a + b ≤ c is equivalent to a ≤ c − b. then (β) implies that a + b ≤ (c − b) + b.. Again let c = b − a. go from 0 to a. To locate the point a + c on the number line. But a + c = b.
(β) If a ≤ b and c ≤ d.
87
. The point of final destination is then a + c. then we have to show that b > a.
Therefore (−3)(−7) > (−3)(−4). ( ) If a < 0 and b ≤ c. then ab ≥ ac. which is what we set out to prove. The way to explain (δ) in general is to appeal to (α). therefore ac−ab ≤ 0. Equivalently. we let a = −a so that a > 0. we have (−3)2 > (−3)5. b = −7. and c = −4. we obtain ac ≤ ab. So −(3 · 5) is further away from 0 than −(3 · 2). Now −(3 · 2) is the mirror reflection of 3 · 2 with respect to 0. Thus if we let d = c − b. But ac − ab = a(c − b) and since a > 0 and c − b ≥ 0 (because by hypothesis b ≤ c). This is the most mysterious among the basic facts concerning inequalities. but less so if they are both negative. Hence ac − ab ≥ 0 and therefore ab ≤ ac. while −(3 · 5) is the mirror reflection of 3 · 5 with respect to 0. Since c ≥ b. although 2 < 5. 88
. Let us prove directly why.(δ) If a > 0 and b ≤ c. Example 1 Exhibit all the numbers x on the number line which satisfy (5 − 2x) + 12 > 4 − (3x − 5). Thus assume b ≤ c and a < 0 are given. From 2 < 5. This is obvious if both b and c are positive. −a d ≤ 0. by (γ). We will instead rely on a skillful use of (α) and prove ( ) in one stroke. then ab ≤ ac. We will make frequent use of (α) – ( ) in the succeeding discussion. Let us illustrate by letting a = −3. (α) implies that c − b ≥ 0. Since (−3)(−7) = 21 and (−3)(−4) = 12. we see that a(c − b) ≥ 0 by the definition of fraction multiplication. we have 21 > 12. Using (γ). −(3 · 2) = (−3)2 and −(3 · 5) = (−3)5. However. Now −a d = −a (c − b) = (−a )(c − b) = a(c − b) = ac − ab. Here is one simple application. this line of reasoning would require the separate considerations of three cases: (i) 0 < b ≤ c. For the proof of the general case. d ≥ 0 and therefore a d ≥ 0. Therefore the inequality (−3)2 > (−3)5 is clear from the following picture:
−(3 · 5)
−(3 · 2)
0
(3 · 2)
(3 · 5)
The proof of (δ) can be obtained in a similar manner if 0 < b ≤ c. and (iii) b ≤ c ≤ 0. (ii) b ≤ 0 ≤ c. It can get tedious. Proving ab ≤ ac then becomes proving ac − ab ≥ 0. we get 3 · 2 < 3 · 5.
This means all the x's which satisfy (5 − 2x) + 12 > 4 − (3x − 5) are exactly the same as those which satisfy x > −8. which.
−8
0
Now we are finally in a position to deal with task (i) above. one simply isolates the variable x in the inequality.e. Thus we see that (5 − 2x) + 12 > 4 − (3x − 5) is equivalent to the inequality x > −8. b. The graph of ax + by > c is defined in like manner. Note that the concept of the graph of an inequality or of a collection of inequalities is usually used without an explicit definition in textbooks and professional development materials. i. The graph of a collection of inequalities is by definition the set of all the points which satisfy each of the inequalities in the collection. in the sense of transposing all the x's to one side of the inequality by making repeated use of (γ). is equivalent to 5 − 2x + 3x > −3.. y ) in the plane whose coordinates x and y satisfy this inequality. and Example 1 is usually expressed as: graph (5−2x)+12 > 4−(3x−5) on the number line. and we will use this notation below. which therefore can be represented by the thickened semi-infinite line segment below. x > −8.. Now 4−(3x−5)−12 = −8−(3x−5) = −8−3x+5 = −3−3x. what does the graph of a collection of inequalities look like in the plane? Formally.The set of all these x's is called the graph of (5 − 2x) + 12 > 4 − (3x − 5) on the number line.e. As in the case of solving linear equations. ax + by ≥ c. i. When this happens.
89
. Thus (5−2x)+12 > 4−(3x−5) is equivalent to (5−2x) > 4−(3x−5)−12. by (γ) again. c are given constants) is the set of all the points (x . So (5 − 2x) + 12 > 4 − (3x − 5) is equivalent to 5 − 2x > −3 − 3x. no mathematical reasoning is possible in any discussion concerning graphs of linear inequalities. and is equivalent to −2x + 3x > −3 − 5. It is customary in mathematics to denote the graph of an inequality such as ax + by ≥ c by the notation {ax + by ≥ c}. i. It follows that the graph of a collection of inequalities is the intersection of all the graphs of the individual inequalities. the graph of a linear inequality of two variables ax + by ≥ c (where a. Please be sure to teach your students this definition. Similarly.e. one defines the graphs of ax + by ≤ c and ax + by < c..
First assume L is not horizontal. If L is vertical. 0). and X + . Similarly. Pictorially. L+ (resp. and those to the right of xL . L. A little thought will also reveal that. the half-plane L− of L is by definition all the points P so that the line passing through P and parallel to L meets the x-axis at X − . then L+ (resp. Therefore. If xL is the origin O. we have to discuss the separation property of lines in the plane. called half-planes. L− . This point separates the x-axis into two parts: those points to the left of xL . L− ) is the set of all the points P in the plane so that the xintercepts of the line passing through P and parallel to L is bigger than xL (resp. we have invoked the Parallel Postulate throughout the foregoing discussion. the set consisting of the single point xL . the number xL is not the point (xL . which is the x-coordinate of the point of intersection of the line with the x-axis. we identify xL with (xL . It is clear that the three subsets of the plane. implicitly. 0). Strictly speaking. to be denoted by X − . negative). The half-plane L+ of L is by definition all the points P so that the line passing through P and parallel to L meets the x-axis at X + . respectively.To describe the graph of an inequality. to be denoted by X + . Then L meets the x-axis at a point (xL . 0). and L+ are disjoint and that their union is the whole plane. We are going to show that every line L in the plane separates the plane into two parts. we may rephrase the definitions of L+ and L− in the following equivalent way: L+ (resp.
L
5 5 55 5 5 P 5 L− 5 5 5 r 5 5 P5 r 5 5 5 5 5 5 r 5 55 r r5 5 5 5 5 5 xL 5
L+
X−
X+
Recall that every non-horizontal line has an x-intercept. b) so that the x-coodinate a is positive (resp. Note that the x-axis is the union of X − . then X − and X + are of course the usual negative and positive x-axes. L− ) in this case is 90
. smaller than xL ). but by common practice. L− ) is nothing other than all the points (a. and they are defined as follows.
In that case. The right and left half-planes in the case of a vertical line are now replaced by the upper half-plane L+ and the lower half-plane L− . L L− xL L+ x
On the other hand. The half-planes L+ and L− do not include L. left) of the L. then the linear equation becomes y = c and L is horizontal.literally the part of the plane to the right (resp. We first dispose of two extreme cases. but we shall see presently that there is a need to consider half-planes together with L itself. c are given constants). if L is horizontal.) The reason for our interest in half-planes is that they allow us to identify the graphs {ax + by > c} and {ax + by < c} in terms of the half-planes of the line L defined by ax + by = c (a. We will refer to L+ ∪L and L− ∪ L as the two closed half-planes of L. for the following reason. we will refer to L+ and L− as the two open half-planes of L for emphasis. we call this L+ (resp. the point of intersection of L with the y-axis. we simply replace the x-axis by the y-axis. as shown. y L+ yL L− The occasion will arise when more precision regarding half-planes is needed. b. If there is any fear of confusion. The closed half-planes are not disjoint as they have L in common. The consideration of the b 91 L
. however. left half-plane) of the vertical line L. (The terminology of "open" and "closed" is standard in mathematics. Then the point xL would be replaced by yL . then of course none of these considerations apply. L− ) the right half-plane (resp. For this reason. If a = 0.
and c. An entirely analogous discussion then shows that in this case. One more general comment on Theorem 1 before beginning the proof. So assuming a = 0. b Similarly. There are. y ) of L+ and we check whether the inequality ax + by > c is true or not. b. {ax + by > c}. c Now if b = 0. the graph {ax + by > c} is the right half-plane and the graph {ax + by < c} is the left half-plane. By Theorem 1. we know
92
. while for a < 0. y ) does not belong to ax + by > c so that {ax + by > c} cannot be L+ . Similarly. The proof of this theorem contains all the key ingredients needed for the solution of the Manufacturing Problem. which is the upper half-plane of L. after all. and we have {ax + by > c} = L+ . Indeed. On the other hand. b > 0 and b < 0. This theorem shows why the concept of a closed half-plane is relevant. Take a point (x . What we have seen are special cases of the following general theorem. — in the form of Theorem 1bis below. for a > 0. the graph {ax + by < c} in this case is just the lower half-plane. If it is not. the graph {ax + by < c} in this case is the upper half-plane. But since Theorem 1 guarantees that {ax + by > c} must be a half-plane. and the graph {ax + by > c} is just {y > c }. we conclude that {ax + by > c} = L− . if b < 0. The statement of this theorem gives the impression of sloppiness: if we focus on. there is no need for such a general statement. then the equation ax + by = c becomes x = a and its graph L is therefore vertical. b b which is the lower half-plane of L. ax + by < c) is one of the two half-planes L+ and L− of the line defined by ax + by = c. the inequality ax + by > c is equivalent to y > c (by b (δ)). suppose we want to know the graph of the weak inequality ax + by ≤ c. then the theorem does not seem to specify which of the two half-planes of the line L defined by ax + by = c is equal to {ax + by > c}. If it is. it can be done very simply. Theorem 1 Given constants a. y ) belongs to {ax + by > c} by definition. only two half-planes of L. so that the graph {ax + by > c} is now {y < c }. — but in practice. if b > 0.linear inequalities themselves separates into two cases. Such explicit information will in fact come with the proof. the graph {ax + by > c} is the left half-plane and the graph {ax + by < c} is the right half-plane. on account of the difference between assertions (δ) and ( ) above. and if it is a matter of deciding which of the two is {ax + by > c}. as follows. for example. still with a = 0. the graph of ax + by > c (respectively. then (x . then the inequality ax + by > c is now equivalent to y < c by virtue of ( ). then (x .
Then the graph {ax + by > c} is the half-plane L+ . b = 0. we may further assume that a > 0. Suppose a < 0 in the inequality ax + by > c. Now. which for convenience will be referred to as Theorem 1bis: Let L be the line defined by ax + by = c. it therefore suffices to prove the following assertion. Suppose we are trying to determine the graphs of −3x + y < 9 and −3x + y > 9. Let L be the line defined by −3x + y = 9. Therefore the graph of ax + by ≤ c is simply the closed half-plane L+ ∪ L. we obtain (−a)x − by < −c. Let us give an illustration of how to put Theorem 1bis to use in a concrete case. under the same assumptions. for the following reason. we may equivalently consider the graphs { (−a)x − by < −c} and { (−a)x − by > −c} instead. But of course the coefficient (−a) of x is now positive and the considerations in the case of a > 0 now apply. we may henceforth assume that a = 0 and b = 0. {ax + by < c} = { (−a)x − by > −c} Thus instead of considering the graphs {ax + by > c} and {ax + by < c}. Because −3x + y = 9 is equivalent to 3x − y = −9. and a > 0. To prove Theorem 1. so that {ax + by > c} = { (−a)x − by < −c} Similarly. b. As mentioned above. which is L. Theorem 1 in the form of Theorem 1bis carries the explicit information about which half-plane of L is the graph of (for example) {ax + by > c}. Then we apply Theorem 1bis to obtain: {3x − y > −9} = L+ {3x − y < −9} = L− as shown: 93
. Multiplying both sides of the inequality by (−1) and applying (δ). and the graph {ax + by < c} is the half-plane L− . Having disposed of the special cases of Theorem 1 when either a = 0 or b = 0. let us give the proof of Theorem 1. Therefore we shall henceforth assume that a > 0. where a. L is also the line defined by 3x − y = −9. It follows that the graph of ax+by ≤ c is just L+ together with the graph of ax+by = c.{ax + by < c} is one of L+ and L− . Let us say for definiteness that {ax + by < c} = L+ . c are given constants. Without loss of generality.
and (B) every point of L+ is a point of {ax + by > c}. we get {−3x + y < 9} = L+ {−3x + y > 9} = L−
We now return to the proof of Theorem 1bis.y
¢ ¢ ¢
L 9 L+
L
−
¢ ¢ ¢ ¢ ¢
−3 ¢
¢ ¢ ¢
O
x
In terms of the original question of what the graphs of −3x + y < 9 and −3x + y > 9 are. we use ( ). We will refer to it as (∗). where a > 0. Therefore. will prove to be useful: (∗) A line L defined by ax + by = c with a = 0 and b = 0 has slope − a b c and x-intercept a . (The proof of {ax + by < c} = L− is entirely similar and will be left as an exercise. µ= c a
To prove {ax + by > c} = L+ . We note that. proved in Section 4 (Fact (iv) near the end). So 3x − y > −9 is equivalent to −3x + y < 9 and 3x − y < −9 is equivalent to −3x + y > 9. by (∗).
94
. the x-intercept of L. The following fact. y ) so that the line parallel to L and passing through (x . we have to prove (A) every point of {ax + by > c} is a point of L+ . So given a linear equation ax + by = c with graph L. we have to prove that {ax + by > c} is equal to L+ .) Recall that L+ is by definition all the points (x . y ) has x-intercept bigger than µ.
Using (∗) once more. Then by the definition of L+ . µ > µ. as desired. Therefore a ax +by c > a . and 1 1 = 0. and let be the line passing through (x . we tacitly assumed that. whereas we know a · a = 1. and let have an x-intercept of µ . We conclude (by the Parallel Postulate) that is just . But a > 0 implies 1 > 0. Now as in the c proof of part (A). y ) is obviously a solution of ax + by = c so that also passes through (x . y ) belongs to {ax + by > c}. It cannot be negative either because if it were. The slope of is also − a b (by (∗) again). is defined by ax + by = c . Next we prove (B).e. we have to show µ > µ. we have already seen that µ = a . We begin by determining the equation of . We have completed the proof of Theorem 1bis. If a is not positive. because a > 0. then we must prove that it is also a point of {ax + by > c}. so a has to be positive after all. a > 0. by Theorem 1 of Section 6. then a · a = 0 · a = 0.
The following examples illustrate how to make use of Theorem 1. of is ax + by c = a a Now (x . i.We first prove (A). ax + by > c. by (∗). so that is parallel to L. A precise 1 proof of this fact goes as follows. Let b c = ax + by .e. then it is either 0 or 1 negative.. y ) and parallel to L. we know a · a = 1. (x . µ . we have µ > µ. Let (x . i. we must show that ax + by > c. a we get ax + by > c.
95
. then the product a · a would be the product of a negative number and a positive number (using the 1 assumption a > 0) and is therefore negative. y ) and parallel to L. we see that the x-intercept. a 1 1 (ax + by ) > c a a and this means ax + by c > a a c Since µ = a . Moreover. 1 and 1 is positive. so by definition. µ is the x-intercept of L. y ) is in L+ . also Theorem 1. Suppose (x . It suffices to show that neither possibility is admissible. and therewith. We have now showed that neither possibility can hold. If µ is the x-intercept of . µ =
1 In the proof of (A). we can prove µ = ax +by .. so that by (δ). where as before. But again. Let be the line passing through (x . y ) be a point of {ax + by > c}. and let be the line defined by ax + by = c . The slope of L is − a . a cannot 1 1 be 0 because if it were. y ). Multiplying both sides of the last inequality by a and making use of (δ).
From the picture. y L
L− 5 2 5 − 3 O
L+
x
Since L does not pass through the origin O. Theorem 1bis is directly applicable. Since it is not true that 3 · 0 − 2 · 0 < −5. O is not in {3x − 2y < −5}. {3x − 2y < −5} is a half-plane of L that does not contain O. Thus by Theorem 1. defined by −x − 2y = 4 is the following:
+
The graph of the line
rr
−4 rr r
O
rr rr
rr
−
−2 r rr
r
96
. If we ask for the graph of 3x − 2y ≤ −5. Example 3 Graph −x − 2y < 4 in the plane. So we know right away that {3x − 2y < −5} = L− . {3x − 2y < −5} has to be L− . then either O is in {3x − 2y < −5} or it is not. Of course there is no harm if you remember exactly what Theorem 1bis says and use that to decide which half-plane of L is {3x − 2y < −5}. as before. then we would get the closed half-plane L− ∪ L. Because the coefficient 3 of x in 3x − 2y = −5 is positive.
The line L defined by 3x − 2y = −5 is shown below.Example 2
Graph 3x − 2y < −5 in the plane.
The graph of −x − 2y < 4 and −2x + 3y > 0 is therefore the intersection L− ∩ + . Example 4 −2x + 3y > 0. and we also have (−2)(−3) + 3 · 0 > 0.We first check to see if the origin O belongs to {−x−2y < 4}: the inequality −0−2·0 < 4 being true. {−x − 2y < 4} is a half-plane of that contains O. Again.
97
. Find the graph of the pair of inequalities −x − 2y < 4 and
This example asks for the intersection of the graphs of the individual inequalities. Therefore by Theorem 1. Therefore by Theorem 1. Now by ( ). Thus {−x−2y < 4} = + as before. we see that O lies in + . the inequality x+2y > −4 is equivalent to −x−2y < 4. we can also directly appeal to Theorem 1bis. so that Theorem 1bis implies that {x + 2y > −4} = + . If we ask for the graph of −x−2y ≤ 4. It remains to determine the graph of −2x + 3y > 0. So {−x − 2y < 4} has to be + . then we would get the closed half-plane + ∪ . The equation −x−2y = 4 is the same as x + 2y = −4. which is the dotted region below without the two semi-infinite boundary line segments. We already know the graph of −x − 2y < 4 from Example 3. so (−3. Let L be the line defined by −2x + 3y = 0. Then the picture is the following: L L
−
−3 r −2 O
L+
The point (−3. Looking at the picture. we conclude that {−x − 2y < 4} = + . 0) is also in {−2x + 3y > 0}. 0) is in L− . {−2x + 3y > 0} = L− . O is in {−x − 2y < 4}.
−2x + 3y ≥ 0. to understand the behavior of the profit function P = 125x + 185y of the Manufacturing Problem. the graph would tend to be a polygon with the edges included. and is the same dotted region as above plus the two semi-infinite boundary line segments. i. and is therefore the following dotted triangular region together with the three edges. The graph of the two weak inequalities −x − 2y ≤ 4 and −2x + 3y ≥ 0 is now the intersection of the closed half-planes L− ∪ L and + ∪ . We can illustrate this phenomenon by elaborating on Example 4. In applications such as the Manufacturing Problem of this section. The graph of the three weak inequalities −x − 2y ≤ 4.e.q q rq r
q q q q r r
q q q q rq r
q q q q
q rq r rr
q q q L q q q q q q O r
q q q q
r
r
r
r
The graph in Example 4 is an "infinite region" in a sense that is self-explanatory (although "infinite" in this context can be precisely described in advanced mathematics).. Call this region S. and y ≤ 0
is the intersection of S with the closed lower half-plane of the x-axis. however. we have all the needed information to tackle task (ii) above. L
r rr q r rrq q q q q rr q O r q q rr rr rr r rr r
r
At this point. Recall that we have introduced the region R as the graph of the following four
98
.
y) in the region R. y0 ) of the Manufacturing Problem where the coordinates x0 and y0 are whole numbers.inequalities: x≥0 y≥0 x + y ≤ 50 75x + 165y ≤ 6000 Sometimes we say R is a region defined by these inequalities.e. it makes sense to talk about the profit at a point (x. y0 ) is called a maximum point of the profit function. y ) (which is 125x + 185y ). as a mathematical problem. in the sense that if (x . it makes perfect sense to see. we are able to consider the profit function P = 125x + 185y for any (x. Part 99
. then the profit P at (x0 .. In other words. we come up with a solution (x0 .27 A games and 13 B games. y0 ) of R the profit function P = 125x + 185y attains (or achieves) its maximum in R. what the value of 125x + 185y is so long as at the end. there is no such thing as the profit derived from 1. However. So by freeing x and y from the constraint of being only whole numbers. 125x + 185y) only if x and y are both whole numbers. for 7 any two numbers x and y regardless of whether or not they are whole numbers. 125x0 + 185y0 ≥ 125x + 185y The point (x0 . y) (i. After all. y ) is any other point of R. y0 ) (which is 125x0 + 185y0 ) is at least as big as the profit at (x . This graph is the dotted region below: y
d d d d d x + y = 50 —— d —— q —— d ——d q q —— — d — d—— d q d —— q 75x —— + 165y = 6000 q R d —— q q d —— q q —— d —— d — d O d
x
Our reformulation of the Manufacturing Problem now asks at which point (x0 . We pause to point out that in the context of the Manufacturing Problem.
where the coefficients a and b are the same in both L and P . y0 ) is a maximum point of P in R. and consider the general expression P = ax + by + e. i. ax + by = c − e. This is because if (x . y ) is a point on L. with a = 0 or b = 0. and symbolically we write this set as {P = c }. then by the very definition of L. y ) satisfies ax + by = c − e and by the definition of L. This is so because if P assigns c to (x . ap + bq + e = c . so that ax + by + e = c . and a point in R being a maximum point of P in R are defined similarly. then the line L consists of all the points to which P assigns the value c + e. e be three fixed numbers. then (x . To see this.
100
. To continue the discussion. Accordingly. we let a. Our first observation is that the line L defined by ax + by = c − e. With such a linear function P = ax + by + e understood. For example. (x . and the extension of the concept of "profit" to include 125x + 185y for any numbers x and y is a good example of the needed abstraction for the solution of many problems in algebra. b. Thus (x . y ) is a point to which P assigns the value c . y ) is a point on the line L. then ax + by + e = c . take a point (p. q) the value c . so that ax + by = c − e. we show first of all that P assigns to each point of L the value c . Then the concepts of a linear function attaining (or achieving) a maximum in R.of learning algebra includes learning when to take an abstract approach to a problem. y ) is already in L. y ) to which P assigns the same value c . which is the statement that P assigns to (x . which assigns to any point (x . y ) of R. y ) the value c . which is ax + by + e = c in disguise. we say P = ax + by + e achieves a maximum at a point (x0 . we always have ax0 + by0 + e ≥ ax + by + e. Naturally. Up to this point. We call such a P a linear function of two variables. we have only looked at a line L as the graph of a linear equation. Given a fixed number c .e. y ) in the plane the value ax + by + e.. then we may rephrase the above information in terms of our more familiar notation as follows: given a line ax + by = c and a linear function P = ax + by + e. such as ax + by = c. the corresponding concepts for a minimum in place of a maximum is similarly defined. including the Manufacturing Problem. Next we show that if (x . And of course. one refers to all the points to which P assigns this same value c as a level set of P . Now we get a different perspective: we may also look at L as the totality of all the points to which the linear function P = ax + by + e assigns the value where c + e. consists of all the points (x . q) and let P assign to (p. y0 ) in R if for any point (x . we will see that there is no advantage to gain by studying the specific profit function P = 1215x + 185y instead of something more general. (x0 . y ). If we let c = c − e in the preceding paragraph.
Y ) to which P assigns a value > c . By Theorem 1bis. and leave that relating to L− to an exercise. if a linear function P = ax + by + e achieve its maximum in R at a point (p. which is then the statement that P assigns to (x . Y ) must belong to L+ . Since L is the graph of ax + by = c. y ) a value > c . y ) belongs to {ax + by > c}.
101
. We can now translate Theorem 1bis into the language of the level set of a linear function P = ax + by + e. y0 ). and by item (γ) on inequalities. the level set of P that contains (x0 .Observe that. then (X. if P assigns the value c to a point (x0 . which is {ax + by > c − e}. Then the half-plane L+ consists of all the points to which P assigns values > c . y ) be a point of L+ . y0 ). Let (x0 . Theorem 2 immediately leads to our first conclusion concerning where a linear function P = ax + by + e can achieve its maximum in a region R. we have ax + by + e > c . q) of R. ax + by > c − e. y0 ). q). Again by (γ). First we show that P assigns to every point of L+ a value > c . we see that (X. Y ) belongs to L+ . If on the other hand P achieves its minimum in R at (p. This then completes the proof. where c = c − e. y0 ) be an arbitrary point in the plane and let P assign to it the value c . q). Let L be the line which is the level set of P containing (x0 . We finish the proof by showing that if there is any point (X. Thus we are given that aX + bY + e > c . Theorem 2 Given P = ax + by + e with a > 0. then R lies completely in the closed half-plane L− ∪ L of the level set L of P containing (p. We may therefore summarize our findings as follows: If a linear function P = ax + by + e assigns the value c to a point (x0 . by definition. where c = c − e. Let (x . we get aX + bY > c − e = c. then the level set of P containing (x0 . then R lies completely in the closed half-plane L+ ∪ L. We will concentrate on the statement relating to L+ . With a > 0. The proof of Theorem 2 is very simple indeed. Therefore. y0 ) is exactly {P = c }. and the half-plane L− consists of all the points to which P assigns values < c . y0 ) is the graph of ax + by = c. Theorem 1bis implies that (x .
q)e r e e e e e e e e e e e e e e e e e er er e e e e
L+
R
x
The statement about minimum is proved in the same way. and the coefficient of x in Q is now positive. which is then precisely the statement that Q achieves its minimum in R at (p. then P assigns to (s. Now suppose P achieves its maximum in R at (p. q). y ) in R. we consider instead the linear function Q = −(ax + by + e). We want to show that this is impossible. q). q) because: ap + bq + e ≥ ax + by + e for all (x . t) of L+ . It is instructive to give a proof of this assertion which makes use of the preceding one. we have: If a < 0. but this means Q achieves its minimum in R at (p. Then Q = (−a)x + (−b)y − e. t) a value exceeding the value it assigns to (p. If on the other hand P achieves its minimum at (p. q). q). t) e e e e e (p. We know P achieves its maximum in R at (p.This is because. and yet R contains a point (s . if a linear function P = ax + by + e achieve its maximum in R at a point (p.
102
. q). q) of R. q). then R lies completely in the closed half-plane L− ∪ L. Because the leading coefficient a is now negative rather than positive. by Theorem 2. We can therefore apply to Q what we have learned thus far. and therefore P does not achieve its maximum at (p. t ) of L− . as shown. as shown: L
e e e e e e e er(s. by item ( ) on inequalities. In a similar manner. if R contains a point (s. then R lies completely in the closed half-plane L+ ∪ L of the level set L of P containing (p. q). y ) in R means −(ap + bq + e) ≤ −(ax + by + e) for all (x .
then R lies completely in one of the two closed half-planes of L. a linear function P = ax + by + e cannot achieve a maximum or a minimum at an interior point of a region R. Let us call a point (p. The statement about minimum is proved in the same way.L
e e e e e e e (s . q). q) e e e e e e e e R e e e e e e e e er e r e e e e
L−
x
Let P assign to (p. q) would be horizontal. By item ( ) on inequalities again. First. in all the arguments. q) the value c a usual. q). q) of R.10 In particular then. respectively.
It may help for the understanding of this statement to recall that the definition of L− depends only on the fact that L is a line and has nothing to do amy linear functions P or Q. In particular. t ) a value smaller than −c .
10
103
. although we have always assumed in the above arguments that a > 0 or a < 0.. we have as + bt + e > c . Second. and this contradicts the given data that P achieves its maximum in R at (p. −(as +bt +e) < −c . q) is in R and is disjoint from the boundary of R. Then L is the level set {Q = −c } of Q. we know that L− consists of all the points to which Q assigns values < −c . Hence no such (s . t ) exists. Then: Theorem 3 If a linear function P = ax + by + e achieves a maximum or a minimum at a point (p. i. the case of a = 0 in P = ax + by + e (which is not excluded in Theorem 3) can be easily taken care of by noting that resulting level set L passing through a maximum point (p.e. Therefore. q) an interior point of R if (p. Then we simply replace the erstwhile L+ and L− by the upper and lower half-planes of L. where L is the line which is the level set of P passing through (p. Q assigns to (s . In view of what we have already discussed. only two minor comments on this theorem are called for. We can summarize the foregoing discussion as follows. t )er e e e e e e e r(p.
consider {x > 0} ∩ {y > 0}. so that this region does not contain its boundary. it merely asserts that if there is a maxiIf the region is the intersection of open half-planes.
11
104
. Theorem 3 has an important refinement. For regions which are intersections of closed half-planes such as the one arising from the Manufacturing Problem. then any line L passing through (p. q) would have the property that both of its open half-planes contain points of R. minimum) at a corner. the boundary of the region would not be part of the region.. For example.the reason for the last statement of Theorem 3 is that if (p. then it does so at a corner. q) is an interior point of a region R. y
d d d d d x + y = 50 —t —— d —— q — d ——d q q —— t d —— d—— d q d —— q 75x —— + 165y = 6000 q R d —— q q d —— q q —— d —— t dt — O d d
x
The refinement in question is: Theorem 4 Let a region R be an intersection of closed half-planes.11 The points of intersection of these boundary line segments are called corners. If a linear function P = ax + by + e achieves its maximum or minimum in R. This region is the first quadrant but without the two (positive) coordinate axes. Please note what the theorem does not say. It does not say that P only achieves its maximum (resp. What is important for our purpose about intersections of closed half-planes is that their boundary consists of straight line segments and that the boundary is part of the region. The corners of the region arising from the Manufacturing Problem are indicated with a dot in the picture below.
The case of a = 0 is simple and can be left to the reader.. q) of R. If (p. q) is already a corner and L contains no point of R except the corner. there would be nothing to prove. minimum point) in R for P . Suppose P achieves its maximum at a point (p. q) is not a corner. there can only be three posibilities for this to happen: either (p. the case of a minimum is entirely similar. Consider a typical such region:
3 33 f £ £ £ f f f £ f £ @ @ f @ £ @@@
x We will prove the theorem for the case a > 0. t) 3t L− 33 ff £ f (p. Because the boundary of R consists of line segments. q) 3t L− 33 ff £ f f £ £ f £ f £ f @@@f £ @@ @ f f
L
f f (s. t) and (u.mum point (resp. v) of the edge 105
. or (p. then the fact that L = {P = c } means P assigns to the corners (s. Let L be the level set {P = c } passing through (p. minimum point). q) is a corner but L also contains an edge of the boundary. q) is already a corner. The case of a < 0 is similar and will be left as an exercise (but see the proof Theorem 3). q). R must lie in the closed half-plane L− ∪ L. or (p. f t q) £ £ f £ f (u. £ f t v) @@@f £ @@@ f f
x
x
x
If (p. By Theorem 3.
L
t t (p. q) is not a corner but lies inside an edge of the boundary and then L contains this edge. as in the case on the right.. as shown. q) 3t L− 33tt f £ ft ft £ £ f t £ f t £ f @@@ t £ @@ @ t t t
L
f f (p. then a corner would already serve as a maximum point (resp. Proof We will deal with the case of a maximum. as in the case on the left and the case in the middle.
q) (and contained in L) the same value c . then it does so at a corner. as shown: y
q(x . Nothing in the statement of the theorem precludes the possibility that P achieves a maximum somewhere else too so that.. Consider for example the linear function P = x + y in the region R0 which is the graph of the pair of inequalities x ≥ 0 and y ≥ 0.
106
. the case on the right in no way contradicts Theorem 4. (u. The solution of the Manufacturing Problem is now relatively simple. but in fact (as we have already pointed out) what it says is merely that there is a corner at which P achieves a maximum. In the context of ordinary communication. as we know from the WYSIWYG characteristic of mathematics. t) (or for that matter. v)). Indeed. While this fact is fairly believable. a linear function will always achieve a maximum in R at a corner. if R is a finite region (in the sense of being contained inside some circle) which is the intersection of closed half-planes. its proof requires some advanced mathematics. then any linear function will always achieve a maximum in R.containing (p. the first quadrant including the positive coordinate axes. Theorem 4 would seem to imply that P achieves a maximum in R at a corner but nowhere else. However. this assertion is not even true. and therefore P also achieves its maximum in R at the corner (s. The case on the right points to the subtlety of the mathematical language. y ) increases without bound as (x . This proves Theorem 4. y
)
d
kd
d
{P = k}
d d d d k
O
x
Clearly the value P assigns to a point (x . A further comment is that one has to pay attention to the careful wording of Theorem 4: it says that if P achieves a maximum in R. but it does not say in any R that is the intersection of closed half-planes. and so P cannot achieve a maxiumum in R0 . i. y ) moves up in the direction of the upper right corner.e.
25). (0.2 y = 80. we get x0 = y0 = 25. So (x0 . In this case. 0).We are looking at the profit function P = 125x + 185y on the graph R of the following weak inequalities: x≥0 y≥0 x + y ≤ 50 75x + 165y ≤ 6000 We have already seen a picture of R:
1
y
d
50 d
d d
2
d
d d ——t d 4 —— 36 11 —— q — d ——d q —— (x0 . 1. (50. this R is called the feasibility region of the problem. The four corners of R are (0. (x0 . . y0 ). 50 107
. try out all possible profits with n A games and 50 − n B games for n = 0. R is a quadrilateral together with its interior bounded by the positive x. y0 ) = (25. . 36 4 ) 11
where every point is there for the obvious reason except for (x0 . . the line 1 defined by x + y = 50 and the line 2 defined by 75x + 165y = 6000. 2. y0 ). for example.and y-axes. y0 ) q dt —— d — d q d —— q —— q R d —— q q d —— q q —— d —— dt t 80— O d 50
x
In the context of the Manufacturing Problem. Now the profit in Manufacturing Problem must have a maximum. . which is the point of intersection of 1 and 2 and is therefore the solution (by Section 6) of the simultaneous system: x + y = 50 75x + 165y = 6000 Solving this system in the standard way (but note that a simplification can be achieved by reducing the second equation to x + 2. 0).
e. 0). then one of the nearby points (inside the feasibility region) with whole number coordinates. Thus the maximum point of the profit function is (25. This is because there is always the 4 possibility that if (0. 36 11 ) had to be accounted for even if 4 it makes no sense to talk about 36 11 B games. respectively. 25) is in the feasibility region of the problem. 25). which is precisely the statement about the manufacturing cost of these 50 games being at most $6000. In other situations. (0. 36) and (1. Graph the following inequalities in the plane: 5 3 ≤ 2 2x + 4y −2x + 3y ≤ 12 1 x−y ≤ 5 3
3. 7750. In 4 our case. consider the solution of 25 each of A and B games. 0). (a) Graph the inequality 3 x − (2 + 7x) ≥ (6 + x) − (1 − 2 x) on the number line. But the corner (0. 0. 36 11 ) is a maximum point.) To give some intuitive content to this way of approaching the problem. It remains to round off the preceding discussion by tying up a loose end. EXERCISES
2 1 1. may be a maximum point among those points with whole number coordinates. 35). (0. (50. one would need to check such a possibility. with x = 25 and y = 25. 0) in search of 4 the maximum possible profit. It is obvious that we could have left out any consideration of the corner (0. 4.. 36 11 ) are. i. Why did we not check to see if the $6000 manufacturing budget is enough to cover the production of 25 A games and 25 B games? This is because (25. 6250. The profits at 3 4 (0. so that in particular it is in the half-plane of 75x + 165y ≤ 6000. (See Problem 10 in the following Exercises. (b) Graph the inequality 2 − 1 x ≥ 1 x + 1 on the number line. (75 × 25) + (165 × 25) ≤ 6000.and the maximum is among these. and 6725 11 . 36 11 ) is not large enough. the fact that {ax+by < c} = L− .. it could be that 36 B games or 1 A game and 35 B games produces the maximum profit. Therefore. this did not happen because even the profit at (0. 5 2 5 6 2. 108
. Theorem 4 therefore tells us where to look for the maximum point: check the profit at each of the four corners above. 25). (25. Give a proof of the half of Theorem 2 concerning L− . i. and the solution is: manufacture exactly 25 A games and 25 B games to make the maximum profit of $7750. Give a proof of the other half of Theorem 1.e. Thus.
u. In some school mathematics and physics textbooks. and 300 mgs of protein.e.
8
Exponents and Absolute Value
So far we have dealt with linear equations.u. The maximum distance of the object from the sun. linear inequalities.. total dollar intake) assuming he can sell everything he produces? Use a scientific calculator.
13 12
110
. Linear objects are important because they are the basic building blocks of mathematics. 50 mgs of iron.) iron (mg) protein (mg) 50 4. How many 100 bushels of Crop A and Crop B should he produce in order to maximize his revenue (i. this law is stated using "mean distance" in place of "major axis". (c) He has to complete the harvesting of the crops in 200 hours.2 15
A mountain climber wants to bring enough of both items for her trip so that she would get at least 2700 calories. and the same amount of Crop B requires only 1 hour. It takes $125 to produce 100 bushels of Crop A and $95 to produce the same amount of Crop B. and linear functions of two variables. but life is often not linear. 1500 i.80.20 and each unit of Item B costs $2.(a) He has 150 acres of land. How many units of each should she buy so that the total cost is minimum and her nutritional requirements are met? Use a scientific calculator. Suppose business transactions are done only in terms of 100 bushels. It takes 1 acre to produce 100 bushels of Crop A but 2 acres to produce the same amount of Crop B. The nutritional values of a basic unit of two food items are tabulated below: A B calorie 156 112 vitamin C (i. 11. and that is an error. A good example is Kepler's famous Third Law governing the motion of an object around the sun: the square of the period12 divided by the cube of the so-called semi-major axis of the elliptic orbit13 is a fixed constant no
The time it takes the object to complete a revolution around the sun. (b) He has a capital of $9500. of vitamin C.1 13 200 1. Each 100 bushels of Crop A require 4 hours of work. Suppose each unit of Item A costs $4.
For example.. such as T 3 and even x−3/5 . in some vague sense. In symbols. by definition. As to the triviality of their proofs. n are positive integers.e. and m. . (α3 )5 = (ααα)5 = (ααα)(ααα)(ααα)(ααα)(ααα) = α5×3 = α3×5 111
.1) αm αn = αm+n (8. α. β. Here are the most basic facts concerning exponents: (8. this means there is a number C so that. Let us start from the beginning. Pluto). Unless stated to the contrary. T2 =C D3 Thus if the object is far from the sun compared with the earth (e. will stand for positive integers.1) says that..g.g. One also speaks of αn as raising α to the n-th power. What this means is that. . k. trivial to prove and "fun" to use due to their simplicity.. By multiplying both sides with D3 . These three facts are. These are the most basic nonlinear quantities. γ will always stand for positive numbers. any asteroid). α2 = αα. as well as the absolute value of a number. if T is the period and D is the semi-major axis of an object revolving around the sun. Recall that. For example. to progress further into mathematics.2) (αm )n = αmn (8.3) (αβ)m = αm β m where m. n. exponents are additive under multiplication. any meteor. (8. there is no doubt of that. we would have to deal with powers of numbers. any planet. . let us agree that for the remainder of this section.matter what the object may be (e. |x|. then it would take much more than a year for tht object to complete a revolution around the sun. simultaneously. α4 = αααα. i. and in general. αn = αα · · · α (n times)
The positive integer n is called the exponent or power of αn . we can rewrite this equality as T 2 − CD3 = 0 You can see that this is not a linear equation in T and D. α3 = ααα.
This heuristic argument is the same if 5 is replaced by any positive number α: if the equality α1 α0 = α1+0 makes sense and is correct. First. This doesn't matter to us because we are merely trying to make the right guess about what 50 ought to be.3) should still hold even when m. Multiplying both sides by 5 gives 1 ·5·50 = 1 ·5. such as it is.0)–(8. The boldness of this experiment therefore lies in the fact that we are assuming that we know the very thing we are trying to find out. What should α0 mean? What about fractions in the exponents? In fact.1)–(8.1) even when n is 0.The general proof of (8. we know what we want out of it: (8.3) are to remain meaningful and continue to be valid even when one or more of the exponents is allowed to be 0. 5 5 so that 50 = 1. As we have emphasized. So with this understood.3).1) with n = 0 gives: 51 · 50 = 51+0
1 In other words. We further assume that this object 50 that we know nothing about satisfies (8. the version of (8. is exactly the argument we gave above. then it is necessary to define 50 = 1. The presentation of the definition of α0 in some books gives the impression that one can prove α0 = 1. we now define the 0-th power of a positive numbex α to be α0 = 1
We pause to make a point. Right now we do not know what 50 means but for once will pretend that we do. and we are not claiming to be proving anything. The "proof". Having established the desirability of (8. we should have 5·50 = 5. To recapitulate: if (8.2) is almost identical.1)–(8. What it does is to give us confidence that we have probably made the correct definition to ensure the 112
. On the basis of this heuristic argument. or n is allowed to be 0. mathematics only deals with concepts with precise definitions. Therefore we shall perform a bold experiment by throwing out all the usual rules in mathematics. we may ask why something so good should be restricted to positive integer values of m and n. and the same is true of the proofs of the other two identities. It is therefore necessary to underscore once more the fact that the preceding heuristic argument is not a proof of why α0 = 1. what about rational numbers? We now approach these three questions systematically. what could 50 mean? While we do not know what meaning to give 50 . so that necessarily α0 = 1. then αα0 = α.
A good mathematics education sometimes has the beneficial effect of making you stop and think about things that you may have taken for granted all along. For example. So what gives us the confidence that there is a "square root of 2"? This is where mathematical knowledge is helpful by providing the answer we need.)
1 1 1 1
Note the emphasis throughout on the positivity of α and γ. and gain new understanding in the process.g. so that no matter how many decimal digits one writes down. in case α > 0.3) even when one or both of m. e.1421352 = 1. n is 0. that not only square roots.validity of (8. (Recall that γ n = γγ · · · γ (n times). Fractional exponent is next.1)–(8.3) again. then there is no number on the number line whose square is a negative number. There is a theorem. it will just be an approximation. we learn what it actually is and how to directly estimate it. something most of us have been familiar with since perhaps primary school. may yet be another such case. but any so-called n-th roots exist and are unique. by definition.4142135 . .1)–(8. We start with the simplest case: what could α 2 mean? We will be guided by (8. (Do you know why?) Moreover. How do we know that there is a number whose square is exactly 2? One can rattle off 1.99999091405925. proved in advanced courses. This is why we have to specify the positivity of γ in the preceding paragraph. Then given a positive number α. You recognize this as saying γ is the square root of 2. a positive number γ is said to be a positive n-root of α if γ n = α. 2 and −2. This is because if α = −2. α = 4. and (8. The "square root of 2". Precisely. namely. this says γ should be a number so that γ 2 = 2. A case in point for most of us is our first serious encounter with the number π.2) suggests that (α 2 )2 = α 2 2 = α If we write γ for α 2 . there is one and only one positive number γ so that γ n = α. let n be a positive integer. . 1. but one may also be aware that the decimal expansion of the square root of 2 is non-repeating. Theorem Given a positive number α and a positive integer n.. as that number.
113
.
Then the theorem that resolves all doubts in this context is the following. there will at least two numbers whose square is 2.
which we have just defined. 4 3 means the cube root of 4 raised to the fifth power.3) as a guide for the correct definition of rational exponents. for any positive integers m and n. Henceforth. if the analogue of (8. we must have (α 2 )2 = α 2 ·2 = α1 = α 1 for any positive α.. which says that there is at most one such γ. assuming that the analogue of (8. The third root of α is traditionally called its cube root. we feel sufficiently confident to define the n -th power of a positive number α to be √ 1 α n = the positive n-th root n α of α
3
The next question is: what should 2 2 mean? Again.1)–(8. by convention. and never −2. α 2 must be the positive square root of α. Note that the case n = 2 is distinguished and the notation for the positive square root √ √ √ is α rather than the more elaborate 2 α. By Theorem 1. more simply √ as the n-root of α. one message of the theorem is that there is such a thing as the positive square root of 2. Thus for any positive integer 1 n. the cube of 2 2 . This suggests that.2) for rational exponents is valid.It is to be remarked that the uniqueness part of the theorem. we shall refer to the γ in the theorem as the positive n-th root of α and. Please remember that n α is always √ positive. for any positive integer n. in general. The same 1 heuristic argument would yield the fact that. In any case. α n = (α n )m For example.e.2) is valid in general. 1 1 Thus. i. We are now in a position to resume our discussion of rational exponents. is actually not too difficult to prove. We will continue to use (8. we must have 2 2 = 2 2 ·3 = (2 2 )3 = 2 2 · 2 2 · 2 2
1 3 1 1 1 1 1
i. we define α n = αn αn · · · αn In other words. and for any positive number α. if there is no fear of confusion.. 43 = 43 · 43 · 43 · 43 · 43
5 1 1 1 1 1 5 m 1 m 1 1 1
( m times)
114
. The standard notation for the positive n-th root of α is n α. (α n )n = α. Therefore 4 = 2. but we will postpone this proof so as not to interrupt our discussion. 1 and therefore α n has to be the positive n-th root of α.e.
6) are. and for any rational numbers r and s.e.4) αr αs = αr+s (8.. α−A = 1 αA 4− 3 =
5 5 5 5
1
5
We have now finished defining αr for any positive number α and for any rational number r. we now claim that for any positive number α and β.We are almost through with the definitions! There is only one more to go: negative 5 exponents. the following are valid: (8. so if we insist that (8. positive rational number) A. Our more modest task here is to at least get a partial understanding of these laws for rational exponents. unsuitable for school mathematics.1)–(8.1) be valid for rational exponents.4)–(8.5) (αr )s = αrs (8.1) for rational exponents to obtain: 5 5 5 5 4− 3 · 4 3 = 4− 3 + 3 = 40 = 1 Thus we get 4− 3 · 4 3 = 1 Multiplying both sides by 1 43 we get 43 5 There is nothing special about either the number 4 or the exponent − 3 . rational or irrational. Nevertheless.6) are 115
.4)–(8. What should 4− 3 mean? We appeal to the analogue of (8. These proofs are as unpleasant as the proofs of (8. we should define for any positive number α and for any fraction (i. students should at least have a glimpse of why (8. and prove these laws in one fell swoop for all numbers r and s using sophisticated reasoning. our goal is rather to generalize (8.6) (αβ)r = αr β r These are the laws of exponents.1)–(8. But our work has barely begun! We do not want to get a general definition of αr just to satisfy our idle curiosity. In advanced courses. With this in mind. it is shown how to define αr for any number r.3) to rational exponents. especially grade eight. without a doubt.3) are trivial. The complete proofs of (8.
either because it is not as central or not as comprehensive as others already designated as theorems. because they will be using these laws often and it is altogether a bad idea if they are total strangers to any basic tools they use. We suggest first of all to check a few simple cases. then for any positive integer n. so 64 3 = (64 3 )2 = 42 = 3 64 2 · 64 3 = 8 · 16 = 128
1 2
1
2
1
On the other hand. Proof Suppose α < β. (In the tradition of Euclid. a lemma is a theorem that is deemed to be of a lower status. αn < β n . such as the following: 1 3 1 5− 2 · 5− 4 = 5− 4 1 2 (11) (52 ) 3 = 5 3 1 1 1 (αβ) n = α n β n when the last equality is for all positive numbers α and β and for all positive integer n. Also 64 = 43 .6).) Lemma 1 If two positive numbers α and β satisfy α < β. Thus
7 6
=
1 2
+ 2 . all prospective teachers should at least go through the complete proof of at least one of these laws. It is sometimes a very subjective judgment whether something should be a lemma or a theorem. so also 64 6 = (64 6 )7 = 27 = 128 and we are done. so 64 2 = 8.4)–(8. Let us consider the problem of how to give students a glimpse of the reasoning behind the laws of exponents. then they should at least know whereof they speak. Now 64 = 82 . We also suggest giving a proof of a special case of each of (8. 64 = 26 .true. For example: we check 2 7 1 64 2 · 64 3 = 64 6 Note that 16. then (because α > 0). because if they will be telling their students that these proof are too difficult. The proofs of all three items in (11) make use of the following useful lemma. multiplying both sides by α gives α2 < αβ 116
7 1
. On the other hand.
As to the deduction of Corollary 1 from Lemma 1. Now we multiply both sides of α < β by β to get αβ < β 2 Combining the preceding two displayed inequalities. we get α3 < β 3 . but 3 > (−4). then α < β. This completes the proof. Multiplying both sides of α2 < β 2 by α. we get α4 < αβ 3 Now multiply both sides of the just-proven inequality αβ 2 < β 3 by β and we get αβ 3 < β 4 Combining the last two displayed inequalities. we make use of what is called the trichotomy law among numbers. which states that for any two numbers a and b. For example. we get α2 < β 2 . 32 < (−4)2 . if αn < β n for some positive integer n. we get α3 < αβ 2 Now we make use of the just-proven inequality αβ < β 2 : multiply both sides by β to get αβ 2 < β 3 Combining the preceding two displayed inequalities. or a < b.(see item (δ) of Section 7). we want to show next that α4 < β 4 . let us note first of all what it does not say: it does not say that if any two numbers α and β satisfy αn < β n for some positive integer n. Corollary 1 For two positive numbers α and β. one and only one of the three possibilities holds: either a = b. Multiplying both sides of α3 < β 3 by α. prove that α3 < β 3 . The following two corollaries (immediately drawn conclusions) of Lemma 1 are of independent interest. we will eventually get to αn < β n no matter what n is. so it is a curious fact that it is also a corollary of Lemma 1. Before explaining why. This is of course the converse of Lemma 1. we get α4 > β 4 . Continuing one more step. or a > b. 117 Next we will
. then α < β. In this way. So the truth of Corollary 1 depends critically on the positivity of both α and β.
+nk (Incidentally, we need the explicit formula m + k = m n here, and the use of this n formula in this setting may be as good a reason as any as to why one should not follow the common method of defining fraction addition in terms of the LCM of the denominators n and .) By Corollary 2, it suffices to prove
which is exactly (a). The proof of case (c) is simpler. By (12), we know αn α = αn+
m k m k
By the cross-multiplication algorithm, this is equivalent to 1 α α which is of course the same as α− n α− = α− n − The proof of (8.4) is complete. Remark We have chosen to give a complete proof of (8.4) because until one goes through such a proof, one doesn't know what it means that the proofs of (8.4)–(8.6) are unpleasant. Now in mathematics, when things get unpleasant, 90% of the time it is because they are not done "the right way". In this case, the right way is to prove (8.4)– (8.6) for all number r, s all at once without restricting oneself to only rational values of r and s. Such a proof is achieved by appealing to basic properties of the exponential and logarithmic functions and using differentiation. This will be done in advanced courses. For our task at hand, it remains to emphasize that you should feel free to use (8.4)–(8.6) for all values of r and s and worry about their proofs later. 122
m k m k m n k
=
1 α
m +k n
We will put (8.4)–(8.6) to use in the next section. Before we leave the discussion of exponents, we would like to expand on the meaning of an expression as defined in Section 1. Recall that an expression (or number expression) is simply a collection of numbers x, y, etc. connected by the four arithmetic operations. Now that exponents are available, we can add to the meaning of an expression by defining it to mean a collection of numbers x, y, etc. which are connected by the four arithmetic operations and the use of rational exponents. In this context, an expression is also called an algebraic expression. Thus, the following is an algebraic expression: x−3 + {(yz)2 + 5} 4 − (
3
xy 5 ) z
We now introduce the concept of absolute value of a number. For any number x, the absolute value |x| of x is by definition: |x| = the distance of x from 0 Thus |x| = x −x if x ≥ 0 if x < 0
For example, | − 7 | = 7 , |1728| = 1728, and | − 62.9| = 62.9. 6 6 Note in particular that |x| ≥ 0 for every number x, and |x| = 0 is equivalent to x = 0. The relevance of absolute value in the context of exponents lies in the following alternate definition of the absolute value |x| of a number x: |x| = √ x2
The fact that this is equivalent to the original definition is easy to see. A question often asked is, "why bother with absolute value?" A superficial answer is that very often one wishes to say that a number x is "close" to another number, and the concept of absolute value allows us to express this "closeness" in a simple manner. For example, a common expression is that "today's temperature is about 60 (degrees Fahrenheit)". It means the temperature of today, say x degrees, is at most 5 degree more than 60 or 5 degrees less than 60. Thus, 123
suppose x satisfies the Second Expression. then |x − 60| ≤ 5. then 60 − x ≤ 5. If x ≥ 60. We want to show that they are equivalent. |x − c| is the distance between x and c on the number line. then we will show it satisfies the Second Expression. we have |x − 60| ≤ 5 no matter what x is. as shown: c x x c
124
.if x ≥ 60. then |x − 60| = x − 60. Thus we see that x satisfies the First Expression. and we isolate the two key points. The idea behind the preceding discussion has a wider application. and then the case x ≤ c. let us refer to these two ways of expressing "today's temperature is about 60" as the First Expression and Second Expression. Altogether. We now show that one could express this much more simply as: |x − 60| ≤ 5 For convenience of discussion. then |x − 60| = 60 − x and again we know 60 − x ≤ 5 in this case. Conversely. respectively. then |x − 60| = x − 60 and therefore x − 60 ≤ 5. then |x − 60| = 60 − x and the fact that |x − 60| ≤ 5 now becomes the statement that 60 − x ≤ 5. So suppose x satisfies the First Expression. Then the first is that for any two numbers x and c. by first considering the case that x ≥ c. On the number line. If x ≥ 60. then x − 60 ≤ 5. This is easily seen. and if x < 60. If x < 60. Define the distance between two numbers on the number line as the length of the segment joining them. Thus also |x − 60| ≤ 5. x < 60. If on the other hand. So |x − 60| ≤ 5 in this case. as in the preceding discussion. and we know x − 60 ≤ 5 from the First Expression. an x satisfying the First Expression must stay within the thickened line segment below:
'
x 60
E
55
65
Now consider |x − 60|.
c be arbitrary numbers and let be a positive number. A second example is the inequality. But then this is exactly the content of the double inequality c − ≤ x ≤ c + . also valid for all numbers x and y: |2xy| ≤ x2 + y 2 125
. b. Then |x − c| ≤ is equivalent to the double inequality c − ≤ x ≤ c + .
'
x c
E
c−
c+
We now give a more satisfactory answer to the question of "why bother with absolute value?" It is because in mathematics the distance of a number x (regardless of whether x is positive or negative) from 0 is very often an issue of intense interest. A second key point is to point out that an inequality involving absolute value can be expressed directly in terms of ordinary inequalities.e.. i. We introduce a new notation for this purpose: if three numbers a. then allows for precise computations with this symbol to estimate this distance. We give two elementary examples to illustrate this kind of computation.Thus to say "x is within units of c". and can move up only as far as c + . we can simply say: |x − c| ≤ . c satisfy the following inequalities: a ≤ b. Having a symbol to denote this distance. The latter means that x can move down the number line only as far as c − . We have seen that the inequality |x − c| ≤ is the statement that the distance of x from c is at most . The proof is best achieved through the use of a picture of the number line. |x|. and b≤c
then we abbreviate these inequalities into a double inequality: a≤b≤c We now prove: Lemma 3 Let x. The first is the inequality |x + y| ≤ |x| + |y| for all number x and y.
most of the important inequalities are those involving absolute values. Notice that the second inequality actually compares two absolute values. 2 2
EXERCISES
126
. Problem 6 (iv) of the following Exercises). From the picture. i.
r −3 1 2
−3
r 1 −2 2
we see that the graph is the union of two semi-infinite segments: the segment to the left 1 1 of −3 2 and including −3 1 . For example. |2xy| ≤ |x2 + y 2 | One gains an appreciation of the importance of absolute values if one experiments with ways of formulating inequalities like these without having absolute values at one's disposal. In Example 1 of Section 7. with the only difference that we now make use of absolute value. we graphed an inequality on the number line. which in turn is the same as 1 |3 + x| > 2 (cf.. but nothing further to the left on the 2 number line. the inequality |2xy| ≤ |x2 + y 2 | says this does not happen because the product 2|xy| must satisfy |2xy| ≤ 1 so that the product xy can be as far from 0 as − 1 .Imagine how clumsy these inequalities would be if the concept of absolute value were not available. or even −150. Which carries more information? The inequality 2xy ≤ x2 + y 2 does not exclude the possibility that two numbers x and y satisfy x2 + y 2 = 1 but their product xy is −2. However. compare an inequality such as 2xy ≤ x2 + y 2 with the preceding one.
1 1 The inequality is the same as 2 |6 + 2x| > 2 · 1.e. Since |3 + x| = |x − (−3)|. Example Graph |6 + 2x| > 1 on the number line. or −3. In advanced mathematics. This means we have to find all 1 the points x so that its distance from −3 is bigger than 2 . 1 the inequality is therefore the same as |x − (−3)| > 2 . Here is more of the same. and the segment to the right of −2 1 and including −2 2 .
(ii) 3 1 1 |x| − 4 < 3 . 7. 2. (iii) 9 − |3x − 1| < 4. and for any positive number α. but without using (8. Verify each of the following about absolute values of arbitrary numbers x. Given two similar triangles as shown:
£e £ e £ e
a£
£e £ e £ e e e e £ £ £ £
£
e e e e e e e
a££
£ £
If the ratio of the area of the smaller triangle to the area of the bigger triangle is s.5). Do the same to 1 2 1 117649 2 · 117649 6 = 117649 3 m m 1 1 4. 8. Graph on the number line each of the following: (i) |x| − 14 > −8. Prove that for all rational numbers r. what a is the ratio a in terms of s ? 9. Prove: |2xy| ≤ x2 + y 2 for all numbers x and y. Explain why there is no number on the number line whose square is a negative number. (ii) −|x| ≤ x ≤ |x|.1. s and t. (v) |6x + 1| + 2 1 < 5. 10. (iii) |x + y| ≤ |x| + |y|. y: (i) | − x| = |x|. 5. 4
127
. Prove using only the definition of α n . Prove that for any rational number r. (iv) |2x + 5 | ≥ 5 . Verify the equality 729 2 · 729 3 = 729 6 by direct computation. Derive a formula which gives the amount of money in an account at the end of n years if the account has an initial deposit of P dollars and an annual interest rate of x percent. α−r = 1 αr
6. Recall that an annual interest rate of x percent means that an account of P dollars earns at the end of one full year an amount of x 100 P
dollars. (iv) |xy| = |x| · |y|. (αr β s )t = αrt β st . 1 5 1 3. that (α n ) m = α n .
For the moment. We note for emphasis that F (x) is always ≥ 0 for any x. if F is the function F : {all numbers} → {all numbers} which assigns to each number its square. heart. Functions are needed to describe the various processes and phenomena that occur in the human and the natural worlds. yet. then f : A → B is the correct notation to capsulize this information. Both worlds are in a state of change. to these common cases. the kind of functions one encounters are usually those from a set of numbers or a set of points in the plane to some other set of numbers. Suppose you have just 128
. then we write f (a) = b For example. we want to give a general discussion and would not want to limit ourselves. If G is the function G : {a deck of cards} → {club. If the function is denoted by f . We will give three examples. so if we write instead F : {all numbers} → {all numbers ≥ 0} then it would also be correct. then F can be succinctly given as F (x) = x2 for each number x.. However. etc. and functions are the tools to describe change. First we want to say a few words about why we study functions. spade} which assigns to each card its suit. the function f is often denoted generically by f (x). a precise prescription) that assigns to each element of A an element of B. then what G does to each card would be difficult to describe in symbols.e. In a course on introductory algebra. such as G(King of diamonds) = diamond G(Two of spades)= spade G(Queen of hearts) = heart. however. If f assigns the element b of B to an element a of the set A. One can illustrate by giving some examples.9
Functions and Their Graphs
A function from a set A to a set B is a rule (i. when A and B are understood. diamond.
Thus f is a function which assigns to every number t ≥ 0 another number.e. we see that 4. Here then is an alternative: denote by f (t) the temperature of the coffee (in degrees Fahrenheit) t minutes after the coffee is brewed. In this case. the temperature is 137◦ F.8 4. namely. it is not a single number. it is 165◦ . In another minute. the temperature of the coffee t minutes after it is brewed. The following table then summarizes every piece of information in the last paragraph: t 0 1 2 3 4 5 f (t) 195 180 165 153 143 135
Of course. respectively. Finally afer five minutes. but an infinite collection of numbers..8 f (t) 172 159 148 145 139 137
Reading from the second table.5 3. it drops to 135◦ and it becomes borderline drinkable. A minute later.8 minutes (i. For example: t 1. Now this verbal description of the change in temperature is cumbersome and totally unsuitable for scientific investigations. the temperature drops to 180◦ .5 3. i. "the temperature" is a number that depends on the time of the measurement. it is important to render this information into symbolic form so that we can compute with it. As we have emphasized all along.5 2.brewed a cup of coffee and are waiting for it to cool down. So you watch out for the temperature of the coffee. 4 minutes and 48 seconds) after the coffee is brewed.e. each corresponding to a particular time of measurement. one can get much more information about the temperature of the coffee by measuring it at shorter time intervals. in succeeding minutes it is 153◦ and 143◦ . Let us say it is 195◦ (Fahrenheit) at the beginning.5 4.
129
..
He manages to get home at 43 minutes after his departure and it takes him only about a minute to get the necessary document. As a final example. so that after 10 minutes he only travels two and a half miles.5 13 7. clearly one number won't get the job done because this distance depends on the time when the distance is measured. consider the problem of the temperature of the city of Berkeley on a certain day.5 0
We can see that he has to start his trip slowly probably because of city traffic. he turns around as the values of f (27) and f (28) and those of subsequent minutes show that he is driving away from the airport. Around the 26-th minute after he leaves home. Even a skeletal description of this function in terms of a few values of t can tell a story. strictly 130
. not trivial considering the traffic condition these days. He has a few minutes to spare. If we want to see how far he is from the airport. as for instance: t 0 5 10 15 20 25 26 27 28 f (t) 25 24 22.
Consider a second example: a man drives to the airport which is 25 miles away. In general. Then he speeds a bit as he makes it to the airport in 23 minutes (67 − 44 = 23). He plans to leave his house two hours before departure time. F assigns to each number t ≥ 0 another number which is his distance in miles from the airport at time t. Our experience with the coffee problem suggests that we make use of a function F for this description. He forgets to bring his photo-ID (a guess!). such that F (t) = his distance (in miles) from the airport t minutes after he leaves his house Thus F (0) = 25. To say that Berkeley is 67◦ (Fahrenheit) makes no sense.5 21 16 10 9 10 11 t 30 35 40 43 44 45 55 60 67 f (t) 13 19 24 25 25 24.f : {all number ≥ 0} → {all numbers} f (t) = the temperature of the coffee t minutes after it is brewed.
we need two more numbers which may be thought of as the idealized x and y coordinates. t) = the temperature of Berkeley. a scientifically usable description of the temperature of Berkeley would make use of a function T .e. in the Manufacturing Problem of Section 7. |y| ≤ 5 (miles). y) These are among the simplest examples of how functions naturally arise. (Compare the discussion of the term "variable" given at the beginning of Section 2. y. these are examples of function T . Therefore.speaking. then 0 ≤ t ≤ 24. without holding forth on the philosophical implications of what a "variable" is. i. and translations. Transformations of the plane. And where is the temperature measured: at the top of the hill (about 1500 feet high). y) = (x − 2. Is the temperature taken in the early dawn or in the afternoon? In Berkeley. See Lemma 3 of Section 8. this could mean a 25◦ difference. This is as it should be. or by the Bay? The difference here could be another 15◦ . y. t hours past midnight at a spot specified by the x and y coordinates Incidentally. including rotations. the translation T which moves every point of the plane 2 units to the left horizontally is precisely given by: T (x.) Quite apart from the description of change. are functions which assign to each point of the plane another point of the plane. To specify the geographic location. For example. y. then14 F : S → {all numbers} F (x. It may not have escaped your attention that we give F this name as an afterthought. and 0 ≤ t ≤ 24 (hours). if S is the region in 3-space consisting of all ordered triples of numbers (x. downtown. so that. such a function F is said to be a function of three variables. we had to use the profit function P to describe the possible
Notice also the use of absolute value to describe the physical extent of the city. Berkeley being a small city. because three numbers x. It means of course that −5 ≤ x ≤ 5 and −5 ≤ y ≤ 5. t) so that x and y satisfy |x|. functions have already forced their way into our work whether we know it or not.
14
131
. so that T : {the plane} → {the plane} For example. reflections. Of course functions are everywhere as soon as you look around. and t are involved in its definition.. If we start measuring the time t in hours from midnight. 5 miles from the city center in any direction would include everything.
In principle. (Back then we slipped in the nomenclature of a "function" without warning. of course. we can improve on the notation as follows: if the game manufacturer makes x A games and y B games. so that P (28. y). 22) = 125 × 28 + 185 × 22 = 7570. Precisely. y ) in the feasibility region R the number 125x + 185y . H assigns to it the number H(x. we already came across a special case of this concept. y) = 125x + 185y. created out of necessity. let f be a function from a set of numbers A to a set of numbers B. It should be stated that plenty of practice in plotting points on a graph is an essential component in the learning process of getting to know graphs and functions. Then the graph of f is the set of all the points (x. for any point (x. i. f (x)) in the plane. Then P becomes a function P : R → {all numbers} P (x . y ) = 125x + 185y
The purpose of this discussion is to make it plain that the concept of a function is not something artifically concocted for the purpose of giving students a hard time. We already saw in Section 8 that it was to our advantage to expand the role of P by allowing it to assign to any point (x . In Section 7.profits of the game manufacturer in various scenarios. it is a tool. to effectively describe the phenomena around us be they sociological or mathematical. f : A → B. 132
. In general. Thus the profit of selling 28 A games and 22 B games is $7570. we introduce a few related concepts. Rather. So please remember: use the graphing calculator only after you have acquired fluency in plotting points. the set A is infinite. y) = the profit she makes by selling x A games and y B games. Such a function H is called a function of two variables. Before we give some examples. where x is an element of A. Recall that P (x.) Now that we know the language and notation of a function a bit better. It is indispensable. y) in the plane. so that the graph of f is an infinite set as well. then P (x.. Suppose we are given a function H from the plane to the set of all numbers. and therefore of the function itself.e. it is usually the case that plotting a finite number of well-chosen points in the graph is enough to reveal the essential features of the graph. it is impossible to literally get hold of the whole graph of any function. The best way to get to know the concept of a function is to look at many examples of functions and to examine their graphs. However.
3) are on the graph. y ) so that H(x . This fact will be useful below. a linear function of two variables. y) being in the graph of f itself.g. such as g1 (x) = 2x + 3 (thus a = 2. y0 ) and (x0 . 1) (1. f : {all numbers} → {all numbers} Define a function of two variables H0 by H0 (x. Where these concepts of graphs come together is the following. Once these "obvious" points have been plotted.namely. 9)
133
. y) = (x. e. Example 1 A linear function of one variable g is by definition a function from numbers to numbers so that for some fixed constants a and b. The last is of course equivalent to (x. −1) (−1. f (x)).. what are the values of x0 and y0 ? Since all points on the graph of g1 are of the form (x. y) = y − f (x) for all x and y. we get the following points on the graph of g1 : (−2. 3 3 Therefore y0 = 3 and x0 = − 2 . b = 3). 0) and (0. 2x + 3). 1. y) being in the graph of H0 = 0 is equivalent to y − f (x) = 0. This is because (x. 5) (2. 3. letting x = −2. Then {the graph of f } = {the graph of the equation H0 = 0 }. 0) are on the graph. First look at a special case. y ) = 0. Suppose f is a function from numbers to numbers. if (0. which is then equivalent to (x. Imagine that we are approaching the graph of g1 for the first time. The graph of the equation H = 0 is the set of all the points (x . 7) (3. we see that y0 = 2 × 0 + 3 and 2x0 + 3 = 0. which points should we plot? We should certainly find out where the graph crosses the coordinate axes. so that (− 2 . g(x) = ax + b for all numbers x. which is equivalent to y = f (x). we pick some other points in random. −1. 2. The graph of g is therefore the set of all points (x. In other words. ax + b).
we know in addition that this line has slope a and y-intercept b. which is in turn the same as (−a)x + y = b. Then the same reasoning shows that the graph of the linear function of one variable f is the same as the graph of the linear equation of two variables (−a)x + y = b.
134
. This is why the graph of g1 is a straight line. We already know from Section 4 that the latter is a straight line. Example 2 Graph the function s(x) = x2 .e.. The general case is no different: Given f (x) = ax + b. one would say that the graph of g1 is a straight line. and is therefore a line. Thus the graph of the linear function of one variable g1 is the same as the graph of the linear equation of two variables (−a)x + y = b (which is of course the same as the equation (−a)x + y − b = 0). By Section 4. Activity 1 problem above. This turns out to be correct because the graph of g1 is the same as the graph of y = g1 (x).8 q
q
7 q
5 q3 q q −2 −1 q
1 O 1 2 3
If one has to guess on the basis of these seven points. Thus the graph of a linear function of one variable is always a line in the plane. Plot the given points on the graph of the function f in the coffee
Activity 2 Plot the given points on the graph of the function F in the problem of the man driving to the airport. i. y = ax + b.
The graph of s consists of all the points of the form (x, x2 ), where x is arbitrary. Since (−x)2 = x2 , we see that the graph includes both (x, x2 ) and (−x, x2 ), no matter what x may be. The point (0, 0) is the obvious point on the graph. We can put in values of x = ±1, ±2, ±3, ±4 to get the points (±1, 1) Let us also throw in (±0.5, 0.125) (±1.5, 2.25) (±2.5, 6.25) (±3.5, 12.25) (±2, 4) (±3, 9) (±4, 16)
for good measure, and we get the following sequence of points on the graph of s. Note that in order to make the picture manageable, we have shrunk the scale of the y-axis by half.
q q q q q q −4 −2 q 2 q q q q q 2 4 10 q q q 16 14 q q
6
O
It is not difficult to extrapolate from these points to envision the graph as the following curve:
135
This curve is an example of what is called a parabola. Parabolas are discussed more fully in Sections 11 and 12. We emphasize once more that it is the plotting of points on the graph, and the ability to envision the whole graph on the basis of these points, that are our major concern here. The beginning stage of learning about graph is best done being away from a graphing calculator. Example 3
1 Graph the function h given by h(x) = x .
We first address an issue concerning this h that we have not confronted thus far, and it is this. Would it be correct to say that h is a function from all numbers to all numbers? In other words, would it be correct to say that h : {all numbers} → {all numbers} The answer is no, because h cannot assign any number to 0, as infinity is not a number. So the correct statement is that h : {all nonzero numbers} → {all numbers} That said, we start plotting points. Again, there are two obvious points: (1, 1) and 1 (−1, −1). Beyond that we take some random values of x and compute x , and we will
Notice that there are two separate curves here, and they are called the two branches of a hyperbola. Regrettably, we will not pursue the study of hyperbolas here. 1 The plotted points above exhibit a pattern: if 0 < a < b or a < b < 0, then a > 1 . b (In an exercise, you are asked to prove this in general.) This pattern justifies that this limited choice of the points on the graph is enough to reveal the general behavior of the graph: it tells us that as the upper right curve extends to the right end of the positive x-axis, all it does is to get closer and closer to the x-axis, and as it approaches 0 from the positive x-direction, all it does is to get closer and closer to the positive y-axis. A similar statement also applies to the lower left curve. Example 4 Graph the function given by f (x) = √ x.
We note, as in Example 3, that this is not a function from all numbers to all numbers, but f : { all numbers ≥ 0} → { all numbers} 137
8) (−2. G(±3). √ √ then a < b. a function from all numbers to all numbers. G(±4). at the same time. This fact then tells us that there is no need to see more points on the graph because as we go towards the right end of the positive x-axis.g. getting the following points on the graph of G: (−4. Example 5 Graph the function G given by G(x) = x3 − 3x + 6. −48) (1. 4) (−3.. G(0). 4) (3. x > x.The following sequence of points on the graph of f are self-explanatory: 4 3 2 1
q q q qqq q q q q q q q q
q
q
O
1
4
9
16
√ This sequence of points exhibit a different pattern: when 0 < x < 1. the graph simply rises in the same way as it does here for the values of 1 ≤ x ≤ 16.) Note also that if 0 < a < b. (This too will be an exercise. 6)
By compressing the y-axis. G(±1). x < x. and √ when 1 < x. we can exhibit these points as follows:
138
. −12) (2. 58) (0. 24) (−1. we try some obvious numbers. e. G(±2).
This is a cubic polynomial and. 8) (4. Since we have no idea what to expect.
so it is intuitively clear that x0 is between −2. then 0 = G(x0 ) by definition of the graph of G. The roots of a polynomial equation are of great interest in mathematics (see the last two sections on the roots of quadratic polynomials equations). and continue to go up as we go to the right of the x-axis. etc. we already know that there are no further troughs and bumps below x = −4 and above x = 4. 0). If we were less fortunate and the chosen points happen not to reveal the bump and the trough of the graph. 0 3 Such an x0 is called a root of the cubic polynomial equation x − 3x + 6 = 0. G(−2. that the graph of a cubic polynomial can have at most one bump and one trough. It is a known fact. This means x3 − 3x0 + 6 = 0. Notice that the graph has a bump above (roughly) −1.3) = 0. For this reason.
139
. So the graph will continue to go down as we go to the left x-axis. one may try to get a better estimate of this x0 . We have G(−2. Suppose it crosses the x-axis at (x0 .5) = −2..125.31). Therefore we are very fortunate that with the choice of the nine obvious points on this graph. we can get even better estimates of this root.624.3.4 and −2. proved in advanced courses.73. and has a trough above (roughly) 1. By experimenting with G(−2. then we would have to plot more points.60 50 40 30 20 q q q 10 q q 2 3 q
q
−4 −3 −2 −1 q
O 1 −10 −20 −30 −40
4
q
−50 −60
The graph seems to cross the x-axis between −3 and −2.4) = −0. G(−2. since these features may not have appeared yet or may even not exist.
we therefore have: average speed from t1 to t2 = f (t2 ) − f (t1 ) t2 − t1
Average speed therefore measures miles per hour (mph). and this is because for all rational numbers r and s so that r < s. For our purpose. where f (t) = the distance traveled from time 0 to time t Let us say t is measured in hours and f (t) in miles. An object in motion is described by a function f . especially Examples 4–8 and make better sense of that discussion. let us consider the case of motion. we are now in a position to revisit the earlier discussion of rates and constant rates in Section 5. The proof of a more general fact is left as an exercise. or the functions αx for a positive α. Finally. For definiteness. they give a hint of the reason why we want to devote a whole section to the discussion of rational exponents. We do not have the time here to pursue the study of this function H. We adopt the usual convention of letting time 0 be the starting time and that f (0) = 0 by definition. we say the motion has constant speed v mph if the average speed over any time interval is the fixed number 141
. Formally introduce the concept of average speed during the time interval from time t1 to time t2 (0 ≤ t1 < t2 ) as the distance traveled from time t1 to time t2 divided by the length of the time interval (which is of course t2 − t1 ). It is worthwhile to mention nevertheless that these functions are important in mathematics and the sciences. we have 2r < 2s . In terms of f .16 14 12 10 8 6 4 q −4 q q q q q q q −3 −2 −1 q 2 q q 1 q q q q q
q
O
2
3
4
Notice that the points rise steadily to the right. With the concept of a function available.
the fact that the motion has constant speed v mph is equivalent to the fact that f is given by f (t) = vt miles for all t ≥ 0. f (t2 ) − f (t1 ) =v t2 − t1 for any t1 and t2 so that 0 ≤ t1 < t2 . let F be the function so that F (t) is the amount of water (in gallons) coming out of the faucet from time 0 to time t (in minutes). work done at a constant rate. average speed from t1 to t2 = f (t2 ) − f (t1 ) t2 − t1 vt2 − vt1 = t2 − t1 v(t2 − t1 ) = t2 − t1 = v
This then proves the theorem. by definition. in case of water flow from a faucet (let us say). an entirely similar discussion can be given for water flow at a constant rate. Proof First suppose the motion has constant speed v. it would be good to point out that the cancellation of (t2 − t1 ) in the last calculation depends on the fact that it is a nonzero number. For example. we have f (t) = vt for all t. The rate of flow is constant. Theorem Notation as above. which explains why we were so careful to specify that t1 < t2 . For your beginning students in algebra. By letting t1 = 0 and let t2 be arbitrary. suppose f (t) = vt. we get f (t2 ) =v t2 which is the same as f (t2 ) = vt2 Since t2 is arbitrary.v. Conversely. Then by definition. Here is the basic observation. By definition. etc. if there is a fixed number r so that during any time interval from t1 to t2 (0 ≤ t1 < t2 ) the total amount of water coming out of the faucet in this time interval
142
. Naturally.
and m that the number of square feet that each of Joshua. one can also appreciate the importance of being able to transcribe verbal information into symbolic expressions (Section 2). One then proves in exactly the same way that the rate of water flow being a constant r gallons per minute is equivalent to F (t) = rt gallons for all t ≥ 0. respectively. . and 16 hours. One then proves as before that the rate of painting the house being c square feet per minute is equivalent to H(t) = ct square feet. to paint the whole house. We now do a prototypical "rate problem" from the perspective of linear functions. This problem can be done without algebra. if there is a fixed number c so that during any time interval from t1 to t2 (0 ≤ t1 < t2 ) the total number of square feet painted in this time interval divided by the length of the time interval is always equal to c square feet per minute. how long will it take them to do it together? By the constant rate assumption. Li and Manfred paints in t hours are. H(t2 ) − H(t1 ) =c t2 − t1 for all t1 and t2 so that 0 ≤ t1 < t2 . In the ensuing discussion. respectively. It is estimated that. F (t2 ) − F (t1 ) =r t2 − t1 for all t1 and t2 so that 0 ≤ t1 < t2 . 15 hours. Li and Manfred are going to paint a house together. so it is the reasoning behind the solution based on algebra that will be the main interest here. Example 7 Joshua. In other words. we know that there are numbers j. For work such as painting a house. individually.divided by the length of the time interval is always equal to r gallons per minute. by definition. it will take them 18 hours. The rate of painting is constant. J(t) = jt L(t) = t
M (t) = mt 143
. Assuming that each person paints at a constant rate. Thus. let H be the function so that H(t) is the number of square feet painted from time 0 to time t (in minutes).
1 18
1 +
1 15
+
1 16
=4
Note two things. Let A be the the number of square feet of the house that need painting. A second thing of note is that.. and when 1 < x. x < x. Prove that if 0 < a < b or a < b < 0. if one is used to the use of symbols.
144
. 18 t. If they paint A square feet (i. In like manner. If all three paint together. the number b is called the constant term of the function. x > x. A
we get 1 1 1 + + 18 15 16 t0 = 1 32 37
and therefore t0 = hours. In a linear function f (x) = ax + b. It remains to observe that the linear functions that arise from rate problems are linear functions without constant term. when the three work together. then the preceding solution is entirely straightforward and is devoid of subtlety. Therefore all three together paint A A A 1 1 1 t+ t+ t= + + At 18 15 16 18 15 16 square feet in t hours. and m. Compare with any solution that does not use algebra. the whole house) in t0 hours. we see that J(18) = A. b √ √ 2. If b = 0. then we say the function f (x) = ax is without constant term. Such functions are the subject of the next section. Since it take Joshua 18 hours to paint the A house. and 16 t square feet. Li A A A and Manfred paints. respectively. then a > 1 . we get A A = 15 and m = 16 . then 1 1 1 + + At0 = A 18 15 16 Multiplying both sides by
1 . Before we leave this section. EXERCISES
1 1. Prove that when 0 < x < 1. A A A they paint at a constant rate of 18 + 15 + 16 square feet per hour. Thus 18j = A and j = 18 . each of Joshua. 15 t. then in t hours.e.We can determine each of these constants j. we introduce a terminology. The preceding argument shows that.
in how many days can each each do it alone?
145
. A man walks from point A to point B at a constant rate. Suppose Jessica can do a piece of work in 5 days and Jessica and Helena together can do it in 3 days. (it iv) x5 . (iii) x3 − x2 − 4x + 4. then n a < n b. (it iv) 2x3 + 9x2 − 15x − 22. find their speeds. and what are the differences? (Use a scientific calculator.) 8. Together they can do it in 3 11 days. Plot enough points in the graph of each of the following functions to get an accurate picture of the graph: (i) x3 . Assuming that each works at a constant rate. while if 0 < b < 1. It runs 80 miles in the same time that the passenger trains runs 112 miles. 112 miles apart. Assuming constant rate of work. (ii) x3 − 4x2 + x + 6. A train leaves A for B. for any positive interger n. 12. and one hour later a train leaves B for A. A freight train runs 6 miles an hour less than a passenger train.3. at 9 am. (The fact that this is true for n = 2 is used in Example 6. find the rate of each train. If the second train had started at 9 am and the first at 9 : 50 am. Plot enough points in the graph of 4x without the use of a calculator to get an accurate picture of its graph. 9. Prove that if 0 < a < b. ar < as . then for all rational numbers r and s so that r < s. they met at 12 noon. (ii) x4 . Compare the graphs of (i) – (iii) by putting all three in the same coordinate grid if necessary: what are the similarities. Assume that each train runs at a fixed constant speed. br > bs . in how many days can Helena do the work alone? 10. and B can do it in 4 as many 3 5 7 days as C. 6.) √ √ 7. Assuming that both trains run at a constant rate. then it takes him 5 1 minutes more to get to point B than if he 2 walks at the rate of 4 yards per 3 seconds. 13. (Use a scientific calculator. they would also have met at noon. (Use a scientific calculator. Plot enough points in the graph of each of the following functions to get an accurate picture of the graph: (i) x3 − x.) 5.) 4. (ii) 3x . Prove that if a > 1. (iii) x3 + 5. If he walks at the rate of 1 yard per second. (iii) 5x . Plot enough points in the graph of each of the following functions to get an accurate 1 picture of the graph: (i) ( 3 )x . How far is point A from point B? 11. A can do a piece of work in 2 as many days as B.
and c = 2. The classical (traditional) approach to such a problem is to set up a proportion: if w is the number of liters of water 8 people need per day. 83):
A group of 8 people are going camping for three days and need to carry their own water. one has to begin by making explicit the assumption that if f (x) is the amount of water (in liters) x people need each day.5. What is given is that f (5) = 12. How much water should they carry?
This problem is supposed to illustrate the importance of a concept called "proportional reasoning" in middle school mathematics. the problem cannot be done without the additional assumption that each person drinks the same amount of water each day. then 12.5.5 liters is to 5 people as w is to 8 people. It would seem that this concept does not have a precise meaning but has something to do with thinking in terms of multiplication. Now for "proportional reasoning" to be applicable in this case. at least none that is based on what is given in the problem. the definition of f means that c is the amount of water one person needs per day. and the answer to the problem is therefore 3w liters. They read in a guide book that 12.10
Proportional Reasoning (outlined)
Consider the following problem (taken from NCTM Standards. One guess is that "proportional reasoning" means the ability to fluently apply the concept of a linear function without constant term. p. What is noteworthy is that this assumption is missing from the problem. then f (x) is a linear function in x without constant term.5 = 20.) Therefore we may write f (x) = cx for some fixed number c. Thus 5c = 12.5.5 liters are needed for a party of 5 persons for 1 day. see the discussion of the profit function in Section 7. and the answer is 3f (8) = 60 liters.5 w = 5 8 Use this equation to solve for w. so 12. there is no logical reasoning that can be used to explain this approach. (We proceed to allow x to be an arbitrary number rather than just whole numbers. Unfortunately. 146
. Yet as it stands. It follows that f (8) = 8 × 2. Recall the WYSIWYG characteristic of mathematics in this context! A more recent approach is to isolate the concept of "proportional reasoning" and use it as a way to understand problems of this type. Since f (1) = c.
How many inches long will the scarf be at the end of 2 days? Explain how you figured it out. with the availability of the linear function f (x) = cx. Consider another problem (taken from Balanced Assessment for the Mathematics Curriculum):
John's grandfather enjoys knitting. He can knit a scarf 30 inches in 10 hours. Give explanation such as: To knit 27 inches takes 27 ÷ 3 hours.
Think through and do this problem yourself before comparing your solution with the suggested solution in Balanced Assessment. 9 hours. In 2 days he knits 2 × 6 inches. 3 inches. 12 inches. which is as follows: 1. The preceding proportion of 12. so for arbitrary x1 and x2 . by division: 10 ÷ 2. by division: 30 ÷ 10.Incidentally. How many inches can he knit in 1 hour? 2. 4. How many days will it take Grandpa to knit a scarf 30 inches long? 3. we will always have x1 = f (x2 ) . we can at last understand what "setting up a proportion" is all about. 2. We know that f (x) = c no x f (x1 ) matter what x may be. He always knits for 2 hours each day. 4. x2 because both quotients are equal to c. Give explanation such as: In one day he knits 3 × 2 = 6 inches. 5 days. 1. How many hours will it take Grandpa to knit a scarf 27 inches long? Explain your reasoning.5 = w is now seen 5 8 to be nothing other than the statement that f (8) f (5) = 5 8 Therefore one can begin to make sense of this venerable device of "setting up a proportion" from the standpoint of a linear function without constant term. 3.
147
.
The answers to the four parts are then. Thus if we write g(t) = t for some constant . if g(t) is the number of inches grandfather can knit in t hours. 1 person. For the case at hand. The first problem is a so-called discrete problem: there is a natural unit for people. we make one more comment on the proposed solution to Part 4 in Balanced Assessment. and = 3. then from g(10) = 30. if h has the same meaning as before and we are given g(h) = 27. What must be made explicit is the fact that. students are asked to develop conceptual understanding of a given topic and yet the topic is not taught in a manner that meets the minimum requirements of mathematics. To drive home the point of the need to understand linear functions without constant term in all such "proportional reasoning" problems. There 148
. we get 10 = 30.As a problem in mathematics. and finally. then g(t) is a linear function without constant term. we know that g(t) is always equal to the constant no matter what t may be. which results in t0 = 9. students are asked to develop an understanding of a multiplicative process but they cannot do so because the given assumptions are not sufficient to support the needed reasoning. Note that there is a real difference between these two problems. A little reflection would reveal that this method of solution is merely a repackaging of the proportion 30 27 = 10 h where h is the number of hours it takes grandfather to knit 27 inches. t In particular. 10 = 5. the 2 value of t0 so that f (t0 ) = 27. Think of WYSIWYG again. namely. From our vantage point. how can one begin to think about such a problem? Where to begin? This problem is symptomatic of a generic failure of mathematics education at the moment. this problem is not doable as it stands because what is given cannot support any kind of logical reasoning for its solution. then we have g(10) g(h) = . namely. in succession: f (1) = 3. with the exception of Part 2. The missing assumption is that grandfather knits at a constant rate. 10 h and this is exactly the proportion implicitly used in Balanced Assessment to solve the problem. Without this assumption. The given data is that g(10) = 30. f (4) = 12.
Mathematics education should not engage in the practice of not telling students that there is such a linear function without constant term. in textbooks and in the classroom. there is no reason why they should be linear. for example. a quadratic function. but rather a matter of being able to clearly and unambiguously articulate.edu/∼wu/. Without an explicit assumption. we say f is a quadratic polynomial function. We must never forget the WYSIWYG characteristic of mathematics. The graph of such a function is called a parabola. the knitting problem is a standard rate problem in that there is no natural unit for time. and in a way that is grade appropriate.27 persons. The lack of a natural unit in this case makes the knitting problem harder. Any time interval can be broken down into smaller time intervals. what a linear function is and when a linear function is being assumed in word problems.
11
Quadratic Functions and Their Graphs
If a function f from all numbers to all numbers is given by f (x) = ax2 + bx + c for some given numbers a. This is the difference between the discrete and the continuous.is no such thing as 1.berkeley. b. Mathematics is not concerned with making wild guesses about a hidden agenda. or more commonly. As we have seen in the preceding two examples. It is not a matter of what kind of conceptual understanding students should have about proportional reasoning. students have no way of doing the problems except by guessing. Further discussion of the role of proportional reasoning in the school mathematics curriculum is given at the end of Section 4 in What is so difficult about the preparation of mathematics teachers?. and then make judgement about their conceptual understanding (or lack thereof) because they cannot guess that there is such a linear function. It makes no sense whatsoever to demand that students have the conceptual understanding that those functions are linear. 149
. In one way or another. students must be given the information that a certain function is a linear function without constant term. untill we make explicit the fact that the functions f and g are linear functions without constant terms. On the other hand. See the discussion at the end of the preceding section. and c. because this is part of the basic assumption of constant rate.
Recall that a set S in the plane is said to be symmetric with respect to a line L if for every point Q in S. Since x2 is always positive.The main theme of this section is quadratic functions and their graphs. then f (x) < 0 for |x| large. We will use a symmetry property of the graph of f to cut down our effort by half. Given a quadratic function f (x) = ax2 + bx + c as above: (A) |f (x)| becomes arbitrarily large if |x| gets sufficiently large. f (x) ∼ a|x2 | when |x| is sufficiently large. for (A): "f (x) > R for a given large positive number R. Write f (x) = a+ c b + 2 x x x2
1 Now if |x| is sufficiently large. Consequently. If we use 150
. If a < 0. If the concept of absolute value were not at our disposal. In a suggestive notation. then |x| would be sufficiently small. and (A) also follows from the same expression for f (x) when |x| is sufficiently large. We have come across the simplest quadratic function f0 (x) = x2 in Example 2 of Section 9. for example. If we use ∼ to indicate "approximately equal to". we would have to write something like the following. At the end of this section. when |x| is sufficiently large. we shall prove that all parabolas are similar (in the sense of the composition of a congruence followed by a dilation) to the graph of f0 . and therefore so are b b | x | and | xc2 |. we will paraphrase the combined statements of (A) and (B) by writing:
|x|→∞
lim f (x) =
∞
for a > 0 for a < 0
|x|→∞
lim f (x) = − ∞
Notice how we have to use absolute value to express both (A) and (B) succinctly. then we have f (x) ∼ ax2 when |x| is sufficiently large. This proves (B). We begin with two general observations.
It remains to find out how f (x) behaves when x is not very large. (B) If a > 0. or f (x) < −R for a large positive number R if the distance of x from 0 is sufficiently large"
The reason for (A) is as follows. the reflection R across L maps Q to a point R(Q) which also lies in S. then f (x) > 0 for |x| large. (a+ x + xc2 ) would be very close to a.
So suppose f (x) = a(x − p)2 + q. S ∩ L+ or S ∩ L− . then p is the number at which f achieves a maximum. f (p + k) = f (p − k) = ak 2 + q This is easy: f (p − k) = a((p − k) − p)2 + q = ak 2 + q = a((p + k) − p)2 + q = f (p + k) 151
. then the study of S itself reduces to a study of one of the two halves. We also say that x0 is a zero of the function f is f (x0 ) = 0. If we write a(x − p)2 + q = ax2 − 2apx + (ap2 + q).g. Similarly. we say f achieves a maximum at a point β if f (x) ≤ f (β) for any number x. a quadratic function). As usual. then to say that f can be written as f (x) = a(x − p)2 + q is to say that the coefficients b and c in f (x) = ax2 + bx + c can be expressed in terms of two numbers p and q as b = −2ap and c = ap2 + q
In the second part. The first part assumes that the quadratic function can be written in the form f (x) = a(x − p)2 + q for some numbers p and q. Let us examine what this means. We observe that for any number k. and the graph of f is symmetric with respect to the vertical line defined by x = p. and if a < 0. if a > 0.. To describe this symmetry more precisely. Given a function f from all numbers to all numbers (e. there is a unique number p so that. then p is the number at which f achieves a minimum. This explains the interest in such a symmetry. If there is such a symmetry. We say f achieves a minimum at a point β if f (x) ≥ f (β) for any number x. S being symmetric with respect to L means the part of S in the half-plane L+ is congruent to the part of S in the other half-plane L− . Theorem 1 For any quadratic function f (x) = ax2 + bx + c. The following is the main theorem about quadratic functions. f can have at most two zeros. we introduce a concept already used informally in Section 7. we give a detailed proof of this theorem because the ideas of the proof contain important information about quadratic functions in general. we will show that in fact every quadratic function can be written in the form f (x) = a(x − p)2 + q. Furthermore.the terminology of Section 7. We will break up the proof into two parts.
f (p + k)) because we have seen that f (p − k) = f (p + k). The following picture assumes that k is positive. those (x.We now use the fact that f (p + k) = f (p − k) to show that the graph of f is symmetric with respect to the vertical line L defined by x = p. Indeed..e. L
(p − k. so that
152
. f (x)) where x ≤ p) can be obtained by reflection. but by the symmetry of the graph of f with respect to L (i. that is. we have proved our claim of symmetry with respect to L. then f (p + k) decreases with k so that f (p + k) is biggest when k = 0. f (p − k)) is on the graph of f . then f (p + k) increases with k so that f (p + k) is smallest when k = 0. f (x)) where x ≥ p. By symmetry with respect to L. they remain true for all x. f (p − k)). By symmetry with respect to L. this symmetry says it suffices to study (x. y ) r
p−k
q
p
p+k
Now if a point (p − k. the rest (i. Thus f has one zero at p + k0 among the numbers p + k for k ≥ 0. f (p − k0 ) is also zero. and since p is the only minimum of f . Finally. the line x = p). we see that f (p + k0 ) = 0 for some k0 > 0. for a > 0. we can write x = p + k. and if a < 0. The preceding assertions is about the behavior of f (x) for x ≥ p. f (p + k)) is a point on the graph of f (the set of all the points of the form (x. then f (p + k) = ak 2 + q > 0 and there would be no zero for f when x ≥ p. y ) r (p + k. where k ≥ 0. To study the graph of f . Now if x ≥ p.. Thus if a > 0.e. then under the reflection R. by definition). f attains its maximum only at x = p. Moreover. there is exactly one zero for f . which is equal to (p + k. y ) to the point (p − k. we have f (p + k) = ak 2 + q. But from (A) and (B) above. Since (p + k. y ). if q > 0. the reflection R with respect to L maps a point (p + k. If q = 0. if q < 0. f attains its minimum only at x = p. Because f (p + k) increases with k. where k and y are both arbitrary. and from the above discussion. we know that f (p + k) becomes large positive when k is large. f (x)). that is. this point is moved to (p + k. f has no zero in this case. then f (p) = q < 0. then f (p) = 0.
If on the other hand q < 0. we digress to interpret the information from the preceding discussion in terms of the shape and the location of the graph of f (x) = a(x − p)2 + q relative to the coordinate axes. 153
. and q > 0. 0) (which is of course (x0 . first assume a > 0. q = 0. and the graph would touch the x-axis at exactly (p. 0). Thus the graph of f crosses the x-axis at two points exactly. so that x0 is a zero of f . then f (p) = 0. f has exactly two zeros in this case. Does the graph cross the x-axis?
Remark Before continuing with the proof of Theorem 1. Finally.) If the graph of f intersects the x-axis at (x0 . ak 2 + q) where k is arbitrary. So the proof of Theorem 1 is complete in case f (x) = a(x − p)2 + q. if q > 0. which consists of all the points (p + k. then (x0 . Thus the whole graph of f would be above the x-axis and it would not intersect the x-axis. Our interest in the intersection of the graph of f with the x-axis naturally coincides with our interest in locating the zeros of the polynomial function f (x) = ax2 + bx + c. Conversely. because every quadratic function will be shown presently to be expressible in the form of a(x − p)2 + q. Thus locating the points of intersection of the graph of f with the x-axis is equivalent to locating the zeros of f . Therefore the graph of f goes from below the x-axis at p to above the x-axis when |x| is very large. 0) being on the graph means 0 = f (x0 ). one zero. 0). if f (x0 ) = 0. this means the graph of f also rises as we go to the left. according to whether q < 0. then f (p + k) = ak 2 + q > q > 0 for every k. the graph rises as we go to the right. and therefore also once in L− because of the symmetry with respect to L. Moreover. Then we have seen that f (p + k) with k > 0 gets bigger as k gets bigger. then the point (x0 . (A zero of the polynomial ax2 + bx + c is also called a root of the polynomial equation ax2 + bx + c = 0. and two zeros.f has another zero among the numbers p − k for k > 0. In a similar manner.
1 Activity Plot points on the graph of f (x) = 4(x − 2)2 + 3 which are symmetric with respect to the line x = 2. then f has no zero. In terms of the graph of f . we can show that if a < 0. then f (p) = q < 0 whereas lim|x|→∞ f (x) = ∞. once in L+ because the graph rises as we go to the right. With L as the line x = p and f (x) = a(x − p)2 + q. As mentioned above. respectively. Altogether. Because of the symmetry with repsect to L. if q = 0. f (x0 ))) is on both the graph of f and the x-axis and is therefore a point of the intersection of the graph with the x-axis. what we are going to say is valid for all quadratic functions.
What we have established is that. Furthermore. the graph of f will now have points both above and below the x-axis and will therefore intersect the axis at precisely two points (the reason for "two" is the same as before). Still with f (x) = a(x − p)2 + q. the whole graph would be below the x-axis and f would have no zero. now assume a < 0. q = 0 in the middle. and the point (p. and q > 0 on the right:
p
q q
p
q
p
q q
In the case of a > 0. then because f (p) = q. with the case of q < 0 on the left. because f (p + k) = ak 2 + q and a < 0 so that f gets more and more negative as k goes to the right of the x-axis. then the graph touches the x-axis at exactly (p. The same discussion can be carried through. f (p)) is still called the vertex of the parabola. and q > 0. it is now the highest point on the parabola when a < 0. The vertex is the lowest point of the parabola in this case. or no zero. If q > 0. in case a > 0 in f (x) = a(x − p)2 + q. two zeros. and the graph of f goes down as we go to the right of the x-axis. By symmetry with respect to L. 0). these three cases lead to the following kinds of graphs. but now f achieves a maximum at p and.We will discuss the zeros of a quadratic function a bit more in the next section and we will see that these zeros provide critical information in applications. if q < 0. depending on whether q = 0. respectively. f may have one zero. q < 0. and the y-coordinate is the minumum value of f . Typically. The three possibilites are given in the following pictures:
q q q q q
p
p
p
154
. Precisely. If q = 0. we call f (p) (which equals q) the minimum value of f . the x-coordinate of the vertex is the point at which f achieves its minimum. The point (p. the graph of f also goes down as we go to the left of the x-axis. f (p)) is then called the vertex of the parabola.
Be that as it may. Write b f (x) = a(x2 + x) + c a and compare it with what we want. we now get to work.15 Given f (x) = ax2 + bx + c. Thus a = −2p. set q = c − 4a = 4ac−b . the equality of all the corresponding b coefficients. we have to set 4a +q = c. We first do it algebraically. the proof is not too difficult.
155
. One should first work out some concrete examples such as f (x) = 2x2 + 6x − 7 or f (x) = 3x2 − 4x + 1 to get some feeling for the problem before embarking on the general case. 16 This fact can be proven in general. here is the general argument. and in the case of a < 0 as a down parabola. at least heuristically. we want to find numbers p and q so that f (x) = a(x − p)2 + q. b which would suggest that we let p = − 2a . 2 2 b or what is the same thing. Formally.16 In particular. We try that and get: a(x − p)2 + q = a(x + b b 2 b2 b2 ) + q = a(x2 + x + 2 ) + q = ax2 + bx + ( + q) 2a a 4a 4a
2
b If we hope to make this last expression equal to = ax2 +bx+c.e. i. we must have.It is common to refer to the graph of a quadratic function in the case of a > 0 as a up parabola. let b 4ac − b2 p=− and q = 2a 4a Then:
a(x − p)2 + q = a x +
15
b 2a
2
+
4ac − b2 4a
You will see that the identities (x + y)2 = x2 + 2xy + y 2 and (x − y)2 = x2 − 2xy + y 2 of Section 1 play a crucial role in the subsequent argument.. which is f (x) = a(x − p)2 + q = a(x2 − 2px + p2 ) + q For these two expressions to be equal for all x. See the Exercises in the next section. the two coefficients of x must be equal. but for quadratic functions. We can now return to the proof of the second part of Theorem 1. b a(x2 + x) + c = a(x2 − 2px + p2 ) + q a for all x. With this heuristic reasoning in the 4a background.
and the maximum or minimum value of f at p is q. and the maximum value f (− 2a ) is
and achieves
b b its minimum at − 2a if a > 0. But we have the explicit values of p and q given above in terms of the coefficients a. and the minimum value f (− 2a ) is 4ac−b2 . 4a
b b its maximum at − 2a if a < 0. looking back. depending on whether a < 0 or a > 0. 4ac−b ). b. and we go through both carefully. we now know that for the given f (x) = ax2 + bx + c. from the proof of the first part of Theorem 1.b b2 4ac − b2 = a x + x+ 2 + a 4a 4a 2 2 b 4ac − b = ax2 + bx + + 4a 4a = ax2 + bx + c
2
It follows that with these values for p and q. There are two comments on this proof that are of equal importance. the first one is that. 4a
Furthermore. f (x) = a(x − p)2 + q Then the proof of the first part of Theorem 1 now takes over to completely prove the theorem. With f (x) = ax2 + bx + c understood. we can see that the most critical step in the proof of the second part of Theorem 1. we know that p is the point at which f achieves its maximum or minimum. the quadratic function can be rewritten as b f (x) = a x + 2a
2
+
4ac − b2 4a
b It goes without saying that the vertex of the graph of f has coordinates (− 2a . one that allows us to reduce the most general 156
. 4a
2
A second comment is that. and c of f . Theorem 2 The quadratic function f (x) = ax2 + bx + c achieves
4ac−b2 . We can therefore summarize these conclusions in the following important theorem about quadratic functions.
157
. then we would immediately recognize that ax2 + bx + c b2 b could be put in the form of a(x − p)2 + q with p = − 2a and q = c − 4a . 2
17 No.
b = a x+ 2a
+ c−
Such an identity has wide applicability in mathematics and should be stated more generally. the Babylonians did not state completing the square as identity (13). and what the name "completing the square" means.17 Writing x2 + βx as β x2 + 2 ·x 2 and remembering that β and x are numbers.C. we have x + βx =
2
β x+ 2
2
−
β 2
2
(13)
In this form. is the identity that b x + x= a
2
b x+ 2a
2
−
b 2a
2
Indeed. because it is a simple computation to see that b ax2 + bx + c = a x2 + x + c a = a x+ b 2a
2 2
−
b2 4a2
+c b2 4a . as follows: for any number β. one would be led into thinking that it can be expressed as the difference of two squares. but if they had the symbolic notation that we have today. we see that this expression is exactly the area of the following figure consisting of a square with one side of length x and two rectangles each with sides of length β and x. identity (13) is known under the name of completing the square. What is more interesting is how. if we knew this identity. Its validity is not in doubt if you start with the right side and simply expand it and simplify to arrive at the left side.quadratic function to the manageable special case of a(x − p)2 + q. starting with the left side x2 + βx. they might have. Both questions are answered simultaneously if we retrace the steps of the Babylonians who made extensive use of such an identity some 38 centuries ago (around 1800 B.).
we get immediately identity (13). the total area is now the area of the big square with a side of length x + β . With this dotted 2 2 square added to the original figure. it would be natural to complete it to a bigger square with a side of length x + β by adding the small dotted square at the lower right corner: 2 x
β 2
x
β 2
p p p pp p p p p
Now the dotted square has a side of length β .x
β 2
x
β 2
Looking at this picture. which is by definition the graph of the quadratic function given by f1 (x) = x2 158
. so its area is ( β )2 . 2 We conclude this section by fulfilling the promise made at the beginning of the section: we are going to prove that the graph of any quadratic function is similar to the standard parabola G1 . Thus: 2 (x + βx) +
2
β 2
2
=
β x+ 2
2
Transposing ( β )2 to the right.
The final reason. You may have gotten the message by now that these notes consider learning precise definitions to be a critical component of a mathematics education. Part 1 is proved as follows. y) = (kx. and in fact the main one. The first is that this is a surprising result. In this case. Moreover. k x2 ) = ( { k x}. and there is no more convincing illustration of this message than this proof. then D( k x. is that the proof deepens your understanding of the precise definition of the graph of a function. k { k x}2 ). are similar. k 2 t2 ) = (kt. k x2 ) = (x. which are not rectilinear figures. It is quite clear that this theorem is not part of the standard algebra curriculum. and even mentioning something like this in an algebra class could be inspirational or intriguing to students. Indeed. where x is arbitrary. then a reflection followed by a translation would map G to a Gk of Part 1. Part 1 and 2 together prove that the graph of any quadratic function is similar to the standard parabola. rotations. (kt)2 ) which shows that D(Gk ) is part of G1 . We break the proof up into two parts: Part 1: Let Gk be the graph of the function fk (x) = kx2 where k > 0. ky) maps Gk to the standard parabola G0 . whether it be Algebra I or Algebra II. x2 ). instead we are forced to use the definition of similarity in terms of dilation. x2 ) for all x. if (x. kx2 ). So why spend time on it here? There are three reasons. kt2 ) of Gk . then the dilation D defined by D(x. kt2 ) = (kt. then a translation would map G to a Gk of Part 1. x2 ) is an arbitrary point 1 1 1 1 1 1 of G1 . Part 2: Let G be the graph of a quadratic function f (x) = ax2 + bx + c. For an arbitrary point (t. Gk consists of all points of the form (x. If a > 0. Because a similarity is by definition a rigid motion (a composition of translations. There are no line segments to measure and no angles to compare. By definition. while G1 consists of all points of the form (x. we have to prove that two parabolas. the proof hinges on exactly what the graph of the function fk (x) = kx2 is. we have D(t.
A few comments before the proof. which shows 159
. or reflections) followed by a dilation. On the other hand. A second reason is that the proof of the theorem shows why it is essential that we know a precise definition of similarity.for all x. ( k x. if a < 0.
(−a)(x − p)2 − q) } is exactly the graph of the function g given by g(x) = (−a)(x − p)2 − q. a(x − p)2 + q) of G is mapped by R to (x. and that a < 0. Because a < 0. f (x)) = (x. −q) to (0. In other words.
(p. which is the graph of f = a(x − p)2 + q. 160
. (−a) > 0. q) q
Let R be the reflection with respect to the x-axis. Observe that T (p. Observe that the collection of all these point { (x. where γ > 0 so that R(G) is an up parabola. Thus D(Gk ) is all of G1 . 0). We will prove that after a reflection and a translation.
For Part 2. Write γ for (−a) for simplicity. If the vertex of G is (p. q). is of the form (x. k x2 ) belongs to Gk . (−a)(x − p)2 − q). The points (x. Then R maps a point (x. γ(x−p)2 ). This proves Part 1. 0). y + q) for all (x. −q)
Notice that the vertex of R(G) is (p. γ(x−p)2 −q+q) = (x−p. then by Theorem 2. −y).1 1 that ( k x. and we see that the reflected parabola R(G) is the graph of g given by g(x) = γ(x − p)2 − q. as shown:
q
(p. The translation T is described algebraically by T (x. γ(x − p)2 − q) of the reflected parabola R(G) are now translated to (x−p. Next we will use a translation to move (p. Therefore each point (x. let the graph G of a quadratic function be given. Now a point on G. y). −q). G is a down parabola. −q) = (0. G is the graph of the quadratic function f (x) = a(x − p)2 + q. a(x − p)2 + q). We are going to assume the worst case scenario so that both p and q are nonzero. G is a Gk for some k. y) = (x − p. y) to (x.
given ax2 + bx + c = 0. We want to solve this equation.. From Theorem 1 of Section 11. y) = (x − p.T (x. we know that there are at most two zeros of f . we start from the beginning and assume nothing about quadratic functions.e. solving this equation is the same as locating the zeros of the quadratic function f (x) = −x2 + 90x − 180. This is known as a quadratic equation in the variable x. Suppose there is a root s. determine all the numbers x0 so that x0 (90 − x0 ) = 180. and this problem illustrates how such equations arise naturally without the intervention of a quadratic function. −x2 + 90x0 − 180 = 0. Such an x0 is also called a 0 2 root of the polynomial equation −x + 90x − 180 = 0. or. Our study of the quadratic function therefore covers more ground than the solving of quadratic equations. Thus an understanding of the quadratic function itself gives a comprehensive understanding of the nature of its zeros. y − q). Explain why this is true. By completing the square. we now approach the quadratic polynomial ax2 + bx + c purely algebraically. then we have an equation in x.
12
The Quadratic Formula and Applications (outlined)
Consider the problem: A rectangle has a perimeter 180 linear units and an area of 1800 area units. nevertheless. x(90 − x) = 180. and therefore the quadratic equation has at most two roots. because it gives a different perspective on the zeros of the quadratic function. without considering its graph. From the perspective of the last section. in studying a quadratic equation by itelf without reference to the associated quadratic function. we get b s+ 2a
2
= 162
b2 − 4ac 4a2
. and show again that it has at most two zeros. Thus. There is value. i. The proof of Theorem 1 in facts gives an algorithm (completing the square) for locating these zeros if they exist. namely. For this reason. What are its dimensions? If the length of one side is x linear units.
such as those listed below. "multiple" would have to mean "multiplication by another polynomial". then (ax + β)(x − r) would match ax2 +bx+c up to the coefficients of x2 and x. But how close is "close"? We would obviously welcome a multiple of (x − r) that matches ax2 + bx + c coefficient-by-coefficient for as many coeficients as possible. this "multiple" has degree either 1 or 0. (We say (x − r) divides a polynomial p(x) if p(x) factors as p(x) = q(x)(x − r) for some polynomial q(x). What it says 164
. but through its many substantial consequences. i. and the remainder is determined by the quotient once the quotient is clearly defined. (b) If r1 and r2 are the roots of ax2 + bx + c = 0. This is what γ is all about. (a) A number r is a root of ax2 + bx + c = 0 if and only if (x − r) divides ax2 + bx + c. we review the division-with-remainder for whole numbers. So the remainder ax2 + bx + c − (ax + β)(x − r) in this case is just a number (or a constant. Theorem 2 can appear to be deceptively simple (or perhaps plain opaque). ax2 + bx + c = a(x − r1 )(x − r2 ) (This identity may seem too obvious to you because some may even regard this as the definition of the roots r1 and r2 . but in this case. and γ the remainder.e. When we say "59 divided by 7 has quotient 8 and remainder 3". among all whole number multiples of 7. then for all x.. we are making a statement about multiplication. It turns out that if we let β = ar + b. and it misses 59 by 3 (3 = 59 − (8 × 7)). we look for a "multiple" of (x − r) that is "closest" to ax2 + bx + c. The linear polynomial is called the quotient of the division. namely. Analogously. one gains a better understanding of it.Theorem 2 (ax + β) so that
Given ax2 + bx + c and a number r. and since ax2 + bx + c has degree 2. If this terminology seems strange. But it is not obvious. the one that is closest to 59 but not exceeding it is the 8th multiple (which is of course 56). a and b. Thus to say (x − r) divides ax2 + bx + c means precisely that γ = 0 in Theorem 2). The quotient is therefore this multiple. Since the degrees add when polynomials are multiplied. there is a linear polynomial ax2 + bx + c = (ax + β)(x − r) + γ
for some constant γ and for all x. which is the common terminology).
Example If an object is thrown from a height of h meters from the ground with an initial velocity of v0 m/sec.) (d) If a quadratic polynomial ax2 + bx + c can be factored into a product of linear polynomials. but also more sophisticated proof of this fact by a judicious use of (a).is that. we get t0 = 2.) (c) Let r1 and r2 be the roots of x2 + bx + c = 0. which is 20 10 meters. approximately.2 seconds. and then explcitly compute a(x − r1 )(x − r2 ) to show that it is equal to a(r1 )2 + br1 + c. the QF provides a factorization for all trinomials. if r1 and r2 are any two numbers which satisfy a(r1 )2 +br1 +c = 0 and a(r2 )2 + br2 + c = 0. Thus solving −4. where r1 and r2 are the roots of ax2 +bx+c = 0. One way is to get explicit expressions of r1 and r2 in terms of a. Here is a typical example of how all this knowledge about quadratic functions and quadratic equations is put to use in solving word problems. b. Such a striking statement undoubtedly demands an explanation. (In short. and c by appealing to QF. It is also possible to obtain a more conceptual. then the factorization is a(x − r1 )(x − r2 ).)
Activity Factor 10x2 − 13x − 30. and when does it hit the ground? The highest point above the ground is the maximum of the quadratic function f (t) = −4.
165
. then b = −(r1 + r2 ) and c = r1 r2 (This needs (b) and Problem 2 in the following Exercises. what is the highest point of the object above the ground. then its distance f (t) above the ground t seconds after it is thrown (in meters) is f (t) = −4. The object hits the ground after 49 t0 seconds if f (t0 ) = 0. when does it get there. then the equality ax2 + bx + c = a(x − r1 )(x − r2 ) is valid for all x.9t2 + 2t + 20 = 0.9t2 + v0 t + h (This follows from Newton's second law.9t2 + 2t + 20.) Now if h = 20 meters and v0 = 2 m/sec.
traveling at 6 mph faster. reaches the end of the same run 20 minutes later than the first train. Find the time of the run of each train. how many hours will it take to fill the tank by each faucet separately?
167
. If the water flows from the faucets at a constant rate. A tank can be filled by the larger of two faucets in 5 hours less time than by the smaller one. A train makes a run of 120 miles.) 15. It is filled by them both together in 6 hours. A second train starts one hour later and.14. (Assume that both trains make the run at a constant rate. | 677.169 | 1 |
has been designed to teach algorithmic approaches to solving engineering problems. Focus is on developing common algorithmic patterns and how to use them to solve complex problems. Engineering applications requiring use of algebra,More...
This title has been designed to teach algorithmic approaches to solving engineering problems. Focus is on developing common algorithmic patterns and how to use them to solve complex problems. Engineering applications requiring use of algebra, calculus, and physics | 677.169 | 1 |
3
ALEKS ES UN PROGRAMA DE TUTOREO CON BASE EN LA RED CIBERNETICA. ESTE PROGRAMA TIENE COMO META EL ESTRUCTURAR Y CONSTRUIR LAS MAESTRIAS NECESARIAS PARA TENER EXITOS EN LAS MATEMATICAS. What is ALEKS? What is ALEKS? ¿QUE ES ALEKS? ALEKS IS A WEB BASED TUTORING PROGRAM THAT FOCUSES ON STRUCTURING AND BUILDING THE NECESSARY SKILLS FOR SUCCESS IN MATHEMATICS ALEKS IS A WEB BASED TUTORING PROGRAM THAT FOCUSES ON STRUCTURING AND BUILDING THE NECESSARY SKILLS FOR SUCCESS IN MATHEMATICS
7
TO LOGIN TO ALEKS, STUDENTS MUST VISIT AND USE THE LOG IN NAME AND PASSWORD PROVIDED TO THEM BY THEIR TEACHER. PARAR ENTRAR A ALEKS, EL ESTUDIANTE DEBE VISIT Y USAR EL NOMBRE DE ENTRADA Y LA CONTRA SENADA PREVISTO POR SU MAESTRO
8
. ONCE LOGGED ON, UNA VEZ QUE EL ESTUDIANTE A ENTRADO, THE STUDENT WILL SEE A PIE WHICH TELLS THEM THEIR AREAS OF STRENGHT AND AREAS TO BUILD ON THE STUDENT WILL SEE A PIE WHICH TELLS THEM THEIR AREAS OF STRENGHT AND AREAS TO BUILD ON MIRARA SU GRAFICO, EL CUAL INDICA SUS AREAS DE MAESTRIA, Y CUALES DEBEN AUN FORTALECER. MIRARA SU GRAFICO, EL CUAL INDICA SUS AREAS DE MAESTRIA, Y CUALES DEBEN AUN FORTALECER.
15
IN CONCLUSION, ALEKS CAN BE USED IN SCHOOL, AT HOME OR THE PUBLIC LIBRARY. STUDENTS ARE EXPECTED TO WORK ON THIS DAILY IN ORDER TO SEE RESULTS. EN CONCLUSION, ALEKS PUEDE USARSE EN LA ESCUELA, EN CASA O LA BIBLIOTECA PUBLICA. SE ESPERA QUE LOS ESTUDIANTES USEN ALEKS PARA VER UN CAMBIO POSITIVO. | 677.169 | 1 |
Affordable Access
Abstract
Publisher Summary The purpose of this chapter is to begin the study of the theory of linear operators, which are basic to differential equations. This chapter presents an outline of the necessary facts about vector spaces. It is very useful to be able to treat vectors (and later, operators) as objects independent of any particular coordinate system. Henceforth this chapter shall use the term vector space to mean "subspace of a Cartesian space." An element of a vector space will be called a vector (also a point). To distinguish them from vectors, real numbers are called scalar.
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Maths Essentials for MBA Success (edX)
Discover and acquire the fundamental maths skills that you will need to use while studying an MBA program, from algebra to differentiation and geometric series.
Planning to study for an MBA but unsure of your basic maths skills? All MBA programs require some maths, particularly on quantitative subjects such as Accounting, Economics and Finance.
In this mathematics course, you will learn the fundamental business math skills needed to succeed in your MBA study. These math skills will also give you an edge in the workplace enabling you to apply greater analytical skill to your decision making.
You will learn how to evaluate and manipulate the types of formulae that appear in an accounting syllabus, how to perform the calculus required to solve optimization problems in economics and how to apply the concept of geometric series to solving finance-related problems such as calculating compound interest payments.
This course assumes no prior knowledge of business maths, concepts are explained clearly and regular activities give you the opportunity to practice your skills and improve your confidence.
What you'll learn:
- How to evaluate, manipulate and graph the types of linear equations that appear throughout an MBA syllabus
- How to evaluate and manipulate the complex relationships that appear (for example, those that are represented as quadratic or exponential equations)
- How calculus is used to solve optimization problems, such as finding the quantity of sales that maximizes a firm's profit
- How tools relating to mathematical series and sequences are adopted on an MBA program to calculate concepts such as net present value and compound interestAn introduction to probabilistic models, including random processes and the basic elements of statistical inference. The world is full of uncertainty: accidents, storms, unruly financial markets, noisy communications. The world is also full of data. Probabilistic modeling and the related field of statistical inference are the keys to analyzing data and making scientifically sound predictions | 677.169 | 1 |
McGraw-Hill's Conquering the New GRE Math by Robert E. Moyer
Book Description
Be ready for the mathematics sections of the GRE General Test - scheduled to be revised in August 2011. "McGraw-Hill's Conquering the New GRE Math" offers you intensive review for every kind of GRE math question. Within each topic, solved problems of gradually increasing difficulty help you build your problem-solving skills. Exercises show how each math concept is tested on the GRE. Full-length GRE math sections provide practice with questions just like those on the real test. It includes features such as: complete coverage of the new math question types scheduled to be introduced in August 2011; intensive drill and practice to improve your math skills to get into the graduate program of your choice; sample GRE math questions build your test-taking confidence; and, expertise from an author who specializes in providing instruction to students whose math skills are weak or rusty.
This title includes topics such as: The GRE Quantitative Reasoning Section; The Math You Need to Review; How the Questions Are Asked; GRE Quantitative Comparison; GRE Problem-solving (Multiple-choice); GRE Data Interpretation; GRE Numeric Entry Questions; GRE Mathematics Review; Number Properties; Arithmetic Computation; Algebra; Geometry; GRE Math Practice Tests; GRE Math Practice Test 1; GRE Math Practice Test 2; and, GRE Math Practice Test 3.
Buy McGraw-Hill's Conquering the New GRE Math book by Robert E. Moyer from Australia's Online Bookstore, Boomerang Books.
Includes more than 600 fully solved problems, examples, and practice exercises to sharpen your problem-solving skills. This book gives you: 618 solved problems to reinforce knowledge; concise explanations of all trigonometry concepts; and updates that reflect the course scope and sequences, with coverage of periodic functionsand curve graphing.
This math-refresher gives students in-depth practice and intensive review for every type of math problem on the GMAT. The book provides a comprehensive overview of every test topic through sample exams and answered problems that increase in difficulty--helping to build conceptual and problem-solving skills.
Author Biography - Robert E. Moyer
Robert E. Moyer, Ph.D., is currently an associate professor of mathematics at Southwest Minnesota State University in Marshall, Minnesota. The coauthor of four bestselling Schaum's Outlines in mathematics, he has also written math questions for the GED and the ASVAB. From 1978 to 1990 he worked with the Georgia Assessment Program that administered the K-12 testing program in Georgia and the Georgia Teacher Certification program, writing test questions and conducting workshops for teachers who needed help passing the mathematics test for high school teachers or the mathematics portions of the elementary and middle grades teacher tests | 677.169 | 1 |
College Algebra Exam Review 16 - 26 1. ALGEBRAIC THEMES...
26 1. ALGEBRAIC THEMES Although the multiplicative structure of the integers is subordinate to the additive structure, many of the most interesting properties of the in-tegers have to do with divisibility , factorization, and prime numbers. Of course, these concepts are already familiar to you from school mathemat-ics, so the emphasis in this section will be more on a systematic, logical development of the material, rather than on exploration of unknown terri-tory. The main goal will be to demonstrate that every natural numbers has a unique factorization as a product of prime numbers; this is trickier than one might expect, the uniqueness being the difficult part. On the way, we will, of course, be practicing with logical argument, and we will have an introduction to computational issues: How do we actually compute some abstractly defined quantity? Let's begin with a definition of divisibility. We say that an integer a divides an integer b (or that b is divisible by a ) if there is an integer
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This note was uploaded on 12/15/2011 for the course MAC 1105 taught by Professor Everage during the Fall '08 term at FSU. | 677.169 | 1 |
Algebraic Expressions and Equations
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This is the opening lesson (or homework) for the unit on Algebraic Expressions and Equations for middle school math. It is an Excel format to enable editing. Each question include a strategy guide to help the students complete it correctly. It's aligned to the Common Core. An answer key is | 677.169 | 1 |
Mastering Mathematics 11-16 for Edexcel GCSE focuses on strands of learning within the new national curriculum to improve progression throughout secondary mathematics, offering a seamless five year progression.
Mastering Mathematics 11-16 for OCR GCSE focuses on strands of learning within the new national curriculum to improve progression throughout secondary mathematics, offering a seamless five year progression.
Mastering Mathematics 11-16 for WJEC GCSE focuses on strands of learning within the new national curriculum to improve progression throughout secondary mathematics, offering a seamless five year progression. | 677.169 | 1 |
Enumerative combinatorics deals with finite sets and their cardinalities. In other words, a typical problem of enumerative combinatorics is to find the number of ways a certain pattern can be formed.
In the...
Enumerative combinatorics deals with finite sets and their cardinalities. In other words, a typical problem of...
A very beautiful classical theory on field extensions of a certain type (Galois extensions) initiated by Galois in the 19th century. Explains, in particular, why it is not possible to solve an equation of degree...
A very beautiful classical theory on field extensions of a certain type (Galois extensions) initiated by Galois in the...
This course covers mathematical topics in trigonometry. Trigonometry is the study of triangle angles and lengths, but trigonometric functions have far reaching applications beyond simple studies of triangles....
This course covers mathematical topics in trigonometry. Trigonometry is the study of triangle angles and lengths, but...
Calculus Two: Sequences and Series is an introduction to sequences, infinite series, convergence tests, and Taylor series. The course emphasizes not just getting answers, but asking the question "why is this true?"
Calculus Two: Sequences and Series is an introduction to sequences, infinite series, convergence tests, and Taylor...
Learn how to think the way mathematicians do - a powerful cognitive process developed over thousands of years.
Learn how to think the way mathematicians do - a powerful cognitive process developed over thousands of years.
Calculus is about the very large, the very small, and how things change. The surprise is that something seemingly so abstract ends up explaining the real world. Calculus plays a starring role in the biological,...
Calculus is about the very large, the very small, and how things change. The surprise is that something seemingly so...
This is a course about the Fibonacci numbers, the golden ratio, and their intimate relationship. In this course, we learn the origin of the Fibonacci numbers and the golden ratio, and derive a formula to...
This is a course about the Fibonacci numbers, the golden ratio, and their intimate relationship. In this course, we...
This course covers mathematical topics in college algebra, with an emphasis on functions. The course is designed to help prepare students to enroll for a first semester course in single variable calculus.
This course covers mathematical topics in college algebra, with an emphasis on functions. The course is designed to... | 677.169 | 1 |
Synopses & Reviews
Publisher Comments
"""A First Course in Analysis"" is a new approach to the elements of real analysis for post-calculus students. The text outlines the evolution of number systems, then proceeds to introduce the fundamental ideas of completeness, compactness and continuity via a plausible heuristic investigation of the problem of extreme values. The author then presents the standard properties of continuous functions, along with a continuity-focused development of the elementary functions. The book concludes with chapters on the foundations of calculus: differentiation, integration, and infinite series. Applications and looks-ahead to more advanced analysis are pointed out. This text can be adapted not only to courses of different lengths and emphasis, but also to different modes of classroom presentation. In particular, it complements a strictly deductive presentation, that begins, say, with the ordered field axioms."
Review
"This book is a very enjoyable introduction into Analysis. Considering the treatment of the material it should be emphasized that a very valuable feature is that a lot of interesting questions are raised in such a way that definitions, theorems, and proofs are motivated for the readers. This volume is very useful for students and lectures even for bright high school students interested in the introductory theory of functions. ACTA SCIENTIARUM MATHEMATICARUM"
Synopsis
The first course in Analysis, which follows calculus, along with other courses, such as differential equations and elementary linear algebra, in the curricu lum, presents special pedagogical challenges. There is a change of stress from computational manipulation to "proof. " Indeed, the course can become more a course in Logic than one in Analysis. Many students, caught short by a weak command of the means of mathematical discourse and unsure of what is expected of them, what "the game" is, suffer bouts of a kind of mental paralysis. This text attempts to address these problems in several ways: First, we have attempted to define "the game" as that of "inquiry," by using a form of exposition that begins with a question and proceeds to analyze, ultimately to answer it, bringing in definitions, arguments, conjectures, exam ples, etc., as they arise naturally in the course of a narrative discussion of the question. (The true, historical narrative is too convoluted to serve for first explanations, so no attempt at historical accuracy has been made; our narra tives are completely contrived. ) Second, we have kept the logic informal, especially in the course of preliminary speculative discussions, where common sense and plausibility tempered by mild skepticism-serve to energize the inquiry."
Synopsis
This text on advanced calculus discusses such topics as number systems, the extreme value problem, continuous functions, differentiation, integration and infinite series. The reader will find the focus of attention shifted from the learning and applying of computational techniques to careful reasoning from hypothesis to conclusion. The book is intended both for a terminal course and as preparation for more advanced studies in mathematics, science, engineering and computation. | 677.169 | 1 |
Orlando Meetings: Presentation Summary
This is the summary of a presentation given at the Joint Mathematics
Meetings, January 10-13, 1996, Orlando, Florida.
A course in mathematics and technology for prospective teachers
Several national mathematics organizations have recommended the
incorporation of appropriate computer technology into classrooms at all
levels. A corollary of these recommendations is that prospective teachers
of mathematics must thoroughly learn the appropriate uses, capabilities,
and limitations of these devices, so that they can effectively manage the
resultant classroom environment involving greater emphasis on student
activity and discovery.
The Texas A & M Department of Mathematics has approached the challenge of
training future secondary-school teachers in the use of various
technologies by instituting
a course in which students use TI-82 calculators, The Geometer's
Sketchpad, and Maple to solve a variety of problems involving
precalculus mathematics, discrete mathematics, calculus, geometry, linear
algebra, and statistics. The prerequisite is the completion of a
senior-level mathematics course, so that the course comes late in
students' careers, and they can concentrate on the use of the technology as
a problem-solving tool rather than attempt to learn substantial amounts
of new mathematical content simultaneously. This is not a teaching
methods course, but one in which students learn appropriate uses for the
technology they will use in their teaching careers, in
essentially the same type of active problem-solving setting that their
future students will encounter. This is accomplished through a number of
carefully
chosen and challenging problem-solving experiences involving and extending
mathematics learned in previous courses. Students also explore open-ended
"project" problems of particular interest to them. Thus, the course
provides
a "capstone" experience in problem-solving as well as a mathematical
content review for the future mathematics teacher. | 677.169 | 1 |
Courses How does US maths and physics courses compare with british A-level?
Just a matter of interest, but how do the US physics and mathematics courses for students before university (The final year before university) compare with the British A-level courses? what is the typical content of the US courses at this age? For example, what is included in "precalculus", and so on.
Precalculus as I was taught it includes more on trigonometric functions, trigonometric identities, a little formal proof, some geometry, a little stuff about polynomials. It basically finished covering all math up to the SAT II Mathematics IIC test (see though there's no guarantee that a given student will learn all of it.
High school calculus as I learned it covered basic integration and differentiation in 1 variable, some techniques of integration, limits, some series and sequences, finding volumes of solids of revolution, and implicit differentiation/rates of change problems. It basically covered all math up to the AP Calculus AB or BC test (BC being the more advanced but some schools only teaching AB). See
High school physics as I learned it was one year of non-calculus-based mechanics and wave motion.
This is purely from my own experience in High School (I'm a 12th grade student currently) and that of people I've spoken to in other states and in different school systems.
Precalculus- At my school, this course is basically half a year of review over trignometry, functions, logarithms, exponential functions, basic algebra. In the halfway through the year, my teacher began teaching limits with epsilon-delta. The class spent a good time doing limits this way and studying limits in general. We moved on to Derivatives using the limit definition, proved many basic derivatives (e^x, sin(x), cos(x), etc.). The teacher proceded to teach derivatives, applications of derivatives, and anti-differentiation. We did no actual definite integration.
AP Calculus AB: This course is typically offered at my school. It covers a normal Calculus I course (I believe) starting with limits and continuity, continuing through with derivatives, applications of derivatives, integration, applications of integration (Volume via shell and cylinder methods.)
AP Calculus BC: This course covers a standard Calculus I and II course. It continues where AB Calculus left off and explores more applications of integration (Work, Arc Length, Surface Area), explores techniques of integration, does infinite series, power series, polar and parametric and basic vector derivatives and things of that sort.
Many High Schools in the US offer a system where a student can take classes at nearby universities. One of my friends opted to do this instead of taking large numbers of AP classes. He is currently in Multivariable Calculus and Differential Equations (I also go to DE, mostly for education but also because it is entertaining.)
Furthermore, some schools actually offer Linear Algebra, although I doubt it is on par with university level Linear Algebra.
I've spoken with many students from other states; they basically do the same AP Calculus work, but less proofs for basic derivatives and their precalculus lacked calculus. That's, from what I've seen, how the US does High School math. | 677.169 | 1 |
arson IS student success. INTERMEDIATE ALGEBRA owes its success to the hallmark features for which the Larson team is known: learning by example, a straightforward and accessible writing style, emphasis on visualization through the use of graphs to reinforce algebraic and numeric solutions and to interpret data, and comprehensive exercise sets. These pedagogical features are carefully coordinated to ensure that students are better able to make connections between mathematical concepts and understand the content. With a bright, appealing design, the new Fifth Edition builds on the Larson tradition of guided learning by incorporating a comprehensive range of student success materials to help develop students' proficiency and conceptual understanding of algebra. The text also continues coverage and integration of geometry in examples and exercises. | 677.169 | 1 |
ized N a concise presentation of the basic theory of numerical methods that offers numerous examples, problems and computer programs. It can be used alone or in a course with heavy computer emphasis or as a supplement to the author's basic text, Numerical Methods in Engineering Practice. | 677.169 | 1 |
Online Algebra Resources
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Great interactive websites offer students manipulatives for students to practice basic algebraic skills like solving for variables and balancing equations.
This would be perfect to use at a math center if you have access to laptops or iPads and would also be a good resource to send home with students who are struggling so they could do some extra | 677.169 | 1 |
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Unformatted text preview: Functions Mariusz Bajger COMP2781/8781 School of Computer Science, Engineering and Mathematics May 6, 2011 1/5 Reading and Exercises Reading Epp, Chapter 7 Exercises Sec. 7.1, 7.2 and 7.3 (use as you need; you are supposed to know this material from MATH1121 or elsewhere), here we focus on Section 7.4 (all in blue exercises are recommended5 General concepts - revision I It is expected that you know basic properties of functions from MATH1121 including graphs of elementary functions like log , exp and polynomials. I Chapter 1 (Epp) gives a gentle introduction to functions, more is covered in Chapter 7.1 and 7.2 - read as much as you need. I Ensure that you understand the definition of function, equality of functions, one-to-one and onto functions, inverse functions, logarithmic and exponential functions,...
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This note was uploaded on 06/12/2011 for the course MATH 103 taught by Professor Wouters during the Spring '08 term at Wisc Oshkosh. | 677.169 | 1 |
Randall - Accounting A Level and AS Level
Randall
CIE Accounting: AS Level and A Level meets the requirements of the Cambridge International Examinations (CIE) syllabuses for AS Level and A Level Accounting, and is endorsed by CIE for use with these examinations. Suitable for one year AS courses and for two year A level courses, the book helps develop the required computational, accounting, problem-solving and written skills. Theoretical and practical aspects of all topics are covered, with teaching points being illustrated by fully worked examples and exercises to test students' understanding. Answers to the exercises are in the appendix. Features include examination hints, multi-choice questions at the end of each chapter, answers in the appendix, additional exercises from past papers, worked examples, diagrams and charts to illustrate teaching points and a grid to distinguish between AS topics and A Level topics.
Comprehensive CIE syllabus coverage
Combination of theory and practice
The book has been produced by experienced teachers and lecturers, with senior examining roles
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The National Curriculum and the Scottish Guidelines emphasise the importance of mental calculation in mathematics. This is the first of four small books that give practice in applying basic mathemat...
The National Curriculum and the Scottish Guidelines emphasise the importance of mental calculations in mathematics. This is the second of four small books for junior age children, that give practice...
The National Curriculum and the Scottish Guidelines emphasise the importance of mental calculations in mathematics. This is the third of four small books for junior age children, that give practice ...
The National Curriculum and the Scottish Guidelines emphasise the importance of mental calculations in mathematics. This is the last of four small books for junior age children, that give practice i...
It is hard to imagine a classroom where mental mathematics is not used every day. However, setting aside a regular time for mental mathematics ensures that children have plenty of experience and pra...
It is hard to imagine a classroom where mental mathematics is not used every day. However, setting aside a regular time for mental mathematics ensures that children have plenty of experience and pra...
It is hard to imagine an infant classroom where mental mathematics is not used every day. However, setting aside a regular time for mental mathematics ensures that children have plenty of experience...
This invaluable little book highlights the real mistakes that students make in the exam - and shows how to avoid them. Based on analysis of thousands of exam scripts, each unit targets a key problem... | 677.169 | 1 |
BANGALORE, India - March 6, 2014 - PRLog -- Lanika announces a major new release of Maplesoft's flagship product, Maple™, the mathematical computing software for education and research in mathematics, engineering, and the sciences. With Maple 18, Maplesoft offers enhanced tools for developing interactive applications and quizzes, together with additional features to enrich and streamline the student experience.
Maple supports the easy creation of interactive Math Apps for use in the classroom and through The Möbius Project, an initiative from Maplesoft that supports the creation, sharing, and grading of Math Apps. With Maple 18, instructors can take advantage of increased flexibility in the one-step Math App creation tool to quickly create even more complex applications, and easily create randomly generated quizzes for their students.
Maple 18 also offers a host of improvements to enhance the student experience, so they can learn important concepts quickly and easily.
· New Clickable Math™ tools include a dedicated calculus palette, improved context-sensitive menus for student-learning, and over 75 new Math Apps for math, biology, chemistry, and engineering.
· A new statistics package designed specifically to help teachers and students with an introductory course in statistics offers formulas and visualizations of statistical quantities, hypothesis testing, and interactive exploration.
Other important enhancements were made in visualization, mathematics, and the user interface:
· Extensive updates to visualization include a new package for quickly creating fractals, custom plot shading and image textures, background images on plots, and new visualization tools for signal processing and time series analysis.
· The many mathematical computation improvements include updates to physics, time series analysis, polynomial operations, dynamic systems, and graph theory.
· A powerful new search tool provides instant access to all the relevant help pages, tasks, assistants, tutors, and Math Apps in Maple and in The Möbius Project, with previews available so customers can easily select the exact resource they need.
"Maple is used in so many different ways by our customers, from a student's first exploration of a new concept, to instructors creating learning materials to engage and evaluate their class, to researchers working on large, complex projects," says Dr. Laurent Bernardin, Executive Vice-President and Chief Scientist at Maplesoft. "Maple 18 offers important new advantages for all these customers. With advancements in application development tools, specialized tools for student learning, and enhancements in its computation engine, Maple 18 helps customers reach their education and research goals more quickly than ever before."
Maple 18 is part of a broader release of Maplesoft's product suite, including Maple add-ons and e-books and a new version of MapleSim, the advanced system-level modeling and simulation tool based on Maple's mathematical engine. MapleSim 6.4 and all its toolboxes and connectivity add-ons have been updated to take advantage of the enhancements to Maple 18's mathematical engine. Other improvements include more powerful tools for creating custom components, performance enhancements, a significantly expanded MapleSim Control Design Toolbox, and enhancements to the model generators for Simulink® and FMI.
Maple 18 is available in English, with built-in language extensions available for several languages, including French, Traditional Chinese, Simplified Chinese, Greek, and Brazilian Portuguese. A Japanese version of Maple 18 will be available in April.
Lanika is provider of technical computing software & high-end hardware tools for engineers and scientists in industry, government and education. The Company partners with reputed principals developing industry leading solutions that help a wide base of clients throughout the Indian sub-continent solve the toughest engineering problems.
Lanika Solutions product offerings and support reflects the philosophy that given great tools, clients can simplify development, increase productivity, and dramatically reduce time to market. Company's suites of technical products help clients to quickly solve practical problems within the framework of the premier products and services provided.
Maplesoft™, a subsidiary of Cybernet Systems Co., Ltd. in Japan, has over 25 years of experience developing products for technical education and research, offering a solution that applies to every aspect of academic life. Its product suite reflects the philosophy that given great tools, people can do great things.
Maplesoft's core technology is the world's most advanced symbolic computation engine, which is the foundation for all of its products, including Maple™, the technical computing and documentation environment; MapleSim™, the high-performance, multi-domain modeling and simulation tool for physical systems; and Maple T.A.™, a web-based system for creating and assessing online tests and assignments.
Maplesoft also introduced a fundamental shift in technical education through its Clickable Math™ and Clickable Engineering™ initiatives. The idea behind this shift is to create technology that will allow students and teachers to focus on the concepts, not the tool. These initiatives deliver powerful mathematics through visual, interactive point-and-click methods in Maple, while the intuitive physical modeling environment of MapleSim helps teachers to quickly demonstrate the connection between modeling concepts and the underlying mathematical theory.
Over 90% of advanced research institutions and universities worldwide, including MIT, Stanford, Oxford, the NASA Jet Propulsion Laboratory, and the U.S. Department of Energy, have adopted Maplesoft solutions to enhance their education and research activities. In industry Maplesoft's customers include Ford, Toyota, NASA, Canadian Space Agency, Motorola, and DreamWorks, covering sectors such as automotive, aerospace, electronics, defense, consumer products, and entertainment. | 677.169 | 1 |
BCcampus OpenEd Resources
Learning about, and experiencing, open educational practicesThu, 23 Feb 2017 17:29:15 +0000en-UShourly1 new textbooks in the BC Open Textbook Collection
09 Feb 2017 22:20:03 +0000 of today, we now have 173 open textbooks in 8 main and 36 secondary subject areas in the BC Open Textbook Collection. New titles are being adapted and adopted often – here are the 5 latest open textbooks that we've added to the collection since the beginning of the year.
Description: Prealgebra is designed to meet scope and sequence requirements for a one-semester prealgebra course. The text introduces the fundamental concepts of algebra while addressing the needs of students with diverse backgrounds and learning styles. Each topic builds upon previously developed material to demonstrate the cohesiveness and structure of mathematics. Prealgebra follows a nontraditional approach in its presentation of content. The beginning, in particular, is presented as a sequence of small steps so that students gain confidence in their ability to succeed in the course. The order of topics was carefully planned to emphasize the logical progression throughout the course and to facilitate a thorough understanding of each concept. As new ideas are presented, they are explicitly related to previous topics.
Description: The book examines the underlying principles that guide effective teaching in an age when everyone, and in particular the students we are teaching, are using technology. A framework for making decisions about your teaching is provided, while understanding that every subject is different, and every instructor has something unique and special to bring to their teaching. The book enables teachers and instructors to help students develop the knowledge and skills they will need in a digital age: not so much the IT skills, but the thinking and attitudes to learning that will bring them success.
Description: Greek and Latin Roots: Part I – Part I will try to impart some skill in the recognition and proper use of words derived from Latin Latin, and does not involve the grammatical study of this language—except for a few basic features of noun and verb formation that will help students to understand the Latin legacy in English. Although there will be some attention paid to the historical interaction of Latin Greek and Latin Roots: Part II – Part II will try to impart some skill in the recognition and proper use of words derived from Greek Greek, and does not involve the grammatical study of this language—except for a few basic features of noun and verb formation that will help students to understand the Greek legacy in English. All students will be asked to learn the Greek alphabet. This skill is not absolutely essential for a general knowledge of Greek roots in English. However, it will help students understand a number of otherwise puzzling features of spelling and usage. Although there will be some attention paid to the historical interaction of Greek residents—both inheritors of ancient Indigenous Knowledge and wisdom, and newcomers who can experience the engagement, joy and promise of science instilled with a sense of place. This book takes a step forward toward preserving and actively using the knowledge, stories, and lessons for today and future generations, and with it a worldview that informs everyday attitudes toward the earth. Knowing Home: Braiding Indigenous Science with Western Science is far more than a set of research papers or curriculum studies. The project outputs include both, but they are incorporated into a theoretical structure that can provide the methodological basis for future efforts that attempt to develop culturally responsive Indigenous Science curricula in home places. It is not just one or two angels to organize, but multiple interwoven approaches and cases that give this project its exceptional importance. Thus, the project outputs have been organized into two books. Book 1 provides an overview of why traditional knowledge and wisdom should be included in the science curriculum, a window into the science and technologies of the Indigenous peoples who live in Northwestern North America, Indigenous worldview, culturally responsive teaching strategies and curriculum models, and evaluative techniques. It is intended that the rich examples and cases, combined with the resources listed in the appendices, will enable teachers and students to explore Indigenous Science examples in the classroom; and in addition, support the development of culturally appropriate curriculum projects.Faculty and Students working together to create OER: the Open Case Studies Project at UBC
This is #2 in a series of webinars – Open Education Stories from Across B.C.
Presentation Description
The Open Case Studies project at the University of British Columbia brings together faculty and students to create and edit case studies as open educational resources. We have started by focusing on courses that have to do with sustainability because there are many courses across our campus that address issues in sustainability in some way and that use case studies to do so.
However, we are interested in any courses that use case studies and that want to make them open and available for others to revise and reuse. We have a small number of case studies so far, but hope the numbers will grow in the coming years as more faculty and students become involved. So far all of our cases are from courses at UBC, but we are investigating ways to include contributions from people around BC or beyond. Christina Hendricks, the lead faculty member on the project, will discuss how the project has worked so far, how courses have involved students in reading, adding to, or creating case studies, and how others can get involved.
Christina Hendricks is a Professor of Teaching in Philosophy at the University of British Columbia-Vancouver, who is also an advocate for open education generally, including the use and creation of open textbooks and other open educational resources. From 2014-2015 she was a Faculty Fellow with the BCcampus Open Textbook project, and from 2015-2017 she was an OER Research Fellow with the Open Education Group.
View other webinars in the series:
Privacy Note]]> change with the BCcampus Open Education Advocacy and Research Fellows
31 Jan 2017 15:41:17 +0000 is an honour to announce the three instructors who have joined the BCcampus OpenEd team to fulfill a one-year term as the 2017/18 BCcampus Open Education Advocacy and Research Fellows.
The 2017/18 BCcampus Open Education Advocacy and Research Fellows are Jennifer Barker, Biology Instructor at Douglas College, Ken Jeffery, Digital Arts Instructor at BCIT, and George Veletsianos, Associate Professor at Royal Roads University. This group will help raise awareness of open educational practices through advocacy to conduct, present, and publish research on open educational practices at B.C. post-secondary institutions.
As mentioned in the November post, Dr. Rajiv Jhangiani has been selected to be the Senior Open Education Advocacy and Research Fellow. Alongside the BCcampus OpenEd team, Dr. Jhangiani will lead the incoming BCcampus Open Education Advocacy and Research Fellows.
Get to know the 2017/18 BCcampus Open Education Advocacy and Research Fellows:
Jennifer Barker
After completing a BSc and MSc in Zoology at Toronto, a PhD in Neuroscience at UBC, and finally a neuroscience postdoc in Belgium, three years ago I left the research stream of academia and returned to B.C. to take up a teaching position in Biology at Douglas College. I've discovered that I enjoy teaching so much that I haven't looked back since!
I hope to help raise awareness among students and colleagues of both the availability and the value of Open Textbooks and Open Educational Resources in general, and to encourage their adoption within my department, institution, and province. I also hope to be able to participate in the collection of empirical evidence, and its presentation to colleagues as well as students, that the adoption of Open Educational Resources in our classrooms can help our students succeed by alleviating the enormous financial burden many of them experience, without compromising the quality of the education they receive.
Ken Jeffery
Ken Jeffery is an instructor in the Digital Arts Department at British Columbia Institute of Technology. He holds a Master's Degree in Learning and Technology from Royal Roads University, where he researched strategies for effectively implementing social media in the classroom. He is co-chair of the open education working group at BCIT, and a co-author of the open textbook Graphic Design and Print Production Fundamentals. With over 18 years of prior experience in printed communications, and a strong background in typography and design, Ken has become a keen advocate for open educational practices in higher education.
"One of the challenges I see that still impedes the growth of open education is that of translating visibility into action. As support grows stronger, the message needs to be clear. There are so many benefits to going open, but sharing successes (and lessons learned) with a wider audience can still be difficult. Marketing the "idea" of open education seems easy, but delivering a compelling call to action is equally as important. With traditional academic publishing models being constantly challenged, it is more important than ever to expand our advocacy to as many educational avenues as possible. I hope to be able to contribute by inspiring others to take action.
Rajiv Jhangiani
Dr. Rajiv Jhangiani is the University Teaching Fellow in Open Studies and a Psychology Instructor at Kwantlen Polytechnic University, where he conducts research in open education and the scholarship of teaching and learning. A recipient of the Robert E. Knox Master Teacher Award from UBC and the Dean of Arts Teaching Excellence award at KPU, Dr. Jhangiani serves as the Senior Open Education Advocacy and Research Fellow with BCcampus, an Associate Editor of Psychology Learning and Teaching, and a faculty workshop facilitator with the Open Textbook Network. Dr. Jhangiani has revised two open textbooks—for Research Methods and Social Psychology—and advocates for the adoption of open educational and science practices. His books include A Compendium of Scales for Use in the Scholarship of Teaching and Learning (2015) and Open: The Philosophy and Practices that are Revolutionizing Education and Science (forthcoming, Ubiquity Press).
I am looking forward to working closely with Jennifer, Ken, and George to design and conduct rigorous and relevant research about open educational practices in British Columbia. In addition to supporting their development as open education researchers, I expect that the data we collect will yield practical insights that will prove invaluable to faculty, students, administrators, and policymakers who are interested in harnessing the benefits of open education. By the end of this year, I hope that we will have succeeded in substantially increasing awareness of and capacity for open educational practices within our disciplines and across B.C. institutions.
George Veletsianos
Dr. George Veletsianos is Canada Research Chair in Innovative Learning and Technology and an Associate Professor at Royal Roads University. His research aims to understand and improve teaching and learning in emerging digital environments. He achieves this by examining the practices and experiences of learners, educators, and scholars with/in online learning, social media, and open education.
I am hoping to make better sense of the open education landscape in B.C., and by so doing, support provincial efforts to improve teaching and learning. I am also hoping to collaborate on a variety of research endeavors with colleagues across the province and worldwide to advance our evidence-based understanding open practices.
]]> Continuum of Openness: Highlighting the RRU MALAT program
26 Jan 2017 16:41:03 +0000 is full of acronyms, as we're sure you all know, and you're probably wondering what the RRU MALAT program is all about so let us introduce you to the Royal Roads University Master of Arts in Learning and Technology (RRU MALAT) program. Over the past 2 years, this program has undergone a significant redesign including extensive consultations with various stakeholder groups. A core value and key principal of the program is openness, but not just in the form of open textbooks, but by applying the principles of networked learning, open pedagogy and digital mindset, with students creating and evaluating multiple digital learning environments.
As a leader and advocate for open education, BCcampus is excited to see programs that embrace open practices and embody an open philosophy, such as the RRU MALAT program.
WhileTo find out more about this fascinating program and of course, its connection to open education, we were eager to catch up withElizabeth Childs, Associate Professor and Program Head of the MALAT program in the School of Education and Technology at Royal Roads University to ask her a few questions. Here's what she had to say:
Can you describe how the program is changing as a result of openness as a core value?
We began by looking at the five Rs of openness (reuse, revise, remix, redistribute and retain) and how they might support what we were trying to achieve in the program redesign. At Royal Roads, we try to think beyond the independent, standalone courses, creating a synergy across all courses. After the 5-year program review, we were able to step back and look at the courses and the program systematically to see how they all work together and how we could design it along the continuum of openness which lead us to where the program is now. It has changed in a number of ways, such as:
Program delivery. We will be delivering the program in WordPress, which means that the content, activities, blog posts, assignments, etc. will be open and publicly available which is a first for Royal Roads.
Digital footprint. Students will gain and maintain their own WordPress site throughout the program, allowing them to take an active and participatory role in the wider education community. They will be cultivating their digital identity and presence as they grow to be seen as a valuable contributor in this space.
Readings and resources. First and foremost, we aim to use open resources, journals, and peer-reviewed articles.
Personalization of the learning experience. There are a variety of changes in this realm, making the program customizable by the students. The learner will create their own learning plan, drawing on content and experiences from outside of the university and bring it back to the program – enhancing the overall learning outcome.
Networks. The strength of the cohort model is great, but in today's connected world, a cohort is small. We need to focus on the network that students can build, cultivate and contribute to throughout the program to cross that traditional boundary, taking this network with them when they graduate. Lifelong learning includes a network to learn from and collaborate with.
Why is it important to the program to include open as a core value?
It might sound silly, but at a base level, openness is about sharing, kindness, generosity and optimism. As people who work in designing learning environments and experiences, this is inherent in all that we do – design things to then make those things better.
The RRU Teaching and Learning Model speaks to authentic experiential and inquiry-based learning, outcomes-based learning, enhanced learning through technology so when we lined these characteristics up with principles of openness and open pedagogy broadly defined there was a direct dovetail.
During the program redesign, Royal Roads put design documents on Moodle and pushed them out via social media for others to weigh in, offer value, feedback, etc. Clint Lalonde, Manager, Educational Technologies, BCcampus took this opportunity to contribute and collaborate as an RRU MALAT Alumni, but also as a lifelong learner in educational technology.
Notable quote:
"The RRU MALAT program will help develop educators who not only understand that learning in an open, networked world is different, but that will also help them maximize the potential of open, networked learning. This is an education program that, first and foremost, fundamentally acknowledges the unique learning affordances of the Internet." – Clint Lalonde, Manager, Educational Technologies, BCcampus
Webinar series:
Jan 31st webinar recording: Open Education Stories from Across B.C
This webinar focused on open education stories from across British Columbia, and explored a case study on the makings of the RRU MALAT program.
Faculty Experiences Adopting Openness as a Core Value
This is #1 in a series of webinars – Open Education Stories from Across B.C.
Presentation Description
This presentation reports on the experiences faculty members designing and developing a Master's degree in Learning and Technology when they adopted openness as a core value and key design principle.
A growing body of evidence suggests that adoption of open educational resources (OER), and especially open textbooks, leads to lower costs for students without having negative impacts on academic outcomes (e.g., Hilton, in press; Wiley, Williams, DeMarte, & Hilton, 2016). WhileThe MA program that we will present represents a case study for the open community. In this degree, students contribute meaningfully to digital learning networks and communities in the field. The degree prepares students to work in the creation and evaluation of digital learning environments and apply theoretical and practical knowledge to critically analyze learning innovations and assess their impact on organizations and society. Openness is central to the achievement of this program goal. Openness was adopted as a program value predicated on the philosophical stance that open practices lead to collaboration and the development of a digital mindset that values sharing and cultivates networked learning.
In this case study of an MA program, open practices are evidenced at the course level in the design of the online experience; through the use of "renewable assignments"; authentic assessment opportunities, in resource curation and, through online facilitation approaches. At the program level, it manifests itself in the intentional open spaces that have been designed into the program which require student identification and completion of personalized areas of inquiry. It is also evidenced through the adoption of open pedagogy as a design principle that informs design and delivery decisions at the course and program level
Initial tensions such as coming to a common understanding of openness and what openness can be within the constraints of an institution; how openness supports or detracts from online community; the role of openness in the creation of safe learning environments and, ways to support adjunct faculty for designing for openness will be highlighted in this presentation. This MA program is currently under development so by November additional strategies that have been used to develop a common understanding around openness and open pedagogy at the faculty and institutional level will also be discussed. In addition, various supports that have been used in working with adjunct faculty in course design with openness as a core value will be shared.
View other webinars in the series:In 2004, BCcampus initiated the project to implement a repository for sharing content developed through the Online Program Development Fund (OPDF). The concept of a learning resource repository was developed in a series of working groups where faculty members from BC post-secondary institutions played a key advisory role.
In April 2006, BCcampus launched SOL*R as a service for British Columbia public post-secondary educators. Since that time, hundreds of online learning resources have been contributed to SOL*R for sharing and reuse.
SOL*R is a repository service provided by BCcampus that allows educators to access FREE online learning resources. It facilitates sharing, discovery, reuse, and remixing of a growing collection of content created by BC post-secondary educators.
Screen shot of the SOL*R Repository
SOL*R includes learning resources from a wide variety of disciplines and subject areas. Resources range from open textbooks, individual learning activities and tools, all the way to full programs.
Evolution of Content
SOL*R was initially seeded with content funded by the OPDF. As part of an OPDF project, developers were required to license their learning resources openly and then contribute them to SOL*R. While initial content was provided by the OPDF, all online learning resources are welcome and SOL*R now contains non-OPDF resources as well.
Moving Forward
In September 2016 BCcampus received notification from Equella, the robust engine behind SOL*R, that Equella would be entering into a Sustain mode- i.e. there would not be further development or technical support provided for the Equella system. Due to the number of resources supported by BCcampus it quickly became evident that an alternative solution for a repository engine was required.
Since September 2016 BCcampus has been engaged in the review of a variety of open source repository platforms. The initial review included identifying the required functional specifications for the new open source repository.
Based on those specifications the task was then to identify which open source repositories best met our needs This resulted in some excellent and very informative conversations with collegues Scott Leslie at the BC Libraries Co-op (the original SOLR guru and implementer, and Sunni Nishimura and Brandon Weigel of BC ELN. Their organizations have embarked upon similar projects and their expertise was invaluable.
As a result of the conversations and research, BCcampus will be migrating the SOL*R repository to DSpace . DSpace is currently being used in a number of BC institutions including UBC and UVic. More information about DSpace can be found on the website. Our decision to move to DSpace is based on a number of factors, most importantly these include- heavily documented materials and resources, a robust community of developers; and the fact that it is in alignment with BCcampus's API requirements.
DSpace
Timeline of activity
While BCcampus is at the initial stages of setting up a DSpace instance and establishing a technical process for migrating over 3,000 open educational resources from SOL*R to DSpace, we will keep the communication open and transparent about the process of the Repository Replacement Project.
The repository will officially launch mid September 2017.
As the process continues the BCcampus team is available for questions and comments you might have as one of our users and community members. Please direct all questions to Amanda Coolidge (acoolidge.at.bccampus.dot.ca)
]]>New OER/Ancillary Resource: Open Case Studies from UBC
Wed, 21 Dec 2016 15:31:52 +0000 very happy to announce that one of our OER Grant recipients, Christina Hendricks from UBC, has completed the "Open Case Studies" website. This project will continue to evolve as more open case studies are added to the site over the years.
Below are a few links to individual case studies, as ancillary resources, to some of the books in our collection:
Would you like to help create and publish an Open Textbook, "An Introduction to Philosophy," and at the same time contribute to developing a new, collaborative model for publishing Open Textbooks?"An Introduction to Philosophy", in development by lead author Christina Hendricks (University of British Columbia), is one of a handful of Open Textbook projects being undertaken with the support of the Rebus Community for Open Textbook Creation*.
We're looking for people to volunteer time, with a variety of tasks envisaged, from chapter contributions, to proofreading and a host of other things. There will be space for everyone from tenured faculty to students to the general public to participate in the project, though chapters will be written by faculty or graduate students (preferably PhD students) in Philosophy.Interested in helping? You can let us know by doing the following:
Or, if you have more questions you could send an email to: hugh@rebus.foundation .
]]>Announcement: Senior Open Education Advocacy and Research Fellow
17 Nov 2016 18:29:37 +0000 News! The BCcampus Open Education Team is making a change to the Faculty Fellows program. As of January 2017 the Faculty Fellows program will be known as the Open Education Advocacy and Research Group. The group will include a lead, Senior Open Education Advocacy and Research Fellow and three Advocacy and Research Fellows. The term of the 2017 Open Education Advocacy and Research Fellowship Program will be January 2017-December 2017.
We are pleased to announce that the Senior Open Education Advocacy and Research Fellow for January 2017-December 2017 is Dr. Rajiv Jhangiani.
Dr. Rajiv Jhangiani
Dr. Jhangiani is a Psychology faculty member at Kwantlen Polytechnic Unviersity as well as the Open Studies Teaching Fellow at KPU. Dr. Jhangiani is a passionate advocate for Open Education both provincially and internationally. His contribution to research in the field of Open Education is extensive and we look forward to his continued work in this field as he works with and leads the Open Education Advocacy and Research Fellowship Program.
The Senior Open Education Advocacy and Research Fellow is responsible for the following:
Lead a team of faculty researchers (3) in the development of Open Education Research, e.g. 1 cumulative study conducted by the entire team. The research should relate to the educational outcomes that take place when OER are substituted for traditional learning materials and should be specifically focussed on the BC context.
Organize a one-day face-to-face or virtual meeting with all successful candidates in January to discuss deliverables, potential research question(s), etc.
Support team of researchers in the research stages.
Mentor team of researchers and conduct monthly meetings to ensure success of research
Mentor and support team in the submission of research findings to one open access journal
Mentor and support team in publishing blog posts to the BCcampus Open Education blog
Mentor team members on open education advocacy and best practices in communicating Open Education messaging in BC institutions.
Hire and supervise open education research graduate assistants, as needed for support of research data and findings
For B.C. faculty interested in applying to be one of the three Open Education Advocacy and Research Fellows we will be putting out a call for applications in the next couple of weeks. Stay Tuned!!
Should you have questions about the program please email: Amanda Coolidge, Senior Manager of Open Education (acoolidge@bccampus.ca)
The 2016 Open Education conference, in Richmond, Virginia, is wrapping up today. As always, this event generates and inspires new ideas for me and my colleagues at BCcampus. This year's gathering was both familiar and different; it was a mixture of the old and new.
New People
I've noticed that there are more folks new to open textbooks and OER at this year's event. This is encouraging because it means that this idea of creating and providing open educational resources at the institutional and faculty level is catching on. And it's becoming important enough that colleges and universities, institutes and schools are spending the money to send their instructors and staff to this conference to learn.
Someone new to OER and open education hasn't yet acquired a foundation of knowledge on this topic so they, of course, ask a lot of questions. This is a wonderful opportunity for those of us well versed in OER to teach and, as a result, clarify and refine our understanding of open education.
While attending this conference, I've been reminded that these new people can teach us old timers a few things. Those with new eyes see OER from a fresh perspective. When looking at OER support materials, for example, "new eyes" see the gaps or confusion in our instructions and procedures. New eyes, and minds, often draw conclusions about the OER world that the rest of us haven't thought of. New OER converts are the "canaries in the coal mine" as they move through the OER maze trying to find their way as they challenge the rest of us on what's missing or unclear, and help us do better work.
People with new eyes can be the innovators of this innovative movement, without knowing it. I met a professor from a Midwest community college on my flight here who described a very comprehensive OER training program for faculty that she's spearheaded and now leads at her institution. Not only does this program educate her college's instructors about OER, but shows them how to create a plan to adopt, adapt, and/or author OER for their classroom.
Wow!
Reinventing the Wheel
It's gratifying to hear about all of the projects people are working on. However, it does distress me to learn that the same or similar work is taking place at different institutions, without awareness between the participants, that this is happening.
We often talk of not "reinventing the wheel" in regards to the creation of OER. Less often do we discuss applying this concept to our OER projects and plans. For instance, at BCcampus I'm responsible for maintaining the B.C. Open Textbook Collection and was very interested to learn what the Open Textbook Network Library manager and OpenSUNY manager were doing. I learned that we had some similar items on our work agenda. It seemed a shame that each of us was spending precious time and energy on tackling the same issues in isolation. So, we have started a discussion around how we can help each other lighten the load by sharing our individual ideas.
The spreadsheet, detail-oriented part of me wishes there was an easier way to systematically track all of these efforts.
The OER World Map has a goal "…to provide the most complete and comprehensible picture of the global Open Educational Resources (OER) movement. Information is collected on people, organizations, services, projects and events related to OER. " Contributions to this tracker is, however, voluntary and therefore incomplete.
While I will certainly refer to this map often, I will continue to hunt down colleagues who are doing similar work. If open education is about sharing and collaboration, I want to make sure that I apply this philosophy to my work plan. In other words, I want to make sure that I'm not doing more than I need to if someone else has already done the work.
Seeing your Face
I've been working remotely for 30 years. This started out with sitting in my home office as a writer and connecting with people by phone, then — years later — fax, and finally through email. Today, online work is standard for many in the workforce. It certainly is the method I use to communicate with a broad array of individuals that I've never meet face-to-face.
At each Open Education conference, I get to put a face and hand shake to the email addresses I've been writing to for months. In my humble opinion, all of life — including work — is based on relationships. So getting to meet my at-a-distance colleagues, look them in the eye, share a meal and drink, talk about our families and lives outside of work, deepens that professional relationship and makes this work more effective and enjoyable.
Inspiration and the 5Rs
Lastly, these conferences are inspiring: nothing new here. They generate in my mind, and the minds of my BCcampus colleagues, new ideas and the fuel to continue this work throughout the year. I love to pull out business cards I've collected during the three days of OpenEd and remember whom I've talked with, and who has sparked in me a new idea, filled me with admiration for the work they've done, and moved me to seek out new answers to the challenges we face in British Columbia and Canada.
A conference like OpenEd is, to me, a time to recharge my OER battery and see what else I can learn. It is a remixing of ideas provoked through conversation and presentations. And once remixed, concepts about OER are often revised. And then retained, and reused throughout the year. Finally, when the following year's Open Education conference arrives, we get to redistribute our ideas and accomplishments from the past 12 months. While, of course, looking for new eyes, old faces, and more inspiration. | 677.169 | 1 |
Archimedes & The Law of The Lever
Informazione importanti
Corso
Online
Durata: 6 Weeks
Quando: Flessible
DescrizioneTeacher Education
Mathematics
Maths
Law
Teachers
Programma
This is a Professional Development course for high school maths and science teachers developed at the School of Mathematics and Statistics, Faculty of Science, UNSW.
This purely online course will introduce teachers to Archimedes' law of the lever, centres of mass, expected values of probability distributions, barycentric coordinates, and interesting applications via online videos.
There will also be fun activities and challenges that will develop and deepen your mathematical understanding, and give you ideas and materials for engaging your students.
You will be able to interact with fellow educators from around the country. | 677.169 | 1 |
Algebra - Functions, Tables, and Graphs Worksheet
Word Document File
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0.03 MB | 2 pages
PRODUCT DESCRIPTION
This worksheet shows the relationship between functions, tables, and graphs. Students are given one piece of information and asked to find the other two pieces. Graphs are provided when graphing is required | 677.169 | 1 |
Please contact your nearest Dymocks store to confirm availability
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Discrete mathematics has now established its place in most undergraduate mathematics courses. This textbook provides a concise, readable and accessible introduction to a number of topics in this area, such as enumeration, graph theory, Latin squares and designs. It is aimed at second-year undergraduate mathematics students, and provides them with many of the basic techniques, ideas and results. It contains many worked examples, and each chapter ends with a large number of exercises, with hints or solutions provided for most of them. As well as including standard topics such as binomial coefficients, recurrence, the inclusion-exclusion principle, trees, Hamiltonian and Eulerian graphs, Latin squares and finite projective planes, the text also includes material on the mnage problem, magic squares, Catalan and Stirling numbers, and tournament schedules. | 677.169 | 1 |
Pa-ra-boo-las (Quadratic Ghosts)
PDF (Acrobat) Document File
Be sure that you have an application to open this file type before downloading and/or purchasing.
0.43 MB | 11 pages
PRODUCT DESCRIPTION
Challenge your students to create pa-ra-boo-las, ghosts made entirely out of parabolas. This classroom tested activity will help students better understand the parameters of quadratic polynomials and how they relate to their graphs. The 11 page pdf includes directions, examples, and the link to the permanent online applet. If you enjoy this activity, visit my store, Natual Numbers - The Healthier Choice | 677.169 | 1 |
Sue McClure
Sue McClure is a lecturer in the School of Mathematical and Statistical Sciences at Arizona State University. Educated at Ball State University, Purdue University, and Indiana University, Sue has acquired years of experience teaching courses ranging from high school mathematics to college calculus. Her efforts in the Mathematics Department at Angola High School helped rank the school as one of Indiana's finest high schools, and her interest in educational technologies has led Sue to explore and integrate personalized learning through adaptive mathematics and online education into her courses at Arizona State University.
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Learn the basics of Algebra while preparing for future courses in Calculus through this credit-eligible college level math course. In this college level Algebra course, you will learn to apply algebraic reasoning to solve problems effectively. You'll develop skills in linear and quadratic functions, general polynomial functions, rational functions, and exponential and logarithmic functions. You will also study systems of linear equations. This course will emphasize problem-solving techniques, specifically by means of discussing concepts in each of | 677.169 | 1 |
Calculators in the new A level Mathematics (from 2017)
Ofqual's subject-level conditions and requirements for Mathematics and Further Mathematics state that calculators used must include the following features:
an iterative function
the ability to compute summary statistics and access probabilities from
standard statistical distributions
the ability to perform calculations with matrices up to at least order 3 x 3 (FM only)
Approved graphical calculators will be permitted in the examinations. Approved graphical calculators include all of the requirements listed above and can be a useful tool for you and your students that will help you to embed the use of technology into teaching and learning, as required by the new mathematics A levels. They can help students with some questions on the examination.
Further guidance about use of calculators, along with advice about suitable models, will appear on this page. | 677.169 | 1 |
Product Description:
This book is designed for students studying Information Systems, especially Computer Programming or Computer electronics at a two-year college. The text, due to its content and organization is to be used in a two-academic quarter sequence or in a one-semester course. This textbook has an Intuitive Approach; each topic is carefully explained and illustrated with appropriate examples. The textbook also has plenty of Chapter Exercises for the student. This provides more than adequate opportunity to practice and master each concept. This textbook by Lance has been Mohican Publishing Company's top-selling book for the past five years. *A Solutions Manual is also available. Table of Contents: Part I - Chapter 1: Mathematics: A Necessary Tool for Data Processing and Computer Programming; Chapter 2: Computer Arithmetic; Chapter 3: Algorithms, Flowcharts, Pseudocode, and Decision Tables; Chapter 4: Decimal and Nondecimal Numeration Systems; Chapter 5: Sets; Chapter 6: Intermediate Algebra. Table of Contents: Part 2: Chapter 7: Functions; Chapter 8: Logic and Computer Programming; Chapter 9: Boolean Algebra; Chapter 10: Systems of Equations; Chapter 11: Arrays, Matrices and Determinants; Chapter 12: Linear Programming; Chapter 13: Statistics and Probability, Contents: A - Appendix, B - Appendix, C - Appendix - ASCII (American Standard Code for Information Interchange Table); Glossary, Solutions of Odd-Numbered Exercises
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REVIEWS for Computer | 677.169 | 1 |
Linear Algebra
Recommended School Credit: One full year of high school credit equal to or greater than an AP class or one semester of college credit
Course Code: LIN
Course Description
Description
Linear Algebra is an online and individually-paced course equivalent to a first-year college linear algebra course. This course covers the entire syllabus from the Johns Hopkins semester-based, in-person Linear Algebra course, plus several additional topics. Computer based interactives, homeworks and quizzes help to reinforce concepts taught in the class. Projects covering advanced applications will introduce students to mathematical typesetting with LaTeX. Online course materials supplement the required textbook.
Each student is assigned to a CTY instructor to help them during their course. Students can contact their instructor via email with any questions or concerns at any time. Live one-on-one online review sessions can be scheduled as well to prepare for the graded assessments, which include quizzes, homework, midterm exams, and a cumulative final. Instructors use virtual classroom software allowing video, voice, text, screen sharing and whiteboard interaction.
Topics include:
Linear Equations
Matrix Algebra
Determinants
Vector Spaces
Eigenvalues
Orthogonality
Least Squares
Symmetric Matrices
Quadratic Forms
For a detailed list of topics, click the "List of Topics" tab.
Materials Needed
A textbook purchase is required for this course.
Linear Algebra and its Applications, 4th edition, by David C. Lay.
ISBN 13: 978-0-321-38517-8
List of Topics
Upon successful completion of the course, students will be able to demonstrate mastery over the following topics:
Linear Equations in Linear Algebra
Systems of Linear Equations
Row Reduction and Echelon Form
Vector Equations
The Matrix Equation Ax->=b->
Solution Sets of Linear Systems
Applications of Linear Systems
Linear Independence
Introduction to Linear Transformations
The Matrix of a Linear Transformation
Matrix Algebra
Matrix Operations
The Inverse of a Matrix
Characterizations of Invertible Matrices
Subspaces of Rn
Dimension and Rank
Determinants
Introduction to Determinants
Properties of Determinants
Cramer's Rule, Volume, and Linear Transformations
Eigenvalues and Eigenvectors
Eigenvalues and Eigenvectors
The Characteristic Equation
Diagonalization
Eigenvalues and Linear Transformations
Complex Eigenvalues
Discrete Dynamical Systems
Vector Spaces
Vector Spaces and Subspaces
Null Spaces, Column Spaces, and Linear Transformations
Linearly Independent Sets; Bases
Coordinate Systems
The Dimension of a Vector Space
Rank
Change of Basis
Applications to Difference Equations
Applications to Markov Chains
Orthogonality and Least Squares
Inner Products Length, and Orthogonality
Orthogonal Sets
Orthogonal Projections
The Gram-Schmidt Process
Least-Squares Problem
Inner Product Spaces
Symmetric Matrices and Quadratic Forms
Technical Requirements
This course requires a properly maintained computer with high-speed internet access and an up-to-date web browser (such as Chrome or Firefox). The student must be able to communicate with the instructor via email. Visit the Technical Requirements and Support page for more details.
This course uses an online virtual classroom for discussions with the instructor. The classroom works on standard computers with the Adobe Flash plugin, and also tablets or handhelds that support the Adobe Connect Mobile app. | 677.169 | 1 |
2
Evaluation Assignments –4 each worth 10% –Late assignments will be accepted up to 24 hrs late with a penalty of 25% –You are encouraged to discuss the problems with other students however, the actual write up must be an individual effort –You must be able to reproduce any solution that you submit. The penalty for cheating ranges from a zero on the assignment to suspension from the university
5
Course Expectations Expectations of the lecturer Give clear, organized lectures Assign fair, challenging assignments that ensure that you, the student, understand the material Be available for help in office hours Help every student achieve their goals in the course (this requires your help!)
6
Course Expectations Expectations of the student Attend lectures and participate Bring course notes to class Review lecture notes after each class, not just before the exam Complete homework fully, neatly and independently Have respect for your classmates and lecturers
7
Course Expectations What does neatly mean? Staple sheets Write legibly (if you are incapable of this skill, please type) Your work should be of a quality that you would feel comfortable giving to your boss in a work environment.
8
Discrete Mathematics Who needs it? Anyone in computer science or a mathematical science Why? In CS we need to be able to speak precisely without ambiguity analyze problems and formulate solutions apply the concepts associated with probability, graph theory and counting theory.
13
How Do I Become Good At This Stuff? Same way that you become a good hockey player, calculus student, violinist… practice, practice, practice...
14
14 Weighted Job Scheduling Problem Given a set of jobs where each job has a value or weight jobs have a start time and duration Schedule the jobs so that the total value is maximized scheduled jobs do not overlap Q. What are some examples of this type of problem? A. printer queue, airplanes on runway, server use, scheduling surgeries…
15
15 Some Examples Time
16
Some Examples Time
17
17 Finding an Algorithm Q. What is an algorithm? A. A step-by-step procedure for solving a problem or accomplishing some end especially by a computer. * Lets list some potential algorithms to try: Earliest start time. * Merrian-Webster Time
20
20 So...Whats the Answer? Use a mathematical tool called Dynamic Programming. You can read about it in a text such as Algorithm Design by Kleinberg and Tardos. Or visit this link: programming-2x2.pdf Or take CSCC73.
21
21 A Simpler Problem Interval Scheduling Problem Given a set of jobs where jobs have a start time and finish time Schedule the jobs so that the number of jobs scheduled is maximized jobs do not overlap Notation J : The set of jobs j i : The i th job s i : The start time of the i th job f i : The finish time of the i th job
27
27 Is It Correct? Q. How do we know our algorithm is correct? A. Prove it! One common proof technique is Proof by Contradiction Idea Play Devils Advocate Assume our solution is not the best This means there is a better solution B Show that our algorithms solution is as good as B by making B equivalent to our solution.
28
28 The Proof Let S be the schedule our algorithm creates. S = (s 1, s 2, s 3,..., s n ) Let B be a better solution. B = (b 1, b 2, b 3,..., b m ) Q. What do we know about m and n? A. m >= n Q. What do we know about b 1 and s 1 ? A. The finish time for s 1 is less than or equal to the finish time for b 1. in order of non- decreasing finish time
29
29 The Proof S = (s 1, s 2, s 3,..., s n ) B = (b 1, b 2, b 3,..., b m ) Q. If the finish time for s 1 is less then or equal to the finish time for b 1, what can we say about B* = (s 1, b 2, b 3,..., b m )? A. |B*| = |B|. So B* is as good a solution as B but is more similar to S. Q. How should the proof finish? A. Repeat the argument with S = (s 2, s 3,…, s n ) and B = (b 2, b 3,…, b m ). Q. Why does this work? A. Induction - more on this later.
30
30 Challenge Another Scheduling Problem Given a set of employees, want to set up a meeting that everyone can attend. Each person has a calendar which says whether they are available for any given time slot during the day. Give an efficient algorithm to schedule the meeting so that everyone can attend (if possible). | 677.169 | 1 |
Week 16: Graphs Chapters 30 and 31. We have already studies special cases of graphs: linked lists and trees Definitions : A graph is a collection of vertices and edges that connect the vertices in pairs. Edges with direction are directed ; graphs with direction are digraphs Undirected edges go in both directions. A path is a sequence of edges connecting two vertices A simple path repeats no vertex A cycle is a graph where the first vertex = the last vertex Weights are values on edges (i.e. miles if a map, voltage if a wiring diagram, etc.) Task 1: Do #1 on page 794 and check your answer in the appendix. More definitions : A connected graph has a path from / to every vertex. A complete graph has edges connecting every pair of vertices. (there will be pow(v,2) edges needed) Two vertices are adjacent (neighbors) if they are connected by an edge A tree is a graph with no cycles A forest is a collection of trees. Task 2:
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Summary and Info
This is a practical anthology of some of the best elementary problems in different branches of mathematics. They are selected for their aesthetic appeal as well as their instructional value, and are organized to highlight the most common problem-solving techniques encountered in undergraduate mathematics. Readers learn important principles and broad strategies for coping with the experience of solving problems, while tackling specific cases on their own. The material is classroom tested and has been found particularly helpful for students preparing for the Putnam exam. For easy reference, the problems are arranged by subject.
More About the Author
Andrea Scutellari or Andrea da Viadana (16th-century) was an Italian painter of the Renaissance period. | 677.169 | 1 |
Walter Rudin:
Mathematics Series This is an advanced text for the one- or two-semester course in analysis taught primarily to math, science, computer science, and electrical engineering majors at the junior, senior or graduate level. Real and Complex Analysis: This is an advanced text for the one- or two-semester course in analysis taught primarily to math, science, computer science, and electrical engineering majors at the junior, senior or graduate level., Mcgraw-Hill Education Ltd | 677.169 | 1 |
Product Description:
Alice Kaseberg's respected Intermediate Algebra: Everyday Explorations, Fourth Edition, helps students build confidence in algebra. This text's popularity is attributable to the author's use of guided discovery, explorations, and problem solving, all of which help students learn new concepts and strengthen their skill retention. Known for an informal, interactive style that makes algebra more accessible to students while maintaining a high level of mathematical accuracy, Intermediate Algebra includes a host of teaching and learning tools that work together for maximum flexibility and a high student success rate. With the Fourth Edition, instructors have access to an Instructor's Annotated Edition that provides additional examples, as well as a robust Instructor's Resource Manual, algorithmic computerized testing, and an extensive online homework system.
REVIEWS for Intermediate Algebra | 677.169 | 1 |
Advanced Higher Maths
The aim of Advanced Higher Mathematics is to extend further students' mathematical experience in Pure Maths, as well as providing an opportunity to study some Applied Maths. It is particularly useful for students who may go on to study degree courses that involve Mathematics.
Course content
Unit 1
Unit 2
Unit 3
Algebra
Further Differentiation
Vectors
Differentation
Further Integration
Matrix Algebra
Integration
Complex Numbers
Futher Sequences and Series
Properties of Functions
Sequences and Series
Ordinary differential Equations
Matrices
Number Theory and methods of Proof
Further Number Theory and Proofs
Assessment
To gain the overall award students must pass written assessments in all three units of the course and an external assessment. There is a single SQA external exam paper. A prelim, which is of the same form as the SQA external exam, takes place under exam conditions in February.
Advanced Higher Notes
Revision
Homework
Past Papers and Solutions
SQA Past Papers
SQA Solutions | 677.169 | 1 |
Class 11 Physics/Maths Ebook Download|NCERT Solutions Pdf
Physics and Maths is an important and scoring subjects.This page contains Class 11 Physics and Maths Ebook Download along
with NCERT Solutions Pdf .
We also have a page dedicated to physics class 11
and class 11 maths
where you can find easy to follow notes , assignments, worksheets and much more.
Our aim through this page is to provide all the required material for students success in examination and better understanding of mathematics
and physics as a whole. Hope you like them and do not forget to like , social share
and comment at the end of the page. | 677.169 | 1 |
Online math classes
ALISON provides free online math courses to help you gain more knowledge in algebra, geometry and statistics. Start ALISON online math programs today. Free online math classes can help you master the basics without having to struggle through complicated textbooks or pay for a tutor. Check out this collection of the. NetMath is a distance learning program that offers online math courses for college credit. Our mission is to bring the academic resources from one of the nation's top. Students who searched for Where to Find Free Math Courses Online found the following information relevant and useful.
Online math classes
Free online math classes can help you master the basics without having to struggle through complicated textbooks or pay for a tutor. Check out this collection of the. An undergraduate degree in mathematics provides an excellent basis for graduate work in mathematics or computer science, or for employment in such mathematics. Save money & earn college credits with the affordable, self-paced online college math courses like Algebra, Calculus, Statistics & more.
Study.com has engaging online math courses in pre-algebra, algebra, geometry, statistics, calculus, and more! Our self-paced video lessons can help you study for. NetMath is a distance learning program that offers online math courses for college credit. Our mission is to bring the academic resources from one of the nation's top. Online math courses in geometry, algebra, basic math, calculus and statistics for adult learners, highschool and college students.
Students who searched for Where to Find Free Math Courses Online found the following information relevant and useful. Free online math courses. Explore, create, and track courses from the world's top universities. | 677.169 | 1 |
$210Partial differential equations (PDEs) are used to describe a large variety of physical phemena, from fluid flow to electromagnetic fields, and are indispensable to such disparate fields as aircraft simulation and computer graphics. While most existing texts on PDEs deal with either analytical or numerical aspects of PDEs, this invative and comprehensive textbook features a unique approach that integrates analysis and numerical solution methods and includes a third component - modeling - to address real-life problems. The authors believe that modeling can be learned only by doing; hence a separate chapter containing 16 user-friendly case studies of elliptic, parabolic, and hyperbolic equations is included, and numerous exercises are included in all other chapters. | 677.169 | 1 |
Dayo
★★★★☆
21 February 2017
Cool....Maths rocks! It's just what I needed; a refresher!
David El
★★★★★
30 August 2016
Wow...! Really Good book for revising the basics
Description
Mathematical skills are an unavoidable prerequisite for beginners of a study at a university. This book covers all major subjects from a secondary school education in mathematics which are required for studying successfully. Among others, it deals with mathematical foundations, real numbers, equations and inequalities, geometry, sequences, functions, differentation, integration, vectors and probability theory. The book presents the major facts from each subject and contains many worked examples so that the reader gets practice in solving mathematical problems. At the end of each chapter, exercises from the particular subjects are given. The book will provide the reader with the required mathematical knowledge and tools for a contemporary study in economics, engineering, computer science and other disciplines.
Content
Some Mathematical Foundations
Real Numbers and Arithmetic Operations
Equations
Inequalities
Trigonometry and Goniometric Equations
Analytic Geometry in the Plane
Sequences and Partial Sums
Functions
Differentiation
Integration
Vectors
Combinatorics, Probability Theory and Statistics
This website uses cookies to improve user experience. By using our website you consent to all cookies in accordance with EU regulation. | 677.169 | 1 |
The Effects of Using the TI-92 Calculator To Enhance Junior High Students' Performance in and Attitude toward Geometry.
Ryan, Walter F.
This study extended over three months and involved 344 subjects. It compared the performance and attitudes of seventh-grade students taught geometry using the TI-92 calculator and a supplementary activity manual written to be used with the TI-92 calculator with the performance and attitudes of students using the traditional approach. On a comprehensive common post study geometry examination, the students who were taught geometry using the TI-92 calculator had significantly higher scores than those taught by traditional means. The students who were taught geometry using the TI-92 calculator had a significantly more positive attitude toward the study of geometry than students taught by traditional means. (Contains 14 references.) (Author/WRM) | 677.169 | 1 |
Providing an introduction to mathematical analysis as it applies to economic theory and econometrics, this book bridges the gap that has separated the teaching of basic mathematics for economics and the increasingly advanced mathematics demanded in economics research today. Dean Corbae, Maxwell B. Stinchcombe, and Juraj Zeman equip students with the knowledge of real and functional analysis and measure theory they need to read and do research in economic and econometric theory | 677.169 | 1 |
Functions
Assessment
• Final exam: 55% (May 2010) • Mid-term: 30% • Continuous Assessment: 15% Webwork (computer based
system)
• Homeworks in each tutorial but not part of assessment • Exam questions are based on the homeworks and the
Webwork
.Preliminaries
Chapter 1.
population modelling. . finance.
growth and decay
• Introduced by Isaac Newton (1642-1727) (and
independently by Gottfried Wilhelm Leibniz (1646-1716)) in his study of planetary motion
• Initially used primarily in physics and engineering • calculus now used in all physical sciences. and much more.Preliminaries
Chapter 1. motion.
ecology. .Functions
What is Calculus?
• Calculus is the quantitative science of change. biology and medicine.
.
you should have some understanding of what the calculus is and how it is used. By the end of the course.Preliminaries
Chapter 1.
.Functions
In this course you will get an introduction to some of the basic ideas and techniques of the Calculus. You should have also mastered some of the basic computations and rules of the calculus.
A precise definition is:
• Specify the input set A.
.Functions
What is a function?
Informal definition: A FUNCTION assigns to each element of an input set a unique element of an output set. • Specify the output set B. Read f (x) as "f of x". also called the domain. • For each x ∈ A we have to define an element f (x) in the
set B. f (x) is called the function value of f at x.Preliminaries
Chapter 1.
f (Moe) = Chocolate.
. f (Larry) = Vanilla. define the output set by B = {Chocolate. Define the domain (input set) by A = {Curly. Now define a function f : A → B (read: a function f from A to B) by f (Curly) = Chocolate.Functions
An example
A function from a set A to a set B can be given by explicitly providing a unique element of B for each element of A.
Here we think that Chocolate and Vanilla are the favourite ice-cream flavours. Larry}. Vanilla}.Preliminaries
Chapter 1. Moe.
Each student in A is asked what he perfers: chocolate or vanilla. We define a function f : A → B by f (x) := favourite ice-cream flavour of student x ∈ A. If we are dealing with a very big set (e. Vanilla}.Preliminaries
Chapter 1. the natural numbers) then this is the only way one can define a function.Functions
Favourite ice-cream flavours for students in MATH10070
Now let us define A as the set of all MATH10070 students.
. Here we define f (x) by means of a variable x.g. and the output set B = {Chocolate.
. and they seem to scare people!
If you see an x or a y in mathematics it is just standing in for an unspecified number.Functions
Variables
Letters are used a lot in mathematics. instead of full names.
This is a lot like using "he" or "she" in English.Preliminaries
Chapter 1.
division.
.Functions
6 + 2 × 3 + 4 =?
There are four basic arithmetic operations – multiplication.
Z
First multiplication and division.
So 6 + 2 × 3 + 4 equals 16. then addition and subtraction. addition and subtraction and there is a precedence in the order which these are to be carried out.Preliminaries
Chapter 1.
.Preliminaries
Chapter 1. Note that 3. Can write it as 3(4).
.Functions
Laziness.4 is a decimal.
We don't often use the multiplication sign ×. But can't shorten 3 × 4 to 34. The multiplication x × y is usually abbreviated to x · y or just xy .. | 677.169 | 1 |
ReadingGuide - Reading Guide for Ghahramani: Fundamentals...
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Unformatted text preview: Reading Guide for Ghahramani: Fundamentals of Probability I list here the parts of the textbook that directly support the material that I will have delivered in the lectures by Feb. 25. Sections 1.1, 1.2. From beginning to on page 3 to page 5, up to and including the paragraph after Example 1.6. Read the rest of this section, if the notation of set theory is unfamiliar. Comment: I write A B where the book writes AB . Section 1.3. Page 12. Definition (Probability Axioms). The examples on pages 15 17. Comment: These examples illustrate basic patters of notation that all probabilists use. All of your work (quizzes, homework and tests) should conform to these usages. Section 1.4. Page 18-19: Theorems 1.4 and 1.6. Sections 1.5, 1.6, 1.7. OMIT for now. Sections 2.1, 2.2. This is not about probability, but the principles discussed here have direct applications to probability: a) number of elements in the product of several sets (Theorems 2.1 and 2.2), b) number of subsets of a set (Theorem 2.3), and c)the idea of a tree diagram. the product of several sets (Theorems 2....
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This note was uploaded on 11/29/2011 for the course MATH 3355 taught by Professor Britt during the Spring '08 term at LSU. | 677.169 | 1 |
This course is an introduction to the theory of numbers.
Topics will include prime numbers, divisibility, congruences,
powers of an integer, quadratic reciprocity, greater integer
function, and diophantine functions.
Students will have experiences with recognition, representation, and
communication of relations found in naturally occurring situations.
The program can be customized to focus on areas of science or business
through free electives.
Medaille College has a long-standing reputation in Western New
York for producing some of the area's most effective teachers, and the
Mathematics program is closely aligned with the Medaille's degree in
Adolescent/Secondary (7-12) Education. The program emphasizes active
experiential learning, both inside and outside the classroom, through
frequent field observations and practice over four years, and students in
the Adolescent/Secondary (7-12) Education program may also complete
the Mathematics degree by taking at most 6 additional credits.
The current trend toward a technology- based economy places a growing
demand on mathematic and scientific skills, and as the world becomes
increasingly complex, the ability to collect relevant data, process it, and
make decisions based upon the results has increasing value. Mathematics
at Medaille provides students with a flexible background in these skills,
which can be applied to a variety of career and graduate school choices.
Many fields – from science to business to education - have growing
technological needs which mathematics training will help our students
fulfill.
PARENT/GUEST NAME:
Mathematics is more than just numbers: students will gain
experience in how to recognize, represent, and communicate
relations and patterns found in naturally occurring situations.
Medaille's Mathematics degree prepares learners with the
conceptual mathematical knowledge and the ability to apply
that knowledge to a variety of situations.
PARENT/GUEST NAME:
TOTAL NUMBER OF GUESTS*:
*additional guests are welcome
Please RSVP by Friday, October 5, 2007.
Please check the events you will attend
Friday, October 12, 2007
Bachelor of Science
� Student for a Day
� Environmental Sustainability;
Bringing the "Academic" to Life!
Mathematics Overview
Bachelor of Science Degree
As the world becomes increasingly complex, information skills become increasingly vital. The ability to collect
relevant data, process it, and make decisions based upon
the results is a common theme in industrialized society.
An aspect of the Mathematics program is to provide an
opportunity to enhance these information skills by covering the core areas of mathematics.
The curriculum is tailored to complement the degree
program in Adolescent Teaching: Mathematics 7-12 with
Special Education. Students in this degree program may
also complete the Mathematics Program by taking at
most 6 extra credits.
Credit Distribution
General Education Component
ENG 110 College Writing II (3)
ENG 200 Analytical Writing (3)
GEN 110 Introduction to Critical Thought (3)
GEN 220 A Global Perspective on Colonial North America (3)
GEN 230 Creative Expressions (3)
GEN 240 Scientific Discovery (3)
GEN 410 Baccalaureate Capstone I (3)
GEN 411 Baccalaureate Capstone II (3)
SPE 130 Fundamentals of Public Speaking (3)
MAT 251 Calculus I (3)
MAT 381 Statistics and Probability I (3)
Minimum General Education Credit Hours 33 | 677.169 | 1 |
Holt algebra 1 homework help
I have viewed and agree to the Terms of Use and Privacy Policy required. I am located outside US and understand HMH may share my information with its licensed. Pearson Prentice Hall and our other respected imprints provide educational materials, technologies, assessments and related services across the secondary curriculum. I have viewed and agree to the Terms of Use and Privacy Policy required. I am located outside US and understand HMH may share my information with its licensed. McDougal-Littell Algebra 2 Help with a Personal Math Teacher. Get the exact McDougal-Littell Algebra 2 help you need by entering the page number of your.
Thousands of users are using our software to conquer their algebra homework. Here are some of their experiences. Algebra 2 textbooks online are sought after by many algebra 2 students each year. The enormous popularity that algebra 2 textbooks online have has only gained. Algebra 2 textbooks online are sought after by many algebra 2 students each year. The enormous popularity that algebra 2 textbooks online have has only gained. Solutions in Holt Algebra 1 Homework and Practice Workbook (9780030466373. My.HRW.com is website that caters to both students and educators and helps both come together to make learning much easier. At "My HRW", students with a Holt or. We would like to show you a description here but the site won't allow us. Get the Remind app. Students and parents use the Remind app to get updates from their teachers. Visit one of the app stores and download the app to log in and view.
Holt algebra 1 homework help
View Your Algebra 2 Answers Now. Free. Browse the books below to find your textbook and get your solutions now. Table 1. Acculturation vs. Language Learning; Acculturatio. Language Learnin. Stage 1. The new culture is almost inaccessible. Frustration is constant. My.HRW.com is website that caters to both students and educators and helps both come together to make learning much easier. At "My HRW", students with a Holt or. Jefferson Middle School Algebra 1 Syllabus Welcome to the 8th grade at Jefferson Middle School – your last but very important year before High School.
ClassZone Book Finder. Follow these simple steps to find online resources for your book. Table 1. Acculturation vs. Language Learning; Acculturatio. Language Learnin. Stage 1. The new culture is almost inaccessible. Frustration is constant. View Your Algebra 2 Answers Now. Free. Browse the books below to find your textbook and get your solutions now. Get the Remind app. Students and parents use the Remind app to get updates from their teachers. Visit one of the app stores and download the app to log in and view.
Math Homework Help. Need math homework help? MathHelp.com's online math lessons are matched to your exact textbook and page number! Get started by. Solutions in Holt Algebra 1 Homework and Practice Workbook (9780030466373. McDougal-Littell Algebra 2 Help with a Personal Math Teacher. Get the exact McDougal-Littell Algebra 2 help you need by entering the page number of your. Activation Code: - Example: 1234567-10 or 12345678-100. When you click the Continue button, you will be asked to sign into ClassZone or to create a. Standards Documents • High School Mathematics Standards • Coordinate Algebra and Algebra I Crosswalk • Analytic Geometry and Geometry Crosswalk.
You have been redirected to Sadlier Connect from one of our product URLs where you used to access additional materials to support your Sadlier program.
Activation Code: - Example: 1234567-10 or 12345678-100. When you click the Continue button, you will be asked to sign into ClassZone or to create a. Thousands of users are using our software to conquer their algebra homework. Here are some of their experiences | 677.169 | 1 |
At the end of this course, students should be able to apply functions and equations to an contextual situation and mathematically model it to make appropriate inferences and conclusions based on their knowledge of different mathematical representations.
The course starts by providing a general overview of different families of functions so students can identify the key features that distinguish different families of functions in a variety of representations and then determine in which situations a given family is best for modeling. Each subsequent unit gives students opportunity to develop more depth of knowledge with a thorough understanding of how to use each function family including transformations, finding roots, understanding the effects of restricting domain, range and number sets, inverses, solving for key features including maxima and minima, roots, intercepts and end behavior and how to apply these tools to real life and abstract situations. This course culminates by allowing students to build on this knowledge and have further opportunity for real life application in probability and prediction and statistics and inference from data. | 677.169 | 1 |
80547625683
Educational Level:
Elementary/High School
Author:
Steck-Vaughn Company
Age Level:
Sixth Grade
Publication Year:
20110000
ISBN:
9780547625683
Detailed item info
Synopsis
This very practical series will help adolescents and adults alike to understand mathematics as it relates to their everyday lives. Each book covers basic math concepts and skills before exploring the more specific topics. Clear explanations are followed by ample practice. Each section also has a pretest, a section review, and posttest | 677.169 | 1 |
Career Academy Blended Mathematics
Career Academy Blended Mathematics (CABM is a series of exciting in–depth, multi–day projects that align with geometric, algebraic, and trigonometric topics found in every high school mathematics class.
CABM engages students in math with career-focused activities across a wide-range of career opportunities from Pasty Chef to NASCAR Racer. The career-focused themes equip students with the critical thinking abilities to prosper beyond high school and correspond with national, state and district mathematics standards.
Students work together planning to build a staircase in CABM's Staircase Design lesson | 677.169 | 1 |
Infinite Sequence and Series
From the Basic of Sequences and series you will become Master in Convergent and Divergent series and Much more ...
4The Calculus II Course on INFINITE SEQUENCE AND SERIES assume that you've got a working knowledge Calculus I, including Limits, Derivatives, and Integration (up to basic substitution). It is also assumed that you have a fairly good knowledge of Trig. Several topics rely heavily on trig and knowledge of trig functions.
Description
This course is not only virtual learning material . You can ask any questions related to Infinite Sequence and series after enrolling in this course.
Here is a listing and brief description of the topics in this course.
A brief discussion of sequences. The basic terminology and convergence of sequences.
Quick look about monotonic and bounded sequences.
Series The Basics we will discuss some of the basics of infinite series.
Series Convergence/Divergence about the convergence/divergence of a series
Series Special Series Geometric Series, Telescoping Series, and Harmonic Series in this section.
Integral Test to determine if a series converges or diverges.
Comparison Test/Limit Comparison Test to determine if a series converges or diverges.
Alternating Series Test Using the Alternating Series Test to determine if a series converges or diverges.
Absolute Convergence A brief discussion on absolute convergence and how it differs from convergence.
Ratio Test Using the Ratio Test to determine if a series converges or diverges.
Root Test Using the Root Test to determine if a series converges or diverges.
Strategy for Series A set of general guidelines to use when deciding which test to use.
Estimating the Value of a Series estimating the value of an infinite series.
Power Series An introduction to power series and some of the basic concepts.
This course Contains 90% of Video tutorials which include practice section. Master in Sequence and series for your Calculus Course!
If you can understand my pronunciation of English in PROMO VIDEO then you can understand all lecture. So Why are you waiting? Click the button "TAKE THIS COURSE".
How much content is included in the course?
Course is updated on Weekly basis and you will get more content in next week.
Hurry! The price will be increasing soon.
What will you know, or be able to do, after completing the course?
Benefits of this course are:
After completing this course you will be Master in solving Sequences and series problem of any type.
Graduated in Electrical engineering from University of Rajasthan in year 2006, Taught over 1000 students of High school after graduation and still counting, worked as Yoga trainer and trained more than 700 person, Worked as an online tutor for mathematics. Provide coaching to students who prepare India's engineering entrance examination, competitive examination. Published 2 books on calculus, algebra. Prepare Exam papers and question bank based on Entrance exam/test prep after high school, Provide homework help through various freelancing websites, Build strong relationship with students globally to meet their subject requirement and help them to improve their grade and develop their interest in understanding of mathematics.
Worked as an Online tutor with reputed tutoring agency of India.
Enhanced knowledge of many students in the area of Calculus, Algebra, Probability, Statistics, Geometry, Vedic Mathematics, Trigonometry, and applications of mathematics in Economics, Engineering.
Producing and publishing Video on mathematics for students of all over India and Global students.
As a yoga trainer Conduct yoga session to aware people about respiration system, deep breathing through pranayama, asthang yoga of patanjali, give them Indian method of treatment of any disease through Ayurveda.
Through Yoga Session and training I reduced weight of many people, and cure obesity, constipation, acidity, and many common disease caused by inactive life style.
I am Full time Producer and Publisher of Video tutorials of self learning content. | 677.169 | 1 |
9780534351847
05343518299.99
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Summary
Authors Cheney and Kincaid show students of science and engineering the modern computer's potential for solving numerical problems and gives them ample opportunity to hone their skills in programming and problem solving. The text helps students learn about errors that inevitably accompany scientific computing and arms them with methods for detecting, predicting, and controlling these errors. In this edition a discussion of how to locate codes for numerical algorithms on the World Wide Web has been added. A new section on iterative methods for solving large systems of linear equations has also been added.
A less scholarly approach and a different menu of topics sets this book apart from the authors' highly regarded NUMERICAL ANALYSIS: MATHEMATICS OF SCIENTIFIC COMPUTING, SECOND EDITION. | 677.169 | 1 |
This is my 5th year teaching Math at McCollum High School. I grew up in Southern California and went to Loyola Marymount University in Los Angeles, California where I earned a Bachelors Degree in Electrical Engineering. After college I was commissioned as an officer in the United States Air Force, where I proudly served for 22 years before retiring in 2012. I had always planned on becoming an educator following my military career and I'm glad that I finally found my way to McCollum High in San Antonio, Texas. Go Cowboys!
We have an exciting year ahead of us as we set out to learn and master the subject of Algebra 1. Our class will be challenging yet fun as students will learn through exploration and discovery. By working hard and persevering, we will be prepared to apply principles of Algebra to other high school math and science classes, to future college studies, and in everyday life scenarios.
I can be contacted at robert.schlegel@harlandale.net. I am always available before and after school to answer students' questions and to provide assistance. I look forward to personally getting to know each student and helping prepare for a future of success.
Sincerely, Mr. Rob Schlegel
Google Classroom - We will be using Google Classroom site to post important information such as course syllabus, assignments, homework reminders and tips, and other helpful resources.
Students can access our Google Classroom account from the following website using their myhisd.net login provided to them during the first week of school:
or through Google Classroom I-Phone and Android smartphone apps.
Period - Google Classroom Code
2nd Period - uopvt4y
3rd Period - weyeKu0
6th Period -69k2em6 8th Period - uqanefd
9th Period -fm9rzpk
Tutoring Sessions available after school most days
Monday (4:15 - 5:00pm, Room 209)
Weds/Thurs (4:15 -6:00pm, Room 218)
Supply List
1. Pencils and erasers – All work should be done in pencil (we all make mistakes and erasers help keep our work neat).
2. Interactive notebook – One will be provide during first week for our Algebra 1 Journal.
3. TI-84 graphing calculators will be provided in class. See for Wabbit Emu Mac/PC/android application to use at home.
Most other supplies will be provided in class.
Syllabus -
Course syllabus will be sent some with students during first week of school for parent/guardians to sign and return.
Textbook: Students can check out a textbook (student responsible for returning it at end of year) or may access the textbook online (access instructions will be provided soon and included on Google Classroom). | 677.169 | 1 |
PDF (Acrobat) Document File
Be sure that you have an application to open this file type before downloading and/or purchasing.
0.43 MB | 5 pages
PRODUCT DESCRIPTION
This is the third (in a series of 9) worksheet I created to teach the ideas between linear and exponential functions.
Topics covered in this worksheet are: geometric and arithmetic sequences, geometric and arithmetic sequences in context, comparing a linear and an exponential function (graphs) and deciding which has the biggest rate of change over different intervals | 677.169 | 1 |
Check Your Delivery Options
In Part I, I give you an overview of Calculus II, plus a review of more foundational math concepts. Chapter 1 introduces the definite integral, a mathematical statement that expresses area. I show you how to formulate and think about an area problem by using the notation of calculus. I also introduce you to the Riemann sum equation for the integral, which provides the definition of the definite integral as a limit. Beyond that, I give you an overview of the entire book . Chapter 2 gives you a need-to-know refresher on Pre-Calculus and Calculus I. Chapter 3 introduces the indefinite integral as a more general and often more useful way to think about the definite integral.Part II focuses on a variety of ways to solve indefinite integrals.Part III discusses a variety of intermediate topics, after you have the basics of integration under your belt.In Part IV, I introduce the infinite series — that is, the sum of an infinite number of terms.Part VI includes a few top-ten lists on a variety of calculusrelated topics. | 677.169 | 1 |
Description
To the Teacher. This book is designed to introduce a student to some of the important ideas of algebraic topology by emphasizing the re- lations of these ideas with other areas of mathematics. Rather than choosing one point of view of modem topology (homotopy theory, simplicial complexes, singular theory, axiomatic homology, differ- ential topology, etc.), we concentrate our attention on concrete prob- lems in low dimensions, introducing only as much algebraic machin-- ogists-without, we hope, discouraging budding topologists. We also feel that this approach is in better harmony with the historical devel- opment of the subject. What would we like a student to know after a first course in to- pology (assuming we reject the answer: half of what one would like the student to know after a second course in topology)?
Our answers to this have guided the choice of material, which includes: under- standing the relation between homology and integration, first on plane domains, later on Riemann surfaces and in higher dimensions; wind- ing numbers and degrees of mappings, fixed-point theorems; appli- cations such as the Jordan curve theorem, invariance of domain; in- dices of vector fields and Euler characteristics; fundamental groups | 677.169 | 1 |
Check Your Delivery Options
Use of mathematical tools in economics helps in systematic understanding of relationships and derivation of certain results which would be more complex through verbal logic. The mathematical approach can be considered as a quick mode of transportation from a set of postulates to a set of conclusion. The purpose of this book is to give an introduction to basic tools of mathematics for use in economic theory and econometrics. The author believes that this book serves as elementary study materials for the students of economics as it is presumed that the readers are acquainted with basic calculus, matrix algebra and economic theory.This book can be used as reference material for the student of M.S. Agricultural Economics. | 677.169 | 1 |
Developmental Mathematics
In addition to supporting the mission, goals, and student learning outcomes of the FCC Mathematics Department, Developmental Mathematics courses at FCC focus on building foundational mathematical skills in an active learning environment. Developmental Mathematics courses emphasize preparation for credit math classes and development of lifelong learning strategies.
If you need more information or have any questions, call the FCC Mathematics Department at 301.629.7835 to speak with Ms. Kylena Cross or ask an advisor.
Frequently Asked Questions
Where do I start?
To begin your math studies at FCC, you must take the Math Placement Test.
You may be exempt from taking the Math Placement Test. Visit the Testing Center FAQ page to find out whether you are exempt.
You may take the Math Placement Test twice before enrolling in a math course. If you are not satisfied with your math placement after the first attempt, then you may re-test. However, once you begin taking any developmental math class, you cannot take the Math Placement Test.
To prepare for the Math Placement Test, you can access the official web-based study app and free accuplacer study questions on the College Board site. If you would like assistance on any of these problems, then feel free to visit the STEM center during regular business hours for walk-in tutoring.
For test information, times and locations, or to schedule an appointment, call the Testing Center at 301.846.2522.
What's on the math placement test?
Learn more about placement testing, how to prepare, and view sample test questions here.
What courses do I need to take?
Developmental Mathematics has three levels of math courses, some of which have been combined with credit courses to help you complete your degree in a timely manner. If you are ready to take MA82, then you MUST meet with an advisor, as MA103A or MA206A could be an option depending on your degree program.
Come to the STEM Learning Center in B212 and get help from the STEM learning assistants and math tutors. The Learning Center is open to all students currently enrolled in a math class at FCC. You may study in the Learning Center and use the resources available there. Math books of every level and graphing calculator manuals are also available. Reference material must stay in B212. | 677.169 | 1 |
Download Presentation
Algebra for Precalculus and Calculus Students Mindful Manipognition After a Unit on Number Systems:
There are 10 kinds of mathematicians. Those who can think in binary and those who can't...
NO recognition
Mistakes My Students Truly Made Last Month in Calc II
Mistakes My Students Truly Made Last Month in Calc II
Mistakes My Students Truly Made Last Month in Calc II
Issues My Students Routinely Have
Not knowing when to distribute (when to, UGH, "FOIL".) GET RID OF FOIL!!
Pattern recognition
Reading tables of integrals
Copying what I write rather than what the problem solving strategy is
Refusing to think
Mistakes My Students Routinely Make
Everything is commutative
Everything is distributive
Things behave the way we want them to as in: (fg)'=f 'g' since it "works" with sums
Everything is linear: f(a+b) = f(a)+f(b)
Canceling
Fractions
I'm ANGRY. Who Is doing the teaching, and what is being taught??
Mistakes with Cancelling My Students Routinely Make (A Joke)
Mistakes with Fractions My Students Routinely Make (A Joke)
The chef instructs his apprentice:
"You take two thirds of water, one third of cream, one third of broth..."
The apprentice:
"But that makes four thirds already!"
"Well -just take a larger pot!"
Mistakes People Make (NOT A Joke)
Item costs $200
There is a 40% discount
Then…….
Mistakes with Commutativity Young Students Make (Not a Joke)
Teacher: "Who can tell me what 7 times 6 is?"
Student: "It's 42!"
Teacher: "Very good! - And who can tell me what 6 times 7 is?"
Same student: "It's 24!"
UNFORTUNATELY,
Students truly believe, and indeed expect, that if they come to class, they will receive a passing grade and, more likely than not, the grade will be a C or better.
Students believe that coming to class is their sum responsibility as students.
The vast majority of students do no homework, ever.
Students have come to expect long review sessions before an exam, and that the exam will contain only the problems covered in the review, with altered constants.
UNFORTUNATELY,
Mediocrity is now acceptable and even encouraged.
Partial credit for mostly incorrect work encourages mediocrity.
Having a cursory knowledge of a topic is considered sufficient.
Multi-step reasoning is an unfair expectation.
UNFORTUNATELY,
Students do not have a number sense.
Therefore, they do not have an algebra sense,
Nor do they have the sense to ask "Does the answer seem reasonable?"
The Fall of Rome
"No student left behind" has produced expectations aimed at the lowest common denominator. Skills have become so weak that examinations have had to be rescaled.
Making students and parents feel good is more important than setting realistic goals and limits.
The concept of "on time" is very loosely applied.
Better: Each child to his/her maximum potential
I'm Angry!
If a student does poorly on an exam, it is deemed the fault of the exam.
If a student fails to live up to expectations, it is deemed the fault of the teacher.
A college degree today is significantly different from such a degree of 25 years ago.
Many experienced, well-trained teachers are being replaced by faculty who are weaker.
A non-trivial number of high school mathematics teachers and/or supervisors:
do not have an undergraduate degree that has the word "mathematics" in it, or,
do not have an graduate degree that has the word "mathematics" in it, or,
have neither an undergraduate nor graduate degree that has the word "mathematics" in it.
What to do?
We are urged to have higher retention rates!
I, for one, chose to be part of the cure.
No student of mine will ever be asked by a professor "Who did you have in the previous course?"
Visit
~cheifp/Itseemstome1.htm
Which of the following three questions would you deem most important for your algebra students?
What algebraic law would you use to solve the equation 3(x + 5) = 27?
For what value of x is equal to zero?
Use an algebraic expression to prove or disprove that the sum of any three consecutive integers is a multiple of three.
In the following expressions, a and x are positive. What is the effect of increasing a on the value of each expression?
Without solving, decide whether the following equations have a positive solution, a negative solution, a zero solution, or no solution.
A car trip costs $1.50 per fifteen miles for gas and 30¢ per mile for other expenses, plus $20 for car rental. The total cost for the trip is given by theequation
Explain what each term represents in terms of the trip.
What units for cost and distance are being used?
Is the equationlinear?
The number of people who attend a concert is 160 – p when the price of a ticket is $p.
What is the practical interpretation of the 160?
Why is it reasonable that the p term has a negative sign?
The number of people who attend a movie at ticket price $p is 175 – p. If tickets are the same price, does the concert or the movie draw a larger audience?
The number of people who attend a dance recital at ticket price $p is 160 – 2p. If tickets are the same price, does the concert or the dance recital draw a larger audience?
A company uses two different sized trucks to deliver sand. The first truck can transport x cubic yards and the second y cubic yards. The first truck makes S trips to the job site and the second makes T trips. What do the following expressions represent in practical terms?
You plan to drive 300 miles at 55 miles per hour, stopping for a two-hour rest. You want to know t, the number of hours the journey will take. Which of the following equations would you use?
Precalculus: Fostering a deeper understanding
What is it?
For whom?
What topics do we need in such a course?
Calculus bound?
Liberal arts?
Understanding the information a graph imparts
Match each graph with a description of its zeros
Two real solutions
One real solution
c. No real solutions
Function Notation
If f (x) = 5 then f (2) = 10
If f (x) = 5 then 2f (x) = 10
If f (x) = 6 then f (2x) = 12
f (a + b) = f (a) + f (b)
Reading and Interpreting Graphs
If f (x) = h(x) – g(x) then the y-intercept of f is at
a. 0 b. 1 c. 2 d. 3 e. none of these
g(x)
h(x)
Reading and Interpreting Graphs
If f (x) = h(x) – g(x) then the x-intercept of f is at
a. 0 b. 1 c. 2 d. 3 e. none of these
g(x)
h(x)
Deeper Understanding of Symbols
In the equation 3x + 2y = 9, if x increases by 2 then y
a. Increases by 2
b. Decreases by 6
c. Increases by 3
d. Increases by 6
e. Decreases by 3
Understanding Exponential Growth
Match the equation to the description
a. Astronomical growth
b. A 6% increasing growth rate
c. A 6% decay rate
d. Quickly approaches zero
How to Solve it?Classify each equation as
Solvable using the exponential function.
Solvable using the log function
Not solvable using the log function or the exponential function but a solution exists
No real number is a solution
Reading Tables
Which table shows an increasing function?
Reading Tables
Which of the following tables shows data from an invertible function?
The Relationship Between Equations and Their Graphs
Which of the following are possible equations for the graph below?
d
a b c
In Closing:
Where are we, as a mathematics community, going?
What are we going to do about it?
Be an ambassador for common sense
Talk to anyone you can about the sorry state of education in the US and in particular, about improving mathematical reasoning | 677.169 | 1 |
9780534378912
05343789200.00
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Summary
Discover the many ways mathematics is relevant to your life with MATHEMATICS: A PRACTICAL ODYSSEY and its accompanying online resources. You'll master problem solving skills in such areas as calculating interest and understanding voting systems and come to recognize the relevance of mathematics and to appreciate its human aspect. Included with your purchase is access to the ThomsonNOW, an online tutorial that allows you to work with real math notation in real time, with unlimited practice problems, instant analysis and feedback, and streaming video to illustrate key concepts and Personal Tutor with SMARTHINKING a live, online mathematics tutor.
Table of Contents
1 LOGIC
1
(53)
Deductive vs. Inductive Reasoning
2
(10)
Problem Solving
2
(1)
Deductive Reasoning
3
(1)
Deductive Reasoning and Venn Diagrams
4
(1)
Historical Note: Aristotle
5
(4)
Inductive Reasoning
9
(1)
Exercises 1.1
10
(2)
Symbolic Logic
12
(9)
Statements
12
(1)
Compound Statements and Logical Connectives
13
(1)
The Negation ≈p
14
(1)
Historical Note: Gottfried Wilhelm Leibniz
15
(1)
The Conjunction p q
16
(1)
The Disjunction p q
16
(1)
The Conditional p → q
17
(2)
Exercises 1.2
19
(2)
Truth Tables
21
(13)
The Negation ≈p
21
(1)
The Conjunction p q
21
(1)
The Disjunction p q
22
(3)
The Conditional p &rarrl q
25
(3)
Equivalent Expressions
28
(1)
Historical Note: George Boole
29
(2)
De Morgan's Laws
31
(2)
Exercises 1.3
33
(1)
More on Conditionals
34
(7)
Variations of a Conditional
34
(1)
Equivalent Conditionals
35
(2)
The ``Only If' Connective
37
(1)
The Biconditional p ↔ q
38
(1)
Exercises 1.4
39
(2)
Analyzing Arguments
41
(13)
Valid Arguments
41
(3)
Tautologies
44
(2)
Historical Note: Charles Lutwidge Dodgson
46
(3)
Exercises 1.5
49
(2)
Chapter 1 Review
51
(3)
Sets and Counting
54
(58)
Sets and Set Operations
55
(12)
Notation
55
(2)
Universal Set and Subsets
57
(1)
Intersection of Sets
58
(1)
Mutually Exclusive Sets
58
(1)
Union of Sets
58
(3)
Complement of a Set
61
(1)
Historical Note: John Venn
62
(1)
Shading Venn Diagrams
63
(1)
Set Theory and Logic (prerequisite: Chapter 1)
63
(1)
Exercises 2.1
64
(3)
Applications of Venn Diagrams
67
(11)
Surveys
67
(4)
De Morgan's Laws
71
(2)
Historical Note: Augustus De Morgan
73
(1)
Exercises 2.2
74
(4)
Introduction to Combinatorics
78
(8)
The Fundamental Principle of Counting
78
(3)
Factorials
81
(2)
Exercises 2.3
83
(3)
Permutations and Combinations
86
(13)
With versus Without Replacement
86
(1)
Permutations
86
(3)
Combinations
89
(6)
Historical Note: Chu Shi-chieh
95
(1)
Exercises 2.4
96
(3)
Infinite Sets
99
(13)
One-to-One Correspondence
99
(2)
Historical Note: Georg Cantor
101
(1)
Countable Sets
102
(3)
Uncountable Sets
105
(1)
Points on a Line
106
(2)
Exercises 2.5
108
(2)
Chapter 2 Review
110
(2)
Probability
112
(75)
History of Probability
113
(7)
Roulette
114
(1)
Dice and Craps
115
(1)
Cards
116
(1)
Historical Note: Blaise Pascal
117
(1)
Historical Note: Gerolamo Cardano
118
(1)
Exercises 3.1
119
(1)
Basic Terms of Probability
120
(15)
Finding Probabilities and Odds
122
(2)
Relative Frequency versus Probability
124
(1)
Mendel's Use of Probabilities
125
(1)
Historical Note: Gregor Johann Mendel
126
(2)
Probabilities in Genetics
128
(1)
Genetic Screening
129
(2)
Historical Note: Nancy Wexler
131
(1)
Exercises 3.2
132
(3)
Basic Rules of Probability
135
(9)
Mutually Exclusive Events
136
(1)
Pair-of-Dice Probabilities
136
(3)
More Probability Rules
139
(1)
Probabilities and Venn Diagrams
140
(1)
Exercises 3.3
140
(3)
Fractions on a Graphing Calculator
143
(1)
Combinatorics and Probability
144
(10)
Lotteries
146
(2)
Historical Note: Lotteries
148
(1)
Keno
149
(1)
Cards
150
(2)
Exercises 3.4
152
(2)
Expected Value
154
(7)
Why the House Wins
156
(1)
Decision Theory
157
(1)
Betting Strategies
158
(1)
Exercises 3.5
158
(3)
Conditional Probability
161
(11)
Probabilities and Polls
161
(3)
The Product Rule
164
(1)
Tree Diagrams
165
(4)
Exercises 3.6
169
(3)
Independence; Trees in Genetics
172
(15)
Dependent and Independent Events
172
(3)
Product Rule for Independent Events
175
(1)
Trees in Medicine and Genetics
176
(3)
Hair Color
179
(2)
Exercises 3.7
181
(4)
Chapter 3 Review
185
(2)
Statistics
187
(101)
Population, Sample, and Data
188
(24)
Population versus Sample
188
(1)
Frequency Distributions
188
(2)
Grouped Data
190
(2)
Histograms
192
(2)
Histograms and Relative Frequency Density
194
(3)
Pie Charts
197
(1)
Exercises 4.1
198
(10)
Histograms on a Graphing Calculator
208
(4)
Measures of Central Tendency
212
(10)
The Mean
212
(5)
The Median
217
(1)
The Mode
218
(1)
Exercises 4.2
219
(3)
Measures of Dispersion
222
(15)
Deviations
223
(1)
Variance and Standard Deviation
224
(2)
Alternate Methods for Finding Variance
226
(4)
Exercises 4.3
230
(4)
Measures of Central Tendency and Dispersion on a Graphing Calculator
234
(3)
The Normal Distribution
237
(19)
Discrete versus Continuous Variables
238
(1)
Normal Distributions
238
(2)
Probability, Area, and Normal Distributions
240
(2)
Historical Note: Carl Friedrich Gauss
242
(1)
The Standard Normal Distribution
243
(6)
Converting to the Standard Normal
249
(5)
Exercises 4.4
254
(2)
Polls and Margin of Error
256
(13)
Sampling and Inferential Statistics
256
(1)
Sample Proportion versus Population Proportion
257
(2)
Margin of Error
259
(1)
Historical Note: George H. Gallup
260
(6)
Exercises 4.5
266
(3)
Linear Regression
269
(19)
Linear Trends and Line of Best Fit
270
(3)
Coefficient of Linear Correlation
273
(4)
Exercises 4.6
277
(4)
Linear Regression on a Graphing Calculator
281
(3)
Chapter 4 Review
284
(4)
Finance
288
(80)
Simple Interest
289
(9)
National Debt
292
(1)
Add-On Interest
293
(1)
Credit Card Finance Charge
294
(1)
Historical Note: Credit Card History
295
(1)
Exercises 5.1
296
(2)
Compound Interest
298
(75)
Annual Yield
304
(3)
Exercises 5.2
307
(2)
Doubling Time on a Graphing Calculator
309
(64)
Annuities
373
Calculating Short-Term Annuities
313
(3)
Calculating Long-Term Annuities
316
(1)
Tax-Deferred Annuities
317
(2)
Sinking Funds
319
(1)
Present Value of an Annuity
320
(2)
Exercises 5.3
322
(3)
Annuities on a Graphing Calculator
325
(1)
Amortized Loans
326
(21)
Amortization Schedules
327
(5)
Finding an Unpaid Balance
332
(3)
Exercises 5.4
335
(5)
Amortization Schedules on a Computer
340
(7)
Annual Percentage Rate on a Graphing Calculator
347
(9)
Historical Note: The Truth in Lending Act
349
(1)
Finance Charges
349
(3)
Estimating Prepaid Finance Charges
352
(2)
Exercises 5.5
354
(2)
Payout Annuities
356
(12)
Calculating Short-Term Payout Annuities
356
(1)
Comparing Payout Annuities and Savings Annuities
357
(1)
Calculating Long-Term Payout Annuities
358
(2)
Payout Annuities with Inflation
360
(2)
Exercises 5.6
362
(2)
Chapter 5 Review
364
(4)
Geometry
368
(92)
Perimeter and Area
369
(14)
Polygons
369
(4)
Heron's Formula for the Area of a Triangle
373
(3)
Right Triangles
376
(1)
Circles
377
(2)
Exercises 6.1
379
(4)
Volume and Surface Area
383
(12)
Problem Solving
383
(2)
Surface Area
385
(2)
Spheres
387
(2)
Cones and Pyramids
389
(2)
Exercises 6.2
391
(4)
Egyptian Geometry
395
(11)
Units of Measurement
396
(1)
Empirical Geometry (If It Works, Use It)
397
(2)
The Great Pyramid of Cheops
399
(1)
The Rhind Papyrus
400
(2)
Pi and the Area of a Circle
402
(2)
Exercises 6.3
404
(2)
The Greeks
406
(12)
Thales of Miletus
407
(1)
Pythagoras of Samos
408
(2)
Euclid of Alexandria
410
(1)
Deductive Proof
411
(1)
Historical Note: Archimedes of Syracuse
412
(2)
Archimedes of Syracuse'
414
(1)
Exercises 6.4
415
(3)
Right Triangle Trigonometry
418
(14)
Angle Measurement
419
(1)
Special Triangles
420
(1)
Trigonometric Ratios
421
(4)
Using a Calculator
425
(1)
Finding an Acute Angle
426
(2)
Angles of Elevation and Depression
428
(1)
Exercises 6.5
429
(3)
Conic Sections and Analytic Geometry
432
(13)
Historical Note: Hypatia
433
(1)
Conic Sections
433
(1)
The Circle
434
(1)
The Parabola
435
(3)
The Ellipse
438
(2)
The Hyperbola
440
(4)
Exercises 6.6
444
(1)
Non-Euclidean Geometry
445
(15)
The Parallel Postulate
445
(1)
Girolamo Saccheri
446
(1)
Carl Friedrich Gauss
447
(1)
Janos Bolyai
447
(1)
Nikolai Lobachevsky
448
(1)
Bernhard Riemann
449
(1)
Geometric Models
449
(2)
A Comparison of Triangles
451
(1)
Geometry in Art
452
(3)
Exercises 6.7
455
(1)
Chapter 6 Review
456
(4)
Matrices and Markov Chains
460
(59)
Review of Matrices
467
(6)
Terminology and Notation
461
(1)
Matrix Multiplication
462
(3)
Properties of Matrix Multiplication
465
(1)
Historical Note: Arthur Cayley and James Joseph Sylvester
466
(2)
Identity Matrices
468
(1)
Exercises 7.0
469
(1)
Matrix Multiplication on a Graphing Calculator
469
(4)
Markov Chains
473
(14)
Transition Matrices
474
(1)
Probability Matrices
475
(1)
Using Markov Chains to Predict the Future
476
(6)
Historical Note: Andrei Andreevich Markov
482
(1)
Exercises 7.1
483
(4)
Systems of Linear Equations
487
(21)
Linear Equations
487
(1)
Systems of Equations
487
(3)
Solving Systems: The Elimination Method
490
(1)
Solving Systems: The Gauss-Jordan Method
491
(8)
Exercises 7.2
499
(2)
Technology and the Row Operations
501
(1)
The Row Operations on a Graphing Calculator
502
The Row Operations and Amortrix
000
(507)
Augmented Identity Matrices
507
(1)
Long-Range Predictions with Markov Chains
508
(11)
Exercises 7.3
514
(3)
Chapter 7 Review
517
(2)
Linear Programming
519
(61)
Review of Linear Inequalities
521
(12)
Systems of Linear Inequalities
524
(3)
Finding Corner Points
527
(2)
Exercises 8.0
529
(1)
Graphing Linear Inequalities on a Graphing Calculator
529
(4)
The Geometry of Linear Programming
533
(16)
Creating a Model
534
(3)
Analyzing the Model
537
(2)
Some Graphing Tips
539
(1)
Why the Corner Principle Works
539
(5)
Historical Note: George Dantzig
544
(2)
Exercises 8.1
546
(3)
Introduction to the Simplex Method
549
(11)
Comparison of the Gauss-Jordan and Simplex Methods
555
(1)
Karmarkar's New Method
556
(1)
Historical Note: Hindu Mathematics
557
(1)
Exercises 8.2
558
(2)
The Simplex Method: Complete Problems
560
(20)
Pivoting with the Simplex Method
561
(2)
When to Stop Pivoting
563
(2)
Why the Simplex Method Works
565
(6)
Exercises 8.3
571
(2)
Technology and the Simplex Method
573
(5)
Chapter 8 Review
578
(2)
Exponentials and Logarithmic Functions
580
(74)
Review of Exponentials and Logarithms
581
(10)
Functions
581
(1)
Exponential Functions
581
(1)
Rational and Irrational Numbers
582
(1)
The Natural Exponential Function
583
(2)
Logarithms
585
(1)
Historical Note; Leonhard Euler
586
(1)
Common Logarithms
586
(3)
The Natural Logarithm Function
589
(2)
Exercises 9.0A
591
(1)
Review of Properties of Logarithms
591
(13)
The Inverse Properties
591
(2)
Solving Exponential Equations
593
(2)
The Exponent-Becomes-Multiplier Property
595
(1)
Historical Note: John Napier
596
(2)
The Division-Becomes-Subtraction Property
598
(2)
The Multiplication-Becomes-Addition Property
600
(1)
Solving Logarithmic Equations
601
(1)
Exercises 9.0B
602
(2)
Exponential Growth
604
(16)
Delta Notation
604
(3)
The Exponential Model
607
(6)
Exponential Growth and Compound Interest (for those who have read Chapter 5: Finance)
613
(2)
Exercises 9.1
615
(5)
Exponential Decay
620
(16)
Half-Life
623
(3)
Relative Decay Rate
626
(3)
Historical Note: Marie Curie
629
(1)
Radiocarbon Dating
629
(3)
Historical Note: Willard Frank Libby
632
(2)
Exercises 9.2
634
(2)
Logarithmic Scales
636
(18)
Earthquakes
636
(3)
The Richter Scale
639
(5)
Earthquake Magnitude and Energy
644
(1)
The Decibel Scale
645
(4)
Historical Note: Alexander Graham Bell
649
(1)
Exercises 9.3
650
(2)
Chapter 9 Review
652
(2)
Calculus
654
Review of Ratios, Parabolas, and Functions
655
Ratios, Proportions, and Rates
655
Delta Notation
656
Similar Triangles
657
Parabolas
660
Secant Lines and Their Slopes
661
Locating the Vertex
662
Functions
664
Exercises 10.0
666
The Antecedents of Calculus
668
Ancient Mathematics
668
Greek Mathematics
669
Hindu Mathematics
671
Arabic Mathematics
672
Historical Note: Arabic Mathematics
676
Mathematics during the Middle Ages
677
Mathematics during the Renaissance
677
Historical Note: Rene Descartes
681
Exercises 10.1
682
Four Problems
684
Find the Distance Traveled by a Falling Object
686
Find the Trajectory of a Cannonball
689
Historical Note: Galileo Galilei
690
Find the Line Tangent to a Given Curve
693
Find the Area of Any Shape
695
Exercises 10.2
697
Newton and Tangent Lines
700
Cauchy's Reformulation of Newton's Method
702
Historical Note: Isaac Newton
705
Exercises 10.3
708
Newton on Falling Objects and the Derivative
709
Average Speed and Instantaneous Speed
709
Newton on Gravity
709
The Derivative
712
Interpreting a Derivative
716
Exercises 10.4
718
The Trajectory of a Cannonball
720
The Motion Due to the Explosion
721
The Motion Due to Gravity
722
The Equation for the Cannonball's Motion
722
The Generalized Equation for the Cannonball's Motion (for those who have read Section 6.5) | 677.169 | 1 |
Algebra Writing for deeper understanding
Word Document File
Be sure that you have an application to open this file type before downloading and/or purchasing.
1.39 MB | 3 pages
PRODUCT DESCRIPTION
This is a tool used to help students not only solve algebraic word problems, but explain how they came to their solutions. An example has been included to explain how your students need to use this tool.You can replace the example problem with any word problem giving your students trouble | 677.169 | 1 |
Introduction to Algebra - PowerPoint with activities for students
Presentation (Powerpoint) File
Be sure that you have an application to open this file type before downloading and/or purchasing.
1.99 MB | 40 pages
PRODUCT DESCRIPTION
This Presentation was designed around Unit 7 of the Year 8 C2C Maths Queensland Curriculum. The PowerPoint introduces students to Algebra and using algebraic expressions. This presentation comes with questions for the students to work on as you proceed through the PowerPoint. It has a hyperlink to a quiz that the students can then do to reinforce their understanding of how to expand algebraic equations.
I also have a specific PowerPoint Quiz titled 'Intro to Algebra Quiz' that is available on my site.
I have designed the Quiz to follow directly on from this PowerPoint and it is actually more appropriate to have students do this Quiz before you guide them to the hyperlinked Quiz on the internet.
The Presentation provides good visual steps rather than students needing to rely on the teacher just talking. Presentation designed to provide students with visual examples. Caters for students who are visual learners or those with Hearing Impairments and Auditory Processing Disorders. This Presentation was designed to cater for Year 8 Maths Queensland Curriculum but could be used for any year level from year 6 to Year 9.
(Australia, Queensland, Mathematics, Special Education, Year 8, C2C) PowerPoint Queen Australia | 677.169 | 1 |
In this role, you will ability to choose the right mathematical methods or formulas to solve a problem. Ability to work with mathematical concepts such as probability and statistical inference and fundamentals of plane and solid geometry and trigonometry. Ability to apply concepts such as fractions, percentages, ratios and proportions to practical situations | 677.169 | 1 |
A Beginner's Guide to Discrete Mathematics
Description
This introduction to discrete mathematics is aimed at freshmen and sophomores in mathematics and computer science. It begins with a survey of number systems and elementary set theory before moving on to treat data structures, counting, probability, relations and functions, graph theory, matrices, number theory and cryptography. The end of each section contains problem sets with selected solutions, and good examples occur throughout the text. | 677.169 | 1 |
This book has been designed for the use of honours and postgraduate students of various Indian universities. It will also be found useful by the students preparing for various competitive examinations.Do not start this book with an unreasonable fear. There are no mysteries in Mathematics. It is all simple and honest reasoning explained step by step which anybody can follow with a little effort and concentration. Often a student has difficulty in following a mathematical explanation only because the author skips steps which he assumes the students to be familiar with. If the student fails to recount the missing steps, he may be faced with a gap in the reasoning and the author's conclusion may become mysterious to him.
Questions asked in recent papers of GATE and various university examinations have been inserted at appropriate places. This enriched inclusion of solved examples and variety of new exercises at the end of each article and chapter makes this book more useful to the reader.
Audience of the Book :
[For BA, B.Sc. and Honours (Mathematics and Physics), M.A., M.Sc. (Mathematics and Physics), B.E. Students of Various Universities and for I.A.S., P.C.S., A.M.I.E. GATE, C.S.I.R. U.G.C. NET and Various Competitive Examinations]
Key Features:
The following major changes have been made in the present edition:
Almost all the chapters have been rewritten so that in the present form, the reader will not find any difficulty in understanding the subject matter.
The matter of the previous edition has been re-organised so that now each topic gets its proper place in the book.
More solved examples have been added so that the reader may gain confidence in the techniques of solving problems.
References to the latest papers of various universities and I.A.S. examination have been made at proper places.
Errors and omissions of the previous edition have been corrected
Table of Contents:
PART-I: ELEMENTARY DIFFERENTIAL EQUATIONS
• Differential Equations, their Formation and Solutions
• Equations of First Order and First Degree
• Trajectories
• Equations of the First Order but not of the First Degree Singular Solutions and Extraneous Loci | 677.169 | 1 |
Calculators at the Secondary Level
Scientific Calculators are needed for all levels of math prior to Algebra 2. Once a student is entering into Algebra 2, a graphing calculator is required.
The calculator you purchase now will be used in all subsequent math courses and science courses including those at the college level. There are many good graphing calculators on the market. We have found that the TI-Nspire CX, TI-83+ or TI-84+ made by Texas Instruments are good quality calculators that hold up well to student use and best suit the needs of our program. (The TI-84+ is a newer version of the TI-83+.) Our teachers will demonstrate procedures using one of these three calculators. We have found that students who use one of these models can easily follow class demonstrations and will not be distracted while trying to figure out how to execute a similar operation on a calculator with a different operating system. Some students or parents may feel that if the TI-83+ or TI-84+ is good, a TI-86, 89 or 92 is better. This IS NOT the case. The TI-86 is designed for a different purpose and lacks some of the programming that is frequently used in math classes. The TI-92 is not allowed on SAT and AP tests. And the TI-89 is more expensive because it contains programming that is not necessary until students reach much higher level courses.
We understand that graphing calculators are very costly. We do have TI-83 calculators to loan to students for the year. The student is responsible for returning the calculator to his/her teacher at the end of the year. | 677.169 | 1 |
Classes for Students Entering 10-12th Grades
Summer 2017 Information Coming Soon!
ACT Math Preparation
Description: This course will review the math concepts that are tested on the ACT exam
Secondary Math 2 (Full Year Course)
Description: This course will cover the entire Secondary Math 2 core and is designed for students who have already completed Secondary Math 1 math and would like to accelerate or for students who have taken Secondary Math 2 2 Quarter 1
Description: This course will cover the first quarter of Secondary Math 2 core and is designed for students who have taken Secondary Math 2 and need remediation.
Secondary Math 2 Quarter 2
Description: This course will cover the second quarter of Secondary Math 2 core and is designed for students who have taken Secondary Math 2 and need remediation.
Secondary Math 2 Quarter 3
Description: This course will cover the third quarter of Secondary Math 2 core and is designed for students who have taken Secondary Math 2 and need remediation.
Secondary Math 2 Quarter 4
Description: This course will cover the fourth quarter of Secondary Math 2 core and is designed for students who have taken Secondary Math 2 and need remediation.
Secondary Math 2 Honors Topics Only
Description: This course is for students who would like to transition to the honors track. Topics will include: writing radical expressions in equivalent forms, analyzing and solving problems using polynomials, complex numbers (conjugates, representations, multiplying and dividing, finding the distance between points, etc.), inverse matrices, trigonometric identities, solving problems involving trigonometry, conics, and deriving volume formulas for solid figures.
Secondary Math 3 (Full Year Course)
Description: This course will cover the entire Secondary Math 3 core and is designed for students who have already completed Secondary Math 2 math and would like to accelerate or for students who have taken Secondary Math 3 3 Quarter 1
Description: This course will cover the first quarter of Secondary Math 3 core and is designed for students who have taken Secondary Math 3 and need remediation.
Secondary Math 3 Quarter 2
Description: This course will cover the second quarter of Secondary Math 3 core and is designed for students who have taken Secondary Math 3 and need remediation.
Secondary Math 3 Quarter 3
Description: This course will cover the third quarter of Secondary Math 3 core and is designed for students who have taken Secondary Math 3 and need remediation.
Secondary Math 3 Quarter 4
Description: This course will cover the fourth quarter of Secondary Math 3 core and is designed for students who have taken Secondary Math 3 and need remediation.
Secondary Math 3 Honors Topics Only
Description: This course is for students who would like to transition to the honors track. Topics will include: graphing rational functions, composing functions, inverse functions, even and odd symmetry and periodicity of trigonometric functions, inverse trigonometric functions, polar coordinates, parametric equations, and finite arithmetic and infinite geometric series. | 677.169 | 1 |
Expanding Logarithms Task Cards plus HW
PDF (Acrobat) Document File
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0.68 MB | 19 pages
PRODUCT DESCRIPTION
Properties of Logarithms: Expanding Logarithms. This activity is designed for Algebra 2, PreCalculus, or College Algebra students.
•Task Cards: The cards are designed to reinforce the concepts taught in class. There are 24 cards. There are also four blank cards for you to personalize problems you may have emphasized in your class. The problems include all bases, numerical, literal, base 10 and base e. Included in the bundle is a Master List of the questions contained on the Task Cards.
•Handouts I have included are two handouts with 16 questions each similar to the problems on the task cards. These can be used as HW, assessments, or enrichment.
If you teach PreCalculus or Calculus, please have a look at •Calculus Survival Kit, an incredible 70+ page resource for PreCalculus and Calculus teachers75. | 677.169 | 1 |
This site contains a collection of fully developed high school curriculum modules that use the Internet in significant ways. ...
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This site contains a collection of fully developed high school curriculum modules that use the Internet in significant ways. There are currently 15 modules in Mathematics and 6 modules in Science; also, there are approximately two dozen additional modules that have been created by instructors and/or Education students.The learning modules here are web-based, technology intensive lessons focusing on mathematics and science in an applied context. They have been developed for teachers, by teachers, aligned with the Illinois State Learning Standards and the National Council for Teachers of Mathematics (NCTM) Standards. Some of the lessons are designed to last over several days, some only for a class periodQuadratic Functions contains two applets that allow the user to change the coefficients of a quadratic equation and observe...
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Quadratic Functions contains two applets that allow the user to change the coefficients of a quadratic equation and observe the change in the corresponding graph. One applet uses the standard form of a quadratic equation to investigate the role of ?a?, ?b?, and ?c?? and the second uses the standard form for a parabola to study the role of ?a?, ?h?, and ?k?? Quadratic Functions to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material Quadratic Functions
Select this link to open drop down to add material Quadratic Mathematics--Algebra to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material S.O.S. Mathematics--Algebra
Select this link to open drop down to add material S.O.S. Mathematics--Al Wave Geometry to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material Sine Wave Geometry
Select this link to open drop down to add material Sine Wave Ge Standard and General Form of a Parabola to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material The Standard and General Form of a Parabola
Select this link to open drop down to add material The Standard and General Form of a Parabola Gauss to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material Visual Gauss
Select this link to open drop down to add material Visual Gauss Scoop to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material Math Scoop
Select this link to open drop down to add material Math Scoop to your Bookmark Collection or Course ePortfolio
Mathway is a mathematics problem solving tool where students can select their math course - Basic Math, Pre-Algebra, Algebra,...
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Mathway is a mathematics problem solving tool where students can select their math course - Basic Math, Pre-Algebra, Algebra, Trigonometry, PreCalculus, Calculus or Statistics and enter a problem. The computer solves the problem and shows the steps for the solution. It also has a worksheet generatorway to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material Mathway
Select this link to open drop down to add material Mathway to your Bookmark Collection or Course ePortfolio
Old Egyptian Fractions at MathCats(Web and android version)Old Egyptian Math Cats knew fractions like 1/2 or 1/4 (one piece...
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Old Egyptian Fractions at MathCats(Web and android version)Old Egyptian Math Cats knew fractions like 1/2 or 1/4 (one piece of a pie).But to make fractions like 3/4, they had to add pieces of pies like 1/2 + 1/4 = 3/4.Old Egyptian Math cats never repeated the same fraction when adding.They never wrote: 1/4 + 1/4 + 1/4 = 3/4How it works:Choose puzzles from the list on the top. ( * = easier, **** = very hard.)Add 2 or 3 fraction pieces below.After you find one solution, the puzzle is marked "Solved." Can you find more solutions? (Click to see them listed on the bottom.)There's the android version in the mirror site link Old Egyptian Fractions to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material Old Egyptian Fractions
Select this link to open drop down to add material Old Egyptian Fractions to your Bookmark Collection or Course ePortfolio | 677.169 | 1 |
9780764123825
076412380.01
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Summary
Updated to reflect the most recent Advanced Placement exams in Calculus, this manual presents four practice exams in Calculus AB and four more in Calculus BC, all with questions answered and explained. Extensive review sections offer brush-ups in functions and their graphs, derivatives and integrals, differential equations, and sequences and series. Additional features include test-taking tips and guidelines for using a graphing calculator. Review material includes multiple-choice questions, free-response questions, and many applications problems.
Table of Contents
Preface
p. vii
Introduction
p. viii
The Courses
p. viii
Comparison of Calculus AB and Calculus BC Courses
p. viii
The "New" AP Calculus Courses
p. viii
Topics That May Be Tested on the Calculus AB Exam
p. ix
Topics That May Be Tested on the Calculus BC Exam
p. x
The Examinations
p. xi
The Graphing Calculator: Using Your Graphing Calculator on the AP Exam | 677.169 | 1 |
Publisher Description
The Windows desktop version combines the test banks and has more game features than does the hand held versions. With the desktop, there are options for either single or multiple players; an answer box for displaying the correct answer; the ability to modify the games test bank (over 580 questions); an electronic book for reviewing basic algebraic principles, and more. The demo version has a limited number of questions in the test bank. | 677.169 | 1 |
Exponential Functions
Resources for Exponential Functions that have been developed for
the Algebra for All project are provided below. Use the comments
area at the bottom of page to disucss these resources or share
additional online resources related to Exponential Functions.
Student Access:
You do not need to go to the AfA Social Network to access these
activities. Students and teachers can access these resources
directly by going straight to the urls provided. Simply copy and
paste the links given below and provide those links to your
students. Click here for more detailed instructions.
Exponential Explorations
Description: In this activity students explore
exponential relationships through two concrete examples. Through
paper-folding, and rolling dice, students make connections between
abstract exponential equations and concrete situations that are
modeled by exponential equations
Exponential Dice Roller Applet: In this applet students
create a model of exponential decay by rolling (up to) 98 dice and
taking out dice of different values. By varying which dice are kept
and how many dice are used, students can connect parts of
exponential equations with a concrete model.
Exponential Stations Graphing Applets: Each of these
applets provides an opportunity to graph a set of four exponential
function, providing an opportunity to compare and contrast the
features of each graph and learn about the different parts of an
exponential equation. | 677.169 | 1 |
College Algebra
Browse related Subjects ...
Read More addition, the book includes many real-world examples that show you how mathematics is used to model in fields like engineering, business, physics, chemistry, and biology.
Read Less 6e, Instructor's Review Copy Item may show signs of shelf wear. Pages may include limited notes and highlighting. Includes supplemental or companion materials if applicable. Access codes may or may not work. Connecting readers since 1972. Customer service is our top priority.
Brand new. [(Annotated) Instructor's Edition] [AIE has the same contents but with extra notes, answers for teachers; perfect for home study or review] [No ancillary materials included unless specified]
Fair. 6th edition. Pages are unmarked. Cover is worn and covered and used bookstore stickers. NOT pretty! Unless noted, used texts do NOT include supplements such as CDs & codes. Orders packed carefully and shipped daily with tracking # emailed to you. Canadian and international orders welcomed | 677.169 | 1 |
About this product
Description
Description
Organized for easy reference and crucial practice, this title offers coverage of all the essential topics presented as 500 AP-style questions with detailed answer explanations. 5 Steps to a 5: 500 AP Calculus AB/BC Questions to Kw by Test Day is tailored to meet your study needs - whether This title features: 500 AP-style questions and answers referenced to core AP materials; review explanations for right and wrong answers; additional online practice; close simulations of the real AP exams; updated material that reflects the latest tests; and, online practice exercises.
Author Biography
Cynthia Johnson has over 20 years' experience in writing and editing test-prep guides for America's leading educational publishers. Her team of writers and editors have also produced thousands of test items for ETS and for Harcourt Assessment. | 677.169 | 1 |
How to Approach an SAT Math Problem
197 Downloads
Word Document File
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0.04 MB | 2 pages
PRODUCT DESCRIPTION
The SAT Test Prep market is flooded with so much material and methods of "beating" the SATs that students are often overwhelmed by it all. They often find it difficult to take away from the dense Test Prep guides what they actually do need to know to succeed in the Math portion of the SAT. Teachers often lack the time (we never have enough, do we?) to put together, in one handout, a summary of what they know their students need to know. This 11-point summary is an ideal handout for your student studying for the SATs or PSATs. It touches upon the wide variety of SAT Math questions covered in the current SATs and outlines simple and easy-to-understand approaches to these questions. These methods will help a student successfully tackle most SAT Math problems and build their confidence in their abilities. The more they use these approaches, the more they will succeed.
My name is Vidya Forrester, and I have been teaching high school Math for 13 years. I have taught SAT test prep for 10 of those years | 677.169 | 1 |
Description: This is a text that covers the standard topics in a sophomore-level course in discrete mathematics: logic, sets, proof techniques, basic number theory, functions, relations, and elementary combinatorics, with an emphasis on motivation. It explains and clarifies the unwritten conventions in mathematics, and guides the students through a detailed discussion on how a proof is revised from its draft to a final polished form. | 677.169 | 1 |
Differential Equations and Linear Agebra thorough introduction to the basics of differential equations and linear algebra with a carefully balanced and sound integration of the two topics. Flexible in format, it explains concepts clearly and logically without sacrificing level orMore...
For a thorough introduction to the basics of differential equations and linear algebra with a carefully balanced and sound integration of the two topics. Flexible in format, it explains concepts clearly and logically without sacrificing level or rigor and supports material with a vast array of problems of varying levels from which students/instructors can choose.
First-Order Differential Equations
Second-Order Linear Differential Equations
Matrices and Systems of Linear Algebraic Equations
Determinants
Vector Spaces
Linear Transformations and the Eigenvalue/Eigenvector Problem
Linear Differential Equations of Ordern
Systems of Differential Equations
The Laplace Transform and Some Elementary Applications
Series Solutions to Differential Equations
A Review of Complex Numbers
A Review of Partial Fractions
A Review of Integration Techniques
An Existence and Uniqueness Theorem for First-Order Differential | 677.169 | 1 |
... more...
Recently teachers and researchers have become interested in students? mental computation skills & strategies. Many curriculums expect mental computation to be taught and many also specify particular methods to focus on. Most of these expectations are based on beliefs rather than evidence. This book makes a valuable contribution to developing the evidence... more...
A self-contained, elementary introduction to wavelet theory and applications Exploring the growing relevance of wavelets in the field of mathematics, Wavelet Theory: An Elementary Approach with Applications provides an introduction to the topic, detailing the fundamental concepts and presenting its major impacts in the world beyond academia. Drawing... more...
Use a practical approach to teaching mathematics that integrates proven literacy strategies for effective instruction. This professional resource will help to maximize the impact of instruction through the use of whole-class instruction, small-group instruction, and Math Workshop. Incorporate ideas for using ongoing assessment to guide your instruction... more...
Written specifically for K–12 mathematics teachers, this resource provides the "nuts and bolts" of differentiation. Presented in an easy-to-implement format, this handy notebook is designed to facilitate the understanding and process of writing differentiated lessons to accommodate all readiness levels, learning styles, and interests.... more...
Low Attainers in Primary Mathematics focuses on data from students in Singapore schools. It is widely acknowledged that students from Singapore do well in mathematics in international studies. This book provides readers with a glimpse of students from Singapore who are at the other end of the ability spectrum. The book is based on a study that explored... more...
Penrose is back, and ready to usher young readers along as he encounters more amazing mathematical ideas in a sequence of adventure tales. At once demystifying and challenging, the book gives readers visuals to consider and things to do as they along with Penrose discover mathematical "rep-tiles"; meet x, the mathematical actor; find out when one... more... | 677.169 | 1 |
Provides a number of additional challenging problems for students to solve that are drawn from real-world applications. Problems are keyed to each chapter and are designed to highlight and emphasize key concepts.
"synopsis" may belong to another edition of this title.
About the Author:
Allan G. Bluman is Professor of Mathematics at Community College of Allegheny County, near Pittsburgh. For the McKeesport and New Kensington Campuses of Pennsylvania State University, he has taught teacher-certification and graduate education statistics courses. Prior to his college teaching, he taught mathematics at a junior high school. Professor Bluman received his B.S. from California State College in California, Penn.; his M.Ed. from the University of Pittsburgh; and, in 1971, his Ed.D., also from the University of Pittsburgh. His major field of study was mathematics education. In addition to Elementary Statistics: A Step by Step Approach, Third Edition, and Elementary Statistics: A Brief Version, the author has published several professional articles and the Modern Math Fun Book (Cuisenaire Publishing Company). He has spoken and presided at national and local mathematics conferences and has served as newsletter editor for the Pennsylvania State Mathematics Association of Two-Year Colleges. He is a member of the American Statistical Association, the National Council of Teachers of Mathematics, and the Mathematics Council of Western Pennsylvania. Al Bluman is married and has two children. His hobbies include writing, bicycling, and swimming. | 677.169 | 1 |
Scatterplots
Compressed Zip File
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5.05 MB | 13 pages
PRODUCT DESCRIPTION
This is an Algebra 1 Common Core Lesson on Scatterplots, trend line, causation and correlation. After a few definitions and teacher led examples, students will work alone or with a partner to practice. There is a link to a public Kahoot game at the end of the lesson. (Note: Kahoot was not created by Math Masters | 677.169 | 1 |
Friends Who Are Going
Friends Attending
Friends Attending
Friends Attending
Description
KS5: Developing Mathematical Thinking in Post 16 Students
Appropriate for teachers of A level Mathematics, this course highlights the need for strong mathematical thought processes in students in order that they are able to solve a variety of mathematical problems. It aims to provide teachers with the skills and strategies needed to effectively prepare their students for the problem-solving requirements of their final examination. Teachers will be encouraged to integrate a variety of problem-solving activities into their A level mathematics classrooms in order to enrich the experience of students in mathematics lessons and give them the opportunity to develop the strong mathematical thought processes needed to successfully progress to a university mathematics course.
A range of problem-solving materials designed for post 16 students will be introduced and assessed. | 677.169 | 1 |
Mathematical functions are relationships that assign each member of one set (domain) to a unique member of another set (range), and the relationship is recognizable across representations.
Numbers, measures, expressions, equations, and inequalities can represent mathematical situations and structures in many equivalent forms.
Numerical measures describe the center and spread of numerical data.
Patterns exhibit relationships that can be extended, described, and generalized.
Relations and functions are mathematical relationships that can be represented and analyzed using words, tables, graphs, and equations.
Some questions can be answered by collecting, representing, and analyzing data, and the question to be answered determines the data to be collected, how best to collect it, and how best to represent it.
The set of real numbers has infinite subsets including the sets of whole numbers, integers, rational, and irrational numbers.
There are some mathematical relationships that are always true and these relationships are used as the rules of arithmetic and algebra and are useful for writing equivalent forms of expressions and solving equations and inequalities.
Understand and apply the Pythagorean Theorem to find distances between points in a coordinate plane and to analyze polygons and polyhedra.
Use fundamental facts about distances and angles to describe and analyze figures and situations in 2- and 3-dimensional spaces and to solve problems including those with multiple steps.
Use linear functions, linear equations, and linear inequalities to represent, analyze, and solve a variety of problems.
Use the appropriate graphical data representation and extend understanding of the influence of scale in data interpretation.
Objectives
In this unit, students will learn to use equations and inequalities to model real-world situations and to solve basic linear equations and inequalities, including the following forms/techniques. Students will:
isolate a variable using inverse operations.
combine like terms.
balance variables on both sides of an equation or inequality.
use the distributive property.
write an equation in slope-intercept form.
Essential Questions
How do we recognize when it is appropriate to use a linear model to represent a real-world situation and what are the benefits of using a linear model to answer questions about the situation?
Divide both sides by −15 or division property of inequality (Multiplying by −1/15 also acceptable, with explanation of Multiply both sides by −1/15 or multiplication property of inequality)
(Also note: The student must reverse the inequality sign to get full credit.)
x < 8 Simplify
Points
Description
2
Correctly and completely solves the inequality
Demonstrates thorough understanding of finding the LCD, the distributive property, the addition/subtraction property of inequalities, and the multiplication/division property of inequalities
Supports each step with an explanation or identification of the correct property.
1
Correctly solves the inequality, but the answer may be incomplete and does not show all the steps or shows all the correct procedures but includes one calculation error
Demonstrates partial understanding of finding the LCD, the distributive property, the addition/subtraction property of inequalities, or the multiplication/division property of inequalities
Attempts to support each step with an explanation or identification of the correct property
0
Makes no attempt at solving or incorrectly attempts to solve the inequality
Demonstrates no understanding of finding the LCD, the distributive property, the addition/subtraction property of inequalities, or the multiplication/division property of inequalities
Does not give any explanation or property identification
Performance Assessment:
Eighth Grade Talent Show/Fundraiser:
The Eighth Grade Student Council in your school has been given permission to use the gymnasium to hold a Talent Show to raise funds for the local food shelf. As part of the planning group, you have the chance to help.
Set a goal of how much money you would like to raise. Think of different ways to make this money at the event, including ticket sales and refreshments. Do you want to charge the same for all ages or have separate prices? If you are using different prices, estimate the fraction of the total that will be in the different age brackets.
After deciding upon the ticket pricing and other items to be sold, list these different amounts. Use p for the number of people attending, and if some amounts involve only some of the attendees (for example, a portion of the attendees will purchase a soda, a portion of the attendees will be under 10 years old, etc.), estimate these amounts and represent them as fractions of the total p.
Research and determine any expenses involved with the event. These could include making programs, purchasing refreshments to sell, and producing posters and flyers to advertise the event.
Using the amounts from problem 2, write an equation that can be used to determine the number of people attending (p) needed to make your goal amount of money. Be sure to include expenses in this equation.
Does your answer in problem 4 sound reasonable? If it does not seem possible to get that many attendees, what other solution(s) could be used? Show how a new solution would change your equation.
Write a plan that your group will submit to the principal. Include the estimates and equation that you wrote.
Performance Assessment Scoring Rubric:
Points
Description
4
Responds completely with detailed explanation
Contains no math/calculation errors
Demonstrates complete understanding of how to model a real-world situation in mathematical terms
Shows complete understanding of the questions, mathematical ideas, and processes
Goes beyond what is required by the problem, shows creativity
3
Responds completely with clear explanation
Contains no major math errors or conceptual/procedural errors
Demonstrates understanding of how to model a real-world situation in mathematical terms
Shows substantial understanding of the questions, mathematical ideas, and processes
Meets the problem requirements
2
Responds unclearly or has some parts missing.
Contains several minor errors or one or more serious math errors or conceptual/procedural errors
Demonstrates some understanding of how to model real-world situations
Shows some understanding of the problem
Partially meets the problem requirements
1
Misses key points and/or sections
Contains major math errors or serious conceptual/procedural errors
Demonstrates lack of understanding of how to model real-world situations
Shows lack of understanding of the problem
Does not meet the problem requirements
0
Fails to complete or incorrectly completes most sections
Contains major math errors or serious conceptual/procedural errors
Demonstrates complete lack of understanding of how to model real-world situations | 677.169 | 1 |
Preview — Pre-Calculus for Dummies
by Krystle Rose Forseth
Pre-Calculus for Dummies
a handle on all of the concepts -- not just the number crunching -- and understand how to perform all pre-calc tasks, from graphing to tackling proofs. You'll also get a new appreciation for how these concepts are used in the real world, and find out that getting a decent grade in precalc isn't as impossible as you thought. Discover how to:
Apply the major theorems and formulas Understand quadratic, square-root, absolute value, cubic, and cube-root functions Graph trig functions like a pro Flip-flop with inverse functions Find trig values on the unit circle Work with trig identities and advanced identities Tackle analytic geometry Solve oblique triangles with the laws of sines and cosines Use polar coordinates to express points on a plane Identify function limits and continuity Place vectors on a coordinate plane. Rotate and shift conic section forms Solve systems with mingling and matrices If "the fun and easy way to learn pre-calc" seems like an oxymoron to you, order Pre-Calculus For Dummies today, and get ready to be surprised!...more | 677.169 | 1 |
Translating Polynomials
PDF (Acrobat) Document File
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0.06 MB | 1 pages
PRODUCT DESCRIPTION
This online generated activity presents students with their own individual problems and are assessed for free by MATHguide.com. Two different levels of interactive problems are presented to check for understanding of the translation of quadratics (parabolas).
Student creativity is also required to graph larger concepts of polynomials, like zeros, factors, maximum, and minimum points | 677.169 | 1 |
Description: 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 | 677.169 | 1 |
...
Show More explains how to read scientific literature and understand its different language style, terminology, data and argument analysis. And conflicting interpretations. Next, the book tackles how to take data from a range of sources to construct your own perspective on a scientific issue & how to present it in both written & oral forms. Lastly it assists with one of the most fundamental of all tools for any branch of science- an understanding of the basic laws of mathematics. Key topics include an introduction to tertiary study, e-learning & assessment; reading scientific literature, thinking & arguing critically; writing, research and presenting; and quantitative methods including fundamental mathematics, calculus, problem solving and statistics. Focused on step-by-step skills development, this book aims to help readers become better students who are more likely to succeed | 677.169 | 1 |
This manual is intended to provide students, teachers, counselors, and other school personnel with a set of practical instruments and procedures designed to measure the generalizable mathematics skills of students in secondary vocational programs. Three types of assessment instruments and procedures are presented and discussed: (1) student self-ratings, (2) teacher ratings, and (3) performance tests. The manual is organized around five major sections. The first section describes the list of generalizable mathematics skill areas and skills. The second, third, and fourth sections describe the student self-rating instrument, teacher rating instrument, and performance test, respectively. These three sections provide information regarding the assessment procedures, including instrument development, item analysis, reliability, validity, administration, scoring, and uses. The fifth section describes how the student self-ratings and teacher ratings can be used in combination with the performance test assessment instrument and how the scores may be used and interpreted. The appendixes present the list of generalizable skills color charts; additional important mathematics skills; the student self-rating, teacher rating, and performance test instruments; skill profiles; and the performance test answer key. The information contained in this manual and the information obtained from the assessments can be used in individualized and group planning, assessment, curriculum development, instruction, and evaluation activities. (KC) | 677.169 | 1 |
80375771
Educational Level:
Elementary/High School
Author:
Harold Torrance
Age Level:
Sixth Grade
Publication Year:
2011
ISBN:
1580375774
Detailed item info
Synopsis
Make math matter to students in all grades using Math Tutor: Pre-Algebra Skills! This 80-page book provides step-by-step instructions of the most common math concepts and includes practice exercises, reviews, and vocabulary definitions. The book covers factoring, positive and negative numbers, order of operations, variables, exponents, and formulas such as perimeter, area, and volume. It aligns with state, national, and Canadian provincial standards., The Math Tutor series provides step-by-step instruction in the most common math concepts needed by students of all ages. Included are practice exercises, reviews, and vocabulary definitions. Math Tutor: Pre-Algebra covers factoring, positive and negative numbers, order of operations, variables, exponents, and various formulas such as perimeter, area, and volume. Correlated to state, national, and Canadian provincial standards. 80 pages | 677.169 | 1 |
Showing 1 to 3 of 5
Dr. Figgers help me to the best of my ability to understand algebra. If she hadn't been there i don't think I would've passed.
Course highlights:
How to study, and gettingmy work done on time.
Hours per week:
9-11 hours
Advice for students:
Study, and do all work and extra credit
Course Term:Fall 2016
Professor:Dr. Figgers
Course Required?Yes
Course Tags:Math-heavyGo to Office HoursA Few Big Assignments
Jul 27, 2016
| Would recommend.
Pretty easy, overall.
Course Overview:
I would recommend this course, not because of the professor, rather than the amount of educated students who came together to help reteach lesson that were discussed in class. The method of teaching was a contradiction to learning, however, that never stopped me or my classmates. By teaching each other we were able to create a stronger atmosphere in class. While maintaining this close nit attitude and understanding were able to comprehend what he was trying to teach while at the same time teaching him new ways of figuring out the problem as well.
Course highlights:
I learned that one mind is very strong, however a collective of different minds and learning styles is better. The highlights would be when we taught him a lesson within the third week of class, after he attempted to teach us. Another highlight is when we created a Group chat that had schedule library meetings and notes. I learned that multiple learning styles are always benefits the people who don't speak up when their lost or confused. The course in a whole was very interesting especially the Math Labs.
Hours per week:
3-5 hours
Advice for students:
Find your craft and what works for you then teach it to somebody else. Be the voice that ask's the question, because somebody may have the exact question you do. Make sure to figure out the hard workers over the people who float through class. Express any thoughts to your professor, they will help you. Put forth the effort and you will pass the class. | 677.169 | 1 |
Your enquiries
A mathematical way to think about biologyBiology
Mathematics
Probability
Academics
Stochastic systems
Course programme
A mathematical way to think about biology comes to life in this lavishly illustrated video book. After completing these videos, students will be better prepared to collaborate in physical sciences-biology research. These lessons demonstrate a physical sciences perspective: training intuition by deriving equations from graphical illustrations.
"Excellent site for both basic and advanced lessons on applying mathematics to biology." -Tweeted by the U.S. National Cancer Institute's Office of Physical Sciences Oncology
What are the requirements?
Algebra
Exposure to calculus (there is an appendix for students interested in review)
What am I going to get from this course?
Over 134 lectures and 15.5 hours of content!
Apply physical sciences perspectives to biological research
Be able to teach yourself quantitative biology
Be able to communicate with mathematical and physical scientists
What is the target audience?
Undergraduate students
Graduate students
Postdoctoral scholars
Lab managers
Funding agency program staff
Principal investigators and grant writers
Citizen scientists
Patient advocates
Lifelong learners
Integrative Cancer Biology Program members
Physical Sciences Oncology Network members
National Centers for Systems Biology members
A mathematical way to think about biologyUdemy
Free
Do you see something that is not right in this course? Let us know if there are any mistakes and you will help users like yourself. | 677.169 | 1 |
Math and science for Drupal
Easily include maths formulas, calculations and graphics in your web pages within Drupal.
Create mathematical formulas with a visual editor (WYSIWYG)
Perform all your math calculations and plots in 2D and 3D
Drupal can be used with different HTML editors like CKEditor or TinyMCE. WIRIS plugin enhances your text editor with new icons to edit math equations with WIRIS editor and include calculations and graphics with WIRIS CAS . | 677.169 | 1 |
Strong mathematics performance in the middle grades is more important than ever--and teachers entering the field need to prepare for this endeavor in new and innovative ways. This new approach introduces some basic concepts of number theory and modern algebra that underlie middle grade arithmetic and algebra, with a focus on collaborative learning combined with extensive in-class and out-of-class assignments. Gives both pre-service and in-service teachers a fundamental understanding of the key mathematical ideas that they will be teaching, so that in turn they can help their students learn important mathematics. Directly connects college-level abstract algebra and number theory to standards-based middle grade mathematics curricula. Gives specific examples from middle-grade curricular materials to show readers the direct connections between the mathematics they are learning and the mathematics they will be teaching. Focuses on the mathematics in new reform materials. Offers Classroom Problems and Classroom Discussions that focus on discovery and collaborative learning. A useful reference for teachers of middle-grades mathematics.
REVIEWS for Algebra Connections | 677.169 | 1 |
9780070537446Discrete Mathematics and Its Applications
Designed for undergraduate courses in Discrete Mathematics found in either the Mathematics or Computer Science Departments, this text offers a strong view of the core topics and a wide variety of optional ones. This flexible organization allows the text to fit into a wide array of course outlines. Topics presented using five central themes - mathematical reasoning, combinatorial analysis, discrete structures; algorithms are treated as a central theme of the course and described in both English and Pascal-like pseudocode; and combinatorics are extensively covered including counting techniques and graph theory | 677.169 | 1 |
Get this great activity designed to help your Algebra 2 students understand permutations, combinations, theoretical and experimental probability, independent and dependent events, two-way tables, conditional probability, and compound events at the end of the unit on Probability. There are 12 task cards in the activity. | 677.169 | 1 |
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